\documentclass[11pt]{book} %PAPIR - US TRADE \paperwidth 15.24cm \paperheight 22.86cm %TEKST \textwidth 11.9cm \textheight 19.4cm \oddsidemargin=-0.5cm \evensidemargin=-1.2cm \topmargin=-15mm \headheight=13.86pt %\usepackage[slovene]{babel} \usepackage[english]{babel} %\usepackage[cp1250]{inputenc} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{amsmath} \usepackage{color} \usepackage{amsfonts} \usepackage{makeidx} \usepackage{calc} \usepackage{gclc} %\usepackage[dvips]{hyperref} \usepackage{amssymb} \usepackage[dvips]{graphicx} \usepackage{fancyhdr} %za slike \usepackage{caption} \DeclareCaptionFormat{empty} \def\contentsname{Content} \makeindex \newcommand{\ch}{\mathop {\mathrm{ch}}} \newcommand{\sh}{\mathop {\mathrm{sh}}} \newcommand{\tgh}{\mathop {\mathrm{th}}} \newcommand{\tg}{\mathop {\mathrm{tg}}} \newcommand{\ctg}{\mathop {\mathrm{ctg}}} \newcommand{\arctg}{\mathop {\mathrm{arctg}}} \newcommand{\arctgh}{\mathop {\mathrm{arcth}}} \def\indexname{Index} \definecolor{green1}{rgb}{0,0.5,0} \definecolor{viol}{rgb}{0.5,0,0.5} \definecolor{viol1}{rgb}{0.2,0,0.9} \definecolor{viol3}{rgb}{0.3,0,0.6} \definecolor{viol4}{rgb}{0.6,0,0.6} \definecolor{grey}{rgb}{0.5,0.5,0.5} \def\qed{$\hfill\Box$} \newcommand{\kdokaz}{\color{red}\qed\vspace*{2mm}\normalcolor} \newcommand{\res}[1]{\color{green1}\textit{#1}\normalcolor} \newtheorem{izrek}{Theorem}[section] \newtheorem{lema}{Lemma}[section] \newtheorem{definicija}{Definition}[section] \newtheorem{aksiom}{Axiom}[section] \newtheorem{zgled}{Exercise}[section] \newtheorem{naloga}{Problem} \newtheorem{trditev}{Proposition}[section] \newtheorem{postulat}{Postulate} \newtheorem{ekv}{E} %BARVA \newcommand{\pojem}[1]{\color{viol4}\textit{#1}\normalcolor} %\newcommand{\pojemFN}[1]{\textit{#1}} \newcommand{\blema}{\color{blue}\begin{lema}} \newcommand{\elema}{\end{lema}\normalcolor} \newcommand{\bizrek}{\color{blue}\begin{izrek}} \newcommand{\eizrek}{\end{izrek}\normalcolor} \newcommand{\bdefinicija}{\begin{definicija}} \newcommand{\edefinicija}{\end{definicija}} \newcommand{\baksiom}{\color{viol3}\begin{aksiom}} \newcommand{\eaksiom}{\end{aksiom}\normalcolor} \newcommand{\bzgled}{\color{green1}\begin{zgled}} \newcommand{\ezgled}{\end{zgled}\normalcolor} \newcommand{\bnaloga}{\color{red}\begin{naloga}} \newcommand{\enaloga}{\end{naloga}\normalcolor} \newcommand{\btrditev}{\color{blue}\begin{trditev}} \newcommand{\etrditev}{\end{trditev}\normalcolor} \newcommand{\del}[1]{\chapter{#1}} \newcommand{\poglavje}[1]{\section{#1}} %\newcommand{\naloge}[1]{\color{red}\section*{#1}\normalcolor} \newcommand{\naloge}[1]{\section{#1}} \newcommand{\ppoglavje}[1]{\subsection{#1}} \setlength\arraycolsep{2pt} \author{Milan Mitrovi\'c} \title{\textsl{\Huge{\textbf{Euclidean Plane Geometry}}}} \date{} %_________________________________________________________________________________________ \begin{document} \pagestyle{fancy} \lhead[\thepage]{\textsl{\nouppercase{\rightmark}}} \rhead[\textsl{\nouppercase{\leftmark}}]{\thepage} \cfoot[]{} \vspace*{-12mm} \hspace*{24mm} \textsl{\Huge{\textbf{Euclidean }}}\\ \hspace*{24mm} \textsl{\Huge{\textbf{Plane}}}\\ \hspace*{24mm} \textsl{\Huge{\textbf{Geometry}}} \vspace*{8mm}\hspace*{60mm}Milan Mitrovi\'c % \normalcolor \vspace*{0mm} \hspace*{-21mm} \input{sl.NASL2A4.pic} %\color{viol1} \hspace*{48mm}Sevnica \hspace*{49mm} 2013 % \normalcolor %slikaNova0-1-1 %\includegraphics[width=120mm]{slikaNaslov.pdf} \setcounter{section}{0} \thispagestyle{empty} \newpage %\pagecolor{white} %\color{viol1} \vspace*{17mm} \hspace*{77mm} \textit{To Boris and Jasmina} %\normalcolor %slikaNova0-1-1 %\includegraphics[width=120mm]{slikaNaslov.pdf} \thispagestyle{empty} \newpage %________________________________________________________________________ %PREDGOVOR %{\hypertarget{Vsebina}\tableofcontents} %\printindex \thispagestyle{empty} %_______________________________________________________________________ \chapter*{Preface} \thispagestyle{empty} \thispagestyle{empty} The present book is the result of the experience I gained as a professor in teaching geometry at the Mathematical Gymnasium in Belgrade for many years and preparing students in Slovenia for the International Mathematical Olympiad (IMO). Formally, the substance is presented in such a way that it does not rely on prior knowledge of geometry. In the book, we will deal only with planar Euclidean geometry - all definitions and statements refer to the plane. The first two chapters deal with the history and axiomatic design of geometry. The consequences of the axioms of incidence, congruence and parallelism are discussed in detail, while in the other two groups (axioms of order and continuity) the consequences are mostly not proven. Chapters three and four deal with the relation of the congruence of figures, the use of the triangle congruence theorems, and a circle. In the fifth chapter, a vectors are defined. Thales's theorem of proportion is proven. Chapter six deals with isometries and their use. Their classification has been performed. Chapters 7 and 8 deal with similarity transformations, figure similarity relation, and area of figures. The ninth chapter presents the inversion. At the end of each chapter (except the introductory one) are exercises. Solutions and instructions can be found in the last, tenth chapter. The book contains 341 theorems, 247 examples and 418 solved problems (28 of them from the IMO). In this sense, the book in front of you is at the same time a preparing guide for the IMO. For some well-known theorems and problems, are given brief historical remarks. That can help high school and college students to better understand the development of geometry over the centuries. A lot of help in writing the book was selflessly offered to me by Prof. Roman Drstvenšek, who read the manuscript in its entirety. With his professional and linguistic comments, he made a great contribution to the final version of the book. In that work, he was assisted by Prof. Ana Kretič Mamič. I kindly thank both of them for the effort and time they have generously devoted to this book. I would especially like to thank Prof. Kristjan Kocbek, who read the partial manuscript and with his critical remarks contributed to significant improvement of the book. %I sincerely thank prof. Gordana Kenda Kozinc, who read and proofread the introductory chapter, and thanks also to prof. Alenka Brilej, who helped Roman with the language test with quite a few useful tips. I also thank Prof. Dr Predrag Janiči\'c, who wrote the wonderful software package \textit {GCLC} for \ LaTeX {}. Almost all pictures in this book were made with this package. Last but not least, I would like to thank the students of the Bežigrad Grammar School, the 1st Grammar School in Celje and the Brežice Grammar School for their inspiration and support. Students from these schools attended the renewed course of geometry which I have already taken before years in Belgrade. \thispagestyle{empty} \vspace*{12mm} Sevnica, December 2013 \hfill Milan Mitrovi\'c %\newpage %________________________________________________________________________ \tableofcontents %\thispagestyle{empty} %\newpage % DEL 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - %________________________________________________________________________________ % O DEDUKTIVNI IN INDUKTIVNI METODI %________________________________________________________________________________ \del{Introduction} \label{pogUVOD} %________________________________________________________________________________ \poglavje{Deductive and Inductive Method} \label{odd1DEDUKT} We learn a lot of geometric concepts in elementary school, such as: triangle, circle, right angle, etc. Later we also learn some propositions: propositions about the congruence of triangles, Pythagoras' and Tales' proposition. At the beginning we do not prove the propositions, but we verify the facts based on several individual examples. This way of reasoning is called inductive method. Inductive method (lat. inductio -- introduction) is thus a way of reasoning, in which we come to general conclusions from individual examples. Later we start proving individual propositions. Through these proofs we first encounter the so-called deductive way of reasoning, or deduction. Deduction (lat. deductio -- deduction) is a way of reasoning, in which we come from general to individual conclusions. The idea of this method is thus to deduce a general conclusion by proving it and then use it in individual examples. Since we cannot verify all examples using inductive method, because their number is usually infinite, we can also come to wrong conclusions using this method. With deductive method we always get correct conclusions, if the assumptions we use in the proof are correct. Let us analyze both of these methods using the following example. We try to come to the conclusion: \btrditev \label{TalesUvod} The diameter of a circle subtends a right angle to any point on the circle. \etrditev \begin{figure}[!htb] \centering \input{sl.1.2.1.6.pic} \caption{} \label{sl.sl.1.2.1.6.pic} \end{figure} %slikaNova0-1-1 %\includegraphics[width=50mm]{slikaNova0-1-1.pdf} If we used inductive method, we would verify whether this proposition is true in some individual examples; for example in the case when the apex of the angle is in the center of the circle and similar (Figure \ref{sl.sl.1.2.1.6.pic}). If we only deduced the general conclusion from these individual examples, of course we could not be sure that the proposition is not true in some example we did not verify. We will now use the deductive method. Let $AB$ be the radius of the circle with center $O$ and $L$ any point on this circle, different from points $A$ and $B$ (Figure \ref{sl.sl.1.2.1.6.pic}). We will prove that the angle $ALB$ is a right angle. Because $OA\cong OB\cong OL$, it follows that the triangles $AOL$ and $BOL$ are isosceles, therefore $\angle ALO\cong\angle LAO=\alpha$ and $\angle BLO\cong\angle LBO=\beta$. Then $\angle ALB=\alpha+\beta$. The sum of the interior angles in triangle $ALB$ is equal to $180^0$, therefore $2\alpha+2\beta=180^0$. From this it follows: $$\angle ALB=\alpha+\beta=90^0$$ We notice that in the case of using the deductive method or in the proof of the statement, we have not considered a certain point $L$ on the circle, but an arbitrary point (in general position). This means that the statement is valid for every point on the circle (except $A$ and $B$), if the proof is correct, of course. But is the proof correct? In this proof, we used the following two statements: \btrditev If two sides in a triangle are congruent, then the angles opposite the congruent sides are congruent angles. \etrditev \btrditev The sum of the interior angles of a triangle is equal to $180^0$. \etrditev We also used concepts such as: isosceles triangle, angle congruence; in the statement itself, we also used the concepts: diameter, circle, angle over the diameter and right angle. In order to be sure that the statement we proved is true, we must be sure that the two statements we used in the proof are also true. In our case, we assume that we have already proved the aforementioned two statements and that we have introduced all the concepts mentioned. It is clear that this problem arises with every statement - even with the two on which we relied in the proof. This requires a certain systematization of the entire geometry. The question arises as to how to start if we again refer to previously proven statements in the proof of each statement. This process could then continue indefinitely. Thus we come to the need for initial statements - \index{aksiomi} \pojem{aksiomih}. The same applies to concepts - we need t. i. \pojem{začetni pojmi}.\index{začetni pojmi} In this way, each geometry (there can be more of them), which we consider, depends on the choice of initial concepts and axioms. We call this approach to building a geometry \pojem{sintetični postopek}, and we call the geometry itself \pojem{sintetična geometrija}\index{geometrija!sinteti\v{c}na}. %________________________________________________________________________________ \poglavje{Basic Terms and Basic Theorems} \label{odd1POJMI} In some theory (like geometry) we introduce every new concept with a \index{definition} \pojem{definition}, which describes this concept with the help of some initial or already defined concepts. The connections between concepts and their appropriate properties are given by statements, which we call \pojem{theory statements}. As we have already mentioned, we call the initial statements \index{aksiomi} \pojem{aksiomi}, the statements derived from them are \pojem{izreki} of this theory. Formally, \pojem{proof} \index{dokaz izreka} of some statement $\tau$ is a sequence of statements, which logically follow one from the other, each of which is either an axiom or a statement derived from the axioms (izrek), and the last one in this sequence is precisely the statement $\tau$. Although the choice of axioms is not uniquely determined, it cannot be arbitrary. When making this choice we must be careful not to lead to contradictory statements or to a contradiction. This means that, for a given choice of axioms, there is no such statement that both the statement and its negation are izreki in this theory. We also need enough axioms so that we can determine, for every statement that we can formulate in this theory, whether it is true or not. This means that either the statement or its negation is an izrek in this theory. For a system of axioms that satisfies the first requirement, we say that it is \index{sistem aksiomov!neprotisloven} \pojem{neprotisloven}, for one that satisfies the second requirement, we say that it is \index{sistem aksiomov!popoln} \pojem{popoln}. When choosing axioms, there is also a third requirement -- that the system of axioms is \index{sistem aksiomov!minimalen} \pojem{minimalen}, which means that none of the axioms can be derived from the others. We mention that the last requirement is not as important as the first two. We must also add that we do not build Euclidean geometry independently from algebra and logic. We will use concepts such as set, function, relation with properties that apply to them. We will also use so-called rules of inference, such as the method of contradiction. For mathematical disciplines that we use in this way to build geometry, we say that they are \pojem{predpostavljene teorije}. %________________________________________________________________________________ \poglavje{A Brief Historical Overview of the Development of Geometry} \label{odd1ZGOD} People started dealing with geometry in early history. At first, it was only the observation of characteristic shapes, such as a circle or a square. Based on the drawings found on the walls of old caves, we conclude that people in prehistory were interested in the symmetry of shapes. In further development, man was discovering various properties of geometric shapes. This was due to practical needs, e.g. measuring the area of ​​land - which is also how the word "geometry" came about. In this period, geometry developed as an inductive science. This means that geometric propositions were coming from experiences - by means of measurements and checking on individual examples. In this sense, geometry was developed by all ancient civilizations: Chinese, Indian and especially Egyptian. In Egypt, geometry developed mainly as a study of measurements. Because the Nile river often flooded, the land had to be measured very often. In addition, the knowledge of geometry was used in construction. They knew e.g. the formula for calculating the volume of a pyramid and a truncated pyramid, although they came to it empirically. So geometry was for the Egyptians primarily a pragmatic discipline. The oldest records date back to approximately 1700 BC. Geometry of three-dimensional space was not discussed as much as in Egypt. There is not as much data on Chinese geometry as on Egyptian, although we know that it was also very developed. In the oldest preserved records we find a description of the calculation of the volume of a prism, pyramid, cylinder, cone, truncated pyramid and truncated cone. Indian geometry is much younger than the previous three. It dates back to approximately the 5th century BC. In it we already see the first attempts at proof. Later it developed parallel to Greek geometry. The turning point in the development of geometry took place in Ancient Greece. This is when the deductive method was first used in geometry. The first geometric proofs are associated with Tales\footnote{The Ancient Greek philosopher and mathematician \textit{Tales} \index{Tales} from Miletus (640--546 BC).}. We connect his name with the well-known statement about the proportionality of segments in parallel lines. He also proved the statement that the angles above the diameter of a circle are right, even though this claim was known without proof to the Babylonians 1000 years earlier. This way of developing geometry was continued by other Ancient Greek philosophers, of which Pythagoras\footnote{The Ancient Greek philosopher and mathematician \textit{Pythagoras} \index{Pythagoras} from the island of Samos (ca. 580--490 BC).} was one of the most important. Of course, his \index{statement!Pythagorean}\textit{Pythagorean statement} is famous. However, the Egyptians knew this statement as a fact 3000 years BC (maybe the statement was known even before that), but Pythagoras gave the first known proof. Archimedes\footnote{The Ancient Greek philosopher and mathematician \textit{Archimedes} \index{Archimedes} from Syracuse (287--212 BC).} was the first to present a theoretical calculation of the number $\pi$, by considering the inscribed and circumscribed polygons with $96$ sides. The statements about the congruence of triangles were also proven. With the rapid progress of geometry, reflected in the large number of proven statements, the need for systematization and, with that, the need for axioms, became apparent. The need for axioms was first described by Plato\footnote{The Ancient Greek philosopher and mathematician \textit{Plato} \index{Plato} (427--347 BC).} and Aristotle\footnote{The Ancient Greek philosopher and mathematician \textit{Aristotle} \index{Aristotle} from Athens (384--322 BC).}. Plato is also known in mathematics for his research of regular polygons: the tetrahedron, the cube, the octahedron, the dodecahedron and the icosahedron, which is why we also call them Platonic solids after him. One of the first attempts at an axiomatic approach to geometry - and the only one that has been preserved from that time - was made by the most famous mathematician of that time, Euclid, Plato's student, in his well-known work \textit{The Elements}, which consists of 13 books. In it, he systematized all the existing knowledge of geometry. He divided the initial statements into axioms and so-called postulates, of which the latter are purely geometric content (today we also call them axioms). \textit{The Elements} became one of the most important and influential books in the history of mathematics. The geometry, which he developed in this way, with minor unimportant changes, is the one that is taught in schools today. Proofs, such as the one about the central and peripheral angle, have been preserved in practically unchanged form. We give the postulates as Euclid gave them (Figure \ref{sl.sl.1.3.1.9.pic}): \color{viol3} \begin{postulat} We can draw a straight line from any point to any point. \end{postulat} \begin{postulat} We can produce a finite straight line continuously in a straight line. \end{postulat} \begin{postulat} We can describe a circle with any center and distance. \end{postulat} \begin{postulat} All right angles are equal to one another. \end{postulat} \begin{postulat} If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the straight lines, if produced indefinitely, will meet on that side on which the angles are less that two right angles. \end{postulat} \normalcolor \begin{figure}[!htb] \centering \input{sl.1.3.1.9.pic} \caption{} \label{sl.sl.1.3.1.9.pic} \end{figure} %slikaNova1-3-3 %\includegraphics[width=100mm]{slikaNova1-3-3.pdf} However, the system of axioms that Euclid gave was not perfect. In some proofs, he took certain parts as obvious and did not prove them. Of course, we should not be too critical, because this was a revolutionary work for those times. \textit{The Elements} were an example and inspiration for mathematicians for centuries and laid the foundations for the further development of geometry to this day. The last axiom, i.e. the \index{aksiom!fifth Euclid's axiom} \pojem{fifth Euclid's axiom}, was particularly important for the further development of geometry. The problem of its independence from the other axioms was open for the next 2000 years! The last in a series of great ancient Greek mathematicians were Apolonius\footnote{The Greek mathematician \textit{Apolonius} \index{Apolonius} from Perga (262--190 BC).}, Menelaus\footnote{The Greek mathematician \textit{Menelaus} \index{Menelaus} from Alexandria (ca. 70--130).} and Pappus\footnote{The Greek mathematician \textit{Pappus} \index{Pappus} from Alexandria (ca. 290--350).}. In his book \textit{On the Cutting of a Cone}, Apolonius defined the ellipse, parabola and hyperbola as the intersections of a plane and a circular (infinite) cone. This allowed him to consider their properties simultaneously, which was quite a modern approach for the time. Menelaus and Pappus proved certain theorems which didn't become relevant until the 19th century with the development of projective geometry. So the ideas of these three mathematicians were quite modern and in a way we can say that they were on the brink of discovering the first non-Euclidean geometry. After the final fall of the Old Greece under the Roman Empire, the period of the glorious ancient Greek geometry came to an end. Although the Old Romans took over a large part of the ancient Greek culture and built roads, aqueducts and so on, it is interesting that they never really showed much interest in the ancient theoretical mathematics. So their contribution to the development of geometry is very modest. An important role in the further development of geometry was taken over by the Arabs. First of all, we should say that all the works of the Ancient Greeks including Euclid's \textit{Elements} are known to us today because they were translated and thus preserved by the Arabs. After the foundation of Baghdad in 762, in the next 100 years they translated most of the works of ancient Greek and Indian mathematics. They also made a synthesis of ancient Greek mostly geometric and Indian mostly algebraic approach. We mention that the word itself \pojem{algebra} is of Arabic origin. In addition, the Arabs continued the development of \pojem{trigonometry}, which was designed by the Ancient Greeks. A. R. al-Biruni\footnote{The Arab mathematician \textit{ A. R. al-Biruni} \index{al-Biruni, A. R.} (973--1048).} proved the now known \pojem{sine theorem}. The development of geometry in Europe began in the 12th century, when Arabic and Jewish mathematicians brought their knowledge to Spain and Sicily. (Euclid's \textit{Elements} were translated from Arabic into Latin around 1200); but Europe did not experience its true flowering until the 16th century. In the Middle Ages, mathematics developed very slowly. In the Middle Ages, Western European mathematicians only learned the Greek geometric heritage from Arabic translations, but this process was not quick. When this knowledge was accumulated and social and political conditions changed, a new era began in the development of geometry. The first new results were given by Italian mathematicians of that time, who paid a lot of attention to constructions using a ruler and a compass. As we mentioned earlier, Euclid's fifth axiom had a very big impact on the further development of geometry. Because of its formulation, which is not as simple as the previous axioms, and also because of its importance, many mathematicians of that time thought that it did not need to be considered as an axiom, but it could be proven as a theorem with the other axioms. If we read Euclid's other initial propositions, it is really true that the fifth axiom is more complex. The problem of the independence of the fifth axiom from the others occupied many mathematicians in the following centuries. Until the second half of the 19th century the problem was not solved. In many attempts to prove the fifth axiom from the others, propositions were used, whose proof was omitted. It was later shown that these propositions could not even be proven from the other axioms if the fifth axiom was omitted from the list. Just as they follow from the fifth axiom, these propositions also follow from the fifth axiom (of course using the other axioms). We therefore call them \pojem{equivalents of the fifth Euclidean axiom}. We give some examples of these equivalents (Figure \ref{sl.sl.1.3.1.9a.pic}): \color{blue} \begin{ekv} If $ ABCD $ is a quadrilateral with two right angles on the side $BC$ and the sides $AB$ and $CD$ are congurent, then the two remain angles of this quadrilateral are also right angles.\footnote{This equivalent was set by Italian mathematician \index{Saccheri, G. G.} \textit{G. G. Saccheri} (1667--1733).} \end{ekv} \begin{ekv} A line perpendicular to one arm of an acute angle intersects his other arm. \end{ekv} \begin{ekv} Every triangle can be circumscribed. \end{ekv} \begin{ekv} If three angles of a quadrilateral are right angles, then the fourth angle is also a right angle.\footnote{\index{Lambert, J. H.}\textit{J. H. Lambert} (1728--1777), French mathematician.} \end{ekv} \begin{ekv} The sum of the interior angles in every triangle is $180^0$.\footnote{\index{Legendre, A. M.} \textit{A. M. Legendre} (1752--1833), French mathematician.} \end{ekv} \begin{ekv} For any given line $p$ and point $A$ not on $p$, in the plane containing both line $p$ and point $A$ there is just one line through point $A$ that do not intersect line $p$\footnote{\index{Playfair, J.} \textit{J. Playfair} (1748--1819), Scottish mathematician.}. \end{ekv} \normalcolor \begin{figure}[!htb] \centering \input{sl.1.3.1.9a.pic} \caption{} \label{sl.sl.1.3.1.9a.pic} \end{figure} So mathematicians were able to prove the fifth Euclidean axiom with the help of each of these assertions, but eventually it turned out that none of them could be proved without the fifth axiom. Therefore, these assertions, as we have already mentioned, are equivalent to the fifth axiom. Today, Playfair's equivalent is most commonly used, which was later added to Euclid's axioms instead of the fifth axiom. But how did mathematicians find out that the fifth Euclidean axiom could not be derived from the other axioms? Just the fact that they were not able to prove it did not mean that it was not possible. The answer to this question came at the end of the 19th century and, as we will see, brought much more to the development of geometry than just the fact of the unprovability of the fifth axiom. The next breakthrough in the development of geometry was the discovery \index{geometrija!neevklidska}\pojem{non-Euclidean geometries} in the 19th century. For the beginning of this development we count N.~I.~Lobačevskega\footnote{\index{Lobačevski, N. I.}\textit{N. I. Lobačevski} (1792--1856), Russian mathematician.}. He also dealt with the problem of the independence of the fifth Euclidean axiom. Based on its negation or the negation of Playfair's equivalent statement, Lobačevski postulated that through a point that does not lie on a straight line, there are at least two straight lines that do not intersect with that line and are coplanar. In an attempt to come to a contradiction (thus the fifth axiom would be proven), he built a whole sequence of new statements. One of them, for example, is that the sum of the interior angles of a triangle is always less than the extended angle. But none of these statements were in contradiction with the other axioms, if of course we exclude the fifth Euclidean axiom from the list. From this he got the idea that it is possible to build a completely new geometry that is non-contradictory and based on all Euclidean axioms except the fifth, which we replace with its negation. Today we call this geometry \index{geometrija!hiperbolična} \pojem{hyperbolic geometry} or \pojem{Lobačevski geometry}. Independently of Lobačevski, J. Bolyai\footnote{\index{Bolyai, J.} \textit{J. Bolyai} (1802--1860), Hungarian mathematician.}. As it often happens, the ideas of Lobačevski during his lifetime unfortunately were not accepted. The complete confirmation of these ideas or the proof of the non-contradiction of this new geometry was at the end of the 19th century, that is, only after the death of Lobačevski, presented by A. Poincar\'{e}\footnote{\index{Poincar\'{e}, J. H.} \textit{J. H. Poincar\'{e}} (1854--1912), French mathematician.}. Poincar\'{e} built a model on the basis of which he showed that a possible contradiction of Lobačevski geometry would at the same time be a contradiction of Euclidean geometry. Later, the discovery of other non-Euclidean geometries followed. Although at the end of the 19th and the beginning of the 20th century the system of Euclid's geometry axioms was already almost completely built, the first correct and complete system was given by D. Hilbert\footnote{\index{Hilbert, D.} \textit{D. Hilbert} (1862--1943), German mathematician.} in his famous book \textit{The Foundations of Geometry}, published in 1899. We use a very similar system of axioms in an almost unchanged form even today. Parallel to the research of the fifth Euclid's axiom and the development of non-Euclidean geometries, other important methods have also emerged in the study of geometry. As early as around 1637, R. Descartes\footnote{\index{Descartes, R.} \textit{R. Descartes} (1596--1650), French mathematician.} in his book \textit{Geometry} showed that every point in a plane can be described by a suitable pair of two real numbers and similarly in space as a triple of three real numbers. He connected this with the concept of a coordinate representation of the dependence of one quantity (function) on another (variable), which was known earlier. Today, we call such a way of determining points in space after him \pojem{Cartesian coordinate system}. Lines and planes can then be described as sets of solutions of appropriate linear equations, where the unknowns are coordinates of points. Thus, under the influence of the ideas of F. Vi\'{e}te\footnote{\index{Vi\'{e}te, F.} \textit{F. Vi\'{e}te} (1540--1603), French mathematician.}, Descartes and P. Fermat\footnote{\index{Fermat, P.} \textit{P. Fermat} (1601--1665), French mathematician.}, the two very important mathematical disciplines began to develop - first \index{geometry!analytical} \pojem{analytical geometry}, then \index{linear algebra} \pojem{linear algebra}, which represent the connection between algebra and geometry. The further development of these two disciplines allowed the development of \index{geometry!multidimensional} \pojem{multidimensional geometry}, in which spaces of dimensions greater than three can be considered, because in algebra there are no such limitations as we have in the geometric perception of space. Thus, we can define the so-called \pojem{polytope} - objects of multidimensional space, which are the analogy of two-dimensional polygons and three-dimensional polyhedrons. Later, the discovery of other non-Euclidean geometries followed. In the 19th century, the so-called \index{geometrija!projektivna}\pojem{projective geometry} developed, but its development was not axiomatic like in hyperbolic geometry; the appropriate system of axioms was set only later. In this geometry, there are no lines in the plane that do not intersect. One of the first motives for the beginning of the development of projective geometry originates from painting or from the desire to transfer the feeling of three-dimensional space into a plane. Already in very early painting, we encounter a very important property--that in the picture, parallel lines are represented as lines that intersect. In the 15th century, Italian artists were very interested in the geometry of space. The theory of perspective was first considered by F. Brunellechi\footnote{\index{Brunellechi, F.} \textit{F. Brunellechi} (1377--1446), Italian architect.} in 1425. His work was continued by L. B. Alberti\footnote{\index{Alberti, L. B.} \textit{L. B. Alberti} (1404--1472), Italian mathematician and painter.} and A. D\"{u}rer\footnote{\index{D\"{u}rer, A.} \textit{A. D\"{u}rer} (1471--1528), German painter.}. Alberti's book from 1435 represents the first presentation of central projection. For the beginning of the development of projective geometry as a mathematical discipline, we consider the period when J. Kepler\footnote{\index{Kepler, J.} \textit{J. Kepler} (1571--1630), German astronomer.} and G. Desargues\footnote{\index{Desargues, G.} \textit{G. Desargues} (1591--1661), French architect.} independently introduced the concept of points at infinity. Kepler showed that a parabola has two foci, one of which is a point at infinity. Desargues was writing in 1639: ‘‘Two parallel lines have a common endpoint at an infinite distance.’’ In 1636, he wrote a book on perspective, and in 1639, he wrote about cones. The famous \textit{Desargues' Theorem} was published in 1648. With the further development of projective geometry we connect French mathematics. The genius B. Pascal\footnote{\index{Pascal, B.} \textit{B. Pascal} (1623--1662), French philosopher and mathematician.} was already as a sixteen year old when he proved an important theorem about cones, which we today call \textit{Pascal's theorem}. This theorem, which is one of the basic theorems of projective geometry, was published in 1640. G. Monge\footnote{\index{Monge, G.} \textit{G. Monge} (1746--1818), French mathematician.} was among the first mathematicians who we can consider a specialist; he is in fact the first true geometer. He developed \pojem{descriptive geometry} as a special discipline. In his research in descriptive geometry we find many ideas of projective geometry. The most original Monge's student was J. V. Poncelet\footnote{\index{Poncelet, J. V.} \textit{J. V. Poncelet} (1788--1867), French mathematician.}. Although Pappus\footnote{\index{Pappus} \textit{Pappus of Alexandria} (3rd century), Greek mathematician.} discovered the first projective theorems, Poncelet with a completely projective way of reasoning proved them only in the 19th century. In 1822 Poncelet published his famous "Treatise on the projective properties of figures", in which all the important concepts characteristic for projective geometry appear: harmonic quadruple, perspectivity, projectivity, involution, etc. Poncelet introduced a line at infinity for all planes that are parallel to a given plane. Poncelet and J. D. Gergonne\footnote{\index{Gergonne, J. D.} \textit{J. D. Gergonne} (1771--1859), French mathematician.} independently from each other studied duality in projective geometry, C.~J.~Brianchon\footnote{\index{Brianchon, C. J.} \textit{C. J. Brianchon} (1783--1864), French mathematician.} however proved a theorem which is dual to Pascal's theorem. M.~Chasles\footnote{\index{Chasles, M.} \textit{M. Chasles} (1793--1880), French mathematician.} was the last of the great school of French projective geometers of that time. A typical representative of so-called pure geometry (today we would say synthetic geometry) was J. Steiner\footnote{\index{Steiner, J.} \textit{J. Steiner} (1796--1863), Swiss mathematician.}. Steiner developed projective geometry very systematically, from perspective to projectivity and then to conics. In the middle of the 19th century, German mathematicians took over the lead in the development of projective geometry. They advocated a synthetic approach h geometry. All mathematicians until then had designed projective geometry based on Euclidean metric geometry -- by adding points at infinity. But C.~G.~C.~Staudt \footnote{\index{Staudt, K. G. C.} \textit{C. G. C. Staudt} (1798--1867), German mathematician.} was the first to try to make it independent and design it only on incidence axioms, without the help of metric. This led to the abolition of the difference between points at infinity and ordinary points or the transition from extended Euclidean to projective space. F. Klein\footnote{\index{Klein, F. C.} \textit{F. C. Klein} (1849--1925), German mathematician.} set the projective geometry on algebraic foundations in 1871 with the help of so-called \pojem{homogeneous coordinates}, which were discovered independently of each other in 1827, by K. W. Feuerbach\footnote{\index{Feuerbach, K. W.} \textit{K. W. Feuerbach} (1800--1834), German mathematician.} and A. F. M\"{o}bius\footnote{\index{M\"{o}bius, A. F.} \textit{A. F. M\"{o}bius} (1790--1868), German mathematician.}. A. Cayley\footnote{\index{Cayley, A.} \textit{A. Cayley} (1821--1895), English mathematician.} and Klein found the use of projective geometry in other non-Euclidean geometries. They discovered the model of hyperbolic geometry and models of other geometries in projective. The first to completely axiomatically design projective geometry, were G. Fano\footnote{\index{Fano, G.} \textit{G. Fano} (1871--1952), Italian mathematician.} in 1892 and M. Pieri\footnote{\index{Pieri, M.} \textit{M. Pieri} (1860--1913), Italian mathematician.} in 1899. Because of its relative simplicity, the development of classical (synthetic) projective geometry was almost completed by the end of the 19th century. Its development today continues within the framework of other theories -- especially in algebra and algebraic geometry as $n$-dimensional projective geometry. G. F. B. Riemann\footnote{\index{Riemann, G. F. B.} \textit{G. F. B. Riemann} (1828--1866), German mathematician.} defined the space of arbitrary dimension in his book \textit{On the hypotheses which lie at the foundations of geometry}, which is not always of constant curvature. After him, we today call it the \pojem{Riemann metric space}\index{Riemann spaces}. Euclidean geometry is then obtained as a special case: if the curvature is constant and equal to 0; hyperbolic geometry is obtained if we choose that the curvature is constant and negative. If the curvature is constant and positive, we obtain the so- called \index{geometry!elliptic} \pojem{elliptic geometry}. The latter geometry is actually projective geometry, if we add a metric to it. This research was also the beginning of the development of a new discipline in mathematics, namely the \index{geometry!differential} \pojem{differential geometry}. If we think about non-Euclidean geometries, it may seem strange to us that in mathematics there can be considered more different theories, such as Euclidean geometry and hyperbolic geometry, which are in contradiction with each other. For modern mathematics, it is most important that both geometries are determined by systems of axioms, which are (each for itself) consistent and complete. To the question of which of these two geometries is valid, it is pointless to seek an answer within mathematics. This is because it depends on which axioms we have chosen. Such a question would be the same as the question of which axioms are valid. But the axioms are assumed by definition without proof. Of course, we can ask the question of what the geometry of space is in the physical sense and how we can describe it with axioms." To answer this question, a physical interpretation of the basic geometric concepts is needed. For example, it is most natural to interpret a line as a ray of light. In this sense, physical space is not Euclidean. It is not even determined by hyperbolic geometry. With the advent of Einstein's\footnote{\index{Einstein, A.} \textit{A. Einstein} (1879--1955), famous German physicist.} theory of relativity at the beginning of the 20th century, it turned out that in space of cosmic dimensions it is more convenient to use non-Euclidean geometry with variable curvature (Riemann metric spaces!). We can say that the geometry of the universe is locally different in each point, depending on the proximity and size of some mass. Einstein's theory also tells us that space and time are interrelated and that time (which is of course surprising) does not flow evenly in each point of the universe. In connection with the mentioned connection of space and time, the so-called \pojem{Minkowsky space of four dimensions}\footnote{This space was discovered by \index{Minkowsky, H.} \textit{H. Minkowsky} (1864--1909), German mathematician.} is important. Since the 20s of the 20th century and the development of the theory of the primordial soup, we know that the universe is not static and that it is expanding. Its destiny depends on what geometry globally describes it best. But we still do not know for sure what the shape of the universe is or what its destiny is. We do not even know if the universe is finite or infinite. As S. Hawking writes in his famous popular book from 1988 A Brief History of Time (\cite{KratkaZgodCasa}), the universe may even be finite and unlimited. The latter seems paradoxical, although we can imagine it if, instead of three-dimensional, we imagine a "two-dimensional universe". So the beings of this two-dimensional universe could once find out that their universe is actually not a plane but a sphere that is finite but unlimited. The sphere is part of three-dimensional space. So we can theoretically imagine the universe as a three-dimensional sphere in four-dimensional space. The three-dimensional sphere is one of the 3-manifolds. In this case, the geometry of the universe would be globally elliptical. Although the question of the shape and destiny of the universe is a question of theoretical physics and cosmology, we see how modern geometry (non-Euclidean geometry, geometry of multi-dimensional spaces, etc.) is closely related to this problem (\cite{Oblika}). It is important to understand that geometry (and every other mathematical discipline) develops and is treated as an abstract discipline in which physical interpretations are only inspiration - in this sense, only axioms and initial concepts remain on which we then build mathematical theory. At the end, we mention another one of the most important geometers of the 20th century H. S. M. Coxeter\footnote{\index{Coxeter, H. S. M.} \textit{H. S. M. Coxeter} (1907--2003), Canadian mathematician. One of the greatest geometers of the 20th century.}. Coxeter further researched polytopes in arbitrary dimensions, primarily regular polytopes. In addition, he was mostly occupied with groups of isometries in hyperbolic geometry and with multidimensional hyperbolic geometry. It needs to be added that with everything that we have said, the development of geometry is far from over. Quite the opposite -- contrary to the usual perception -- geometry and mathematics in general are now developing even faster than ever before. Even today, there are in mathematics (in geometry in particular from the non-Euclidean geometries) many problems that are still unresolved. % DEL 2 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - %________________________________________________________________________________ % AKSIOMI RAVNINSKE EVKLIDSKE GEOMETRIJE %________________________________________________________________________________ \del{Axioms of Planar Euclidean Geometry} \label{pogAKS} In what follows, we will illustrate the axiomatic design of planar Euclidean geometry. We will list the initial concepts and initial statements - axioms, and then derive some new concepts and statements. We mention that we have chosen the axioms of the plane, because in this book we will only deal with the geometry of the Euclidean plane. Let $\mathcal{S}$ be a non-empty set. Its elements are called \index{point} \pojem{points} and we denote them with $A, B, C, \ldots$ Certain subsets of the set $\mathcal{S}$ are called \index{line} \pojem{lines} and we denote them with $a, b, c, \ldots$ The set $\mathcal{S}$ (the set of all points) is also called the \index{plane} \pojem{plane}. In addition to these basic concepts, there are also two relations on the set $\mathcal{S}$. The first is the \index{relation!$\mathcal{B}$} \pojem{relation $\mathcal{B}$} and it applies to three points. The fact that points $A$, $B$ and $C$ are in this relation, we will denote with $\mathcal{B}(A,B,C)$ and read: Point $B$ is between points $A$ and $C$. The second is the \index{relation!compatibility of point pairs} \pojem{relation of compatibility of point pairs}; the fact that pairs of points $A, B$ and $C, D$ are in this relation, we will denote with $(A,B) \cong (C,D)$ and read: The pair of points $(A,B)$ is compatible with the pair of points $(C,D)$. With the help of the aforementioned basic concepts, we can also define the following derived concepts: If point $A$ belongs to line $p$ ($A\in p$), or line $p$ contains point $A$ ($p\ni A$), we will say that point $A$\index{relation!lies on a line} \pojem{lies on} line $p$, or that line $p$ \index{relation!goes through a point}\pojem{goes through} point $A$. For three or more points we say that they are \index{collinear points}\pojem{collinear}, if they lie on the same line, otherwise they are \index{non-collinear points}\pojem{non-collinear}. Two different lines \pojem{intersect}, if their intersection (the intersection of two subsets) is not an empty set. We call their intersection the \index{intersection of two lines} \pojem{intersection} of two lines. Any non-empty subset $\Phi$ of the set $\mathcal{S}$ ($\Phi\subset\mathcal{S}$) is called a \index{figure} \pojem{figure}. We say that figures $\Phi_1$ and $\Phi_2$ \index{figures!coincide}\pojem{coincide} (or they are \index{figures!identical}\pojem{identical}), if $\Phi_1=\Phi_2$. Now we will also list some basic theorems - axioms. By their nature, they are divided into five groups: \begin{enumerate} \item incidence axioms (three axioms), \item ordering axioms (four axioms), \item congruence axioms (four axioms), \item continuity axiom (one axiom), \item Playfair's axiom (one axiom). \end{enumerate} %________________________________________________________________________________ \poglavje{Incidence Axioms} \label{odd2AKSINC} Because lines as basic notions represent certain sets of points, we can consider appropriate relations between elements and sets of points and lines: $\in$ and $\ni$ - relations we also call \index{relacija!incidencije}\pojem{relations of incidence}. These axioms describe just the basic properties of these relations (Figure \ref{sl.aks.2.1.1.pic}): \vspace*{3mm} \baksiom \label{AksI1} For every pair of distinct points $A$ and $B$ there is exactly one line $p$ such that $A$ and $B$ lie on $p$. \eaksiom \baksiom \label{AksI2} For every line there exist at least two distinct points such that both lie on. \eaksiom \baksiom \label{AksI3} There exist three points that do not all lie on any one line. \eaksiom \vspace*{3mm} \begin{figure}[!htb] \centering \input{sl.aks.2.1.1.pic} \caption{} \label{sl.aks.2.1.1.pic} \end{figure} From the first two axioms \ref{AksI1} and \ref{AksI2} it follows that each line is determined by its two distinct points. Therefore the line $p$, which is determined by the points $A$ and $B$, we also call the line $AB$. From the first axiom \ref{AksI1} it follows that the intersection of two lines that intersect is one point. If, for example, two lines had one more common point, according to this axiom they would coincide (they would be identical), but in the definition of lines that intersect, we required that they are different. The fact that the lines $p$ and $q$ intersect in the point $A$, we will write $p\cap q=\{A\}$ or shorter $p\cap q=A$ (Figure \ref{sl.aks.2.1.2.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.1.2.pic} \caption{} \label{sl.aks.2.1.2.pic} \end{figure} The third axiom \ref{AksI3} can also be expressed as follows: There exist at least three points that are not collinear. So we deduced the first consequences of the incidence axioms; for simplicity, we did not express them in the form of propositions. These are almost all the consequences that arise from the first group of axioms. Because of this, geometry, which is based solely on the axioms of incidence, is too simple. In it, we could only prove the existence of three points and three lines. So we need new axioms. %________________________________________________________________________________ \poglavje{Ordering Axioms} \label{odd2AKSURJ} The axioms in this group describe the basic characteristics of the relation $\mathcal{B}$, which we listed as the basic concept. \vspace*{3mm} \baksiom \label{AksII1} If $\mathcal{B} (A, B, C)$, then $A$, $B$ and $C$ are three distinct collinear points, and also $\mathcal{B} (C, B, A)$ (Figure \ref{sl.aks.2.2.1.pic}). \eaksiom \begin{figure}[!htb] \centering \input{sl.aks.2.2.1.pic} \caption{} \label{sl.aks.2.2.1.pic} \end{figure} \baksiom \label{AksII2} If $A$, $B$ and $C$ are three distinct collinear points, exactly one of the relations holds: $\mathcal{B}(A,B,C)$, $\mathcal{B}(A,C,B)$, $\mathcal{B}(C,A,B)$ (Figure \ref{sl.aks.2.2.2.pic}). \eaksiom \begin{figure}[!htb] \centering \input{sl.aks.2.2.2.pic} \caption{} \label{sl.aks.2.2.2.pic} \end{figure} \baksiom \label{AksII3} Given a pair of distinct points $A$ and $B$ there is a point $C$ on line $AB$, so that is $\mathcal{B}(A,B,C)$ (Figure \ref{sl.aks.2.2.3.pic}). \eaksiom \vspace*{-1mm} \begin{figure}[!htb] \centering \input{sl.aks.2.2.3.pic} \caption{} \label{sl.aks.2.2.3.pic} \end{figure} \baksiom \label{AksPascheva}\index{aksiom!Paschev} (Pasch's\footnote{\index{Pasch, M.} \textit{M. Pasch} (1843--1930), German mathematician, who introduced the concept of ordering points in his 'Lectures on Modern Geometry' from 1882. These axioms were later supplemented by Italian mathematician \index{Peano, G.} \textit{G. Peano} (1858--1932), in 'Principles of Geometry', then by German mathematician \index{Hilbert, D.}\textit{D. Hilbert} (1862--1943) in his famous book 'Foundations of Geometry' from 1899.} axiom) Let $A$, $B$ and $C$ be three noncollinear points and $l$ be a line that does not contain point $A$. If there is a point $P$ on $l$ that is $\mathcal{B}(B,P,C)$ then either $l$ contains a point $Q$ that is $\mathcal{B}(A,Q,C)$ or $l$ contains a point $R$ that is $\mathcal{B}(A,R,B)$ (Figure \ref{sl.aks.2.2.4.pic}). \eaksiom \begin{figure}[!htb] \centering \input{sl.aks.2.2.4.pic} \caption{} \label{sl.aks.2.2.4.pic} \end{figure} In the previous axiom, we did not particularly emphasize that the line $l$ lies in the plane $ABC$, because we are building a plane Euclidean geometry, where all points lie in the same plane. At this point we will not derive all the consequences of the ordering axioms. The formal derivation of all the facts is not so simple and would take up a lot of space. Most of the proofs can be found in \cite{Lucic}. We prove the first consequence of the ordering axioms. \bizrek \label{izrekAksUrACB} Given a pair of distinct points $A$ and $B$ there is a point $C$, so that is $\mathcal{B}(A,C,B)$. \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.2.5.pic} \caption{} \label{sl.aks.2.2.5.pic} \end{figure} \textbf{\textit{Proof.}} By Axiom \ref{AksI1} there exists exactly one line that goes through points $A$ and $B$ - we'll mark it with $AB$. By Axiom \ref{AksI3} there are at least three non-linear points. Therefore, there is at least one point outside of line $AB$ - we'll mark it with $D$ (Figure \ref{sl.aks.2.2.5.pic}). Next, by Axiom \ref{AksII3} there is such a point $E$, that $\mathcal{B}(B,D,E)$ is true, and then such a point $F$, that $\mathcal{B}(A,E,F)$ is true. $A$, $B$ and $E$ are non-linear points, because otherwise point $D$ would lie on line $AB$ (Axiom \ref{AksI1}). Line $FD$ does not go through point $A$, because by Axiom \ref{AksI1} points $F$, $D$, $A$ and $E$ would be linear, and so would point $B$. But that is not possible, because it would imply that point $D$ lies on line $AB$. We'll now use Pasch's Axiom \ref{AksPascheva} on points $A$, $B$ and $E$ and line $FD$. Line $FD$ intersects line $EB$ in such a point $D$, that $\mathcal{B}(B,D,E)$ is true, and therefore it intersects either line $AE$ in such a point $F$, that $\mathcal{B}(A,F,E)$ is true, or line $AB$ in such a point $C$, that $\mathcal{B}(A,C,B)$ is true. But since $\mathcal{B}(A,E,F)$ is already true, by Axiom \ref{AksII2} $\mathcal{B}(A,F,E)$ cannot be true as well. Therefore line $FD$ intersects line $AB$ in such a point $C$, for which $\mathcal{B}(A,C,B)$ is true. \kdokaz Relation $\mathcal{B}$ and the order axioms related to it, allow us to define new concepts. Let $A$ and $B$ be any two different points. \index{distance!open}\pojem{Open distance} $AB$ with notation $(AB)$ is the set of all points $X$, for which $\mathcal{B}(A,X,C)$ is true (Figure \ref{sl.aks.2.2.6.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.6.pic} \caption{} \label{sl.aks.2.2.6.pic} \end{figure} If we add points $A$ and $B$ to an open line segment $AB$, we get a \pojem{line segment} (or a \pojem{closed line segment}) $AB$, which we also denote with $[AB]$. Points $A$ and $B$ are its \pojem{endpoints}, and all other points on it are \pojem{interior points} of the line segment (Figure \ref{sl.aks.2.2.6.pic}). More formally: a line segment (or a closed line segment) is the union of an open line segment and the set $\{A,B\}$ or $[AB]=(AB)\cup \{A,B\}$. Similarly, we define a \pojem{half-open line segment}: $(AB]=(AB)\cup \{B\}$, or $[AB)=(AB)\cup \{A\}$ (Figure \ref{sl.aks.2.2.6a.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.6a.pic} \caption{} \label{sl.aks.2.2.6a.pic} \end{figure} From Axiom \ref{AksII1} it follows immediately that line segments $AB$ and $BA$ are the same. From the same axiom it also follows that line segment $AB$ is a subset of the line $AB$. Therefore, we say that line segment $AB$ \pojem{lies on the line} $AB$, and we call the line $AB$ the \pojem{line supporting the line segment} $AB$ (Figure \ref{sl.aks.2.2.6b.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.6b.pic} \caption{} \label{sl.aks.2.2.6b.pic} \end{figure} By Theorem \ref{izrekAksUrACB}, line segment $AB$ has, besides its endpoints $A$ and $B$, at least one more point $C_1$. In this way, we can get an infinite sequence of points $C_1$, $C_2$, ..., for which $\mathcal{B}(A, C_n, C_{n-1})$ is true ($n\in \{2,3,\cdots\}$) (Figure \ref{sl.aks.2.2.6c.pic}). At this point, we will not formally prove the fact that all points in the sequence are different and that all of them lie on line segment $AB$. From this statement it follows that every line segment (and consequently every line) has infinitely many points. \begin{figure}[!htb] \centering \input{sl.aks.2.2.6c.pic} \caption{} \label{sl.aks.2.2.6c.pic} \end{figure} Let's define the relation $\mathcal{B}$, which relates to more than three collinear points. We say that $\mathcal{B}(A_1,A_2,\ldots,A_n)$ ($n\in\{4,5,\ldots\}$), if for every $k\in\{1,2,\ldots,n-2\}$ it holds that $\mathcal{B}(A_k,A_{k+1},A_{k+2})$ (Figure \ref{sl.aks.2.2.6d.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.6d.pic} \caption{} \label{sl.aks.2.2.6d.pic} \end{figure} Let $S$ be a point that lies on the line $p$. On the set $p\setminus \{S\}$ (all points of the line $p$ without the point $S$) we define two relations. We say that the points $A$ and $B$ ($A,B\in p\setminus \{S\}$) \index{relacija!na različnih straneh točke} \pojem{on different sides of the point} $S$ (which we denote by $A,B\div S$), if $B(A,S,B)$, otherwise the points $A$ and $B$ ($A,B\in p\setminus \{S\}$) \index{relacija!na isti strani točke} \pojem{on the same side of the point} $S$ (which we denote by $A,B\ddot{-} S$). So for points $A,B\in p\setminus \{S\}$ it holds that $A,B\ddot{-} S$, if it is not $A,B\div S$ (Figure \ref{sl.aks.2.2.7.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.7.pic} \caption{} \label{sl.aks.2.2.7.pic} \end{figure} Let $A$ and $B$ be two different points. The set of all such points $X$, for which $B,X\ddot{-} A$ including the point $A$, we call \index{poltrak}\pojem{poltrak} $AB$ with \pojem{starting point} or \pojem{origin} $A$. The line $AB$ is \index{nosilka!poltraka} \pojem{the carrier of the poltrak} $AB$ (Figure \ref{sl.aks.2.2.8.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.8.pic} \caption{} \label{sl.aks.2.2.8.pic} \end{figure} From the definition itself it follows that the poltrak is a subset of its carrier or that it lies on its carrier. From the relation $B,X\ddot{-} A$ it follows that $B$, $X$ and $A$ are collinear points, so the point $X$ lies on the line $AB$. We will not prove other important properties of lines and line segments that we will use later. Let's mention some of these properties. If $C$ is the an interior point of the line segment $AB$, then that line segment can be expressed as the union the line segments $AC$ and $CB$. \bizrek \label{izrekAksIIDaljica} If $C$ is an interior point of the line segment $AB$, then that line segment can be expressed as the union the line segments $AC$ and $CB$ (Figure \ref{sl.aks.2.2.9.pic}). \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.2.9.pic} \caption{} \label{sl.aks.2.2.9.pic} \end{figure} \bizrek \label{izrekAksIIPoltrak} Each point lying on the line determines exactly two rays on it. The union of these rays is equal to that line (Figure \ref{sl.aks.2.2.9.pic}). \eizrek The proof of the previous statement is based on the fact that the relation $\ddot{-} A$ is equivalent to the relation that has two classes. Each of the classes is a suitable open ray. The rays from the previous statement, which are determined by the same initial point on the line, are called \index{poltrak!komplementarni}\pojem{complementary rays}. The concepts of line and ray allow us to define new concepts. Let $A_1$, $A_2$, ... $A_n$ be such points in the plane that no three of them are collinear. The union of the lines $A_1A_2$, $A_2A_3$,... $A_{n-1}A_n$ is called \index{lomljenka} \pojem{broken line} $A_1A_2\cdots A_n$ or \index{poligonska črta}\pojem{polygonal line} $A_1A_2\cdots A_n$ (Figure \ref{sl.aks.2.2.10.pic}). The points $A_1$, $A_2$, ... $A_n$ are \index{oglišče!lomljenke} \pojem{vertices of the broken line}, the lines $A_1A_2$, $A_2A_3$,... $A_{n-1}A_n$ are \index{stranica!lomljenke} \pojem{sides of the broken line}. The sides of the broken line with a common vertex are \index{sosednji stranici!lomljenke} \pojem{adjacent sides of the broken line}. \begin{figure}[!htb] \centering \input{sl.aks.2.2.10.pic} \caption{} \label{sl.aks.2.2.10.pic} \end{figure} If the sides of a broken line do not have common points, except for adjacent sides that have a common vertex, such a broken line is called a \index{lomljenka!enostavna} \pojem{simple broken line} (Figure \ref{sl.aks.2.2.10a.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.10a.pic} \caption{} \label{sl.aks.2.2.10a.pic} \end{figure} A broken line $A_1A_2\cdots A_nA_{n+1}$, for which $A_{n+1}=A_1$ and $A_n$, $A_1$ and $A_2$ are non-linear points, is called a \index{lomljenka!sklenjena} \pojem{closed broken line} $A_1A_2\cdots A_n$ (Figure \ref{sl.aks.2.2.10b.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.10b.pic} \caption{} \label{sl.aks.2.2.10b.pic} \end{figure} We will be particularly interested in \pojem{simple closed broken lines} (Figure \ref{sl.aks.2.2.10b.pic}). Let $p$ and $q$ be two semi-lines with a common starting point $O$ (Figure \ref{sl.aks.2.2.10c.pic}). The union of these two semi-lines is called a \index{kotna lomljenka} \pojem{angular broken line} $pq$ (or $pOq$). \begin{figure}[!htb] \centering \input{sl.aks.2.2.10c.pic} \caption{} \label{sl.aks.2.2.10c.pic} \end{figure} For a figure $\Phi$ we say that it is \index{lik!konveksen}\pojem{convex}, if for any two of its points $A,B\in \Phi$ the distance $AB$ is a subset of this figure or if the following is true (Figure \ref{sl.aks.2.2.10d.pic}): $$(\forall A)(\forall B)\hspace*{1mm} (A,B\in \Phi \Rightarrow [AB]\subseteq \Phi).$$ For a figure that is not convex, we say that it is \index{lik!nekonveksen}\pojem{non-convex} (Figure \ref{sl.aks.2.2.10d.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.10d.pic} \caption{} \label{sl.aks.2.2.10d.pic} \end{figure} It follows directly from the definition that a straight line is a convex figure. As a result of the axioms of this group, it can be proved that a distance and a semi-line are convex figures. We say that a figure $\Phi$ is \index{figure!connected}\pojem{connected}, if for every two of its points $A,B\in \Phi$ there exists a broken line $AT_1T_2\cdots T_nB$, which is a subset of this figure, or if the following is true (Figure \ref{sl.aks.2.2.10e.pic}): $$(\forall A\in \Phi)(\forall B\in \Phi)(\exists T_1,T_2,\cdots , T_n)\hspace*{1mm} AT_1T_2\cdots T_nB\subseteq \Phi.$$ \begin{figure}[!htb] \centering \input{sl.aks.2.2.10e.pic} \caption{} \label{sl.aks.2.2.10e.pic} \end{figure} A figure that is not connected, we call \index{figure!not connected}\pojem{not connected}. It is clear that every convex figure is also connected. For the broken line it is enough to take the distance $AB$. The converse is of course not true. There are figures that are connected, but not convex, which we will discover later. Now we will define two relations that are analogous to the relations $\ddot{-} S$ and $\div S$. Let $p$ be a line that lies in the plane $\alpha$ (because we axiomatically build only the Euclidean geometry of the plane, in fact all the points that exist to us are in this plane). On the set $\alpha\setminus p$ (all points except the points of the line $p$) we define two relations. We say that the points $A$ and $B$ ($A,B\in \alpha\setminus p$) are \index{relation!on different sides of the line} \pojem{on different sides of the line} $p$ (which we denote by $A,B\div p$), if the line $AB$ has a common point with the line $p$, otherwise the points $A$ and $B$ ($A,B\in \alpha\setminus p$) are \index{relation!on the same side of the line} \pojem{on the same side of the line} $p$ (which we denote by $A,B\ddot{-} p$). So for points $A,B\in \alpha\setminus p$ it is $A,B\ddot{-} p$, if it is not $A,B\div p$ (Figure \ref{sl.aks.2.2.11.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.11.pic} \caption{} \label{sl.aks.2.2.11.pic} \end{figure} Let $A$ be a point that does not lie on the line $p$. The set of all such points $X$, for which $A,X\ddot{-} p$, is called \index{polravnina!odprta}\pojem{open half-line} $pA$. The union of the open half-lines $pA$ and the line $p$ is the \index{polravnina!zaprta}\pojem{closed half-line} or just \index{polravnina}\pojem{half-line} $pA$. The line $p$ is the \index{rob!polravnine} \pojem{edge} of this half-line (Figure \ref{sl.aks.2.2.11a.pic}). If the points $B$ and $C$ lie on the edge $p$ of the half-line $pA$, we will call this half-line the half-line $BCA$. In addition, we will denote half-lines by Greek letters $\alpha$, $\beta$, $\gamma$,... \begin{figure}[!htb] \centering \input{sl.aks.2.2.11a.pic} \caption{} \label{sl.aks.2.2.11a.pic} \end{figure} Similarly to the case of a segment, it can be shown (as a consequence of the axioms of this group), that each line $p$ in the plane determines two half-lines $\alpha$ and $\alpha'$, which have the line $p$ as an edge (Figure \ref{sl.aks.2.2.11a.pic}). We say that in this case $\alpha$ and $\alpha'$ are \index{polravnina!komplementarna}\pojem{complementary half-lines}. It turns out that the union of two complementary half-lines is the whole plane. Similarly to the case of a segment, the proof of these statements is based on the fact that the relation $\ddot{-} p$ is equivalent to the relation with two classes. Each of the classes is the appropriate open half-line. Let $pq$ or $pOq$ be an angle. Define a new relation on the set of all points of the plane except the points that lie on the angle. We say that the points $A$ and $B$ are on the same side of the angle $pq$ (which we denote by $A,B\ddot{-} pq$), if there exists an angle $AT_1T_2\cdots T_nB$, which does not intersect the angle $pq$ or does not have any common points with it (Figure \ref{sl.aks.2.2.12.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.12.pic} \caption{} \label{sl.aks.2.2.12.pic} \end{figure} The relation $\ddot{-} pq$ is also an equivalence relation that has two classes. The union of each of these two classes with the angular bracket $pq$ is called \index{kot}\pojem{the angle} $pq$, which is denoted by $\angle pq$, or $\angle pOq$. The angular bracket therefore determines two angles. We will soon resolve the dilemma of which angle is meant by the notation $\angle pOq$. The segments $p$ and $q$ are \index{krak!kota}\pojem{the sides of the angle} and the point $O$ \index{vrh kota}\pojem{the vertex of the angle}. If $P\in p$ and $Q\in q$ are points that lie on the sides of the angle $pOq$ and differ from its vertex $O$, we will also call the angle $POQ$ and denote it by $\angle POQ$ (Figure \ref{sl.aks.2.2.12a.pic}). If we know which angle it is, we will denote it by its vertex: $\angle O$. We will also denote angles by Greek letters $\alpha$, $\beta$, $\gamma$,... All points of the angle $pOq$, which do not lie on either of the sides $p$ and $q$, are called \index{notranje točke!kota} \pojem{internal points of the angle}, the set of all these points is called \index{notranjost!kota}\pojem{the interior of the angle}. It is clear that these are points of the appropriate class determined by the relation $\ddot{-} pq$. The points of the other class are \index{zunanje!točke kota}\pojem{external points of the angle}, the whole class is \index{zunanjost!kota}\pojem{the exterior of the angle}. The points that lie on the sides, or on the angular bracket $pOq$, which determines the angle $pOq$, are \index{robne točke!kota}\pojem{the boundary points of the angle}, the whole bracket is \index{rob!kota}\pojem{the boundary of the angle}. \begin{figure}[!htb] \centering \input{sl.aks.2.2.12a.pic} \caption{} \label{sl.aks.2.2.12a.pic} \end{figure} If the sides of the angle are complementary segments, such an angle is called \index{kot!iztegnjeni}\pojem{an extended angle} (Figure \ref{sl.aks.2.2.12b.pic}). As a set of points, this angle is essentially the same as the half-line with the boundary that is the carrier of both sides. \begin{figure}[!htb] \centering \input{sl.aks.2.2.12b.pic} \caption{} \label{sl.aks.2.2.12b.pic} \end{figure} If the angular bisector $pOq$ does not determine the extended angle or is not equal to the line, it turns out that $pOq$ determines two angles that represent a convex and a concave shape - we call them the \index{kot!konveksen}\pojem{convex (protruding) angle} and the \index{kot!nekonveksen}\pojem{concave (indented) angle}. We will omit the formal proof of this fact here. Unless otherwise stated, we will always mean the convex angle under the label $\angle pOq$ (or $\angle pq$ or $\angle POQ$). In this sense, it is clear from the definition of the angle that (convex) angle $pOq$ and $qOp$ represent the same angle. The angle $pOq$ and $qOr$, which have a common leg $q$, which is also their intersection (as a set of points), are \index{kot!sosednji}\pojem{adjacent angles} (Figure \ref{sl.aks.2.2.12c.pic}). If in addition the segments $p$ and $r$ are complementary (or determine the extended angle), we say that $pOq$ and $qOr$ are \index{kota!sokota}\pojem{sokota} (Figure \ref{sl.aks.2.2.12c.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.12c.pic} \caption{} \label{sl.aks.2.2.12c.pic} \end{figure} The angle $pOq$ and $rOs$ are \index{kota!sovršna}\pojem{sovršna kota}, if $p$ and $r$ or $q$ and $s$ are a pair of complementary (supplementary) segments (Figure \ref{sl.aks.2.2.12d.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.12d.pic} \caption{} \label{sl.aks.2.2.12d.pic} \end{figure} Let $A_1A_2\cdots A_n$ ($n\in \{3,4,5,\cdots\}$) be a simple closed curve. Similarly to the angular bisector, on the set of all points in the plane except for the points that lie on the curve $A_1A_2\cdots A_n$, we can define the following relation: we say that the points $B$ and $C$ are on the same side of the simple closed curve $A_1A_2\cdots A_n$ (which we denote by $B,C\ddot{-} A_1A_2\cdots A_n$), if there exists a curve $BT_1T_2\cdots T_nC$, which does not intersect the simple closed curve $A_1A_2\cdots A_n$ or does not have any common points with it (Figure \ref{sl.aks.2.2.13.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.13.pic} \caption{} \label{sl.aks.2.2.13.pic} \end{figure} It can also be proven in this case that it is an equivalence relation with two classes - for one class there is a line that lies entirely within it, but for the other class there is no such line. The union of the class that does not contain any line (intuitively - the one that is limited) and the simple closed broken line, $A_1A_2\cdots A_n$ is called \index{polygon}\pojem{polygon} $A_1A_2\cdots A_n$ or \index{$n$-gon}\pojem{$n$-gon} $A_1A_2\cdots A_n$ (Figure \ref{sl.aks.2.2.13a.pic}). All points of the aforementioned class that do not contain any line are called \index{internal points!of a polygon} \pojem{internal points of a polygon}, the entire class is \index{interior!of a polygon}\pojem{interior of a polygon}. The points of the other class are \index{external!points of a polygon}\pojem{external points of a polygon}, the entire class is \index{exterior!of a polygon}\pojem{exterior of a polygon}. The points that lie on the broken line $A_1A_2\cdots A_n$, which determines the polygon, are \index{vertices!of a polygon}\pojem{vertices of a polygon}, the entire broken line is \index{border!of a polygon}\pojem{border of a polygon}. \begin{figure}[!htb] \centering \input{sl.aks.2.2.13a.pic} \caption{} \label{sl.aks.2.2.13a.pic} \end{figure} An equivalent statement holds true for the internal points of a polygon. We will state the statement without proof. \bizrek A point $N$ is an interior point of a polygon if and only if any ray from the point $N$, that does not contain the vertices of the polygon, intersects an odd number of sides of the polygon (Figure \ref{sl.aks.2.2.13b.pic}). \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.2.13b.pic} \caption{} \label{sl.aks.2.2.13b.pic} \end{figure} The points $A_1$, $A_2$,..., $A_n$ (the vertices of the broken line) are the \index{oglišče!večkotnika}\pojem{vertices of the polygon}, the segments $A_1A_2$, $A_2A_3$, ... $A_{n-1}A_n$, $A_nA_1$ are the \index{stranica!večkotnika}\pojem{sides of the polygon}. The lines $A_1A_2$, $A_2A_3$, ... $A_{n-1}A_n$, $A_nA_1$ are the \index{nosilka!stranice}\pojem{supports of the sides} $A_1A_2$, $A_2A_3$, ... $A_{n-1}A_n$, $A_nA_1$. The sides that contain a common vertex are the \index{stranica!sosednja}\pojem{adjacent sides}, otherwise the sides are \index{stranica!nesosednja}\pojem{non-adjacent}. If the vertices are at the same time the endpoints of the same side, we say that the vertices are \index{oglišče!sosednje}\pojem{adjacent}, otherwise the vertices are \index{oglišče!nesosednje}\pojem{non-adjacent}. It is clear from the definition that each vertex has exactly two adjacent vertices. Similarly, each side has exactly two adjacent sides. The segment determined by two non-adjacent vertices is called the \index{diagonala!večkotnika}\pojem{diagonal of the polygon} (Figure \ref{sl.aks.2.2.13c.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.13c.pic} \caption{} \label{sl.aks.2.2.13c.pic} \end{figure} The following statement is true for the diagonals of the polygon. \bizrek The number of diagonals of an $n$-gon is $\frac{n(n-3)}{2}$. \eizrek We will give the proof of this statement in section \ref{odd3Helly}, where we will separately consider the combinatorial properties of sets of points in the plane. Let's define the angles of a polygon. Let $O$ be any vertex of a polygon, and $P$ and $Q$ be its two adjacent vertices. The line segments $OP$ and $OQ$ are denoted by $p$ and $q$. In this case, the angular bisector $pOq$ determines two angles. The angle for which it is true that every line segment with initial point $O$, which belongs to this angle and does not contain any other vertices of the polygon, intersects the edge of the polygon except at point $O$ in an odd number of points, is called the \index{angle!inner polygon}\pojem{inner angle of the polygon} or, more briefly, the \index{angle!inner}\pojem{angle of the polygon} at vertex $O$ (Figure \ref{sl.aks.2.2.13d.pic}). If the inner angle of the polygon is convex, its supplement is called the \index{angle!outer polygon}\pojem{outer angle of the polygon} (Figure \ref{sl.aks.2.2.13d.pic}). The angles of the polygon are \index{angle!adjacent}\pojem{adjacent angles}, if their vertices are adjacent vertices of the polygon, otherwise the angles are \index{angle!non-adjacent}\pojem{non-adjacent}. \begin{figure}[!htb] \centering \input{sl.aks.2.2.13d.pic} \caption{} \label{sl.aks.2.2.13d.pic} \end{figure} The most simple $n$-gon and at the same time one of the most used shapes in the geometry of a plane is the case $n=3$ - \index{trikotnik}\pojem{trikotnik}. In the case of the triangle $ABC$ (we will mark it with $\triangle ABC$), the points $A$, $B$ and $C$ are its \pojem{oglišča}, and the lines $AB$, $BC$ and $CA$ are its \index{stranica!trikotnika}\pojem{stranice} (Figure \ref{sl.aks.2.2.14.pic}). Obviously, every two sides of the triangle are adjacent. The same goes for every two vertices. The triangle therefore has no diagonal. We say that the vertex $A$ ($B$ and $C$) or the angle $BAC$ ($ABC$ and $ACB$) is the \index{oglišče!nasprotno trikotnika}\pojem{nasprotno oglišče} or the \index{kot!nasprotni trikotnika}\pojem{nasprotni kot} of the side $BC$ ($AC$ and $AB$) of the triangle $ABC$. And also the side $BC$ ($AC$ and $AB$) is the \index{stranica!nasprotna trikotnika}\pojem{nasprotna stranica} of the vertex $A$ ($B$ and $C$) or the angle $BAC$ ($ABC$ and $ACB$) of the triangle $ABC$. The angles (internal) of the triangle $ABC$ at the vertices $A$, $B$ and $C$ are often marked with $\alpha$, $\beta$ and $\gamma$, the corresponding external ones with $\alpha_1$, $\beta_1$ and $\gamma_1$. \begin{figure}[!htb] \centering \input{sl.aks.2.2.14.pic} \caption{} \label{sl.aks.2.2.14.pic} \end{figure} The triangle $ABC$ can also be defined as the intersection of the half-planes $ABC$, $ACB$ and $BCA$. We will not prove the equivalence of these two definitions here. Pasch's axiom in terms of triangles can now be expressed in a shorter form: If a line in the plane of a triangle intersects one of its sides and does not pass through any of its vertices, then intersects exactly one more side of this triangle \bizrek \label{PaschIzrek} If a line, not passing through any vertex of a triangle, intersects one side of the triangle then the line intersects exactly one more side of the triangle (Figure \ref{sl.aks.2.2.14a.pic}). \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.2.14a.pic} \caption{} \label{sl.aks.2.2.14a.pic} \end{figure} In the case of $n=4$ for the $n$-gon we get \index{štirikotnik}\pojem{štirikotnik}. Because each vertex of the quadrilateral has exactly one non-adjacent vertex, we will also call this vertex the \index{oglišče!nasprotno štirikotnika}\pojem{opposite vertex} of the quadrilateral. Similarly, the non-adjacent sides will be the \index{stranica!nasprotna štirikotnika}\pojem{opposite sides} of the quadrilateral. The diagonal of the quadrilateral is therefore determined by the opposite vertices. From the definition itself, it is clear that the quadrilateral has two diagonals\index{diagonala!štirikotnika} (Figure \ref{sl.aks.2.2.14b.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.14b.pic} \caption{} \label{sl.aks.2.2.14b.pic} \end{figure} For the non-adjacent angle of the quadrilateral, we also say that they are the \index{kot!nasprotni štirikotnika}\pojem{opposite angles} of the quadrilateral. The angles (internal) of the quadrilateral $ABCD$ at the vertices $A$, $B$, $C$ and $D$ are usually denoted by $\alpha$, $\beta$, $\gamma$ and $\delta$, and the corresponding external (those that exist) by $\alpha_1$, $\beta_1$, $\gamma_1$ and $\delta_1$ (Figure \ref{sl.aks.2.2.14c.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.14c.pic} \caption{} \label{sl.aks.2.2.14c.pic} \end{figure} As a result of the axioms of order, we will also introduce the concepts of orientation of a triangle and orientation of an angle in this section. The triangle $ABC$, for which the vertices are a triple $(A,B,C)$, is called the \index{orientacija!trikotnika} \pojem{oriented triangle}. We say that the oriented triangles $ABC$ and $BCA'$ are of the \pojem{same orientation}, if $A,A'\ddot{-} BC$, and of the opposite orientation, if $A,A'\div BC$ (Figure \ref{sl.aks.2.2.15.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.15.pic} \caption{} \label{sl.aks.2.2.15.pic} \end{figure} When we talk about the orientation of two triangles, from now on we will always mean an oriented triangle (we will often omit the word oriented). Triangles $ABC$ and $A'B'C'$ are of the \pojem{same orientation} or are \pojem{equally oriented}, if there exists such a sequence of triangles: $\triangle ABC=\triangle P_1P_2P_3$, $\triangle P_2P_3P_4$, $\triangle P_3P_4P_5$, ..., $\triangle P_{n-2}P_{n-1}P_n=\triangle A'B'C'$, that in this sequence the number of changes in orientation of two adjacent triangles is even (Figure \ref{sl.aks.2.2.15a.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.15a.pic} \caption{} \label{sl.aks.2.2.15a.pic} \end{figure} It can be proven that the relation of the same orientation of two triangles is equivalent to a relation that has two classes. For two triangles that are not in the same class, we say that they are of \pojem{different orientation} or are \pojem{differently oriented}. Each of the two classes determines the \index{orientation!plane}\pojem{orientation of the plane}. We call them the \pojem{positive orientation} and the \pojem{negative orientation}. For the sake of easier understanding, let us agree that the orientation that corresponds to the direction of the rotation of the hour hand is negative, and the opposite is the positive orientation (Figure \ref{sl.aks.2.2.15b.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.2.15b.pic} \caption{} \label{sl.aks.2.2.15b.pic} \end{figure} Let us also define the orientation of angles. Angles $ASB$ and $A'S'B'$, neither of which is a right angle, are of the \pojem{same orientation}, if: \begin{itemize} \item both are convex or both are concave, and triangles $ASB$ and $A'S'B'$ are of the same orientation (Figure \ref{sl.aks.2.2.16.pic}), \item one angle is convex and the other is concave, and triangles $ASB$ and $A'S'B'$ are of opposite orientation (Figure \ref{sl.aks.2.2.16c.pic}). \end{itemize} \begin{figure}[!htb] \centering \input{sl.aks.2.2.16.pic} \caption{} \label{sl.aks.2.2.16.pic} \end{figure} \begin{figure}[!htb] \centering \input{sl.aks.2.2.16c.pic} \caption{} \label{sl.aks.2.2.16c.pic} \end{figure} If $\angle ASB$ is an obtuse angle, $\angle A'S'B'$ is a convex angle, then the angles $ASB$ and $A'S'B'$ have the same orientation, if there is a point $C$ inside the angle $ASB$, such that the angles $ASC$ and $A'S'B'$ have the same orientation (Figure \ref{sl.aks.2.2.16d.pic}). We do the same if the angle $A'S'B'$ is obtuse or if both angles $ASB$ and $A'S'B'$ are obtuse. \begin{figure}[!htb] \centering \input{sl.aks.2.2.16d.pic} \caption{} \label{sl.aks.2.2.16d.pic} \end{figure} It turns out that the relation of the same orientation of angles is an equivalent relation that has two classes. In this case, the positive orientation of the angle represents the class in which the triangle $ASB$ has a negative orientation for the convex angle $ASB$ from this class (Figure \ref{sl.aks.2.2.16a.pic}). In this sense, the angles $ASB$ and $BSA$ are oriented in the opposite direction. \index{orientation!angle} \pojem{Oriented angle} $ASB$ we will denote with $\measuredangle ASB$. \begin{figure}[!htb] \centering \input{sl.aks.2.2.16a.pic} \caption{} \label{sl.aks.2.2.16a.pic} \end{figure} \begin{figure}[!htb] \centering \input{sl.aks.2.2.16b.pic} \caption{} \label{sl.aks.2.2.16b.pic} \end{figure} If $C$ is an arbitrary point that does not lie on the edge of the angle $ASB$, we will define the sum of the oriented angles $\measuredangle ASC$ and $\measuredangle CSB$ (Figure \ref{sl.aks.2.2.16b.pic}): \begin{eqnarray} \measuredangle ASC+\measuredangle CSB = \measuredangle ASB. \label{orientKotVsota} \end{eqnarray} %________________________________________________________________________________ \poglavje{Congruence Axioms} \label{odd2AKSSKL} The following axioms are needed to introduce the concept and properties of the congruence of figures. With the previous axioms, we could introduce and consider the concepts: distance, segment, angle, polygon, ... but not the concepts related to congruence: circle, right angle, congruence of triangles,~... The intuitive idea of the compatibility of shapes that we used in elementary school is associated with the movement that the first shape transforms into the other. We will now use this idea to formally define the concept of compatibility and its properties. We will first start with the basic, already mentioned, concept of compatibility of pairs of points $(A,B)\cong (C,D)$ (Figure \ref{sl.aks.2.3.1.pic}) and formally define the concept of ‘‘movement’’. \begin{figure}[!htb] \centering \input{sl.aks.2.3.1.pic} \caption{} \label{sl.aks.2.3.1.pic} \end{figure} Using the compatibility of pairs of points, we first define the compatibility of an $n$-tuple of points. We say that two $n$-tuples of points are compatible (Figure \ref{sl.aks.2.3.2.pic}) or $$(A_1 , A_2,\ldots ,A_n ) \cong ( A'_1 , A'_2 ,\ldots , A'_n ),$$ if: $(A_i,A_j)\cong (A'_i,A'_j)$ for every $i,j\in \{1,2,\ldots, n\}$. \begin{figure}[!htb] \centering \input{sl.aks.2.3.2.pic} \caption{} \label{sl.aks.2.3.2.pic} \end{figure} A bijective mapping of a plane into a plane $\mathcal{I}:\mathcal{S}\rightarrow \mathcal{S}$ is \index{izometrija}\pojem{izometrija} or \pojem{izometrijska transformacija}, if it preserves the relation of compatibility of pairs of points (Figure \ref{sl.aks.2.3.3.pic}) or if for every two points $A$ and $B$ it holds: $$(\mathcal{I}(A),\mathcal{I}(B))\cong (A,B).$$ \begin{figure}[!htb] \centering \input{sl.aks.2.3.3.pic} \caption{} \label{sl.aks.2.3.3.pic} \end{figure} With the following axioms, we will introduce the properties of the newly defined mapping. \vspace*{3mm} \baksiom \label{aksIII1} Isometries preserve the relation $\mathcal{B}$ (Figure \ref{sl.aks.2.3.4.pic}), which means that for every isometry $\mathcal{I} $ holds: $$\mathcal{I}: A, B,C\mapsto A',B',C'\hspace*{2mm}\wedge \hspace*{2mm} \mathcal{B}(A,B,C) \hspace*{1mm}\Rightarrow\hspace*{1mm} \mathcal{B}(A',B',C').$$ \eaksiom \begin{figure}[!htb] \centering \input{sl.aks.2.3.4.pic} \caption{} \label{sl.aks.2.3.4.pic} \end{figure} \baksiom \label{aksIII2} If $ABC$ and $A'B'C'$ are two half-planes (Figure \ref {asl.aks.2.3.5.pic}), then there is a single isometry $\mathcal{I}$, which maps: \begin{itemize} \item point $A$ to point $A'$, \item half-line $AB$ in half-line $A'B'$, \item half-plane $ABC$ to half-plane $A'B'C'$. \end{itemize} If $(A,B)\cong (A',B')$ holds, then it is $\mathcal{I}(B)=B'$. \\ If in addition $(A,B,C)\cong (A',B',C')$ also holds, then it is $\mathcal{I}(B)=B'$ and $\mathcal{I}(C)=C'$. \eaksiom \begin{figure}[!htb] \centering \input{sl.aks.2.3.5.pic} \caption{} \label{asl.aks.2.3.5.pic} \end{figure} \baksiom \label{aksIII3} For every two points $A$ and $B$ there exists isometry such that holds $$\mathcal{I}: A, B\mapsto B,A.$$ If $(S,A)\cong (S,B)$ and $S\in AB$, then for each isometry $\mathcal{I}$ with this property holds $\mathcal{I}(S)=S$ (Figure \ref{sl.aks.2.3.6.pic}). \eaksiom \begin{figure}[!htb] \centering \input{sl.aks.2.3.6.pic} \caption{} \label{sl.aks.2.3.6.pic} \end{figure} \baksiom \label{aksIII4} The set of all isometries with respect to the composition of mappings form a group, which means that: \begin{itemize} \item composition of two isometries $\mathcal{I}_2\circ \mathcal{I}_1$ is isometry, \item identity map $\mathcal{E}$ is isometry, \item if $\mathcal{I}$ is isometry, then its inverse transformation $\mathcal{I}^{-1}$ is also isometry. \end{itemize} \eaksiom \vspace*{3mm} We mention that in the structure of a group the property of associativity is also required, i.e. $\mathcal{I}_1\circ (\mathcal{I}_2\circ \mathcal{I}_3)= (\mathcal{I}_1\circ \mathcal{I}_2)\circ \mathcal{I}_3$ (for any isometries $\mathcal{I}_1$, $\mathcal{I}_2$ and $\mathcal{I}_3$), which is automatically fulfilled in the operation of the composition of functions. We also mention the \pojem{identity} \index{identity} $\mathcal{E}$ from the previous axiom of the mapping, for which $\mathcal{E}(A)=A$ for every point on the plane. The mapping $\mathcal{I}^{-1}$ is the \pojem{inverse mapping} for the isometry $\mathcal{I}$, if $\mathcal{I}^{-1}\circ \mathcal{I} =\mathcal{I}\circ\mathcal{I}^{-1}=\mathcal{E}$. According to the previous axiom, the identity and the inverse mapping are therefore also isometries of every isometry. We prove the first consequences of the compatibility axioms. First, we will consider the following properties of isometries. \bizrek \label{izrekIzoB} Isometry maps a line to a line, a line segment to a line segment, a ray to a ray, a half-plane to a half-plane, an angle to an angle and an $n$-gon to an $n$-gon. \eizrek \textbf{\textit{Proof.}} According to axiom \ref{aksIII1}, isometries preserve the relation $\mathcal{B}$. Therefore, all points of the line $AB$ are mapped by isometry $I$ into points that lie on the line $A'B'$, where $A'=\mathcal{I}(A)$ and $B'=\mathcal{I}(B)$. Since the inverse mapping $\mathcal{I}^{-1}$ is also an isometry (axiom \ref{aksIII4}), each point of the line $A'B'$ is the image of some point that lies on the line $AB$. So with isometry $\mathcal{I}$ the line $AB$ is mapped into the line $A'B'$. We defined the other shapes from the statement using the relation $\mathcal{B}$, so the proof is similar to the one for the line. \kdokaz From the proof of the previous statement it follows that the endpoints of the line $AB$ are mapped by isometry into the endpoints of the image $A'B'$. In a similar way, the starting point of the ray is mapped into the starting point of the ray, the edge of the half-plane into the edge of the half-plane, the vertex of the angle into the vertex of the angle and the vertices of the polygon into the vertices of the polygon. Isometries are defined as bijective maps, which preserve the congruence of pairs of points. But does it also hold true, that for every congruent pair of points there is an isometry, which maps the first pair into the second one? Let us answer this question with the following theorem. \bizrek \label{izrekAB} If $(A,B)\cong (A',B')$, then there is an isometry $\mathcal{I}$, which maps the points $A$ and $B$ to the points $A'$ and $B'$, i.e.: $$\mathcal{I}: A, B\mapsto A',B'.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.3.7.pic} \caption{} \label{sl.aks.2.3.7.pic} \end{figure} \textbf{\textit{Proof.}} Let $C$ be a point, which does not lie on the line $AB$, and $C'$ be a point, which does not lie on the line $A'B'$ (Figure \ref{sl.aks.2.3.7.pic}). By Axiom \ref{aksIII2} there is a single isometry $\mathcal{I}$, which maps the point $A$ into the point $A'$, the segment $AB$ into the segment $A'B'$ and the plane $ABC$ into the plane $A'B'C'$. Because of $(A,B)\cong (A',B')$ from the same Axiom \ref{aksIII2}, it follows that $\mathcal{I}(B)=B'$. \kdokaz The proof of the following theorem, which will later be stated in a different form as the first theorem about the congruence of triangles, is similar. \bizrek \label{IizrekABC} Let $(A,B,C)$ and $(A',B',C')$ be triplets of non-collinear points such that $$(A,B,C)\cong (A',B',C'),$$ then there is a single isometry $\mathcal{I}$, that maps the points $A$, $B$ and $C$ into the points $A'$, $B'$ and $C'$, i.e.: $$\mathcal{I}: A, B,C\mapsto A',B',C'.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.3.5.pic} \caption{} \label{sl.aks.2.3.5.pic} \end{figure} \textbf{\textit{Proof.}} By Axiom \ref{aksIII2} there exists a single isometry $\mathcal{I}$, that maps point $A$ to point $A'$, line segment $AB$ to line segment $A'B'$ and plane $ABC$ to plane $A'B'C'$ (Figure \ref{sl.aks.2.3.5.pic}). Because, by the assumption $(A,B,C)\cong (A',B',C')$ from the same Axiom \ref{aksIII2}, it follows that $\mathcal{I}(B)=B'$ and $\mathcal{I}(C)=C'$. It is necessary to prove that $\mathcal{I}$ is the only such isometry. Assume that there exists such an isometry $\mathcal{\widehat{I}}$, that $\mathcal{\widehat{I}}: A, B,C\mapsto A',B',C'$. By Theorem \ref{izrekIzoB} isometry $\mathcal{\widehat{I}}$ also maps line segment $AB$ to line segment $A'B'$ and plane $ABC$ to plane $A'B'C'$. From Axiom \ref{aksIII2} it follows that $\mathcal{\widehat{I}}=\mathcal{I}$. \kdokaz A direct consequence is the following theorem. \bizrek \label{IizrekABCident} Let $A$, $B$ and $C$ be three non-collinear points, then the identity map $\mathcal{E}$ is the only isometry that maps points $A$, $B$, and $C$ to the same points $A$, $B$ and $C$. \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.3.5a.pic} \caption{} \label{sl.aks.2.3.5a.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.aks.2.3.5a.pic}) First, the identical mapping $\mathcal{E}$, that maps points $A$, $B$ and $C$ to the same points $A$, $B$ and $C$, is an isometry by Axiom \ref{aksIII4}. From the previous theorem \ref{IizrekABC} it follows that such an isometry is unique. \kdokaz For point $A$ we say that it is \index{point!fixed} \pojem{fixed point} (or \index{point!invariant} \pojem{invariant point}) of isometry $\mathcal{I}$, if $\mathcal{I}(A)=A$. The previous theorem tells us that the only isometries that have three fixed non-collinear points are identities. We will discuss isometries in more detail in chapter \ref{pogIZO}, but here we will use them primarily to help us introduce the concept of congruence of figures. Two figures $\Phi$ and $\Phi'$ are \index{figures!congruent}\poem{congruent} (we will write $\Phi\cong \Phi'$), if there exists an isometry $I$, that transforms figure $\Phi$ into figure $\Phi'$. A direct consequence of axiom \ref{aksIII4} is the following proposition. \bizrek Congruence of figures is an equivalence relation. \label{sklRelEkv} \eizrek \textbf{\textit{Proof.}} \textit{Reflexivity.} For every figure $\Phi$ it holds that $\Phi \cong \Phi$, because the identity transformation $\mathcal{E}$ is an isometry (axiom \ref{aksIII4}) and $\mathcal{E}:\Phi\rightarrow\Phi$. \textit{Symmetry.} From $\Phi \cong \Phi_1$ it follows that there exists an isometry $\mathcal{I}$, that transforms figure $\Phi$ into figure $\Phi_1$. The inverse transformation $\mathcal{I}^{-1}$, which is an isometry according to axiom \ref{aksIII4}, transforms figure $\Phi_1$ into figure $\Phi$, so $\Phi_1 \cong \Phi$ holds. \textit{Transitivity.} From $\Phi \cong \Phi_1$ and $\Phi_1 \cong \Phi_2$ it follows that there exist such isometries $\mathcal{I}$ and $\mathcal{I}'$, that $\mathcal{I}:\Phi\rightarrow\Phi_1$ and $\mathcal{I}':\Phi_1\rightarrow\Phi_2$ hold. Then the composition $\mathcal{I}'\circ\mathcal{I}$, which is an isometry according to axiom \ref{aksIII4}, transforms figure $\Phi$ into figure $\Phi_2$, so $\Phi \cong \Phi_2$ holds. \kdokaz The concept of congruence of figures also applies to lines. We have intuitively associated the congruence of lines with the congruence of pairs of points. Now we will prove the equivalence of both relations. \bizrek \label{izrek(A,B)} $AB \cong A'B' \Leftrightarrow (A,B)\cong (A',B')$ \eizrek \textbf{\textit{Proof.}} ($\Rightarrow$) If $(A,B)\cong (A',B')$, according to proposition \ref{izrekAB} there exists an isometry $\mathcal{I}$, that transforms points $A$ and $B$ into points $A'$ and $B'$. From proposition \ref{izrekIzoB} it follows that isometry $\mathcal{I}$ transforms line $AB$ into line $A'B'$ or $AB \cong A'B'$ holds. ($\Leftarrow$) If $AB \cong A'B'$, there exists an isometry $\mathcal{I}$, which maps the line segment $AB$ to the line segment $A'B'$. By the consequence of Theorem \ref{izrekIzoB}, the endpoint of the line segment is mapped to the endpoint of the line segment. This means that either $\mathcal{I}:A,B\mapsto A',B'$ or $\mathcal{I}:A,B\mapsto B',A'$ holds. From the first relation it follows that $(A,B)\cong (A',B')$ and from the second that $(A,B)\cong (B',A')$. But from the second example we also get $(A,B)\cong (A',B')$, which is a consequence of Axioms \ref{aksIII3} and \ref{aksIII4}. \kdokaz Because of the previous theorem, in the following we will always write the relation $(A,B)\cong (A',B')$ instead of $AB\cong A'B'$. \bizrek \label{ABnaPoltrakCX} For each line segment $AB$ and each ray $CX$, there is exactly one point $D$ on the ray $CX$ that $AB\cong CD$ holds. \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.3.5b.pic} \caption{} \label{sl.aks.2.3.5b.pic} \end{figure} \textbf{\textit{Proof.}} Let $P$ be a point that does not lie on the line $AB$ and $Q$ be a point that does not lie on the line $CX$ (Figure \ref{sl.aks.2.3.5b.pic}). By Axiom \ref{aksIII2}, there is only one isometry $\mathcal{I}$, which maps the point $A$ to the point $C$, the line segment $AB$ to the line segment $CX$ and the half-plane $ABP$ to the half-plane $CXQ$. Let $D=\mathcal{I}(C)$, then $AB \cong CD$ holds. We assume that on the line segment $CX$ there is another point $\widehat{D}$, for which $AB \cong C\widehat{D}$ holds. Because the line segments $CX$ and $CD$ are congruent, and the isometry $\mathcal{I}$ maps the point $A$ to the point $C$, the line segment $AB$ to the line segment $CD$ and the half-plane $ABP$ to the half-plane $CDQ$, from Axiom \ref{aksIII2} it follows that $\mathcal{I}(C)=\widehat{D}$ or $\widehat{D}=D$. \kdokaz \bizrek \label{izomEnaC'} Let $A$, $B$, $C$ be three non-collinear points and $A'$, $B'$ points of the edge of a half-plane $\pi$ such that $AB \cong A'B'$. Then there is exactly one point $C'$ in the half-plane $\pi$ such that $AC \cong A'C'$ and $BC \cong B'C'$. \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.3.11a.pic} \caption{} \label{sl.aks.2.3.11a.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.aks.2.3.11a.pic}) By Axiom \ref{aksIII2} there is only one isometry $\mathcal{I}$, which maps the point $A$ into the point $A'$, the segment $AB$ into the segment $A'B'$ and the plane $ABC$ into the plane $\pi$ and holds $\mathcal{I}(B)=B'$. Let $C'=\mathcal{I}(C)$, then it holds $AC \cong A'C'$ and $BC \cong B'C'$. We assume that there is such a point $\widehat{C}'$, which lies in the plane $\pi$ and holds $AC \cong A'\widehat{C}'$ and $BC \cong B'\widehat{C}'$. Because $AB \cong A'B'$, by Theorem \ref{IizrekABC} there is only one isometry $\mathcal{\widehat{I}}$, which maps the points $A$, $B$ and $C$ into the points $A'$, $B'$ and $\widehat{C}'$. But this also maps the segment $AB$ into the segment $A'B'$ and the plane $ABC$ into the plane $A'B'\widehat{C}'=\pi$. By Axiom \ref{aksIII2} is $\mathcal{\widehat{I}}=\mathcal{I}$ and therefore also $\widehat{C}'=\mathcal{\widehat{I}}(C)=\mathcal{I}(C)=C'$. \kdokaz \bizrek \label{izoABAB} If $\mathcal{I}$ is an isometry that maps a points $A$ and $B$ into the same points $A$ and $B$ (i.e. $\mathcal{I}(A)=A$ and $\mathcal{I}(B)=B$), then it also holds for each point $X$ on the line $AB$ (i.e. $\mathcal{I}(X)=X$). \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.3.8.pic} \caption{} \label{sl.aks.2.3.8.pic} \end{figure} \textbf{\textit{Proof.}} Let $X$ be an arbitrary point on the line $AB$. Without loss of generality, we assume that the point $X$ lies on the segment $AB$ (Figure \ref{sl.aks.2.3.8.pic}). We prove that $\mathcal{I}(X)=X$. Let $P$ be a point that does not lie on the line $AB$ and $P'=\mathcal{I}(P)$. The isometry $\mathcal{I}$ maps the point $A$ to the point $A$, the segment $AB$ to the segment $AB$ (or the segment $AX$ to the segment $AX$) and the half-plane $ABP$ to the half-plane $ABP'$ (or the half-plane $AXP$ to the half-plane $AXP'$). By Axiom \ref{aksIII2} from $AX\cong AX$ it follows that $\mathcal{I}(X)=X$. \kdokaz We introduce new concepts related to distances. We say that the line $EF$ \index{vsota!daljic}\pojem{vsota daljic} $AB$ and $CD$, which we denote $EF=AB+CD$, if there exists such a point $P$ on the line $EF$, that $AB \cong EP$ and $CD \cong PF$ (Figure \ref{sl.aks.2.3.9.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.3.9.pic} \caption{} \label{sl.aks.2.3.9.pic} \end{figure} The line $EF$ is \index{razlika!daljic}\pojem{razlika daljic} $AB$ and $CD$, which we denote $EF=AB-CD$, if $AB=EF+CD$ (Figure \ref{sl.aks.2.3.9.pic}). In a similar way, we can also define multiplication of a line by a natural and a positive rational number. For the lines $AB$ and $CD$ it is $AB=n\cdot CD$ ($n\in \mathbb{N}$), if there exist such points $X_1$, $X_2$,..., $X_{n-1}$, that $\mathcal{B}(X_1,X_2,\ldots,X_{n-1})$ and $AX_1 \cong X_1X_2 \cong X_{n-1}B \cong CD$ (Figure \ref{sl.aks.2.3.10.pic}). In this case, it is also $CD=\frac{1}{n}\cdot AB$. At this point, we will not formally prove the fact that for every line $PQ$ and every natural number $n$ there exists a line $AB$, for which $AB=n\cdot PQ$, and a line $CD$, for which $CD=\frac{1}{n}\cdot PQ$. \begin{figure}[!htb] \centering \input{sl.aks.2.3.10.pic} \caption{} \label{sl.aks.2.3.10.pic} \end{figure} We introduce multiplication of a line segment with a positive rational number in the following way. For $q=\frac{n}{m} \in \mathbb{Q^+}$ we have: $$q\cdot AB=\frac{n}{m}\cdot AB = n\cdot\left(\frac{1}{m}\cdot AB\right)$$ If for a point $P$ on the line segment $AB$ it holds that $AP=\frac{n}{m}\cdot PB$, we say that the point $P$ \pojem{divides} the line segment $AB$ in the \index{razmerje} \pojem{ratio} $n:m$, which we write as $AP:PB=n:m$. The line segment $AB$ is \index{relacija!urejenosti daljic}\pojem{longer} than the line segment $CD$, which we denote $AB>CD$, if there exists such a point $P\neq B$ on the line segment $AB$, that it holds $CD \cong AP$ (Figure \ref{sl.aks.2.3.11.pic}). In this case we also say that the line segment $CD$ is \pojem{shorter} than the line segment $AB$ (notation $CDCD$, $AB \angle cd$), if there exists a segment $s=SX$ in the angle $ab$, such that $\angle as \cong \angle cd$ (Figure \ref{sl.aks.2.3.17.pic}). In this case, the angle $cd$ is also \pojem{smaller} than the angle $ab$ ($\angle cd< \angle ab$). It is not difficult to prove that for two angles $ab$ and $cd$ one of the relations holds: $\angle ab > \angle cd$, $\angle ab < \angle cd$ or $\angle ab \cong \angle cd$. \begin{figure}[!htb] \centering \input{sl.aks.2.3.17.pic} \caption{} \label{sl.aks.2.3.17.pic} \end{figure} The angles are \index{angles!supplementary}\pojem{supplementary}, if their sum is equal to the straight angle (Figure \ref{sl.aks.2.3.18.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.3.18.pic} \caption{} \label{sl.aks.2.3.18.pic} \end{figure} The segment $s=SX$ is the \index{bisector of an angle}\pojem{bisector of the angle} $\angle pSq=\alpha$ (Figure \ref{sl.aks.2.3.19.pic}), if it lies in this angle and it holds $\angle ps \cong \angle sq$. The carrier of this bisector is the \index{simetrala!kota}\pojem{simetrala kota} $pSq$ (the line $s_{\alpha}$). \begin{figure}[!htb] \centering \input{sl.aks.2.3.19.pic} \caption{} \label{sl.aks.2.3.19.pic} \end{figure} Similarly to the center of a line, the following statement holds for the bisector of an angle. \bizrek \label{izrekSimetralaKota} An angle has exactly one bisector. %(oz. eno samo simetralo). \eizrek \textbf{\textit{Proof.}} Let $\alpha=pSq$ be an arbitrary angle, $P$ an arbitrary point, which lies on the segment $Sp$ ($P\neq S$) and $Q$ a point, which lies on the segment $Sq$ and it holds $SP\cong SQ$. \begin{figure}[!htb] \centering \input{sl.aks.2.3.20.pic} \caption{} \label{sl.aks.2.3.20.pic} \end{figure} Let the angle $\alpha$ be the extended angle (Figure \ref{sl.aks.2.3.20.pic}), which determines the line $\pi$. Let $A$ be any point on it. By the statement \ref{izomEnaC'} in the line $\pi$ there is only one point $B$, so that $(P,Q,A)\cong (Q,P,B)$. From the statement \ref{IizrekABC} it follows that there is only one izometry $\mathcal{I}$, which maps points $P$, $Q$ and $A$ into points $Q$, $P$ and $B$. Let $\mathcal{I}(B)=\widehat{A}$. Because $(Q,P,B)\cong(P,Q,\widehat{A})$, by the statement \ref{izomEnaC'} $\widehat{A}=A$. Therefore: $$\mathcal{I}:P,Q,A,B\mapsto Q,P,B,A.$$ Therefore, the centers $S$ and $L$ of the lines $PQ$ and $AB$ map into each other (axiom \ref{aksIII4}), which then also holds for the line segment $s=SL$ and every point on it (statement \ref{izoABAB}). Therefore, the izometry $\mathcal{I}$ maps the line segment $pSs$ into the line segment $sSq$, so the line segment $pSs\cong sSq$ or the line segment $s$ is the bisector of the angle $pSq$. We will prove that $s$ is the only bisector of the angle $\alpha$. Let $\widehat{s}=S\widehat{L}$ be a line segment that lies in the angle $\alpha$ and $pS\widehat{s}\cong \widehat{s}Sq$. Then there is an izometry $\mathcal{\widehat{I}}$, which maps the angle $pS\widehat{s}$ into the angle $\widehat{s}Sq$. This izometry maps the point $S$ into the point $S$, the line segment $p$ into the line segment $q$ and the line $\pi$ into the line $\pi$, so by the axiom \ref{aksIII2} $\mathcal{\widehat{I}}=\mathcal{I}$. Therefore $\mathcal{I}(\widehat{s})= \mathcal{\widehat{I}}(\widehat{s})=\widehat{s}$. If $\widehat{L} \notin s$, the izometry $\mathcal{I}$ maps three non-collinear points $S$, $L$ and $\widehat{L}$ into itself and is the identical mapping (statement \ref{IizrekABCident}), which is not possible. Therefore $\widehat{L} \in s$ or $\widehat{s}=s$. If $\alpha$ is an unextended convex angle (Figure \ref{sl.aks.2.3.20.pic}), then points $S$, $P$ and $Q$ are nonlinear, so according to Theorem \ref{IizrekABC} there is only one isometry $\mathcal{I}$, which maps points $P$, $S$ and $Q$ to points $Q$, $S$ and $P$. With $L$ we denote the center of the line $PQ$. By the Axiom \ref{aksIII3} we have $\mathcal{I}(L)=L$. Then also all points of the ray $s=SL$ are mapped to themselves (Theorem \ref{izoABAB}). As $\alpha$ is a convex angle, this means that the point $L$ and then also the ray $s$ lie within this angle. Therefore the isometry $\mathcal{I}$ maps angle $pSs$ to angle $sSq$, so $pSs\cong sSq$ or the ray $s$ is the bisector of angle $pSq$. Similarly as in the previous example we prove that angle $\alpha$ has no other bisectors. If $\alpha$ is a non-convex angle, the bisector is obtained as the complementary (supplementary) ray of the ray $s$. \kdokaz We prove two theorems, which relate to right angles and perfect angles.\index{angle!right} \index{angle!perfect} \bizrek The adjacent supplementary angles of two congruent angles are also congruent. \label{sokota} \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.3.20a.pic} \caption{} \label{sl.aks.2.3.20a.pic} \end{figure} \textbf{\textit{Proof.}} Let $\alpha'=\angle P'OQ$ and $\alpha_1'=\angle P_1'O_1Q_1$ be the supplement angles of two adjacent angles $\alpha=\angle POQ$ and $\alpha_1=\angle P_1O_1Q_1$ (Figure \ref{sl.aks.2.3.20a.pic}). By Axiom \ref{aksIII2}, there exists a single isometry $\mathcal{I}$, which maps point $O$ to point $O_1$, line segment $OP$ to line segment $O_1P_1$, and line $POQ$ to line $P_1O_1Q_1$. Let $Q_2=\mathcal{I}(Q)$. Then $\angle P_1O_1Q_2\cong \angle POQ$. Isometry $\mathcal{I}$ maps line $POQ$ to line $P_1O_1Q_1$, so point $Q_2$ (and also line segment $O_1Q_2$) lies on line $P_1O_1Q_1$. Since, by assumption, $\angle POQ\cong\angle P_1O_1Q_1$, by \ref{KotNaPoltrak} Theorem, $OQ_1$ and $OQ_2$ represent the same line segment. Therefore, point $Q_2$ lies on line segment $O_1Q_1$. Let $P_2'=\mathcal{I}(P')$. Since isometries map line segments to line segments (Theorem \ref{izrekIzoB}), point $P_2'$ lies on line segment $O_1P_1'$. From $\mathcal{I}:P',O,Q\mapsto P_2',O_1,Q_2$ it follows that isometry $\mathcal{I}$ maps angle $P'OQ$ to angle $P_2'O_1Q_2$ (Theorem \ref{izrekIzoB}), so $\angle P'OQ\cong \angle P_2'O_1Q_2=\angle P_1'O_1Q_1$. \kdokaz \bizrek \label{sovrsnaSkladna} Vertical angles are congruent. \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.3.20b.pic} \caption{} \label{sl.aks.2.3.20b.pic} \end{figure} \textbf{\textit{Proof.}} Let $\alpha=\angle POQ$ and $\alpha'=\angle P'OQ'$ be two adjacent angles, where points $P$, $O$, $P'$ (or $Q$, $O$, $Q'$) are collinear (Figure \ref{sl.aks.2.3.20b.pic}). Angle $\beta=\angle QOP'$ is the supplement of both angles $\alpha$ and $\alpha'$. Since $\beta\cong\beta$, by the previous Theorem \ref{sokota}, $\alpha\cong\alpha'$. \kdokaz \bizrek \label{sredZrcObstoj} For each point $S$ there exists an isometry $\mathcal{I}$ such that $\mathcal{I}(S)=S$. In addition, for each point $X\neq S$ the following holds:\\ if $\mathcal{I}(X)=X'$, then $S$ is the midpoint of the line segment $XX'$. \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.3.20c.pic} \caption{} \label{sl.aks.2.3.20c.pic} \end{figure} \textbf{\textit{Proof.}} Let $P$ be an arbitrary point different from $S$ (Figure \ref{sl.aks.2.3.20c.pic}). By Axiom \ref{AksII3}, there exists a point $Q$ on the line $SP$ such that $\mathcal{B}(P,S,Q)$. We mark the half-planes determined by the edge $SP$ with $\alpha$ and $\alpha'$. By Axiom \ref{aksIII2}, there exists (one and only one) isometry $\mathcal{I}$, which maps the point $S$ to the point $S$, the line segment $SP$ to the line segment $SQ$, and the half-plane $\alpha$ to the half-plane $\alpha'$. We mark the line $SP$ with $p$. The point $P'=\mathcal{I}(P)$ lies on the line segment $SQ$ or on the line $p$. Therefore, since $\mathcal{I}:S,P \mapsto S,P'$, the line $SP$ is mapped to the line $SP'$ by Axiom \ref{aksIII1}, i.e. $\mathcal{I}:p\rightarrow p$. The image of the half-plane $\alpha'$ with the edge $p$ is therefore a half-plane with the same edge (Proposition \ref{izrekIzoB}). This half-plane cannot be $\alpha'$, since the isometry $\mathcal{I}$ is a bijective mapping and it maps the half-plane $\alpha$ to the half-plane $\alpha'$. Therefore, $\mathcal{I}:\alpha'\rightarrow \alpha$. Now it is clear that, without loss of generality, it is enough to carry out the proof only for points that lie in the half-plane $\alpha$ (without the edge or only the line segment $SP$). Let $X\in \alpha\setminus p$ and $X'=\mathcal{I}(X)$. We immediately see that $X'\in \alpha'\setminus p$. By Axiom \ref{AksII3} there exists on the line $SX$ such a point $X_1$, that $\mathcal{B}(X,S,X_1)$ is true. Because $\angle PSX$ and $\angle P'SX_1$ are perfect angles, by Theorem \ref{sovrsnaSkladna} they are also compatible. But from $\mathcal{I}:S,P,X \mapsto S,P',X'$ it follows that $\angle PSX \cong \angle P'SX'$. Therefore $\angle P'SX_1\cong \angle P'SX'$ is true (Theorem \ref{sklRelEkv})), so by Theorem \ref{KotNaPoltrak} the line segments $SX_1$ and $SX'$ are identical. This means that the point $X'$ lies on the line segment $SX_1$ or $\mathcal{B}(X,S,X')$ is true. Because of $\mathcal{I}:S,X \mapsto S,X'$ it is also true that $SX\cong SX'$, so by definition the point $S$ is the center of the line $XX'$. Let in the end $Y$ be an arbitrary point on the line segment $SP$, which is different from the point $S$, and $Y'=\mathcal{I}(Y)$. The point $Y'$ lies on the line segment $SQ$, so $\mathcal{B}(Y,S,Y')$ is true. Because of $\mathcal{I}:S,Y \mapsto S,Y'$ it is also true that $SY\cong SY'$, so by definition the point $S$ is the center of the line $YY'$. \kdokaz In section \ref{odd6SredZrc} we will discuss the isometry, which is mentioned in the previous Theorem \ref{sredZrcObstoj}, in more detail. Let us define new types of angles. A \index{kot!ostri}\pojem{acute angle} is a \index{kot!pravi} \pojem{right angle} or a \index{kot!topi}\pojem{obtuse angle}, if it is smaller, equal to or greater than its supplement (Figure \ref{sl.aks.2.3.21.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.3.21.pic} \caption{} \label{sl.aks.2.3.21.pic} \end{figure} From the definition it follows that acute (or obtuse) angles are those convex angles, which are smaller (or greater) than a right angle. From Theorem \ref{izrekSimetralaKota} it follows that a right angle exists, since the bisector of an extended angle divides it into two compatible supplements. It is not difficult to prove that every two right angles are compatible and that an angle, which is compatible with a right angle, is also a right angle. If the sum of two angles is a right angle, we say that the angles are \pojem{complementary} (Figure \ref{sl.aks.2.3.22.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.3.22.pic} \caption{} \label{sl.aks.2.3.22.pic} \end{figure} We will now introduce an extremely important relation between lines. If the lines $p$ and $q$ contain the segments of a right angle, we say that $p$ and $q$ are \pojem{perpendicular}, which we denote $p \perp q$ (Figure \ref{sl.aks.2.3.23.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.3.23.pic} \caption{} \label{sl.aks.2.3.23.pic} \end{figure} From the definition itself it is clear that the perpendicularity is a symmetric relation, i.e. from $p \perp q$ it follows that $q \perp p$. If $p \perp q$ and $p \cap q=S$, we say that the line $p$ is \pojem{perpendicular} to the line $q$ at the point $S$ or that $p$ is a \pojem{perpendicular} of the line $q$ at this point. The following theorem is the most important theorem that characterizes the relation of perpendicularity. \bizrek \label{enaSamaPravokotnica} For each point $A$ and each line $p$, there is a unique line $n$ going through the point $A$, which is perpendicular on the line $p$. \eizrek \textbf{\textit{Proof.}} Assume that point $A$ does not lie on line $p$. Let $B$ and $C$ be any points that lie on line $p$ (Figure \ref{sl.aks.2.3.24.pic}). We denote the plane $BCA$ with $\pi$, and the complementary plane with $\pi_1$. By izreku \ref{izomEnaC'}, there exists only one point $A_1\in \pi_1$, for which $(A,B,C) \cong (A_1,B,C)$. From izreku \ref{IizrekABC} it follows that there exists only one izometrija $\mathcal{I}$, which maps points $A$, $B$ and $C$ into points $A_1$, $B$ and $C$. We denote the line $AA_1$ with $n$. Because $A$ and $A_1$ are on different sides of line $p$, line $n$ intersects line $p$ in some point $S$. From $\mathcal{I}:B,C \mapsto B,C$ it follows that $\mathcal{I}(S)=S$ (izrek \ref{izoABAB}). Therefore, izometrija $\mathcal{I}$ maps angle $ASB$ into angle $A_1SB$. It follows that $\angle ASB$ and $\angle A_1SB$ are complementary angles, therefore they are right angles. Therefore, $n \perp p$. \begin{figure}[!htb] \centering \input{sl.aks.2.3.24.pic} \caption{} \label{sl.aks.2.3.24.pic} \end{figure} We will prove that $n$ is the only perpendicular to line $p$ through point $A$. Let $\widehat{n}$ be a line, for which $A\in \widehat{n}$ and $\widehat{n} \perp p$. Let point $\widehat{S}$ be the intersection of lines $\widehat{n}$ and $p$. By the assumption, $\angle A\widehat{S}B$ is a right angle and is compatible with its complementary angle $\angle B\widehat{S}A_2$ ($A_2$ is such a point that $\mathcal{B}(A,\widehat{S},A_2)$), which is also a right angle. From $\mathcal{I}:B,C \mapsto B,C$ it follows $\mathcal{I}(\widehat{S})=\widehat{S}$ (statement \ref{izoABAB}). Therefore, the isometry $\mathcal{I}$ maps the angle $A\widehat{S}B$ into the angle $A_1\widehat{S}B$. It follows that $\angle A\widehat{S}B$ and $\angle A_1\widehat{S}B$ are congruent, so $\angle A_1\widehat{S}B$ is a right angle. Therefore, the angle $A_1\widehat{S}B$ and $A_2\widehat{S}B$ are right angles and are therefore congruent. It follows that the line segments $\widehat{S}A_1$ and $\widehat{S}A_2$ are the same, so $A_1 \in \widehat{S}A_2=\widehat{n}$ or $\widehat{n}=AA_1=n$. In the case when the point $A$ lies on the line $p$, the rectangle $n$ is the symmetry of the corresponding extended angle (statement \ref{izrekSimetralaKota}). \kdokaz The previous statement has the following important consequence - the existence of pairs of disjoint lines in the plane - or those that do not have common points. This is the content of the following two statements. \bizrek \label{absolGeom1} Let $p$ and $q$ be a lines perpendicular on a line $PQ$ in the points $P$ and $Q$. Then the lines $p$ and $q$ do not have a common points i.e. $p\cap q=\emptyset$. \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.3.25b.pic} \caption{} \label{sl.aks.2.3.25b.pic} \end{figure} \textbf{\textit{Proof.}} The statement is a direct consequence of the previous statement \ref{enaSamaPravokotnica}. Namely, if the lines $p$ and $q$ intersected at some point $S$, we would have two rectangles on the line $PQ$ from the point $S$ (Figure \ref{sl.aks.2.3.25b.pic}), which is in contradiction with the aforementioned statement. \kdokaz \bizrek \label{absolGeom} If $A$ is a point that does not lie on a line $p$, then there exists at least one line (in the same plane) passing through the point $A$ and not intersecting the line $p$ (Figure \ref{sl.aks.2.3.25a.pic}). \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.3.25a.pic} \caption{} \label{sl.aks.2.3.25a.pic} \end{figure} \textbf{\textit{Proof.}} By the statement \ref{enaSamaPravokotnica} there exists (exactly one) rectangle $n$ of the line $p$, which goes through the point $A$. We mark with $A'$ the intersection of the lines $p$ and $n$. From the same statement it follows that there exists another rectangle $q$ of the line $n$ in the point $A$. By the previous statement \ref{absolGeom1} the line $q$ goes through the point $A$ and doesn't have any common points with the line $p$. \kdokaz The point $A'$ is the \index{node}\pojem{node} or the \index{orthogonal projection}\pojem{orthogonal projection} of the point $A$ on the line $p$, if the rectangle of the line $p$ through the point $A$ intersects the line in the point $A'$. We will mark it with $A'=pr_{\perp p}(A)$ (Figure \ref{sl.aks.2.3.25.pic}). From the previous statement it follows that for every point and line there exists only one orthogonal projection. \begin{figure}[!htb] \centering \input{sl.aks.2.3.25.pic} \caption{} \label{sl.aks.2.3.25.pic} \end{figure} The line, which goes through the center $S$ of the distance $AB$ and is perpendicular to the line $AB$, is called the \index{symmetry!of a line}\pojem{symmetry of the line} $AB$ and we mark it with $s_{AB}$ (Figure \ref{sl.aks.2.3.26.pic}). The properties of the symmetry of a line we will discuss in the next chapter. \begin{figure}[!htb] \centering \input{sl.aks.2.3.26.pic} \caption{} \label{sl.aks.2.3.26.pic} \end{figure} We say that the point $A$ is \index{symmetry!with respect to a line}\pojem{symmetric} to the point $B$ with respect to the line $s$, if the line $s$ is the symmetry of the line $AB$. The symmetry with respect to a line (as a mapping - i.e. the basic mirroring) we will discuss in more detail in the section \ref{odd6OsnZrc}. Let $S$ be a point and $AB$ a line. The set of all points $X$, for which it holds that $SX \cong AB$, is called the \index{circle}\pojem{circle} with \index{center!of a circle}\pojem{center} $S$ and \index{radius of a circle}\pojem{radius} $AB$; we mark it with $k(S,AB)$ (Figure \ref{sl.aks.2.3.27.pic}) i.e.: $$k(S,AB)=\{X;\hspace*{1mm}SX \cong AB\}.$$ \begin{figure}[!htb] \centering \input{sl.aks.2.3.27.pic} \caption{} \label{sl.aks.2.3.27.pic} \end{figure} Of course, a circle is a set of points in a plane, because in this book we only consider planar geometry (all points and all figures belong to the same plane). From the definition it is clear that for the radius we can choose any distance that is consistent with the distance $AB$, that is, any distance $SP$, where $P$ is any point on the circle. Since the radius is not tied to a specific distance, we usually denote it with a small letter $r$. So we can also write the circle like this: $$k(S,r)=\{X;\hspace*{1mm}SX \cong r\}.$$ The set $$\{X;\hspace*{1mm}SX \leq r\}$$ we call the \index{krog}\pojem{circle} with center $S$ and radius $r$ (Figure \ref{sl.aks.2.3.28.pic}) and denote it with $\mathcal{K}(S,r)$. The set $$\{X;\hspace*{1mm}SX < r\}$$ is the \index{notranjost!kroga} \pojem{interior of the circle} $\mathcal{K}(S, r)$, and its points are \pojem{interior points of the circle}. This means that the circle is actually the union of its interior and the corresponding circle. The set $$\{X;\hspace*{1mm}SX > r\}$$ we call the \index{zunanjost!kroga}\pojem{exterior of the circle} $\mathcal{K}(S, r)$ and its points \pojem{exterior points of the circle}. For practical reasons, we will call the interior of the circle $\mathcal{K}(S, r)$ the \index{notranjost!krožnice}\pojem{interior} of the corresponding circle $k(S,r)$, the exterior of the circle $\mathcal{K}(S, r)$ the \index{zunanjost!krožnice}\pojem{exterior} of the corresponding circle $k(S,r)$. We define the \pojem{interior} and \pojem{exterior} points of the circle in the same way. \begin{figure}[!htb] \centering \input{sl.aks.2.3.28.pic} \caption{} \label{sl.aks.2.3.28.pic} \end{figure} If $P$ and $Q$ are two points on the circle $k(S, r)$, the distance $PQ$ is called the \index{tetiva krožnice} \pojem{tetiva} of the circle. If the tether contains the center of the circle, it is called \index{premer krožnice}\pojem{premer} or \index{diameter krožnice}\pojem{diameter} of the circle (Figure \ref{sl.aks.2.3.29.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.3.29.pic} \caption{} \label{sl.aks.2.3.29.pic} \end{figure} We prove the following statement. \bizrek \label{premerInS} The centre $S$ of the circle $k(S, r)$ is at the same time the midpoint of each diameter of that circle. \eizrek \textbf{\textit{Proof.}} \begin{figure}[!htb] \centering \input{sl.aks.2.3.30.pic} \caption{} \label{sl.aks.2.3.30.pic} \end{figure} If $PQ$ is the diameter of the circle $k(S, r)$, the points $P$ and $Q$ lie on the circle, which means: $SP \cong SQ \cong r$ (Figure \ref{sl.aks.2.3.30.pic}). Because the point $S$ lies on the line $PQ$, it follows that the point $S$ is the center of this line. \kdokaz From the previous statement it follows that the diameter is equal to two radii, because: $PQ = PS + SQ = 2\cdot PS = 2\cdot r$. This means that all diameters of a circle are consistent with each other. Let $P$ and $Q$ be any two points on the circle $k(S, r)$. The intersection of the circle $k$ with one of the planes (in the plane of this circle) with the edge $s=PQ$ is called the \index{krožni!lok} \pojem{krožni lok} $PQ$ (or shorter \pojem{lok}) with the endpoints $P$ and $Q$. Thus, each tether $PQ$ on a circle $k$ determines two arcs. Assume that the center $S$ does not lie on the edge of the plane that generates the circular arc. If this plane contains the center $S$ of the circle $k$, it is a \pojem{veliki lok} $PQ$, otherwise it is a \pojem{mali lok} $PQ$. But if the center $S$ is on the edge $PQ$ of the plane, then each of the two arcs $PQ$ \index{polkrožnica}\pojem{polkrožnica} $PQ$ (Figure \ref{sl.skk.4.2.1.pic}). \begin{figure}[!htb] \centering \input{sl.skk.4.2.1.pic} \caption{} \label{sl.skk.4.2.1.pic} \end{figure} Since the locus is not uniquely determined by its endpoints, we must also know at least one point on the circle that belongs or does not belong to this locus. In a similar way, we define certain parts of the circle. Let $P$ and $Q$ be arbitrary points on the circle $k(S, r)$. The intersection of the circle $\mathcal{K}(S, r)$ with one of the planes (in the plane of this circle) with the edge $s=PQ$ is called \index{krožni!odsek} \pojem{circular segment}. So each chord $PQ$ on some circle $k(S, r)$ determines on the circle $\mathcal{K}(S, r)$ two circular segments. Assume that the center $S$ does not lie on the edge of the plane that generates the circular segment. If this plane contains the center $S$ of the circle $k$, it is a \pojem{major circular segment} $PQ$, otherwise it is a \pojem{minor circular segment} $PQ$. If the center $S$ is on the edge $PQ$ of the plane, then each of the two circular segments \index{polkrog}\pojem{arc} is an \pojem{arc}. From the definition it is clear that the edge of the circular segment is the union of the chord $PQ$ and the corresponding arc (Figure \ref{sl.skk.4.2.1b.pic}). \begin{figure}[!htb] \centering \input{sl.skk.4.2.1b.pic} \caption{} \label{sl.skk.4.2.1b.pic} \end{figure} We will also define one more term related to the circle. Let $P$ and $Q$ be arbitrary points on the circle $k(S, r)$. The intersection of the circle $\mathcal{K}(S, r)$ with one of the angles $PSQ$ is called \index{krožni!izsek} \pojem{circular sector}. Again we have two circular sectors. If the angle $PSQ$ is an obtuse angle, we get two arcs, otherwise a convex and a concave circular sector, depending on whether the angle $PSQ$ is convex or concave (Figure \ref{sl.skk.4.2.1c.pic}). \begin{figure}[!htb] \centering \input{sl.skk.4.2.1c.pic} \caption{} \label{sl.skk.4.2.1c.pic} \end{figure} %________________________________________________________________________________ \poglavje{Continuity Axiom} \label{odd2AKSZVE} We have already learned in elementary school, when introducing the numerical line and the coordinate system, that it is possible to establish a connection in which each point of a line corresponds to a certain real number, and vice versa, each real number can be assigned a point that lies on that line. This is related to the following axiom. \baksiom \label{aksDed}\index{aksiom!Dedekind's} (Dedekind's\footnote{\index{Dedekind, R.} \textit{R. Dedekind} (1831--1916), German mathematician.} axiom) Suppose that all points on open line segment $AB$ are divided into the union of two nonempty disjoint sets $\mathcal{U}$ and $\mathcal{V}$ such that no point of $\mathcal{U}$ is between two points of $\mathcal{V}$ and vice versa: no point of $\mathcal{V}$ is between two points of $\mathcal{U}$. Then there is a unique point $C$ on open line segment $AB$ such that $B(A',C,B')$ for any two points $A'\in \mathcal{U}\setminus{C}$ and $B'\in \mathcal{V} \setminus {C}$ (Figure \ref{sl.aks.2.4.1.pic}). \eaksiom \begin{figure}[!htb] \centering \input{sl.aks.2.4.1.pic} \caption{} \label{sl.aks.2.4.1.pic} \end{figure} Let us state without proof two important consequences of the axiom of continuity\footnote{Until the 19th century, mathematicians did not feel the need to prove these two statements, or the need to introduce the axiom of continuity. Even \index{Evklid} Euclid from Alexandria (3rd century BC) in his famous work 'Elements', does not prove that a certain circle intersects.}. \bizrek \label{DedPoslKrozPrem} Let $k$ be a circle and $P$ a point inside that circle. Then any line $p$ passing through the point $P$ and the circle $k$ has exactly two common points (Figure \ref{sl.aks.2.4.2.pic}). \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.4.2.pic} \caption{} \label{sl.aks.2.4.2.pic} \end{figure} \bizrek \label{DedPoslKrozKroz} If $k$ and $l$ are circles such that $l$ contains at least one point inside and one point outside $k$, then the circles has exactly two points (Figure \ref{sl.aks.2.4.3.pic}). \eizrek \poglavje{The basics of Geometry} \label{osn9Geom} We have already learned in elementary school, when introducing the numerical line and the coordinate system, that it is possible to establish a connection in which each point of a line corresponds to a certain real number, and vice versa, each real number can be assigned a point that lies on that line. This is related to the following axiom. \baksiom \label{aksDed}\index{aksiom!Dedekind's} (Dedekind's\footnote{\index{Dedekind, R.} \textit{R. Dedekind} (1831--1916), German mathematician.} axiom) Suppose that all points on open line segment $AB$ are divided into the union of two nonempty disjoint sets $\mathcal{U}$ and $\mathcal{V}$ such that no point of $\mathcal{U}$ is between two points of $\mathcal{V}$ and vice versa: no point of $\mathcal{V}$ is between two points of $\mathcal{U}$. Then there is a unique point $C$ on open line segment $AB$ such that $B(A',C,B')$ for any two points $A'\in \mathcal{U}\setminus{C}$ and $B'\in \mathcal{V} \setminus {C}$ (Figure \ref{sl.aks.2.4.1.pic}). \eaksiom \begin{figure}[!htb] \centering \input{sl.aks.2.4.1.pic} \caption{} \label{sl.aks.2.4.1.pic} \end{figure} Let us state without proof two important consequences of the axiom of continuity\footnote{Until the 19th century, mathematicians did not feel the need to prove these two statements, or the need to introduce the axiom of continuity. Even \index{Evklid} Euclid from Alexandria (3rd century BC) in his famous work 'Elements', does not prove that a certain circle intersects.}. \bizrek \label{DedPoslKrozPrem} Let $k$ be a circle and $P$ a point inside that circle. Then any line $p$ passing through the point $P$ and the circle $k$ has exactly two common points (Figure \ref{sl.aks.2.4.2.pic}). \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.4.2.pic} \caption{} \label{sl.aks.2.4.2.pic} \end{figure} \bizrek \label{DedPoslKrozKroz} If $k$ and $l$ are circles such that $l$ contains at least one point inside and one point outside $k$, then the circles has exactly two points (Figure \ref{sl.aks.2.4.3.pic}). \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.4.3.pic} \caption{} \label{sl.aks.2.4.3.pic} \end{figure} Dedekind's axiom is used in a few different ways to establish the set of real numbers. This is reminiscent of the already mentioned connection between the set of points on a line and the set of real numbers. We have already defined the multiplication operation of line segment $AB$ by any positive rational number $q$. Now we can extend the concept of multiplication to any positive real number $\lambda$. The definition of line segment $\lambda\cdot AB$, ($\lambda \in \mathbb{R}^+$), which we will not formally derive here, is associated with two sets of points on line segment $CD$: \begin{eqnarray*} && \{X\in CD;\hspace*{1mm}CX=q\cdot AB,\hspace*{1mm}q\leq\lambda,\hspace*{1mm}q\in \mathbb{Q}^+ \} \hspace*{1mm}\textrm{ in}\\ && \{X\in CD;\hspace*{1mm}CX=q\cdot AB,\hspace*{1mm}q>\lambda,\hspace*{1mm}q\in \mathbb{Q}^+ \} \end{eqnarray*} and Dedekind's axiom \ref{aksDed}. With the help of the axiom of continuity \ref{aksDed} we can also introduce the concepts of measuring line segments and angles. When measuring line segments we will assign each line segment $AB$ a positive real number $\textsl{m}(AB)$ in the following way. Let $\mathcal{D}$ be the set of all line segments and $\mathbb{R}^+$ be the set of all positive real numbers. The mapping $\textsl{m}:\mathcal{D}\rightarrow\mathbb{R}^+$, which satisfies the following properties: \begin{itemize} \item $(\exists A_0B_0\in\mathcal{D})\hspace*{1mm}\textsl{m}(A_0B_0)=1$, \item $(\forall AB, CD\in\mathcal{D})\hspace*{1mm}(AB\cong CD \Rightarrow\textsl{m}(AB)=\textsl{m}(CD))$, \item $(\forall AB, CD, EF\in\mathcal{D})\hspace*{1mm}(AB+CD=EF\Rightarrow \textsl{m}(AB)+\textsl{m}(CD)=\textsl{m}(EF))$, \end{itemize} is called the \index{dolžina!daljice}\pojem{length of a line segment} or the \index{mera!daljice}\pojem{measure of a line segment}, and the triple $\textsl{M}=(\mathcal{D},\mathbb{R}^+,\textsl{m})$ is called the \index{sistem merjenja!daljic}\pojem{system of measuring line segments}. The length of the line $AB$ (or $\textsl{m}(AB)$) will usually be denoted by $|AB|$. It is intuitively clear that there are an infinite number of measuring systems, which depend on the choice of the unit length line $A_0B_0$ - the one that has a length of 1, or $\textsl{m}(A_0B_0)=1$. In one measuring system, the length of any line $AB$ is represented by a positive real number $x$, for which $AB=x\cdot A_0B_0$. So $\textsl{m}(AB)=x$ exactly when $AB=x\cdot A_0B_0$ (Figure \ref{sl.aks.2.4.4.pic}). Now it is clear why we need the axiom of connectivity - without it, we would have problems with the definition of the length of the diagonal of a square with a unit length side (of length 1)\footnote{The ancient Greeks always represented the length as a rational number, so they called the side and the diagonal of the square \pojem{incomparable lines}.}. \begin{figure}[!htb] \centering \input{sl.aks.2.4.4.pic} \caption{} \label{sl.aks.2.4.4.pic} \end{figure} \bizrek \label{meraDalj1} For any measuring system of lengths it holds: (\textit{i}) $AB|AB|$. \begin{figure}[!htb] \centering \input{sl.aks.2.4.6.pic} \caption{} \label{sl.aks.2.4.6.pic} \end{figure} (\textit{ii}) Suppose that $AB\not\cong CD$. Without loss of generality, let $AB\alpha$ and $\gamma'>\beta$. The relations are a direct consequence of the proven inequality $\alpha+\beta=\gamma'$ from the previous \ref{zunanjiNotrNotr}. \kdokaz Based on the internal angles, we can consider three types of triangles. We have already proven that in any triangle the sum of the internal angles is equal to $180^0$ (the statement \ref{VsotKotTrik}). This means that at most one of these angles is a right angle or a reflex angle, or at least two are acute. So we have three possibilities (Figure \ref{sl.aks.2.6.4a.pic}): \begin{itemize} \item The triangle is \index{trikotnik!ostrokotni} \pojem{acute-angled}, if all of its internal angles are acute. \item The triangle is \index{trikotnik!pravokotni}\pojem{right-angled}, if it has one internal angle that is a right angle. The side opposite the right angle of a right-angled triangle is called \index{hipotenuza}\pojem{the hypotenuse}, the other two sides are \index{kateta}\pojem{the catheti}. Because the sum of the internal angles in any triangle is equal to $180^0$, the angles at the hypotenuse of a right-angled triangle are complementary. \item The triangle is \index{trikotnik!topokotni}\pojem{obtuse-angled}, if it has one internal angle that is obtuse. The other two internal angles are then acute. \end{itemize} \begin{figure}[!htb] \centering \input{sl.aks.2.6.4a.pic} \caption{} \label{sl.aks.2.6.4a.pic} \end{figure} %slika 33.1 We will now prove a statement that applies to any $n$-gon. \bizrek \label{VsotKotVeck} The sum of the interior angles of any $n$-gon is equal to $(n - 2) \cdot 180^0$. \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.5.5c.pic} \caption{} \label{sl.aks.2.5.5c.pic} \end{figure} \textbf{\textit{Proof.}} First, let's assume that $A_1A_2\ldots A_n$ is a convex $n $-gon (Figure \ref{sl.aks.2.5.5c.pic}). Its $n-3$ diagonals $A_1A_3$, $A_1A_4$, ... $A_1A_{n-1}$ divide this polygon into $n-2$ triangles $\triangle_1$, $\triangle_2$, ..., $\triangle_{n-2}$. Because each internal angle of the $n $-gon $A_1A_2\ldots A_n$ is divided into the appropriate internal angles of the aforementioned triangles, the sum of all internal angles $n $-gon $A_1A_2\ldots A_n$ is equal to the sum of all angles of triangles $\triangle_1$, $\triangle_2$, ..., $\triangle_{n-2}$. By Theorem \ref{VsotKotTrik} in the end this sum is equal to exactly $(n - 2) \cdot 180^0$. At this point we will not prove the fact that even in the case of a non-convex $n $-gon $A_1A_2\ldots A_n$ it can be divided into $n-2$ triangles. \kdokaz \bizrek \label{VsotKotVeckZuna} The sum of the exterior angles of any $n$-gon is equal to $360^0$. \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.5.5b.pic} \caption{} \label{sl.aks.2.5.5b.pic} \end{figure} \textbf{\textit{Proof.}} We mark with $\alpha_1$, $\alpha_2$,..., $\alpha_n$ the internal and $\alpha'_1$, $\alpha'_2$, ..., $\alpha'_n$ the corresponding external angles at the vertices $A_1$, $A_2$,...,$A_n$ of a convex $n $-gon $A_1A_2\ldots A_n$. (Figure \ref{sl.aks.2.5.5b.pic}). For the appropriate internal and external angle we have: \begin{eqnarray*} & & \alpha_1+\alpha'_1=180^0\\ & & \alpha_2+\alpha'_2=180^0\\ & & \vdots\\ & & \alpha_n+\alpha'_n=180^0 \end{eqnarray*} If we add all the equalities and take into account the proven equality from the previous theorem \ref{VsotKotVeck} $$\alpha_1+\alpha_2+\cdots + \alpha_n=(n - 2) \cdot 180^0,$$ we get $(n - 2) \cdot 180^0+\alpha'_1+\alpha'_2+\cdots + \alpha'_n=n\cdot 180^0$ or $$\alpha'_1+\alpha'_2+\cdots + \alpha'_n=2 \cdot 180^0=360^0,$$ which was to be proven. \kdokaz From the statement \ref{VsotKotVeck} it follows that the sum of all the internal angles of an arbitrary quadrilateral is equal to $360^0$, and from the statement \ref{VsotKotVeckZuna} it also follows that the sum of all the external angles of a convex quadrilateral is equal to $360^0$ (Figure \ref{sl.aks.2.5.5a.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.5.5a.pic} \caption{} \label{sl.aks.2.5.5a.pic} \end{figure} Because a triangle (as the intersection of three planes) is a convex figure, the sum of all the external angles of an arbitrary triangle is equal to $360^0$ (Figure \ref{sl.aks.2.5.5d.pic}). \begin{figure}[!htb] \centering \input{sl.aks.2.5.5d.pic} \caption{} \label{sl.aks.2.5.5d.pic} \end{figure} A similar claim, based on the statement \ref{KotaVzporKraki}, which refers to an angle with parallel sides, also applies to an angle with perpendicular sides. \bizrek \label{KotaPravokKraki} Angles with perpendicular sides are either congruent or supplementary. \index{kota!s pravokotnimi kraki} \eizrek \begin{figure}[!htb] \centering \input{sl.aks.2.5.5.pic} \caption{} \label{sl.aks.2.5.5.pic} \end{figure} \textbf{\textit{Proof.}} Let $\angle aSb$ and $\angle a'S'b'$ be such angles that $a\perp a'$ and $b\perp b'$ (Figure \ref{sl.aks.2.5.5.pic}). Let $A$ be the intersection of the sides of $a$ and $a'$, and $B$ be the intersection of the sides of $b$ and $b'$. In the quadrilateral $SAS'B$, the internal angles at the vertices $A$ and $B$ each measure $90^0$. According to the statement \ref{VsotKotVeck}, the sum of all the internal angles of this quadrilateral is equal to $360^0$. Therefore, the internal angle $BSA$ and $AS'B$ of this quadrilateral measure together $180^0$, which means that two angles, determined by the sides of $a$ and $b$ or $a'$ and $b'$, are supplementary. However, if we replace one of these angles with its supplement, the corresponding angles are congruent. \kdokaz %________________________________________________________________________________ \naloge{Exercises} \begin{enumerate} \item Let $P$, $Q$ and $R$ be the internal points of the sides of a triangle $ABC$. Prove that $P$, $Q$ and $R$ are non-collinear. \item Let $P$ and $Q$ be points on sides $BC$ and $AC$ of triangle $ABC$ and at the same time different from its vertices. Prove that the line segments $AP$ and $BQ$ intersect in one point. \item Points $P$, $Q$ and $R$ are located in order on sides $BC$, $AC$ and $AB$ of triangle $ABC$ and are different from its vertices. Prove that the line segments $AP$ and $QR$ intersect in one point. \item A line $p$, which lies in the plane of the quadrilateral, intersects its diagonal $AC$ and does not pass through any vertex of this quadrilateral. Prove that the line $p$ intersects exactly two sides of this quadrilateral. \item Prove that the half-line is a convex figure. \item Prove that the intersection of two convex figures is a convex figure. \item Prove that any triangle is a convex figure. \item If $\mathcal{B}(A,B,C)$ and $\mathcal{B}(D,A,C)$, then $\mathcal{B}(B,A,D)$ is also true. Prove it. \item Let $A$, $B$, $C$ and $D$ be such collinear points that $\neg\mathcal{B}(B,A,C)$ and $\neg\mathcal{B}(B,A,D)$ are true. Prove that $\neg\mathcal{B}(C,A,D)$ is also true. \item Let $A_1A_2\ldots,A_{2k+1}$ be an arbitrary polygon with an even number of vertices. Prove that there is no line that intersects all of its sides. \item If izometry $\mathcal{I}$ maps figure $\Phi_1$ and $\Phi_2$ into figure $\Phi'_1$ and $\Phi'_2$, then the intersection $\Phi_1\cap\Phi_2$ is mapped by this izometry into the intersection $\Phi'_1\cap\Phi'_2$. Prove it. \item Prove that any two strips of a plane are compatible with each other. \item Prove that any two lines of a plane are compatible with each other. \item Let $k$ and $k'$ be two circles of a plane with centers $O$ and $O'$ and radii $AB$ and $A'B'$. Prove the equivalence: $k\cong k' \Leftrightarrow AB\cong A'B'$. \item Let $\mathcal{I}$ be a non-identical izometry of a plane with two fixed points $A$ and $B$. If $p$ is a line of this plane, which is parallel to the line $AB$ and $A\notin p$. Prove that there are no fixed points of izometry $\mathcal{I}$ on the line $p$. \item Let $S$ be the only fixed point of the isometry $\mathcal{I}$ in some plane. Prove that if this isometry maps the line $p$ to itself, then $S\in p$. \item Prove that any two lines of a plane either intersect or are parallel. \item If a line in a plane intersects one of the two parallel lines of the same plane, then it also intersects the other parallel line. Prove. \item Prove that every isometry maps a parallel line to a parallel line. \item Let $p$, $q$ and $r$ be such lines of a plane that $p\parallel q$ and $r\perp p$. Prove that $r\perp q$. \item Prove that a convex $n$-gon cannot have more than three acute angles. \end{enumerate} % DEL 3 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - %________________________________________________________________________________ % SKLADNOST TRIKOTNIKOV. VEČKOTNIKI %________________________________________________________________________________ \del{Congruence. Triangles and Polygons} \label{pogSKL} %________________________________________________________________________________ \poglavje{Triangle Congruence Theorems} \label{odd3IzrSkl} From the general definition of the congruence of figures it follows that two triangles are congruent if there exists an isometry that maps the first triangle to the second. It is clear that from the congruence of two triangles follows the congruence of the corresponding sides and interior angles. We are interested in the inverse problem: When does the congruence of some of the corresponding sides and angles imply the congruence of two triangles? This is expressed by the following \index{statement!about the congruence of triangles}\emph{triangle congruence theorems}\footnote{The first, third and fourth triangle congruence theorems are attributed to \index{Pythagoras} \textit{Pythagoras of Samos} (6th century BC), while the second theorem is assumed to have been known to \index{Thales} \textit{Thales of Miletus} (7th--6th century BC). All four are mentioned by \index{Euclid} \textit{Euclid of Alexandria} (3rd century BC) in the first book of his \textit{Elements}.}: \bizrek \label{SSS} (\textit{SSS}) Triangles are congruent if three sides of one triangle are congruent to the corresponding sides of the other triangle, i.e. (Figure \ref{sl.skl.3.1.1.pic}): \begin{eqnarray*} \left. \begin{array}{l} AB \cong A'B'\\ BC \cong B'C'\\ AC \cong A'C' \end{array} \right\}\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC \cong \triangle A'B'C' \end{eqnarray*} \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.1.1.pic} \caption{} \label{sl.skl.3.1.1.pic} \end{figure} \textbf{\textit{Proof.}} From the congruence of sides by \ref{izrek(A,B)} there is also $(A,B) \cong (A',B')$, $(B,C) \cong (B',C')$ and $(A,C) \cong (A',C')$ or $(A,B,C) \cong (A',B',C')$. From \ref{IizrekABC} it follows that there is an isometry $\mathcal{I}$, which maps points $A$, $B$ and $C$ to points $A'$, $B'$ and $C'$. It maps triangle $ABC$ to triangle $A'B'C'$, which is a consequence of \ref{izrekIzoB}. So triangles $ABC$ and $A'B'C'$ are congruent. \kdokaz \bizrek \label{SKS} (\textit{SAS}) Triangles are congruent if two pairs of sides and the included angle of one triangle are congruent to the corresponding sides and angle of the other triangle, i.e. (Figure \ref{sl.skl.3.1.2.pic}): \begin{eqnarray*} \left. \begin{array}{l} AB \cong A'B'\\ AC \cong A'C'\\ \angle BAC \cong \angle B'A'C' \end{array} \right\}\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC \cong \triangle A'B'C' \end{eqnarray*} \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.1.2.pic} \caption{} \label{sl.skl.3.1.2.pic} \end{figure} \textbf{\textit{Proof.}} Because angles $BAC$ and $B'A'C'$ are congruent, there exists an isometry $\mathcal{I}$, which maps angle $BAC$ to angle $B'A'C'$. This isometry maps point $A$ to point $A'$, and segments $AB$ and $AC$ to $A'B'$ and $A'C'$. Let $\mathcal{I}(B)=\widehat{B}'$ and $\mathcal{I}(C)=\widehat{C}'$. From this it follows that $AB \cong A'\widehat{B}'$ and $AC \cong A'\widehat{C}'$, but according to \ref{ABnaPoltrakCX} we have that $\widehat{B}'=B'$ and $\widehat{C}'=C'$. So the isometry $\mathcal{I}$ maps points $A$, $B$ and $C$ to points $A'$, $B'$ and $C'$, or triangle $ABC$ to triangle $A'B'C'$, which means that triangles $ABC$ and $A'B'C'$ are congruent. \kdokaz \bizrek \label{KSK} (\textit{ASA}) Triangles are congruent if two pairs of angles and the included side of one triangle are congruent to the corresponding angles and side of the other triangle i.e. (Figure \ref{sl.skl.3.1.3.pic}): \begin{eqnarray*} \left. \begin{array}{l} AB \cong A'B'\\ \angle BAC \cong \angle B'A'C'\\ \angle ABC \cong \angle A'B'C' \end{array} \right\}\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC \cong \triangle A'B'C' \end{eqnarray*} \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.1.3.pic} \caption{} \label{sl.skl.3.1.3.pic} \end{figure} \textbf{\textit{Proof.}} From axiom \ref{aksIII2} it follows that there exists an isometry $\mathcal{I}$, that maps point $A$ to point $A'$, line segment $AB$ to line segment $A'B'$ and plane $ABC$ to plane $A'B'C'$. Because $AB \cong A'B'$, from the same axiom it follows $\mathcal{I}(B)=B'$. Let $\mathcal{I}(C)=\widehat{C}'$. Then $\angle BAC \cong \angle B'A'\widehat{C}'$ and $\angle ABC \cong \angle A'B'\widehat{C}'$. Because according to the assumption also $\angle BAC \cong \angle B'A'C'$ and $\angle ABC \cong \angle A'B'C'$, from \ref{KotNaPoltrak} it follows that line segments $A'\widehat{C}'$ and $A'\widehat{C}'$ (or $B'C'$ and $B'\widehat{C}'$) are equal. Therefore $\widehat{C}'=A'\widehat{C}'\cap A'\widehat{C}'=A'C'\cap A'C'=C'$. So $\mathcal{I}:A,B,C \mapsto A',B',C'$, therefore triangles $ABC$ and $A'B'C'$ are congruent. \kdokaz We will omit the proof of the fourth statement about congruent triangles. \bizrek \label{SSK} (\textit{SSA}) Triangles are congruent if two pairs of sides and the angle opposite to the longer side of one triangle are congruent to the corresponding sides and angle of the other triangle. (Figure \ref{sl.skl.3.1.4.pic}). \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.1.4.pic} \caption{} \label{sl.skl.3.1.4.pic} \end{figure} One of the most important consequences of theorems about the congruence of triangles is the following statement. \bizrek \label{enakokraki} If two sides of a triangle are congruent, then angles opposite those sides are congruent.\\ And vice versa:\\ If angles opposite those sides are congruent, then two sides of a triangle are congruent. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.1.5.pic} \caption{} \label{sl.skl.3.1.5.pic} \end{figure} \textbf{\textit{Proof.}} Let $ABC$ be such a triangle that $AB \cong AC$ (Figure \ref{sl.skl.3.1.5.pic}). Because $AC \cong AB$ and $BC \cong CB$ still hold, from the \textit{SSS} theorem it follows that the triangles $ABC$ and $ACB$ are congruent (these two triangles have different orientations). Therefore, $\angle ABC \cong \angle CBA$. In the same way, we could have proven the converse statement. In this case, we would have used the \textit{ASA} theorem. \kdokaz A triangle (as is the triangle $ABC$ from the previous theorem), which has at least two sides congruent, is called \index{trikotnik!enakokraki}\pojem{an isosceles triangle}. Each of the two congruent sides is a \index{krak!enakokrakega trikotnika}\pojem{arm}, and the third side is \index{osnovnica!enakokrakega trikotnika}\pojem{the base} of this triangle. So, according to the previous theorem, the angles at the base of an isosceles triangle are congruent. And vice versa - if two angles of a triangle are congruent, then this triangle is isosceles. \bzgled Let the $E$ and $F$ be a points lies on the line containing the hypotenuse $AB$ of a perpendicular triangle $ABC$ and let $B(E,A,B,F)$, $EA\cong AC$ and $FB\cong BC$. What is the measure of the angle $ACB$? \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.1.6.pic} \caption{} \label{sl.skl.3.1.6.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.skl.3.1.6.pic}) The internal angle of the triangle $ABC$ at the vertices $A$ and $B$ we denote by $\alpha$ and $\beta$. The triangle $EAC$ and $CBF$ are isosceles triangles with the bases $CE$ and $CF$, so $\angle CEA \cong \angle ACE$ and $\angle CFB \cong \angle BFC$ (statement \ref{enakokraki}). The $\alpha$ is the external angle of the triangle $EAC$, so by the statement \ref{zunanjiNotrNotr}: $\alpha = 2\angle ECA$ or $\angle ECA = \frac{1}{2} \alpha$. Similarly, $\angle FCB = \frac{1}{2} \beta$. Therefore: \begin{eqnarray*} \angle ECF&=&\angle ECA+\angle ACB+\angle BCF=\\ &=&\frac{1}{2} \cdot\alpha+90^0+ \frac{1}{2} \cdot\beta=90^0+ \frac{1}{2} \cdot\left(\alpha+\beta\right)=\\ &=&90^0+ \frac{1}{2}\cdot 90^0=135^0, \end{eqnarray*} or $\angle ECF=135^0$. \kdokaz A triangle in which all sides are equal is called \index{trikotnik!enakostranični}\pojem{an equilateral triangle} (Figure \ref{sl.skl.3.1.7.pic}), which is a special case of an isosceles triangle. Therefore, from the above statement it follows that all the internal angles of an equilateral triangle are equal. Since their sum is equal to $180^0$, each of these angles measures $60^0$. The converse is also true: if at least two angles of a triangle are equal to $60^0$ (and therefore also the third one), then this triangle is equilateral. \begin{figure}[!htb] \centering \input{sl.skl.3.1.7.pic} \caption{} \label{sl.skl.3.1.7.pic} \end{figure} \bzgled Let $ABC$ be an equilateral triangle and $P$, $Q$ and $R$ points such that $\mathcal{B}(A,B,R)$, $\mathcal{B}(B,C,Q)$, $\mathcal{B}(C,A,P)$ and $BR\cong CQ\cong AP$. Prove that $PQR$ is also an equilateral triangle. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.1.8.pic} \caption{} \label{sl.skl.3.1.8.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.1.8.pic}) From the given conditions, it follows first that $AR\cong BQ\cong CP$. The triangle $ABC$ is an equilateral triangle, so all three internal angles measure $60^0$. From this it follows that $\angle PAR\cong \angle RBQ\cong QCP$. By the \textit{SAS} theorem, the triangles $PAR$, $RBQ$ and $QCP$ are similar and $PR\cong RQ\cong QP$. This means that $PQR$ is also an equilateral triangle. \kdokaz In the previous chapter we defined the perpendicular bisector of a line segment as the line that is perpendicular to the line segment and goes through its center. Now we prove an equivalent definition of the perpendicular bisector, which will be very important in the sequel. \index{perpendicular bisector} \bizrek \label{simetrala} The perpendicular bisector of a line segment $AB$ is the set of all points $X$ that are equidistant from its endpoints, i.e. $AX \cong BX$. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.1.9.pic} \caption{} \label{sl.skl.3.1.9.pic} \end{figure} \textbf{\textit{Proof.}} Let $s$ be a line that is the perpendicular bisector of the line segment $AB$ in some plane. By definition, $s$ is perpendicular to the line $AB$ through its center - the point $S$. We denote by $\mathcal{M}$ the set of all points $X$ in this plane, for which $AX \cong BX$. It is necessary to prove that $s =\mathcal{M}$. We will prove this by two inclusions (Figure \ref{sl.skl.3.1.9.pic}). ($s\subseteq \mathcal{M}$). Let $X \in s$. We will prove that then $X \in \mathcal{M}$ is true. From the relations $AS \cong BS$, $XS \cong XS$ and $\angle ASX \cong \angle BSX = 90^0$ it follows that the triangles $ASX$ and $BSX$ are congruent (theorem \ref{SAS}). Therefore $AX \cong BX$ or $X \in \mathcal{M}$. ($\mathcal{M}\subseteq s$). Now let $X \in \mathcal{M}$. We will prove that $X \in s$ is true. From $X \in \mathcal{M}$ it follows that $AX \cong BX$. Now from $AS \cong BS$, $XS \cong XS$ and $AX \cong BX$ it follows that the triangles $ASX$ and $BSX$ are congruent (theorem \ref{SSS}). Therefore the angles $ASX$ and $BSX$ are congruent and the angle between the lines is both a right angle. This means that the line $XS$ is perpendicular to the line segment $AB$ in its center. Therefore, the line $XS$ is the perpendicular bisector of $s$ or $X \in s$. \kdokaz The next problem is an example of multiple use of the theorem of an isosceles triangle (\ref{enakokraki}). \bnaloga\footnote{42. IMO USA - 2001, Problem 5.} In a triangle $ABC$, let $AP$ bisect $\angle BAC$, with $P$ on $BC$, and let $BQ$ bisect $\angle ABC$, with $Q$ on $CA$. It is known that $\angle BAC=60^0$ and that $|AB|+|BP|=|AQ|+|QB|$. What are the possible measures of interior angles of triangle $ABC$? \enaloga \begin{figure}[!htb] \centering \input{sl.skk.4.9.IMO1.pic} \caption{} \label{sl.skk.4.9.IMO1.pic} \end{figure} \textbf{\textit{Solution.}} Let's mark the interior angles of triangle $ABC$ with $\alpha=60^0$, $\beta$ and $\gamma$. Let $D$ and $E$ be such points that $BD\cong BP$, $\mathcal{B}(A,B,D)$, $QE \cong QB$ and $\mathcal{B}(A,Q,E)$ (Figure \ref{sl.skk.4.9.IMO1.pic}). From these conditions it follows that $DBP$ and $BQE$ are isosceles triangles with bases $DP$ and $BE$. From the given condition $|AB|+|BP|=|AQ|+|QB|$ it also follows that $AD\cong AE$, which means that $ADE$ is also an isosceles triangle with base $DE$. Since $DBP$ is an isosceles triangle, from the statements \ref{enakokraki} and \ref{zunanjiNotrNotr} it follows: $\angle BDP\cong \angle BPD =\frac{1}{2}\angle ABC=\frac{1}{2}\beta$. Since $BQE$ is also an isosceles triangle, $\angle QBE\cong\angle QEB$ follows. From the congruence of triangles $ADP$ and $AEP$ (the statement \textit{SAS} \ref{SKS}) it follows that $\angle ADP\cong \angle AEP$ and $PD\cong PE$. If we connect the proven relations, it holds: \begin{eqnarray*} && \angle AEP\cong \angle BDP =\frac{1}{2}\beta\cong \angle QBP\\ &&\textrm{ and } \angle AEB\cong\angle QBE \end{eqnarray*} Let's first assume that $QB>QC$ or $\mathcal{B}(Q,C,E)$ holds. In this case: \begin{eqnarray*} \angle PEB &=&\angle AEB-\angle AEP=\\ &=&\angle QBE-\angle QBP=\\ &=&\angle PBE. \end{eqnarray*} This means that $PBE$ is an isosceles triangle with the base $BE$ or $PE\cong PB$. But from the already proven $PE\cong PD$ and the assumption $PB\cong BD$ it follows that $PD\cong PB\cong BD$, therefore $BDP$ is an equilateral triangle. From this it follows that $\beta=2\angle BDP=2\cdot 60^0=120^0$, or $\alpha+\beta=60^0+120^0=180^0$, which is not possible (the statement \ref{VsotKotTrik}). Therefore the relation $QB>QC$ is not possible. In a similar way, the relation $QB>QC$ leads to a contradiction. This means that only $QB\cong QC$ is possible. In this case $C=E$ and $\gamma=\angle ACB=\angle AEB\cong AEP=\frac{1}{2}\beta$ holds. From $\alpha+\beta+\gamma=180^0$, it follows that $60^0+\beta+\frac{1}{2}\beta=180^0$ or $\beta=80^0$. We have shown that from the conditions of the task it follows that $\beta=80^0$. So the only possible solution is $\beta=80^0$. It is still necessary to show that $\beta=80^0$ is a solution, or that from $\alpha=60^0$, $\beta=80^0$ it follows that $|AB|+|BP|=|AQ|+|QB|$. First, from $\angle QCB=\gamma=\frac{1}{2}\beta=40^0=\angle QBC$ it follows (from the statement \ref{enakokraki}) $QC\cong QB$ or $|AQ|+|QB|=|AQ|+|QC|=|AC|$. If we define the point $D$ in the same way as in the first part, we again get $\angle ADP\cong \angle BPD=\frac{1}{2}\angle ABC=40^0=\angle ACB=\angle ACP$. This means that the triangles $ADP$ and $ACP$ are congruent (from the statement \textit{ASA} \ref{KSK}) or $AD\cong AC$. Therefore, in the end: $$|AB|+|BP|=|AB|+|BD|=|AD|=|AC|=|AQ|+|QB|,$$ which had to be proven. \kdokaz %________________________________________________________________________________ \poglavje{Constructions in Geometry} \label{odd3NacrtNaloge} At the next example we will describe the so-called design tasks. \bzgled \label{načrt1odd3} Two congruent line segments $AB$ and $A'B'$ in a plane are given. Construct a point $C$ such that $\triangle ABC \cong \triangle A'B'C$. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.1.10.pic} \caption{} \label{sl.skl.3.1.10.pic} \end{figure} \textbf{\textit{Solution.}} We assume that $C$ is a point in the plane of the segments, for which $\triangle ABC \cong \triangle A'B'C$. Then $AC\cong A'C$ and $BC\cong B'C$ or the point $C$ lies on the perpendiculars of the segments $AA'$ and $BB'$ (from the statement \ref{simetrala}). This fact allows us to construct (Figure \ref{sl.skl.3.1.10.pic}). We draw the perpendiculars of the segments $AA'$ and $BB'$. We get the point $C$ as their intersection. We prove that $C$ is the desired point or that it satisfies the conditions of the task. By assumption, $AB\cong A'B'$ already. Because we got the point $C$ as the intersection of the perpendiculars of the segments $AA'$ and $BB'$, it is $AC\cong A'C$ and $BC\cong B'C$. From the statement \ref{SSS} (SSS) it follows that the triangles $ABC$ and $A'B'C$ are congruent. The task has a solution (one) exactly when the lines of symmetry $AA'$ and $BB'$ intersect, or when the lines $AA'$ and $BB'$ are not parallel. \kdokaz The previous example is therefore called the \index{task!design} \pojem{design task}, in which for the given data it is necessary to plan or construct a new element or figure, which in relation to the given data satisfies certain conditions. The \pojem{planning} or \index{construction} \pojem{construction} means the use of a ruler and a compass or the use of \index{construction!elementary} \pojem{elementary construction}:\label{elementarneKonstrukcije} \begin{itemize} \item for the given points $A$ and $B$ we draw: \begin{itemize} \item the line $AB$, \item the line segment $AB$, \item the midpoint $AB$; \end{itemize} \item we draw the circle $k$: \begin{itemize} \item with the center $S$, which goes through the given point $A$, \item with the center $S$ and the radius, which is consistent with the given line segment; \end{itemize} \item we draw the circular arc with the given center and radius, \item we draw the intersection (or intersections): \begin{itemize} \item of two lines, \item of a line and a circle, \item of two circles. \end{itemize} \end{itemize} The solution to the design task (draw the figure $\Phi$, which satisfies the conditions $\mathcal{A}$) is formally composed of four steps: \begin{itemize} \item \textit{analysis} - in which we assume that the figure $\Phi$ is already designed and satisfies the conditions $\mathcal{A}$, then we look for new conditions $\mathcal{B}$, which the figure satisfies. These follow from the conditions $\mathcal{A}$ and are more favorable for the construction of the figure $\Phi$. We prove $\mathcal{A}\Rightarrow \mathcal{B}$. \item \textit{construction} - we plan the figure $\Phi'$, which satisfies the conditions $\mathcal{B}$. We exactly describe the course of the design. \item \textit{proof} - we prove that $\Phi' = \Phi$ or $\mathcal{B}\Rightarrow \mathcal{A}$. \item \textit{discussion} - we investigate the number of task solutions, depending on the conditions $\mathcal{A}$. \end{itemize} \bzgled \label{konstrTrik1} Line segments $a$, $l$ and an angle $\alpha$ are given. Construct a triangle $ABC$, such that the side $BC$ is congruent to the line segment $a$, the sum of the sides $AB + AC$ equal to the line segment $l$ and the interior angle $BAC$ congruent to the angle $\alpha$ ($a$, $b+c$, $\alpha$). \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.1.10a.pic} \caption{} \label{sl.skl.3.1.10a.pic} \end{figure} \textbf{\textit{Analysis.}} Let $ABC$ be a triangle, where $BC \cong a$, the sum of $AB+AC$ is equal to the given line segment $l$ and $\angle BAC\cong \alpha$ (Figure \ref{sl.skl.3.1.10a.pic}). Let $D$ be a point on the line segment $BA$, such that $AD\cong AC$ and points $B$ and $D$ are on different sides of point $A$. Therefore $BD=BA+AD=AB+AC=l$ or $BD\cong l$. Triangle $ACD$ is isosceles, so the angles $ADC$ and $ACD$ are congruent by Theorem \ref{enakokraki}. Because they are also the interior angles of triangle $CAD$, both are equal to half of the exterior angle $BAC$ of this triangle (Theorem \ref{zunanjiNotrNotr}), or $\angle BDC=\angle ADC\cong \angle ACD=\frac{1}{2}\angle BAC=\frac{1}{2}\alpha$. This allows us to construct triangle $BCD$. \textbf{\textit{Construction.}} First, let's construct triangle $BCD$, where $\angle BDC=\frac{1}{2}\alpha$, $BC\cong a$ and $BD\cong l$, then point $A$ as the intersection of the line segment $CD$'s perpendicular with line segment $BD$. We will prove that $ABC$ is the desired triangle. \textbf{\textit{Proof.}} First, $BC\cong a$ by construction. By construction, point $A$ lies on the perpendicular of line segment $CD$, so $AD\cong AC$ (Theorem \ref{simetrala}). Therefore triangle $CAD$ is isosceles with the base $CD$, so it is (Theorem \ref{enakokraki}) also $\angle ADC\cong \angle ACD$. Because of this, (Theorem \ref{zunanjiNotrNotr}) $\angle BAC = 2 \cdot \angle BDC= 2\cdot\frac{1}{2}\alpha=\alpha$. From $AD\cong AC$ it follows $AB + AC = AB + AD = BD \cong l$ . \textbf{\textit{Discussion.}} The task has a solution (namely one or two) exactly when the line segment $DC$ intersects the circle $k(B,a)$ and the perpendicular of line segment $CD$ intersects line segment $BD$. \kdokaz In the future, we will not carry out all the steps in every design task. In most cases, we will only do the first step and thus indicate the course of the solution. %________________________________________________________________________________ \poglavje{Triangle inequality} \label{odd3NeenTrik} First, we will prove two important expressions that are a consequence of theorems about the congruence of triangles. \bizrek \label{vecstrveckot} One side of a triangle is longer than another side of a triangle if and only if the measure of the angle opposite the longer side is greater than the angle opposite the shorter side. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.2.1.pic} \caption{} \label{sl.skl.3.2.1.pic} \end{figure} \textbf{\textit{Proof.}} Let $ABC$ be a triangle in which $AC > AB$ (Figure \ref{sl.skl.3.2.1.pic}). We prove that then $\angle ABC > \angle ACB$. Because $AC > AB$, there is such a point $B'$ between points $A$ and $C$, for which $AB \cong AB'$. Then the triangle $BAB’$ is isosceles and $\angle ABB' \cong \angle AB'B$ (theorem \ref{enakokraki}). The segment $BB'$ is inside the angle $ABC$, so $\angle ABC > \angle ABB '$. Then $\angle AB'B$ is the external angle of the triangle $BCB'$. By theorem \ref{zunanjiNotrNotrVecji} this angle is greater than its adjacent internal angle $B'CB$. If we use what has been proven so far, we get: $$\angle ABC > \angle ABB ' \cong \angle AB'B > \angle B'CB \cong \angle ACB.$$ Therefore $\angle ABC > \angle ACB$. In a similar way, we prove that the converse is also true. \kdokaz If we denote the lengths of the sides $BC$, $AC$ and $AB$ of the triangle $ABC$ with $a$, $b$ and $c$, and the measures of the opposite angles at the vertices $A$, $B$ and $C$ with $\alpha$, $\beta$ and $\gamma$, we can write the previous theorem in the form: $$a > b \Leftrightarrow \alpha > \beta,$$ the theorem about the isosceles triangle \ref{enakokraki} in the form: $$a = b \Leftrightarrow \alpha = \beta.$$ This means that both expressions $a-b$ and $\alpha-\beta$ are both positive, both negative or both equal to zero. Thus we have proven the following property: \bzgled \label{vecstrveckotAlgeb} For each triangle $ABC$ is: $$(a-b)(\alpha-\beta)\geq 0, \hspace*{4mm} (b-c)(\beta-\gamma)\geq 0, \hspace*{4mm} (c-a)(\gamma-\alpha)\geq 0$$ \ezgled The following proposition follows directly from \ref{vecstrveckot}: \bizrek \label{vecstrveckotHipot} The hypotenuse of a right-angled triangle is longer than its two legs. The longest side of an obtuse triangle is the one opposite to the obtuse angle. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.2.2.pic} \caption{} \label{sl.skl.3.2.2.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.2.2.pic}) The sum of the inner angles of a triangle is equal to $180^0$. Therefore, in a right-angled triangle, the largest angle is a right angle. The hypotenuse of a right-angled triangle is the longest side of this triangle according to the previous proposition. We can similarly prove this for an obtuse triangle. \kdokaz A line $AA'$ is the \index{height!of a triangle}\pojem{height} of a triangle $ABC$, if $AA'\perp BC$ and $A'\in BC$. The latter of the two relations means that the point $A'$ lies on the line $BC$, but not necessarily on the line segment $BC$. The relation $\mathcal{B}(B,A',C)$ is valid exactly when both of the internal angles at the vertices $B$ and $C$ are acute (Figure \ref{sl.skl.3.2.3.pic}). This is a consequence of the theorem about the sum of the internal angles of any triangle (theorem \ref{VsotKotTrik}). In the case that $\angle ABC\geq 90^0$ and $\mathcal{B}(B,A',C)$, the sum of the internal angles in the triangle $ABA'$ would be greater than $180^0$. So the height of a triangle is not always inside the triangle. In a right triangle, the heights from the two vertices with acute angles are equal to the corresponding catheti. The heights from the vertices $A$, $B$ and $C$ are usually denoted by $v_a$, $v_b$ and $v_c$. From the previous theorem \ref{vecstrveckotHipot} it follows that the length of the height of any triangle is less than or equal to the length of the opposite side of that triangle, e.g.: $v_a\leq b$, $v_a\leq c$, ... \begin{figure}[!htb] \centering \input{sl.skl.3.2.3.pic} \caption{} \label{sl.skl.3.2.3.pic} \end{figure} Now we will solve a design problem in which the height of a triangle is given as data. \bzgled Construct a triangle $ABC$ such that the sides $AB$, $AC$ and the altitude from the vertex $B$ are congruent to the three given line segments $c$, $b$ and $v_b$. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.2.4a.pic} \caption{} \label{sl.skl.3.2.4a.pic} \end{figure} \textbf{\textit{Analysis.}} Let $ABC$ be a triangle for which $AB\cong c$, $AC\cong b$ and $AD\cong v_b$ (where $BD$ is the height of this triangle from the vertex $B$). In the right triangle $ABD$ therefore the known hypotenuse $AB\cong c$ and the cathetus $AD\cong v_b$ are known, which means that we can design it. The third vertex $C$ of the triangle $ABC$ lies on the line $AD$ (Figure \ref{sl.skl.3.2.4a.pic}). \textbf{\textit{Construction.}} Let's first draw a rectangular triangle $ABD$ (with conditions: $AB \cong c$, $\angle ADB = 90^0$ and $BD \cong v_b$). On the line $AD$ then determine such a point $C$, so that $AC \cong b$. Prove that $ABC$ is the desired triangle. \textbf{\textit{Proof.}} First, $AB \cong c$ and $AC \cong b$ already by construction. Since $\angle ADB = 90^0$, $BD$ is the height of the triangle $ABC$ from the vertex $B$ and is consistent with the distance $v_b$ by construction. \textbf{\textit{Discussion.}} The task has a solution exactly when it is possible construction of the triangle $ABD$ or $hb \leq c$. In the construction of point $C$ there are two possibilities - on different sides of point $A$, which means that we have two solutions for triangle $ABC$. In the case $hb \cong c$, the solutions are a right triangle and a congruent triangle. \kdokaz \bzgled If $v_a$, $v_b$ and $v_c$ are altitudes corresponding to the sides $a$, $b$ and $c$ of a triangle, then: $$\frac{v_a}{b+c}+\frac{v_b}{a+c}+\frac{v_c}{a+b}<\frac{3}{2}.$$ \ezgled \textbf{\textit{Proof.}} By adding the inequalities $v_a \leq b$, $v_a \leq c$ we get $2v_a \leq b + c$ or $\frac{v_a}{b + c} \leq \frac{1}{2}$. Similarly, we get $\frac{v_b}{a + c} \leq \frac{1}{2}$ and $\frac{v_c}{a + b} \leq \frac{1}{2}$. Since all inequalities cannot be true at the same time, by adding we get the desired inequality. \kdokaz The next property of the triangle will be called \index{triangle inequality} \pojem{triangle inequality}. \bizrek \label{neenaktrik} The sum of any two sides of a triangle is greater than the third side. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.2.4.pic} \caption{} \label{sl.skl.3.2.4.pic} \end{figure} \textbf{\textit{Proof.}} Let $ABC$ be an arbitrary triangle. We will prove that $AB + AC > BC$. We mark a point $D$ so that $\mathcal{B}(B,A,D)$ and $AD \cong AC$ (Figure \ref{sl.skl.3.2.4.pic}). By the \ref{enakokraki} theorem ($\triangle CAD$ is an isosceles triangle with the base $CD$), we also have that $\angle BDC=\angle ADC \cong \angle ACD$. The line segment $CA$ is inside the angle $DCB$, so $\angle ACD < \angle DCB$. Then $\angle BDC < \angle DCB$ as well. From \ref{vecstrveckot} (referring to the triangle $BCD$) it follows that: $$BC < BD = AB + AD = AB + AC,$$ which was to be proven. \kdokaz From the triangle inequality we obtain a criterion for the existence of such a triangle, that its sides are consistent with three given line segments. \bzgled Let $a$, $b$ and $c$ be three line segments. A triangle with sides $a$, $b$ and $c$ exist if and only if: $$b + c > a,\hspace*{2mm} a + c > b \hspace*{1mm}\textrm{ in }\hspace*{1mm} a + b > c.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.2.5.pic} \caption{} \label{sl.skl.3.2.5.pic} \end{figure} \textbf{\textit{Proof.}} If such a triangle exists, then the three relations are direct consequences of the triangle inequality \ref{neenaktrik}. So we assume that all three relations are true. Without loss of generality, let $a$ be the longest side of this triangle (it is enough that it is not shorter than any other side) and let $B$ and $C$ be any points for which $BC \cong a$ (Figure \ref{sl.skl.3.2.5.pic}). Because $b + c > a$, this means that the circles $k(B,c)$ and $k(C,b)$ intersect in some point $A$ (a consequence of Dedekind's axiom - \ref{DedPoslKrozKroz} theorem, because each of them contains the inner points of the other), which is not on the line segment $BC$. The triangle $ABC$ is then the desired triangle. \kdokaz If we know which of the three sides is the longest, it is enough to check only one inequality, as the other two are automatically fulfilled. The proof of the previous statement can also be used for the following, equivalent criterion. \bzgled \label{neenaktrik1} Let $a$, $b$ and $c$ be three line segments, such that $a \geq b,c$. A triangle with sides $a$, $b$ and $c$ exist if and only if $b + c > a$. \ezgled For example, we can determine that a triangle exists with sides that have lengths 7, 5 and 3 (because $5+3>7$), but a triangle with sides that have lengths 9, 6 and 2 does not exist (because $6+2$ is not greater than 9). Let's look at some consequences of the previous statements. \bzgled If $X$ is an arbitrary point of the side $BC$ of a triangle $ABC$, then: $$AX < AB + AC.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.2.6.pic} \caption{} \label{sl.skl.3.2.6.pic} \end{figure} \textbf{\textit{Proof.}} If we use the triangle inequality for the triangles $ABX$ and $AXC$ (Figure \ref{sl.skl.3.2.6.pic}), we get: $$AX < AB + BX \hspace*{1mm} \textrm{ and }\hspace*{1mm} AX < AC + CX.$$ By adding these two inequalities and using the triangle inequality for the triangle $ABC$, we get: $$2AX < AB + AC + BC < 2(AB + AC),$$ which was to be proven. \kdokaz %% !!! Dosegel magično stran - 100!!! Wow Bravo!!! The next inequality is a generalization of the previous one. In this sense, the previous statement is its consequence and it was not necessary to prove it separately. \bzgled Let $X$ be an arbitrary point of the side $BC$, different from the vertices $B$ and $C$ of a triangle $ABC$. Then the line segment $AX$ is shorter than at least one of the two line segments $AB$ and $AC$ i.e.: $$AX < \max\{AB, AC\}.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.2.7.pic} \caption{} \label{sl.skl.3.2.7.pic} \end{figure} \textbf{\textit{Proof.}} Because for the point $X$ it holds that $\mathcal{B}(B, X, C)$, one of the sides $AXB$ and $AXC$ is not acute. Without loss of generality, let it be $AXC$ (Figure \ref{sl.skl.3.2.7.pic}). Then it is the largest angle in the triangle $AXB$, which means that \\ $AX < AC$ (statement \ref{vecstrveckot}). Similarly, if the angle $AXB$ is not acute, it holds that $AX < AB$. \kdokaz We will especially consider the case of the distance $AX$, if the point $X$ is from the previous two statements the center of the side $BC$ (Figure \ref{sl.skl.3.2.8.pic}). Such a distance, which is determined by the vertex and the center of the opposite side of the triangle, is called \index{težiščnica trikotnika} \pojem{težiščnica} trikotnika. Težiščnice, ki ustrezajo ogliščem $A$, $B$ in $C$ trikotnika $ABC$, običajno označujemo s $t_a$, $t_b$ in $t_c$. Zadnji dve trditvi lahko uporabimo tudi za težiščnice. Toda za težiščnice velja še dodatna lastnost, ki jo bomo sedaj dokazali. \begin{figure}[!htb] \centering \input{sl.skl.3.2.8.pic} \caption{} \label{sl.skl.3.2.8.pic} \end{figure} \bzgled \label{neenTezisZgl} If $a$, $b$, $c$ are the sides and $t_a$ the corresponding median of a triangle $ABC$, then: $$\frac{b+c-a}{2} AB$ and $AA_1 + A_1C > AC$ or: $$t_a+\frac{a}{2}>c \hspace*{2mm} \textrm{ in } \hspace*{2mm} t_a+\frac{a}{2}>b.$$ If we add these two inequalities, we get $\frac{b+c-a}{2} AD = 2AA_1 = 2t_a,$$ which had to be proven. \kdokaz We will show some more examples of the use of the triangle inequality. \bzgled Let $M$ be an arbitrary point of the bisector of the exterior angle at the vertex $C$ of a triangle $ABC$. Then $$MA + MB \geq CA + CB.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.2.10.pic} \caption{} \label{sl.skl.3.2.10.pic} \end{figure} \textbf{\textit{Proof.}} Let $D$ be such a point on the line segment $BC$, that $AC \cong CD$ and $\mathcal{B}(A,C,D)$ (Figure \ref{sl.skl.3.2.10.pic}). The triangles $ACM$ and $DCM$ are congruent, by statement \textit{SAS} \ref{SKS} ($AC \cong DC$, $CM \cong CM$, $\angle ACM \cong \angle DCM$). Therefore $MA \cong MD$. If we use the triangle inequality now, we get: $$MA + MB = MD + MB \geq BD = DC + CB = CA + CB.$$ Of course, the equality holds when the points $B$, $M$ and $D$ are collinear or $M = C$. \kdokaz \bzgled If the bisector of the interior angle at the vertex $A$ of a triangle $ABC$ intersects the side $BC$ in the point $E$, then $$AB > BE \hspace*{2mm} \textrm{ in }\hspace*{2mm} AC > CE .$$ \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.2.11.pic} \caption{} \label{sl.skl.3.2.11.pic} \end{figure} \textbf{\textit{Proof.}} Because $\mathcal{B}(B,E,C)$, the angle $AEB$ is the external angle of triangle $AEC$ (Figure \ref{sl.skl.3.2.11.pic}). Therefore $\angle BEA > \angle EAC \cong \angle BAE$ (statement \ref{zunanjiNotrNotrVecji}). Opposite the larger angle in triangle $BAE$ is the larger side, or $AB > BE$ (statement \ref{vecstrveckot}). Similarly, we prove that the other of the two relations is also true. \kdokaz \bzgled \label{zgled3.2.9} If $M$ is the interior point of a triangle $ABC$, then \\ $BA + AC > BM + MC.$ \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.2.12.pic} \caption{} \label{sl.skl.3.2.12.pic} \end{figure} \textbf{\textit{Proof.}} Let $N$ be the intersection of lines $BM$ and $CA$ (Figure \ref{sl.skl.3.2.12.pic}). Because $M$ is an interior point of triangle $ABC$, $\mathcal{B}(B,M,N)$ and $\mathcal{B}(A,N,C)$ are true. If we now use the triangle inequality (statement \ref{neenaktrik}) twice, we get: \begin{eqnarray*} \hspace*{-4mm}BM + MC &<& BM + (MN + NC) = (BM + MN) + NC = BN + NC\\ \hspace*{-4mm}&<& (BA + AN) + NC = BA + (AN + NC) = BA + AC. \end{eqnarray*} Let's define two new concepts. The sum of all sides of a polygon is called its \index{obseg!večkotnika} \pojem{obseg}. Half of this sum is the \pojem{polobseg} of this polygon. \bzgled If $M$ is the interior point and $s$ the semiperimeter of a triangle $ABC$, then $$s < AM + BM + CM < 2s.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.2.13.pic} \caption{} \label{sl.skl.3.2.13.pic} \end{figure} \textbf{\textit{Proof.}} We first get the inequality, if we use the triangle inequality for the triangles $MAB$, $MBC$ and $MCA$ and then add them up. The second inequality is obtained, if we use the previous statement (Example \ref{zgled3.2.9}) and add the corresponding inequalities (Figure \ref{sl.skl.3.2.13.pic}). \kdokaz \bzgled In each convex pentagon there exist three diagonals, which are congruent to the sides of a triangle. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.2.14.pic} \caption{} \label{sl.skl.3.2.14.pic} \end{figure} \textbf{\textit{Proof.}} Let $AD$ be the longest diagonal of the pentagon $ABCDE$ (not shorter than any other diagonal). We prove that $AD$, $AC$ and $BD$ are the desired diagonals, i.e. those for which there exists a triangle, whose sides are congruent to these diagonals (Figure \ref{sl.skl.3.2.14.pic}). Because $AD\geq AC$ and $AD\geq BD$, it is enough to prove (Example \ref{neenaktrik1}), that $AC + BD > AD$. The pentagon $ABCDE$ is convex, so its diagonals $AC$ and $BD$ intersect in some point $S$. Then: $$AC + BD > AS + SD > AD,$$ which was to be proven. \kdokaz The following consequence of the theorems about the congruence of triangles is very important. We will also need the triangle inequality in this proof. \bizrek \label{SkladTrikLema} Let $ABC$ and $A'B'C'$ triangles such that $AB \cong A'B'$ and $AC \cong A'C'$. Then $BC > B'C'$ if and only if $\angle BAC > \angle B' A'C'$ i.e. $$BC > B'C' \Leftrightarrow \angle BAC > \angle B' A'C'.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.2.15.pic} \caption{} \label{sl.skl.3.2.15.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.2.15.pic}) ($\Leftarrow$) Let $\angle BAC > \angle B' A'C'$. Then there exists within the angle $BAC$ such a line segment $l$, that $\angle BA,l \cong \angle B'A'C'$. With $C''$ we mark the point of the line segment $l$, for which $AC'' \cong A'C'$. Then both triangle $ABC''$ and $A'B'C'$ are congruent and $BC'' \cong B'C'$. It is enough to prove that $BC > BC''$. If $C''$ lies on the side $BC$, it is trivially fulfilled. We assume that the point $C''$ does not lie on the side $BC$. Let the point $E$ be the intersection of the perpendicular bisector of the angle $CAC''$ and the side $BC$. By the \textit{SAS} theorem, the triangles $ACE$ and $AC''E$ are congruent, therefore $CE \cong C''E$. Now it is: $$BC = BE + EC = BE + EC'' \hspace{0.1mm} > BC'' = B'C'.$$ ($\Rightarrow$) Let $BC > B'C'$. The relation $\angle BAC \cong \angle B' A'C'$ is not true, because then (by the \textit{SAS} theorem) the triangle $ABC$ and $A'B'C'$ would be congruent and then also $BC \cong B'C'$. If it would be true that $\angle BAC < \angle B' A'C'$, then from what has already been proven it would follow that $BC < B'C'$. Therefore $\angle BAC > \angle B' A'C'$. \kdokaz \bizrek \label{neenakIzlLin} If $A_1A_2\ldots A_n$ ($n\in \mathbb{N}$, $n\geq 3$) is polygonal chain, then $$|A_1A_2|+|A_2A_2|+\cdots +|A_{n-1}A_n|\geq |A_1A_n|.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.2.16.pic} \caption{} \label{sl.skl.3.2.16.pic} \end{figure} \textbf{\textit{Proof.}} We will carry out the proof by induction over $n$ (Figure \ref{sl.skl.3.2.16.pic}). In the case $n=3$ we get the triangle inequality - izrek \ref{neenaktrik}. Let's assume that the inequality is true for $n=k$ ($k\in \mathbb{N}$, $k> 3$) or $|A_1A_2|+|A_2A_2|+\cdots +|A_{k-1}A_k|\geq |A_1A_k|.$ We will prove that the inequality is also true for $n=k+1$ or $|A_1A_2|+|A_2A_2|+\cdots +|A_kA_{k+1}|\geq |A_1A_{k+1}|.$ If we first use the induction assumption, and then the triangle inequality, we get: \begin{eqnarray*} && |A_1A_2|+|A_2A_2|+\cdots +|A_{k-1}A_k|+|A_kA_{k+1}|\geq\\ && \geq|A_1A_k|+|A_kA_{k+1}|\geq |A_1A_{k+1}|, \end{eqnarray*} which is what needed to be proven. \kdokaz We will now prove another inequality that is true in any triangle. \bzgled If $a$, $b$, $c$ are the sides and $\alpha$, $\beta$, $\gamma$ the opposite interior angles of a triangle, then $$60^0\leq \frac{a\alpha+b\beta +c\gamma}{a+b+c} < 90^0.$$ \ezgled \textbf{\textit{Proof.}} We will prove each of the inequalities separately. In doing so, we will use the statement about the sum of the interior angles of a triangle (statement \ref{VsotKotTrik}). First, we will prove the second inequality: \begin{eqnarray*} \frac{a\alpha+b\beta +c\gamma}{a+b+c} < 90^0 &\Leftrightarrow& a\alpha+b\beta +c\gamma - 90^0(a+b+c)<0\\ &\Leftrightarrow& a(\alpha-90^0)+b(\beta-90^0) +c(\gamma-90^0)<0\\ &\Leftrightarrow& a(180^0-2\alpha)+b(180^0-2\beta) +c(180^0-2\gamma)>0\\ &\Leftrightarrow& a(\beta+\gamma-\alpha)+b(\alpha+\gamma-\beta) + c(\alpha+\beta-\gamma)>0\\ &\Leftrightarrow& \alpha(b+c-a)+\beta(a+c-b) + \gamma(a+b-c)>0 \end{eqnarray*} The last inequality is fulfilled because, according to the triangle inequality (statement \ref{neenaktrik}), $b+c-a>0$, $a+c-b>0$ and $a+b-c>0$ hold. We will now prove the first inequality: \begin{eqnarray*} && \frac{a\alpha+b\beta +c\gamma}{a+b+c} \geq 60^0\Leftrightarrow\\ &\Leftrightarrow& a\alpha+b\beta +c\gamma - 60^0(a+b+c)\geq 0\\ &\Leftrightarrow& a(\alpha-60^0)+b(\beta-60^0) +c(\gamma-60^0)\geq 0\\ &\Leftrightarrow& a(3\alpha-180^0)+b(3\beta-180^0) +c(3\gamma-180^0)\geq 0\\ &\Leftrightarrow& a(2\alpha-\beta-\gamma)+b(2\beta-\alpha-\gamma) + c(2\gamma-\alpha-\beta)\geq 0\\ &\Leftrightarrow& a(\alpha-\beta+\alpha-\gamma)+b(\beta-\alpha+\beta-\gamma) + c(\gamma-\alpha+\gamma-\beta)\geq 0\\ &\Leftrightarrow& a(\alpha-\beta)+a(\alpha-\gamma)+ b(\beta-\alpha)+b(\beta-\gamma) + c(\gamma-\alpha)+c(\gamma-\beta)\geq 0\\ &\Leftrightarrow& (a-b)(\alpha-\beta)+(a-c)(\alpha-\gamma)+ (b-c)(\beta-\gamma) \geq 0 \end{eqnarray*} The last inequality is a consequence of statement \ref{vecstrveckotAlgeb}. \kdokaz The next statement will be the motivation for defining the distance of a point from a line. \bizrek Let $A'=pr_{\perp p}(A)$ be the foot of the perpendicular from a point $A$ on a line $p$. If $X\in p$ and $X\neq A'$, then $AX>AA'$. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.2.17.pic} \caption{} \label{sl.skl.3.2.17.pic} \end{figure} \textbf{\textit{Proof.}} By definition, $AA'\perp p$ (Figure \ref{sl.skl.3.2.17.pic}), which means that $AA'X$ is a right angled triangle with hypotenuse $AX$. From \ref{vecstrveckotHipot} it follows that $AX>AA'$. \kdokaz If $A'=pr_{\perp p}(A)$, we say that the length of the line $AA'$ \index{distance!point from a line} \pojem{distance of point $A$ from line $p$}. We denote it by $d(A,p)$. So $d(A,p)=|AA'|$. %________________________________________________________________________________ \poglavje{Circle and Line} \label{odd3KrozPrem} In the following we will deal with a circle and the mutual position of a circle and a line. We prove first a property of the diameter of a circle, which is a simple consequence of the triangle inequality. \bizrek \label{premerNajdTetiva} The longest chord of a circle is its diameter. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.3.1.pic} \caption{} \label{sl.skl.3.3.1.pic} \end{figure} \textbf{\textit{Proof.}} Let $AB$ be any chord of a circle that is not a diameter, and $C$ a point on the line $AS$, for which $CS\cong SA$ and $\mathcal{B}(A,S,C)$ (Figure \ref{sl.skl.3.3.1.pic}). Then point $C$ lies on the circle $k$ and $AC$ is its diameter. We have already shown (a consequence of \ref{premerInS}) that all diameters of a circle are mutually congruent. So it is enough to show that $AC>AB$. This follows from the triangle inequality (triangle $ASB$). It holds: $$AC=AS+SC=AS+SB>AB,$$ which was to be proved. \kdokaz It follows another property of a chord of a circle, as a consequence of \ref{vecstrveckotHipot}. \bzgled \label{tetivaNotrTocke} Every point that lies on a chord of a circle, except its endpoints, is an interior point of the circle. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.3.2.pic} \caption{} \label{sl.skl.3.3.2.pic} \end{figure} \textbf{\textit{Proof.}} Let $X$ be an inner point of the segment $AB$ with shorter sides on the circle $k(S,r)$ (Figure \ref{sl.skl.3.3.2.pic}). The angle $AXS$ and $BXS$ are adjacent angles, which means that they are not both acute angles. Without loss of generality, assume that the angle $BXS$ is not an acute angle. Then in the triangle $SXB$ the side $SB$ is the longest (by \ref{vecstrveckotHipot}), which means that: $$SX ST = r.$$ Therefore, none of the points $T_1$ ($T_1 \neq T$), lying on the line $PT$, lies on the circle $k$. This means that the line $PT$ is a tangent of this circle. \kdokaz From the proof of the previous statement ($\Leftarrow$) it follows that all points lying on the tangent of the circle (except for its point of contact), are external points of this circle. With this property we will prove the following statement. \bzgled \label{tangKrozEnaStr} All points of a circle are on the one side of its tangent. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.3.7.pic} \caption{} \label{sl.skl.3.3.7.pic} \end{figure} \textbf{\textit{Proof.}} Let $T$ be the point of tangency of the circle $k(S, r)$ and its tangent $t$ (Figure \ref{sl.skl.3.3.7.pic}). The tangent $t$ divides the plane in which $k$ and $t$ lie into two half-planes. The half-plane in which the point $S$ lies, we denote by $\alpha_1$, the other half-plane by $\alpha_2$. We prove that all points of the circle $k$ lie in the half-plane $\alpha_1$. Let $X$ be an arbitrary point of the half-plane $\alpha_2$. Since the points $S$ and $X$ are on different sides of the line $t$, it follows that the open line segment $SX$ intersects at some point $Y$. Then we have: $$SX = SY + YX > SY \geq ST = r,$$ which means that the point $X$ does not lie on the circle $k$ and is its external point. Therefore, none of the points of the half-plane $\alpha_2$ lies on the circle $k$, that is, all of them are in the half-plane $\alpha_1$ with the edge $t$. \kdokaz A direct consequence of the statement \ref{TangPogoj} is also that in each point of the circle we can draw only one tangent. If $X$ is an internal point of the circle $k(S, r)$, then no tangent passes through this point, since all lines through $X$ are intersecting, which is a consequence of Dedekind's axiom (statement \ref{DedPoslKrozPrem}). Later (statement \ref{tangentiKroznice}) we will find that through each external point of the circle we can draw exactly two tangents. For the time being, we prove the following statement (the reader will remember that this is a statement that we have already considered at the beginning of the introductory chapter - statement \ref{TalesUvod}). \bizrek \label{TalesovIzrKroz} \index{izrek!Talesov za krožnico} Thales' theorem for a circle\footnote{Starogrški filozof in matematik \textit{Tales} \index{Tales} iz Mileta (640--546 pr. n. š.) ni prvi, ki je odkril to trditev. Kot empirično dejstvo so jo poznali že stari Egipčani in Babilonci. Izrek imenujemo po Talesu, ki ga je prvi dokazal. V dokazu je uporabljal lastnosti enakokrakih trikotnikov in dejstvo, da je vsota notranjih kotov trikotnika enaka vsoti dveh pravih kotov. Torej je dokaz enak temu, ki ga bomo izpeljali tukaj.}:\\ Let $AB$ be a diameter of a circle $k$. Then for any point $X$ of this circle different from $A$ and $B$ ($X\in k$ in $X\neq A$ in $X\neq B$) is $\angle AXB=90^0$ \index{izrek!Talesov za krožnico} \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.3.8.pic} \caption{} \label{sl.skl.3.3.8.pic} \end{figure} \textbf{\textit{Proof.}} Naj bo $O$ središče krožnice $k$ (Figure \ref{sl.skl.3.3.8.pic}). Ker $A,B,X\in k$, je $OA\cong OB\cong OX$. Torej sta trikotnika $AOX$ in $BOX$ enakokraka, zato je (izrek \ref{enakokraki}): $\angle AXO\cong\angle XAO=\alpha$ in $\angle BXO\cong\angle XBO=\beta$. Tedaj je $\angle AXB=\alpha+\beta$. Vsota notranjih kotov v trikotniku $AXB$ je enaka $180^0$ (izrek \ref{VsotKotTrik}), torej je $2\alpha+2\beta=180^0$. Iz tega sledi $$\angle AXB=\alpha+\beta=90^0,$$ kar je bilo treba dokazati. \kdokaz Dokažimo tudi obratno trditev. \bizrek \label{TalesovIzrKrozObrat} If $A$, $B$ and $X$ are non-collinear points such that $\angle AXB=90^0$, then the point $X$ lies on a circle with the diameter $AB$. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.3.9.pic} \caption{} \label{sl.skl.3.3.9.pic} \end{figure} \textbf{\textit{Proof.}} Let $O$ be the center of the line $AB$ and $k$ the circle with center $O$ and radius $OA$ or diameter $AB$ (Figure \ref{sl.skl.3.3.9.pic}). From $\angle AXB=90^0$ it follows from the theorem \ref{VsotKotTrik}: \begin{eqnarray} \angle XAB+ \angle XBA = 90^0 \label{relacija336} \end{eqnarray} We prove $X\in k$. We assume the contrary, i.e. that the point $X$ does not lie on the circle $k$. In this case $OX\neq OA$. Let $X_1$ be the point on the segment $OX$, for which $OX_1\cong OA$ (theorem \ref{ABnaPoltrakCX}). This means that the point $X_1$ lies on the circle $k$ and by the theorem \ref{TalesovIzrKroz} we have $\angle AX_1B=90^0$. By our assumption $OX\neq OA$ it is clear that $X\neq X_1$. We will consider two possibilities: \textit{1)} Let $OX_1OX$ or $\mathcal{B}(O,X,X_1)$. Similarly as in the first case we get: $$\angle X_1AB+ \angle X_1BA>\angle XAB+ \angle XBA = 90^0.$$ In this case the sum of the angles in the triangle $AX_1B$ is greater than $180^0$, which by the theorem \ref{VsotKotTrik} is not possible. It follows that $OX=OX_1$ or $X\in k$. \kdokaz Use the previous theorem for the design of tangents. \bzgled \label{tangKrozKonstr} Let $A$ be an exterior point of a circle $k(S,r)$. Construct all tangents of this circle passing through the point $A$. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.3.10.pic} \caption{} \label{sl.skl.3.3.10.pic} \end{figure} \textbf{\textit{Solution.}} Let $l$ be a circle with diameter $SA$ (Figure \ref{sl.skl.3.3.10.pic}). Because $S$ is an interior, $A$ is an exterior point of the given circle $k$, the circle $k$ and $l$ have exactly two common points $T_1$ and $T_2$ (statement \ref{DedPoslKrozKroz}). By Tales' statement \ref{TalesovIzrKroz}, $\angle ST_1A\cong \angle ST_2A=90^0$. Because $ST_1$ and $ST_2$ are radii of the circle $k$, $AT_1$ and $AT_2$ are tangents of the circle $k$ through the point $A$ (statement \ref{TangPogoj}). We prove that $AT_1$ and $AT_2$ are the only tangents of the circle $k$ from the point $A$. If $AT$ is a tangent from the point $A$, which the circle $k$ touches in the point $T$, by statement \ref{TangPogoj} $\angle ATS=90^0$. This means that the point $T$ lies on the circle $l$ (statement \ref{TalesovIzrKrozObrat}) or $T\in k\cap l$. Therefore $T$ is one of the points $T_1$ and $T_2$, so $AT_1$ and $AT_2$ are the only tangents of the circle $k$ from the point $A$. \kdokaz The following statement follows from the previous construction. \bizrek \label{tangentiKroznice} If $V$ is an exterior point of a circle $k(S,r)$, then there are exactly two tangents of the circle $k$ through the point $V$. \eizrek We prove some more properties of a tangent of a circle. \bzgled \label{TangOdsek} If $VA$ and $VB$ are tangents of a circle $k(S,r)$ in a points $A$ and $B$ of the circle, then the centre $S$ lies on the bisector of the angle $AVB$ and $VA \cong VB$. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.3.11.pic} \caption{} \label{sl.skl.3.3.11.pic} \end{figure} \textbf{\textit{Proof.}} From izrek \ref{TangPogoj} it follows: $VA \perp AS$ and $VB \perp BS$ (Figure \ref{sl.skl.3.3.11.pic}). Therefore $ASV$ and $BSV$ are right-angled triangles with a shared hypotenuse $SV$. Because $SA \cong SB = r$, these two triangles are congruent (izrek \textit{SSA} \ref{SSK}). Therefore the angles $AVS$ and $BVS$ are also congruent, which means that the line $VS$ is the bisector of angle $AVB$. From the congruence of these two triangles it also follows that $VA \cong VB$. \kdokaz The converse is also true. \bzgled \label{SimKotaKraka} If a point $S$ lies on the bisector of a convex angle, then it is the centre of a circle touching both sides of this angle. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.3.12.pic} \caption{} \label{sl.skl.3.3.12.pic} \end{figure} \textbf{\textit{Proof.}} Let $A$ and $B$ be the right-angled projections of point $S$ on the sides of given angle with the vertex $V$ (Figure \ref{sl.skl.3.3.12.pic}). Triangles $ASV$ and $BSV$ are congruent (izrek \textit{ASA} \ref{KSK}), because they have a shared side $VS$ and two pairs of congruent angles - from $\angle AVS\cong \angle BVS$ and $\angle SAV\cong \angle SBV=90^0$ it follows that $\angle ASV\cong \angle BSV$. Therefore $SA\cong SB$ and $k(S,SA)$ is the desired circle. The sides of given angle are tangents to the circle by izrek \ref{TangPogoj}. \kdokaz Now we will prove another criterion for the mutual position of a line and a circle in a plane. \bizrek \label{TangSekMimobKrit} Let $P$ be the foot of the perpendicular from the centre of a circle $k(S,r)$ on a line $p$ (lying in the plane of the circle). Then the line $p$ is: (i) secant, if and only if $SP < r$; (ii) tangent, if and only if $SP \cong r$; (iii) non-intersecting line, if and only if $SP> r$. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.3.13.pic} \caption{} \label{sl.skl.3.3.13.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.3.13.pic}) The statement (ii) follows directly from the criterion for tangency (the statement \ref{TangPogoj}). (i) In the proof of the direct direction of equivalence we use the fact that the hypotenuse of a right angled triangle is longer than either of the two shorter sides (the statement \ref{vecstrveckotHipot}). If $A$ and $B$ are the intersections of the secant $p$ and the circle $k$, then $SA$ is the hypotenuse of the right angled triangle $ASP$ and it holds: $r \cong SA > SP$. In the proof of the inverse direction of equivalence we use the consequence of Dedekind's axiom (the statement \ref{DedPoslKrozPrem}). Because in this case $P$ is an inner point of this circle, each straight line of this plane that goes through the point $P$ is a secant of the circle $k$. (iii) It follows from the proven (i) and (ii). Because if $SP > r$, then neither $SP < r$ nor $SP \cong r$. From the equivalences (i) and (ii) it follows that the straight line $p$ is neither a secant nor a tangent. Therefore $p$ is a parallel of the circle $k$. In the same way we prove the inverse direction of equivalence. \kdokaz %________________________________________________________________________________ \poglavje{Quadrilaterals} \label{odd3Stirik} In section \ref{odd2AKSURJ} we introduced the concept of a quadrilateral as a polygon with four sides and four vertices. We defined the concepts of adjacent and opposite sides, adjacent and opposite angles and diagonal. To a quadrilateral $ABCD$ with the lengths of its sides $AB$, $BC$, $CD$ and $DA$ we usually assign the letters $a$, $b$, $c$ and $d$, and to the lengths of its diagonals $AC$ and $BD$ the letters $e$ and $f$ (Figure \ref{sl.skl.3.4.1.pic}). \begin{figure}[!htb] \centering \input{sl.skl.3.4.1.pic} \caption{} \label{sl.skl.3.4.1.pic} \end{figure} In the same section, we introduced the concepts of internal and external angles of a quadrilateral. We also mentioned that the internal angles at vertices $A$, $B$, $C$ and $D$ of a quadrilateral $ABCD$ are usually denoted by $\alpha$, $\beta$, $\gamma$ and $\delta$, and its external angles by $\alpha'$, $\beta'$, $\gamma'$ and $\delta'$. We proved (as a consequence of the general statement \ref{VsotKotVeck}) that the sum of all four internal angles of an arbitrary quadrilateral is equal to $360^0$ (Figure \ref{sl.skl.3.4.1.pic}). The sum of external angles is also equal to $360^0$ (in a convex quadrilateral). So: \begin{eqnarray*} \alpha+\beta+\gamma+\delta=360^0,\\ \alpha'+\beta'+\gamma'+\delta'=360^0 \end{eqnarray*} Let us also add that we call the internal angle \pojem{adjacent} or \pojem{opposite}, if the corresponding vertices are adjacent or opposite. Now we will consider some types of quadrilaterals in more detail. A quadrilateral $ABCD$ is a \pojem{trapezoid}, if $AB\parallel CD$ (Figure \ref{sl.skl.3.4.2.pic}). Sides $AB$ and $CD$ are the \pojem{bases}, $BC$ and $AD$ are the \pojem{legs} of this trapezoid. The line $PQ$ ($P\in AB$, $Q\in CD$ and $PQ\perp AB$) is the \pojem{height} of the trapezoid. We often denote it by $v$. A trapezoid is \pojem{isosceles}, if $BC \cong AD$ and $BC \not\parallel AD$, or \pojem{right}, if at least one of the internal angles is a right angle. \begin{figure}[!htb] \centering \input{sl.skl.3.4.2.pic} \caption{} \label{sl.skl.3.4.2.pic} \end{figure} Two internal angles at the same leg of a trapezoid are supplementary, because they are angles with parallel legs (statement \ref{KotiTransverzala}). The supplementarity of these angles is also a sufficient condition for a quadrilateral to be a trapezoid. It follows from this that a right trapezoid has at least two right internal angles. We get another group of quadrilaterals as a special type of trapezoids. These are \pojem{parallelograms}. They can be defined in different ways. We will choose one, and prove that the others are equivalent. Quadrilateral $ABCD$ is a \index{paralelogram} \pojem{parallelogram}, if $AB \parallel CD$ and $AD \parallel BC$ (Figure \ref{sl.skl.3.4.3.pic}). Line $PQ$ ($P\in AB$, $Q\in CD$ and $PQ\perp AB$) and $MN$ ($M\in BC$, $N\in AD$ and $MN\perp BC$) are the \index{višina!paralelograma}\pojem{heights of the parallelogram}. We often denote them with $v_a$ and $v_b$. \begin{figure}[!htb] \centering \input{sl.skl.3.4.3.pic} \caption{} \label{sl.skl.3.4.3.pic} \end{figure} A parallelogram is therefore a quadrilateral that has two pairs of parallel sides. The term congruence is not used in the definition of a parallelogram. We can also consider parallelograms (and trapezoids) in so-called \index{geometrija!afina} \pojem{affine geometry}. This is a geometry that is based on all the axioms of Euclidean geometry, with the exception of the third group of axioms - the axioms of congruence. We will now prove the aforementioned equivalents for the definition of a parallelogram. \bizrek \label{paralelogram} Let $ABCD$ be a convex quadrilateral. Then the following statements are equivalent: \begin{enumerate} \item The quadrilateral $ABCD$ is a parallelogram. \item Any two adjacent interior angles of the quadrilateral $ABCD$ are supplementary. \item Any two opposite interior angles of the quadrilateral $ABCD$ are congruent. \item $AB \parallel CD$ and $AB \cong CD$\footnote{This equivalent in a slightly different form is given by \index{Euclid} \textit{Euclid of Alexandria} (3rd century BC) in the first book of his 'Elements'.}. \item $AB \cong CD$ and $AD \cong BC$. \item The diagonals of the quadrilateral $ABCD$ bisect each other, i.e. line segments $AC$ and $BD$ have a common midpoint. \end{enumerate} \eizrek \textbf{\textit{Proof.}} It is enough to prove the equivalence of all the statements $(1)-(6)$. To avoid proving all equivalences (two implications - for example, with the statement (1), which would give 10 implications in total), we will simplify the proof a little, so that we prove implications according to the following scheme. \vspace*{5mm} \hspace*{25mm} $\begin{array}{ccccccc} \textit{(1)} & \Leftarrow & \textit{(2)} & \Leftarrow & \textit{(3)} & & \\ \Downarrow & & & & \Uparrow & & \\ \textit{(4)} & & \Rightarrow & & \textit{(5)} & \Leftrightarrow & \textit{(6)} \end{array}$ \vspace*{5mm} As we can see, this is enough because the implication $\textit{(1)} \Rightarrow \textit{(2)}$ follows directly from: $\textit{(1)}\Rightarrow \textit{(4)}\Rightarrow \textit{(5)}\Rightarrow \textit{(3)} \Rightarrow \textit{(2)}$. Let's mark with $\alpha$, $\beta$, $\gamma$ and $\delta$ the internal angles at the vertices $A$, $B$, $C$ and $D$ of the quadrilateral $ABCD$. The quadrilateral $ABCD$ is convex, which means that its diagonals intersect in some point $S$. $\textit{(2)}\Rightarrow \textit{(1)}$. Let the angles $\alpha$ and $\beta$ be complementary (Figure \ref{sl.skl.3.4.4.pic}). Then the angles at the transversal $AB$ are congruent to the angles $AD$ and $BC$, so $AD\parallel BC$ (by Theorem \ref{KotiTransverzala}). Similarly, from the complementarity of the angles $\beta$ and $\gamma$ it follows that $AB\parallel CD$. Therefore, the quadrilateral $ABCD$ is a parallelogram. \begin{figure}[!htb] \centering \input{sl.skl.3.4.4.pic} \caption{} \label{sl.skl.3.4.4.pic} \end{figure} $\textit{(3)}\Rightarrow\textit{(2)}$. Let $\alpha =\gamma$ and $\beta =\delta$ (Figure \ref{sl.skl.3.4.4.pic}). Since $\alpha + \beta +\gamma +\delta = 360°$ (the sum of all internal angles in a quadrilateral is $360°$ - by Theorem \ref{VsotKotVeck}), it follows that $\alpha + \beta =180°$ and $\beta +\gamma = 180°$. \begin{figure}[!htb] \centering \input{sl.skl.3.4.4a.pic} \caption{} \label{sl.skl.3.4.4a.pic} \end{figure} $\textit{(1)}\Rightarrow\textit{(4)}$. Let the quadrilateral $ABCD$ be a parallelogram, i.e. let $AB \parallel CD$ and $AD \parallel BC$ (Figure \ref{sl.skl.3.4.4a.pic}). We will prove that then also $AB \cong CD$. The line $AC$ is a transversal of the parallels $AB$ and $CD$, which means that the angles $CAB$ and $ACD$ are alternate angles at this transversal and are therefore congruent. Similarly, from the parallelism of the lines $AD$ and $BC$ it follows that the angles $ACB$ and $CAD$ are congruent. Since $AC \cong AC$, the triangles $ACB$ and $CAD$ are congruent (by Theorem \ref{KSK} - \textit{ASA}). Therefore, $AB \cong CD$. \begin{figure}[!htb] \centering \input{sl.skl.3.4.4b.pic} \caption{} \label{sl.skl.3.4.4b.pic} \end{figure} $\textit{(4)}\Rightarrow \textit{(5)}$. Let $ABCD$ be such a quadrilateral that $AB \parallel CD$ and $AB \cong CD$ (Figure \ref{sl.skl.3.4.4b.pic}). We will prove that then also $AD \cong BC$. The line $AC$ is a transversal of the parallels $AB$ and $CD$, which means that the angles $CAB$ and $ACD$ are alternate angles at this transversal and are therefore congruent. Since $AC \cong AC$, the triangles $ACB$ and $CAD$ are congruent (by Theorem \ref{SKS} - \textit{SAS}). Therefore, $BC \cong AD$. \begin{figure}[!htb] \centering \input{sl.skl.3.4.4c.pic} \caption{} \label{sl.skl.3.4.4c.pic} \end{figure} $\textit{(5)}\Rightarrow \textit{(3)}$. Let $ABCD$ be a quadrilateral such that $AB \cong CD$ and $AD \cong BC$ (Figure \ref{sl.skl.3.4.4c.pic}). We prove that then $\beta =\delta$ and $\alpha =\gamma$. Because $AC \cong AC$, the triangles $ACB$ and $CAD$ are congruent (\ref{SSS} - \textit{SSS}). It follows that $\angle ABC \cong \angle CDA$ or $\beta =\delta$. In a similar way we prove that $\alpha =\gamma$. \begin{figure}[!htb] \centering \input{sl.skl.3.4.4d.pic} \caption{} \label{sl.skl.3.4.4d.pic} \end{figure} $\textit{(5)}\Leftrightarrow \textit{(6)}$. Let $ABCD$ be a quadrilateral such that $AB \cong CD$ and $AD \cong BC$ (Figure \ref{sl.skl.3.4.4d.pic}). We prove that the point $S$ is the common center of its diagonals $AC$ and $BD$. Because $AC \cong AC$, the triangle $ACB$ is congruent to the triangle $CAD$ (\ref{SSS} - \textit{SSS}). Therefore $\angle ACB \cong \angle CAD$ or $\angle SCB \cong \angle SAD$. From the congruence of the right angles $CSB$ and $ASD$, it follows that the angles $SBC$ and $SDA$ are also congruent. Because $AD \cong BC$, it follows that $\triangle CSB \cong \triangle ASD$ (\ref{KSK} - \textit{ASA}). Therefore $SB \cong SD$ and $SC \cong SA$ or the point $S$ is the common center of the diagonals $AC$ and $BD$. \begin{figure}[!htb] \centering \input{sl.skl.3.4.4e.pic} \caption{} \label{sl.skl.3.4.4e.pic} \end{figure} Conversely, let $S$ be the common center of the diagonals $AC$ and $BD$ (Figure \ref{sl.skl.3.4.4e.pic}). Then $SB \cong SD$ and $SC \cong SA$. The right angles $CSB$ and $ASD$ are also congruent, so $\triangle CSB \cong \triangle ASD$ (\ref{SKS} - \textit{SAS}). It follows that $AD \cong BC$. In a similar way we prove that $AB \cong CD$. \kdokaz We recommend to the reader to prove the previous statement by using a similar scheme. This will be a good exercise in using theorems about the congruence of triangles. Let us now define a type of quadrilateral for which it will be shown that they are special cases of parallelograms (Figure \ref{sl.skl.3.4.5.pic}). \begin{figure}[!htb] \centering \input{sl.skl.3.4.5.pic} \caption{} \label{sl.skl.3.4.5.pic} \end{figure} A quadrilateral with all sides congruent is called a \index{rhombus} \pojem{rhombus}. A quadrilateral with all interior angles congruent (and therefore equal to $90^0$, because their sum is $360^0$) is a \index{rectangle} \pojem{rectangle}. A quadrilateral with all sides congruent and all interior angles congruent (and equal to $90^0$) is called a \index{square} \pojem{square}. It is not difficult to prove that each of these quadrilaterals is also a parallelogram. This is a direct consequence of the previous statement. The rhombus is a parallelogram due to $\textit{(5)}\Rightarrow\textit{(1)}$; the rectangle is a parallelogram due to $\textit{(2)}\Rightarrow\textit{(1)}$ (or $\textit{(3)}\Rightarrow\textit{(1)}$). For the square it is clear that it is also a rectangle and a rhombus, so it is also a parallelogram. From the previous statement \ref{paralelogram} - equivalent (\textit{5}) it follows that a parallelogram with two adjacent sides congruent is a rhombus. Similarly, according to the same statement from the equivalents \textit{(2)} and \textit{(3)} it follows that a parallelogram with at least one right angle is a rectangle. The next statement provides additional criteria when a parallelogram is also a rhombus, a rectangle or a square. This statement refers to diagonals. In a parallelogram, the diagonals always intersect, but in a rhombus, a rectangle or a square we will have additional properties. \bizrek \label{RombPravKvadr} $ $ (Figure \ref{sl.skl.3.4.5a.pic}) a) A parallelogram is a rhombus if and only if their diagonals are perpendicular. b) A parallelogram is a rectangle if and only if their diagonals are congruent. c) A parallelogram is a square if and only if their diagonals are perpendicular and congruent. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.4.5a.pic} \caption{} \label{sl.skl.3.4.5a.pic} \end{figure} \textbf{\textit{Proof.}} Let $ABCD$ be a parallelogram and $S$ the intersection of its diagonals $AC$ and $BD$. According to the previous statement \ref{paralelogram}, $S$ is their common center. From the same statement it also follows that $AB \cong CD$ and $AD \cong BC$. \begin{figure}[!htb] \centering \input{sl.skl.3.4.6.pic} \caption{} \label{sl.skl.3.4.6.pic} \end{figure} \textit{a)} (Figure \ref{sl.skl.3.4.6.pic}) If $ABCD$ is a rhombus, all sides are congruent. Therefore the triangles $ABS$ and $ADS$ are congruent (statement \ref{SSS} - \textit{SSS}). Then the angles $ASB$ and $ASD$ are also congruent and are (as the supplement of the angle) both right angles. This means that the diagonals are perpendicular. If the diagonals of the parallelogram $ABCD$ are perpendicular, the triangles $ABS$ and $ADS$ are congruent (statement \ref{SKS} - \textit{SAS}). Therefore the sides $AB$ and $AD$ are congruent. In a similar way, we prove that all sides of this parallelogram are congruent, which means that the parallelogram is a rhombus. \begin{figure}[!htb] \centering \input{sl.skl.3.4.6a.pic} \caption{} \label{sl.skl.3.4.6a.pic} \end{figure} \textit{b)} (Figure \ref{sl.skl.3.4.6a.pic}) If $ABCD$ is a rectangle, all interior angles are congruent and are right angles. Then the triangles $ABC$ and $DCB$ are congruent (statement \ref{SKS} - \textit{SAS}). Therefore $AC \cong DB$. If in the parallelogram $ABCD$ it holds that $AC \cong DB$, the triangles $ABC$ and $DCB$ are congruent (statement \ref{SSS} - \textit{SSS}). From this it follows that the interior angles at the vertices $B$ and $D$ of the parallelogram $ABCD$ are congruent. According to the previous statement \ref{paralelogram} they are supplementary, which means that they are both right angles. Similarly, all angles of this parallelogram are right angles, which means that the parallelogram is a rectangle. \textit{c)} A parallelogram is a square if and only if it is a rhombus and a rectangle at the same time. The latter is equivalent to the fact that the diagonals are perpendicular and congruent, which follows from the proven (\textit{a.} and \textit{b.}). \kdokaz \begin{figure}[!htb] \centering \input{sl.skl.3.4.7.pic} \caption{} \label{sl.skl.3.4.7.pic} \end{figure} Since the diagonals of a rectangle are perpendicular and intersect, there is a circle that contains all the vertices of this rectangle (Figure \ref{sl.skl.3.4.7.pic}). This is called the \index{circumscribed circle!rectangle} \pojem{circumscribed circle of the rectangle}. Its center is the intersection of its diagonals. If the point $S$ is the intersection of the diagonals of the rectangle $ABCD$, then from the previous equation \ref{RombPravKvadr} it follows: $$SA \cong SC \cong SB \cong SD.$$ The radius of this circle is equal to half the diagonal of the rectangle. Since the square is a special type of rectangle, it also has a circumscribed circle. We will now prove an important property of isosceles trapezoids. \bizrek \label{trapezEnakokraki} Interior base angles of an isosceles trapezium are congruent. The diagonals of an isosceles trapezium are congruent line segments. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.4.8.pic} \caption{} \label{sl.skl.3.4.8.pic} \end{figure} \textbf{\textit{Proof.}} Let $ABCD$ be an isosceles trapezoid with the base $AB$ (Figure \ref{sl.skl.3.4.8.pic}). Without loss of generality, assume that $AB>CD$. Let $C'$ and $D'$ be the orthogonal projections of the vertices $C$ and $D$ onto the line $AB$. The quadrilateral $D'C'CD$ is a parallelogram with a right angle, so it is a rectangle. From the fact that $D'C'CD$ is a parallelogram, it follows that $CC'\cong DD'$ (equation \ref{paralelogram}). Since $\angle CC'B\cong \angle DD'A=90^0$, the triangles $CC'B$ and $DD'A$ are congruent (the \textit{SSA} equation \ref{SSK}), so $\beta=\angle CBC'\cong \angle DAD'=\alpha$. We will now prove that the diagonals $AC$ and $BD$ are congruent. This follows from the congruence of the triangles $ABC$ and $BAD$ (the \textit{SAS} equation \ref{SKS}). \kdokaz \bzgled Let $ABCD$, $AEBK$ and $CEFG$ be equally oriented squares in a plane. Then $B$, $D$ and $F$ are collinear points and the point $B$ is a midpoint of the line segment $DF$. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.4.9.pic} \caption{} \label{sl.skl.3.4.9.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.4.9.pic}) The triangles $CAE$ and $FBE$ are congruent, because $CE\cong FE$, $AE\cong BE$ and $\angle AEC=90^0-\angle CEB=\angle BEF$ (statement \ref{SKS} - \textit{SAS}). Therefore $\angle EBF$ is a right angle and the points $D$, $B$ and $F$ are collinear. From the congruence of these two triangles it also follows that $BF\cong AC\cong BD$. \kdokaz \begin{figure}[!htb] \centering \input{sl.skl.3.4.10.pic} \caption{} \label{sl.skl.3.4.10.pic} \end{figure} Except for trapezoids and parallelograms, we will define one more group of quadrilaterals. A quadrilateral $ABCD$ is \index{deltoid}\pojem{deltoid}, if its diagonals are perpendicular and one of the diagonals bisects the other (Figure \ref{sl.skl.3.4.10.pic}). The following statement is related to the deltoid and is equivalent to its definition. \bzgled A quadrilateral is a deltoid if and only if it has two pairs of congruent adjacent sides. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.4.11.pic} \caption{} \label{sl.skl.3.4.11.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.4.11.pic}) Let the quadrilateral $ABCD$ be a deltoid. Then the diagonals $AC$ and $BD$ are perpendicular and one of the diagonals bisects the other. Without loss of generality, let the diagonal $BD$ bisect the diagonal $AC$. It follows that the right triangles $ABS$ and $CBS$ are congruent (statement \ref{SKS} - \textit{SAS}). Then $AB \cong CB$. From the congruence of the triangles $ADS$ and $CDS$ it follows that $AD \cong CD$. Let $ABCD$ be a quadrilateral, in which $AB \cong CB$ and $AD \cong CD$ hold. The triangles $ABD$ and $CBD$ are congruent (statement \ref{SSS} - \textit{SSS}), so the angles $ADS$ and $CDS$ are also congruent. From this it follows that the triangles $ADS$ and $CDS$ are also congruent (statement \ref{SKS} - \textit{SAS}). Therefore $S$ is the center of the diagonal $AC$, and the angles $DSA$ and $DSC$ are right angles, because they are congruent with the angles. \kdokaz \bzgled Let $k_1(S_1,r)$, $k_2(S_2,r)$, $k_3(S_3,r)$ be congruent circles and $k_1\cap k_2=\{B,A_3\}$, $k_2\cap k_3=\{B,A_1\}$, $k_3\cap k_1=\{B,A_2\}$. Prove that the lines $S_1A_1$, $S_2A_2$ and $S_3A_3$ intersect at a single point . \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.2.12.pic} \caption{} \label{sl.skk.4.2.12.pic} \end{figure} \textbf{\textit{Proof.}} We will prove that the lines $O_1A_1$, $O_2A_2$ and $O_3A_3$ have the same center, or that the corresponding quadrilaterals are parallelograms (Figure \ref{sl.skk.4.2.12.pic}). Because $k_1$, $k_2$ and $k_3$ are congruent circles, the quadrilaterals $O_1A_2O_3B$ and $O_2A_1O_3B$ are rhombuses. Because of this, the lines $O_1A_2$ and $O_2A_1$ are congruent and parallel, which means that the quadrilateral $O_1A_2A_1O_2$ is a parallelogram (statement \ref{paralelogram}). From the same statement it follows that its diagonals $O_1A_1$ and $O_2A_2$ have a common center. In a similar way we prove that the lines $O_2A_2$ and $O_3A_3$ have a common center, which means that this is also true for all three lines $O_1A_1$, $O_2A_2$ and $O_3A_3$ at the same time. \kdokaz \bzgled Construct a rectangle $ABCD$ if its diagonals and the difference of its sides are congruent with the two given line segments $d$ and $l$. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.4.10a.pic} \caption{} \label{sl.skl.3.4.10a.pic} \end{figure} \textbf{\textit{Solution.}} Let $ABCD$ be a rectangle, where $AC\cong d$ and $AB-BC=l$ (Figure \ref{sl.skl.3.4.10a.pic}). Let $E$ be a point on side $AB$ such that $EB\cong BC$. In this case, $AE=AC-EB=AC-BC=l$. Because $EBC$ is an isosceles right triangle, $\angle CEB\cong\angle ECB=45^0$ (\ref{enakokraki} and \ref{VsotKotTrik}) or $\angle AEC=135^0$. This allows us to first construct the triangle $AEC$ ($AC\cong d$, $\angle AEC=135^0$ and $AE\cong l$), and then the rectangle $ABCD$. \kdokaz \bzgled Construct a triangle with given $b$, $c$ and $t_a$ (sides $AC$, $AB$ and triangle median $AA_1$). \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.4.10b.pic} \caption{} \label{sl.skl.3.4.10b.pic} \end{figure} \textbf{\textit{Solution.}} Let $ABC$ be a triangle, where $AC\cong b$, $AB\cong c$ and $AA_1\cong t_a$, where $A_1$ is the center of line segment $BC$ (Figure \ref{sl.skl.3.4.10b.pic}). Let $D$ be a point on line segment $AA_1$ such that $DA_1\cong AA_1$ and $\mathcal{B}(A, A_1,D)$. This means that $A_1$ is the common center of line segments $BC$ and $AD$, so by \ref{paralelogram} quadrilateral $ABDC$ is a parallelogram. By the same \ref{paralelogram}, $CD\cong AB\cong c$. This allows us to first construct the triangle $ADC$ ($AC\cong b$, $CD\cong c$ and $AD=2t_a$), and then point $A$. \kdokaz We will now introduce a shorter form of data entry for designing triangles. Similarly, as we had in the previous task with the notation: $b$, $c$, $t_a$, for the elements of the triangle $ABC$ we will usually use the following labels: \begin{itemize} \item $a$, $b$, $c$ - sides, \item $\alpha$, $\beta$, $\gamma$ - internal angles, \item $v_a$, $v_b$, $v_c$ - altitudes, \item $t_a$, $t_b$, $t_c$ - centroids, \item $l_a$, $l_b$, $l_c$ - distances that are determined by the vertex and the intersection of the internal angle's symmetry line at that vertex with the opposite side; \item $s$ - semi-perimeter ($s=\frac{a+b+c}{2}$), \item $R$ - radius of the circumscribed circle (see section \ref{odd3ZnamTock}), \item $r$ - radius of the inscribed circle (see section \ref{odd3ZnamTock}), \item $r_a$, $r_b$, $r_c$ - radii of the excircles (see section \ref{odd4Pricrt}). \end{itemize} \bzgled Construct a trapezium if its sides are congruent with the four given line segments $a$, $b$, $c$ and $d$. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.4.10c.pic} \caption{} \label{sl.skl.3.4.10c.pic} \end{figure} \textbf{\textit{Solution.}} Without loss of generality, we will first assume that $a\geq c$. Let $ABCD$ be a trapezium, in which sides $AB\cong a$, $BC\cong b$, $CD\cong c$ and $DA\cong d$ (Figure \ref{sl.skl.3.4.10c.pic}). In this case, $AB\geq CD$, so on the side $AB$ there exists a point $E$, such that $AE\cong CD$. Because $AB\parallel CD$, by \ref{paralelogram} the quadrilateral $AECD$ is a parallelogram, so by the same theorem $CE\cong DA\cong d$. It also holds that $EB=AB-AE=AB-CD=a-c$. This allows us to construct the triangle $EBC$ ($EB=a-c$, $BC\cong c$ and $CE\cong d$), and then the vertices $A$ and $D$ (from the condition $AE\cong CD\cong c$). \kdokaz %________________________________________________________________________________ \poglavje{Regular Polygons}\label{odd3PravilniVeck} The concept of a square fits into the general definition of a new type of polygons. A polygon is \index{pravilni!večkotniki} \pojem{regular}, if all of its sides are congruent and all of its interior angles are congruent (Figure \ref{sl.skl.3.5.1.pic}). \begin{figure}[!htb] \centering \input{sl.skl.3.5.1.pic} \caption{} \label{sl.skl.3.5.1.pic} \end{figure} A square is therefore a regular quadrilateral. An equilateral triangle \index{trikotnik!pravilni}\pojem{regular triangle} is also a regular polygon. This is due to the fact that, in an equilateral triangle, all angles are also equal. We have already established that the sum of the interior angles of any $n$-gon is equal to $(n - 2) \cdot 180^0$ (Theorem \ref{VsotKotVeck}). Because in a regular $n$-gon, all interior angles are congruent, we can calculate the interior angle by dividing the sum of all angles by the number $n$. Thus, we have proved the following theorem (Figure \ref{sl.skl.3.5.2.pic}). \bizrek \label{pravVeckNotrKot} The measure of each interior angle of a regular $n$-gon is: $$\frac{(n - 2)\cdot 180^0}{n}.$$ \eizrek Thus, the interior angle of a regular triangle measures $60^0$, the interior angle of a regular quadrilateral measures $90^0$, the interior angle of a regular pentagon measures $108^0$, the interior angle of a regular hexagon measures $120^0$, ... \begin{figure}[!htb] \centering \input{sl.skl.3.5.2.pic} \caption{} \label{sl.skl.3.5.2.pic} \end{figure} We shall now prove two important properties of regular polygons. \bizrek \label{sredOcrtaneKrozVeck} For each regular polygon, there exists a circle passing through each of its vertices. \eizrek \textbf{\textit{Proof.}} Let $A_1A_2\ldots A_n$ be a regular $n$-gon (Figure \ref{sl.skl.3.5.2.pic}). Then all sides are congruent and all internal angles are congruent and equal to $\frac{(n - 2)\cdot 180^0}{n}$. Let $s_1$ and $s_2$ be the lines of symmetry of sides $A_1A_2$ and $A_2A_3$ of this polygon and let $S$ be their intersection point. From \ref{simetrala} it follows that $SA_1 \cong SA_2$ and $SA_2 \cong SA_3$ or: $$SA_1 \cong SA_2 \cong SA_2 \cong SA_3.$$ It follows that the triangle $A_1SA_2$ and $A_2SA_3$ are congruent (from \ref{SSS} - \textit{SSS}). Then the angles $SA_1A_2$, $SA_2A_1$, $SA_2A_3$ and $SA_3A_2$ are congruent. From $\angle SA_2A_1 \cong \angle SA_2A_3$ it follows that both angles are equal to half of the internal angle of the polygon or $\frac{\alpha}{2}=\frac{(n-2)\cdot 180^0}{2n}$. Therefore: $$\angle SA_3A_4 =\alpha - \frac{\alpha}{2}=\frac{\alpha}{2} = \angle SA_3A_2.$$ Thus the triangle $A_2SA_3$ and $A_3SA_4$ are congruent (from \ref{SKS} - \textit{SAS}). Because of this, $SA_3 \cong SA_4$ or: $$SA_1 \cong SA_2 \cong SA_2 \cong SA_3\cong SA_4.$$ If we continue this process, we get: $$SA_1 \cong SA_2 \cong SA_2 \cdots \cong SA_n,$$ which means that the point $S$ is the center of the circle $k(S, SA_1)$, which contains all its vertices. \kdokaz The circle from the previous theorem is called the \index{circumscribed circle!regular polygon} \pojem{circumscribed circle of a regular polygon}. From the proof of the previous theorem it is clear that its center lies at the intersection of the lines of symmetry of all its sides. We prove the following theorem in an analogous way. \bizrek \label{sredVcrtaneKrozVeck} For each regular polygon, there exists a circle touching each of its sides. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.5.3.pic} \caption{} \label{sl.skl.3.5.3.pic} \end{figure} \textbf{\textit{Proof.}} Let $A_1A_2\ldots A_n$ be a regular $n$-gon (Figure \ref{sl.skl.3.5.3.pic}). Define the point $S$ as in the proof of the previous statement. We have shown that: $SA_1 \cong SA_2 \cong SA_2 \cdots \cong SA_n$. From this, by statement \ref{SSS} - \textit{SSS}, it follows that the congruence of the isosceles triangles: $$\triangle A_1SA_2 \cong \triangle A_2 SA_3 \cong \cdots \cong \triangle A_{n-1}SA_n \cong \triangle A_nSA_1.$$ Because of this, all angles at the bases of these triangles are also congruent. Therefore, the lines $SA_1$, $SA_2$,..., $SA_n$ are the altitudes of the polygon $A_1A_2\ldots A_n$. Let $P_1$, $P_2$,..., $P_n$ be the points of intersection of these altitudes with the polygon $A_1A_2\ldots A_n$. From the congruence of the triangles $\triangle A_1SP_1$, $\triangle A_2SP_1$, $\triangle A_2SP_2$, ..., $\triangle A_1SP_n$ (statements \ref{SSK} and \ref{KSK}), it follows that the segments $SP_1$, $SP_2$,..., $SP_n$ are congruent. By statement \ref{TangPogoj}, the circle $k(S, SP_1)$ touches all sides of the polygon $A_1A_2\ldots A_n$. \kdokaz The circle from the previous statement is called the \index{circumscribed circle!of a regular polygon} \pojem{circumscribed circle of a regular polygon}. From the proof of this statement, it is clear that the center of the circumscribed circle of a regular polygon lies at the intersection of the altitudes of all its internal angles. From the proof it is also clear that the points at which this circle touches the sides of the regular polygon are also the centers of these sides. The center of the circumscribed and inscribed circle is the same point and is therefore also called the \index{center!of a regular polygon}\pojem{center of a regular polygon}. We have also seen that all triangles, determined by the center of the regular $n$-gon and by its sides, are isosceles and all congruent. The radius of the circumscribed and inscribed circle of the $n$-gon is equal to the leg or. the height of each of these triangles. The angles at the top of these triangles are also congruent and because there are a total of $n$ (the same as the sides of the $n$-gon), each of them measures (Figure \ref{sl.skl.3.5.4.pic}): $$\varphi = \frac{360^0}{n}.$$ \begin{figure}[!htb] \centering \input{sl.skl.3.5.4.pic} \caption{} \label{sl.skl.3.5.4.pic} \end{figure} For a regular hexagon, or for $n = 6$, it holds: $$\varphi = \frac{360^0}{6}=60^0.$$ This means that the aforementioned triangles are regular. Therefore, a regular hexagon is composed of six regular triangles (Figure \ref{sl.skl.3.5.4.pic}). In the following we will consider the properties of regular $n$-gons. Let $n$ be an even number and $k = \frac{n}{2}+1$. We say that $A_k$ is the \pojem{opposite vertex} of vertex $A_1$ of the regular $n$-gon $A_1A_2\ldots A_n$ (Figure \ref{sl.skl.3.5.5.pic}). Analogously, $A_2$ and $A_{k+1}$, $A_3$ and $A_{k+2}$, ... , $A_{k-1}$ and $A_n$ are opposite vertices of this $n$-gon. Similarly, the sides $A_1A_2$ and $A_kA_{k+1}$, ... , $A_{k-1} A_k$ and $A_nA_1$ of the polygon $A_1A_2\ldots A_n$ are \pojem{opposite sides}. We notice that it holds: $$\angle A_1SA_k=\frac{n}{2}\varphi = \frac{n}{2}\cdot \frac{360^0}{n}=180^0,$$ which means that the diagonal $A_1A_k$ of this $n$-gon contains its center. Therefore, this diagonal represents the diameter of the circumscribed circle. Analogously, this holds for all diagonals determined by opposite vertices. Because of this, we call such diagonals \index{velika diagonala pravilnega $n$-kotnika} \pojem{major diagonals} of a regular $n$-gon. The radius of the circumscribed circle is equal to half the major diagonal. \begin{figure}[!htb] \centering \input{sl.skl.3.5.5.pic} \caption{} \label{sl.skl.3.5.5.pic} \end{figure} It can be proven in a similar way that for an even number $n$ of centers of the opposite sides of a regular $n$-gon $A_1A_2\ldots A_n$ determine the diameters of the inscribed circle of this $n$-gon. If we consider the previous labels, we get: \begin{eqnarray*} \angle P_1SP_k&=&\angle P_1SA_2 + \angle A_2SA_{k-1} + \angle A_{k-1}SP_k\\ &=& \frac{\varphi}{2}+\frac{n-2}{2}\cdot \varphi+\frac{\varphi}{2} =\frac{n}{2}\cdot\varphi =180^0. \end{eqnarray*} The distances that are determined by a pair of centers of opposite sides of an $n$-gon $A_1A_2\ldots A_n$ or the distances $P_1P_k$, $P_2P_{k+1}$, ... , $P_{k-1}P_n$, are called the \pojem{heights} \index{height!of a regular $n$-gon} of this $n$-gon. The radius of the inscribed circle is equal to half the height. So every regular $n$-gon, where $n$ is an even number, contains $\frac{n}{2}$ big diagonals (equal to the diameter of the circumscribed circle) and $\frac{n}{2}$ heights (equal to the diameter of the inscribed circle). Each of them goes through the center of this $n$-gon. Let $n$ be an odd number now (Figure \ref{sl.skl.3.5.6.pic}) and $k=\frac{n+1}{2}+1$. Then: \begin{eqnarray*} \angle P_1SA_k=\angle P_1SA_2 + \angle A_2SA_k= \frac{\varphi}{2}+\frac{n-1}{2}\cdot \varphi =\frac{n}{2}\cdot\varphi =180^0. \end{eqnarray*} \begin{figure}[!htb] \centering \input{sl.skl.3.5.6.pic} \caption{} \label{sl.skl.3.5.6.pic} \end{figure} This means that the distance $P_1A_k$ contains the center $S$ of a regular $n$-gon $A_1A_2\ldots A_n$. This distance is called the \index{height!of a regular $n$-gon} \pojem{height} of this $n$-gon, the side $A_1A_2$ and the point $A_k$ are \pojem{opposite}. We define the remaining $n$ heights and $n$ pairs of opposite sides and points in a similar way. In a similar way we can prove that the other heights of this $n$-gon also contain its center. In a right (isosceles) triangle, we therefore have three heights, which intersect in its center (Figure \ref{sl.skl.3.5.7.pic}). If this is true for any triangle, we will find out later \begin{figure}[!htb] \centering \input{sl.skl.3.5.7.pic} \caption{} \label{sl.skl.3.5.7.pic} \end{figure} In a square, its diagonals are also the main diagonals and intersect in its center (Figure \ref{sl.skl.3.5.7.pic}). The heights of the square are consistent with its side, which is not difficult to prove. We have already mentioned that a regular hexagon is composed of six triangles, which intersect in its center. We will consider a regular hexagon in more detail later. We will also prove some properties of a regular pentagon, heptagon, nonagon and dodecagon. The problem of designing regular $n$-gons for any number $n$ will be particularly interesting. We will now prove an interesting property of a regular nonagon. \bzgled If $a$ is a side and $d$ and $e$ are the shortest and longest diagonal of a regular nonagon ($9$-gon), then $e - d = a$. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.5.8.pic} \caption{} \label{sl.skl.3.5.8.pic} \end{figure} \textbf{\textit{Proof.}} Let $d = CE$ and $e = BF$ be the shortest and longest diagonals of a regular nonagon $ABCDEFGHI$ with side $a$ and $P$ the intersection of lines $BC$ and $FE$ (Figure \ref{sl.skl.3.5.8.pic}). The internal angle of this nonagon measures $\angle CDE=\frac{9-2}{9}\cdot 180^0=140^0$, so $\angle ECD = \angle CED = 20^0$. From this it follows $\angle BCE = \angle FEC = 120^0$, i.e.: $$\angle ECP = \angle CEP = 60^0.$$ Therefore, triangle $CPE$ is regular. Since $CB = EF=a$ and $\angle BPF \cong \angle CPE = 60^0$, triangle $BPF$ is also regular. Therefore: $$e = BF = BP = BC + CP = BC + CE = a + d,$$ which was to be proved. \kdokaz %%________________________________________________________________________________ \poglavje{Midsegment of Triangle} \label{odd3SrednTrik} We will now look at a very important property of a triangle, which we will use often. Let $P$ and $Q$ be the centers of the sides $AB$ and $AC$ of the triangle $ABC$. The distance $PQ$ is called the \index{midsegment!of a triangle} \pojem{midsegment of the triangle} $ABC$, which corresponds to the side $BC$ (Figure \ref{sl.skl.3.6.1.pic}). We also say that $PQ$ is the midsegment of the triangle $ABC$ with the base $BC$. We prove the basic property that relates to the midsegment. \begin{figure}[!htb] \centering \input{sl.skl.3.6.1.pic} \caption{} \label{sl.skl.3.6.1.pic} \end{figure} \bizrek \label{srednjicaTrik} Let $PQ$ be the midsegment of a triangle $ABC$ corresponding to the side $BC$. Then: $$ PQ = \frac{1}{2} BC\hspace*{2mm} \textrm{ in } \hspace*{2mm} PQ \parallel BC.$$ \eizrek \textbf{\textit{Proof.}} Let $R$ be a point such that $PQ \cong QR$ and $\mathcal{B}(P,Q,R)$ (Figure \ref{sl.skl.3.6.1.pic}). The line segments $AC$ and $PR$ have a common center, so the quadrilateral $APCR$ is a parallelogram (\ref{paralelogram}). Therefore, the line segments $AP$ and $RC$ are congruent and parallel. The point $P$ is the center of the line segment $AB$, so the line segments $PB$ and $RC$ are congruent and parallel. This means that the quadrilateral $PBCR$ is a parallelogram. It follows that the line segments $BC$ and $PR$ are congruent and parallel. The final conclusion follows from the fact that the point $Q$ is the center of the line segment $PR$. \kdokaz \bzgled Let $AB$ and $A'B'$ be congruent line segments, $C$ and $D$ the midpoints of the line segments $AA'$ and $BB'$. Suppose that $CD =\frac{1}{2} AB$. What is a measure of the angle between the lines $AB$ and $A'B'$? \ezgled \textbf{\textit{Solution.}} Let point $S$ be the center of line $A'B$ (Figure \ref{sl.skl.3.6.2.pic}). Lines $CS$ and $DS$ are the medians of triangles $A'AB$ and $BA'B'$, so: $$CS = \frac{1}{2}AB = CD = \frac{1}{2}A'B'= DS,$$ or $SCD$ is an equilateral triangle. Angles $\angle AB,A'B'$ and $\angle CSD$ have corresponding sides. Therefore: $\angle AB, A'B' \cong \angle CSD = 60^0$. \kdokaz The next consequence of \ref{srednjicaTrik} applies to a trapezium. \bizrek \label{srednjTrapez} Let $P$ and $Q$ be the midpoints of legs $BC$ and $DA$ of a trapezium $ABCD$. Suppose that $M$ and $N$ are the midpoints of the diagonals $AC$ and $BD$ of that trapezium. Then the points $M$ and $N$ lie on the line $PQ$, which is parallel to the bases $AB$ in $CD$ of the trapezium, and also: $$PQ = \frac{1}{2}( AB + CD) \hspace*{2mm} \textrm{ in } \hspace*{2mm} MN=\frac{1}{2}( AB - CD).$$ \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.6.3.pic} \caption{} \label{sl.skl.3.6.3.pic} \end{figure} \textbf{\textit{Proof.}} The lines $PN$, $NQ$ and $PM$ are in turn medians of the triangles $DAC$, $ACB$ and $ADB$ for the corresponding bases $DC$, $AB$ and $AB$ (Figure \ref{sl.skl.3.6.3.pic}). Because of this, all three lines $PN$, $NQ$ and $PM$ are parallel to the bases $CD$ and $AB$. Since through each point (first $N$, then $P$) there is only one parallel to the line $AB$ (Playfair's\footnote{\index{Playfair, J.}\textit{J. Playfair} (1748--1819), Scottish mathematician.} axiom \ref{Playfair}), the points $P$, $N$, $M$ and $Q$ are collinear. It also holds (from the statement \ref{srednjicaTrik}): $PN = \frac{1}{2} CD$ and $NQ = PM = \frac{1}{2} AB$. From this it follows: \begin{eqnarray*} PQ&=& PN+NQ=\frac{1}{2}CD+ \frac{1}{2}AB=\frac{1}{2}\left(AB+CD\right)\\ NM&=& PM-PN=\frac{1}{2}AB- \frac{1}{2}CD=\frac{1}{2}\left(AB-CD\right), \end{eqnarray*} which was to be proven. \kdokaz The line $PQ$ from the previous statement is called \index{srednjica!trapeza} \pojem{median of the trapezoid}. The following statements apply to an arbitrary quadrilateral. \bizrek \label{Varignon} Let $ABCD$ be an arbitrary quadrilateral and $P$, $Q$, $K$ and $L$ the midpoints of the sides $AB$, $CD$, $BC$ and $AD$, respectively. Then the quadrilateral $PKQL$ is a parallelogram (so-called \index{paralelogram!Varignonov} Varignon\footnote{\index{Varignon, P.} \textit{P. Varignon} (1654--1722), French mathematician, who was the first to prove this property. However, the statement was not published until after his death in 1731. Given the simplicity, it is quite surprising that this statement "waited" so long to be discovered.} parallelogram). \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.6.4.pic} \caption{} \label{sl.skl.3.6.4.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.6.4.pic}) The lines $PK$ and $LQ$ are the medians of the triangles $ABC$ and $ADC$ for the same base $AC$, so they are congruent and parallel. Therefore, the quadrilateral $PKQL$ is a parallelogram. \kdokaz In a special case, Varignon's parallelogram can even be a rectangle, rhombus or square. When is this possible? This question gives us an idea for the next statement. We know that a parallelogram is a rectangle if it has at least one internal angle right. The other (equivalent) condition is that it has congruent diagonals. A similar treatment can be used for rhombus and square. \bzgled \label{VarignonPoslPravRomb} Let $ABCD$ be an arbitrary quadrilateral and $P$, $K$, $Q$ and $L$ the midpoints of the sides $AB$, $BC$, $CD$ and $DA$, respectively. Then: a) $AC \perp BD \Leftrightarrow PQ \cong KL$; b) $AC \cong BD \Leftrightarrow PQ \perp KL$. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.6.5.pic} \caption{} \label{sl.skl.3.6.5.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.6.5.pic}) From the previous statement \ref{Varignon} it follows that the quadrilateral $PKQL$ is always a parallelogram – Varignon's parallelogram. The lines $PL$ and $PK$ are the medians of the triangles $ABD$ and $ABC$ for the bases $AD$ and $AC$. Therefore, we have: $PL= \frac{1}{2}BD$ and $PL \parallel BD$ and $PK= \frac{1}{2}AC$ and $PK \parallel AC$. Therefore, we have: a) $AC \perp BD \Leftrightarrow PL \perp PK \Leftrightarrow PKQL \textrm{ rectangle} \Leftrightarrow PQ \cong KL$; b) $AC \cong BD \Leftrightarrow PL \cong PK \Leftrightarrow PKQL \textrm{ rhombus } \Leftrightarrow PQ \perp KL$. \kdokaz If Varignon's parallelogram is a square, all four conditions from the previous equivalences are fulfilled, or in this case: $AB \perp CD$, $AB \cong CD$, $PQ \perp KL$ and $PQ \cong KL$. Let's look at one more use of the property of Varignon's parallelogram. \bzgled \label{VagnanPosl} Let $ABCD$ be an arbitrary quadrilateral. If $P$, $Q$, $K$, $L$, $M$ and $N$ are the midpoints of the line segments $AB$, $CD$, $BC$, $AD$, $AC$ and $BD$, respectively, then the line segments $PQ$, $KL$ and $MN$ have a common midpoint. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.6.6.pic} \caption{} \label{sl.skl.3.6.6.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.6.6.pic}) Štirikotnik $PKQL$ je Varignonov paralelogram (izrek \ref{Varignon}). Na podoben način dokazujemo, da je tudi štirikotnik $LNKM$ paralelogram (srednjice trikotnikov $ADB$ in $ACB$). Paralelograma $PKQL$ in $LPKQ$ imata skupno diagonalo $LK$. Ker se diagonali poljubnega paralelograma razpolavljata (izrek \ref{paralelogram}), imajo daljice $LK$, $PQ$ in $MN$ skupno središče. \kdokaz Točko iz prejšnjega primera, v kateri se daljice sekajo, imenujemo \index{težišče!štirikotnika} \pojem{težišče štirikotnika}. Več o tem bomo povedali v razdelku \ref{odd5TezVeck}. Naslednja trditev je lep primer kombiniranja neenakosti trikotnika in srednjice trikotnika. Trditev je pravzaprav posplošitev izreka o srednjici trapeza \ref{srednjTrapez}. \bizrek If $P$ and $Q$ are the midpoints of the sides $AB$ and $CD$ of an arbitrary quadrilateral $ABCD$, then: $$PQ \leq \frac{1}{2}\left( BC + AD\right).$$ \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.6.7.pic} \caption{} \label{sl.skl.3.6.7.pic} \end{figure} \textbf{\textit{Proof.}} Let $S$ be the center of the diagonal $AC$ of the quadrilateral $ABCD$ (Figure \ref{sl.skl.3.6.7.pic}). If we use the theorem about the median of a triangle (\ref{srednjicaTrik}) and the triangle inequality (\ref{neenaktrik}), we get: $$BC + AD = 2PS + 2SQ = 2(PS + SQ) \geq 2PQ.$$ The equality holds when the points $P$, $S$ and $Q$ are collinear, i.e. when the quadrilateral $ABCD$ is a trapezoid with the base $BC$. \kdokaz We mention that the inequality from the previous example also holds when the points $A$, $B$, $C$ and $D$ are not in the same plane, i.e. when the $ABCD$ is a \pojem{tetrahedron}. \bzgled \label{TezisceSredisceZgled} Let $P$ be the midpoint of the median $AA_1$ of a triangle $ABC$ and $Q$ the intersection of the side $AC$ and the line $BP$. Determine the ratios $AQ :QC$ and $BP : PQ$. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.6.8.pic} \caption{} \label{sl.skl.3.6.8.pic} \end{figure} \textbf{\textit{Solution.}} Let $R$ be the center of the line segment $QC$ (Figure \ref{sl.skl.3.6.8.pic}). The line segment $A_1R$ is the median of the triangle $BCQ$ for the base $BQ$, so (by the theorem \ref{srednjicaTrik}) $BQ = 2A_1R$ and $BQ\parallel A_1R$. From this parallelism and the definition of the point $P$ it follows that $PQ$ is the median of the triangle $AA_1R$ for the base $A_1R$, so (by the theorem \ref{srednjicaTrik} and Playfair's axiom \ref{Playfair}) the point $Q$ is the center of the line segment $AR$ and it holds $A_1R = 2PQ$. Therefore: $AQ \cong QR \cong RC$ i.e. $AQ:QC=1:2$. In the end it is also $BQ=2A_1R=4PQ$ i.e. $BP:PQ=3:1$. \kdokaz \bnaloga\footnote{19. IMO Yugoslavia - 1977, Problem 1.} Equilateral triangles $ABP$, $BCL$, $CDM$, $DAN$ are constructed inside the square $ABCD$. Prove that the midpoints of the segments $LM$, $MN$, $NP$, $PL$, $AN$, $LB$, $BP$, $CM$, $CL$, $DN$, $DM$ in $AP$ are the twelve vertices of a regular dodecagon ($12$-gon). \enaloga \begin{figure}[!htb] \centering \input{sl.skl.3.6.IMO1.pic} \caption{} \label{sl.skl.3.6.IMO1.pic} \end{figure} \textbf{\textit{Solution.}} We mark with $a$ the length of the side and with $O$ the center of the square $ABCD$ (Figure \ref{sl.skl.3.6.IMO1.pic}). We first prove that the quadrilateral $MNPL$ is also a square with the same center $O$. Because $ABP$, $BCL$, $CDM$ and $DAN$ are all right triangles, the diagonals $MP$ and $LN$ of the quadrilateral $MNPL$ are on the similitudes of the sides $AB$ and $BC$ of the square $ABCD$. From this it follows that $MP \perp LN$. Since $d(M,AB)=d(P,CD)=a-v$ (where $v$ is the length of the height of the aforementioned right triangles), it also holds that $OM\cong OP$. Similarly, $OL\cong ON$, which means that the quadrilateral $MNPL$ is really a square with the same center~$O$. We now prove that $LAM$ is a right triangle. Because $AB\cong AD\cong BL\cong DM=a$ and $\angle LBA\cong\angle MDA =90^0-60^0=30^0$, the triangles $LBA$ and $MDA$ are similar (by the \textit{SAS} \ref{SKS}). This means that $LA\cong MA$ and $\angle DAL= 90^0-\angle LAB=15^0$. Similarly, $\angle BAM=15^0$ or $\angle LAM = 90^0-2\cdot 15^0=60^0$. Therefore, $LAM$ is a right triangle, so the side of the square $MNPL$ has length $b=|LM|=|LA|$. Let's mark the center of the line $LM$ with $S$. The centers of the sides of the square $MNPL$ lie on the circle $k(O,\frac{b}{2})$, which is the inscribed circle of this square. We will prove that the point $T$ - the center of the line $AN$ - also lies on this circle. The line $OT$ is the median of the triangle $LAN$ for the base $LA$, so $OT\parallel LA$ and $|OT|=\frac{1}{2}|LA|=\frac{b}{2}$. Therefore, the point $T$ and, analogously, all the points defined in the problem of the $12$-gon lie on the circle $k(O,\frac{b}{2})$. We will also prove that the aforementioned $12$-gon is regular. Without loss of generality, it is enough to prove that $\angle SOT=\frac{360^0}{12}=30^0$. But from the already proven fact $OT\parallel LA$ it follows that $\angle SOT\cong \angle LAS=\frac{1}{2}\angle LAM=30^0$. \kdokaz %________________________________________________________________________________ \poglavje{Triangle Centers} \label{odd3ZnamTock} We will continue our research with a triangle - the most simple polygon, which is at the same time a figure that has unexpectedly many interesting properties. We will discuss some of them in this section, some of them later, when we will deal with other concepts, such as isometries and similarity. Now we will consider four \index{značilne točke trikotnika}\pojem{značilne točke trikotnika}\footnote{These four points are mentioned by the Ancient Greeks, although they (especially the centroid) were probably known long before that.} and their use in quadrilaterals and polygons. We will start with the first among the characteristic points, which is related to the medians of the triangle. We have already defined the median in chapter \ref{odd3NeenTrik}. \bizrek \label{tezisce} The medians of a triangle intersect at one point. That point divides the medians in the ratio $2:1$ (from the vertex to the midpoint of the opposite side). \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.7.1.pic} \caption{} \label{sl.skl.3.7.1.pic} \end{figure} \textbf{\textit{Proof.}} Let $AA_1$, $BB_1$ and $CC_1$ be the altitudes of the triangle $ABC$ (Figure \ref{sl.skl.3.7.1.pic}). Because of Pasch's axiom, \ref{AksPascheva} with respect to the triangle $BCB_1$ and the line $AA_1$, the line $AA_1$ intersects the line segment $BB_1$. Similarly, the line $BB_1$ intersects the line $AA_1$, which means that the altitudes $AA_1$ and $BB_1$ intersect at some point $T$. Let $A_2$ and $B_2$ be the midpoints of the line segments $AT$ and $BT$. The line segments $A_1B_1$ and $A_2B_2$ are the medians of the triangles $ABC$ and $ABT$ with the same base $AB$. This means that the line segments $A_1B_1$ and $A_2B_2$ are parallel and equal to half of the side $AB$. Therefore, the quadrilateral $B_2A_1B_1A_2$ is a parallelogram (statement \ref{paralelogram}), which means that its diagonals $A_1A_2$ and $B_1B_2$ intersect at their common center $T$. So it holds: $A_1T\cong TA_2 \cong A_2A$ and $B_1T\cong TB_2 \cong B_2B$ or $AT:TA_1=2:1$ and $BT:TB_1=2:1$ and $$A_1T=\frac{1}{3}A_1A \textrm{ and } B_1T = \frac{1}{3}B_1B.$$ In the same way we prove that the altitudes $AA_1$ and $CC_1$ intersect at some point $T'$, for which it holds $AT':T'A_1=2:1$ and $CT':T'C_1=2:1$ or: $$A_1T'=\frac{1}{3}A_1A \textrm{ and } C_1T' = \frac{1}{3}C_1C.$$ This means that $T$ and $T'$ are points on the line segment $A_1A$, for which it holds $A_1T\cong A_1T'=\frac{1}{3}A_1A$, so according to \ref{ABnaPoltrakCX} $T = T'$, which means that the altitudes $AA_1$, $BB_1$ and $CC_1$ intersect at the point $T$ and it holds $AT:TA_1=BT:TB_1=CT:TC_1=2:1$. \kdokaz The point from the previous statement, in which all altitudes of the triangle intersect, is called the \index{težišče!trikotnika}\pojem{center of the triangle}. The center of a triangle in the physical sense represents the point that is the center of mass of that triangle. This will be even more clear when we in section \ref{odd8PloTrik} prove the fact that the center divides the triangle into triangles with the same area. In the next chapter \ref{pogVEKT} (section \ref{odd5TezVeck}) we will consider the center of any polygon. In section \ref{odd3PravilniVeck} we found that for any regular polygon there exist circumscribed (which contains all its vertices) and inscribed (which touches all its sides) circle with the same center. This property is also transferred to regular or equilateral triangles. But how is it with any triangle? We will prove that there exist the mentioned circles for any triangle, just that in the general case they have different centers. \bizrek \label{SredOcrtaneKrozn} The perpendicular bisectors of the sides of any triangle intersect at a single point, which is the centre of a circle containing all its vertices. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.7.2.pic} \caption{} \label{sl.skl.3.7.2.pic} \end{figure} \textbf{\textit{Proof.}} Let $p$, $q$ and $r$ be the perpendicular bisectors of sides $BC$, $AC$ and $AB$ of triangle $ABC$ (Figure \ref{sl.skl.3.7.2.pic}). The perpendicular bisectors $p$ and $q$ are not parallel (because in that case by Playfair's axiom \ref{Playfair} the lines $BC$ and $AC$ would be parallel too) and they intersect in some point $O$. Because this point lies on the perpendicular bisectors $p$ and $q$ of sides $BC$ and $AC$, we have $OB \cong OC$ and $OC \cong OA$. From this follows $OA \cong OB$, which means that the point $O$ also lies on the perpendicular bisector $r$ of the line $AB$. So the perpendicular bisectors $p$, $q$ and $r$ intersect in one point. Because $OA \cong OB \cong OC$, the point $O$ is the center of the circle $k(O,OA)$, which contains all vertices of the triangle $ABC$. \kdokaz The circle from the previous theorem, which contains all the vertices of the triangle, is called the \index{circumscribed circle!triangle} \pojem{circumscribed circle of the triangle}, and its center is the \index{center!circumscribed circle!triangle} \pojem{center of the circumscribed circle of the triangle}. \bizrek \label{SredVcrtaneKrozn} The bisectors of the interior angles of any triangle intersect at a single point, which is the centre of a circle touching all its sides. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.7.3.pic} \caption{} \label{sl.skl.3.7.3.pic} \end{figure} \textbf{\textit{Proof.}} Let $p$, $q$, and $r$ be the angle bisectors of the angles at the vertices $A$, $B$, and $C$ of the triangle $ABC$ (Figure \ref{sl.skl.3.7.3.pic}). We shall prove that the bisectors $p$ and $q$ are not parallel. Otherwise, by Theorem \ref{KotiTransverzala}, the sum of the halves of the angles at the vertices $A$ and $B$ would be equal to $180°$, which, by Theorem \ref{VsotKotTrik}, is not possible. Therefore, $p$ and $q$ intersect at some point $S$. Since the point $S$ lies on the angle bisectors of the angles at the vertices $A$ and $B$, it is equidistant from the sides $AC$ and $AB$ (Theorem \ref{SimKotaKraka}). From this it follows that the point $S$ is also equidistant from the sides $BA$ and $BC$, which means that it lies on the bisector $r$ of the angle at the vertex $C$. Therefore, the angle bisectors $p$, $q$, and $r$ intersect at the point $S$. With $P$, $Q$ in $R$ we denote the orthogonal projections of point $S$ onto the sides $BC$, $CA$ and $AB$. Because $\frac{1}{2}\angle CBA<90^0$ and $\frac{1}{2}\angle BCA<90^0$, it follows that $\mathcal{B}(B,P,C)$. Similarly, $\mathcal{B}(C,Q,A)$ and $\mathcal{B}(A,R,B)$ also hold. Because of the already proven properties of point $S$, it follows that $SP \cong SQ \cong SR$. Therefore, the point $S$ is the center of the circle $l$, which passes through points $P$, $Q$ and $R$. Because of the perpendicularity of the radii $SP$, $SQ$ and $SR$ to the respective sides, they are tangent to the circle $l$. Because $\mathcal{B}(B,P,C)$, $\mathcal{B}(C,Q,A)$ and $\mathcal{B}(A,R,B)$ hold, the circle $l$ is tangent to all sides of the triangle $ABC$. \kdokaz The circle from the previous statement, which is tangent to all sides of the triangle, is called \index{včrtana krožnica!trikotnika} \pojem{the inscribed circle of the triangle}, and its center \index{središče!včcrtane krožnice!trikotnika} \pojem{the center of the inscribed circle of the triangle}. There is one more of the four aforementioned characteristic points of the triangle. It is related to the altitudes of the triangle. \bizrek \label{VisinskaTocka} The lines containing the altitudes of a triangle intersect at a single point. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.7.4.pic} \caption{} \label{sl.skl.3.7.4.pic} \end{figure} \textbf{\textit{Proof.}} Let $p$, $q$ and $r$ be the altitude lines $AA'$, $BB'$ and $CC'$ of the triangle $ABC$ (Figure \ref{sl.skl.3.7.4.pic}). We denote by $a$, $b$ and $c$ the lines that are perpendicular to the respective heights in points $A$, $B$ and $C$. Because the lines $a$, $b$ and $c$ are parallel to the sides of the triangle $ABC$, each two of them intersect. We denote by $P$, $Q$ and $R$ the intersections of the lines $b$ and $c$, $a$ and $c$, and $a$ and $b$, in order. The quadrilateral $ABCQ$ and $RBCA$ are parallelograms, which means that $RA \cong BC \cong AQ$, or the point $A$ is the center of the line $RQ$. The line $AA'$ is therefore the perpendicular bisector of the side $RQ$ of the triangle $PQR$. Similarly, $BB'$ and $CC'$ are the perpendicular bisectors of the sides $PR$ and $PQ$ of the same triangle. By the theorem \ref{SredOcrtaneKrozn}, the perpendicular bisectors $AA'$, $BB'$ and $CC'$ of the triangle $PQR$ intersect in some point $V$. The point $V$ is therefore the intersection of the altitude lines $AA'$, $BB'$ and $CC'$ of the triangle $ABC$. \kdokaz The point from the previous theorem, in which the altitude lines intersect, is called \index{višinska točka trikotnika} \pojem{the altitude point of the triangle}. The triangle $A'B'C'$, which is determined by the altitudes of the triangle $ABC$, is called \index{trikotnik!pedalni} \pojem{the pedal triangle} of the triangle $ABC$. We have found that every triangle has four characteristic points, namely: the centroid, the center of the circumscribed circle, the center of the inscribed circle and the altitude point. But these are not the only characteristic points of the triangle. We will mention some of them later. For a point in the plane of the triangle in general, we say that it is its \pojem{characteristic point}, if its definition is symmetrical with respect to the vertices of this triangle. \begin{figure}[!htb] \centering \input{sl.skl.3.7.5.pic} \caption{} \label{sl.skl.3.7.5.pic} \end{figure} It is clear that in any triangle the four characteristic points differ (Figure \ref{sl.skl.3.7.5.pic}). In an isosceles triangle, the centroid, the altitude, the line of symmetry of the base, and the line of symmetry of the internal angle opposite the base have the same carrier. If $A_1$ is the center of the base $BC$ of the isosceles triangle $ABC$, the triangles $ABA_1$ and $ACA_1$ are congruent, which means that the angle at the vertex $A_1$ is a right angle and the angles $BAA_1$ and $CAA_1$ are congruent. Therefore, the distance $AA_1$ is both the centroid and the altitude, and the line $AA_1$ is both the line of symmetry of the side $BC$ and the line of symmetry of the internal angle at the vertex $A$ of the triangle $ABC$. It follows that all four characteristic points of this triangle lie on one line $AA_1$ (Figure \ref{sl.skl.3.7.6.pic}). If we use the already proven property of an isosceles triangle for an equilateral triangle, we find that all the appropriate centroids, altitudes, side lines of symmetry, and internal angle lines of symmetry have the same carrier. This means that in an equilateral triangle all four characteristic points coincide (Figure \ref{sl.skl.3.7.6.pic}). This is actually already defined (section \ref{odd3PravilniVeck}) as the center of this equilateral (or regular) triangle. \begin{figure}[!htb] \centering \input{sl.skl.3.7.6.pic} \caption{} \label{sl.skl.3.7.6.pic} \end{figure} We will also show the position of the characteristic points with respect to the type of triangle. \begin{figure}[!htb] \centering \input{sl.skl.3.7.7.pic} \caption{} \label{sl.skl.3.7.7.pic} \end{figure} The centroids are always inside the triangle. Therefore, the center is an internal point of every triangle (Figure \ref{sl.skl.3.7.7.pic}). The same conclusion applies to the center of the inscribed circle (Figure \ref{sl.skl.3.7.7s.pic}). \begin{figure}[!htb] \centering \input{sl.skl.3.7.7s.pic} \caption{} \label{sl.skl.3.7.7s.pic} \end{figure} In an acute triangle, the vertices of the perpendiculars from its vertices lie on the sides of this triangle, which means (Pasch's axiom \ref{AksPascheva}), that its altitudes intersect in the interior. So in an acute triangle, the altitude point lies in its interior (Figure \ref{sl.skl.3.7.7v.pic}). In a right-angled triangle, the altitude point is the vertex at the right angle. This is because its catheti are at the same time altitudes of the triangle. The altitude point of an obtuse triangle lies in its exterior, because not all of the altitudes are in its interior. The corresponding vertices belong to the sides' supports, not to the sides themselves. \begin{figure}[!htb] \centering \input{sl.skl.3.7.7v.pic} \caption{} \label{sl.skl.3.7.7v.pic} \end{figure} The centre of the circumscribed circle is an interior or exterior point of the triangle, depending on whether the triangle is acute or obtuse (Figure \ref{sl.skl.3.7.7o.pic}). We will omit the formal proof of this fact. We will only prove the following statement, which refers to right-angled triangles. \begin{figure}[!htb] \centering \input{sl.skl.3.7.7o.pic} \caption{} \label{sl.skl.3.7.7o.pic} \end{figure} \bizrek The circumcentre of a right-angled triangle is at the same time the midpoint of its hypotenuse. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.7.8.pic} \caption{} \label{sl.skl.3.7.8.pic} \end{figure} \textbf{\textit{Proof.}} We mark with $O$ the centre of the hypotenuse $AB$ and with $P$ the centre of the cathetus $AC$ of the right-angled triangle $ABC$ (Figure \ref{sl.skl.3.7.8.pic}). The distance $OP$ is the median of this triangle, which corresponds to the cathetus $BC$, so $OP\parallel BC$. From this it follows that $OP \perp AC$. Therefore, the triangles $OPC$ and $OPA$ are congruent (the statement \textit{SAS} \ref{SKS}) and then $OC \cong OA$. Since $OB \cong OA$ as well, the point $O$ is the centre of the circumscribed circle of this triangle. \kdokaz The line $OC$ from the previous theorem is the median of the triangle. This means that the median of a right angled triangle is equal to the radius of the inscribed circle of that triangle, and also to half of its hypotenuse. The previous theorem is also connected to Thales’ theorem for a circle \ref{TalesovIzrKroz} and its converse \ref{TalesovIzrKrozObrat}. We will now state all of these theorems in one theorem. \bizrek Thales’ theorem for a circle (several forms - Figure \ref{sl.skl.3.7.9.pic}): \index{theorem!Thales’ for a circle} \label{TalesovIzrKroz2} \begin{enumerate} \item The circumcentre of a right-angled triangle is at the same time the midpoint of its hypotenuse. \item If $t_c$ is the median of a right-angled triangle for its hypotenuse $c$ and $R$ the circumradius of that triangle, then $R=t_c=\frac{c}{2}$. \item If $AB$ is a diameter of a circle $k$, then for any point $X\in k$ ($X\neq A$ and $X\neq B$) is $\angle AXB=90^0$. \item If $A$, $B$ in $X$ are three non-collinear points, such that $\angle AXB=90^0$, then the point $X$ lies on a circle with the diameter $AB$. \end{enumerate} \index{theorem!Thales’ for a circle} \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.7.9.pic} \caption{} \label{sl.skl.3.7.9.pic} \end{figure} Knowing the characteristic points of a triangle and the properties of a median of a triangle allows us to prove various other properties of both triangles and quadrilaterals and $n$-gons. \bzgled Let $CD$ be the altitude at the hypotenuse $AB$ of a right-angled triangle $ABC$. If $M$ and $N$ are the midpoints of the line segments $CD$ and $BD$, then $AM \perp CN$. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.7.10.pic} \caption{} \label{sl.skl.3.7.10.pic} \end{figure} \textbf{\textit{Proof.}} The line $NM$ is the median of the triangle $BCD$, so by the \ref{srednjicaTrik} $NM \parallel BC$ (Figure \ref{sl.skl.3.7.10.pic}). Because the angle at the vertex $C$ is a right angle, $NM \perp AC$ as well. Because of this, the line $NM$ is the altitude of the triangle $ANC$. Since $CD$ is also the altitude of this triangle, $M$ is its height point. Therefore, the line $AM$ is the altitude of the third triangle of this triangle and $AM \perp CN$ applies. \kdokaz \bzgled If $P$ and $Q$ are the midpoints of the sides $BC$ and $CD$ of a parallelogram $ABCD$, then the lines $AP$ and $AQ$ divide the diagonal $BD$ of this parallelogram into three congruent line segments. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.7.11.pic} \caption{} \label{sl.skl.3.7.11.pic} \end{figure} \textbf{\textit{Proof.}} We mark with $E$ and $F$ the intersections of the lines $AP$ and $AQ$ with the diagonal $BD$ of the parallelogram $ABCD$ and with $S$ the intersection of its diagonals $AC$ and $BD$ (Figure \ref{sl.skl.3.7.11.pic}). The diagonals of the parallelogram are divided (from \ref{paralelogram}), so the point $S$ is the common center of the lines $AC$ and $BD$. This means that the points $E$ and $F$ are the centers of the triangles $ACB$ and $ACD$, so they divide the altitudes $SB$ and $SD$ of these triangles in the ratio $2:1$ (from \ref{tezisce}). Therefore: \begin{eqnarray*} BE &=& \frac{2}{3}BS = \frac{2}{3}DS = FD,\\ EF&=& ES+ SF= \frac{1}{3}SB+ \frac{1}{3}SD= \frac{1}{3}(SB +SD)=\frac{1}{3} BD, \end{eqnarray*} which is what we wanted to prove. \kdokaz \bzgled Let $BAKL$ and $ACPQ$ be positively oriented squares in the same plane. Prove that the lines $BP$ and $CL$ intersect at a point lying on the line containing the altitude $AA'$ of the triangle $ABC$. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.7.12.pic} \caption{} \label{sl.skl.3.7.12.pic} \end{figure} \textbf{\textit{Proof.}} Let $AA'$ be the altitude of the triangle $ABC$ (Figure \ref{sl.skl.3.7.12.pic}). We mark with $X$ and $Y$ the intersections of the line $AA'$ with the rectangles on the line $CL$ through the point $B$ and on the line $BP$ through the point $C$. We prove that $X = Y$. The triangle $BLC$ and $ABX$ are similar according to the \textit{ASA} theorem \ref{KSK} because: $BL \cong AB$, $\angle BLC\cong\angle ABX$ and $\angle BCL\cong\angle AXB$ (the angle with perpendicular sides - theorem \ref{KotaPravokKraki}). Therefore $AX \cong BC$. Similarly, the triangle $CPB$ and $ACY$ are similar, so $AY \cong BC$. Therefore $AX \cong AY$ or $X = Y$. This means that the lines $AA'$, $BP$ and $CL$ are the altitudes of the triangle $XBC$, so they intersect at one point. \kdokaz \bzgled \label{zgledPravokotnik} Let $K$ be the midpoint of the side $CD$ of a rectangle $ABCD$. A point $L$ is the foot of the perpendicular from the vertex $B$ on the diagonal $AC$ and $S$ is the midpoint of the line segment $AL$. Prove that $\angle KSB$ is a right angle. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.7.13.pic} \caption{} \label{sl.skl.3.7.13.pic} \end{figure} \textbf{\textit{Proof.}} Let $V$ be the center of the line $BL$ (Figure \ref{sl.skl.3.7.13.pic}). The line $SV$ is the median of the triangle $ABL$ for the base $AB$, so $SV\parallel AB$ and $SV =\frac{1}{2} AB$ (statement \ref{srednjicaTrik}). From the first relation and $BC\perp AB$ it follows that $SV\perp BC$ (statement \ref{KotiTransverzala}). This means that $BL$ and $SV$ are the altitude of the triangle $CSB$. Therefore, $V$ is the altitude point of this triangle, so $CV$ is the altitude of its third altitude (statement \ref{VisinskaTocka}) or $CV\perp SB$ is true. From $SV\parallel AB$ and $SV =\frac{1}{2} AB=KC$ it follows that the quadrilateral $SVCK$ is a parallelogram or $CV\parallel SK$ is true. From this and $CV\perp SB$ it finally follows (statement \ref{KotiTransverzala}) $SK\perp SB$, so $\angle KSB$ is a right angle. \kdokaz \bzgled Let $AP$, $BQ$ and $CR$ be the altitude, the median and the bisector of the angle $ACB$ ($R\in AB$) of a triangle $ABC$. Prove that if the triangle $PQR$ is regular, then the triangle $ABC$ is also regular. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.7.14.pic} \caption{} \label{sl.skl.3.7.14.pic} \end{figure} \textbf{\textit{Proof.}} Let $PQR$ be a right triangle or $PQ\cong QR\cong RP$ (Figure \ref{sl.skl.3.7.14.pic}). The point $Q$ is the center of the hypotenuse $AC$ of the right triangle $APC$, so $QA \cong QC \cong QP$ (statement \ref{TalesovIzrKroz2}. Because in this case $QR\cong QC\cong QP$, from the same statement it follows that $\angle ARC$ is a right angle. From the similarity of the triangles $ACR$ and $BCR$ (statement \ref{KSK}) we get that the point $R$ is the center of the side $AB$ and that $AC\cong BC$ is true. Because the point $R$ is the center of the hypotenuse $AB$ of the right triangle $APB$, $AB = 2RP = 2PQ = 2AQ = AC$. Therefore, $AB\cong AC\cong BC$ is true, which means that $ABC$ is a right triangle. \kdokaz It is not difficult to prove that a triangle is equilateral if and only if the corresponding medians are concurrent. The same is true for altitudes. But is something similar true for so-called angle bisectors? The line segments $BB'$ and $CC'$, where $BB'$ and $CC'$ are angle bisectors of the triangle $ABC$ and $B'\in AC$ and $C'\in AB$, are called \index{angle bisector} \pojem{angle bisectors}. Angle bisectors are denoted by $l_a$, $l_b$ and $l_c$. The aforementioned statement is also true in this case, but the proof is not so simple. This is the subject of the following well-known theorem. \bizrek \index{theorem!Steiner-Lemus} (Steiner-Lehmus\footnote{\textit{D. C. L. Lehmus} (1780--1863),\index{Lehmus, D. C. L.} French mathematician, who in 1840 sent this, at first glance simple statement, to the famous Swiss geometer \index{Steiner, J.} \textit{J. Steiner} (1796--1863), who derived a very extensive proof of this theorem. Then followed several different solutions to this problem and one of them was published in 1908 by the French mathematician \index{Poincar\'{e}, J. H.} \textit{J. H. Poincar\'{e}} (1854--1912).}) Let $BB'$ ($B' \in AC$) and $CC'$ ($C' \in AB$) be the bisectors of the interior angles of a triangle $ABC$. Then: $$AB \cong AC \Leftrightarrow BB'\cong CC'.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.7.15.pic} \caption{} \label{sl.skl.3.7.15.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.7.15.pic}) ($\Rightarrow$) From $AB \cong AC$ it follows that $\angle ABC \cong \angle ACB$ (theorem \ref{enakokraki}) or $\angle B'BC \cong \angle C'CB$. By the \textit{ASA} theorem \ref{KSK}, the triangles $B'BC$ and $C'CB$ are congruent, so $BB'\cong CC'$. ($\Leftarrow$) Let $BB'\cong CC'$. We assume that $AB\not\cong AC$. Without loss of generality let $AB < AC$. In this case $\angle ACB < \angle ABC$ (by \ref{vecstrveckot}) or $\angle ACC'< \angle ABB'$. This means that inside the angle $ABB'$ there is a segment $p$ with endpoint $B$, which intersects the side $AC$ in such a point $D$, that both $\mathcal{B}(A,D,B')$ and $\angle DBB'\cong \angle ACC'$ are true. In the triangle $BCD$ is $\angle ACB < \angle DBC$ and because of that also $BD < CD$ (by \ref{vecstrveckot}). Therefore there is such a point $E$, which is between the points $C$ and $D$, so that $BD \cong CE$. By the \textit{SAS} \ref{SKS} the triangles $BDB'$ and $CEC'$ are similar, therefore the angles $BDB'$ and $CEC'$ are similar. We prove that this is not possible. Because of Pasch's axiom \ref{AksPascheva} (used for the triangle $AC'E$ and the line $BD$) the line $BD$ intersects the line $C'E$ in some point $S$. In the triangle $SDE$ is the angle $SEC$ (or the angle $CEC'$) external and by \ref{zunanjiNotrNotrVecji} it can not be similar to the adjacent internal angle $SDE$ (or the angle $BDB'$). This means that the assumption $AB < AC$ (analogously $AB > AC$) is not possible. Therefore $AB\cong AC$. \kdokaz \bzgled Let $A_1$ be the midpoint of the side $BC$ of a triangle $ABC$. Calculate the measure of the angle $AA_1C$, if $\angle BAC=45^0$ and $\angle ABC=30^0$. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.7.1a.pic} \caption{} \label{sl.skl.3.7.1a.pic} \end{figure} From $\angle CC'B = 90^0$ it follows first $\angle C'CB=60^0$, then that the point $C'$ lies on the circle above the diameter $CB$ and with the center $A_1$ (by \ref{TalesovIzrKroz}), so $A_1C'\cong A_1C\cong A_1B$. Therefore, the triangle $CC'A_1$ is isosceles, or by \ref{enakokraki} it holds $\angle CC'A_1 \cong\angle C'CB=60^0$. This means that the triangle $CC'A_1$ is equilateral and $C'C\cong C'A_1$. From the fact that $AC'C$ is an isosceles triangle ($\angle CAC'=\angle ACC'=45^0$), it follows that $AC'\cong C'C$. If we connect this with the previous relation, we get $AC'\cong C'A_1$, which means that the triangle $AC'A_1$ is also isosceles. Therefore, it is (by \ref{enakokraki} and \ref{zunanjiNotrNotr}): $$\angle C'A_1A\cong\angle C'AA_1=\frac{1}{2}\angle A_1C'B= \frac{1}{2}\angle C'BA_1=\frac{1}{2}\cdot 30^0=15^0.$$ In the end there is also: $$\angle AA_1C=\angle C'A_1C-\angle C'A_1A =60^0-15^0=45^0,$$ which needed to be calculated. \kdokaz \bzgled Let $P$ be the midpoint of the side $BC$ of an isosceles triangle $ABC$ and $Q$ the foot of the perpendicular from the point $P$ on the leg $AC$ of that triangle. Let $S$ be the midpoint of the line segment $PQ$. Prove that $AS \perp BQ$. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.7.16.pic} \caption{} \label{sl.skl.3.7.16.pic} \end{figure} \textbf{\textit{Proof.}} From the congruence of the triangles $ABP$ and $ACP$ (from the statement \textit{SSS} \ref{SSS}) it follows that the congruence of the sides $APB$ and $APC$ or $AP\perp BC$ (Figure \ref{sl.skl.3.7.16.pic}). We mark with $R$ the center of the line $QC$. The line $SR$ is the median of the triangle $QPC$ for the base $PC$, so according to the statement \ref{srednjicaTrik} $SR\parallel CP$. From this and $AP\perp BC$ it follows that $SR\perp AP$. So $S$ is the altitude point of the triangle $APR$, so according to the statement \ref{VisinskaTocka} also $AS\perp PR$. But the line $PR$ is the median of the triangle $BQC$ for the base $BQ$, so $PR\parallel BQ$ (from the statement \ref{srednjicaTrik}). From $AS\perp PR$ and $PR\parallel BQ$ we get $AS\perp BQ$. \kdokaz \bzgled \label{kotBSC} If $S$ is the incentre and $\alpha$, $\beta$, $\gamma$ the interior angles at the vertices $A$, $B$, $C$ of a triangle $ABC$, then $$\angle BSC=90^0+\frac{1}{2}\cdot\alpha.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.7.1c.pic} \caption{} \label{sl.skl.3.7.1c.pic} \end{figure} \textbf{\textit{Proof.}} According to the statement \ref{SredVcrtaneKrozn} the simetrals of the internal angles of the triangle $ABC$ intersect in the center of the inscribed circle - in the point $S$ (Figure \ref{sl.skl.3.7.1c.pic}). So $\angle SBC =\frac{1}{2}\cdot \beta$ and $\angle SCB =\frac{1}{2}\cdot \gamma$. Because according to the statement \ref{VsotKotTrik} in every triangle the sum of the internal angles is equal to $180^0$, it follows: $$\angle BSC = 180^0-\frac{1}{2}\cdot\left( \beta+ \gamma\right)=180^0-\frac{1}{2}\cdot\left( 180^0- \alpha\right)=90^0+\frac{1}{2}\cdot\alpha,$$ which had to be proven. \kdokaz \bzgled Construct a triangle with given $a$, $t_a$, $R$ (see the labels in section \ref{odd3Stirik}). \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.7.16a.pic} \caption{} \label{sl.skl.3.7.16a.pic} \end{figure} \textbf{\textit{Proof.}} The construction can be carried out by first drawing the circumscribed circle $k(O,R)$, choosing an arbitrary point $B\in k$, planning the cord $BC\cong a$ of the circle, the center $A_1$ of the cord $BC$ and finally the point $A$ as the intersection of the circles $k(O,R)$ and $k_1(A_1,t_a)$ (Figure \ref{sl.skl.3.7.16a.pic}). It is clear that the task of the solution is exactly when $a\leq 2R$ and the intersection of the circles $k(O,R)$ and $k_1(A_1,t_a)$ is not an empty set. The number of solutions in this case depends on the number of intersections of the circles $k(O,R)$ and $k_1(A_1,t_a)$. \kdokaz \bzgled Construct a right-angled triangle if the hypotenuse and the altitude to that hypotenuse are congruent to the given line segments $c$ and $v_c$. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.7.16b.pic} \caption{} \label{sl.skl.3.7.16b.pic} \end{figure} \textbf{\textit{Analysis.}} Let $ABC$ be a right-angled triangle with a right angle at the vertex $C$, in which the hypotenuse $AB$ and the altitude $CC'$ are congruent to the line segments $c$ and $v_c$ (Figure \ref{sl.skl.3.7.16b.pic}). By the \ref{TalesovIzrKroz2} theorem, the center $O$ of the hypotenuse $AB$ is also the center of the circumscribed circle of the triangle $ABC$. Therefore, the vertex $C$ lies on the circle $k$ with diameter $AB$. Since $CC'\cong v_c$, the vertex $C$ also lies on the parallel $p$ to the line $AB$ at a distance $v_c$. The point $C$ is then the intersection of this parallel and the circle $k$. \textbf{\textit{Construction.}} First, we plan the line $AB$, which is congruent to the given line segment $c$, then the center $O$ of the line $AB$ and the circle $k(O,OA)$. Then we plan the parallel $p$ to the line $AB$ at a distance $v_c$. One of the intersections of the line $p$ and the circle $k(O,OA)$ is denoted by $C$. We prove that $ABC$ is the desired triangle. \textbf{\textit{Proof.}} By construction, point $C$ lies on the circle with radius $AB$, so by \ref{TalesovIzrKroz2} $\angle ACB=90^0$, which means that $ABC$ is a right triangle with hypotenuse $AB$. By construction, it is consistent with the distance $c$. Let $CC'$ be the altitude of the triangle $ABC$. By construction, point $C$ lies on the line $p$, which is from the line $AB$ at a distance $v_c$, so also $|CC'|=d(C,AB)=d(p,AB)$ or $CC'\cong v_c$. \textbf{\textit{Discussion.}} The number of solutions is dependent on the number of intersections of the line $p$ and the circle $k(O,OA)$. \kdokaz \bnaloga\footnote{2. IMO Romania - 1960, Problem 4.} Construct triangle ABC, given $v_a$, $v_b$ (the altitudes from $A$ and $B$) and $t_a$, the median from vertex $A$. \enaloga \begin{figure}[!htb] \centering \input{sl.skl.3.7.IMO1.pic} \caption{} \label{sl.skl.3.7.IMO1.pic} \end{figure} \textbf{\textit{Solution.}} Let $ABC$ be a triangle, such that $AA' \cong v_a$ and $BB' \cong v_b$ are its altitudes and $AA_1\cong t_a$ is its median (Figure \ref{sl.skl.3.7.IMO1.pic}). We mark with $A'_1$ the orthogonal projection of point $A_1$ on the line $AC$. The distance $A_1A'_1$ is the median of the triangle $BB'C$ for the base $BB'$, so by \ref{srednjicaTrik}: $$|A_1A'_1|=\frac{1}{2}\cdot|BB'|=\frac{1}{2}\cdot v_b \hspace*{1mm} \textrm{ in } \hspace*{1mm} A_1A'_1 \parallel BB'.$$ From this it follows that the lines $A_1A'_1$ and $AC$ are perpendicular in point $A'_1$, thus the line $AC$ is tangent to the circle $k(A_1,\frac{1}{2} v_b)$ \ref{TangPogoj}. The proven properties allow us to construct it. First, we can plan the rectangular triangle $AA'A_1$ ($AA'\cong v_a$, $AA_1\cong t_a$ and $\angle AA'A_1 = 90^0$), then the circle $k(A_1,\frac{1}{2} v_b)$. From the point $A$ we plan the tangents to the circle $k(A_1,\frac{1}{2} v_b)$. The intersection of one of the tangents with the line $A'A_1$ is denoted by $C$. In the end, we plan such a point $B$, that $BA_1 \cong CA_1$ and $\mathcal{B}(C,A_1,B)$. We prove that the triangle $ABC$ satisfies the given conditions. From the construction it is $AA'\cong v_a$ the height and $AA_1 \cong t_a$ the centroid (because $A_1$ is the center of the line $BC$) of the triangle $ABC$. Let $BB'$ be the height of this triangle. We prove that $BB' \cong v_b$. The line $AC$ is by construction the tangent of the circle $k(A_1,\frac{1}{2} v_b)$. Their point of contact is denoted by $A'_1$. By the theorem \ref{TangPogoj} the lines $A_1A'_1$ and $AC$ are perpendicular in the point $A'_1$, therefore the line $A_1A'_1$ is the median of the triangle $BB'C$ for the base $BB'$ and it is valid $|BB'|= 2\cdot |A_1A'_1|=2\cdot\frac{1}{2}\cdot v_b=v_b$. The task has no solution when $v_a>t_a$. If $v_a\leq t_a$, the number of solutions depends on the number of tangents, which we can plan from the point $A$ on the circle $k(A_1,\frac{1}{2} v_b)$. In this case, the tangent must intersect the line $A'A_1$. If $\frac{1}{2} v_b4. \label{teselRelHyp} \end{eqnarray} This inequality has infinitely many solutions in the set $\mathbb{N}^2$, which means that we have in hyperbolic geometry infinitely many regular tilings. Two of them are for example $(3,7)$ and $(4,5)$ (Figure \ref{sl.skl.3.9.2H.pic}\footnote{http://math.slu.edu/escher/index.php/Category:Hyperbolic-Tessellations}). In the latter, five squares touch around one vertex. This is possible because in hyperbolic geometry the interior angle of a square is always sharp and is not constant. It turns out that the square with the longer side has a smaller interior angle. It is possible to choose such a side of the square that the interior angle is equal to $\frac{360^0}{5} =72^0$, which just corresponds to the tiling $(4,5)$. \begin{figure}[!htb] \centering \includegraphics[width=0.413\textwidth]{whyptess1.eps}\hspace*{4mm} \includegraphics[width=0.387\textwidth]{whyptess.eps} \caption{} \label{sl.skl.3.9.2H.pic} \end{figure} In elliptic geometry, where the sum of the angles in a triangle is always greater than $180^0$, the aforementioned relation for triangle $OSB$ becomes: $\frac{360^0}{2n}+\frac{360^0}{2m}+90^0>180^0$, or: \begin{eqnarray} (n-2)(m-2)<4. \label{teselRelElipt} \end{eqnarray} This equation has solutions $(3,3)$, $(4,3)$, $(3,4$), $(5,3)$ and $(3,5)$ in the set $\mathbb{N}^2$. Because elliptic geometry is realized as a model on a sphere, these solutions represent tessellations of the sphere with spherical polygons. The sides of these polygons are arcs of great circles of the sphere. If in Euclidean space with distances we connect the appropriate vertices of these tessellations, we get the so-called \index{pravilni!poliedri} \pojem{regular polyhedra} (Figure \ref{sl.skl.3.9.2E.pic}\footnote{http://www.upc.edu/ea-smi/personal/claudi/web3d/}): \pojem{regular tetrahedron}, \pojem{cube} (or \pojem{regular hexahedron}), \pojem{regular octahedron}, \pojem{regular dodecahedron} and \pojem{regular icohedron}. For example, $(4,3)$ would represent a cube, in which three squares (regular 4-gon) meet at one point. \begin{figure}[!htb] \centering \includegraphics[bb=0 0 11cm 6cm]{wpoliedri.eps} \caption{} \label{sl.skl.3.9.2E.pic} \end{figure} Let's go back to the Euclidean plane. If we allow the possibility that in covering the plane we use more (finitely many) types of regular polygons or more types of cells that are arranged the same way at each vertex, then in addition to the three regular tessellations from the statement \ref{pravilnaTlakovanja} there are eight so-called \index{tlakovanja!Arhimedova} \pojem{Archimedean tessellations} \footnote{The proof of this statement was carried out by the German astronomer, mathematician and physicist \index{Kepler, J.} \textit{J. Kepler} (1571--1630).} (Figure \ref{sl.skl.3.9.2A.pic}\footnote{http://commons.wikimedia.org/wiki/File\%3AArchimedean-Lattice.png}). \begin{figure}[!htb] \centering \includegraphics[bb=0 0 10cm 7.7cm]{Archimedean.eps} \caption{} \label{sl.skl.3.9.2A.pic} \end{figure} In addition to regular and Archimedean tilings, there are also other tilings with polygons that are not regular. The simplest example is the tiling with congruent parallelograms (Figure \ref{sl.skl.3.9.3a.pic}), which we get if we deform the regular tiling $(4,4)$ so that instead of squares in one vertex, four parallelograms meet. This tiling is determined by the grid of two sets of parallel lines. \begin{figure}[!htb] \centering \input{sl.skl.3.9.3aa.pic} \input{sl.skl.3.9.3aaa.pic} \caption{} \label{sl.skl.3.9.3a.pic} \end{figure} \begin{figure}[!htb] \centering \input{sl.skl.3.9.3bb.pic} \input{sl.skl.3.9.3cc.pic} \caption{} \label{sl.skl.3.9.3bc.pic} \end{figure} If we divide all parallelograms into two triangles with diagonals that have the same direction, we get a tiling of the plane with congruent triangles (Figure \ref{sl.skl.3.9.3a.pic}). The triangle (basic cell) can be arbitrary, because two such (congruent) triangles can always be connected by a common side into a parallelogram and thus get a tiling with parallelograms. Tiling with arbitrary congruent triangles is a generalization of the regular tiling $(3,6)$ with regular triangles. %\vspace*{-1mm} As special cases of tiling with parallelograms, we get tiling with rectangles and tiling with rhombuses (Figure \ref{sl.skl.3.9.3bc.pic}). We will prove an even more general statement, which may not be so obvious. It is possible that there is also a tiling with arbitrary congruent quadrilaterals. %\vspace*{-1mm} \bzgled Let $ABCD$ be an arbitrary quadrilateral. Prove that it is possible to tessellate the plane with the cell $ABCD$ so that each vertex is surrounded by four such quadrilaterals. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.9.4.pic} \caption{} \label{sl.skl.3.9.4.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.9.4.pic}) Let $ABCD$ be an arbitrary quadrilateral in the plane with internal angles $\alpha$, $\beta$, $\gamma$ and $\delta$. Points $O$ and $S$ are the centers of its sides $AB$ and $BC$. With central reflections $\mathcal{S}_O$ and $\mathcal{S}_S$ (for the definition of central reflection see section \ref{odd6SredZrc}) the quadrilateral $ABCD$ is mapped into quadrilateral $BAC_1D_1$ and $A_2CBD_2$. Here $\angle ABD_1\cong\alpha$ and $CBD_2\cong\gamma$, therefore $\angle D_1BD_2\cong\delta$. Because $BD_1\cong AD$ and $BD_2\cong CD$, there exists a point $E$, such that quadrilaterals $D_1ED_2B$ and $ABCD$ are congruent. So around point $B$ four quadrilaterals intersect, all of which are congruent to quadrilateral $ABCD$. We can continue the process of tiling the plane, if we use central symmetry with respect to the centers of sides of newly formed quadrilaterals. \kdokaz Therefore there exist tilings of the plane with any triangle and any quadrilateral. It is clear that for any pentagon, hexagon,... this property does not hold. For a regular hexagon there exists a regular tiling $(6,3)$. If in the previous statement we put together two appropriate adjacent quadrilaterals, we get a tiling of the plane with congruent hexagons, which are not necessarily regular, but are always centrally symmetric. %_______________________________________________________________________________ \poglavje{Sets of Points in a Plane. Sylvester's Problem} \label{odd3Silvester} In this section we will investigate problems related to sets of points in the plane and the lines determined by these points. At the beginning we will consider some consequences of the first two groups of axioms (incidence and order). First we will define new concepts. Let $\mathfrak{T}$ be a set of $n$ ($n>2$) points in the plane. With $\mathcal{P}(\mathfrak{T})$ we denote the set of all lines, each of which goes through at least two points from the set $\mathfrak{T}$ (Figure \ref{sl.skl.3.10.1.pic}). Because the set $\mathfrak{T}$ contains at least two points, from the axioms of incidence it follows that the set $\mathcal{P}(\mathfrak{T})$ is not empty. The question arises, how many lines are in the set $\mathcal{P}(\mathfrak{T})$. \begin{figure}[!htb] \centering \input{sl.skl.3.10.1.pic} \caption{} \label{sl.skl.3.10.1.pic} \end{figure} \bizrek \label{stevPremic} Let $\mathfrak{T}$ be a set of $n$ ($n>2$) points in the plane such that no three of them are collinear. Then the number of lines of the set $\mathcal{P}(\mathfrak{T})$ is equal to $$\frac{n(n-1)}{2}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.10.2.pic} \caption{} \label{sl.skl.3.10.2.pic} \end{figure} \textbf{\textit{Proof.}} Through each of the $n$ points from the set $\mathfrak{T}$ there are exactly $n -1$ lines from the set $\mathcal{P}(\mathfrak{T})$. Because we count each line in this way twice (Figure \ref{sl.skl.3.10.2.pic}), we have to divide by 2. So there are exactly $\frac{n(n-1)}{2}$ lines in the set $\mathcal{P}(\mathfrak{T})$. \kdokaz The previous statement can also be solved in the following way: through the first point there are $n -1$ lines, through the second point $n - 2$ lines (one less, because the line determined by these two points is not counted twice), $n - 3$ lines through the third point and so on until one line through the penultimate point. This is a total of $(n -1) + (n - 2) +\cdots+1$ lines. Of course, this is again equal to $\frac{n(n-1)}{2}$. If we take $n-1= k$, we get the known formula for the sum of the first $k$ natural numbers: $$1+ 2+ \cdots + k=\frac{k(k+1)}{2}.$$ From the previous statement we can derive the formula for the number of diagonals of an arbitrary $n$-gon. \bizrek If $D_n$ is the number of diagonals of an arbitrary $n$-gon, then $$D_n=\frac{n( n-3)}{2}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.10.3.pic} \caption{} \label{sl.skl.3.10.3.pic} \end{figure} \textbf{\textit{Proof.}} We mark with $\mathfrak{O}$ the set of all vertices of an arbitrary $n$-gon. The number of diagonals is equal to the number of all lines from the set $\mathcal{P}(\mathfrak{O})$ (statement \ref{stevPremic}) decreased by the number of its sides (Figure \ref{sl.skl.3.10.3.pic}). Therefore: $$D_n=\frac{n(n- 1)}{2}-n=\frac{n^2-3n}{2}=\frac{n(n-3)}{2},$$ which was to be proven. \kdokaz We mention that the formula for the number of diagonals of an $n$-gon could also be derived directly - with a similar treatment as in the proof of statement \ref{stevPremic}. From each of the $n$ vertices of an $n$-gon we can draw $n - 3$ diagonals. In this way we count each diagonal twice, so we have to divide by 2 and we get the previous formula. Statement \ref{stevPremic} referred to the number of lines determined by a set of points in a plane that are in such a position that no three of them are collinear. In the next example we will check what happens if we add new conditions. \bzgled Let $\mathfrak{T}$ be a set of $n$ ($n > 2$) points in the plane which are in such a position that $m$ ($m < n$) of them lies on the same line, but otherwise no other three points are collinear. What is the number of lines in the set $\mathcal{P}(\mathfrak{T})$? \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.10.4.pic} \caption{} \label{sl.skl.3.10.4.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.skl.3.10.4.pic}) Without the additional condition, there would be $\frac{n(n -1)}{2}$ lines in the set $\mathcal{P}(\mathfrak{T})$ (statement \ref{stevPremic}). The additional condition in the problem reduces this number by one less than the number of lines determined by the set of $m$ points in general position. This is because otherwise these lines would be counted multiple times. So the number of lines in the set $\mathcal{P}(\mathfrak{T})$ is equal to: $$\frac{(n-1)}{2}-\frac{(m -1)}{2}+1.$$ \kdokaz The next simple example will be an introduction to the very interesting problem of the relationship between the set of points $\mathfrak{T}$ and the set of all lines $\mathcal{P}(\mathfrak{T})$, which this set of points determines. \bzgled Construct nine points lying on ten lines in such a way, that each of those ten lines contains exactly three of these nine points\footnote{\index{Newton, I.} \textit{I. Newton} (1642--1727), znani angleški matematik in fizik, ki je zastavil ta problem v obliki: ‘‘How can you plant 9 trees in a garden with 10 rows and each row containing exactly 3 trees?’’}. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.10.5.pic} \caption{} \label{sl.skl.3.10.5.pic} \end{figure} \textbf{\textit{Proof.}} One of the possibilities is the following. Let $ABCD$ be an arbitrary rectangle, $P$ and $Q$ be the centers of sides $AB$ and $CD$, $S$ be the intersection of the diagonals, $K$ be the intersection of lines $AP$ and $DQ$, and $L$ be the intersection of lines $BP$ and $CQ$ (Figure \ref{sl.skl.3.10.5.pic}). Because the rectangle $ABCD$ is a rectangle, the points $P$, $Q$ and $S$ are collinear. So are the points $K$, $L$ and $S$ (points $K$ and $L$ are the centers of rectangles $AQPD$ and $QBCP$). So we have nine points $A$, $B$, $C$, $D$, $P$, $Q$, $S$, $K$ and $L$, of which three lie on each of the ten lines $AB$, $CD$, $PQ$, $KL$, $AC$, $BD$, $PA$, $PB$, $QC$ and $QD$. \kdokaz If in the previous example we denote the set of nine points with $\mathfrak{T}$, we see that the set of ten lines is not the set $\mathcal{P}(\mathfrak{T})$. In the set $\mathcal{P}(\mathfrak{T})$ we would have sixteen lines - our ten and also the additional lines $AD$, $BC$ $DL$, $AL$, $CK$ and $BK$. But each of these six lines contains only two points. So the condition that they contain exactly three points of the initial set is not fulfilled. It is now logical to ask the following question: Is it possible in a plane to set a finite set of non-collinear points $\mathfrak{T}$ so that each line from the set $\mathcal{P}(\mathfrak{T})$ contains exactly three points from the set $\mathfrak{T}$? It is clear that it is possible if we require that each line from $\mathcal{P}(\mathfrak{T})$ contains exactly two points from $\mathfrak{T}$. The most simple example for this are the vertices of a triangle and its altitudes, or any set of points from the statement \ref{stevPremic}. The aforementioned problem for three points is not so simple, and we will find the answer in the continuation. We mention that the answer is negative. Even more - the answer is negative even if we require that each line from $\mathcal{P}(\mathfrak{T})$ contains at least three points from $\mathfrak{T}$. First, we prove one lemma (an auxiliary statement). \bizrek \label{SylvesterLema} Let $\mathfrak{T}$ be a finite set of points in the plane which do not all lie on the same line and $\mathcal{P}=\mathcal{P}(\mathfrak{T})$. If $T_0\in \mathfrak{T}$ and $p_0\in \mathcal{P}$ ($T_0\notin p_0$) are such that they determine the minimum distance, i.e. $$(\forall T\in \mathfrak{T})(\forall p\in \mathcal{P}) ( T\notin p\Rightarrow d(T,p)\geq d(T_0,p_0)),$$ then the line $p_0$ contains exactly two points from the set $\mathfrak{T}$. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.10.6.pic} \caption{} \label{sl.skl.3.10.6.pic} \end{figure} \textbf{\textit{Proof.}} We assume the opposite. Let the line $p_0$ contain at least three different points $A$, $B$ and $C$ from the set $\mathfrak{T}$ (Figure \ref{sl.skl.3.10.6.pic}). We mark with $T'_0$ the orthogonal projection of the point $T_0$ on the line $p_0$. If the point $T'_0$ differs from the points $A$, $B$ and $C$, then at least two of these three points (let it be $B$ and $C$) are on the line $p_0$ on the same side of the point $T'_0$. Without loss of generality, let $\mathcal{B}(T'_0,B,C)$. Because $T_0,C \in \mathfrak{T}$, then the line $q= CT_0$ ($q \neq p_0$, because $T_0\notin p_0$) belongs to the set $\mathcal{P}$. Let $B'$ be the orthogonal projection of the point $B$ on the line $q$. It is not difficult to prove that in this case it holds: $$d(B,q)=|BB'| BP > PA$, so $\angle APB$ is the largest angle in the triangle $APB$ and is therefore larger than $60^0$. Similarly, $CB > BP,BC$, so $\angle BPC > 60^0$. The same would hold for the angles $CPD$, $DPE$, $EPF$, and $EPA$, but that is not possible, because their sum is always equal to or even greater than $360^0$, regardless of whether $P$ is an inner or outer point of the hexagon $ABCDEF$. Therefore, the point $P$ is connected to no more than five points. \kdokaz \bzgled What is the maximum number of regions in the plane that can be divided by $n$ lines\footnote{Ta problem je rešil švicarski geometer \index{Steiner, J.} \textit{J. Steiner} (1796--1863).}? \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.10.9.pic} \caption{} \label{sl.skl.3.10.9.pic} \end{figure} \textbf{\textit{Solution.}} We will describe the process of designing such $n$ lines (Figure \ref{sl.skl.3.10.9.pic}). If $n = 1$, or if there is only one line, the plane is divided into two areas. Two lines, if they are not parallel, divide the plane into four areas. If we add a third line $p_3$, which is not parallel to them and does not go through their common point, it intersects the initial two lines in two points. These two points divide the line $p_3$ into three parts, each of which is in one of the three previous four areas of the plane. So with line $p_3$ we get three more parts of the plane, a total of seven. We continue the process. If $n -1$ lines divide the plane into $k$ parts, by adding the $n$-th line $p_n$ (which is not parallel to any of the previous $n -1$ lines and does not contain any of their intersections), we first get $n -1$ intersections on that line, then $n$ of its parts or $n$ new areas of the given plane. So the maximum possible number of areas for $n$ lines is equal to: \begin{eqnarray*} 2+2+3+\cdots+n&=&1+1+2+3+\cdots+n=\\ &=&1+\frac{n(n+1)}{2}=\\&=&\frac{n^2+n+2}{2}. \end{eqnarray*} A formal proof of this fact could be derived by mathematical induction. \kdokaz %_______________________________________________________________________________ \poglavje{Helly's Theorem} \label{odd3Helly} The next important statement is a consequence of only the first two groups of axioms or the axioms of incidence and the axioms of order. We will also use it in tasks related to consistency. \bizrek \label{Helly} Let $\Phi_1$, $\Phi_2$, ... , $\Phi_n$ ($n \geq 4$) be convex sets in the plane. If every three of these sets have a common point, then all $n$ sets have a common point \index{izrek!Hellyjev}(Helly's theorem\footnote{Austrian mathematician \index{Helly, E.} \textit{E. Helly} (1884--1943) discovered this statement in the general case of $n$-dimensional space $\mathbb{E}^n$ in 1913, but published it only in 1923. Alternative proofs were meanwhile given by Austrian mathematician \index{Radon, J. K. A.} \textit{J. K. A. Radon} (1887-–1956) in 1921 and Hungarian mathematician \index{Kőnig, D.} \textit{D. Kőnig} (1884–-1944) in 1922.}). \eizrek \begin{figure}[!htb] \centering \hspace*{10mm} \input{sl.skl.3.11.1.pic} \caption{} \label{sl.skl.3.11.1.pic} \end{figure} \textbf{\textit{Proof.}} We will prove this by induction on $n$. \textit{(A)} Let $n = 4$ and (Figure \ref{sl.skl.3.11.1.pic}): \begin{itemize} \item $P_4\in \Phi_1 \cap \Phi_2 \cap \Phi_3$, \item $P_3\in \Phi_1 \cap \Phi_2 \cap \Phi_4$, \item $P_2\in \Phi_1 \cap \Phi_3 \cap \Phi_4$, \item $P_1\in \Phi_2 \cap \Phi_3 \cap \Phi_4$. \end{itemize} We will prove that there exists a point that lies in each of the figures $\Phi_1$, $\Phi_2$, $\Phi_3$ and $\Phi_4$. Based on the mutual position of points $P_1$, $P_2$, $P_3$ and $P_4$ we will consider only two most general cases (the proof in the other cases is similar). \textit{1)} A quadrilateral, determined by points $P_1$, $P_2$, $P_3$ and $P_4$, is non-convex. In this case, one of the points $P_1$, $P_2$, $P_3$ and $P_4$ is an inner point of the triangle determined by the remaining three points. Without loss of generality, let $P_4$ be the inner point of the triangle $P_1P_2P_3$. The vertices of this triangle lie in the shape $\Phi_4$. Because $\Phi_4$ is a convex shape, all sides and inner points of the triangle $P_1P_2P_3$, as well as the point $P_4$, lie in it. In this case, the point $P_4$ is the common point of shapes $\Phi_1$, $\Phi_2$, $\Phi_3$ and $\Phi_4$. \textit{2)} The quadrilateral determined by points $P_1$, $P_2$, $P_3$ and $P_4$ is convex. Without loss of generality, let its diagonals be $P_1P_2$ and $P_3P_4$. Because the quadrilateral is convex, its diagonals intersect in a point $S$. Given shapes are convex, so from $P_1, P_2\in \Phi_3,\Phi_4$ it follows that the diagonal $P_1P_2$ lies entirely in shapes $\Phi_3$ and $\Phi_4$. Analogously, from $P_3, P_4\in \Phi_1,\Phi_2$ it follows that the diagonal $P_3P_4$ lies entirely in shapes $\Phi_1$ and $\Phi_2$. The point $S$, which lies on both diagonals $P_1P_2$ and $P_3P_4$, lies in all four shapes $\Phi_1$, $\Phi_2$, $\Phi_3$ and $\Phi_4$. With this we have proven that the statement is true for $n=4$. \textit{(B)} Let's now assume that the statement is true for $n = k$ ($k\in \mathbb{N}$ and $k>4$). We shall prove that the statement is also true for $n = k +1$. Let $\Phi_1$, $\Phi_2$, $\ldots$ , $\Phi_{k-1}$, $\Phi_k$ and $\Phi_{k+1}$ be such figures that every triplet of these figures has at least one common point. Let $\Phi'=\Phi_k\cap\Phi_{k+1}$. We shall first prove that every triplet of figures $\Phi_1$, $\Phi_2$ ,$\ldots$, $\Phi_{k-1}$, $\Phi'$ has a common point. For triplets of figures from $\Phi_1$, $\Phi_2$, $\ldots$, $\Phi_{k-1}$ this is already fulfilled according to the assumption. Without loss of generality, it is enough to prove that figures $\Phi_1$, $\Phi_2$ and $\Phi'=\Phi_k\cap\Phi_{k+1}$ have a common point. This is true (based on the proven example for $n = 4$), because every triplet of figures from $\Phi_1$, $\Phi_2$, $\Phi_k$ and $\Phi_{k+1}$ has a common point. From the induction assumption (for $n=k$) it follows that figures $\Phi_1$, $\Phi_2$, $\ldots$, $\Phi_{k-1}$, $\Phi'$ have a common point, which at the same time lies in each of figures $\Phi_1$, $\Phi_2$, $\ldots$, $\Phi_{k-1}$, $\Phi_k$, $\Phi_{k+1}$. \kdokaz In the continuation we shall consider some consequences of Helly's theorem. \bzgled Let $\alpha_1$, $\alpha_2$, $\cdots$, $\alpha_n$ ($n > 3$) be half-planes covering a plane $\alpha$. Prove that there are three of these half-planes that also cover the plane $\alpha$. \ezgled \textbf{\textit{Proof.}} Let $\beta_1$, $\beta_2$, $\cdots$, $\beta_n$ be open half-planes, which are determined by the half-planes $\alpha_1$, $\alpha_2$, $\cdots$, $\alpha_n$ as complementary half-planes with respect to the plane $\alpha$ or $\beta_k=\alpha\setminus\alpha_k$, $k\in\{1,2,\ldots,n\}$. For every point $X$ of the plane $\alpha$ and every $k\in\{1,2,\ldots,n\}$ the equivalence holds: $$X\in \alpha_k \Leftrightarrow X\notin \beta_k.$$ Assume the contrary, that none of the three half-planes $\alpha_1$, $\alpha_2$, $\cdots$, $\alpha_n$ covers the plane $\alpha$. This means that for every triple of them there is a point in the plane $\alpha$, which does not lie on any of them, or for every triple of the half-planes $\beta_1$, $\beta_2$, $\cdots$, $\beta_n$ there is a point of the plane $\alpha$, which lies on each of them. Because the half-planes are convex shapes, from Helly's theorem \ref{Helly} it follows that there is a point $X$, which lies on each of the half-planes $\beta_1$, $\beta_2$, $\cdots$, $\beta_n$. This point therefore lies in the plane $\alpha$, but does not lie in any of the half-planes $\alpha_1$, $\alpha_2$, $\cdots$, $\alpha_n$, which is in contradiction with the basic assumption that the half-planes $\alpha_1$, $\alpha_2$, $\cdots$, $\alpha_n$ cover the plane $\alpha$. Therefore, there is at least one triple of the half-planes $\alpha_1$, $\alpha_2$, $\cdots$, $\alpha_n$, which covers the plane $\alpha$. \kdokaz \bzgled \label{lemaJung} If for every three of $n$ ($n > 3$) points of a plane there is such a circle with radius $r$ containing these three points, then it exists a circle of equal radius containing all these $n$ points. \ezgled \begin{figure}[!htb] \centering \input{sl.skl.3.11.2.pic} \caption{} \label{sl.skl.3.11.2.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skl.3.11.2.pic}) Let $A_1$, $A_2$,$\ldots$, $A_n$ be points with given properties. With $\mathcal{K}_i$ ($i\in\{1,2,\ldots,n\}$) we denote circles with centers $A_i$ and with radius $r$. Let $A_p$, $A_q$ and $A_l$ be any points from the set $\{A_1, A_2,\ldots, A_n\}$. By assumption, there exists a circle with radius $r$, which contains these three points. We denote the center of this circle with $O$. From this it follows that $|OA_p|, |OA_q|, |OA_l|\leq r$, which means that point $O$ lies in each of the circles $\mathcal{K}_p$, $\mathcal{K}_q$ and $\mathcal{K}_l$. Therefore, each three of the circles $\mathcal{K}_1$, $\mathcal{K}_2$,$\ldots$, $\mathcal{K}_n$ have at least one common point. Because the circles are convex figures (statement \ref{KrogKonv}), by Helly's theorem there exists a point $S$, which lies in each of the circles $\mathcal{K}_1$, $\mathcal{K}_2$,$\ldots$, $\mathcal{K}_n$. From this it follows that $\mathcal{K}(S, r)$ is the desired circle, since $|SA_1|, |SA_2|,\ldots |SA_n|\leq r$. \kdokaz An interesting consequence of the last statement \ref{lemaJung} will be given in section \ref{odd7Pitagora} (statement \ref{Jung}). %________________________________________________________________________________ \naloge{Exercises} \begin{enumerate} \item Let $S$ be a point, which lies in the angle $pOq$, and to point $A$ and $B$ the orthogonal projection of point $S$ on the sides $p$ and $q$ of this angle. Prove that $SA\cong SB$ if and only when the line $OS$ is the angle bisector of the angle $pOq$. \item Prove that the sum of the diagonals of a convex quadrilateral is greater than the sum of its opposite sides. \item Prove that in every triangle there is at most one side shorter than the corresponding altitude. \item Let $AA_1$ be the altitude of triangle $ABC$. Prove that of the two angles, which altitude $AA_1$ determines with sides $AB$ and $AC$, the larger one is the one, which altitude $AA_1$ determines with the shorter side. \item Let $BB_1$ and $CC_1$ be the altitudes of triangle $ABC$ and $AB \frac{3}{4}(a + b + c)$. \end{enumerate} \item Let $p$ be a line that is parallel to a circle $k$. Prove that all points of this circle are on the same side of the line $p$. \item If the circle $k$ lies in a convex figure $\Phi$, then the circle determined by this circle also lies in this figure. Prove it. \item Let $p$ and $q$ be two different tangents to the circle $k$ that touch it in points $P$ and $Q$. Prove the equivalence: $p \parallel q$ exactly when $AB$ is the diameter of the circle $k$. \item If $AB$ is the chord of the circle $k$, then the intersection of the line $AB$ and the circle determined by the circle $k$ is equal to this chord. Prove it. \item Let $S'$ be the orthogonal projection of the center $S$ of the circle $k$ onto the line $p$. Prove that $S'$ is an external point of this circle exactly when the line $p$ does not intersect the circle. \item Let $V$ be the altitude of the triangle $ABC$, for which $CV \cong AB$. Determine the size of the angle $ACB$. \item Let $CC'$ be the altitude of the right triangle $ABC$ ($\angle ACB = 90^0$). If $O$ and $S$ are the centers of the inscribed circles of the triangles $ACC'$ and $BCC'$, then the altitude of the internal angle $ACB$ is perpendicular to the line $OS$. Prove it. \item Let $ABC$ be a triangle in which $\angle ABC = 15^0$ and $\angle ACB = 30^0$. Let $D$ be such a point of the side $BC$ that $\angle BAD = 90^0$. Prove that $BD = 2AC$. \item Prove that there exists a pentagon that can be covered with such pentagons that are congruent to it. \item Prove that there exists a decagon that can be covered with such decagons that are congruent to it. \item In a plane, each point is painted red or black. Prove that there exists a right triangle that has all its vertices the same color. \item Let $l_1,l_2,\ldots, l_n$ ($n > 3$) be arcs, which all lie on the same circle. The central angle of each arc is at most $180^0$. Prove that there exists a point, which lies on each arc, if every three arcs have at least one common point. %drugi del \item Let $p$ and $q$ be rectangles, which intersect in the point $A$. If $B, B'\in p$, $C, C'\in q$, $AB\cong AC'$, $AB'\cong AC$, $\mathcal{B}(B,A,B')$ and $\mathcal{B}(C,A,C')$, then the rectangle on the line $BC$ through the point $A$ goes through the center of the line $B'C'$. Prove. \item Prove that the altitudes of an inner angle of a rectangle, which is not a square, intersect in points, which are the vertices of a square. \item Prove that the altitudes of an inner angle of a parallelogram, which is not a rhombus, intersect in points, which are the vertices of a rectangle. Prove also that the diagonals of this rectangle are parallel to the sides of the parallelogram and are equal to the difference of the adjacent sides of this parallelogram. \item Prove that the altitudes of two sides of a triangle are perpendicular to each other. \item Let $B'$ and $C'$ be the points of intersection of the altitudes from the vertices $B$ and $C$ of the triangle $ABC$. Prove the equivalence $AB\cong AC \Leftrightarrow BB'\cong CC'$. \item Prove that a triangle is right, if the center of the circle drawn through the triangle and its altitude point coincide. Is a similar statement true for any two characteristic points of this triangle? \item Prove that a right triangle $ABC$ and a right triangle $A'B'C'$ are congruent exactly when they have congruent altitudes $CD$ and $C'D'$, sides $AB$ and $A'B'$ and angle $ACD$ and $A'C'D'$. \item If $ABCD$ is a rectangle and $AQB$ and $APD$ are right triangles with the same orientation, then the line $PQ$ is congruent with the diagonal of this rectangle. Prove. \item Let $BB'$ and $CC'$ be the altitudes of the triangle $ABC$ ($AC>AB$) and $D$ is such a point on the line segment $AB$, that $AD\cong AC$. The point $E$ is the intersection of the line $BB'$ with the line, which goes through the point $D$ and is parallel to the line $AC$. Prove that $BE=CC'-BB'$. \item Let $ABCD$ be a convex quadrilateral, for which it holds that $AB\cong BC\cong CD$ and $AC\perp BD$. Prove that $ABCD$ is a rhombus. \item Let $BC$ be the base of an isosceles triangle $ABC$. If $K$ and $L$ are such points, that $\mathcal{B}(A,K,B)$, $\mathcal{B}(A,C,L)$ and $KB\cong LC$, then the center of the line $KL$ lies on the base $BC$. Prove. \item Let $S$ be the center of the triangle $ABC$ of an inscribed circle. A line, which goes through the point $S$ and is parallel to the side $BC$ of this triangle, intersects the sides $AB$ and $AC$ in succession in the points $M$ and $N$. Prove that $BM+NC=NM$. \item Let $ABCDEFG$ be a convex heptagon. Calculate the sum of the convex angles, which are determined by the broken line $ACEGBDFA$. \item Prove that the centers of the sides and the vertex of an altitude of an arbitrary triangle, in which no two sides are congruent, are the vertices of an isosceles trapezoid. \item Let $ABC$ be a right triangle with a right angle at the vertex $C$. The points $E$ and $F$ shall be the intersections of the internal angle bisectors at the vertices $A$ and $B$ with the opposite sides, $K$ and $L$ shall be the orthogonal projections of the points $E$ and $F$ on the hypotenuse of this triangle. Prove that $\angle LCK=45^0$. \item Let $M$ be the center of the side $CD$ of a square $ABCD$ and $P$ such a point on the diagonal $AC$, for which it holds that $3AP=PC$. Prove that $\angle BPM$ is a right angle. \item Let $P$, $Q$ and $R$ be the centers of the sides $AB$, $BC$ and $CD$ of a parallelogram $ABCD$. The lines $DP$ and $BR$ shall intersect the line $AQ$ in the points $K$ and $L$. Prove that $KL= \frac{2}{5} AQ$. \item Let $D$ be the center of the hypotenuse $AB$ of a right triangle $ABC$ ($AC>BC$). The points $E$ and $F$ shall be the intersections of the angle trisectors of the sides $CA$ and $CB$ with a line, which goes through $D$ and is orthogonal to the line $CD$. The point $M$ shall be the center of the line $EF$. Prove that $CM\perp AB$. \item Let $A_1$ and $C_1$ be the centers of sides $BC$ and $AB$ of triangle $ABC$. The internal angle bisector at point $A$ intersects the line $A_1C_1$ at point $P$. Prove that $\angle APB$ is a right angle. \item Let $P$ and $Q$ be points on sides $BC$ and $CD$ of square $ABCD$, such that the line $PA$ is the angle bisector of angle $BPQ$. Determine the size of the angle $PAQ$. \item Prove that the center of the circumscribed circle lies closest to the longest side of the triangle. \item Prove that the center of the inscribed circle is closest to the vertex of the largest internal angle of the triangle. \item Let $ABCD$ be a convex quadrilateral. Find a point $P$, such that the sum $AP+BP+CP+DP$ is minimal. \item The diagonals $AC$ and $BD$ of the trapezoid $ABCD$ with the base $AB$ intersect at point $O$ and $\angle AOB=60^0$. Points $P$, $Q$ and $R$ are in order the centers of the lines $OA$, $OD$ and $BC$. Prove that $PQR$ is a right triangle. \item Let $P$ be an arbitrary internal point of triangle $ABC$, for which $\angle PBA\cong \angle PCA$. Points $M$ and $L$ are the orthogonal projections of point $P$ on sides $AB$ and $AC$, point $N$ is the center of side $BC$. Prove that $NM\cong NL$\footnote{Predlog za MMO 1982 (SL 9.)).}. \item Let $AD$ be the internal angle bisector at point $A$ ($D\in BC$) of triangle $ABC$ and $E$ a point on side $AB$, such that $\angle BDE\cong\angle BAC$. Prove that $DE\cong DC$. \item Let $O$ be the center of square $ABCD$ and $P$, $Q$ and $R$ points, which divide its perimeter into three equal parts. Prove that the minimum of the sum $|OP|+|OQ|+|OR|$ is achieved, when one of these points is the center of a side of the square. \item There is a given finite number of lines that divide the plane into areas. Prove that the plane can be colored with two colors, so that each area is colored with one color, and adjacent areas are always colored with different colors. \item Draw a triangle $ABC$, if the following data are given (see labels in section \ref{odd3Stirik}): (\textit{a}) $\alpha$, $\beta$, $s$; \hspace*{2mm} (\textit{b}) $a-b$, $c$, $\gamma$; \hspace*{2mm} (\textit{c}) $a$, $\beta-\gamma$, $b-c$; \hspace*{2mm} (\textit{d}) $a$, $\beta-\gamma$, $b+c$; \hspace*{2mm} (\textit{e}) $b$, $c$, $v_a$; \hspace*{2mm} (\textit{f}) $b$, $v_a$, $v_b$; \hspace*{2mm} (\textit{g}) $\alpha$, $v_a$, $v_b$; \hspace*{2mm} (\textit{h}) $c$, $a+b$, $\gamma$; (\textit{i}) $v_a$, $\alpha$, $\beta$; \hspace*{2mm} (\textit{j}) $b$, $a+c$, $v_c$; \hspace*{2mm} (\textit{k}) $b-c$, $v_b$, $\alpha$; \hspace*{2mm} (\textit{l}) $a$, $t_b$, $t_c$;\hspace*{2mm} (\textit{m}) $b$, $c$, $t_a$; \hspace*{2mm} (\textit{n}) $t_a$, $t_b$, $t_c$; \hspace*{2mm} (\textit{o}) $c$, $v_a$, $l_a$; \hspace*{2mm} (\textit{p}) $c$, $v_a$, $t_b$; (\textit{r}) $b$, $l_a$, $\alpha$; \hspace*{2mm} (\textit{s}) $v_a$, $v_b$, $t_a$; \hspace*{2mm} (\textit{t}) $t_a$, $v_b$, $b+c$; \hspace*{2mm} (\textit{u}) $a$, $b$, $\alpha-\beta$; \item Draw an isosceles triangle $ABC$, if the following are given: \begin{enumerate} \item the base and the sum of the leg and the height on the base, \item the circumference and the height on the base, \item both heights, \item the angle at the base and the segment of its altitude, \item the leg and the point of intersection of the corresponding altitude on it, \item the leg and the corresponding altitude. \end{enumerate} \item Draw a right triangle $ABC$ with a right angle at point $C$, if the following data are given: (\textit{a}) $\alpha$, $a+b$, \hspace*{2mm} (\textit{b}) $\alpha$, $a-b$, \hspace*{2mm} (\textit{c}) $a$, $b+c$, (\textit{d}) $c$, $a+b$,\hspace*{2mm} (\textit{e}) $t_a$, $t_c$, \hspace*{2mm} (\textit{f}) $a$, $c-b$, (\textit{g}) $a+v_c$, $\alpha$, \hspace*{2mm} (\textit{h}) $t_c$, $v_c$,\hspace*{2mm} (\textit{i}) $a$, $t_a$,\hspace*{2mm} (\textit{j}) $v_c$, $l_c$. \item Draw a rectangle $ABCD$, if given: \begin{enumerate} \item the diagonal and one side, \item the diagonal and the perimeter, \item one side and the angle between the diagonals, \item the perimeter and the angle between the diagonals. \end{enumerate} \item Draw a rhombus $ABCD$, if given: \begin{enumerate} \item a side and the sum of the diagonals, \item a side and the difference of the diagonals, \item one angle and the sum of the diagonals, \item one angle and the difference of the diagonals. \end{enumerate} \item Draw a parallelogram $ABCD$, if given: \begin{enumerate} \item one side and the diagonals, \item one side and the altitude, \item one diagonal and the altitude, \item side $AB$, the angle at point $A$ and the sum $BC+AC$. \end{enumerate} \item Draw a trapezoid $ABCD$, if given: \begin{enumerate} \item the bases, the leg and the smaller angle that is not adjacent to this leg, \item the bases and the diagonals, \item the bases and the angle at the longer base, \item the sum of the bases, the altitude and the angle at the longer base. \end{enumerate} \item Draw a deltoid $ABCD$, if given: the diagonal $AC$, which lies on the slanted side of the deltoid, $\angle CAD$ and the sum $AD+DC$." \item Draw a quadrilateral $ABCD$, if it is given: \begin{enumerate} \item four sides and one angle, \item four sides and an angle between the opposite sides, \item three sides and an angle at the fourth side, \item the centers of three sides and a distance that is consistent and parallel to the fourth side. \end{enumerate} \end{enumerate} %%% Do tu pregledala tudi Ana. % DEL 4 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - %________________________________________________________________________________ % SKLADNOST TRIKOTNIKOV IN KROŽNICA %________________________________________________________________________________ \del{Congruence and Circle} \label{pogSKK} We have already looked at some of the properties of circles in the previous two chapters - certain properties of the radius, diameter, cord, the relationship between the circle and the line, and the properties of the tangent to the circle. We saw that for every triangle there is an inscribed and circumscribed circle. We proved that for regular polygons and some quadrilaterals (rectangle, square) there is a circumscribed circle, and for some there is also an inscribed circle. In this chapter we will look further into the properties of the circle that are a result of the congruence of triangles. %________________________________________________________________________________ \poglavje{Two Circles} \label{odd4DveKroz} We will carry out a similar analysis of the position of the circles and lines that we did in section \ref{odd3KrozPrem}, but this time with two circles in the same plane. We will assume that the two circles we are dealing with are in the same plane from now on. First, let's define some terms that relate to two circles. The line that goes through the centers of two circles is the \index{centrala dveh krožnic}\pojem{centrala} of those two circles. The distance between the centers of two circles is called the \index{središčna razdalja dveh krožnic} \pojem{središčna razdalja dveh krožnic} (Figure \ref{sl.skk.4.1.1.pic}). \begin{figure}[!htb] \centering \input{sl.skk.4.1.1.pic} \caption{} \label{sl.skk.4.1.1.pic} \end{figure} Circles in the same plane with the same center (their central distance is equal to $0$) we call \index{circles!concentric} \pojem{concentric circles} (Figure \ref{sl.skk.4.1.1.pic}). If concentric circles have at least one common point, the circles are identical (coincide). If $X$ is a common point of concentric circles $k_1(S,r_1)$ and $k_2(S,r_2)$, it holds $|SX|=r_1=r_2$ or $r_1=r_2$, which means that the circles are identical. Concentric circles are either identical or have no common points. Similarly to the circle and the line, here we raise the question of how many common points different circles can have and what is their mutual position. We will begin with the following statement. \bizrek Two different circles lying in the same plane have at most two common points. \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.1.2.pic} \caption{} \label{sl.skk.4.1.2.pic} \end{figure} \textbf{\textit{Proof.}} We assume the opposite. Let $A$, $B$ and $C$ be three different common points of two circles $k(O,r_1)$ and $l(S,r_2)$ (Figure \ref{sl.skk.4.1.2.pic}). These three points are not collinear, which would mean that the line $AB$ intersects the circle (e.g. $k$) in three different points, which according to izrek \ref{KroznPremPresek} is not possible. But if $A$, $B$ and $C$ are non-collinear points, they determine the triangle $ABC$, which according to izrek \ref{SredOcrtaneKrozn} means that $O=S$, or both points are located in the intersection of the perpendiculars of this triangle. The radii are also equal, because $r_1=|OA|=|SA|=r_2$, so the circles $k(O,r_1)$ and $l(S,r_2)$ are identical - both represent the circle circumscribed around the triangle $ABC$. \kdokaz So, two different circles in the same plane can have two common points, one common point, or no common points. In the first case, we say that the \index{circles!intersect} \pojem{circles intersect}, in the second case the \index{circles!touch} \pojem{circles touch} in their \index{touching point!of two circles} \pojem{touching point}, and in the third case they are \index{circles!are non-intersecting}\pojem{non-intersecting circles} (Figure \ref{sl.skk.4.1.3.pic}). \begin{figure}[!htb] \centering \input{sl.skk.4.1.3.pic} \caption{} \label{sl.skk.4.1.3.pic} \end{figure} If the circles do not intersect, then the interior of at least one of these two circles is either in the interior or in the exterior of the other circle. This is a consequence of Theorem \ref{DedPoslKrozKroz}. The line that is determined by the intersection of two circles that intersect is called the \pojem{secant, which is the tangent of their common chord}. In connection with this, we prove the following theorem. \bizrek \label{KroznPresABpravokOS} If two circles intersect at two points $A$ and $B$, then the line containing the centres of the two circles is perpendicular to the line $AB$. \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.1.4.pic} \caption{} \label{sl.skk.4.1.4.pic} \end{figure} \textbf{\textit{Proof.}} Let $A$ and $B$ be the intersection points of the circles $k_1(S_1,r_1)$ and $k_2(S_2,r_2)$ (Figure \ref{sl.skk.4.1.4.pic}). Because $S_1A\cong S_1B\cong r_1$ and $S_2A\cong S_2B\cong r_2$, the line $S_1S_2$ is the perpendicular bisector of the line segment $AB$ (Theorem \ref{simetrala}), so $S_1S_2\perp AB$. \kdokaz We will prove that the circles that touch have a common tangent at their touching point. \bizrek \label{tangSkupnaDotikKrozn} Let $k_1$ and $k_2$ be different circles with centres $S_1$ and $S_2$ touching at a point $T$. Then: (i) $S_1$, $S_2$ and $T$ are collinear points; (ii) the tangent of the circle $k_1$ at the point $T$ is at the same time the tangent of the circle $k_2$ at the same point. \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.1.5.pic} \caption{} \label{sl.skk.4.1.5.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.1.5.pic}). (\textit{i}) We assume that the points $S_1$, $S_2$ and $T$ are not collinear. The line $S_1S_2$ divides the plane in which the circles lie, into two half-planes. The half-plane that contains the point $T$, we denote by $\pi_1$, the other by $\pi_2$. From the statement \ref{izomEnaC'} it follows that in the half-plane $\pi_2$ there exists (one and only one) point $T'$, for which $S_1T'\cong S_1T$ and $S_2T'\cong ST_2$. This would mean that even the point $T'$, which is different from the point $T$, lies on the circles $k_1$ and $k_2$, which is not possible. Therefore, the points $S_1$, $S_2$ and $T$ are collinear. (\textit{ii}) By the statement \ref{TangPogoj} the tangent of the circle $k_1$ at the point $T$ is perpendicular to the radius $S_1T$. Similarly the tangent of the circle $k_2$ at the same point $T$ is perpendicular to the radius $S_2T$. Because of (\textit{i}) the lines $S_1T$ and $S_2T$ coincide, so do both perpendiculars or tangents. \kdokaz By the statement \ref{tangKrozEnaStr} all points of the circle are on the same side of each of its tangents - on the side where its center is. This means that the circles $k_1$ and $k_2$ from the previous statement are either on the same side or on different sides of their common tangent. When $B(S_1,T,S_2)$, the circles are on different sides of their common tangent and we say that the circles $k_1$ and $k_2$ \pojem{touch each other from the outside}. Otherwise the circles are on the same side of this tangent and we say that they \pojem{touch each other from the inside}. In the first case, the interior of one of these two circles is in the exterior of the other, in the second case, however, it is in the interior of the other circle (Figure \ref{sl.skk.4.1.6.pic}). \begin{figure}[!htb] \centering \input{sl.skk.4.1.6.pic} \caption{} \label{sl.skk.4.1.6.pic} \end{figure} When the circles $k_1(S_1,r_1)$ and $k_2(S_2,r_2)$ touch each other from the outside, it follows from the previous statement that $|S_1S_2| = r_1 + r_2$. If the circles touch each other from the inside, then $|S_1S_2| = |r_1 - r_2|$. It is clear that the converse is also true. The condition $|S_1S_2| = r_1 + r_2$ or $|S_1S_2| = |r_1 - r_2|$ is sufficient for the circles to touch each other from the outside or from the inside. In a similar way, we obtain the other criteria for the mutual position of two circles. \begin{figure}[!htb] \centering \input{sl.skk.4.1.7.pic} \caption{} \label{sl.skk.4.1.7.pic} \end{figure} \bizrek Let $k_1(S_1,r_1)$ and $k_2(S_2,r_2)$ ($r_1\geq r_2$) be two circles. Then (Figure \ref{sl.skk.4.1.7.pic}): (i) the circles $k_1(S_1,r_1)$ and $k_2(S_2,r_2)$ are lying outside each other if and only if $|S_1S_2|>r_1+r_2$; (ii) the circles $k_1(S_1,r_1)$ and $k_2(S_2,r_2)$ are touching each other externally if and only if $|S_1S_2|=r_1+r_2$; (iii) the circles $k_1(S_1,r_1)$ and $k_2(S_2,r_2)$ are intersecting each other at two points if and only if $r_1-r_2<|S_1S_2|\omega$. If $\mathcal{B}(A,N,M)$ holds, we can, by using a similar reasoning, conclude that in this case $\angle AMB<\omega$, which means that for no point $M\notin l \cup l'\setminus\{A,B\}$ it holds that $\angle AMB \cong \omega$. \kdokaz From the proof of the previous theorem we also get the following conclusion. \bizrek \label{obodKotGMTZunNotr} Let $AB$ be an arc of a circle $k$, $\omega$ the corresponding circumferential angle of this arc and $M$ a point of the half-plane with the edge $AB$ not containing this arc (Figure \ref{sl.skk.4.2.8.pic}). Then: (i) $\angle AMB >\omega$ if and only if the point $M$ is an interior point of the circle $k$; (ii) $\angle AMB \cong\omega$ if and only if the point $M$ lies on the circle $k$; (iii) $\angle AMB <\omega$ if and only if the point $M$ is an exterior point of the circle $k$. \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.2.8.pic} \caption{} \label{sl.skk.4.2.8.pic} \end{figure} In the following examples we will see the use of the theorem about the peripheral and central angle and its consequences. \bzgled Construct a triangle with given $a$, $\alpha$, $v_a$. \label{konstr_aalphava} \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.3.1d.pic} \caption{} \label{sl.skk.4.3.1d.pic} \end{figure} \textbf{\textit{Analysis.}} The point $A$ lies at the same time at the geometric location of the points from which the line $BC$ is seen at an angle $\alpha$ (the union of two circular arcs - Theorem \ref{ObodKotGMT}) and at the parallel of the line $BC$, which is $v_a$ away from it (Figure \ref{sl.skk.4.3.1d.pic}). So the vertex $A$ is the intersection of this parallel and the aforementioned geometric location of points. \textbf{\textit{Construction.}} First, let's draw the line $BC\cong a$, then the geometric location of points $\mathcal{L}$, from which this line is seen at an angle $\alpha$ (Theorem \ref{ObodKotGMT}). Then let's draw the parallel $p$ of the line $BC$ at a distance $v_a$. With $A$ we mark the intersection of the line $p$ and the aforementioned geometric location of points $\mathcal{L}$. We will prove that the triangle $ABC$ is the desired triangle. \textbf{\textit{Proof.}} By construction, it is clear that $BC\cong a$. By construction, the point $A$ lies on the geometric location of points from which the line $BC$ is seen at an angle $\alpha$, so $BAC\cong\alpha$. The altitude of the triangle $ABC$ from the vertex $A$ is consistent with the line $v_a$, because the point $A$ by construction lies on the line $p$, which is $v_a$ away from the line $BC$. \textbf{\textit{Discussion.}} The necessary condition is of course $\alpha<180^0$. The number of solutions to the task is equal to the number of intersections of the line $p$ and the set $\mathcal{L}$. \kdokaz \bzgled Let $p$, $q$ and $r$ be lines in the plane intersecting at one point and divide this plane into six congruent angles. Suppose that $P$, $Q$ and $R$ are the foots of the perpendiculars from an arbitrary point $X$ of this plane on the lines $p$, $q$ and $r$, respectively. Prove that $PQR$ is a regular triangle. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.2.9.pic} \caption{} \label{sl.skk.4.2.9.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.2.9.pic}). Let $S$ be the intersection of the lines $p$, $q$ and $r$. It is clear that the lines determine the angles $60^0$. Because $\angle XPS \cong \angle XQS \cong \angle XRS = 90^0$, by \ref{TalesovIzrKroz2} the points $S$, $X$, $P$, $Q$ and $R$ lie on the circle $k$ with diameter $SX$. If we use \ref{ObodObodKot} for the appropriate arcs $PQ$ and $QR$, we have $\angle PRQ \cong \angle PSQ = 60^0$ and $\angle QPR \cong \angle QSR = 60^0$. Because all angles are equal to $60^0$, $PQR$ is a regular triangle. \kdokaz Let $k(O,R)$ and $l(S,r)$ ($R = 2r$) be circles touching each other internally and $P$ an arbitrary point on the circle $l$. Which curve is described by the point $P$ if the circle $l$ rolls without slipping around the circle $k$\footnote{This problem was solved by the Polish astronomer \index{Copernicus, N.} \textit{N. Copernicus} (1473--1543). In the general case, when it is not necessarily $R = 2r$, the curve which is described by the point $P$, is called \index{hipocikloida} \pojem{hipocikloida}. In the case of the outer rolling of the circle around the other circle, the curve is called \index{epicikloida} \pojem{epicikloida}, in the case of the rolling of the circle around the straight line, it is called \index{cikloida} \pojem{cikloida}. The cikloida was first investigated by the German mathematician and philosopher \index{Kuzanski, N.} \textit{N. Kuzanski} (1401--1464) and later by the French mathematician and philosopher \index{Mersenne, M.} \textit{M. Mersenne} (1588--1648). It was named by the Italian physicist, mathematician, astronomer and philosopher \index{Galilei, G.} \textit{G. Galilei} (1564--1642) in 1599.}? \begin{figure}[!htb] \centering \input{sl.skk.4.2.10.pic} \caption{} \label{sl.skk.4.2.10.pic} \end{figure} \textbf{\textit{Solution.}} Let $P_0$ be the position of the point $P$ at the moment when it lies on the circle $k$ (Figure \ref{sl.skk.4.2.10.pic}). At the moment when the point $P$ is in position $P_i$, the circle $l$, which is in position $l_i$, touches the circle $k$ at some point $T_i$. Since this is a "movement without slipping", the lengths of the corresponding arcs $P_0T_i$ and $P_iT_i$ of the circles $k$ and $l$ are equal to each other. The radius of the circle $k$ is twice as large as the radius of the circle $l$, so for the corresponding central angles of the aforementioned arcs we have $\angle T_iS_iP_i= 2\angle T_iOP_0$. But $\angle T_iOP_i$ is the corresponding arc angle of the circle $l$ for the same arc $P_iT_i$, so $2\angle T_iOP_i = \angle T_iS_iP_i$ (statement \ref{ObodObodKot}). From the two previous relations we obtain $\angle T_iOP_i= \angle T_iOP_0$, which means that the point $P_i$ is collinear with the points $O$ and $P_0$, so the desired path of the point $P$ represents the diameter $P_0P'_0$ of the circle $k$. \kdokaz \bzgled Let $C$ be the midpoint of an arc $AB$ and $D$ an arbitrary point of this arc other than $C$. Prove that $$AC + BC > AD + BD.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.2.11.pic} \caption{} \label{sl.skk.4.2.11.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.2.11.pic}). Let $C'$ and $D'$ be such points on the line segments $AC$ and $AD$, that: $CC'\cong CB \cong CA$, $DD'\cong DB$, $\mathcal{B}(A,C,C')$ and $\mathcal{B}(A,D,D')$. The triangles $C'CB$ and $D'DB$ are isosceles, therefore by \ref{enakokraki} and \ref{zunanjiNotrNotr}, it follows: $$\angle CC'B \cong \angle CBC' = \frac{1}{2}\angle ACB \hspace*{2mm}\textit{ and }\hspace*{2mm} \angle DD'B \cong \angle DBD' = \frac{1}{2}\angle ADB.$$ Because the angles $ACB$ and $ADB$ are supplementary (by \ref{ObodObodKot}). The angles $AC'B$ and $AD'B$ are supplementary as well, and the points $C'$ and $D'$ lie on the corresponding arc $l'$ over the segment $AB$ (by \ref{ObodKotGMT}). Because $CC'\cong CB \cong CA$, the distance $AC'$ is the diameter of the circle that contains the arc $l'$, therefore the $\angle AD'C'$ is a right angle (by \ref{TalesovIzrKroz2}). The distance $AC'$ is therefore the hypotenuse of the right triangle $AD'C'$ and by \ref{vecstrveckot} it follows: $$AC + CB = AC'> AD'= AD + DB,$$ which was to be proven. \kdokaz \bnaloga\footnote{35. IMO Hong Kong - 1994, Problem 2.} $ABC$ is an isosceles triangle with $BC \cong AC$. Suppose that: (i) $D$ is the midpoint of $AB$ and $E$ is the point on the line $CD$ such that $EA\perp AC$; (ii) $F$ is an arbitrary point on the segment $AB$ different from $A$ and $B$; (iii) $G$ lies on the line $CA$ and $H$ lies on the line $CB$ such that $G$, $F$, $H$ are distinct and collinear. Prove that $EF$ is perpendicular to $HG$ if and only if $GF\cong FH$. \enaloga \begin{figure}[!htb] \centering \input{sl.skk.4.2.IMO1.pic} \caption{} \label{sl.skk.4.2.IMO1.pic} \end{figure} \textbf{\textit{Proof.}} First, from the similarity of the triangles $CAE$ and $CBE$ (by \textit{SAS} \ref{SKS}) it follows that $\angle EBC\cong\angle EAC=90^0$ and $EA\cong EB$ (Figure \ref{sl.skk.4.2.IMO1.pic}). We prove the equivalence $EF\perp HG \Leftrightarrow GF\cong FH$ in both directions. ($\Rightarrow$) Let's assume that the lines $EF$ and $HG$ are perpendicular, i.e. $\angle EFG\cong\angle EFH=90^0$. Let $k$ and $l$ be circles with diameters $EG$ and $EH$. Because $\angle EAG\cong\angle EFG=90^0$ and $\angle EBH\cong\angle EFH=90^0$, by \ref{TalesovIzrKroz} we have $A,F\in k$ and $B,F\in l$. First, from $EA\cong EB$ it follows that $\angle EAB\cong \angle EBA$. From this and \ref{ObodObodKot} it follows: $$\angle EGF\cong\angle EAF\cong\angle EBF\cong\angle EHF.$$ Therefore, the triangle $EGH$ is equilateral, so its height $EF$ is also the altitude (congruence of triangles $EFG$ and $EFH$, \textit{ASA} \ref{KSK}) or $GF\cong FH$. ($\Leftarrow$) Now let $GF\cong FH$, i.e. the point $F$ is the center of the line $GH$. Let $k$ be a circle with diameter $EG$. In addition to the point $A$ we mark with $\widehat{F}$ the other intersection of this circle with the line $AB$. If the circle $k$ touches the line $AC$, it follows that $G=A$ or $F=D$ and $H=B$, so in this case $GF\cong FH$ is already fulfilled. Assume that $\widehat{F}\neq F$. Let $\widehat{H}$ be the intersection of the lines $G\widehat{F}$ and $CB$. Because the point $\widehat{F}$ lies on the circle $k$ with diameter $EG$, $\angle G\widehat{F}E=90^0$, i.e. $E\widehat{F}\perp G\widehat{H}$. Therefore, for the points $G$, $\widehat{F}$ and $\widehat{H}$ the assumptions of the left side of the equivalence are fulfilled, so from the already proven first part of the statement ($\Rightarrow$) it follows that $G\widehat{F}\cong \widehat{F}\widehat{H}$, i.e. the point $\widehat{F}$ is the center of the line $G\widehat{H}$. In the triangle $GH\widehat{H}$ the $\widehat{F}F$ is the median, so $\widehat{F}F\parallel \widehat{H}H$ and $AB\parallel BC$, which is not possible. Thus, the assumption $\widehat{F}\neq F$ disappears, so $\widehat{F}= F$, so $\widehat{H}=H$. In the end, from $E\widehat{F}\perp G\widehat{H}$ it follows that $EF\perp GH$. \kdokaz %________________________________________________________________________________ \poglavje{More About Circumcircle and Incircle of a Triangle} \label{odd4OcrtVcrt} First, we will consider some important points that lie on the circumcircle of a triangle. \bizrek \label{TockaN} The bisector of the side $BC$ and the bisector of the interior angle $BAC$ of a triangle $ABC$ ($AB\neq AC$) intersect at the Circumcircle $l(O,R)$ of that triangle. \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.3.1.pic} \caption{} \label{sl.skk.4.3.1.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.3.1.pic}). Let the point $N$ be one of the intersections of the Circumcircle $l(O,R)$ of the triangle $ABC$ and the bisector of the side $BC$ (such that $A,N\perp BC$). Because the point $N$ lies on the bisector of the side $BC$, it follows that $NB \cong NC$. Therefore, the triangle $BNC$ is an isosceles triangle, so the angles $\angle NBC$ and $\angle NCB$ are congruent angles (by Theorem \ref{enakokraki}). Because the point $N$ also lies on the Circumcircle $l(O,R)$ of the triangle $ABC$, by Theorem \ref{ObodObodKot} it follows that: \begin{eqnarray*} \angle BAN \cong \angle BCN \textrm{ (obodna kota za krajši lok }BN \textrm{) }\\ \angle NAC \cong \angle NBC \textrm{ (obodna kota za krajši lok } CN \textrm{).} \end{eqnarray*} Therefore, $\angle BAN \cong \angle NAC$ or the line $AN$ is the bisector of the angle $BAC$, thus the theorem is proven. \kdokaz The point $N$ from the previous theorem is the center of that arc $BC$ of the Circumcircle $l(O,R)$ of the triangle $ABC$ which does not contain the vertex $A$. We shall now prove another important property of the point $N$. \bizrek \label{TockaN.NBNC} For the point $N$ from the previous theorem is $NB\cong NS\cong NC$, where $S$ is the incentre of the triangle $ABC$. \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.3.2.pic} \caption{} \label{sl.skk.4.3.2.pic} \end{figure} \textbf{\textit{Proof.}} Let's mark with $\alpha$ and $\beta$ the internal angles of the triangle $ABC$ at the vertices $A$ and $B$ (Figure \ref{sl.skk.4.3.2.pic}). $BNS$ is an isosceles triangle (statement \ref{enakokraki}), because the angles at the vertices $B$ and $S$ are equal. If we use the statement \ref{zunanjiNotrNotr} and \ref{ObodObodKot}, we get: \begin{eqnarray*} \angle BSN &=& \angle ABS + \angle BAS =\frac{1}{2}\alpha+\frac{1}{2}\beta,\\ \angle SBN &=& \angle SBC +\angle CBN = \angle SBC + \angle CAN = \frac{1}{2}\beta+\frac{1}{2}\alpha. \end{eqnarray*} Therefore $NB \cong NS$ and similarly $NC \cong NS$. \kdokaz \bizrek \label{TockaNbetagama} Let $AA'$ be the altitude from the vertices $A$ and $AE$ bisector of the interior angle $BAC$ of a triangle $ABC$ ($A',E\in BC$). Suppose that $l(O,R)$ is the circumcircle of that triangle. If $\angle CBA=\beta\geq\angle ACB=\gamma$, then $$\angle A'AE\cong \angle EAO=\frac{1}{2}\left( \beta-\gamma\right).$$ \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.3.1b.pic} \caption{} \label{sl.skk.4.3.1b.pic} \end{figure} \textbf{\textit{Proof.}} Let $N$ be the point defined as in the previous statements (Figure \ref{sl.skk.4.3.1b.pic}). The lines $AA'$ and $ON$ are parallel, because they are both perpendicular to the line $BC$. Because $OA\cong ON=R$, $AON$ is an isosceles triangle. Therefore, first of all: $$\angle A' AE \cong \angle ANO \cong \angle NAO = \angle EAO,$$ and then: $$\angle A'AE=\frac{1}{2}\alpha-\left(90^0-\beta \right)= \frac{1}{2}\alpha-\left(\frac{\alpha+\beta+\gamma}{2}-\beta \right)= \frac{1}{2}\left( \beta-\gamma\right),$$ which was to be proven. \kdokaz \bzgled \label{tockaNtockePQR} Let $P$, $Q$ and $R$ be the midpoints of those arcs $BC$, $AC$ and $AB$ of the circumcircle of a triangle $ABC$ not containing the vertices $A$, $B$ and $C$ of that triangle. If $E$ and $F$ are intersections of the line $QR$ with sides $AB$ and $AC$, respectively and $S$ the incentre of this triangle, then: (i) $AP \perp QR$, (ii) the quadrilateral $AESF$ is a rhombus. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.3.2a.pic} \caption{} \label{sl.skk.4.3.2a.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.3.2a.pic}) (\textit{i}) Let $L$ be the intersection of the lines $AP$ and $QR$. By \ref{TockaN} points $P$, $Q$ and $R$ lie on the simetrals $AS$, $BS$ and $CS$ of the internal angles of the triangle $ABC$ ($S$ is the center of the triangle $ABC$ of the inscribed circle). If we denote with $\alpha$, $\beta$ and $\gamma$ the internal angles of the triangle $ABC$, then due to the similarity of the corresponding external angles (\ref{ObodObodKot}) we get: \begin{eqnarray*} \angle RPL &=& \angle RPA = \angle RCA =\frac{1}{2}\gamma,\\ \angle PRL &=& \angle PRQ = \angle PRC + \angle CRQ = \angle PAC + \angle CBQ = \frac{1}{2}\alpha+ \frac{1}{2}\beta. \end{eqnarray*} Therefore, the sum of the angles in the triangle $PRL$ (\ref{VsotKotTrik}) $180^0 =\frac{1}{2}\alpha+ \frac{1}{2}\beta +\frac{1}{2}\gamma +\angle RLP = 90° + \angle RLP$. So $\angle RLP = 90°$ or $AP \perp QR$. (\textit{ii}) By the statement \ref{TockaN.NBNC} it is $RA \cong RS$, therefore from the similarity of right-angled triangles $ALR$ and $SLR$ (statement \textit{SSA} \ref{SSK}) it follows that the point $L$ is the center of the diagonal $AS$ of the quadrilateral $AESF$. From the similarity of triangles $AEL$ and $AFL$ (statement \textit{ASA} \ref{KSK}) it follows that the point $L$ is also the center of the diagonal $EF$, therefore the $AESF$ is a parallelogram (statement \ref{paralelogram}). Because the diagonals $AS$ and $EF$ are also perpendicular, the quadrilateral $AESF$ is a rhombus (statement \ref{RombPravKvadr}). \kdokaz From the previous statement we directly get a consequence. \bzgled \label{PedalniLemasPQR} Let $P$, $Q$ and $R$ be the midpoints of those arcs $BC$, $AC$ and $AB$ of the circumcircle of a triangle $ABC$ not containing the vertices $A$, $B$ and $C$ of that triangle. Prove that the incentre of the triangle $ABC$ is at the same time orthocentre of the triangle $PQR$ (Figure \ref{sl.skk.4.3.2b.pic}). \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.3.2b.pic} \caption{} \label{sl.skk.4.3.2b.pic} \end{figure} We prove some consequences of statement \ref{ObodObodKot}, which are related to the altitude point. \bizrek \label{TockaV'} Points that are symmetric to the orthocentre of an acute triangle with respect to its sides lie on the circumcircle of this triangle. \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.3.3.pic} \caption{} \label{sl.skk.4.3.3.pic} \end{figure} \textbf{\textit{Proof.}} Let $V$ be the orthocentre of a triangle $ABC$, $l$ the circumcircle of the triangle $ABC$ and $V_a$ the other intersection point of the circle $l$ with the altitude $AA'$ (Figure \ref{sl.skk.4.3.3.pic}). We prove that the point $V_a$ is symmetric to the point $V$ with respect to the line $BC$. It is enough to prove that $VA'\cong V_aA'$. The angles $V_aBC$ and $V_aAC$ are complementary (circumscribed angle for the chord $V_aC$ - izrek \ref{ObodObodKot}), the angles $V_aAC$ and $CBV$ are complementary angles with perpendicular legs (izrek \ref{KotaPravokKraki}). Therefore, the angles $V_aBC$ and $CBV$ are also complementary, and so are the triangles $V_aBA'$ and $VBA'$ or $VA'\cong V_aA'$. A similar statement holds for the other two altitudes. \kdokaz \bizrek \label{TockaV'a} Let $V$ be the orthocentre of a triangle $ABC$. The circumcircles of the triangles $VBC$, $AVB$ and $ACV$ are congruent to the circumcircle of the triangle $ABC$. \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.3.3a.pic} \caption{} \label{sl.skk.4.3.3a.pic} \end{figure} \textbf{\textit{Proof.}} A direct consequence of the previous izrek \ref{TockaV'}, because the three aforementioned circumcircles are symmetric to the circumcircle of the triangle $ABC$ with respect to the altitude of its sides (Figure \ref{sl.skk.4.3.3a.pic}). \bizrek \label{TockaV1} Points that are symmetric to the orthocentre of an acute triangle with respect to the midpoints of its sides lie on the circumcircle of this triangle. \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.3.3b.pic} \caption{} \label{sl.skk.4.3.3b.pic} \end{figure} \textbf{\textit{Proof.}} Let $V$ be the altitude point of the triangle $ABC$, $l$ the circumscribed circle (with the center $O$) of the triangle $ABC$ and $V_{A_1}$ the point that is symmetrical to the point $A$ with respect to the point $O$ (Figure \ref{sl.skk.4.3.3.pic}). From the very definition of the point $V_{A_1}$ it is clear that it lies on the circle $l$. We shall prove that $V_{A_1}$ is symmetrical to the point $V$ with respect to the point $A_1$, which is the center of the line $BC$. Because $AV_{A_1}$ is the diameter of the circle $l$, according to the Theorem \ref{TalesovIzrKroz2} $\angle ACV_{A_1}=90^0$ or $V_{A_1}C\perp AC$. The line $BV$ is the altitude of the triangle $ABC$, therefore $BV\perp AC$. From the last two relations it follows that $V_{A_1}C\parallel BV$. Similarly $V_{A_1}B\parallel CV$. Therefore the quadrilateral $V_{A_1}CVB$ is a parallelogram, thus its diagonals $VV_{A_1}$ and $BC$ have a common center. The center of the line $BC$ is the point $A_1$, which means that $V_{A_1}$ is symmetrical to the point $V$ with respect to the point $A_1$. The point $V_{A_1}$ but, according to the construction, lies on the circle $l$. \kdokaz We shall also prove some consequences of the Theorem \ref{ObodObodKot}, which are connected with the circumscribed circle of a triangle. \bzgled \label{zgledTrikABCocrkrozP} Let $k$ be the circumcircle of a regular triangle $ABC$. If $P$ is an arbitrary point lying on the shorter arc $BC$ of the circle $k$, then $$|PA|=|PB|+|PC|.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.3.4.pic} \caption{} \label{sl.skk.4.3.4.pic} \end{figure} \textbf{\textit{Proof.}} Because $\angle ACP>60^0>\angle PAC$, by the statement \ref{vecstrveckot} $AP>PC$ (Figure \ref{sl.skk.4.3.4.pic}). Therefore, on the line $AP$ there exists such a point $Q$, that $PQ\cong PC$. By the statement \ref{ObodObodKot} $\angle CPQ=\angle CPA\cong\angle CBA=60^0$, which means that $PCQ$ is an isosceles triangle, therefore also $CQ\cong CP$ and $\angle PCQ=60^0$. From this it follows that $\angle ACQ=\angle ACP-60^0=\angle BCP$. By the statement \textit{SAS} (statement \ref{SKS}) the triangles $ACQ$ and $BCP$ are similar, therefore also $AQ\cong BP$. In the end: $|PB|+|PC|=|AQ|+|PQ|=|AP|$. \kdokaz \bzgled Three circles of equal radii $r$ intersect at point $O$. Furthermore each two of them intersect at one more point: $A$, $B$, and $C$. Prove that the radius of the circumcircle of the triangle $ABC$ is also $r$. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.3.5.pic} \caption{} \label{sl.skk.4.3.5.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.3.5.pic}) We mark with $k$, $l$, $j$ and $o$ the circumcircles of the triangles $OBC$, $OAC$, $OAB$ and $ABC$ and with $P$ an arbitrary point of the circle $k$ so that the points $O$ and $P$ are on different sides of the line $BC$. By the assumption the circles $k$, $l$ and $j$ are similar. The angles $BAO$ and $BCO$ are also similar, because they are the angles between the similar circumcircles $k$ and $j$ over the chord $BO$ (statement \ref{SklTetSklObKot2}). Analogously the angles $CAO$ and $CBO$ are similar. Because of this: $$\angle BAC = \angle BAO + \angle CAO = \angle BCO + \angle CBO = 180° - \angle BOC = \angle BPC.$$ Therefore the circles $k$ and $o$ have a similar circumferential angle over the common chord $BC$, therefore they are similar. \kdokaz \bzgled \label{KvadratKonstr4tocke} Construct a square $ABCD$ such that given points $P$, $Q$, $R$ and $S$ lyes on the sides $AB$, $BC$, $CD$ and $DA$ of this square, respectively. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.3.1c.pic} \caption{} \label{sl.skk.4.3.1c.pic} \end{figure} \textbf{\textit{Solution.}} Since $\angle PAS$ and $\angle QCR$ are right angles, points $A$ and $B$ lie on the circles $k$ and $l$ with diameters $PS$ and $QR$ (Figure \ref{sl.skk.4.3.1c.pic}). The carrier of the diagonal $AC$ of the square $ABCD$ is also the symmetry of the internal angles $BAD$ and $BCD$, so it goes through the centers $N$ and $M$ of the corresponding arcs, which are determined by $k$ and $l$ (statement \ref{TockaN}). The construction can therefore be carried out by first planning the circles $k$ and $l$, then the line $NM$, the points $A$ and $C$ and finally the points $B$ and $D$. \kdokaz \bzgled Construct a triangle $v_a$, $t_a$, $l_a$. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.3.1e.pic} \caption{} \label{sl.skk.4.3.1e.pic} \end{figure} \textbf{\textit{Solution.}} Let $ABC$ be a triangle in which the altitude $AA'$, the median $AA_1$ and the section of the symmetry $AE$ of the internal angle $BAC$ are consistent with the distances $v_a$, $t_a$ and $l_a$. With $O$ we mark the center of the triangle $ABC$ of the drawn circle $k$. By statement \ref{TockaN} the lines $AE$ and $OA_1$ intersect in the point $N$, which lies on the circle $k$ (Figure \ref{sl.skk.4.3.1e.pic}). So we can first plan a right triangle $AA'E$ with the leg $v_a$ and the hypotenuse $l_a$ and the point $A_1$ from the condition $AA_1\cong t_a$. Then plan the point $N$ as the intersection of the line $AE$ and the perpendicular of the line $A'E$ through the point $A_1$. The center $O$ is the intersection of the line $A_1N$ and the symmetry of the distance $AN$ (because $AN$ is the chord of the circle $k$). The points $B$ and $C$ are the intersections of the circle $k(O,OA)$ with the line $A'E$. \kdokaz \bzgled Construct a triangle $R$, $r$, $a$. \label{konstr_Rra} \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.3.1a.pic} \caption{} \label{sl.skk.4.3.1a.pic} \end{figure} \textbf{\textit{Solution.}} Let $ABC$ be a triangle such that $BC\cong a$ and $l(O,R)$ and $k(S,r)$ are its circumscribed and inscribed circle (Figure \ref{sl.skk.4.3.1a.pic}). We denote with $\alpha$, $\beta$ and $\gamma$ its internal angles at vertices $A$, $B$ and $C$. By \ref{SredObodKot} we have $\alpha = \angle BAC = \frac{1}{2}\cdot\angle BOC$. From \ref{kotBSC} it follows that $\angle BSC=90^0+\frac{1}{2}\cdot\alpha$. From two relations we obtain the equality $\angle BSC=90^0+\frac{1}{4}\cdot\angle BOC$, which allows the construction. First we draw an isosceles triangle $BOC$ ($BC\cong a$ and $OB\cong OC\cong R$). We obtain point $S$ as one of the intersections of the arc with the chord $BC$ and the external angle $90^0+\frac{1}{4}\cdot\angle BOC$ and the line, which is at a distance $r$ parallel to the line $BC$. Then we draw the inscribed circle $k(S,r)$ and point $A$ as the intersection of the other two tangents of this circle from points $B$ and $C$. \kdokaz \bnaloga\footnote{47. IMO Slovenia - 2006, Problem 1.} Let $ABC$ be a triangle with incentre $I$. A point $P$ in the interior of the triangle satisfies $$\angle PBA + \angle PCA = \angle PBC + \angle PCB.$$ Show that $|AP| \geq |AI|$, and that equality holds if and only if $P = I$. \enaloga \begin{figure}[!htb] \centering \input{sl.skl.4.3.IMO1.pic} \caption{} \label{sl.skl.4.3.IMO1.pic} \end{figure} \textbf{\textit{Proof.}} We mark with $\alpha$, $\beta$ and $\gamma$ the internal angles of the triangle $ABC$ at the vertices $A$, $B$ and $C$ (Figure \ref{sl.skl.4.3.IMO1.pic}). The condition $\angle PBA + \angle PCA = \angle PBC + \angle PCB$ can be rewritten in the form $\beta-\angle PBC + \gamma-\angle PCB = \angle PBC + \angle PCB$ or: $$\angle PBC + \angle PCB=\frac{1}{2}\left( \beta+\gamma\right).$$ From this and the fact that the sum of the internal angles of each of the triangles $BPC$ and $ABC$ is equal to $180^0$ (\izrekref{VsotKotTrik}), it follows: $$\angle BPC =180^0-\frac{1}{2}\left( \beta+\gamma\right)=90^0+ \frac{1}{2} \alpha.$$ But from \kotBSC it follows $\angle BIC =90^0+ \frac{1}{2}\cdot \alpha$, so $\angle BPC\cong \angle BIC$. Therefore, the points $P$ and $I$ lie on the same curve $\mathcal{L}$ with the chord $BC$ and the central angle $90^0+ \frac{1}{2} \alpha$. Let the point $N$ be the intersection of the perpendicular bisector of the side $BC$ and the perpendicular bisector of the internal angle $BAC$ of the triangle $ABC$. By \izrekref{TockaN}, the point $N$ lies on the circumscribed circle of the triangle $ABC$ and $NB\cong NI\cong NC$ (\izrekref{TockaN.NBNC}). This means that $N$ is the center of the curve $\mathcal{L}$, so $NP\cong NI$ or $\angle NIP\cong\angle NPI<90^0$. Because the points $A$, $I$ and $N$ are collinear (they lie on the perpendicular bisector of the internal angle $BAC$), $\angle AIP =180^0-\angle NIP>90^0$. From \izrekref{vecstrveckot} (for the triangle $API$) it now follows that $|AP| \geq |AI|$ and the equality holds exactly when the triangle $API$ is not, i.e. when $P=I$. \kdokaz \bnaloga\footnote{43. IMO United Kingdom - 2002, Problem 2.} $BC$ is a diameter of a circle center $O$. $A$ is any point on the circle with $\angle AOC>60^0$. $EF$ is the chord which is the perpendicular bisector of $AO$. $D$ is the midpoint of the minor arc $AB$. The line through $O$ parallel to $AD$ meets $AC$ at $J$. Show that $J$ is the incenter of triangle $CEF$. \enaloga \begin{figure}[!htb] \centering \input{sl.skk.4.3.IMO4.pic} \caption{} \label{sl.skk.4.3.IMO4.pic} \end{figure} \textbf{\textit{Proof.}} Because the points $E$ and $F$ lie on the line of symmetry $EF$, and also on the circle with center $O$, we have $$AF\cong FO\cong AO\cong EO \cong EA.$$ This means that the quadrilateral $EOFA$ is a rhombus, which is made up of two congruent triangles $AOF$ and $AEO$. Without loss of generality, we assume that $\angle COE>\angle COF$ (Figure \ref{sl.skk.4.3.IMO4.pic}). First, from the condition $\angle AOC>60^0$ it follows that $\angle COF=60^0- \angle AOC>0^0$, so the points $A$ and $F$ are on the same side of the line $BC$. Because $AOC$ is an isosceles triangle with the base $AC$, by the \ref{enakokraki} and \ref{zunanjiNotrNotr} we have $\angle ACO =\frac{1}{2}\angle AOB$. The point $D$ is the center of the arc $BD$, so $\angle AOD\cong\angle DOB$ or $\angle DOB=\frac{1}{2}\angle AOB$. This means that $\angle ACO\cong\angle DOB$ and the lines $AC$ and $DO$ are parallel by \ref{KotiTransverzala}. Because $AD\parallel JO$ by assumption, the quadrilateral $ADOJ$ is a parallelogram, so $AJ\cong OD$. Therefore $$AJ\cong OD\cong OE\cong AF\cong AE.$$ From $AF\cong AE$ it follows that $AJ$ is the line of symmetry of the internal angle at the vertex $C$ of the triangle $CEF$ (\ref{SklTetSklObKot}). Because $AJ\cong AF\cong AE$, by \ref{TockaN.NBNC} the point $J$ is the center of the inscribed circle of this triangle. \kdokaz %________________________________________________________________________________ \poglavje{Cyclic Quadrilateral} \label{odd4Tetivni} We say that a \index{štirikotnik!tetiven}\index{večkotnik!tetiven}\pojem{tetiven}, if there exists a circumscribed circle, or if there is a circle that contains all of its vertices (Figure \ref{sl.skk.4.5.10.pic}). For vertices in this case we say that they are \index{konciklične točke}\pojem{konciklične točke}. We have already seen that every triangle is tetiven (izrek \ref{SredOcrtaneKrozn}) and also that every regular polygon is tetiven (izrek \ref{sredOcrtaneKrozVeck}). On the other hand, it is clear that not all polygons are cyclic. For example, a diamond is a quadrilateral that does not have a circumscribed circle. In this section we will therefore deal with cyclic quadrilaterals. \begin{figure}[!htb] \centering \input{sl.skk.4.5.10.pic} \caption{} \label{sl.skk.4.5.10.pic} \end{figure} Since a square is a regular polygon, it is also a cyclic quadrilateral. It is not difficult to prove that a rectangle is also a type of cyclic quadrilateral - the center of the circumscribed circle is the intersection of its diagonals, which are consistent and bisect each other. But how would we generally determine if a quadrilateral is cyclic? It is clear that in a cyclic quadrilateral (generally also in a polygon) the altitudes of all its sides intersect in one point (Figure \ref{sl.skk.4.5.10.pic}). This condition is sufficient for the quadrilateral to be cyclic, but it is not sufficiently operative in specific cases. For the cyclicity of quadrilaterals there is a necessary and sufficient condition that is more useful. \bizrek \label{TetivniPogoj} A convex quadrilateral is cyclic if and only if its opposite interior angles are supplementary. Thus, if $\alpha$, $\beta$, $\gamma$ and $\delta$ are the interior angles of a convex quadrilateral $ABCD$, it is cyclic if and only if $$\alpha+\gamma=180^0.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.5.11.pic} \caption{} \label{sl.skk.4.5.11.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.5.11.pic}) ($\Rightarrow$) First, let's assume that the quadrilateral $ABCD$ is cyclic. Because it is convex, the vertices $A$ and $C$ are on different sides of the line $BD$. By \ref{ObodObodKotNaspr} $\alpha+\gamma=180^0$. ($\Leftarrow)$ Now let's assume that the opposite angles of the quadrilateral $ABCD$ are supplementary, i.e. $\alpha+\gamma=180^0$. Let $k$ be the circle drawn through the triangle $ABD$. In this case, the fourth vertex $C$ of the line $BD$ is seen under the angle complementary to the angle at the vertex $A$, which means that the point $C$ also lies on the circle $k$ (\ref{ObodKotGMT}). \kdokaz A direct consequence is the following theorem. \bizrek \label{TetivniPogojZunanji} A convex quadrilateral is cyclic if and only if one of its interior angles is congruent to the opposite exterior angle. Thus, if $\alpha$, $\beta$, $\gamma$ and $\delta$ are the interior angles and $\alpha'$, $\beta'$, $\gamma'$ in $\delta'$ the exterior angles of a convex quadrilateral $ABCD$, it is cyclic if and only if $$\alpha\cong\gamma'.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.5.12.pic} \caption{} \label{sl.skk.4.5.12.pic} \end{figure} We use the criterion from \ref{TetivniPogoj} for a parallelogram and a trapezoid. \bizrek \label{paralelogramTetivEnakokr} A parallelogram is cyclic if and only if it is a rectangle. \eizrek \textbf{\textit{Proof.}} Let $\alpha$, $\beta$, $\gamma$ and $\delta$ be the interior angles of the parallelogram $ABCD$ (Figure \ref{sl.skk.4.5.13.pic}). ($\Leftarrow$) If the parallelogram is a rectangle, $\alpha+\gamma=90^0+90^0=180^0$, which means that $ABCD$ is a cyclic quadrilateral (\ref{TetivniPogoj}). ($\Rightarrow$) Let's assume that $ABCD$ is a trapezoidal parallelogram. Because $ABCD$ is a parallelogram, according to Theorem \ref{paralelogram} $\alpha\cong\gamma$. Because it is also a trapezoid, according to Theorem \ref{TetivniPogoj} $\alpha+\gamma=180^0$. So $\alpha\cong\gamma=90^0$, therefore $ABCD$ is a rectangle. \kdokaz \begin{figure}[!htb] \centering \input{sl.skk.4.5.13.pic} \caption{} \label{sl.skk.4.5.13.pic} \end{figure} \bizrek \label{trapezTetivEnakokr} A trapezium is cyclic if and only if it is isosceles. \eizrek \textbf{\textit{Proof.}} Let $ABCD$ be a trapezoid with a base $AB$ and with internal angles $\alpha$, $\beta$, $\gamma$ and $\delta$ (Figure \ref{sl.skk.4.5.13.pic}). In any trapezoid it holds that $\alpha+\delta=180^0$ and $\beta+\gamma=180^0$. ($\Leftarrow$) Let's assume that trapezoid $ABCD$ is isosceles, i.e. $AD \cong BC$. According to Theorem \ref{trapezEnakokraki} in this case $\alpha\cong\beta$. So $\alpha+\gamma=\beta+\gamma=180^0$, therefore according to Theorem \ref{TetivniPogoj} $ABCD$ is a cyclic quadrilateral. ($\Rightarrow$) Let trapezoid $ABCD$ be a cyclic quadrilateral and $k$ its circumscribed circle. The bases $AB$ and $CD$ are parallel chords of this circle, so they have a common perpendicular, which goes through the center $S$ of the circle $k$ and is perpendicular to the chords $AB$ and $CD$. This means that the legs $AD$ and $BC$ are symmetrical with respect to this perpendicular, so they are congruent and trapezoid $ABCD$ is isosceles. \kdokaz Particularly interesting are cyclic quadrilaterals with perpendicular diagonals\footnote{\index{Brahmagupta}\textit{Brahmagupta} (598--660), Indian mathematician, who studied such quadrilaterals.}. The following example applies to such quadrilaterals. \bzgled \label{TetivniLemaBrahm} Suppose that the diagonals of a cyclic quadrilateral $ABCD$ are perpendicular and intersect at a point $S$. Prove that the foot of the perpendicular from the point $S$ on the line $AB$ contains the midpoint of the line $CD$. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.5.14.pic} \caption{} \label{sl.skk.4.5.14.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.5.14.pic}) Let $N$ and $M$ be the intersections of the rectangle with the line $AB$ through the point $S$ with the sides $AB$ and $CD$ of the quadrilateral $ABCD$. Then it holds: \begin{eqnarray*} \angle CDB &\cong& \angle CAB \hspace*{3mm} \textrm{(external angle for the appropriate locus } CB \textrm{ - izrek \ref{ObodObodKot}})\\ &\cong& \angle NSB \hspace*{3mm} \textrm{ (angle with perpendicular arms - izrek \ref{KotaPravokKraki})}\\ &\cong& \angle MSD \hspace*{3mm} \textrm{(perfect angle)} \end{eqnarray*} Because $\angle CDB\cong \angle MSD$, $MD \cong MS$ (izrek \ref{enakokraki}). Similarly, $MC \cong MS$. Therefore, $MD \cong MC$, which means that $M$ is the center of the side $CD$. \kdokaz We will consider one property of cyclic quadrilaterals with perpendicular diagonals in the example \ref{HamiltonPoslTetiv}. \bzgled \label{TetŠtirZgl0} Let $k$ be the circumcircle of a cyclic quadrilateral $ABCD$ and $N$, $M$, $L$ and $P$ the midpoints of those arcs $AB$, $B$C, $CD$ and $AD$ of the circle $k$, not containing the third vertices of this quadrilateral. Prove that $NL\perp PM$. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.5.0.pic} \caption{} \label{sl.skk.4.5.0.pic} \end{figure} \textbf{\textit{Proof.}} Let $S$ be the intersection of the lines $NL$ and $PM$ (Figure \ref{sl.skk.4.5.0.pic}). If we use the izrek \ref{ObodObodKot} and \ref{TockaN} twice, we get: \begin{eqnarray*} \angle PNS &=& \angle PND +\angle DNL = \angle PBD +\angle DBL =\\ &=& \frac{1}{2} \angle ABD +\frac{1}{2}\angle CBD = \frac{1}{2}\angle ABC. \end{eqnarray*} We similarly prove that $\angle NPS = \frac{1}{2}\angle ADC$. Therefore, according to the statement \ref{TetivniPogoj}: $$\angle PNS +\angle NPS = \frac{1}{2} \left(\angle ABC+\angle ADC\right)=90^0.$$ If we use the statement \ref{VsotKotTrik} for the triangle $PSN$, we get $\angle PSN = 90^0$. \kdokaz \bzgled \label{TetivniVcrtana} Let $ABCD$ be a cyclic quadrilateral. Prove that incentres of the triangles $BCD$, $ACD$, $ABD$ and $ABC$ are the vertices of a rectangle. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.5.1.pic} \caption{} \label{sl.skk.4.5.1.pic} \end{figure} \textbf{\textit{Proof.}} We mark with $A_1$, $B_1$, $C_1$ and $D_1$ the incentres of the triangles $BCD$, $ACD$, $ABD$ and $ABC$ and with $N$, $M$, $L$ and $P$ the incentres of those arcs $AB$, $BC$, $CD$ and $AD$ of the cyclic quadrilateral $ABCD$, which do not contain the other vertices of this quadrilateral (Figure \ref{sl.skk.4.5.1.pic}). From the statement \ref{TockaN} it follows that $BL$ and $DM$ are the angle bisectors of the angles $CBD$ and $BDC$, therefore the point $A_1$ is the intersection of the lines $BL$ and $DM$. Similarly, the point $B_1$ is the intersection of the lines $CP$ and $AL$. According to the statement \ref{TockaN.NBNC}, $LC\cong LA_1\cong LB_1\cong LD$, therefore $A_1LB_1$ is an isosceles triangle with the base $A_1B_1$. From the statement \ref{TockaN} it also follows that $LN$ is the angle bisector of the angle $ALB$ or $B_1LA_1$. In an isosceles triangle $A_1LB_1$ the angle bisector of the angle $B_1LA_1$ contains the altitude of this triangle from the point $L$. This means that $LN\perp A_1B_1$ holds. Similarly, $LN\perp C_1D_1$, $PM\perp A_1D_1$ and $PM\perp C_1B_1$ hold. From the previous statement \ref{TetŠtirZgl0} we know that $LN\perp PM$, therefore the quadrilateral $A_1B_1C_1D_1$ is a rectangle. \kdokaz We have already mentioned that a rectangle is a right-angled quadrilateral. Now we will prove an interesting property of rectangles that relates to points that lie on its circumscribed circle. \bzgled Let $P$ be an arbitrary point on the shorter arc $AB$ of the circumcircle of a rectangle $ABCD$. Suppose that $L$ and $M$ are the foots of the perpendiculars from the point $P$ on the diagonals $AC$ and $BD$, respectively. Prove that the length of the line segment $LM$ does not depend on the position of the point $P$. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.5.15.pic} \caption{} \label{sl.skk.4.5.15.pic} \end{figure} \textbf{\textit{Proof.}} Let $O$ be the center of the circle $k$ (Figure \ref{sl.skk.4.5.15.pic}). The quadrilateral $PMOL$ is a right-angled one, because $\angle OLP + \angle OMP =90^0+90^0= 180^0$ (statement \ref{TetivniPogoj}). We mark with $l$ the circumscribed circle of this quadrilateral. Because the angles $OLP$ and $OMP$ are both right, the distance $OP$ (or the radius of the circle $k$) is the radius of the circle $l$. Then $LM$ is the chord of the circle $l$, which belongs to the peripheral angle $\angle LOM =\angle AOB$, which is constant. Regardless of the choice of the point $P$, the distance $LM$ is the chord of the circle with a constant radius $OA$, which belongs to a constant peripheral angle $AOB$ (or the corresponding constant central angle of this circle). The chords, which belong to the corresponding central angles of the corresponding circles, are proportional to each other, so the length of the distance $LM$ does not depend on the position of the point $P$. \kdokaz The properties of the right-angled quadrilateral are often used to prove various properties of triangles. \bzgled \label{PedalniVS} The orthocentre of an acute triangle is the incentre of its \index{trikotnik!pedalni} pedal triangle. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.5.16.pic} \caption{} \label{sl.skk.4.5.16.pic} \end{figure} \textbf{\textit{Proof.}} Let $AA'$, $BB'$ and $CC'$ be the altitudes of the triangle $ABC$, which intersect in the point $V$ of the altitude of this triangle (Figure \ref{sl.skk.4.5.16.pic}). If $A_1$ is the midpoint of the side $BC$, the points $B'$ and $C'$ lie on the circle $k(A_1,A_1B)$ (statement \ref{TalesovIzrKroz2}). Therefore, the quadrilateral $BC'B'C$ is cyclic, so according to \ref{TetivniPogojZunanji} $\angle AC'B'\cong \angle ACB = \gamma$. Similarly, we prove that the quadrilateral $AC'A'C$ is cyclic, so $\angle BC'A'\cong \angle ACB = \gamma$. Therefore, the angles $AC'B'$ and $BC'A'$ are congruent. Because $CC'\perp AB$, the angles $CC'B'$ and $CC'A'$ are also congruent. This means that the line $C'C$ is perpendicular to the angle $A'C'B'$. Similarly, the lines $A'A$ and $B'B$ are perpendicular to the corresponding internal angles of the triangle $A'B'C'$, so the point $V$ is the centre of the triangle $A'B'C'$ of the inscribed circle. \kdokaz From the proof of the previous statement (\ref{PedalniVS}) we can conclude that the angles, which are determined by the sides of the pedal triangle $A'B'C'$ with the sides of the triangle $ABC$, are equal to the corresponding angles of the triangle $ABC$. We will use this fact in the following example. \bzgled \label{PedalniLemaOcrtana} Let $O$ be the circumcentre of a triangle $ABC$. Prove that the lines $OA$, $OB$ and $OC$ are perpendicular to the corresponding sides of the pedal triangle $A'B'C'$. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.5.17.pic} \caption{} \label{sl.skk.4.5.17.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.5.17.pic}). Let $L$ denote the intersection of the lines $OA$ and $B'C'$. It is enough to prove that the internal angle at the vertex $L$ of the triangle $C'LA$ is a right angle. Let us calculate the other two angles of this triangle. From the previous example \ref{PedalniVS} the angle at the vertex $C'$ is equal to $\gamma$. The triangle $AOB$ is isosceles and $\angle AOB=2\gamma$ (statement \ref{SredObodKot}). Therefore (statements \ref{enakokraki} and \ref{VsotKotTrik}) $\angle C' AL=\angle BAO =90^0-\gamma$, which implies that $\angle ALC'=90^0$. \kdokaz A direct consequence of the statement \ref{PedalniVS} and \ref{PedalniLemasPQR} is the following statement. \bzgled \label{PedalniLemasLMN} Let $P$, $Q$ and $R$ be the midpoints of those arcs $BC$, $AC$ and $AB$ of the circumcircle of a triangle $ABC$ not containing the vertices $A$, $B$ and $C$. Suppose that the point $S$ is the incentre of the triangle $ABC$ and $L=SA\cap QR$, $M=SB\cap PR$ and $N=SC\cap PQ$. Then the triangles $LMN$ and $ABC$ have the common incentre. (Figure \ref{sl.skk.4.5.18.pic}). \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.5.18.pic} \caption{} \label{sl.skk.4.5.18.pic} \end{figure} \bzgled \label{Miquelova točka} Let $P$, $Q$ and $R$ be an arbitrary points on the sides $BC$, $AC$ and $AB$ of the triangle $ABC$, respectively. Prove that the circumcircles of triangles $AQR$, $PBR$ and $PQC$ intersect at in one point (so-called \index{točka!Miquelova} \pojem{Miquel point}\color{green1}\footnote{The point is named after the French mathematician \index{Miquel, A.} \textit{A. Miquel} (1816–-1851), who published this statement in 1838 as an article in Liouville's (\index{Liouville, J.}\textit{J. Liouville} (1809–-1882), French mathematician) journal. But, as is often the case in mathematics, Miquel was not the first to prove this statement. Ten years before him, this fact was discovered and published by the famous Swiss mathematician \index{Steiner, J.} \textit{J. Steiner} (1769--1863).}). \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.5.2.pic} \caption{} \label{sl.skk.4.5.2.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.5.2.pic}) We denote with $k_A$, $k_B$ and $k_C$ the circumcircles of triangles $AQR$, $PBR$ and $PQC$ and the internal angles of triangle $ABC$ in order with $\alpha$, $\beta$ and $\gamma$. Let $S$ be the other intersection of the circles $k_B$ and $k_C$ (the proof is similar in the case when $S = P$). Quadrilateral $BPSR$ and $PCQS$ are tangential, so $\angle RSP = 180^0 - \beta$ and $\angle QSP = 180^0 -\gamma$ (statement \ref{TetivniPogoj}). From this it follows that $\angle RSQ = \beta +\gamma$ and then also $\angle RAQ + \angle RSQ =\alpha + \beta +\gamma = 180^0$. Quadrilateral $ARSQ$ is also tangential (statement \ref{TetivniPogoj}) or it has its own circumcircle, which is actually the circle $k_A$, circumscribed to triangle $AQR$. This means that the circles $k_A$, $k_B$ and $k_C$ intersect in point $S$. \kdokaz In this chapter \ref{pogINV} we will prove one generalization of the previous statement (see example \ref{MiquelKroznice}). \bnaloga\footnote{45. IMO Greece - 2004, Problem 1.} Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$, respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $L$. Prove that the circumcircles of the triangles $BML$ and $CNL$ have a common point lying on the side $BC$. \enaloga \begin{figure}[!htb] \centering \input{sl.skk.4.4.IMO1.pic} \caption{} \label{sl.skk.4.4.IMO1.pic} \end{figure} \textbf{\textit{Proof.}} We mark with $E$ the intersection of the angle bisector of $BAC$ with the side $BC$ of the triangle $ABC$ (Figure \ref{sl.skk.4.4.IMO1.pic}). We prove that $E$ is the desired point i.e. that it lies on the circumcircles of both triangles $BML$ and $CNL$. Because from the construction of the points $M$ and $N$ it follows that $OM\cong ON$, the triangle $OMN$ is isosceles. This means that the bisector of $OL$ is also the bisector of the side $MN$ (follows from the similarity of the triangles $MSO$ and $NSO$, where $S$ is the center of the segment $MN$). Therefore, the point $L$ lies on the bisector of the segment $MN$ of the triangle, so by \ref{TockaN} it lies on the circumcircle $k$ of the triangle $AMN$. The condition $AB\neq AC$ tells us that the angle bisectors of $BAC$ and the side $MN$ (or the angle bisector of $MON$) are different, so their intersection is a point. If we use \ref{TetivniPogojZunanji} and \ref{ObodObodKot}, we get: \begin{eqnarray*} \angle BCA &\cong& AMN \cong\angle ALN,\\ \angle ABC &\cong& ANM \cong\angle ALM. \end{eqnarray*} From these relations and \ref{TetivniPogojZunanji} it follows that $NLEC$ and $LMBE$ are tangential quadrilateral. Therefore, the point $E$ lies on the circumcircles of both triangles $BML$ and $CNL$. \kdokaz We say that a \index{tetragon!tangential}\index{polygon!tangential}\pojem{tangential} \index{tetragon!tangential}\index{polygon!tangential}\pojem{tangential}, if there exists an inscribed circle, or if there is such a circle that the normals of all the sides of the polygon are its tangents (Figure \ref{sl.skk.4.6.1.pic}). We have already seen that every triangle is tangential (\ref{SredVcrtaneKrozn}) and also a regular polygon is tangential (\ref{sredVcrtaneKrozVeck}). On the other hand, it is clear that not all polygons are tangential. For example, a rectangle is a tetragon which does not have an inscribed circle. In this section we will focus on tangential tetragons. \begin{figure}[!htb] \centering \input{sl.skk.4.6.1.pic} \caption{} \label{sl.skk.4.6.1.pic} \end{figure} Since a square is a regular polygon, it is also a tangential tetragon. But how would we in general determine whether a tetragon is tangential? It is clear that in a tangential tetragon (in general also in a polygon) the simetrals of all of its internal angles intersect in one point (Figure \ref{sl.skk.4.6.1.pic}). This condition is sufficient for the tangentiality of a polygon. Unfortunately, this condition is not very useful in specific cases. There is a more useful condition that is necessary and sufficient for the tangentiality of tetragons. \bizrek \label{TangentniPogoj} A quadrilateral $ABCD$ is tangential if and only if $$|AB| + |CD| = |BC| + |AD|.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.6.2.pic} \caption{} \label{sl.skk.4.6.2.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.6.2.pic}) ($\Rightarrow$) First, let's assume that the quadrilateral $ABCD$ is tangent and $k$ is its inscribed circle. Let $P$, $Q$, $R$ and $S$ be the points of tangency of sides $AB$, $BC$, $CD$ and $DA$ with the circle $k$. Because the appropriate tangent lines are concurrent (by Theorem \ref{TangOdsek}), it holds: $AP \cong AS$, $BP \cong BQ$, $CQ \cong CR$ and $DR \cong DS$. Therefore \begin{eqnarray*} |AB| + |CD|&=&|AP| + |PB| + |CR| + |RD| \\&=& |AS| + |SD| + |BQ| + |QC|\\&=&|AD| + |BC|. \end{eqnarray*} ($\Leftarrow$) We prove the converse statement. Let's assume that in the quadrilateral $ABCD$ the sums of the pairs of opposite sides are equal, i.e. $|AB| + |CD| = |BC| + |AD|$. There exists a circle $k$, which touches sides $AB$, $BC$ and $DA$ of this quadrilateral (its center is the intersection of the internal angle bisectors at vertices $A$ and $B$ of this quadrilateral). We prove that this circle also touches side $CD$ of the quadrilateral $ABCD$. Let $D'$ be the intersection of the other tangent from point $C$ of the circle $k$ and the line $AD$. Let's assume that $D'\neq D$. Without loss of generality, let $\mathcal{B}(A,D',D)$. Because the quadrilateral $ABCD'$ is tangent to the circle $k$, by the already proven part of the theorem it holds $|AB| + |CD'| = |AD'|+|BC|$. But since by the assumption also $|AB| + |CD| = |AD| + |BC|$, it also holds $|CD|-|CD'| = |DA| - |D'A|=|DD'|$ i.e. $|CD|= |CD'| + |DD'|$. But this is not possible due to the triangle inequality \ref{neenaktrik} (points $C$, $D$ and $D'$ cannot be collinear, because otherwise it would hold $C\in AD$). In a similar way we arrive at a contradiction also in the case when $\mathcal{B}(A,D,D')$. Therefore it holds $D'= D$, thus $ABCD$ is a tangential quadrilateral. \kdokaz The following theorems are a direct consequence of the previous criterion. \bizrek \label{TangDeltoidRomb} A rhombus, a deltoid, and a square are tangential quadrilaterals (Figure \ref{sl.skk.4.6.3.pic}). \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.6.3.pic} \caption{} \label{sl.skk.4.6.3.pic} \end{figure} \bizrek \label{TangParalelogram} A parallelogram is a tangential quadrilateral if and only if it is a rhombus (Figure \ref{sl.skk.4.6.3.pic}). \eizrek In the next two examples we will consider tension and tangent quadrilaterals at the same time. \bzgled Let $k_A$, $k_B$, $k_C$ and $k_D$ circles with centres $A$, $B$, $C$ and $D$, such that two in a row (also $k_A$ and $k_D$) are touching each other externally. Prove that the quadrilateral defined by the touching points of circles is cyclic, and the quadrilateral $ABCD$ is tangential. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.6.4.pic} \caption{} \label{sl.skk.4.6.4.pic} \end{figure} \textbf{\textit{Proof.}} Let $P$, $Q$, $R$ and $S$ be the touching points of the given circles in order, and $p$ and $r$ the common tangents of the corresponding circles at points $P$ and $R$ (Figure \ref{sl.skk.4.6.4.pic}). First, we have: \begin{eqnarray*} |AD| + |BC| &=& |AP| + |PD| + |BR| + |RC| =\\ &=& |AQ| + |SD| + |QB| + |SC| =\\ &=& |AB| + |CD|. \end{eqnarray*} Therefore, $ABCD$ is a tangent quadrilateral (statement \ref{TangentniPogoj}). Tangents $p$ and $r$ divide the internal angle at vertices $P$ and $R$ of the quadrilateral $PQRS$ into angles, each of which is equal to half of the corresponding central angle (statement \ref{ObodKotTang}). The aforementioned central angles are the internal angles of the quadrilateral $ABCD$. We denote them with $\alpha$, $\beta$, $\gamma$ and $\delta$. Therefore, (statement \ref{VsotKotVeck}): \begin{eqnarray*} \angle QPS+ \angle SRQ&=& \angle QP,p+\angle p,PS+ \angle SR,r+\angle r,RQ=\\ &=& \frac{1}{2}\alpha+\frac{1}{2}\delta+\frac{1}{2}\gamma+\frac{1}{2}\beta=\\ &=& \frac{1}{2}\left(\alpha+\delta+\gamma+\beta\right)=\\ &=& \frac{1}{2}\cdot360^0=180^0, \end{eqnarray*} which means that $PQRS$ is a tension quadrilateral. \kdokaz It is clear that the inscribed circle of the quadrilateral $ABCD$ is also the circumscribed circle of the quadrilateral $PQRS$. Because according to the assumption $PAQ$, $QBR$, $RCS$ and $SDP$ are equilateral triangles with bases $PQ$, $QR$, $RS$ and $SP$, the simetrals of the internal angles of the quadrilateral $ABCD$ are also the simetrals of the sides of the quadrilateral $PQRS$ (Figure \ref{sl.skk.4.6.4a.pic}). This is also the second (simpler) way to prove the second part of the previous example - the assertion that $PQRS$ is a tangential quadrilateral. \begin{figure}[!htb] \centering \input{sl.skk.4.6.4a.pic} \caption{} \label{sl.skk.4.6.4a.pic} \end{figure} \bzgled \label{tetivTangLema} Let $L$ be the intersection of the diagonals of a cyclic quadrilateral $ABCD$. Prove that the foots of the perpendiculars from the point $L$ on the sides of this quadrilateral are the vertices of a tangential quadrilateral. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.6.5.pic} \caption{} \label{sl.skk.4.6.5.pic} \end{figure} \textbf{\textit{Proof.}} Let $P$, $Q$, $R$ and $S$ be the perpendicular projections from the point $L$ on the sides $AB$, $BC$, $CD$ and $DA$ of the quadrilateral $ABCD$ (Figure \ref{sl.skk.4.6.5.pic}). Because of the appropriate right angles, $PBQL$ and $APLS$ are tangential quadrilaterals (statement \ref{TetivniPogoj}). According to the assumption, $ABCD$ is also tangential. If we use this, we get the equality of the appropriate external angles (statement \ref{ObodObodKot}). Therefore: $$\angle SPL \cong \angle SAL = \angle DAC \cong \angle DBC = \angle LBQ \cong \angle LPQ.$$ From this it follows that the line $PL$ is the simetral of the internal angle at the vertex $P$ of the quadrilateral $PQRS$. Similarly, the lines $QL$, $RL$ and $SL$ are the simetrals of the other three internal angles of this quadrilateral. Therefore, $L$ is the center of the inscribed circle of the quadrilateral $PQRS$, so this is tangential. \kdokaz We will now prove another interesting property of tangential quadrilaterals. \bzgled Prove that the incircles of triangles $ABC$ and $ACD$ touch each other if and only if $ABCD$ is a tangential quadrilateral. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.6.6.pic} \caption{} \label{sl.skk.4.6.6.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.6.6.pic}). First, we prove some relations that hold for any convex quadrilateral $ABCD$. Let $P$, $Q$ and $X$ be the points in which the inscribed circle $k$ of triangle $ABC$ touches its sides $AB$, $BC$ and $CA$, and $R$, $S$ and $Y$ be the points in which the inscribed circle $l$ of triangle $ACD$ touches its sides $CD$, $DA$ and $AC$. First, it holds (from \ref{TangOdsek}): \begin{eqnarray*} |AX| &=& |AP|=\frac{1}{2}\left(|AX|+|AP|\right)=\frac{1}{2}\left(|AC|-|CX|+|AB|-|BP|\right)\\ &=& \frac{1}{2}\left(|AC|-|CQ|+|AB|-|BQ|\right)= \frac{1}{2}\left(|AC|+|AB|-|BC|\right), \end{eqnarray*} therefore it holds: $$|AX|=\frac{1}{2}\left(|AC|+|AB|-|BC|\right).$$ In the same way, we prove that it also holds: $$|AY|=\frac{1}{2}\left(|AC|+|AD|-|DC|\right).$$ Now we can start with proving the equivalence. The circles $k$ and $l$ touch each other exactly when $X = Y$ or $|AX| = |AY|$. The last equality holds exactly when: $$\frac{1}{2}\left(|AC|+|AB|-|BC|\right)=\frac{1}{2}\left(|AC|+|AD|-|DC|\right)$$ or $|AB| + |DC| = |AD| + |BC|$, which is fulfilled exactly when $ABCD$ is a tangential quadrilateral (from \ref{TangentniPogoj}). \kdokaz The consequence of this is the following claim. \bzgled Let $ABCD$ be a tangential quadrilateral. Then the incircles of the triangles $ABC$ and $ACD$ touch each other if and only if the incircles of the triangles $ABD$ and $CBD$ touch each other (Figure \ref{sl.skk.4.6.6a.pic}). \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.6.6a.pic} \caption{} \label{sl.skk.4.6.6a.pic} \end{figure} \textbf{\textit{Proof.}} The statements that the incircles of the triangles $ABC$ and $ACD$ or the triangles $ABD$ and $CBD$ touch, are equivalent to the statement that the quadrilateral $ABCD$ is tangent. This means that the initial statements are equivalent. \kdokaz %_______________________________________________________________________________ \poglavje{Bicentric Quadrilateral} \label{odd4TetivniTangentni} Some quadrilaterals are both tangential and chordal. We call them \index{štirikotnik!tetivnotangentni}\pojem{tetivnotangentni} or \index{štirikotnik!bicentrični}\pojem{bicentrični} quadrilaterals. Which quadrilaterals are these? The square is certainly one of them. Is it the only one? The answer is negative. The quadrilateral we get from the example \ref{tetivTangLema} is always tangent. In a certain case it will be chordal as well. Namely, the following statement is true. \bizrek \label{tetivTangIzrek} If $L$ is the intersection of the perpendicular diagonals of a cyclic quadrilateral $ABCD$, then the foots of the perpendiculars from the point $L$ on the sides of this quadrilateral are the vertices of a bicentric quadrilateral. \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.7.2.pic} \caption{} \label{sl.skk.4.7.2.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.7.2.pic}) Let's use the same notation as in the example \ref{tetivTangLema}. We have already proven that $PQRS$ is a tangent quadrilateral. We will now prove that it is also a bicentric quadrilateral. From the proof of the statement in the aforementioned example \ref{tetivTangLema} it follows: \begin{eqnarray*} \angle SPQ &=& \angle SPL+\angle LPQ = \angle SAL+\angle LBQ =\\ &=& \angle DAC + \angle DBC = 2\cdot\angle DBC \end{eqnarray*} or $\angle SPQ= 2\cdot\angle DBC$. Similarly, $\angle SRQ= 2\cdot\angle ACB$. Because, by assumption, $AC\perp BD$, $CLB$ is a right angled triangle, therefore: $$\angle SRQ+\angle SRQ=2\cdot(\angle DBC+\angle ACB)=2\cdot 90^0=180^0.$$ By \ref{TetivniPogoj} $PQRS$ is a bicentric quadrilateral. \kdokaz It is not difficult to convince oneself that the converse statement is also true. \bizrek Let $PQRS$ be a bicentric quadrilateral. Suppose that point $L$ is the incentre of this quadrilateral and at the same time the intersection of the diagonals of a cyclic quadrilateral $ABCD$. If $P$, $Q$, $R$ and $S$ the foots of the perpendiculars from the point $L$ on the sides of quadrilateral $ABCD$, then $AC\perp BD$. \eizrek In the next exercise we will see that for three non-collinear points $A$, $B$ and $C$ there is only one point $D$, such that $ABCD$ is a bicentric quadrilateral. \bnaloga\footnote{4. IMO Czechoslovakia - 1962, Problem 5.} On the circle $k$ there are given three distinct points $A$, $B$, $C$. Construct (using only straightedge and compasses) a fourth point $D$ on $k$ such that a circle can be inscribed in the quadrilateral thus obtained. \enaloga \begin{figure}[!htb] \centering \input{sl.skk.4.5.IMO1.pic} \caption{} \label{sl.skk.4.5.IMO1.pic} \end{figure} \textbf{\textit{Solution.}} Without loss of generality we can assume that $AB\geq BC$. Let $D$ be a point that satisfies the conditions of the task, or such that $ABCD$ is a tangent-chord quadrilateral (Figure \ref {sl.skk.4.5.IMO1.pic}). We denote by $a$, $b$, $c$ and $d$ the sides of $AB$, $BC$, $CD$ and $DA$ of this quadrilateral, and by $\alpha$, $\beta$, $\gamma$ and $\delta$ its internal angles at vertices $A$, $B$, $C$ and $D$. From the condition of the tangency of the quadrilateral $ABCD$ (formula \ref {TetivniPogoj}) it follows that $\delta = 180^0 - \beta$, and from its tangency (formula \ref {TangentniPogoj}) that $a + c = b + d$ or $d-c = a-b$. In this way, the task is reduced to the design of the third vertex $D$ of the triangle $ACD$, where the sides $AC$, the angle $\angle ADC = 180^0 - \beta$ and the difference of sides $AD-CD = AB-BC = a-b$ are given. Let $E$ be a point on the line $AD$, for which $DE \cong DC$. Then $AE = AD-DE = AD-CD = a-b$. The triangle $ECD$ is isosceles, so by formula \ref {enakokraki} it follows that $\angle CED \cong \angle DCE$. Therefore, $\angle AEC = 180^0 - \angle DEC = 180^0 - \frac {1}{2} \beta$. This allows us to construct the triangle $ACE$ or the point $E$. First, we plan the point $E$ as the intersection of the arc $l$ (see the construction described in formula \ref {ObodKotGMT}) $180^0 - \frac {1}{2} \angle ABC$) and the circle $j (A, AB-BC)$ (if $AB \cong AC$, we assume $E = A$). The point $D$ can then be designed as the intersection of the strip $AE$ and the similitude $s_{EC}$ of the line $EC$ (in the case $AB \cong AC$, or $E = A$, $D$ is the second intersection of the similitude $s_{EC} = s_{AC}$ with the circle $k$). We prove that the point $D$ satisfies the conditions of the task, or that $ABCD$ is a tangent-chord quadrilateral. First, we will consider the case when $AB>BC$. By construction, the point $D$ lies on the line of symmetry $EC$, so $DE\cong DC$ and also (by statement \ref{enakokraki}) $\angle DEC\cong\angle DCE$. We have drawn the point $E$ so that it lies on the arc $l$ with the string $AC$ and the angular measure $180^0-\frac{1}{2}\angle ABC$, so $\angle AEC=180^0-\frac{1}{2}\angle ABC$. Because, by construction, $\mathcal{B}(A,E,D)$, we have $\angle DCE\cong\angle DEC=\frac{1}{2}\angle ABC$. From the isosceles triangle $EDC$ by statement \ref{VsotKotTrik} it follows that $\angle ADC=\angle EDC=180^0-\angle ABC$. Therefore, $\angle EDC+\angle ABC=180^0$, so by statement \ref{TetivniPogoj} the quadrilateral $ABCD$ is a string quadrilateral, or $D\in k$. We will now prove that the quadrilateral $ABCD$ is a tangent quadrilateral. In the first part of the proof (the string property) we have already seen that $DE\cong DC$. The point $E$ by construction lies on the circle $j(A,|AB-BC|)$, so $|AE|=|AB|-|BC|$. Because $\mathcal{B}(A,E,D)$ also holds, we have $|AD|-|CD|=|AD|-|DE|=|AE|=|AB|-|BC|$. From this it follows that $|AD|+|BC|=|AB|+|CD|$ and by statement \ref{TangPogoj} the quadrilateral $ABCD$ is tangent. If $AB\cong BC$, the point $D$ by construction already lies on the circle $k$. Because both points $B$ and $D$ lie on the line of symmetry $AC$, the quadrilateral $ABCD$ is a deltoid, so it is also tangent (by statement \ref{TangDeltoidRomb}). We will now investigate the number of solutions to the problem. The circle $k(A,AB-AC)$ and the arc $l$ always intersect in one point $E$. Because $ABC<180^0$, we have $\angle AEC=180^0-\frac{1}{2}\beta>90^0$. This means that the line of symmetry $s_{EC}$ always intersects the half-line $AE$ in one point $D$ and $\mathcal{B}(A,E,D)$ holds. This means that the problem always has one and only one solution. \kdokaz %______________________________________________________________________________ \poglavje{Simson Line} \label{odd4Simson} We will first prove the basic statement. \bizrek \label{SimpsPrem} The foots of the perpendiculars from an arbitrary point lying on the circumcircle of a triangle to the lines containing the sides of this triangle are three collinear points. The line containing these points is the so-called \pojem{Simson\footnote{Premico imenujemo po škotskem matematiku \index{Simson, R.} \textit{R. Simsonu} (1687--1768), čeprav je to lastnost prvi objavil škotski matematik \index{Wallace, W.} \textit{W. Wallace} (1768--1843) šele leta 1799.} line}\color{blue}. \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.7.1a.pic} \caption{} \label{sl.skk.4.7.1a.pic} \end{figure} \textbf{\textit{Proof.}} Let $S$ be an arbitrary point of the circumcircle $k$ of the triangle $ABC$ and $P$, $Q$ and $R$ the orthogonal projections of the point $S$ on the lines containing the sides $BC$, $AC$ and $AB$ (Figure \ref{sl.skk.4.7.1a.pic}). Without loss of generality, we assume that $\mathcal{B}(B,P,C)$, $\mathcal{B}(A,Q,C)$ and $\mathcal{B}(A,B,R)$ hold. In this case, the points $Q$ and $R$ are on different sides of the line $BC$, so it is enough to prove $\angle BPR \cong \angle CPQ$. Because of the appropriate right angles and the position of the point $S$, the quadrilaterals $BRSP$, $ABSC$, $ARSQ$ and $SPQC$ are cyclic, so (from izrek \ref{TetivniPogoj}: \begin{eqnarray*} \angle BPR &=& \angle BSR = \angle RSC - \angle BSC=\\ &=& \angle RSC - (180° - \angle BAC) =\\ &=& \angle RSC - \angle RSQ = \angle CSQ = \angle CPQ, \end{eqnarray*} which means that the points $P$, $Q$ and $R$ are collinear. \kdokaz In the following we will consider further interesting properties of Simson's line. Because each point $X$, which lies on the circumcircle of some triangle, determines the Simson line, we will denote this line by $x$. In this way, each triangle determines one mapping $X\mapsto x$. \bzgled \label{SimsZgled1} Let $P$ be an arbitrary point of the circumcircle $k$ of a triangle $ABC$. Suppose that $P_A$ is the intersection of the perpendicular line of the line $BC$ through the point $P$ with the circle $k$. Prove that the line $AP_A$ is parallel to the Simson line $p$ of the triangle at the point $P$. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.7.1b.pic} \caption{} \label{sl.skk.4.7.1b.pic} \end{figure} \textbf{\textit{Proof.}} Let $X$, $Y$ and $Z$ be the orthogonal projections of the point $P$ on the lines $BC$, $AC$ and $AB$ (Figure \ref{sl.skk.4.7.1b.pic}). By izreku \ref{SimpsPrem} the Simson line $p$ is determined by the points $X$, $Y$ and $Z$. Similarly to izreku \ref{SimpsPrem}, the quadrilateral $PYXC$ is a trapezoid, therefore $\angle YXP \cong \angle ACP$. By izreku \ref{ObodObodKot} the angles $ACP$ and $AP_AP$ above the trapezoid $AP$ are supplementary, therefore $\angle YXP \cong \angle AP_AP$ or $XY\parallel AP$ (izrek \ref{KotiTransverzala}). \kdokaz \bzgled \label{SimsZgled2} Let $P$ and $Q$ be arbitrary points lying on the circumcircle $k(O,r)$ of a triangle $ABC$ and $p$ and $q$ their Simson lines. Prove that $$\angle pq = \frac{1}{2}\angle POQ.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.7.1c.pic} \caption{} \label{sl.skk.4.7.1c.pic} \end{figure} \textbf{\textit{Proof.}} Let $X_P$ and $X_Q$ be the feet of the perpendiculars from points $P$ and $Q$ on the line $BC$ and $P_A$ and $Q_A$ the intersections of these perpendiculars with the circle $k$ (Figure \ref{sl.skk.4.7.1c.pic}). The Simson lines $p$ and $q$ are parallel to the lines $AP_A$ and $AQ_A$ (example \ref{SimsZgled1}). Therefore, the angle determined by the lines $p$ and $q$ is equal to the inscribed angle $Q_AAP_A$, which is equal to half of the central angle $Q_AOP_A$ (by statement \ref{SredObodKot}) or half of the angle $QOP$ (because the trapezoid $PP_AQ_AQ$ is isosceles and by statement \ref{trapezTetivEnakokr} it is also equilateral, i.e. $PQ \cong P_AQ_A$). \kdokaz \bzgled \label{SimsZgled3} Let $P$ be an arbitrary point of the circumcircle $k$ of a triangle $ABC$, $p$ its Simson line and $V$ the orthocentre of this triangle. Prove that the line $p$ bisects the line segment $PV$. \ezgled \begin{figure}[!htb] \centering \input{sl.skk.4.7.1d.pic} \caption{} \label{sl.skk.4.7.1d.pic} \end{figure} \textbf{\textit{Proof.}} Let $P_A$ and $X$ be the points defined as in example \ref{SimsZgled1} (Figure \ref{sl.skk.4.7.1d.pic}). Let $V'$ and $P'$ be the points that are symmetric to the points $V$ and $P$ with respect to the line $BC$. The point $V'$ lies on the circumscribed circle $k$ of the triangle $ABC$ (by statement \ref{TockaV'}). Because of the properties of symmetry or the axis of reflection (see subsection \ref{odd6OsnZrc}), statement \ref{KotiTransverzala}, statement \ref{ObodObodKot} and example \ref{SimsZgled1} it holds: $$\angle VP'P\cong\angle V'PP'\cong\angle AV'P \cong\angle AP_AP\cong\angle p,PP'.$$ Therefore, $\angle VP'P \cong \angle p,PP'$, so by statement \ref{KotiTransverzala} the lines $VP'$ and $p$ are parallel. Because the point $X$ is the midpoint of the segment $PP'$, the line $p$ contains the midpoint of the triangle $PVP'$ (by statement \ref{srednjicaTrik}) or the midpoint of its side $PV$. \kdokaz In sections \ref{odd5Hamilton} and \ref{odd7SredRazteg} we will prove two more properties of Simson lines (see \ref{HamiltonSimson} and \ref{SimsEuler}), which are related to Hamilton's theorem or Euler's circle of a triangle. %________________________________________________________________________________ \poglavje{Torricelli Point} \label{odd4Torricelli} In this section we will give another famous point of a triangle. \bizrek \label{izrekTorichelijev} On each side of a triangle $ABC$ the equilateral triangles $BEC$, $CFA$ and $AGB$ are externally erected. Prove: \begin{enumerate} \item $AE$, $BF$ and $CG$ are congruent line segments; \item the lines $AE$, $BF$ and $CG$ intersect at one point (so-called \pojem{Torricelli\footnote{Problem was first posed by French mathematician \index{Fermat, P.} \textit{P. Fermat} (1601--1665) as a challenge to Italian mathematician and physicist \index{Torricelli, E.} \textit{E. Torricelli} (1608--1647). Torricelli's solution was published by his student - Italian mathematician and physicist \textit{V. Viviani} (1622–-1703) - in 1659. We also call this point \index{point!Fermat's}\pojem{Fermat point}.} point} \color{blue}of this triangle) and every two of them determine an angle with measure $60^0$. \end{enumerate} \index{point!Torricelli's} \eizrek \begin{figure}[!htb] \centering \input{sl.skk.4.7.1.pic} \caption{} \label{sl.skk.4.7.1.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.skk.4.7.1.pic}) (\textit{i}) Triangles $AEC$ and $FBC$ are congruent by \textit{SAS} \ref{SKS} theorem ($AC \cong FC$ , $CE \cong CB$ and $\angle ACE \cong \angle FCB = \angle ACB + 60°$), so $AE\cong BF$. Analogously is $AE\cong CG$. (\textit{ii}) Let $k$, $l$ and $j$ be the circumscribed circles of the triangles $BEC$, $CFA$ and $AGB$. We shall first prove that these circles intersect in one point. With $T$ we denote the second intersection point of the circles $k$ and $l$ ($T\neq C$). The quadrilaterals $BECT$ and $CFAT$ are cyclic, therefore (by the statement \ref{TetivniPogoj}) both the angles $BTC$ and $ATC$ measure $120^0$. Thus, also the angle $ATB$ measures $120^0$, which means that the quadrilateral $AGBT$ is cyclic (by the statement \ref{TetivniPogoj}) or that the point $T$ also lies on the circle $j$. We shall prove that each of the lines $AE$, $BF$, $CG$ goes through the point $T$. From the equality of the corresponding circumscribed angles (by the statement \ref{ObodObodKot}) we obtain: \begin{eqnarray*} \angle ATE&=&\angle ATF+\angle FTC+\angle CTE=\\ &=&\angle ACF+\angle FAC+\angle CBE =3\cdot 60^0=180^0. \end{eqnarray*} Thus, $A$, $T$ and $E$ are collinear points, or the point $T$ lies on the line $AE$. Analogously, the point $T$ also lies on the lines $BF$ and $CG$. It is also clear that: $\angle AE,BF\cong\angle ATF\cong\angle ACF=60^0$. \kdokaz In the section \ref{odd9MetrInv} (by the statement \ref{izrekToricheliFerma}) we shall prove another interesting property of the Torricelli's point. %________________________________________________________________________________ \poglavje{Excircles of a Triangle} \label{odd4Pricrt} We have already proved that for any triangle there exist circumscribed and inscribed circle. The first one contains all the vertices of the triangle, the second one touches all its sides. Now we shall show that there also exist circles which touch one side and two lines containing the sides of the triangle. \bizrek The bisector of the interior angle at vertex $A$ and the bisectors of the exterior angles at vertices $B$ and $C$ of a triangle $ABC$ intersect at one point, which is the centre of the circle touching the side $BC$ and the lines containing the sides $AB$ and $AC$. It is so-called \index{pričrtane krožnice trikotnika} \pojem{excircle of the triangle}\color{blue}. \eizrek \begin{figure}[!htb] \centering \input{sl.27.1.94_veliki_zadatak_lema.pic} \caption{} \label{sl.27.1.94_veliki_zadatak_lema.pic} \end{figure} \textbf{\textit{Proof.}} We prove the statement similarly to the inscribed circle of a triangle. The simetrali of the external angles at the vertices $B$ and $C$ are not parallel and they intersect at some point - we mark it with $S_a$ (Figure \ref{sl.27.1.94_veliki_zadatak_lema.pic}). Because the point $S_a$ lies on these two simetrali, it holds $A,S_a\div BC$ and $S_a$ is equally distant from the lines $AB$, $BC$ and $AC$. Therefore, $S_a$ also belongs to the simetral of the internal angle at the vertex $A$ and is the center of the circle that touches the side $BC$ and the lines $AB$ and $AC$. \kdokaz Now we are ready to prove the so-called \index{velika naloga} \pojem{‘‘velika naloga’’}, which is very useful in designing triangles. \bizrek \label{velikaNaloga} Let $P$, $Q$, $R$ be the touching points of the incircle $k(S,r)$ of a triangle $ABC$ with the sides $BC=a$, $AC=b$, $AB=c$ ($b>c$) and $P_i$, $Q_i$, $R_i$ ($i\in \{a,b,c\}$) the touching points of the excircles $k_i(S_i,r_i)$ with lines $BC$, $AC$ in $AB$. Let $l(O,R)$ be the circumcircle of this triangle with the semiperimeter $s=\frac{a+b+c}{2}$, $A_1$ the midpoint of the line segments $BC$, $M$ and $N$, intersections of the line $OA_1$ with the circle $l$ ($N,A\div BC$) and $M’$, $N’$ the foots of the perpendiculars from these points on the line $AB$ (Figure \ref{sl.27.1.94_veliki_zadatak.pic}). Then: \vspace*{2mm} (\textit{i}) $AQ_a\cong AR_a=s$, \hspace*{0.4mm} (\textit{ii}) $AQ\cong AR=s-a$, \hspace*{0.4mm} (\textit{iii}) $QQa\cong RRa\cong a$, % \vspace*{1mm} (\textit{iv}) $PPa=b-c$,\hspace*{1mm} (\textit{v}) $P_bP_c=b+c$, % \vspace*{1mm} (\textit{vi}) $A_1$ is the midpoint of the line segment $PP_a$ in $P_bP_c$, %\vspace*{1mm} (\textit{vii}) $A_1N= \frac{r_a- r}{2}$,\hspace*{2mm} (\textit{viii}) $A_1M= \frac{r_b + r_c }{2} $,\hspace*{1mm} (\textit{ix}) $r_a +r_b +r_c =4R+r$,\footnote{To lastnost trikotnika je leta 1790 odkril francoski matematik \index{L'Huilier, S. A. J.} \textit{S. A. J. L'Huilier} (1750--1840).} %\vspace*{1mm} (\textit{x}) $NN’= \frac{r_a +r}{2}$, \hspace*{1mm} (\textit{xi}) $MM’= \frac{r_b -r_c }{2 }$,\hspace*{1mm} (\textit{xii}) $N’B\cong AM’=\frac{ b - c}{2}$, % \vspace*{1mm} (\textit{xiii}) $AN’\cong BM’= \frac{b + c}{2} $, \hspace*{1mm} (\textit{xiv}) $M’N’\cong b$. \eizrek \begin{figure}[!htb] \centering \input{sl.27.1.94_veliki_zadatak.pic} \caption{} \label{sl.27.1.94_veliki_zadatak.pic} \end{figure} \textbf{\textit{Proof.}} Preden začnemo z dokazovanjem, omenimo, da po izreku \ref{TockaN} točka $N$ leži na simetrali notranjega kota $BAC$ trikotnika $ABC$. (\textit{i}) If we use the equality of tangent lines (statement \ref{TangOdsek}), we get: \begin{eqnarray*} AQ_a&\cong &AR_a= \frac{1}{2}\left(AQ_a+AR_a\right)= \frac{1}{2}\left(AB+BR_a+AC+CQ_a\right)\\ &=& \frac{1}{2}\left(AB+BP_a+AC+CP_a\right)= \frac{1}{2}\left(AB+BC+AC\right)=s. \end{eqnarray*} (\textit{ii}) In a similar way, we get: \begin{eqnarray*} AQ&\cong &AR= \frac{1}{2}\left(AQ+AR\right)= \frac{1}{2}\left(AB-BR+AC-CQ\right)\\ &=& \frac{1}{2}\left(AB-BP+AC-CP\right)= \frac{1}{2}\left(AB+AC-BC\right)=s-a. \end{eqnarray*} (\textit{iii}) $QQ_a\cong AQ_a-AQ=a$. (\textit{iv}) We prove the equality by first calculating:\\ $BP$ and $CP_a\cong CQ_a$. (\textit{v}) $P_bP_c\cong CP_c+BP_b-a=2s-a=b+c$. (\textit{vi}) It follows from $BP\cong CP_a=s-b$. (\textit{vii}) Points $A_1$ and $N$ are the centers of the diagonal of trapezoid $SPS_aP_a$ with bases $SP\cong r$ and $S_aP_a\cong r_a$. The equality follows from statement \ref{srednjTrapez}. (\textit{viii}) Lines $NA$ and $S_cS_b$ are perpendicular (the internal and external angles at point $A$ are symmetrical), so point $M$ lies on line $S_cS_b$. The desired equality follows from the fact that line $A_1M$ is the median of trapezoid $S_cP_cP_bS_b$ (statement \ref{srednjTrapez}). (\textit{ix}) It follows directly from $2\cdot R=NM=NA_1+A_1M$ and (\textit{vii}) and (\textit{viii}). (\textit{x}) It follows from statement \ref{srednjTrapez} and the fact that $NN'$ is the median of trapezoid $SRR_aS_a$. (\textit{xi}) Points $M$ and $M'$ are the centers of the diagonal of trapezoid $R_cS_cR_bS_b$ with bases $R_cS_c\cong r_c$ and $R_bR_b\cong r_b$. The equality follows from statement \ref{srednjTrapez}. (\textit{xii}) Point $N$ is the center of line $RR_a$, so: $$N'B=AN'-AB= \frac{1}{2}\left(AR_a+AR\right)-c= \frac{1}{2}\left(s+(s-a)\right)-c= \frac{1}{2}\left(b-c\right).$$ (\textit{xiii}) and (\textit{xiv}) follow directly from the proven equality (\textit{xii}). \kdokaz %KONSTRUKCIJA (VN) \bzgled Construct a triangle $ABC$, with given: (\textit{a}) $a$, $b-c$, $r$, \hspace*{3mm} (\textit{b}) $b-c$, $r$, $v_b$, \hspace*{3mm} (\textit{c}) $a$, $b+c$, $r$, \hspace*{3mm}(\textit{č}) $R$, $r$, $r_a$. \ezgled \textbf{\textit{Solution.}} For each construction we will use the big task - \ref{velikaNaloga}. We will use the same labels. \begin{figure}[!htb] \centering \input{sl.27.1.94_veliki_zadatak_konstr.pic} \caption{} \label{sl.27.1.94_veliki_zadatak_konstr.pic} \end{figure} \begin{figure}[!htb] \centering \input{sl.27.1.94_veliki_zadatak_konstr2.pic} \caption{} \label{sl.27.1.94_veliki_zadatak_konstr2.pic} \end{figure} (\textit{a}) Because $PP_a=b-c$ and point $A_1$ is the common center of side $BC$ and the line $PP_a$, it also holds that $PA_1 = \frac{1}{2}( b - c)$ (Figure \ref{sl.27.1.94_veliki_zadatak_konstr.pic}). First, we plan the side $BC$, then its center $A_1$, point $P$, the inscribed circle of the triangle, tangents from the vertices $B$ and $C$, and finally the vertices $A$. (\textit{b}) Similarly to the previous example. First, we plan the line $PA_1$, then the inscribed circle of the triangle $ABC$ (Figure \ref{sl.27.1.94_veliki_zadatak_konstr.pic}). We also need to use the condition of the height from the vertex $B$. With $L$ we mark the orthogonal projection from the point $A_1$ onto the line $AC$. The line $A_1L$ is the median of the triangle $CBB'$, so $A_1L =\frac{1}{2}BB'=\frac{1}{2} v_b$. Therefore, the line $AC$ can be constructed as the common tangent of the inscribed circle and the circle $k(A_1, \frac{1}{2}v_b)$. Thus we get the vertex $C$, then the vertices $B$ and $A$. (\textit{c}) We know that $RR_a\cong a$ and $AN'=\frac{1}{2}(b+c)$, and that $N'$ is the center of the line $RR_a$ (Figure \ref{sl.27.1.94_veliki_zadatak_konstr2.pic}). So, from the given data, we first construct the points $A$, $N'$, $R$ and $R_a$, and then also $S$, $N$ and $S_a$. In the end, we draw the dotted and the dashed circle - the sides of the triangle lie on their common tangents. (\textit{č}) Because $A_1N =\frac{1}{2}(r_a -r)$ and $MN = 2R$, we can first plan the points $N$, $A_1$ and $M$, and then also the dotted circle of the triangle $ABC$ and the side $BC$ (Figure \ref{sl.27.1.94_veliki_zadatak_konstr2.pic}). The construction can be finished in two ways. In the first case, we translate the task into a construction of a triangle that we already know: $a$, $R$, $r$ (example \ref{konstr_Rra}), in the second case, we use the equality $RR_a\cong a$. \kdokaz %________________________________________________________________________________ \naloge{Exercises} \begin{enumerate} \item The sides of a triangle are $6$, $7$ and $9$. Let $k_1$, $k_2$ and $k_3$ be the circles with centers in the vertices of this triangle. The circles touch each other so that the circle with the center in the vertex of the smallest angle of the triangle touches the other two circles from the inside, while the remaining two circles touch from the outside. Calculate the lengths of the radii of these three circles. \item Prove that the angle formed by the secants of a circle that intersect each other outside the circle is equal to half the difference of the central angles corresponding to the arcs that lie between the arms of this angle. \item The apex of the angle $\alpha$ is an external point of the circle $k$. Between the arms of this angle, on the circle, there are two arcs, which are in the ratio $3:10$. The larger of these arcs corresponds to the central angle $40^0$. Determine the measure of the angle $\alpha$. \item Prove that the angle formed by the tangents of a circle is equal to half the difference of the central angles corresponding to the arcs that lie between the arms of this angle. \item Let $L$ be the orthogonal projection of an arbitrary point $K$ of the circle $k$ on its tangent $t$ through the point $T\in k$ and $X$ the point that is symmetric to the point $L$ with respect to the line $KT$. Determine the geometric position of the points $X$. \item Let $BB'$ and $CC'$ be the altitudes of the triangle $ABC$ and $t$ the tangent of this triangle at the point $A$. Prove that $B'C'\parallel t$. \item In the right triangle $ABC$ is above the cathetus $AC$ as the diameter drawn circle that intersects the hypotenuse $AB$ at the point $E$. The tangent of this circle at the point $E$ intersects the other cathetus $BC$ at the point $D$. Prove that $BDE$ is an isosceles triangle. \item In the right angle with the vertex $A$ is drawn a circle that touches the sides of this angle at the points $B$ and $C$. Any tangent of this circle intersects the lines $AB$ and $AC$, in order in the points $M$ and $N$ (so that $\mathcal{B}(A,M,B)$). Prove that: $$\frac{1}{3}\left(|AB|+|AC|\right) < |MB|+|NC| < \frac{1}{2}\left(|AB|+|AC|\right).$$ \item Prove that in the right triangle the sum of the sides is equal to the sum of the diameters of the inscribed and drawn circle. \item Let the similitudes of the internal angles of the convex quadrilateral intersect in six different points. Prove that four of these points are the vertices of the pedal quadrilateral. \item Let: $c$ be the length of the hypotenuse, $a$ and $b$ the lengths of the catheti and $r$ the radius of the inscribed circle of the right triangle. Prove that: \begin{enumerate} \item $2r + c \geq 2 \sqrt{ab}$, \item $a + b + c > 8r$. \end{enumerate} \item Let $P$ and $Q$ be the centers of the shorter arcs $AB$ and $AC$ of the regular triangle $ABC$ of the drawn circle. Prove that the sides $AB$ and $AC$ of this triangle divide the chord $PQ$ into three proportional segments. \item Let $k_1$, $k_2$, $k_3$, $k_4$ be four circles, each of which from the outside touches one side and two sides of an arbitrary convex quadrilateral. Prove that the centers of these circles are concircular points. \item Circles $k$ and $l$ touch each other from the outside in point $A$. Points $B$ and $C$ are the points of contact of the common external tangent of these two circles. Prove that $\angle BAC$ is a right angle. \item Let $ABCD$ be a deltoid ($AB\cong AD$ and $CB\cong CD$). Prove: \begin{enumerate} \item $ABCD$ is a tangent quadrilateral, \item $ABCD$ is a parallelogram exactly when $AB\perp BC$. \end{enumerate} \item Circles $k$ and $k_1$ touch each other from the outside in point $T$, where they intersect lines $p$ and $q$. Line $p$ has two more intersections with the circles, $P$ and $P_1$, line $q$ has $Q$ and $Q_1$. Prove that $PQ\parallel P_1Q_1$. \item Let $MN$ be the common tangent of circles $k$ and $l$ ($M$ and $N$ are the points of contact), which intersect in points $A$ and $B$. Calculate the measure of the sum $\angle MAN+\angle MBN$. \item Let $t$ be the tangent of triangle $ABC$ of the circumscribed circle in point $A$. A line parallel to the tangent $t$ intersects sides $AB$ and $AC$ in points $D$ and $E$. Prove that points $B$, $C$, $D$ and $E$ are concyclic. \item Let $D$ and $E$ be any points of the semicircle drawn over diameter $AB$. Let $AD\cap BE= \{F\}$ and $AE\cap BD= \{G\}$. Prove that $FG\perp AB$. \item Let $M$ be a point of circle $k(O,r)$. Determine the geometric location of the centers of all the tangents of this circle that have one endpoint in point $M$. \item Let $M$ and $N$ be points that are symmetric to the vertex $A'$ of altitude $AA'$ of triangle $ABC$ with respect to side $AB$ and $AC$, and let $K$ be the intersection of lines $AB$ and $MN$. Prove that points $A$, $K$, $A'$, $C$ and $N$ are concyclic. \item Let $ABCD$ be a parallelogram, $E$ the altitude point of triangle $ABD$, and $F$ the altitude point of triangle $ABC$. Prove that quadrilateral $CDEF$ is a parallelogram. \item Circles with centers $O_1$ and $O_2$ intersect in points $A$ and $B$. The line $p$, which goes through point $A$, intersects these two circles in points $M_1$ and $M_2$. Prove that $\angle O_1M_1B\cong\angle O_2M_2B$. \item The circle with center $O$ is drawn over the diameter $AB$. Let $C$ and $D$ be such points on the line $AB$, that $CO\cong OD$. Parallel lines through points $C$ and $D$ intersect the circle in points $E$ and $F$. Prove that lines $CE$ and $DF$ are perpendicular to the line $EF$. \item On the string $AB$ of the circle $k$ with center $O$ lies the point $C$, point $D$ is the other intersection of the circle $k$ with the drawn circle of the triangle $ACO$. Prove that $CD\cong CB$. \item Let $AB$ be the transversal of the circle $k$. Lines $AC$ and $BD$ are tangents to the circle $k$ in points $C$ and $D$. Prove that: $$||AC|-|BD||< |AB| < |AC|+|BD|.$$ \item Let $S$ be the intersection of the sides $AD$ and $BC$ of the trapezoid $ABCD$ with the base $AB$. Prove that the drawn circles of the triangles $SAB$ and $SCD$ touch in point $S$. \item Lines $PB$ and $PD$ touch the circle $k(O,r)$ in points $B$ and $D$. Line $PO$ intersects the circle $k$ in points $A$ and $C$ ($\mathcal{B}(P,A,C)$). Prove that the line $BA$ is the angle bisector of the angle $PBD$. \item Quadrilateral $ABCD$ is inscribed in the circle with center $O$. Diagonals $AC$ and $BD$ are perpendicular. Let $M$ be the perpendicular projection of the center $O$ on the line $AD$. Prove that $$|OM|=\frac{1}{2}|BC|.$$ \item Lines $AB$ and $BC$ are adjacent sides of a regular nonagon, which is inscribed in the circle $k$ with center $O$. Point $M$ is the center of the side $AB$, point $N$ is the center of the radius $OX$ of the circle $k$, which is perpendicular to the line $BC$. Prove that $\angle OMN=30^0$. \item Circles $k_1$ and $k_2$ intersect in points $A$ and $B$. Let $p$ be a line that goes through point $A$, circle $k_1$ intersects also in point $C$, circle $k_2$ intersects also in point $D$, and $q$ be a line that goes through point $B$, circle $k_1$ intersects also in point $E$, circle $k_2$ intersects also in point $F$. Prove that $\angle CBD\cong\angle EAF$. \item Circles $k_1$ and $k_2$ intersect in points $A$ and $B$. Draw a line $p$, that goes through point $A$, so that the length of the line $MN$, where $M$ and $N$ are the intersections of line $p$ with circles $k_1$ and $k_2$, is maximal. \item Let $L$ be the orthogonal projection of an arbitrary point $K$ of the circle $k$ on its tangent through the point $T\in k$ and $X$ be the point that is symmetric to the point $L$ with respect to the line $KT$. Determine the geometric position of the points $X$. \item Prove that the string polygon with an even number of vertices, that has all the internal angles congruent, is a regular polygon. \item Two circles touch each other from the inside in the point $A$. The line $AB$ is the diameter of the larger circle, the string $BK$ of the larger circle touches the smaller circle in the point $C$. Prove that the line $AC$ is the bisector of the angle $BAK$. \item Let $BC$ be the string of the circle $k$. Determine the geometric position of the altitude points of all the triangles $ABC$, where $A$ is an arbitrary point that lies on the circle $k$. \item We have a quadrilateral with three acute internal angles. Prove that the longer diagonal goes through the vertex that belongs to the acute angle. \item Let $ABCDEF$ be a string hexagon, $AB\cong DE$ and $BC\cong EF$. Prove that $CD\parallel AF$. \item Let $ABCD$ be a convex quadrilateral, where $\angle ABD=50^0$, $\angle ADB=80^0$, $\angle ACB=40^0$ and $\angle DBC=\angle BDC +30^0$. Calculate the measure of the angle $\angle DBC$. \item Let $M$ be an arbitrary internal point of the angle with the vertex $A$, points $P$ and $Q$ the orthogonal projections of the point $M$ on the sides of this angle, and point $K$ the orthogonal projection of the vertex $A$ on the line $PQ$. Prove that $\angle MAP\cong \angle QAK$. \item In the archery octagon $A_1A_2\ldots A_8$ it holds that $A_1A_2\parallel A_5A_6$, $A_2A_3\parallel A_6A_7$, $A_3A_4\parallel A_7A_8$. Prove that $A_8A_1\cong A_4A_5$. \item A circle intersects each side of a quadrilateral in two points and thus on all sides of the quadrilateral it determines the consistent tautologies. Prove that this quadrilateral is tangent. \item The lengths of the sides of the tangent pentagon $ABCDE$ are natural numbers and at the same time $|AB|=|CD|=1$. The inscribed circle of the pentagon touches the side $BC$ in the point $K$. Calculate the length of the line $BK$. \item Prove that the circle that passes through the adjacent vertices $A$ and $B$ of the regular pentagon $ABCDE$ and its center $O$, also passes through the intersection of its diagonals $AD$ and $BE$. \item Let $H$ be the altitude point of the triangle $ABC$, $l$ the circle above the diameter $AH$ and $P$ and $Q$ the intersections of this circle with the sides $AB$ and $AC$. Prove that the tangents of the circle $k$ through the points $P$ and $Q$ intersect on the side $BC$. \item The circle $l$ touches the circle $k$ from the inside in the point $C$. Let $M$ be any point of the circle $l$ (different from $C$). The tangent of the circle $l$ in the point $M$ intersects the circle $k$ in the points $A$ and $B$. Prove that $\angle ACM \cong \angle MCB$. \item Let $k$ be the inscribed circle of the triangle $ABC$ and $R$ the center of that arc $AB$ of this circle, which does not contain the point $C$. The lines $RP$ and $RQ$ are the tautologies of this circle. The first one is parallel, the second one is perpendicular to the internal angle $\angle BAC$. Prove: \begin{enumerate} \item The line $BQ$ is the internal angle $\angle CBA$, \item The triangle, which is determined by the lines $AB$, $AC$ and $PR$, is a right triangle. \end{enumerate} \item Let $X$ be such an internal point of the triangle $ABC$, that it holds: $\angle BXC =\angle BAC+60^0$, $\angle AXC =\angle ABC+60^0$ and $\angle AXB =\angle AC B+60^0$. Let $P$, $Q$ and $R$ be the other intersections of the lines $AX$, $BX$ and $CX$ with the inscribed circle of the triangle $ABC$. Prove that the triangle $PQR$ is a right triangle. \item Prove that the tangent points of an inscribed circle of a triangle $ABC$ divide its sides into segments of lengths $s-a$, $s-b$ and $s-c$ ($a$, $b$ and $c$ are the lengths of the sides, $s$ is the semi-perimeter of the triangle). \item Circles $k$, $l$ and $j$ touch each other from the outside in non-linear points $A$, $B$ and $C$. Prove that the circumscribed circle of the triangle $ABC$ is perpendicular to the circles $k$, $l$ and $j$. \item Let $ABCD$ be a square with the center of the circumscribed circle in the point $O$. With $E$ we denote the intersection of its diagonals $AC$ and $BD$ and with $F$, $M$ and $N$ the centers of the lines $OE$, $AD$ and $BC$. If $F$, $M$ and $N$ are collinear points, then $AC\perp BD$ or $AB\cong CD$. Prove. \item Draw a triangle $ABC$ (see the labels in section \ref{odd3Stirik}): (\textit{a}) $a$, $\alpha$, $r$, \hspace*{2mm} (\textit{b}) $a$, $\alpha$, $r_a$, \hspace*{2mm} (\textit{c}) $a$, $v_b$, $v_c$, \hspace*{2mm} (\textit{d}) $\alpha$, $v_a$, $s$, \hspace*{2mm} (\textit{e}) $v_a$, $l_a$, $r$, \hspace*{2mm} (\textit{f}) $\alpha$, $v_a$, $l_a$, \hspace*{2mm} (\textit{g}) $\alpha$, $\beta$, $R$, \hspace*{2mm} (\textit{h}) $c$, $r$, $R$, \hspace*{2mm} (\textit{i}) $a$, $v_b$, $R$, \hspace*{2mm} \item Draw a circle $k$ so that: \begin{enumerate} \item it touches two given non-parallel lines $p$ and $q$, and the tangent that determines the touch points is consistent with the given distance $t$, \item its center is the given point $S$, and the given line $p$ determines on it the tangent that is consistent with the given distance $t$, \item it passes through the given points $A$ and $B$, and its center lies on the given circle $l$, \item it has the given radius $r$ and it touches the two given circles $l$ and $j$, \item it touches the line $p$ in the point $P$ and passes through the given point $A$. \end{enumerate} \item Draw a square $ABCD$, if the given vertex $B$ and two points $E$ and $F$ that lie on the sides $AD$ and $CD$ are given. \item Given is a line $CD$ and points $A$ and $B$ ($A,B\notin CD$). On the line $CD$ draw a point $M$, such that $\angle AMC\cong2\angle BMD$. \item Draw a triangle $ABC$ with the following data: (\textit{a}) $v_a$, $t_a$, $\beta-\gamma$, \hspace*{2mm} (\textit{b}) $v_a$, $l_a$, $R$, \hspace*{2mm} (\textit{c}) $R$, $\beta-\gamma$, $t_a$, \hspace*{2mm} (\textit{d}) $R$, $\beta-\gamma$, $v_a$, \hspace*{2mm} (\textit{e}) $R$, $\beta-\gamma$, $a$. \hspace*{2mm} \item On the beam of the side $AB$ of the rectangle $ABCD$ draw a point $E$, from which the sides $AD$ and $DC$ are seen under the same angle. When does the task have a solution? \item In the convex quadrilateral $ABCD$ it holds that $BC\cong CD$. Draw this quadrilateral, if the sides $AB$ and $AD$ and the internal angle at the vertices $B$ and $D$ are given. \item In the given circle $k$ draw a triangle $ABC$, if the vertex $A$, the line $p$, which is parallel to the altitude $AA'$, and the intersection point $B_2$ of the altitude beam $BB'$ and this circle are given. \item Draw a regular triangle $ABC$, if its side $BC$ is congruent to the distance $a$, the altitude beams of the sides $AB$ and $AC$ and the line of symmetry of the internal angle $BAC$ go through the given points $M$, $N$ and $P$ one after another. \item Draw a triangle $ABC$, if the following are given: \begin{enumerate} \item the vertex $A$, the center of the circumscribed circle $O$ and the center of the inscribed circle $S$, \item the center of the circumscribed circle $O$, the center of the inscribed circle $S$ and the center of the escribed circle $S_a$, \item the vertex $A$, the center of the circumscribed circle $O$ and the altitude point $V$, \item the vertices $B$ and $C$ and the internal angle bisector of the internal angle $BAC$, \item the vertex $A$, the center of the circumscribed circle $O$ and the intersection point $E$ of the side $BC$ with the internal angle bisector of the internal angle $BAC$, \item the points $M$, $P$ and $N$, in which the altitude and the centroid from the vertex $A$ and the internal angle bisector of the internal angle $BAC$ intersect the circumscribed circle of the triangle, \item the vertex $A$, the center of the circumscribed circle $O$, the point $N$, in which the internal angle bisector of the internal angle $BAC$ intersects the circumscribed circle of the triangle, and the distance $a$, which is parallel to the side $BC$. \end{enumerate} \item Draw a triangle $ABC$ with the following data: (\textit{a}) $a$, $b$, $\alpha=3\beta$, \hspace*{3mm} (\textit{b}) $t_a$, $t_c$, $v_b$.\hspace*{3mm} \item Through the point $M$, which lies inside the given circle $k$, draw such a cord that the difference of its segments (from the point $M$) is equal to the given distance $a$. \item Draw a triangle $ABC$, if you know: (\textit{a}) $b-c$, $r$, $r_a$, \hspace*{1.8mm} (\textit{b}) $a$, $r$, $r_a$, \hspace*{1.8mm} (\textit{c}) $a$, $r_b+r_c$, $v_a$, \hspace*{1.8mm} (\textit{d}) $b+c$, $r_b$, $r_c$, (\textit{e}) $R$, $r_b$, $r_c$, \hspace*{1.8mm} (\textit{f}) $b$, $R$, $r+r_a$, \hspace*{1.8mm} (\textit{g}) $a$, $v_a$, $r_a-r$, \hspace*{1.8mm} (\textit{h}) $\alpha$, $r$, $b+c$. \item The following are given: the circle $k$, its diameter $AB$ and the point $M\notin k$. With only a straightedge, draw a rectangle from the point $M$ to the line $AB$. \item The given are: square $ABCD$ and such points $M$ and $N$ on sides $BC$ and $CD$, that $\angle MAN=45^0$. With only a straightedge draw a rectangle from point $A$ to line $MN$. \end{enumerate} % DEL 5 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - %________________________________________________________________________________ % VEKTORJI %________________________________________________________________________________ \del{Vectors} \label{pogVEKT} %________________________________________________________________________________ \poglavje{Vector Definition. The Sum of Vectors} \label{odd5DefVekt} Intuitively, a vector is a directed line segment that can be moved parallel\footnote{The concept of a vector was known to the ancient Greeks. The modern concept of vectors, associated with linear algebra and analytic geometry, began to develop in the 19th century as a generalization of complex numbers. In this sense, the English mathematician \index{Hamilton, W. R.}\textit{W. R. Hamilton} (1805--1865) defined the so-called \index{quaternions}\textit{quaternions} $q = w + ix + jy + kz$, $i^2 = j^2 = k^2 = -ijk = -1$ as a generalization of complex numbers in four-dimensional space}. In this sense, the vector $\overrightarrow{AB}$ would represent the entire set of line segments that are consistent, parallel and have the same direction as a given line segment $AB$. We could also write $\overrightarrow{CD}=\overrightarrow{AB}$ for each line segment $CD$ from this set of line segments (Figure \ref{sl.vek.5.1.1.pic}). \begin{figure}[!htb] \centering \input{sl.vek.5.1.1.pic} \caption{} \label{sl.vek.5.1.1.pic} \end{figure} In this way, we get an idea for a formal definition of vectors. First, we introduce the relation $\varrho$ on the set of pairs of points. Let $A$, $B$, $C$ and $D$ be points in the same plane (Figure \ref{sl.vek.5.1.2.pic}). We say that $(A,B)\varrho (C,D)$, if one of the three conditions\footnote{In this way - in \index{geometry!affine}affine geometry (without the axioms of congruence) - the vector was defined by the Serbian mathematician \index{Veljković, M.}\textit{M. Veljković} (1954--2008), professor at the Mathematical Gymnasium in Belgrade.}\index{relation!$\varrho$} is fulfilled: \begin{enumerate} \item The quadrilateral $ABDC$ is a parallelogram, \item There exist points $P$ and $Q$, such that the quadrilaterals $ABQP$ and $CDQP$ are parallelograms, \item $A=B$ and $C=D$. \end{enumerate} \begin{figure}[!htb] \centering \input{sl.vek.5.1.2.pic} \caption{} \label{sl.vek.5.1.2.pic} \end{figure} It is intuitively clear that the second condition needs to be added due to the example when points $A$, $B$, $C$ and $D$ are collinear. The third condition will apply to the so-called vector of zero. From the definition of the relation $\varrho$ itself, we get the following proposition. \bizrek \label{vektRelRo} If $(A,B)\varrho (C,D)$ and $A\neq B$, then the line segments $AB$ and $CD$ are congruent, parallel, and have the same direction. \eizrek \textbf{\textit{Proof.}} Because $A\neq B$, only the first two conditions from the definition remain. In this case, the relations $AB\parallel CD$ and $AB\cong CD$ are direct consequences of the definition of a parallelogram and proposition \ref{paralelogram}. Regarding the orientation of the line segments, we would use the definition: parallel line segments $XY$ and $UV$ are \pojem{equally oriented}, if one of the conditions is fulfilled: \begin{itemize} \item $Y,V\ddot{-} XU$; \item There exist points $S$ and $T$, such that $ST\parallel XY$, $Y,T\ddot{-} XS$ and $V,T\ddot{-} US$. \end{itemize} \kdokaz The next proposition is needed, so that we can define vectors. \bizrek The relation $\varrho$ on the set of pairs of points in the plane is an equivalence relation. \eizrek \textbf{\textit{Proof.}} The reflexivity and symmetry are direct consequences of the definition of the relation $\varrho$. For transitivity, however, it is necessary to check all possible combinations of conditions 1--3 from the definition. \kdokaz \pojem{Vector} \index{vector} is now defined as the class of the equivalence relation $\varrho$: $$\overrightarrow{AB}=[AB]_{\varrho}=\{(X,Y);\hspace*{1mm}(X,Y)\varrho(A,B)\}.$$ \begin{figure}[!htb] \centering \input{sl.vek.5.1.3.pic} \caption{} \label{sl.vek.5.1.3.pic} \end{figure} We will call point $A$ the \index{starting point of a vector}\pojem{starting point}, point $B$ the \index{end point of a vector}\pojem{end point} of the vector $\overrightarrow{AB}$. It is clear that in the case of $(A,B)\varrho (C,D)$, the pairs $(A,B)$ and $(C,D)$ are from the same class, which means that then $\overrightarrow{AB}=\overrightarrow{CD}$. The converse is also true: the relation $\overrightarrow{AB}=\overrightarrow{CD}$ means that $(A,B)\varrho (C,D)$ (Figure \ref{sl.vek.5.1.3.pic}). If it is not necessary to emphasize which representative of the class it is, we will also denote vectors with $\overrightarrow{v}$, $\overrightarrow{u}$, $\overrightarrow{w}$, ... The vector $\overrightarrow{AA}$, which represents the set of coinciding points, we will call the \index{vector!zero}\pojem{zero vector}. The zero vector will also be denoted by $\overrightarrow{0}$ (Figure \ref{sl.vek.5.1.4.pic}). \begin{figure}[!htb] \centering \input{sl.vek.5.1.4.pic} \caption{} \label{sl.vek.5.1.4.pic} \end{figure} We will say that the vector $\overrightarrow{BA}$ is the \index{vector!opposite}\pojem{opposite vector} of the vector $\overrightarrow{AB}$ and denote it by $-\overrightarrow{AB}$. It is clear that the definition of the opposite vector is not dependent on the choice of the representative of the class; the opposite vector is obtained if we exchange the starting and end points of the vector (or of each pair of points of the corresponding class). So if we denote $\overrightarrow{v}=\overrightarrow{AB}$ then $-\overrightarrow{v}=-\overrightarrow{AB}=\overrightarrow{BA}$ (Figure \ref{sl.vek.5.1.4.pic}). We denote the set of all vectors of the plane with $\mathcal{V}$. A direct consequence of \ref{vektRelRo} is the following assertion. \bizrek \label{vektVzpSkl} If $\overrightarrow{AB}=\overrightarrow{CD}$, then the line segments $AB$ and $CD$ are congruent, parallel, and have the same direction. \eizrek Based on the previous statement, we can now correctly define the following concepts: \begin{itemize} \item \index{smer vektorja}\pojem{smer} of vector $\overrightarrow{AB}\neq\overrightarrow{0}$ is determined by any line $p\parallel AB$, \item \index{dolžina!vektorja}\pojem{length} or \index{intenziteta vektorja}\pojem{intensity} of vector $\overrightarrow{AB}$ is $|\overrightarrow{AB}|=|AB|$ ($|\overrightarrow{0}|=0$), \item \index{usmerjenost vektorja}\pojem{orientation} or \index{orientacija!vektorja}\pojem{orientation} of vector $\overrightarrow{AB}\neq\overrightarrow{0}$ is determined by the ordered pair $(A,B)$, where $A$ is the starting point, and $B$ is the end point of this vector. \end{itemize} If vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ have the same direction, we also say that they are \index{vektorja!vzporedna}\pojem{parallel} or \index{vektorja!kolinearna}\pojem{colinear} and denote it with $\overrightarrow{v}\parallel\overrightarrow{u}$ (in this case we also say that $\overrightarrow{0}$ is colinear with every vector), otherwise the vectors are \index{vektorja!nekolinearna}\pojem{non-colinear}. If colinear vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ satisfy $\overrightarrow{v}=\overrightarrow{SA}$ and $\overrightarrow{u}=\overrightarrow{SB}$, we say that vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ are \pojem{equally oriented} (notation $\overrightarrow{u}\rightrightarrows \overrightarrow{v}$), if $A,B\ddot{-} S$, or \pojem{oppositely oriented} (notation $\overrightarrow{u}\rightleftarrows \overrightarrow{v}$), if $A,B\div S$. It is clear that colinear vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are equally oriented exactly when lines $AB$ and $CD$ are equally oriented. From these definitions it directly follows that $|-\overrightarrow{v}|=|\overrightarrow{v}|$ and $-\overrightarrow{v}\parallel\overrightarrow{v}$; vectors $\overrightarrow{v}$ and $-\overrightarrow{v}$ are oppositely oriented (if $\overrightarrow{v}=\overrightarrow{AB}$, then $-\overrightarrow{v}=\overrightarrow{BA}$). The following statement is very important. \bizrek \label{vektABCObst1TockaD} For every three points there $A$, $B$ and $C$ there exists exactly one point $D$, such that $\overrightarrow{AB}=\overrightarrow{CD}$, i.e. $$(\forall A, B, C)(\exists_1 D)\hspace*{1mm}\overrightarrow{AB}=\overrightarrow{CD}.$$ \eizrek \textbf{\textit{Proof.}} We will consider three different options (Figure \ref{sl.vek.5.1.5.pic}). \begin{figure}[!htb] \centering \input{sl.vek.5.1.5.pic} \caption{} \label{sl.vek.5.1.5.pic} \end{figure} \textit{1)} If $A=B$, then $D=C$ is the only point for which $(A,B)\varrho (C,D)$ or $\overrightarrow{CD}=\overrightarrow{AB}=\overrightarrow{0}$. \textit{2)} Let $A\neq B$ and $C\notin AB$. Then $D$ is the fourth vertex of the parallelogram $ABDC$ (which exists and is only one due to Playfair's axiom \ref{Playfair1}) and the only point for which $(A,B)\varrho (C,D)$ or $\overrightarrow{CD}=\overrightarrow{AB}$. \textit{3)} Let $A\neq B$ and $C\in AB$. Let $P$ be any point that does not lie on the line $AB$. By the proven part \textit{2)} there is exactly one point $Q$, for which $\overrightarrow{PQ}=\overrightarrow{AB}$, and then also exactly one point $D$, for which $\overrightarrow{CD}=\overrightarrow{PQ}=\overrightarrow{AB}$. \kdokaz The previous statement can also be written in another way (Figure \ref{sl.vek.5.1.6.pic}). \bizrek \label{vektAvObst1TockaB} For every point $X$ and every vector $\overrightarrow{v}$ there exists exactly one point $Y$, such that $\overrightarrow{XY}=\overrightarrow{v}$, i.e. $$(\forall X)(\forall \overrightarrow{v})(\exists_1 Y)\hspace*{1mm}\overrightarrow{XY}=\overrightarrow{v}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.1.6.pic} \caption{} \label{sl.vek.5.1.6.pic} \end{figure} As it is evident from the proof, for every vector $\overrightarrow{v}$ we can choose an arbitrary representative from its class with an arbitrary starting point. In a similar way, we could have proven that this is also possible regarding the end point. Intuitively, this means that we can move the vector parallel in only one way to its starting or end point. We can therefore put two arbitrary vectors so that they have the same starting point or so that the starting point of the second vector is at the same time the end point of the first vector. This fact allows us to define the sum of two vectors. If $\overrightarrow{v}=\overrightarrow{AB}$ and $\overrightarrow{u}=\overrightarrow{BC}$, the \index{vsota!vektorjev}\pojem{sum of vectors} $\overrightarrow{v}$ and $\overrightarrow{u}$ is the vector $\overrightarrow{v}+\overrightarrow{u}=\overrightarrow{AC}$ (Figure \ref{asl.vek.5.1.7.pic}). So $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ is true. \begin{figure}[!htb] \centering \input{sl.vek.5.1.7.pic} \caption{} \label{asl.vek.5.1.7.pic} \end{figure} It is necessary to prove the correctness of the definition - that the sum of vectors is not dependent on the choice of representatives of two classes ($\overrightarrow{v}$ and $\overrightarrow{u}$) or on the choice of point $A$. \bizrek \label{vektKorektDefSest} If $A$, $B$, $C$, $A'$, $B'$ and $C'$ are points such that $\overrightarrow{AB}=\overrightarrow{A'B'}$ and $\overrightarrow{BC}=\overrightarrow{B'C'}$, then there is also $\overrightarrow{AC}=\overrightarrow{A'C'}$. \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.1.7a.pic} \caption{} \label{sl.vek.5.1.7a.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.1.7.pic}). In the formal proof, we would consider different possibilities regarding the relation $\varrho$. \kdokaz The rule of adding vectors, for which we use the aforementioned equality: \begin{itemize} \item $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$, \end{itemize} We call it the \index{pravilo!trikotniško}\pojem{triangle rule} or the \index{Chaslesova identiteta}\pojem{Chaslesova\footnote{\index{Chasles, M.} \textit{M. Chasles} (1793–-1880), French mathematician.} identity}. The rule is named this way even though it is also valid when $A$, $B$ and $C$ are collinear points (or $\overrightarrow{v}$ and $\overrightarrow{u}$ are collinear vectors); in this case, it is not a triangle (Figure \ref{sl.vek.5.1.8.pic}). \begin{figure}[!htb] \centering \input{sl.vek.5.1.8.pic} \caption{} \label{sl.vek.5.1.8.pic} \end{figure} It turns out that the set of all vectors $\mathcal{V}$ with respect to the operation of adding vectors represents the structure of a so-called \index{group!commutative}\pojem{commutative group} (or \index{group!Abelian}\pojem{Abelian\footnote{\index{Abel, N. H.}\textit{N. H. Abel} (1802--1829), Norwegian mathematician.} group}). The concept of a group (which is not always commutative) was already mentioned in section \ref{odd2AKSSKL} in the context of isometries; we will deal with this structure in more detail in section \ref{odd6Grupe}. The properties of this structure (commutative groups) are given in the statement of the following theorem. \bizrek \label{vektKomGrupa} The ordered pair $(\mathcal{V},+)$ form a commutative group, which means that: \begin{enumerate} \item $(\forall \overrightarrow{v}\in\mathcal{V})(\forall \overrightarrow{v}\in\mathcal{V})\hspace*{1mm} \overrightarrow{v}+\overrightarrow{v}\in\mathcal{V}$ (\index{grupoidnost}\textit{closure}), \item $(\forall \overrightarrow{u}\in\mathcal{V}) (\forall \overrightarrow{v}\in\mathcal{V}) (\forall \overrightarrow{w}\in\mathcal{V}) \hspace*{1mm} \left(\overrightarrow{u}+\overrightarrow{v}\right)+\overrightarrow{w} = \overrightarrow{u}+\left(\overrightarrow{v}+\overrightarrow{w}\right) $ (\index{asociativnost}\textit{associativity}), \item $(\exists \overrightarrow{e}\in\mathcal{V}) (\forall \overrightarrow{v}\in\mathcal{V}) \hspace*{1mm} \overrightarrow{v}+\overrightarrow{e} = \overrightarrow{e}+\overrightarrow{v} $ (\index{nevtralni element}\textit{identity element}), \item $(\forall \overrightarrow{v}\in\mathcal{V}) (\exists \overrightarrow{u}\in\mathcal{V}) \hspace*{1mm} \overrightarrow{v}+\overrightarrow{u} = \overrightarrow{u}+\overrightarrow{v}=\overrightarrow{e} $ (\index{inverzni element}\textit{inverse element}), \item $(\forall \overrightarrow{v}\in\mathcal{V}) (\forall \overrightarrow{u}\in\mathcal{V}) \hspace*{1mm} \overrightarrow{v}+\overrightarrow{u} = \overrightarrow{u}+\overrightarrow{v} $ (\index{komutativnost}\textit{commutativity}). \end{enumerate} \eizrek \textbf{\textit{Proof.}} \textit{1)} The sum of two arbitrary vectors is a vector, which follows from the definition of the sum of two vectors. \textit{2)} Let $\overrightarrow{u}$, $\overrightarrow{v}$ and $\overrightarrow{w}$ be arbitrary vectors and $A$, $B$, $C$ and $D$ such points that $\overrightarrow{u}=\overrightarrow{AB}$, $\overrightarrow{v}=\overrightarrow{BC}$ and $\overrightarrow{w}=\overrightarrow{CD}$ (statement \ref{vektAvObst1TockaB}) (Figure \ref{sl.vek.5.1.9a.pic}). \begin{figure}[!htb] \centering \input{sl.vek.5.1.9a.pic} \caption{} \label{sl.vek.5.1.9a.pic} \end{figure} By the definition of vector addition (and the correctness of this definition - statement \ref{vektKorektDefSest}) it is: \begin{eqnarray*} & & \left(\overrightarrow{u}+\overrightarrow{v}\right)+\overrightarrow{w}= \left(\overrightarrow{AB}+\overrightarrow{BC}\right)+\overrightarrow{CD}= \overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AD}\\ & & \overrightarrow{u}+\left(\overrightarrow{v}+\overrightarrow{w}\right) =\overrightarrow{AB}+\left(\overrightarrow{BC}+\overrightarrow{CD}\right)= \overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AD} \end{eqnarray*} From this it follows that the addition of vectors is associative. \textit{3)} In the case of vector addition, the neutral element is the vector of zero or $\overrightarrow{e}=\overrightarrow{0}$, because for every vector $\overrightarrow{v}=\overrightarrow{AB}$ it is true: \begin{eqnarray*} & & \overrightarrow{v}+\overrightarrow{0}= \overrightarrow{AB}+\overrightarrow{BB}= \overrightarrow{AB}=\overrightarrow{v}\\ & & \overrightarrow{0}+\overrightarrow{v}= \overrightarrow{AA}+\overrightarrow{AB}= \overrightarrow{AB}=\overrightarrow{v}. \end{eqnarray*} \textit{4)} In the case of vector addition, for an arbitrary vector $\overrightarrow{v}=\overrightarrow{AB}$ the inverse element is the opposite vector $\overrightarrow{u}=-\overrightarrow{v}=\overrightarrow{BA}$: \begin{eqnarray*} & & \overrightarrow{v}+\left(-\overrightarrow{v}\right)= \overrightarrow{AB}+\overrightarrow{BA}= \overrightarrow{AA}=\overrightarrow{0}\\ & & -\overrightarrow{v}+\overrightarrow{v}= \overrightarrow{BA}+\overrightarrow{AB}= \overrightarrow{BB}=\overrightarrow{0}. \end{eqnarray*} \textit{5)} Let $v$ and $u$ be any vectors and $A$, $B$ and $C$ such points that $\overrightarrow{v}=\overrightarrow{AB}$ and $\overrightarrow{u}=\overrightarrow{BC}$ (statement \ref{vektAvObst1TockaB}). We will consider two examples (Figure \ref{sl.vek.5.1.9b.pic}). \begin{figure}[!htb] \centering \input{sl.vek.5.1.9b.pic} \caption{} \label{sl.vek.5.1.9b.pic} \end{figure} \textit{a)} We assume that points $A$, $B$ and $C$ are nonlinear or $\overrightarrow{v}$ and $\overrightarrow{u}$ are nonlinear vectors. We mark with $D$ the fourth vertex of the parallelogram $ABCD$. By definition of the relation $\varrho$ it is $(A,B)\varrho (D,C)$ and $(B,C)\varrho (A,D)$, by definition of vectors it then follows that $\overrightarrow{AB}=\overrightarrow{DC}$ and $\overrightarrow{BC}=\overrightarrow{AD}$. So: $$\overrightarrow{v}+\overrightarrow{u}= \overrightarrow{AB}+\overrightarrow{BC}= \overrightarrow{AC}= \overrightarrow{AD}+\overrightarrow{DC}= \overrightarrow{BC}+\overrightarrow{AB} = \overrightarrow{u}+\overrightarrow{v}.$$ \textit{a)} Let $A$, $B$ and $C$ be collinear points or $\overrightarrow{v}$ and $\overrightarrow{u}$ be collinear vectors. Express the vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ as a sum of $\overrightarrow{v}=\overrightarrow{v}_1+\overrightarrow{v}_2$ and $\overrightarrow{u}=\overrightarrow{u}_1+\overrightarrow{u}_2$, so that none of the vectors $\overrightarrow{v}_1$, $\overrightarrow{v}_2$, $\overrightarrow{u}_1$ and $\overrightarrow{u}_2$ are collinear. If we now use what was proven in case \textit{a)}, we get: $$\overrightarrow{v}+\overrightarrow{u}= \overrightarrow{v}_1+\overrightarrow{v}_2+ \overrightarrow{u}_1+\overrightarrow{u}_2= \overrightarrow{u}_1+\overrightarrow{u}_2+ \overrightarrow{v}_1+\overrightarrow{v}_2= \overrightarrow{u}+\overrightarrow{v},$$ which is what needed to be proven. \kdokaz The next statement is a consequence of the previous expression. \bzgled \label{vektABCD_ACBD} For arbitrary points $A$, $B$, $C$ and $D$ is $$\overrightarrow{AB}=\overrightarrow{CD}\Rightarrow \overrightarrow{AC}=\overrightarrow{BD}.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.1.10.pic} \caption{} \label{sl.vek.5.1.10.pic} \end{figure} \textbf{\textit{Proof.}} By the definition of vector addition and the expression \ref{vektKomGrupa} (commutativity) is (Figure \ref{sl.vek.5.1.10.pic}): $\overrightarrow{AC}= \overrightarrow{AB}+ \overrightarrow{BC}= \overrightarrow{CD}+ \overrightarrow{BC}= \overrightarrow{BC}+\overrightarrow{CD}= \overrightarrow{BD}.$ \kdokaz Another consequence of the commutativity of vector addition (expression \ref{vektKomGrupa}) is the following rule of addition for non-collinear vectors. \begin{itemize} \item \textit{For every three non-collinear points $A$, $B$ and $C$ is $\overrightarrow{AB}+\overrightarrow{AC}=\overrightarrow{AD}$ exactly when $ABDC$ is a parallelogram.} \end{itemize} \begin{figure}[!htb] \centering \input{sl.vek.5.1.11.pic} \caption{} \label{sl.vek.5.1.11.pic} \end{figure} This rule is called the \index{pravilo!paralelogramsko}\pojem{paralelogramsko pravilo}\footnote{Paralelogramsko pravilo was probably known to the ancient Greeks. It is assumed that the ancient Greek mathematician and philosopher \index{Aristotel} \textit{Aristotel} (384--322 BC) mentioned it.} (Figure \ref{sl.vek.5.1.11.pic}). The consequence of the associativity from the izrek \ref{vektKomGrupa} is the following rule for the addition of vectors: \begin{itemize} \item $\overrightarrow{A_1A_2}+\overrightarrow{A_2A_3}+\cdots +\overrightarrow{A_{n-1}A_n}=\overrightarrow{A_1A_n}$, \end{itemize} which is called the \index{pravilo!poligonsko}\pojem{poligonsko pravilo} (Figure \ref{asl.vek.5.1.12.pic}). \begin{figure}[!htb] \centering \input{sl.vek.5.1.12.pic} \caption{} \label{asl.vek.5.1.12.pic} \end{figure} The direct consequence of this rule is the following assertion (Figure \ref{sl.vek.5.1.13.pic}). \bzgled For arbitrary points $A_1$, $A_2$,..., $A_n$ in a plane is $$\overrightarrow{A_1A_2}+\overrightarrow{A_2A_3}+\cdots +\overrightarrow{A_{n-1}A_n}+\overrightarrow{A_nA_1}=\overrightarrow{0}.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.1.13.pic} \caption{} \label{sl.vek.5.1.13.pic} \end{figure} With the following izrek we will give an equivalent definition of the center of a line. \bzgled \label{vektSredDalj} A point $S$ is the midpoint of a line segment $AB$ if and only if $\overrightarrow{SA}=-\overrightarrow{SB}$ i.e. $\overrightarrow{SA}+\overrightarrow{SB}=\overrightarrow{0}$. \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.1.14.pic} \caption{} \label{sl.vek.5.1.14.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.1.14.pic}) ($\Leftarrow$) Let's assume that $\overrightarrow{SA}=-\overrightarrow{SB}$ or $\overrightarrow{AS}=\overrightarrow{SB}$. This means that $AS$ and $SB$ are parallel, have the same direction and are consistent lines (statement \ref{vektVzpSkl}) or $SA\cong BS$ and $\mathcal{B}(A,S,B)$. So $S$ is the center of the line $AB$. ($\Rightarrow$) Now let $S$ be the center of the line $AB$. It is enough to prove that $\overrightarrow{SA}=\overrightarrow{BS}$ or $\overrightarrow{BS}=\overrightarrow{SA}$. Let's assume that $\overrightarrow{BS}=\overrightarrow{SA'}$. But in this case $SA'\cong BS$ and $\mathcal{B}(A',S,B)$. By statement \ref{ABnaPoltrakCX} we have $A=A'$, so $\overrightarrow{BS}=\overrightarrow{SA'}=\overrightarrow{SA}$. \kdokaz Let's also define the operation of subtracting two vectors. \pojem{The difference of vectors} \index{razlika!vektorjev} $\overrightarrow{v}$ and $\overrightarrow{u}$ is the sum of vectors $\overrightarrow{v}$ and $-\overrightarrow{u}$ or $$\overrightarrow{v}-\overrightarrow{u}=\overrightarrow{v}+(-\overrightarrow{u}).$$ \bzgled \label{vektOdsev} For arbitrary three points $O$, $A$ and $B$ is $$\overrightarrow{OB}-\overrightarrow{OA}=\overrightarrow{AB}.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.1.15.pic} \caption{} \label{sl.vek.5.1.15.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.1.15.pic}) From the definition of subtraction and addition of vectors and commutativity of addition (statement \ref{vektKomGrupa}) it follows: $$\overrightarrow{OB}-\overrightarrow{OA}= \overrightarrow{OB}+(-\overrightarrow{OA})= \overrightarrow{OB}+\overrightarrow{AO}= \overrightarrow{AO}+\overrightarrow{OB}= \overrightarrow{AB},$$ which had to be proven. \kdokaz \bzgled \label{vektOPi} Suppose that a point $O$ lies on a line $p$ and let $P_1, P_2, \cdots, P_n$ be points lying in the same half-plane $\alpha$ with the edge $p$. If: $$\overrightarrow{OS_n}=\overrightarrow{OP_1}+\overrightarrow{OP_2}+ \cdots+\overrightarrow{OP_n},$$ then the point $S_n$ also lies in the half-plane $\alpha$. \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.2.1a.pic} \caption{} \label{sl.vek.5.2.1a.pic} \end{figure} \textbf{\textit{Solution.}} We will prove the statement by induction according to $n$ (Figure \ref{sl.vek.5.2.1a.pic}). (\textit{i}) For $n=1$ it is clear that $\overrightarrow{OS_1}=\overrightarrow{OP_1}$, and $P_1\in\alpha\Rightarrow S_1=P_1\in\alpha$. (\textit{ii}) We assume that the statement is true for $n=k$. We will prove that it is then also true for $n=k+1$. Let $P_1, P_2, \cdots, P_k, P_{k+1}\in\alpha$. We will prove that then also $S_{k+1}\in \alpha$. It holds: $$\overrightarrow{OS_{k+1}}=\overrightarrow{OP_1}+\overrightarrow{OP_2}+ \cdots+\overrightarrow{OP_k}+\overrightarrow{OP_{k+1}}= \overrightarrow{OS_k}+\overrightarrow{OP_{k+1}}.$$ The point $S_k$ by the induction assumption lies in the half-plane $\alpha$. Because $S_k,P_{k+1}\in \alpha$ and $O\in p$, the angle $\angle S_kOP_{k+1}$ is convex and the whole angle lies in the half-plane $\alpha$. This means that also the point $S_{n+1}$ (as the fourth vertex of the parallelogram $P_{k+1}OP_kS_{k+1}$) lies in the interior of the angle $\angle S_kOP_{k+1}$ or in the half-plane $\alpha$. \kdokaz With the concept of vector addition, we can define the concept of multiplying a vector by an arbitrary natural number: $$1\cdot \overrightarrow{v} =\overrightarrow{v},\hspace*{2mm} (n+1)\cdot \overrightarrow{v}=n\cdot\overrightarrow{v}+\overrightarrow{v},\hspace*{2mm} (n\in \mathbb{N}).$$ We can extend this definition to whole numbers: $0\cdot \overrightarrow{v}=\overrightarrow{0}$ and $-n\cdot \overrightarrow{v}=n\cdot(-\overrightarrow{v})$. It is clear that in this case, the vectors $\overrightarrow{v}$ and $l\cdot \overrightarrow{v}$ ($l\in \mathbb{Z}$) are always collinear and $|l\cdot \overrightarrow{v}|=|l|\cdot |\overrightarrow{v}|$. This gives us an idea for the definition of multiplying a vector by an arbitrary real number $\lambda\in \mathbb{R}$. First, for $\lambda=0$ let us define $0\cdot \overrightarrow{v}=\overrightarrow{0}$. If $\lambda\neq 0$ and $\overrightarrow{v}=\overrightarrow{AB}$, then $\lambda\cdot\overrightarrow{v}=\overrightarrow{AC}$, where $C$ is such a point on the line $AB$ that (Figure \ref{sl.vek.5.2.1.pic}): \begin{itemize} \item $|AC|=|\lambda| \cdot|AB|$, \item $C,B\ddot{-} A$, if $\lambda>0$, \item $C,B\div A$, if $\lambda<0$. \end{itemize} \begin{figure}[!htb] \centering \input{sl.vek.5.2.1.pic} \caption{} \label{sl.vek.5.2.1.pic} \end{figure} It is clear that the point $C$ is uniquely determined for each $\lambda\neq 0$, and the vector $\lambda\cdot \overrightarrow{v}$ does not depend on the choice of point $A$. This means that the definition of multiplying a vector by a real number is correct. From the definition itself it follows that for every vector $\overrightarrow{v}$ and every real number $\lambda\in \mathbb{R}$ it holds that $|\lambda \cdot \overrightarrow{v}|=|\lambda| \cdot |\overrightarrow{v}|$. Multiplying the vector $\overrightarrow{v}$ by the real number $\lambda$ will also be called \index{množenje vektorja s skalarjem}\pojem{multiplying the vector $\overrightarrow{v}$ by the scalar $\lambda$}. Similarly to how we omit the multiplication sign $\cdot$ when multiplying algebraic expressions, we will also omit the multiplication sign $\cdot$ when multiplying a vector by a real number, or rather than writing $\lambda\cdot\overrightarrow{v}$ we will just write $\lambda\overrightarrow{v}$. The next theorem gives us the necessary and sufficient condition for the collinearity of two vectors. \bizrek \label{vektKriterijKolin} Vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ ($\overrightarrow{v},\overrightarrow{u}\neq \overrightarrow{0}$) are collinear if and only if there is such $\lambda\in \mathbb{R}$, that is $\overrightarrow{u}=\lambda\cdot\overrightarrow{v}$. \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.2.2.pic} \caption{} \label{sl.vek.5.2.2.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.2.pic}) Let $\overrightarrow{v}$ and $\overrightarrow{u}$ be any vectors ($\overrightarrow{v},\overrightarrow{u}\neq \overrightarrow{0}$), $A$ any point, and $B$ and $C$ such points that $\overrightarrow{AB}=\overrightarrow{v}$ and $\overrightarrow{AC}=\overrightarrow{u}$ (statement \ref{vektAvObst1TockaB}). ($\Leftarrow$) First, let us assume that $\overrightarrow{u}=\lambda\cdot\overrightarrow{v}$ for some $\lambda\in \mathbb{R}$. By definition, point $C$ lies on the line $AB$, which means that vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$, or $\overrightarrow{v}$ and $\overrightarrow{u}$, are collinear. ($\Rightarrow$) Now let us assume that vectors $\overrightarrow{v}$ and $\overrightarrow{u}$, or $\overrightarrow{AB}$ and $\overrightarrow{AC}$, are collinear. In this case: \begin{eqnarray*} \overrightarrow{u}&=&\overrightarrow{AC}=\frac{|AC|}{|AB|}\cdot \overrightarrow{AB}= \frac{|\overrightarrow{u}|}{|\overrightarrow{v}|}\cdot \overrightarrow{v}, \hspace*{2mm} \textrm{ if } C,B\ddot{-} A;\\ \overrightarrow{u}&=&\overrightarrow{AC}=-\frac{|AC|}{|AB|}\cdot \overrightarrow{AB}= -\frac{|\overrightarrow{u}|}{|\overrightarrow{v}|}\cdot \overrightarrow{v}, \hspace*{2mm} \textrm{ if } C,B\div A, \end{eqnarray*} which is what needed to be proven. \kdokaz If we combine the operation of adding vectors and multiplying a vector by a scalar, we get what is called a linear combination of vectors. More precisely - for any $n$-tuple of vectors $(\overrightarrow{a_1},\overrightarrow{a_2},\ldots,\overrightarrow{a_n})$ and any $n$-tuple of real numbers $(\alpha_1,\alpha_2,\ldots,\alpha_n)\in \mathbb{R}^n$ the vector $$\overrightarrow{v}=\alpha_1\cdot \overrightarrow{a_1}+ \alpha_2\cdot \overrightarrow{a_2} +\cdots +\alpha_n\cdot \overrightarrow{a_n}$$ \index{linearna kombinacija vektorjev}\pojem{linearna kombinacija vektorjev} $\overrightarrow{a_1}$, $\overrightarrow{a_2}$, $\ldots$, $\overrightarrow{a_n}$. \bizrek \label{vektLinKombNicLema} Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two non-collinear non-zero vectors and $(\alpha,\beta)\in \mathbb{R}^2$ two real numbers. If $\alpha\cdot\overrightarrow{a}= \beta\cdot\overrightarrow{b}$, then $\alpha=\beta=0$. \eizrek \textbf{\textit{Proof.}} We assume the opposite - without loss of generality, that $\alpha\neq0$. Then $\overrightarrow{a}= \frac{\beta}{\alpha}\cdot\overrightarrow{b}$, which means (statement \ref{vektKriterijKolin}), that the vector $\overrightarrow{b}$ is collinear with the vector $\overrightarrow{a}$. This is in contradiction with the assumption, so $\alpha=\beta=0$. \kdokaz A direct consequence of the previous statement is the following claim. \bizrek \label{vektLinKombNic} If for the linear combination of two non-collinear non-zero vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ holds $$\alpha\cdot\overrightarrow{a}+ \beta\cdot\overrightarrow{b}=\overrightarrow{0},$$ then $\alpha=\beta=0$. \eizrek \textbf{\textit{Proof.}} If we write the relation $\alpha\cdot\overrightarrow{a}+ \beta\cdot\overrightarrow{b}=\overrightarrow{0}$ in the form $\alpha\cdot\overrightarrow{a}= -\beta\cdot\overrightarrow{b}$, we see that the statement is a direct consequence of the previous expression. \kdokaz The following expression is very important, which refers to the representation of any vector as a linear combination of two nonlinear non-zero vectors. \bizrek \label{vektLinKomb1Razcep} Let $\overrightarrow{a}$ and $\overrightarrow{b}$ be two non-collinear non-zero vectors in the same plane. Each vector $\overrightarrow{v}$ in this plane can be express in a single way as a linear combination of vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, i.e. there is exactly one pair of real numbers $(\alpha,\beta)\in \mathbb{R}^2$, such that $$\overrightarrow{v}=\alpha\cdot\overrightarrow{a}+ \beta\cdot\overrightarrow{b}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.2.3.pic} \caption{} \label{sl.vek.5.2.3.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.3.pic}) Let $O$ be an arbitrary point and $A$, $B$ and $V$ such points that $\overrightarrow{OA}=\overrightarrow{a}$, $\overrightarrow{OB}=\overrightarrow{b}$ and $\overrightarrow{OV}=\overrightarrow{v}$ (expression \ref{vektAvObst1TockaB}). We mark with $A'$ the intersection of the line $OA$ and the parallel of the line $OB$ through the point $V$ and with $B'$ the intersection of the line $OB$ and the parallel of the line $OA$ through the point $V$. So the quadrilateral $OA'VB'$ is a parallelogram. By the expression \ref{vektKriterijKolin} there exist such $\alpha,\beta\in \mathbb{R}$, that $\overrightarrow{OA'}= \alpha\cdot\overrightarrow{OA}= \alpha\cdot\overrightarrow{a}$ and $\overrightarrow{OB'}= \beta\cdot\overrightarrow{OB}= \beta\cdot\overrightarrow{b}$. By the parallelogram rule it is: $$\overrightarrow{v}= \overrightarrow{OV}=\overrightarrow{OA'}+\overrightarrow{OB'}= \alpha\cdot\overrightarrow{a}+\beta\cdot\overrightarrow{b}.$$ We will prove that $(\alpha,\beta)\in \mathbb{R}^2$ is the only such pair. We assume that for $(\alpha',\beta')\in \mathbb{R}^2$ it is true that $\overrightarrow{v}= \alpha'\cdot\overrightarrow{a}+\beta'\cdot\overrightarrow{b}$. From $\overrightarrow{v}= \alpha'\cdot\overrightarrow{a}+\beta'\cdot\overrightarrow{b}= \alpha\cdot\overrightarrow{a}+\beta\cdot\overrightarrow{b}$ it follows that $(\alpha'-\alpha)\cdot\overrightarrow{a}+(\beta'-\beta)\cdot\overrightarrow{b}= \overrightarrow{0}$. According to the previous statement \ref{vektLinKombNic} we have that $\alpha'-\alpha=\beta'-\beta=0$ which means $\alpha'=\alpha$ and $\beta'=\beta$. \kdokaz \bizrek \label{vektVektorskiProstor} The set $\mathcal{V}$ of all vectors in the plane form so-called \index{vector space} vector space \footnote{ The first one to, in a modern way, define the concept of a vector space in 1888 was the Italian mathematician \index{Peano, G.}\textit{G. Peano} (1858–-1932).} over the field $\mathbb{R}$, which means that: \begin{enumerate} \item the ordered pair $(\mathcal{V},+)$ form a commutative group, \item $(\forall \alpha \in \mathbb{R})(\forall \overrightarrow{v}\in \mathcal{V})\hspace*{1mm} \alpha\cdot \overrightarrow{v} \in \mathcal{V}$, \item $(\forall \alpha,\beta \in \mathbb{R})(\forall \overrightarrow{v}\in \mathcal{V})\hspace*{1mm} \alpha\cdot (\beta\cdot\overrightarrow{v})=(\alpha\beta)\cdot \overrightarrow{v}$, \item $(\forall \overrightarrow{v}\in \mathcal{V})\hspace*{1mm} 1\cdot \overrightarrow{v}=\overrightarrow{v}$, \item $(\forall \alpha,\beta \in \mathbb{R})(\forall \overrightarrow{v}\in \mathcal{V})\hspace*{1mm} (\alpha+ \beta)\cdot\overrightarrow{v}=\alpha\cdot \overrightarrow{v}+\beta\cdot \overrightarrow{v}$, \item $(\forall \alpha \in \mathbb{R})(\forall \overrightarrow{v},\overrightarrow{u}\in \mathcal{V})\hspace*{1mm} \alpha\cdot(\overrightarrow{v}+\overrightarrow{u})= \alpha\cdot \overrightarrow{v}+\alpha\cdot \overrightarrow{u}$. \end{enumerate} \eizrek \textbf{\textit{Proof.}} The claim $\textit{1}$ is, in a sense, already proven in \ref{vektKomGrupa}. The claims $\textit{2}-\textit{5}$ are a direct consequence of the definition of multiplication of a vector with a scalar. The proof of claim $\textit{6}$ is not as simple - in the proof we would have to use Dedekind's axiom of continuity \ref{aksDed}. \kdokaz We will use the notation $\mathcal{\overrightarrow{V}}^2$ for the vector space structure on the set of all vectors of the plane $\mathcal{V}$ over the field $\mathbb{R}$. If we were to consider vectors in Euclidean space, we would also get a vector space. But we can also consider a vector space more abstractly as an arbitrary ordered pair $(\mathcal{\overrightarrow{V}},\mathcal{F})$, because $(\mathcal{\overrightarrow{V}},+)$ represents a commutative group, $(\mathcal{F},+,\cdot)$ is a field, and all conditions $\textit{2}-\textit{6}$ from the previous theorem are satisfied. Because every vector in the plane can be expressed as a linear combination of two non-zero vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ of that plane, we say that such two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are called the \index{baza vektorskega prostora}\pojem{baza} (base) of the vector space $\mathcal{\overrightarrow{V}}^2$. If for some vector $\overrightarrow{v}\in \mathcal{V}$ it holds that $\overrightarrow{v}=\alpha\cdot\overrightarrow{a}+ \beta\cdot\overrightarrow{b}$, we say that the pair $(\alpha,\beta)$ represents the \index{koordinate vektorja}\pojem{koordinate} (coordinates) of the vector $\overrightarrow{v}$ in the base $(\overrightarrow{a},\overrightarrow{b})$. From theorem \ref{vektLinKomb1Razcep} it follows that coordinates of every vector are uniquely determined in any base. Therefore, there are infinitely many bases in the vector space $\mathcal{\overrightarrow{V}}^2$, the number of vectors in any base is always 2, so we say that the vector space $\mathcal{\overrightarrow{V}}^2$ has \pojem{dimension} 2 or is \pojem{two-dimensional}. It is clear that the dimension of the vector space $\mathcal{\overrightarrow{V}}^3$, which is determined by the vectors in Euclidean space, is equal to 3. Euclidean space is therefore \pojem{three-dimensional}, each vector in some base is determined by a triplet of real numbers, which represent the coordinates of that vector. In any vector space we determine the base and dimension in a similar way. Because in general the dimension of space is not limited to 3, this allows us to explore \pojem{$n$-dimensional} Euclidean spaces (for any $n\in \mathbb{N}$). The following examples are very useful. \bzgled \label{vektSredOSOAOB} Let $O$, $A$ and $B$ be arbitrary points and $S$ the midpoint of the line segment $AB$. Then $$\overrightarrow{OS}=\frac{1}{2}\cdot\left(\overrightarrow{OA}+ \overrightarrow{OB}\right).$$ \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.2.4.pic} \caption{} \label{sl.vek.5.2.4.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.4.pic}) By \ref{vektSredDalj} we have $\overrightarrow{SA}=-\overrightarrow{SB}$ or $\overrightarrow{AS}=-\overrightarrow{BS}$. By the triangle rule for vector addition we have $\overrightarrow{OS}=\overrightarrow{OA}+\overrightarrow{AS}$ and $\overrightarrow{OS}=\overrightarrow{OB}+\overrightarrow{BS}$. If we add the equalities and take into account $\overrightarrow{AS}=-\overrightarrow{BS}$, we get: $2\cdot\overrightarrow{OS}=\overrightarrow{OA}+\overrightarrow{AS} +\overrightarrow{OB}+\overrightarrow{BS}=\overrightarrow{OA} +\overrightarrow{OB}$ or $\overrightarrow{OS}=\frac{1}{2}\cdot\left(\overrightarrow{OA}+ \overrightarrow{OB}\right)$. \kdokaz \bzgled \label{vektDelitDaljice} Let $O$, $A$ and $B$ be arbitrary points and $P$ a point in the line segment $AB$ such that $|AP|:|PB|=n:m$. Then $$\overrightarrow{OP}=\frac{1}{n+m}\left(m\cdot\overrightarrow{OA}+ n\cdot\overrightarrow{OB}\right).$$ \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.2.5.pic} \caption{} \label{sl.vek.5.2.5.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.5.pic}) First, from $|AP|:|PB|=n:m$ it follows that $|AP|=\frac{n}{n+m}\cdot|AB|$ and $|BP|=\frac{m}{n+m}\cdot|AB|$ or $\overrightarrow{AP}=\frac{n}{n+m}\cdot\overrightarrow{AB}$ and $\overrightarrow{BP}=\frac{m}{n+m}\cdot\overrightarrow{BA}$ (because $\mathcal{B}(A,P,B)$). Then: \begin{eqnarray*} \overrightarrow{OP}&=&\overrightarrow{OA}+\overrightarrow{AP}=\overrightarrow{OA} +\frac{n}{n+m}\cdot\overrightarrow{AB}\hspace*{3mm} \textrm{ in}\\ \overrightarrow{OP}&=&\overrightarrow{OB}+\overrightarrow{BP}=\overrightarrow{OB} +\frac{m}{n+m}\cdot\overrightarrow{BA}. \end{eqnarray*} If we multiply the first equality by $m$, the second by $n$ and then add them, we get: \begin{eqnarray*} (n+m)\cdot\overrightarrow{OP}&=&m\cdot\overrightarrow{OA} +n\cdot\overrightarrow{OB}+\frac{nm}{n+m}\cdot\left(\overrightarrow{AB}+ \overrightarrow{BA}\right)=\\ &=&m\cdot\overrightarrow{OA} +n\cdot\overrightarrow{OB}. \end{eqnarray*} If we divide the obtained equality by $n+m$, we get the desired relation. \kdokaz \bzgled \label{vektParamDaljica} Let $O$, $A$ and $B$ be arbitrary points. A point $X$ lies on the line segment $AB$ if and only if for some scalar $0\leq\lambda\leq1$ is $$\overrightarrow{OX}=(1-\lambda)\cdot\overrightarrow{OA}+ \lambda\cdot\overrightarrow{OB}.$$ \ezgled \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.6.pic}) We first assume that the point $X$ lies on the line $AB$. Then for some $0\leq\lambda\leq1$ it holds that $\overrightarrow{AX}=\lambda\cdot \overrightarrow{AB}$. \begin{eqnarray*} \overrightarrow{OX}&=&\overrightarrow{OA}+\overrightarrow{AX}=\\ &=& \overrightarrow{OA}+\lambda\cdot \overrightarrow{AB}=\\ &=& \overrightarrow{OA}+\lambda\cdot \left(\overrightarrow{OB}-\overrightarrow{OA}\right)=\\ &=& (1-\lambda)\cdot\overrightarrow{OA}+ \lambda\cdot\overrightarrow{OB}. \end{eqnarray*} Assume now that for some $0\leq\lambda\leq1$ we have $\overrightarrow{OX}=(1-\lambda)\cdot\overrightarrow{OA}+ \lambda\cdot\overrightarrow{OB}$. Then: \begin{eqnarray*} \overrightarrow{AX}&=&\overrightarrow{AO}+\overrightarrow{OX}=\\ &=&\overrightarrow{AO}+(1-\lambda)\cdot\overrightarrow{OA}+ \lambda\cdot\overrightarrow{OB}=\\ &=&-\lambda\cdot\overrightarrow{OA}+ \lambda\cdot\overrightarrow{OB}=\\ &=&\lambda\cdot\left(\overrightarrow{AO}+ \overrightarrow{OB}\right)=\\ &=&\lambda\cdot\overrightarrow{AB}. \end{eqnarray*} Therefore, for some $0\leq\lambda\leq1$ we have $\overrightarrow{AX}=\lambda\cdot \overrightarrow{AB}$, which means that the point $X$ lies on the line $AB$. \kdokaz \begin{figure}[!htb] \centering \input{sl.vek.5.2.6.pic} \caption{} \label{sl.vek.5.2.6.pic} \end{figure} A direct consequence is the following theorem. $O$, $A$ in $B$ . A point $X$ lies on the line segment $AB$ if and only if for some \bzgled Let $O$, $A$ and $B$ be arbitrary points. A point $X$ lies on the line segment $AB$ if and only if for some scalars $\alpha, \beta\in [0,1]$ ($\alpha+\beta=1$) is $$\overrightarrow{OX}=\alpha\cdot\overrightarrow{OA}+ \beta\cdot\overrightarrow{OB}.$$ \ezgled In the next theorem we will give the \index{vector equation of a line} \pojem{vector equation of a line}. \bzgled \label{vektParamPremica} Let $O$, $A$ in $B$ be arbitrary points. A point $X$ lies on the line $AB$ if and only if for some scalar $\lambda\in \mathbb{R}$ is $$\overrightarrow{OX}=(1-\lambda)\cdot\overrightarrow{OA}+ \lambda\cdot\overrightarrow{OB}.$$ \ezgled \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.6.pic}) The proof is the same as in the previous statement, only that we use the fact that the point $X$ lies on the line $AB$ exactly when the vectors $\overrightarrow{AX}$ and $\overrightarrow{AB}$ are collinear, or for some $\lambda\in \mathbb{R}$ it holds that $\overrightarrow{AX}=\lambda\cdot \overrightarrow{AB}$ (statement \ref{vektKriterijKolin}). \kdokaz \index{količnik kolinearnih vektorjev}\index{razmerje!kolinearnih vektorjev} If for collinear vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ it holds that $\overrightarrow{v}=\lambda \overrightarrow{u}$ ($\lambda\in\mathbb{R}$) and $\overrightarrow{u}\neq \overrightarrow{0}$, then we can define the \pojem{količnik kolinearnih vektorjev} or the \pojem{razmerje kolinearnih vektorjev} (Figure \ref{sl.vek.5.2.7.pic}): $$\overrightarrow{v}:\overrightarrow{u} =\frac{\overrightarrow{v}}{\overrightarrow{u}}=\lambda.$$ \begin{figure}[!htb] \centering \input{sl.vek.5.2.7.pic} \caption{} \label{sl.vek.5.2.7.pic} \end{figure} From the definition of multiplication of a vector with a real number we get the following statement. \bizrek \label{vektKolicnDolz} Let $\overrightarrow{v}$ and $\overrightarrow{u}$ be collinear vectors and $\overrightarrow{u}\neq \overrightarrow{0}$. Then $$\frac{\overrightarrow{v}}{\overrightarrow{u}}= \left\{ \begin{array}{lll} \frac{|\overrightarrow{v}|}{|\overrightarrow{u}|}, & \textrm{ if } \overrightarrow{u}\rightrightarrows \overrightarrow{v}\\ -\frac{|\overrightarrow{v}|}{|\overrightarrow{u}|}, & \textrm{ if } \overrightarrow{u}\rightleftarrows\overrightarrow{v}\\ 0, & \textrm{ if } \overrightarrow{v}=\overrightarrow{0} \end{array}\right.$$ \eizrek The following statement also refers to the newly defined concept. \bizrek \label{izrekEnaDelitevDaljiceVekt} For any line segment $AB$ and any $\lambda\in\mathbb{R}\setminus\{-1\}$ there exists exactly one such point $P$ on the line $AB$, that is $$\frac{\overrightarrow{AP}}{\overrightarrow{PB}}=\lambda.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.2.8.pic} \caption{} \label{sl.vek.5.2.8.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.2.8.pic}) Let $P$ be such a point that $$\overrightarrow{AP}=\frac{\lambda}{1+\lambda}\cdot\overrightarrow{AB}.$$ Since $\lambda\neq-1$, such a point always exists. Because $$\overrightarrow{PB}=\overrightarrow{PA}+\overrightarrow{AB}= -\frac{\lambda}{1+\lambda}\cdot\overrightarrow{AB}+\overrightarrow{AB}= \frac{1}{1+\lambda}\overrightarrow{AB},$$ it also holds that $\frac{\overrightarrow{AP}}{\overrightarrow{PB}}=\lambda$. If for another point $P'$ it holds that $\frac{\overrightarrow{AP'}}{\overrightarrow{P'B}}=\lambda$, from $\overrightarrow{AP'}=\lambda\cdot\overrightarrow{P'B}$ and $\overrightarrow{P'B}=\overrightarrow{P'A}+\overrightarrow{AB}$ we get $\overrightarrow{AP'}=\frac{\lambda}{1+\lambda}\cdot\overrightarrow{AB}$. Therefore $\overrightarrow{AP'}=\overrightarrow{AP}$ or $\overrightarrow{P'P}=\overrightarrow{P'A}+\overrightarrow{AP}= -\overrightarrow{AP}+\overrightarrow{AP}=\overrightarrow{0}$. So $P'=P$, which means that there is only one point $P$, for which $\frac{\overrightarrow{AP}}{\overrightarrow{PB}}=\lambda$. \kdokaz We say that the point $P$ from the previous statement divides the line segment $AB$ in the \pojem{ratio} $\lambda$. \bzgled Let $ABCD$ be a trapezium with the base $AB$. Calculate the ratio in which the line segment $PD$ divides the diagonal $AC$, if $|AB| = 3\cdot |CD|$ and $P$ is the midpoint of the line segment $AB$. \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.2.9.pic} \caption{} \label{sl.vek.5.2.9.pic} \end{figure} \textbf{\textit{Solution.}} The intersection of the line $DP$ and the diagonal $AC$ is denoted by $S$ (Figure \ref{sl.vek.5.2.9.pic}) and let $\overrightarrow{u}=\overrightarrow{AP}$ and $\overrightarrow{v}=\overrightarrow{AD}$. The vector $\overrightarrow{AS}$ and $\overrightarrow{AC}$ are collinear, so by \ref{vektKriterijKolin} $\overrightarrow{AS}=\lambda\overrightarrow{AC}$, for some $\lambda\in \mathbb{R}$. The vectors $\overrightarrow{PS}$ and $\overrightarrow{PD}$ are also collinear, i.e. for some $\mu\in \mathbb{R}$ we have $\overrightarrow{PS}=\mu\overrightarrow{PD}$. The vector $\overrightarrow{AC}$ and $\overrightarrow{AS}$ are written as a linear combination of the vectors $\overrightarrow{u}$ and $\overrightarrow{v}$: \begin{eqnarray*} \hspace*{-1.8mm} \overrightarrow{AC}&=&\overrightarrow{AD}+\overrightarrow{DC} =\overrightarrow{v}+\frac{2}{3}\overrightarrow{u} =\frac{2}{3}\overrightarrow{u}+\overrightarrow{v},\\ \hspace*{-1.8mm}\overrightarrow{AS}&=&\overrightarrow{AP}+\overrightarrow{PS} =\overrightarrow{u}+\mu\overrightarrow{PD} =\overrightarrow{u}+\mu(-\overrightarrow{u}+\overrightarrow{v}) =(1-\mu)\overrightarrow{u}+\mu\overrightarrow{v}. \end{eqnarray*} Since $\overrightarrow{AS}=\lambda\overrightarrow{AC}$, we get: \begin{eqnarray*} \overrightarrow{AS}&=& \frac{2}{3}\lambda\overrightarrow{u}+\lambda\overrightarrow{v};\\ \overrightarrow{AS}&=& (1-\mu)\overrightarrow{u}+\mu\overrightarrow{v}. \end{eqnarray*} Because $\overrightarrow{u}$ and $\overrightarrow{v}$ are collinear vectors, by \ref{vektLinKomb1Razcep} $\frac{2}{3}\lambda=1-\mu$ and $\lambda=\mu$. If we solve this simple system of equations, we get $\lambda=\mu=\frac{3}{5}$. Therefore $\overrightarrow{AS}=\frac{3}{5}\overrightarrow{AC}$, i.e. $AS:SC=3:2$. \kdokaz %________________________________________________________________________________ \poglavje{Vector Length} \label{odd5DolzVekt} We have already defined the length of a vector in section \ref{odd5DefVekt}. From the definition of multiplication of a vector with a scalar in the previous section \ref{odd5LinKombVekt} we saw that for every vector $\overrightarrow{v}$ and every real number $\lambda$ it holds that $|\lambda \cdot \overrightarrow{v}|=|\lambda| \cdot |\overrightarrow{v}|$. In this section we will discuss some more properties of the length of a vector. First, from the definition of addition of vectors and the triangle inequality we get the following claim. \bizrek \label{neenakTrikVekt} For any two vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ it holds that $$|\overrightarrow{v}+\overrightarrow{u}|\leq |\overrightarrow{v}|+|\overrightarrow{u}|.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.1.7.pic} \caption{} \label{sl.vek.5.1.7.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.1.7.pic}) Let $A$ be an arbitrary point and let $B$ and $C$ be such points that $\overrightarrow{AB}=\overrightarrow{v}$ and $\overrightarrow{BC}=\overrightarrow{u}$ (claim \ref{vektAvObst1TockaB}). From the definition of addition of vectors and the length of a vector and the triangle inequality (claim \ref{neenaktrik}) we get: \begin{eqnarray*} |\overrightarrow{v}+\overrightarrow{u}|&=&|\overrightarrow{AB}+\overrightarrow{BC}|=\\ &=&|\overrightarrow{AC}|=|AC|\leq |AB|+|AC|= |\overrightarrow{AB}|+|\overrightarrow{AC}|=\\ &=&|\overrightarrow{v}|+|\overrightarrow{u}|, \end{eqnarray*} which was to be proven. \kdokaz In a similar way as we can define a fraction, we can also define the product of collinear vectors. If $\overrightarrow{v}$ and $\overrightarrow{u}$ are two collinear vectors, we define the operation \pojem{multiplication of two collinear vectors}\footnote{This is a special case of the so-called \textit{dot product}, which in linear algebra is defined for any two vectors. In Euclidean space, $\overrightarrow{v}\cdot\overrightarrow{u}= |\overrightarrow{v}|\cdot|\overrightarrow{u}|\cdot\cos \angle \overrightarrow{v},\overrightarrow{u}$.}: \begin{eqnarray*} \overrightarrow{v}\cdot \overrightarrow{u}= \left\{ \begin{array}{ll} |\overrightarrow{v}|\cdot|\overrightarrow{u}|, & \overrightarrow{v},\overrightarrow{u}\rightrightarrows ; \\ -|\overrightarrow{v}|\cdot|\overrightarrow{u}|, & \overrightarrow{v},\overrightarrow{u}\rightleftarrows. \end{array} \right. \end{eqnarray*} The result of this operation, which is a real number, is called \index{product of collinear vectors} \pojem{product of collinear vectors}. It is clear from the definition that: \begin{eqnarray} \label{eqnMnozVektDolzina} \overrightarrow{v}\cdot \overrightarrow{v}=|\overrightarrow{v}|^2 \end{eqnarray} It is also clear that for three collinear points $A$, $B$ and $L$ the following equivalence holds (Figure \ref{sl.vek.5.3.1.pic}): \begin{eqnarray} \label{eqnMnozVektRelacijaB} \overrightarrow{LA}\cdot \overrightarrow{LB}<0 \hspace*{1mm}\Leftrightarrow\hspace*{1mm} \mathcal{B}(A,L,B) \hspace*{3mm} \textrm{(}A,B,L \textrm{ are collinear)}. \end{eqnarray} \begin{figure}[!htb] \centering \input{sl.vek.5.3.1.pic} \caption{} \label{sl.vek.5.3.1.pic} \end{figure} \bnaloga\footnote{15. IMO USSR - 1973, Problem 1.} Point $O$ lies on line $g$; $\overrightarrow{OP_1},\overrightarrow{OP_2}, \cdots,\overrightarrow{OP_n}$ are unit vectors such that points $P_1, P_2, \cdots, P_n$ all lie in a plane containing $g$ and on one side of $g$. Prove that if $n$ is odd, $$|\overrightarrow{OP_1}+\overrightarrow{OP_2}+ \cdots+\overrightarrow{OP_n}|\geq 1.$$ \label{OlimpVekt15} \enaloga \begin{figure}[!htb] \centering \input{sl.vek.5.3.IMO1.pic} \caption{} \label{sl.vek.5.3.IMO1.pic} \end{figure} \textbf{\textit{Solution.}} Let $k$ be a unit circle with center at point $O$, which intersects the line $g$ at points $A$ and $B$. All points $P_1, P_2, \cdots, P_n$ lie on the corresponding arc, which on the circle $k$ is determined by points $A$ and $B$. Without loss of generality we can assume that the vectors are labeled so that it holds: $\angle BOP_1\leq \angle BOP_2\leq\cdots\leq \angle BOP_n$ (Figure \ref{sl.vek.5.3.IMO1.pic}). Let $s$ be the bisector of angle $\angle P_1OP_n$ and $g'$ the perpendicular of the line $s$ at point $O$. Because $\angle P_1OP_n$ is a convex angle, the whole lies on the same side of the line $g'$. So are also all points $P_i$ ($i\in \{1,2,\cdots,n\}$ on the same side of the line $g'$, because they lie in this angle. So we can sharpen our assumption from the task, that all points $P_1, P_2, \cdots, P_n$ are on the same side of some line $g$ or $g'$, with the requirement (which follows from this assumption), that the first and last vector of the sequence are symmetric with respect to the perpendicular $s$ of the line $g'$ through point $O$. We will use this fact in the proof. By assumption $n$ is an odd number. Let $n=2k-1$, $k\in \mathbb{N}$. We will prove by induction on $k$. (\textit{i}) If $k=1$ or $n=1$, then $|\overrightarrow{OP_1}|=1$ or $|\overrightarrow{OP_1}|\geq1$ and the statement is fulfilled. (\textit{ii}) Let's assume that the statement is true for $k=l$ or for every sequence $n=2l-1$ of vectors. We will prove that the statement is also true for $k=l+1$ or in the case of a sequence $n=2l+1$ of vectors. Let $\overrightarrow{OP_1},\overrightarrow{OP_2}, \cdots,\overrightarrow{OP_{2l+1}}$ be unit vectors, where $P_1, P_2, \cdots, P_{2l+1}$ are points that lie in the same plane with the edge $g'$ (in the plane $\alpha$) and the vectors $\overrightarrow{OP_1}$ and $\overrightarrow{OP_{2l+1}}$ are symmetrical with respect to the line $s$. From this it follows that the vector $\overrightarrow{OU}=\overrightarrow{OP_1}+\overrightarrow{OP_{2l+1}}$ lies on the line $s$ or $U\in s$. Let $\overrightarrow{OV}=\overrightarrow{OP_2}+\overrightarrow{OP_3}+ \cdots+\overrightarrow{OP_{2l}}$. By the induction assumption $|\overrightarrow{OV}|=|\overrightarrow{OP_2}+\overrightarrow{OP_3}+ \cdots+\overrightarrow{OP_{2l}}|\geq 1$. Therefore: \begin{eqnarray*} |\overrightarrow{OP_1}+\overbrace{\overrightarrow{OP_2}+ \cdots+\overrightarrow{OP_{2l}}}+\overrightarrow{OP_{2l+1}}|= |\overrightarrow{OU}+\overrightarrow{OV}|. \end{eqnarray*} Let $\overrightarrow{OW}=\overrightarrow{OU}+\overrightarrow{OV}$. The quadrilateral $VOUW$ is a parallelogram (or the points $O$, $U$, $V$ and $W$ are collinear). By the example \ref{vektOPi} the points $U$ and $V$ are in the plane $\alpha$. The point $U$ lies on the line $s$, which is the line of symmetry of the obtuse angle, which is determined by the line $g'$ and the point $O$. Therefore $0\leq\angle UOV\leq 90^0$. From the parallelogram $VOUW$ it then follows that $\angle OVW > 90^0$. From the triangle inequality \ref{neenaktrik} (for the triangle $OVW$) it follows that $|\overrightarrow{OW}|>|\overrightarrow{OV}|$. The inequality is also valid in the case when the points $O$, $U$, $V$ and $W$ are collinear. Now we have: \begin{eqnarray*} |\overrightarrow{OP_1}+\overrightarrow{OP_2}+ \cdots+\overrightarrow{OP_{2l}}+\overrightarrow{OP_{2l+1}}|= |\overrightarrow{OU}+\overrightarrow{OV}|= |\overrightarrow{OW}|>|\overrightarrow{OV}|\geq 1, \end{eqnarray*} which had to be proven. \kdokaz %________________________________________________________________________________ \poglavje{Further Use of Vectors} \label{odd5UporabVekt} First, let's look at a statement that is a direct consequence of the definition of vectors (section \ref{odd5DefVekt}). \bizrek \label{vektParalelogram} A quadrilateral $ABCD$ is a parallelogram if and only if $\overrightarrow{AB}=\overrightarrow{DC}$. \eizrek We should also mention that we can already prove the property of the median of a triangle - statement \ref{srednjicaTrik} - now in the following form. \bizrek \label{srednjicaTrikVekt} Let $PQ$ be the midsegment of a triangle $ABC$, corresponding to the side $BC$. Then $$ \overrightarrow{PQ} = \frac{1}{2} \overrightarrow{AB}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.4.1c.pic} \caption{} \label{sl.vek.5.4.1c.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.4.1c.pic}) The statement is a direct consequence of statement \ref{srednjicaTrik} \kdokaz We have seen that we can add vectors according to the parallelogram rule (when they have a common starting point) or according to the triangle rule (when the starting point of the second one is at the end of the first one). Now we will introduce another rule for adding two vectors that are in any position. This is called the \index{pravilo!splošno za seštevanje vektorjev}\pojem{general vector addition rule}. \bizrek \label{vektSestSplosno} If $S_1$ and $S_2$ are the midpoints of line segments $A_1B_1$ and $A_2B_2$, then $$\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}= 2\overrightarrow{S_1S_2}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.4.1b.pic} \caption{} \label{sl.vek.5.4.1b.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.4.1b.pic}) Since $S_1$ and $S_2$ are the midpoints of the lines $A_1B_1$ and $A_2B_2$, $\overrightarrow{S_1A_1}=-\overrightarrow{S_1B_1}$ and $\overrightarrow{S_2A_2}=-\overrightarrow{S_2B_2}$ or $\overrightarrow{S_1A_1}+\overrightarrow{S_1B_1}=\overrightarrow{0}$ and $\overrightarrow{S_2A_2}+\overrightarrow{S_2B_2}=\overrightarrow{0}$. If we decompose the vector $\overrightarrow{S_1S_2}$ in two ways using the polygonal rule for adding vectors, we first get: \begin{eqnarray*} \overrightarrow{S_1S_2}&=&\overrightarrow{S_1A_1}+\overrightarrow{A_1A_2}+ \overrightarrow{A_2S_2}\\ \overrightarrow{S_1S_2}&=&\overrightarrow{S_1B_1}+\overrightarrow{B_1B_2}+ \overrightarrow{B_2S_2}, \end{eqnarray*} then by adding these: \begin{eqnarray*} 2\cdot\overrightarrow{S_1S_2}&=&\overrightarrow{S_1A_1}+\overrightarrow{A_1A_2}+ \overrightarrow{A_2S_2}+\\ &+&\overrightarrow{S_1B_1}+\overrightarrow{B_1B_2}+ \overrightarrow{B_2S_2}=\\ &=& \overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}, \end{eqnarray*} which had to be proven. \kdokaz We will often use the relation from the previous statement in the form: $$\overrightarrow{S_1S_2}= \frac{1}{2}(\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}),$$ which is a generalization of the property of the median of a trapezoid and a triangle. At this point, we will repeat all three rules for adding two vectors. In the first case, we therefore choose representatives of the vectors so that the beginning of the second is at the end of the first, in the second case the vectors have a common starting point, in the third case the representatives are in general position (Figure \ref{sl.vek.5.4.1a.pic}): \begin{itemize} \item \textit{for every three points $A$, $B$ and $C$ it holds that $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ (the triangle rule),} \item \textit{for every three non-collinear points $A$, $B$ and $C$ it holds that $\overrightarrow{AB}+\overrightarrow{AC}=\overrightarrow{AD}$ exactly when $ABDC$ is a parallelogram (the parallelogram rule),} \item \textit{if $S_1$ and $S_2$ are the midpoints of the line segments $A_1B_1$ and $A_2B_2$, then $\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}= 2\overrightarrow{S_1S_2}$ (the general rule).} \end{itemize} \begin{figure}[!htb] \centering \input{sl.vek.5.4.1a.pic} \caption{} \label{sl.vek.5.4.1a.pic} \end{figure} Except for these three, we also have the polygon rule, which is related to the addition of multiple vectors (Figure \ref{sl.vek.5.1.12.pic}): \begin{itemize} \item \textit{$\overrightarrow{A_1A_2}+\overrightarrow{A_2A_3}+\cdots +\overrightarrow{A_{n-1}A_n}=\overrightarrow{A_1A_n}$ (the polygon rule).} \end{itemize} \begin{figure}[!htb] \centering \input{sl.vek.5.1.12.pic} \caption{} \label{sl.vek.5.1.12.pic} \end{figure} \bzgled \label{vektPetkoinikZgled} Points $M$, $N$, $P$ and $Q$ are the midpoints of the sides $AB$, $BC$, $CD$ and $DE$ of a pentagon $ABCDE$. Prove that the line segment $XY$, determined by the midpoints of the line segments $MP$ and $NQ$, is parallel to the line $AE$ and calculate $\frac{\overrightarrow{XY}}{\overrightarrow{AE}}$. \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.4.2.pic} \caption{} \label{sl.vek.5.4.2.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.4.2.pic}) If we use the expressions \ref{vektSestSplosno} and \ref{srednjicaTrikVekt} we get: \begin{eqnarray*} \overrightarrow{XY}&=&\frac{1}{2}\left(\overrightarrow{MQ}+\overrightarrow{PN} \right)=\\ &=&\frac{1}{2}\left(\frac{1}{2}\left(\overrightarrow{AE}+\overrightarrow{BD}\right) +\frac{1}{2}\overrightarrow{DB} \right)=\\ &=&\frac{1}{4}\overrightarrow{AE} \end{eqnarray*} So the vectors $\overrightarrow{XY}$ and $\overrightarrow{AE}$ are collinear and $\frac{\overrightarrow{XY}}{\overrightarrow{AE}}=\frac{1}{4}$. \kdokaz \bzgled Let $O$ be an arbitrary point in the plane of a triangle $ABC$ and $D$ and $E$ points of the sides $AB$ and $BC$ such that $$\frac{\overrightarrow{AD}}{\overrightarrow{DB}}= \frac{\overrightarrow{BE}}{\overrightarrow{EC}}=\frac{m}{n}.$$ Let $F$ be the intersection of the line segments $AE$ and $CD$. Express $\overrightarrow{OF}$ as a function of $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$, $m$ and $n$. \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.4.3.pic} \caption{} \label{sl.vek.5.4.3.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.4.3.pic}) It is enough to express the vector $\overrightarrow{BF}$ as a linear combination of the vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$, because: \begin{eqnarray*} \overrightarrow{OF}&=&\overrightarrow{OB}+\overrightarrow{BF}\\ \overrightarrow{BA}&=&\overrightarrow{OA}-\overrightarrow{OB}\\ \overrightarrow{BC}&=&\overrightarrow{OC}-\overrightarrow{OB}. \end{eqnarray*} By \ref{vektParamPremica} we have: \begin{eqnarray*} \overrightarrow{BF}&=&\lambda\overrightarrow{BD}+(1-\lambda)\overrightarrow{BC}= \lambda\frac{n}{n+m}\overrightarrow{BA}+(1-\lambda)\overrightarrow{BC};\\ \overrightarrow{BF}&=&\mu\overrightarrow{BA}+(1-\mu)\overrightarrow{BE}= \mu\overrightarrow{BA}+(1-\mu)\frac{m}{n+m}\overrightarrow{BC} \end{eqnarray*} for some numbers $\lambda,\mu\in \mathbb{R}$. Because the vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$ are nonlinear, by \ref{vektLinKomb1Razcep} we obtain the system: \begin{eqnarray*} & & \lambda\frac{n}{n+m}=\mu\\ & & 1-\lambda=(1-\mu)\frac{m}{n+m}, \end{eqnarray*} which we solve for $\lambda$ and $\mu$ in terms of $m$ and $n$. It is enough to calculate only $\lambda$ and plug it in $\overrightarrow{BF}= \lambda\frac{n}{n+m}\overrightarrow{BA}+(1-\lambda)\overrightarrow{BC}$. \kdokaz \bzgled Let $A_1$, $A_2$, ..., $A_n$ be points on a line $p$ and $B_1$, $B_2$, ..., $B_n$ points on a line $q$ ($n\geq 3$), such that $$\overrightarrow{A_1A_2}: \overrightarrow{A_2A_3}:\cdots: \overrightarrow{A_{n-1}A_n}= \overrightarrow{B_1B_2}: \overrightarrow{B_2B_3}:\cdots: \overrightarrow{B_{n-1}B_n}.$$ Prove that the midpoints of line segments $A_1B_1$, $A_2B_2$, ..., $A_nB_n$ lie on one line. \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.4.4.pic} \caption{} \label{sl.vek.5.4.4.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.4.4.pic}) It is enough to prove that three arbitrary centers are collinear points. Without loss of generality, we prove only that points $S_1$, $S_2$, $S_3$ are collinear. Let: $$\frac{\overrightarrow{A_1A_2}}{\overrightarrow{A_2A_3}}= \frac{\overrightarrow{B_1B_2}}{\overrightarrow{B_2B_3}}=\lambda.$$ By \ref{vektSestSplosno} is: \begin{eqnarray*} \overrightarrow{S_1S_2}&=&\frac{1}{2}\left(\overrightarrow{A_1A_2}+ \overrightarrow{B_1B_2}\right)=\\ &=&\frac{1}{2}\left(\lambda\overrightarrow{A_2A_3}+ \lambda\overrightarrow{B_2B_3}\right)=\\ &=&\frac{\lambda}{2}\overrightarrow{S_2S_3}, \end{eqnarray*} which means that $\overrightarrow{S_1S_2}$ and $\overrightarrow{S_2S_3}$ are collinear vectors (\ref{vektKriterijKolin}), so $S_1$, $S_2$ and $S_3$ are collinear points. \kdokaz %________________________________________________________________________________ \poglavje{Centroid of a Polygon With Respect to Its Vertices} \label{odd5TezVeck} Now we will generalize the concept of the centroid of a triangle to an arbitrary polygon. In section \ref{odd3ZnamTock} we defined the concept of the centroid of a triangle. The next theorem relates to one additional property of this concept, which is related to the concept of a vector. \bizrek \label{tezTrikVekt} If $T$ is the centroid of a triangle $ABC$, then $$\overrightarrow{TA}+\overrightarrow{TB}+ \overrightarrow{TC}=\overrightarrow{0}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.5.1.pic} \caption{} \label{sl.vek.5.5.1.pic} \end{figure} \textbf{\textit{Proof.}} We mark with $A_1$ the center of the side $BC$ (Figure \ref{sl.vek.5.5.1.pic}). By izreku \ref{vektSredOSOAOB} is $\overrightarrow{TA_1}= \frac{1}{2}\left(\overrightarrow{TB}+\overrightarrow{TC}\right)$, by izreku \ref{tezisce} for the centroid $T$ of the triangle $ABC$ is valid $|AT|:|TA_1|=2:1$ or $\overrightarrow{TA}=-2\cdot\overrightarrow{TA_1}$. So: $$\overrightarrow{TA}+\overrightarrow{TB}+ \overrightarrow{TC}=\overrightarrow{TA}+2\cdot\overrightarrow{TA_1}= \overrightarrow{TA}-\overrightarrow{TA}= \overrightarrow{0},$$ which had to be proven. \kdokaz It is also valid the converse statement: \bizrek \label{tezTrikVektObr} Let $A$, $B$ and $C$ be three non-collinear points. If $X$ is a point such that $$\overrightarrow{XA}+\overrightarrow{XB}+ \overrightarrow{XC}=\overrightarrow{0},$$ then $X$ is the centroid of the triangle $ABC$. \eizrek \textbf{\textit{Proof.}} Let $T$ be the centroid of the triangle. By the previous izreku \ref{tezTrikVekt} is $\overrightarrow{TA}+\overrightarrow{TB}+ \overrightarrow{TC}=\overrightarrow{0}$. By the assumption is also $\overrightarrow{XA}+\overrightarrow{XB}+ \overrightarrow{XC}=\overrightarrow{0}$. If we subtract the two equalities, we get $3 \cdot \overrightarrow{TX}=\overrightarrow{0}$ or $X=T$. \kdokaz The proven property of the centroid of a triangle gives us an idea for the definition of the centroid of an arbitrary polygon (Figure \ref{sl.vek.5.5.2.pic}). \begin{figure}[!htb] \centering \input{sl.vek.5.5.2.pic} \caption{} \label{sl.vek.5.5.2.pic} \end{figure} A point $T$ is the \index{center of mass!polygon}\pojem{center of mass of a polygon $A_1A_2\ldots A_n$ with respect to its vertices}, if the following is true: $$\overrightarrow{TA_1}+\overrightarrow{TA_2}+\cdots +\overrightarrow{TA_n}=\overrightarrow{0}.$$ The previous relation can also be written as: $$\sum_{k=1}^n\overrightarrow{TA_k}=\overrightarrow{0}.$$ We also mention that the center of mass of any figure $\Phi$ in a plane is defined as a point $T$, for which the following is true: $$\sum_{X\in \Phi}\overrightarrow{TX}=\overrightarrow{0}.$$ In the case of our polygon $A_1A_2\ldots A_n$, we actually found the center of mass of a figure that represents the union of all the vertices of this polygon $\{A_1,A_2,\ldots, A_n\}$. That is why we emphasized that this is the center of mass of a polygon with respect to its vertices. In addition, we could talk about the \pojem{center of mass of a polygon with respect to all its points}, which would more accurately match the general definition of the center of mass of any figure. If we consider the center of mass of a figure in a physical sense as - the center of mass - the first variant of the center of mass represents the center of mass, where all the mass is in the vertices and each vertex has the same mass. In the second case, this is the center of mass of the polygon, where the mass is evenly distributed throughout its interior\footnote{\index{Archimedes} \textit{Archimedes of Syracuse} (3rd century BC) was a Greek mathematician who first created the concept of the center of mass, which he used in many of his writings on mechanics, but we can only guess what he had in mind when he considered the center of mass, because none of his surviving writings contain a clear definition of the concept. The center of mass as the center of mass in a physical sense played an important role in Newton's (\index{Newton, I.}\textit{I. Newton} (1643-1727), English physicist and mathematician) mechanics, where larger bodies are often considered as points with a certain mass.}. If we talk about a general polygon, the aforementioned centroids are only equal in any triangle, already in any quadrilateral the centroids differ. In the following we will only consider the centroid of the polygon with respect to its vertices, so we will just call it the \pojem{centroid of the polygon}. First we will consider the centroid of a quadrilateral. \bizrek The centroid of a parallelogram $ABCD$ is its circumcentre $S$, i.e. the intersection of its diagonals. \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.5.3.pic} \caption{} \label{sl.vek.5.5.3.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.5.3.pic}) By izreku \ref{paralelogram} is the point $S$ the common centre of its diagonals $AC$ and $BD$, so from izreku \ref{vektSredDalj} it follows that $\overrightarrow{SA}+\overrightarrow{SC}=\overrightarrow{0}$ and $\overrightarrow{SB}+\overrightarrow{SD}=\overrightarrow{0}$. Therefore it holds: $$\overrightarrow{SA}+\overrightarrow{SB}+ \overrightarrow{SC}+\overrightarrow{SD}=\overrightarrow{0},$$ which means that $S$ is the centroid of the parallelogram $ABCD$. \kdokaz \bizrek \label{vektVarignon} The centroid of a quadrilateral $ABCD$ is the centroid of its Varignon parallelogram (see theorem \ref{Varignon}). \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.5.1a.pic} \caption{} \label{sl.vek.5.5.1a.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.5.1a.pic}) Let $P$, $K$, $Q$ and $L$ be the midpoints of sides $AB$, $BC$, $CD$ and $DA$, or $PKQL$ the Varignon parallelogram of the quadrilateral $ABCD$ (the quadrilateral $PKQL$ is a parallelogram by the statement \ref{Varignon}). According to the previous statement \ref{vektVarignon}, the intersection of diagonals $PQ$ and $LK$ (denoted by $T$) is at the same time the centroid of the parallelogram $PKQL$. Then it is (statements \ref{vektSredOSOAOB}) and \ref{paralelogram}): \begin{eqnarray*} \overrightarrow{TA}+\overrightarrow{TB}+ \overrightarrow{TC}+\overrightarrow{TD}&=& 2\cdot\overrightarrow{TP}+ 2\cdot\overrightarrow{TQ}=\\ &=& 2\cdot\left(\overrightarrow{TP}+ \overrightarrow{TQ}\right)=\\ &=&\overrightarrow{0}, \end{eqnarray*} which means that $T$ is the centroid of the quadrilateral $ABCD$. \kdokaz \bizrek \label{vektTezVeckXT} If $X$ is an arbitrary point and $T$ the centroid of a polygon $A_1A_2\ldots A_n$, then $$\overrightarrow{XT}=\frac{1}{n}\left(\overrightarrow{XA_1} +\overrightarrow{XA_2}+\cdots +\overrightarrow{XA_n}\right).$$ \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.5.4.pic} \caption{} \label{sl.vek.5.5.4.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.5.4.pic}) First, $\overrightarrow{XT}=\overrightarrow{XA_k}+\overrightarrow{A_kT}$ (for every $k\in\{1,2,\ldots,n\}$). If we add all $n$ relations, we get: \begin{eqnarray*} n\cdot\overrightarrow{XT} &=&\sum_{k=1}^{n}\left(\overrightarrow{XA_k}+\overrightarrow{A_kT}\right)=\\ &=&\sum_{k=1}^{n}\overrightarrow{XA_k}+\sum_{k=1}^{n}\overrightarrow{A_kT}=\\ &=& \sum_{k=1}^{n}\overrightarrow{XA_k}-\sum_{k=1}^{n}\overrightarrow{TA_k}=\\ &=& \sum_{k=1}^{n}\overrightarrow{XA_k}-\overrightarrow{0}=\\ &=&\sum_{k=1}^{n}\overrightarrow{XA_k}. \end{eqnarray*} Therefore: $$\overrightarrow{XT}=\frac{1}{n}\cdot\sum_{k=1}^{n}\overrightarrow{XA_k},$$ which was to be proven. \kdokaz As a result of the previous equation, a particularly useful relation holds for the centroid of an arbitrary triangle. \bizrek \label{vektTezTrikXT} If $X$ is an arbitrary point and $T$ the centroid of a triangle $ABC$, then $$\overrightarrow{XT}=\frac{1}{3}\left(\overrightarrow{XA} +\overrightarrow{XB}+\overrightarrow{XC}\right).$$ \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.5.5.pic} \caption{} \label{sl.vek.5.5.5.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.5.5.pic}) A direct consequence of the previous equation for $n=3$ \kdokaz % The process for an n-gon (n-1)-gon ... So far we have had an effective process for determining the centroids of triangles and quadrilaterals. For an arbitrary quadrilateral, we actually do not know whether the centroid exists at all. We will address this issue in the following. We will look for the idea in the following facts. If we consider the distance $AB$ as a degenerate $2$-gon, its centroid is the center of the distance $AB$; we denote it with $T_2$. For the centroid $T_3$ of the triangle $ABC$, it then holds that $\overrightarrow{CT_3}=\frac{2}{3}\cdot \overrightarrow{CT_2}$ (equation \ref{tezisce}). We will generalize this idea in the following equation. \bizrek \label{vektTezVeck} Every polygon $A_1A_2\ldots A_n$ has exactly one centroid. If $T_{n-1}$ is the centroid of the polygon $A_1A_2\ldots A_{n-1}$ and $T_n$ a point such that $$\overrightarrow{A_nT_n}=\frac{n-1}{n}\cdot\overrightarrow{A_nT_{n-1}},$$ then $T_n$ is the centroid of the polygon $A_1A_2\ldots A_n$. \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.5.2a.pic} \caption{} \label{sl.vek.5.5.2a.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.5.2a.pic}) First, from the relation $\overrightarrow{A_nT_n}=\frac{n-1}{n}\cdot\overrightarrow{A_nT_{n-1}}$ it follows that $\overrightarrow{T_nA_n}=-\frac{n-1}{n}\cdot\overrightarrow{A_nT_{n-1}}$ and $\overrightarrow{T_nT_{n-1}}=\frac{1}{n}\cdot\overrightarrow{A_nT_{n-1}}$. Because $T_{n-1}$ is the centroid of the polygon $A_1A_2\ldots A_{n-1}$, by \ref{vektTezVeckXT} it holds that $\overrightarrow{T_nA_1}+\overrightarrow{T_nA_2}+\cdots +\overrightarrow{T_nA_{n-1}}=\left(n-1\right)\cdot \overrightarrow{T_nT_{n-1}}$. So: \begin{eqnarray*} & & \overrightarrow{T_nA_1}+\overrightarrow{T_nA_2}+\cdots +\overrightarrow{T_nA_{n-1}}+\overrightarrow{T_nA_n}=\\ &=&\left(n-1\right)\cdot \overrightarrow{T_nT_{n-1}} +\overrightarrow{T_nA_n}=\\ &=&\frac{n-1}{n}\cdot\overrightarrow{A_nT_{n-1}} -\frac{n-1}{n}\cdot\overrightarrow{A_nT_{n-1}}=\\ &=&\overrightarrow{0}, \end{eqnarray*} which means that $T_n$ is the centroid of the polygon $A_1A_2\ldots A_n$. Assume that the polygon $A_1A_2\ldots A_n$ has another centroid $T'$. But in this case it holds (\ref{vektTezVeckXT}): \begin{eqnarray*} \overrightarrow{0}= \overrightarrow{T'A_1}+\overrightarrow{T'A_2}+\cdots \overrightarrow{T'A_n}= n\cdot\overrightarrow{T'T_n}. \end{eqnarray*} So $\overrightarrow{T'T_n}=\overrightarrow{0}$ or $T'=T_n$, which means that the polygon $A_1A_2\ldots A_n$ has only one centroid. \kdokaz The previous statement \ref{vektTezVeck} allows us to effectively construct the centroid of a polygon $A_1A_2\ldots A_n$, so that we first construct the centroids of polygons $A_1A_2$, $A_1A_2A_3$,$A_1A_2A_3A_4$, ... and finally $A_1A_2\ldots A_n$ in order (Figure \ref{sl.vek.5.5.2a.pic}): \begin{itemize} \item point $T_2$ is the center of the line segment $A_1A_2$, \item $T_3$ is such a point that it holds: $\overrightarrow{A_3T_3}=\frac{2}{3}\cdot \overrightarrow{A_3T_2}$, \item $T_4$ is such a point that it holds: $\overrightarrow{A_4T_4}=\frac{3}{4}\cdot \overrightarrow{A_4T_3}$,\\ $\vdots$ \item $T_n$ is such a point that it holds $\overrightarrow{A_nT_n}=\frac{n-1}{n}\cdot \overrightarrow{A_nT_{n-1}}$. \end{itemize} We get an even easier procedure for determining the centroid of a polygon if we use the relation $$\overrightarrow{XT}=\frac{1}{n}\left(\overrightarrow{XA_1} +\overrightarrow{XA_2}+\cdots +\overrightarrow{XA_n}\right)$$ from statement \ref{vektTezVeckXT}. So for any point $X$ we simply plot the vector $\overrightarrow{XT}$ and get the point $T$. We emphasize that in both cases of the centroid construction of a polygon we need the process of planning a point that divides a given line segment in a certain ratio. We will discuss this process in section \ref{odd5TalesVekt} (see \ref{izrekEnaDelitevDaljice} and \ref{izrekEnaDelitevDaljiceNan}). In the proof of statement \ref{vektTezVeck} we did not use the fact that the points $A_1$, $A_2$, ..., $A_n$ are in the same plane. The statement is also true in the case where $ABCD$ ($n=4$) is a so-called \pojem{tetrahedron} \index{tetrahedron}. The aforementioned point is then called \index{težišče!tetraedra} \pojem{the centroid of the tetrahedron}. It is possible to prove an analogous statement for a tetrahedron; the line segments determined by the vertices of the tetrahedron and the centroids of the opposite faces pass through the centroid of this tetrahedron, which it divides in the ratio $3 :1$. In the general case, if $n \in \mathbb{N}$, we can talk about the so-called \index{simplex}\pojem{simplex} (generalization: point, line, triangle, tetrahedron, ...), which lies in the $(n-1)$-dimensional Euclidean space. \bzgled Let $A$, $B$ and $C$ be the centroids of a triangles $OMN$, $ONP$ and $OPM$, then the centroid $T$ of the triangle $MNP$, the centroid $T_1$ of the triangle $ABC$ and the point $O$ are three collinear points. Furthermore, it is $OT_1:T_1T=2:1$. \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.5.6.pic} \caption{} \label{sl.vek.5.5.6.pic} \end{figure} \textbf{\textit{Proof.}} We mark with $P_1$, $M_1$ and $N_1$ the centroids of the sides $MN$, $NP$ and $PM$ of the triangle $PMN$ (Figure \ref{sl.vek.5.5.6.pic}). If we use the formulas \ref{vektTezTrikXT}, \ref{tezisce} and \ref{vektSredOSOAOB}, we get: \begin{eqnarray*} \overrightarrow{OT_1}&=&\frac{1}{3}\left( \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC} \right)=\\ &=&\frac{1}{3}\left( \frac{2}{3}\overrightarrow{OP_1}+\frac{2}{3}\overrightarrow{OM_1}+ \frac{2}{3}\overrightarrow{ON_1} \right)=\\ &=&\frac{1}{3}\left( \frac{1}{3}\left(\overrightarrow{OM}+\overrightarrow{ON}\right) +\frac{1}{3}\left(\overrightarrow{ON}+\overrightarrow{OP}\right)+ \frac{1}{3}\left(\overrightarrow{OP}+\overrightarrow{OM}\right) \right)=\\ &=&\frac{2}{9}\left( \overrightarrow{OM}+\overrightarrow{ON}+ \overrightarrow{OP} \right)=\\ &=&\frac{2}{3}\overrightarrow{OT} \end{eqnarray*} From $\overrightarrow{OT_1}=\frac{2}{3}\overrightarrow{OT}$ it follows that $\overrightarrow{OT_1}=2\overrightarrow{T_1T}$, which means that the vectors $\overrightarrow{OT_1}$ and $\overrightarrow{T_1T}$ are collinear (also the points $O$, $T_1$ and $T$) and $OT_1:T_1T=2:1$. \kdokaz \bzgled The centroid of a regular $n$-gon $A_1A_2...A_n$ is its centre (i.e. incentre and circumcentre). \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.5.7.pic} \caption{} \label{sl.vek.5.5.7.pic} \end{figure} \textbf{\textit{Proof.}} Let $S$ be the centroid of a regular $n$-gon $A_1A_2...A_n$ (Figure \ref{sl.vek.5.5.7.pic}). It is enough to prove that: $$\overrightarrow{SA_1}+\overrightarrow{SA_2}+ \cdots+\overrightarrow{SA_n}=\overrightarrow{0}.$$ Although in the case when $n$ is an even number, the statement is trivial, we will prove it for a general value of $n$ (even and odd). Assume that $\overrightarrow{SA_1}+\overrightarrow{SA_2}+\cdots+\overrightarrow{SA_n}=\overrightarrow{SX}$, where $X\neq S$. But if we rotate the polygon around the centroid $S$ by an angle $\theta=\frac{360}{n}$ (see the section on rotation \ref{odd6Rotac}), the sum of vectors on the left side of the equality does not change, the result on the right side becomes the vector $\overrightarrow{SX'}$, where $X'$ is the point we get from $X$ with the same rotation. Because the right side of the equality must remain unchanged, we get $\overrightarrow{SX'}=\overrightarrow{SX}$ or $X'=X$. This is possible only in the case when $X=S$ or $\overrightarrow{SX}=\overrightarrow{0}$. \kdokaz %________________________________________________________________________________ \poglavje{Hamilton's Theorem} \label{odd5Hamilton} Sedaj bomo nadaljevali z lastnostmi, ki se nanašajo na značilne točke trikotnika (razdelek \ref{odd3ZnamTock}). \bizrek \label{HamiltonLema} If $O$, $V$ and $A_1$ are the circumcentre, the orthocentre and the midpoint of the side $BC$ of a triangle $ABC$, respectively, then $$\overrightarrow{AV}=2\cdot \overrightarrow{OA_1}.$$ \eizrek \textbf{\textit{Proof.}} Let $B_1$ be the midpoint of the side $AC$ (Figure \ref{sl.vek.5.6.2.pic}). The vectors $\overrightarrow{OA_1}$ and $\overrightarrow{AV}$ are collinear, so $\overrightarrow{OA_1}=\alpha \cdot \overrightarrow{AV}$ for some $\alpha \in \mathbb{R}$ (statement \ref{vektKriterijKolin}). Similarly, for the same reasons $\overrightarrow{OB_1}=\beta \cdot \overrightarrow{BV}$ (or $\overrightarrow{B_1O}=\beta \cdot \overrightarrow{VB}$) for some $\beta \in \mathbb{R}$. By statement \ref{srednjicaTrikVekt} (the median of a triangle) we have: $$\overrightarrow{B_1A_1}=\frac{1}{2}\overrightarrow{AB}= \frac{1}{2}\left(\overrightarrow{AV}+\overrightarrow{VB}\right)= \frac{1}{2}\overrightarrow{AV}+\frac{1}{2}\overrightarrow{VB}.$$ At the same time we also have: $$\overrightarrow{B_1A_1}=\overrightarrow{B_1O}+\overrightarrow{OA_1}= \beta\overrightarrow{VB}+\alpha\overrightarrow{AV}= \alpha\overrightarrow{AV}+\beta\overrightarrow{VB}.$$ Since the vectors $\overrightarrow{AV}$ and $\overrightarrow{VB}$ are non-collinear, from statement \ref{vektLinKomb1Razcep} it follows that $\alpha=\frac{1}{2}$ and $\beta=\frac{1}{2}$. Therefore $\overrightarrow{OA_1}=\alpha \overrightarrow{AV}=\frac{1}{2}\overrightarrow{AV}$ or $\overrightarrow{AV}=2\cdot \overrightarrow{OA_1}$. \kdokaz \begin{figure}[!htb] \centering \input{sl.vek.5.6.2.pic} \caption{} \label{sl.vek.5.6.2.pic} \end{figure} The following statement is very useful. \bizrek \label{Hamilton}\index{izrek!Hamiltonov} (Hamilton's\footnote{\index{Hamilton, W. R.}\textit{W. R. Hamilton} (1805--1865), angleški matematik.} theorem) If $O$ and $V$ are circumcentre and orthocentre of a triangle $ABC$, respectively then $$\overrightarrow{OA}+\overrightarrow{OB} +\overrightarrow{OC}=\overrightarrow{OV}.$$ \eizrek \textbf{\textit{Proof.}} We mark with $A_1$ the centre of the side $BC$ (Figure \ref{sl.vek.5.6.2.pic}). If we use statement \ref{vektSredOSOAOB} and \ref{HamiltonLema}, we get: $$\overrightarrow{OA}+\overrightarrow{OB} +\overrightarrow{OC} =\overrightarrow{OA}+2\cdot \overrightarrow{OA_1}= \overrightarrow{OA}+\overrightarrow{AV}= \overrightarrow{OV},$$ which had to be proven. \kdokaz We will continue with the consequences of the previous two equations. \bzgled \label{HamiltonPoslTetiv} A quadrilateral $ABCD$ is inscribed in a circle with a centre $O$. The diagonals $AC$ and $BD$ are perpendicular. If $M$ is the foot of the perpendicular from the centre $O$ on the line $CD$, then $$|OM|=\frac{1}{2}|AB|.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.6.3a.pic} \caption{} \label{sl.vek.5.6.3a.pic} \end{figure} \textbf{\textit{Proof.}} Let $V$ be the altitude point of the triangle $BCD$ (Figure \ref{sl.vek.5.6.3a.pic}). Because $AC\perp BD$, the point $V$ lies on the diagonal $AC$. By \ref{HamiltonLema}, $\overrightarrow{OM}=\frac{1}{2} \overrightarrow{BV}$, so $|OM|=\frac{1}{2}|BV|$. Because there is also (\ref{ObodObodKot} and \ref{KotaPravokKraki}): $$\angle BAV=\angle BAC\cong\angle BDC\cong\angle AVB,$$ it follows that $BV\cong AB$ (\ref{enakokraki}) or $|OM|=\frac{1}{2}|AB|$. \kdokaz \bzgled Let $V$ be the orthocentre and $O$ the circumcentre of a triangle $ABC$ and $AV\cong AO$. Prove that $\angle BAC=60^0$. \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.6.1a.pic} \caption{} \label{sl.vek.5.6.1a.pic} \end{figure} \textbf{\textit{Proof.}} Let $A_1$ be the center of the side $BC$ of the triangle $ABC$ (Figure \ref{sl.vek.5.6.1a.pic}). By the statement \ref{HamiltonLema} it follows that $|AV|=2\cdot|OA_1|$. Since $OA\cong OC$, it follows that in the right triangle $OA_1C$ it holds that $|OC|=2\cdot|OA_1|$. With $O'$ we mark the point which is symmetrical to the point $O$ with respect to the point $A_1$. From $\triangle OA_1C\cong \triangle O'A_1C$ (the \textit{SAS} statement \ref{SKS}) it follows that $OC\cong O'C\cong OO'$, which means that the $\triangle OO'C$ is an isosceles triangle or $\angle A_1OC=60^0$. From the statement \ref{SredObodKot} and the congruence of the triangles $BOA_1$ and $COA_1$ (the \textit{SSS} statement \ref{SSS}) at the end it follows that $\angle BAC=\frac{1}{2}\angle BOC=\angle A_1OC=60^0$. \kdokaz \bzgled \label{TetivniVisinska} Let $ABCD$ be a cyclic quadrilateral and: $V_A$ the orthocentre of the triangle $BCD$, $V_B$ the orthocentre of the triangle $ACD$, $V_C$ the orthocentre of the triangle $ABD$ and $V_D$ the orthocentre of the triangle $ABC$. Prove that:\\ a) the line segments $AV_A$, $BV_B$, $CV_C$ and $DV_D$ has a common midpoint,\\ b) the quadrilateral $V_AV_BV_CV_D$ is congruent to the quadrilateral $ABCD$. \ezgled \begin{figure}[htp] \centering \input{sl.vek.5.6.3b.pic} \caption{} \label{sl.vek.5.6.3b.pic} \end{figure} \begin{figure}[htp] \centering \input{sl.vek.5.6.3.pic} \caption{} \label{sl.vek.5.6.3.pic} \end{figure} \textbf{\textit{Solution.}} Let $O$ be the center of the circumscribed circle of the cyclic quadrilateral $ABCD$ (Figure \ref{sl.vek.5.6.3.pic}). It is clear that the point $O$ is at the same time the center of the circumscribed circle of the triangles $BCD$, $ACD$, $ABD$ and $ABC$. By Hamilton's statement \ref{Hamilton} it follows: \begin{eqnarray*} \overrightarrow{OV_A}&=&\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}\\ \overrightarrow{OV_B}&=&\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OD} \end{eqnarray*} We then get: \begin{eqnarray*} \overrightarrow{V_BV_A}&=&\overrightarrow{V_BO}+\overrightarrow{OV_A}=\\ &=&\overrightarrow{OV_A}-\overrightarrow{OV_B}=\\ &=&\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD} -(\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OD})=\\ &=&\overrightarrow{OB}-\overrightarrow{OA}=\\ &=&\overrightarrow{AB}. \end{eqnarray*} Therefore $\overrightarrow{V_BV_A}=\overrightarrow{AB}$. By the statement \ref{vektParalelogram} the quadrilateral $ABV_AV_B$ is a parallelogram, by the statement \ref{paralelogram} its diagonals $AV_A$ and $BV_B$ have a common center - we mark it with $S$ (Figure \ref{sl.vek.5.6.3b.pic}). In a similar way each of the pairs of lines $AV_A$ and $CV_C$ or $AV_A$ and $DV_D$ have a common center. Because it is the center of the line $AV_A$, it follows that all four lines $AV_A$, $BV_B$, $CV_C$ and $DV_D$ have a common center - the point $S$. In a similar way as $\overrightarrow{V_BV_A}=\overrightarrow{AB}$ it also follows that $\overrightarrow{V_CV_B}=\overrightarrow{BC}$, $\overrightarrow{V_DV_C}=\overrightarrow{CD}$ and $\overrightarrow{V_DV_A}=\overrightarrow{DA}$. This means that the quadrilateral $V_AV_BV_CV_D$ and $ABCD$ (statement \ref{vektVzpSkl} and \ref{KotaVzporKraki}) have all the congruent sides and interior angles. Therefore $V_AV_BV_CV_D\cong ABCD$. For a formal proof of this we can use the isometry $\mathcal{I}:A,B,C\mapsto V_A,V_B,V_C$ and prove $\mathcal{I}(D)=V_D$. \kdokaz \bzgled \label{HamiltonSimson}\index{premica!Simsonova} Let $ABCD$ be a cyclic quadrilateral and: $a$ is the Simson line with respect to the triangle $BCD$ and the point $A$, $b$ is the Simson line with respect to the triangle $ACD$ and the point $B$, $c$ is the Simson line with respect to the triangle $ABD$ and the point $C$ ter $d$ S is the Simson line with respect to the triangle $ABC$ and the point $D$. Prove that the lines $a$, $b$, $c$ and $d$ intersect at a single point. \ezgled \begin{figure}[htp] \centering \input{sl.vek.5.6.4.pic} \caption{} \label{sl.vek.5.6.4.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.6.4.pic}). The claim is a direct consequence of theorems \ref{SimsZgled3} and \ref{TetivniVisinska} - the lines $a$, $b$, $c$ and $d$ intersect at the point $S$ (from theorem \ref{TetivniVisinska}). \kdokaz %________________________________________________________________________________ \poglavje{Euler Line} \label{odd5EulPrem} We will now prove an important property related to the three characteristic points of a triangle. \bizrek \label{EulerjevaPremica} The circumcentre $O$, the centroid $T$ and the orthocentre $V$ of an arbitrary triangle lies on the same line. Besides that it is $$|OT|:|TV|=1:2.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.7.1.pic} \caption{} \label{sl.vek.5.7.1.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.7.1.pic}) If we use theorem \ref{vektTezTrikXT} and Hamilton's theorem \ref{Hamilton}, we get: $$\overrightarrow{OT}=\frac{1}{3}\left(\overrightarrow{OA} +\overrightarrow{OB}+\overrightarrow{OC}\right)= \frac{1}{3}\overrightarrow{OV}.$$ The vectors $\overrightarrow{OT}$ and $\overrightarrow{OV}$ are therefore collinear and it holds that $\overrightarrow{OT}:\overrightarrow{OV}=1:3$. This means that the points $O$, $T$ and $V$ are collinear and it holds that $|OT|:|TV|=1:2$. \kdokaz The line from the previous theorem, on which three characteristic points lie, is called the \index{line!Euler's} \pojem{Euler line}. In the next theorem we will see the connection between the Euler line and the Euler circle, which we discussed in section \ref{odd3EulKroz}. \bizrek \label{EulerKrozPrem1}\index{circle!Euler's} The centre of Euler of an arbitrary triangle lies on The Euler line of this triangle. Furthermore, it is the midpoint of the line segment determined by the orthocentre and the circumcentre of this triangle. \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.7.2.pic} \caption{} \label{sl.vek.5.7.2.pic} \end{figure} \textbf{\textit{Proof.}} Let $AA'$, $BB'$ and $CC'$ be altitudes, and $A_1$, $B_1$ and $C_1$ be the centres of sides $BC$, $AC$ and $AB$ of triangle $ABC$. We mark the centre of the circumscribed circle with $O$, the point of altitude of this triangle with $V$, and the centres of lines $VA$, $VB$ and $VC$ with $V_A$, $V_B$ and $V_C$ (Figure \ref{sl.vek.5.7.2.pic}). By theorem \ref{EulerKroznica}, points $A'$, $B'$, $C'$, $A_1$, $B_1$, $C_1$, $V_A$, $V_B$ and $V_C$ lie on one circle - i.e. the Euler circle. We mark the centre of this circle with $E$. Because $\angle V_AA'A_1\cong \angle AA'C=90^0$, by theorem \ref{TalesovIzrKroz2} line $V_AA_1$ is the diameter of this circle, or point $E$ is the centre of line $V_AA_1$. According to the statement \ref{HamiltonLema} it holds: $$\overrightarrow{OA_1}=\frac{1}{2}\cdot \overrightarrow{AV}=\overrightarrow{V_AV}.$$ Therefore $\overrightarrow{OA_1}=\overrightarrow{V_AV}$, which means that the quadrilateral $A_1OV_AV$ is a parallelogram (\ref{vektParalelogram}). Its diagonals $VO$ and $V_AA_1$ intersect (\ref{paralelogram}), so the point $E$ is the center of the line $OV$ and lies on Euler's line of the triangle $ABC$ (\ref{EulerjevaPremica}). \kdokaz In section \ref{odd7SredRazteg} (\ref{EulerKroznicaHomot}) we will see the continuation of the previous statement, which refers to Euler's circle. %________________________________________________________________________________ \poglavje{Thales' Theorem - Basic Proportionality Theorem} \label{odd5TalesVekt} We have already seen in section \ref{odd5LinKombVekt} that we can talk about the ratio of two collinear vectors. For collinear vectors $\overrightarrow{v}$ and $\overrightarrow{u}$ ($\overrightarrow{u}\neq \overrightarrow{0}$) we defined their ratio or quotient $$\overrightarrow{v}:\overrightarrow{u} =\frac{\overrightarrow{v}}{\overrightarrow{u}}=\lambda,$$ if for some $\lambda\in\mathbb{R}$ it holds $\overrightarrow{v}=\lambda \overrightarrow{u}$. In a similar way as with numbers, we can define the ratio of two pairs of collinear vectors. If $\overrightarrow{a}$ and $\overrightarrow{b}$ ($\overrightarrow{b}\neq \overrightarrow{0}$) are a pair of collinear vectors or $\overrightarrow{c}$ and $\overrightarrow{d}$ ($\overrightarrow{d}\neq \overrightarrow{0}$) are a pair of collinear vectors, we say that the pairs of vectors \index{sorazmerje kolinearnih vektorjev}\pojem{sorazmerna}, if it holds: $$\frac{\overrightarrow{a}}{\overrightarrow{b}} =\frac{\overrightarrow{c}}{\overrightarrow{d}}.$$ The next very important statement refers to the defined concept of ratio. \bizrek \label{TalesovIzrek}(Thales'\footnote{Starogrški filozof in matematik \textit{Tales} \index{Tales} iz Mileta (640--546 pr. n. š.) je obravnaval sorazmerje ustreznih daljic, ki jih dobimo, če dve premici presekamo z dvema vzporednicama, pri tem pa ni omenjal vektorske oblike.} theorem - Basic Proportionality Theorem)\\ Let $a$, $b$, $p$ and $p'$ be lines in the same plane, and $O=a\cap b$, $A=a\cap p$, $A'=a\cap p'$, $B=b\cap p$ and $B'=b\cap p'$.\\ If $p\parallel p'$, then $$\frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= \frac{\overrightarrow{OB'}}{\overrightarrow{OB}}= \frac{\overrightarrow{A'B'}}{\overrightarrow{AB}}.$$ \index{izrek!Talesov o sorazmerju} \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.8.1.pic} \caption{} \label{sl.vek.5.8.1.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.8.1.pic}) Since by assumption $p\parallel p'$, the vectors $\overrightarrow{A'B'}$ and $\overrightarrow{AB}$ are collinear. By \ref{vektKriterijKolin} $\overrightarrow{A'B'}=\lambda\overrightarrow{AB}$ for some $\lambda\in \mathbb{R}$. In a similar way, from the collinearity of the vectors $\overrightarrow{OA'}$ and $\overrightarrow{OA}$ or $\overrightarrow{OB'}$ and $\overrightarrow{OB}$ it follows that $\overrightarrow{OA'}=\alpha\overrightarrow{OA}$ for some $\alpha\in \mathbb{R}$ or $\overrightarrow{OB'}=\beta\overrightarrow{OB}$ for some $\beta\in \mathbb{R}$. From this we obtain: $$\frac{\overrightarrow{A'B'}}{\overrightarrow{AB}}=\lambda,\hspace*{2mm} \frac{\overrightarrow{OA'}}{\overrightarrow{OA}}=\alpha,\hspace*{2mm} \frac{\overrightarrow{OB'}}{\overrightarrow{OB}}=\beta.$$ It is enough to prove that $\alpha=\beta=\lambda$. If we use the rule for subtracting vectors \ref{vektOdsev} and \ref{vektVektorskiProstor} (point $\textit{6}$), we get: \begin{eqnarray*} \overrightarrow{A'B'}&=&\overrightarrow{OB'}-\overrightarrow{OA'} =\beta\overrightarrow{OB}-\alpha \overrightarrow{OA};\\ \overrightarrow{A'B'}&=&\lambda\overrightarrow{AB}= \lambda\left(\overrightarrow{OB}-\overrightarrow{OA}\right) =\lambda\overrightarrow{OB}-\lambda\overrightarrow{OA}. \end{eqnarray*} Since $\overrightarrow{OA}$ and $\overrightarrow{OB}$ are non-collinear vectors, by \ref{vektLinKomb1Razcep} $\alpha=\beta=\lambda$ or $\frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= \frac{\overrightarrow{OB'}}{\overrightarrow{OB}}= \frac{\overrightarrow{A'B'}}{\overrightarrow{AB}}$. \kdokaz A direct consequence is the Tales theorem in the form of a ratio of distances, which is not in vector form (Figure \ref{sl.vek.5.8.2.pic}). \bizrek \label{TalesovIzrekDolzine} Let $a$, $b$, $p$ and $p'$ be lines in the same plane, and $O=a\cap b$, $A=a\cap p$, $A'=a\cap p'$, $B=b\cap p$ and $B'=b\cap p'$.\\ If $p\parallel p'$, then $$\frac{OA'}{OA}= \frac{OB'}{OB}= \frac{A'B'}{AB}$$ and also $$\frac{OA'}{OB'}= \frac{OA}{OB}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.8.2.pic} \caption{} \label{sl.vek.5.8.2.pic} \end{figure} \textbf{\textit{Proof.}} The claim directly follows from theorems \ref{TalesovIzrek} and \ref{vektKolicnDolz}. \kdokaz We will prove that the converse is also true, i.e. that from the appropriate proportion follows the parallelism of the corresponding lines. (Converse Thales' proportionality theorem)\\ Let $a$, $b$, $p$ and $p'$ be lines in the same plane, and $O=a\cap b$, $A=a\cap p$, $A'=a\cap p'$, $B=b\cap p$ and $B'=b\cap p'$.\\ If $$\frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= \frac{\overrightarrow{OB'}}{\overrightarrow{OB}},$$ then $p\parallel p'$ and also $$\frac{\overrightarrow{A'B'}}{\overrightarrow{AB}}= \frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= \frac{\overrightarrow{OB'}}{\overrightarrow{OB}}.$$ \index{izrek!Talesov obratni o sorazmerju} \textbf{\textit{Proof.}} We mark $\frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= \frac{\overrightarrow{OB'}}{\overrightarrow{OB}}=\lambda$. In this case, first $\overrightarrow{OA'}=\lambda\overrightarrow{OA}$ and $\overrightarrow{OB'}=\lambda\overrightarrow{OB}$. Therefore, (statement \ref{vektOdsev} and \ref{vektVektorskiProstor}): $$\overrightarrow{A'B'}=\overrightarrow{OB'}-\overrightarrow{OA'} =\lambda\overrightarrow{OB}-\lambda\overrightarrow{OA} =\lambda\left(\overrightarrow{OB}-\overrightarrow{OA}\right) =\lambda\overrightarrow{AB}.$$ Since $\overrightarrow{A'B'}=\lambda\overrightarrow{AB}$, according to statement \ref{vektKriterijKolin} vectors $\overrightarrow{A'B'}$ and $\overrightarrow{AB}$ are collinear. This means that $AB\parallel A'B'$ or $p\parallel p'$. Finally, from statement \ref{TalesovIzrek} follows the relation $\frac{\overrightarrow{A'B'}}{\overrightarrow{AB}}= \frac{\overrightarrow{OA'}}{\overrightarrow{OA}}= \frac{\overrightarrow{OB'}}{\overrightarrow{OB}}$. \kdokaz We also mention some consequences of Tales' statement \ref{TalesovIzrek}. \bizrek \label{TalesPosl1} If parallel lines $p_1$, $p_2$, $p_3$ intersect a line $a$ at points $A_1$, $A_2$, $A_3$ and a line $b$ at points $B_1$, $B_2$, $B_3$, then $$\frac{A_1A_2}{B_1B_2}=\frac{A_2A_3}{B_2B_3} \hspace*{1mm} \textrm{ and } \hspace*{1mm} \frac{A_1A_2}{A_2A_3}=\frac{B_1B_2}{B_2B_3}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.8.4.pic} \caption{} \label{sl.vek.5.8.4.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.8.4.pic}) Without loss of generality, let $\mathcal{B}(A_1,A_2,A_3)$ and $\mathcal{B}(B_1,B_2,B_3)$. We mark with $c$ the parallel to the line $b$ through the point $A_1$ and with $C_2$ and $C_3$ the intersections of the line $c$ with the lines $p_2$ and $p_3$. The quadrilateral $B_1B_3C_3A_1$ and $B_2B_3C_3C_2$ are parallelograms, therefore $\overrightarrow{B_1B_3}=\overrightarrow{A_1C_3}$ and $\overrightarrow{B_2B_3}=\overrightarrow{C_2C_3}$. If we use the statement \ref{TalesovIzrekDolzine}, we get: \begin{eqnarray*} \frac{|A_1A_2|}{|A_2A_3|}&=& \frac{|A_1A_3|-|A_2A_3|}{|A_2A_3|}= \frac{|A_1A_3|}{|A_2A_3|}-1=\\ &=&\frac{|A_1C_3|}{|C_2C_3|}-1= \frac{|B_1B_3|}{|B_2B_3|}-1=\\ &=&\frac{|B_1B_3|-|B_2B_3|}{|B_2A_3|}= \frac{|B_1B_2|}{|B_2B_3|}, \end{eqnarray*} or $\frac{A_1A_2}{A_2A_3}=\frac{B_1B_2}{B_2B_3}$ and $\frac{A_1A_2}{B_1B_2}=\frac{A_2A_3}{B_2B_3}$. \kdokaz \bizrek \label{TalesPosl2} If parallel lines $p_1$, $p_2$,..., $p_n$ intersect a line $a$ at points $A_1$, $A_2$,..., $A_n$ and a line $b$ at points $B_1$, $B_2$,..., $B_n$, then $$\frac{A_1A_2}{B_1B_2}=\frac{A_2A_3}{B_2B_3}=\cdots= \frac{A_{n-1}A_n}{B_{n-1}B_n}\hspace*{1mm} \textrm{ and } \hspace*{1mm}$$ $$A_1A_2:A_2A_3:\cdots :A_{n-1}A_n= B_1B_2:B_2B_3:\cdots :B_{n-1}B_n.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.8.5.pic} \caption{} \label{sl.vek.5.8.5.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.8.5.pic}) The statement is a direct consequence of the statement \ref{TalesPosl1}. \kdokaz \bizrek \label{TalesPosl3} Let $p_1$, $p_2$ and $p_3$ be lines that intersect at a point $O$. If $a$ and $b$ are parallel lines that intersect the line $p_1$ at points $A_1$ and $B_1$, the line $p_2$ at points $A_2$ and $B_2$, and the line $p_3$ at points $A_3$ and $B_3$, then $$\frac{A_1A_2}{B_1B_2}=\frac{A_2A_3}{B_2B_3} \hspace*{1mm} \textrm{ and } \hspace*{1mm} \frac{A_1A_2}{A_2A_3}=\frac{B_1B_2}{B_2B_3}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.8.6.pic} \caption{} \label{sl.vek.5.8.6.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.8.6.pic}) By izreku \ref{TalesovIzrekDolzine} we get: $$\frac{A_1A_2}{B_1B_2}= \frac{OA_2}{OB_2}=\frac{A_2A_3}{B_2B_3},$$ which was to be proven. \kdokaz \bizrek \label{TalesPosl4} Let $p_1$, $p_2$,..., $p_n$ be lines that intersect at a point $O$. If $a$ and $b$ are parallel lines that intersect the line $p_1$ at points $A_1$ and $B_1$, the line $p_2$ at points $A_2$ and $B_2$,..., the line $p_n$ at points $A_n$ and $B_n$, then $$\frac{A_1A_2}{B_1B_2}=\frac{A_2A_3}{B_2B_3}=\cdots= \frac{A_{n-1}A_n}{B_{n-1}B_n}\hspace*{1mm} \textrm{ and } \hspace*{1mm}$$ $$A_1A_2:A_2A_3:\cdots :A_{n-1}A_n= B_1B_2:B_2B_3:\cdots :B_{n-1}B_n.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.8.8.pic} \caption{} \label{sl.vek.5.8.8.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.vek.5.8.8.pic}) The statement is a direct consequence of \ref{TalesPosl3}. \kdokaz The following very well-known and useful planning tasks are next. \bzgled \label{izrekEnaDelitevDaljiceNan} \index{delitev daljice!na enake dele} Construct points that divide a line segment $AB$ into $n$ congruent line segments. \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.8.9.pic} \caption{} \label{sl.vek.5.8.9.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.8.9.pic}) Let $X$ be an arbitrary point that does not lie on the line $AB$ and $Q_1$, $Q_2$, ..., $Q_n$ such points on the line segment $AX$, so that $\overrightarrow{AQ_1}=\overrightarrow{Q_1Q_2}=\cdots =\overrightarrow{Q_{n-1}Q_n}$. With $P_1$, $P_2$, ..., $P_{n-1}$ we denote the intersection of the line $AB$ with the parallels of the line $BQ_n$ through the points $Q_1$, $Q_2$, ..., $Q_{n-1}$. We prove that $P_1$, $P_2$, ..., $P_{n-1}$ are the desired points. By \ref{TalesPosl2} we have: $$\frac{AP_1}{AQ_1}=\frac{P_1P_2}{Q_1Q_2}=\dots=\frac{P_{n-1}B}{Q_{n-1}Q_n}.$$ Since, by assumption, $|\overrightarrow{AQ_1}|=|\overrightarrow{Q_1Q_2}|=\cdots =|\overrightarrow{Q_{n-1}Q_n}|$, also $|AP_1|=|P_1P_2|=\cdots =|P_{n-1}B|$. \kdokaz \bzgled \label{izrekEnaDelitevDaljice} \index{delitev daljice!v razmerju} Divide a given line segment $AB$ in the ratio $n:m$ ($n,m\in \mathbb{N}$), i.e. determine such point $T$ on the line $AB$ that $$\frac{\overrightarrow{AT}}{\overrightarrow{TB}}=\frac{n}{m}.$$ Prove that there is the only one solution for such point $T$. \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.8.10.pic} \caption{} \label{sl.vek.5.8.10.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.vek.5.8.10.pic}) By \ref{izrekEnaDelitevDaljiceVekt} there is only one point $T$, for which $\frac{\overrightarrow{AT}}{\overrightarrow{TB}}=\frac{n}{m}$. Now we will describe the process of constructing point $T$. Let $\overrightarrow{v}$ be an arbitrary vector that is not collinear with vector $\overrightarrow{AB}$, and let $P$ and $Q$ be such points that $\overrightarrow{AP}=n\overrightarrow{v}$ and $\overrightarrow{BQ}=-m\overrightarrow{v}$. We denote the intersection of lines $AB$ and $PQ$ by $T$ (the intersection exists because $P,Q\div AB$). Then it holds: \begin{eqnarray*} \frac{\overrightarrow{AT}}{\overrightarrow{TB}}= -\frac{\overrightarrow{TA}}{\overrightarrow{TB}}= -\frac{\overrightarrow{AP}}{\overrightarrow{BQ}}= -\frac{n\overrightarrow{v}}{-m\overrightarrow{v}}=\frac{n}{m}, \end{eqnarray*} which had to be proven. \kdokaz In section \ref{odd7Harm} we will further investigate the question of dividing a line segment into a given ratio. \bzgled \label{vektTrapezZgled} A line parallel to bases of a trapezium intersects its legs and diagonals in four points and determine three line segments. Prove that two of them are congruent. \\ After that, construct a line parallel to the bases of that trapezium such that all three mentioned line segments are congruent. \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.8.11.pic} \caption{} \label{sl.vek.5.8.11.pic} \end{figure} \textbf{\textit{Proof.}} Let $l$ be a line parallel to the bases $AB$ of trapezium $ABCD$, that intersects sides $AD$ and $BC$ as well as diagonals $AC$ and $BD$ in points $M$, $Q$, $N$ and $P$ (Figure \ref{sl.vek.5.8.11.pic}). By Tales' theorem \ref{TalesovIzrekDolzine} it holds: $MN:DC=MA:DA$ or $PQ:DC=BQ:BC$. Because by the consequence \ref{TalesPosl1} of Tales' theorem $MA:AD=BQ:BC$, it also holds $MN:DC=PQ:DC$ or $MN\cong PQ$. If $E$ is the center of the base $AB$ and $N_0$ is the intersection of the lines $DE$ and $AC$, the desired line $l_0$ passes through the point $N_0$ and is parallel to the line $AB$. If $M_0$, $P_0$ and $Q_0$ are the intersections of the line $l_0$ with the lines $AD$, $BD$ and $CB$, then by \ref{TalesPosl3} of Tales' theorem it follows that $M_0N_0\cong N_0P_0$. \kdokaz \bzgled If $r$ is the inradius and $r_a$, $r_b$ and $r_c$ exradii of an arbitrary triangle, then $$\frac{1}{r_a} +\frac{1}{r_b} +\frac{1}{r_c}= \frac{1}{r}.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.vek.5.8.12.pic} \caption{} \label{sl.vek.5.8.12.pic} \end{figure} \textbf{\textit{Proof.}} If we use the labels from the big task \ref{velikaNaloga} (Figure \ref{sl.vek.5.8.12.pic}), from Tales' theorem \ref{TalesovIzrek} it follows that: $$ \frac{r}{r_a} = \frac{SQ}{S_aQ_a} = \frac{AQ}{AQ_a} = \frac{s-a}{s}.$$ Similarly, it is also: $$ \frac{r}{r_b} = \frac{s-b}{s}\hspace*{2mm} \textrm{ and }\hspace*{2mm} \frac{r}{r_c} = \frac{s-c}{s}.$$ By adding the three equalities we get the desired relation. \kdokaz \bizrek \label{velNalTockP'} Suppose that the incircle and the excircle of a triangle $ABC$ touch its side $BC$ in points $P$ and $P_a$. If $PP'$ is a diameter of the incircle $k(S,r)$ ($P'\in k$), then $Pa$, $P'$ and $A$ are collinear points. \eizrek \begin{figure}[!htb] \centering \input{sl.vek.5.8.13.pic} \caption{} \label{sl.vek.5.8.13.pic} \end{figure} \textbf{\textit{Proof.}} Use the labels from the big task \ref{velikaNaloga} (Figure \ref{sl.vek.5.8.13.pic}). Because the distance $PP'$ is a diameter of the inscribed circle, the point $S$ is its center. Let $\widehat{P'}$ be the intersection of the line segment $PS$ with the line $AP_a$. We will prove that $\widehat{P'}=P'$, or that $S\widehat{P'}=r$. From Tales' theorem it follows that: $$\frac{S\widehat{P'}}{S_aP_a} = \frac{AS}{AS_a} =\frac{SQ}{S_aQ_a}.$$ Because $S_a P_a = S_aQ_a = r_a$, it is also true that $S\widehat{P'}=SQ=r$. \kdokaz In a similar way to \ref{velNalTockP'} we also prove the following theorem. \bizrek \label{velNalTockP'1} Suppose that the incircle and the excircle of a triangle $ABC$ touch its side $BC$ in points $P$ and $P_a$. If $P_aP_a'$ is a diameter of the excircle $k_a(S_a,r_a)$ ($P_a'\in k_a$), then $P_a'$, $P$ and $A$ are collinear points (Figure \ref{sl.vek.5.8.14.pic}). \eizrek \begin{figure}[htp] \centering \input{sl.vek.5.8.14.pic} \caption{} \label{sl.vek.5.8.14.pic} \end{figure} %\vspace*{10mm} \bizrek \label{velNalTockP'2} Suppose that the excircles $k_b(S_b,r_b)$ and $k_c(S_c,r_c)$ of a triangle $ABC$ touch the line $BC$ in points $P_b$ and $P_c$. If $P_bP_b'$ and $P_cP_c'$ are diameters of the excircles $k_b$ ($P_b'\in k_b$) and $k_c$ ($P_c'\in k_c$), then $P_c'$, $P_b$ and $A$ are collinear points and also $P_b'$, $P_c$ and $A$ are collinear points (Figure \ref{sl.vek.5.8.15.pic}). \eizrek \begin{figure}[htp] \centering \input{sl.vek.5.8.15.pic} \caption{} \label{sl.vek.5.8.15.pic} \end{figure} The last three statements can be used in the construction of triangles, which we will illustrate in the following task. \bzgled Construct a triangle $ABC$, with given $r$, $b-c$, $t_a$. \ezgled \begin{figure}[htp] \centering \input{sl.vek.5.8.16.pic} \caption{} \label{sl.vek.5.8.16.pic} \end{figure} \textbf{\textit{Solution.}} Let $ABC$ be the desired triangle or the triangle, in which the radius of the inscribed circle is $r$, the centroid $AA_1$ is consistent with $t_a$ and the difference of sides $AC$ and $AB$ is equal to $b-c$. Use the notation from the big task \ref{velikaNaloga} and the formula \ref{velNalTockP'} (Figure \ref{sl.vek.5.8.16.pic}). We know that $PP_a=b-c$, point $A_1$ is the common center of side $BC$ and distance $PP_a$. From the formula \ref{velNalTockP'} it follows that the points $P_a$, $P'$ and $A$ are collinear. So first we construct the right triangle $P'PP_a$, then the circle $k$ with diameter $PP'$ or the inscribed circle of the triangle $ABC$ and the point $A_1$ as the center of the line segment $PP_a$. The point $A$ is the intersection of the line segment $P_aP'$ with the circle $k_1(A_1,t_a)$, points $B$ and $C$ are the intersections of the line $BC$ with the tangents of the circle $k$ from the point $A$. \kdokaz \bzgled Let $A_1$ be the midpoint of the line segment $BC$, $k(S,r)$ the incircle and $AA'$ the altitude of a triangle $ABC$. Suppose that $L$ is the intersection of the lines $A_1S$ and $AA'$. Prove that $AL\cong r$. \ezgled \begin{figure}[htp] \centering \input{sl.vek.5.8.3.pic} \caption{} \label{sl.vek.5.8.3.pic} \end{figure} \textbf{\textit{Solution.}} Let $P$, $Q$ and $R$ be the points of intersection of the circle $k$ with the sides $BC$, $AC$ and $AB$. Let $k(S_a,r_a)$ be the drawn circle that touches the side $BC$ in the point $P_a$, the lines $AB$ and $AC$ in the points $R_a$ and $Q_a$. Let $P'$ be the intersection of the lines $AP_a$ and $PS$ (Figure \ref{sl.vek.5.8.3.pic}). From the statement \ref{velNalTockP'} and the big exercise \ref{velikaNaloga} it follows: \begin{itemize} \item the point $P'$ lies on the circle $k$, \item the point $A_1$ is the center of the segment $PP_a$. \end{itemize} We prove that $AL\cong r$. The segment $SA_1$ is the median of the triangle $PP'P_a$, therefore $SA_1\parallel P'P_a$ or $LS\parallel AP'$. Since $AL\parallel P'S$, the quadrilateral $AP'SL$ is a parallelogram and $AL\cong P'S\cong r$. \kdokaz %________________________________________________________________________________ \naloge{Exercises} \begin{enumerate} %Vsota in razlika vektorjev %_____________________________________ \item Draw any vectors $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ and $\overrightarrow{d}$ so that their sum is equal to: (\textit{a}) one of these four vectors,\\ (\textit{b}) the difference of two of these four vectors. \item Let $ABCDE$ be a pentagon in some plane. Prove that in this plane there exists a pentagon with sides that determine the same vectors as the diagonals of the pentagon $ABCDE$. \item Let $A$, $B$, $C$ and $D$ be any points in a plane. Is it generally true that: (\textit{a}) $\overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AD}+\overrightarrow{BC}$?\\ (\textit{b}) $\overrightarrow{AB}=\overrightarrow{DC}\hspace*{1mm}\Rightarrow \hspace*{1mm} \overrightarrow{AC}+\overrightarrow{BD}=2\overrightarrow{BC}$? \item Given is a segment $AB$. Using only a ruler (constructions in affine geometry) draw a point $C$ so that: (\textit{a}) $\overrightarrow{AC}=-\overrightarrow{AB}$, \hspace*{3mm} (\textit{b}) $\overrightarrow{AC}=5\overrightarrow{AB}$, \hspace*{3mm} (\textit{c}) $\overrightarrow{AC}=-3\overrightarrow{AB}$. \item Let $ABCD$ be a quadrilateral and $O$ any point in the plane of this quadrilateral. Express the vectors of sides and diagonals of this quadrilateral with vectors $\overrightarrow{a}=\overrightarrow{OA}$, $\overrightarrow{b}=\overrightarrow{OB}$, $\overrightarrow{c}=\overrightarrow{OC}$ and $\overrightarrow{d}=\overrightarrow{OD}$. \item Let $ABCD$ be a quadrilateral and $O$ any point in the plane of this quadrilateral. Is the equivalence true that the quadrilateral $ABCD$ is a parallelogram exactly when $\overrightarrow{OA}+\overrightarrow{OC}= \overrightarrow{OB}+\overrightarrow{OD}$? \item Let $ABCD$ be a parallelogram, $S$ the intersection of its diagonals and $M$ any point in the plane of this parallelogram. Prove that: $$\overrightarrow{MS} = \frac{1}{4}\cdot \left( \overrightarrow{MA}+\overrightarrow{MB} +\overrightarrow{MC} + \overrightarrow{MD} \right).$$ \item Let $ABB_1A_2$, $BCC_1B_2$ and $CAA_1C_2$ be parallelograms, which are drawn over the sides of triangle $ABC$. Prove that: $$\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}+ \overrightarrow{C_1C_2}=\overrightarrow{0}.$$ \item The perpendicular lines $p$ and $q$, which intersect in point $M$, intersect the circle $k$ with center $O$ in points $A$, $B$, $C$ and $D$. Prove that: $$\overrightarrow{OA}+ \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = 2\overrightarrow{OM}.$$ \item Let $A$, $B$, $C$ and $D$ be any points in some plane. Can all six lines, which are determined by these points, be oriented so that the sum of the corresponding six vectors is equal to the vector of nothing? \item Let $A_1$, $B_1$ and $C_1$ be the centers of sides $BC$, $AC$ and $AB$ of triangle $ABC$ and $M$ any point. Prove: (\textit{a}) $\overrightarrow{AA_1}+\overrightarrow{BB_1}+ \overrightarrow{CC_1}=\overrightarrow{0}$,\\ (\textit{b}) $\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}= \overrightarrow{MA_1}+\overrightarrow{MB_1}+\overrightarrow{MC_1}$,\\ (\textit{c}) There exists a triangle $PQR$, such that for its vertices it holds:\\ \hspace*{7mm} $\overrightarrow{PQ}=\overrightarrow{CC_1}$, $\overrightarrow{PR}=\overrightarrow{BB_1}$ and $\overrightarrow{RQ}=\overrightarrow{AA_1}$. \item Let $M$, $N$, $P$, $Q$, $R$ and $S$ be in order the centers of the sides of an arbitrary hexagon. Prove that it holds: $$\overrightarrow{MN}+\overrightarrow{PQ}+ \overrightarrow{RS}=\overrightarrow{0}.$$ %Linearna kombinacija vektorjev %_____________________________________ \item Let $ABCDEF$ be a convex hexagon, for which $AB\parallel DE$, and points $K$ and $L$ are the centers of the lines determined by the centers of the remaining pairs of opposite sides. Prove that $K=L$ exactly when $AB\cong DE$. \item Let $P$ and $Q$ be such points of sides $BC$ and $CD$ of a parallelogram $ABCD$, that $BP:PC=2:3$ and $CQ:QD=2:5$. Point $X$ is the intersection of lines $AP$ and $BQ$. Calculate the ratios in which point $X$ divides line $AP$ and $BQ$. \item Let $A$, $B$, $C$ and $D$ be arbitrary points of a plane. Point $E$ is the center of line $AB$, $F$ and $G$ are such points, that $\overrightarrow{EF} = \overrightarrow{BC}$ and $\overrightarrow{EG} = \overrightarrow{AD}$, and $S$ is the center of line $CD$. Prove that $G$, $S$ and $F$ are collinear points. \item Let $K$ and $L$ be such points of side $AD$ and diagonal $AC$ of a parallelogram $ABCD$, that $\frac{\overrightarrow{AK}}{\overrightarrow{KD}}=\frac{1}{3}$ and $\frac{\overrightarrow{AL}}{\overrightarrow{LC}}=\frac{1}{4}$. Prove that $K$, $L$ and $B$ are collinear points. \item Let $X_n$ and $Y_n$ ($n\in \mathbb{N}$) be such points of sides $AB$ and $AC$ of triangle $ABC$, that $\overrightarrow{AX_n}=\frac{1}{n+1}\cdot \overrightarrow{AB}$ and $\overrightarrow{AY_n}=\frac{1}{n}\cdot \overrightarrow{AC}$. Prove that there exists a point, which lies on all lines $X_nY_n$ ($n\in \mathbb{N}$). \item Let $M$, $N$, $P$ and $Q$ be the centers of sides $AB$, $BC$, $CD$ and $DA$ of quadrilateral $ABCD$. Is the following equivalence true, that the quadrilateral $ABCD$ is a parallelogram exactly when: (\textit{a}) $2\overrightarrow{MP}=\overrightarrow{BC}+\overrightarrow{AD}$ and $2\overrightarrow{NQ}=\overrightarrow{BA}+\overrightarrow{CD}$?\\ (\textit{b}) $2\overrightarrow{MP}+2\overrightarrow{NQ}= \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}$? \item Let $E$, $F$ and $G$ be the centers of sides $AB$, $BC$ and $CD$ of parallelogram $ABCD $, lines $BG$ and $DE$ intersect line $AF$ in points $N$ and $M$. Express $\overrightarrow{AF}$, $\overrightarrow{AM}$ and $\overrightarrow{AN}$ as a linear combination of vectors $\overrightarrow{AB}$ and $\overrightarrow{AD}$. Prove that points $M$ and $N$ divide the distance $AF$ in the ratio $2:2:1$. \item Points $K$, $L$, $M$ and $N$ lie on sides $AB$, $BC$, $CD$ and $DA$ of quadrilateral $ABCD$. If the quadrilateral $KLMN$ is a parallelogram and it is true that $$\frac{\overrightarrow{AK}}{\overrightarrow{KB}}= \frac{\overrightarrow{BL}}{\overrightarrow{LC}} =\frac{\overrightarrow{CM}}{\overrightarrow{MD}}= \frac{\overrightarrow{DN}}{\overrightarrow{NA}}=\lambda$$ for some $\lambda\neq\pm 1$, then the quadrilateral $ABCD$ is a parallelogram. Prove it. \item Let $M$ be the center of side $DE$ of regular hexagon $ABCDEF$. Point $N$ is the center of line $AM$, point $P$ is the center of side $BC$. Express $\overrightarrow{NP}$ as a linear combination of vectors $\overrightarrow{AB}$ and $\overrightarrow{AF}$. %The length of a vector %_____________________________________ \item Prove that for any points $A$, $B$ and $C$ it is true that: (\textit{a}) $|\overrightarrow{AC}|\leq|\overrightarrow{AB}|+|\overrightarrow{BC}|$ \hspace*{6mm} (\textit{b}) $|\overrightarrow{AC}|\geq|\overrightarrow{AB}|-|\overrightarrow{BC}|$\\ Under which conditions is the equality true? \item Let $M$ and $N$ be points that lie on the lines $AD$ and $BC$, respectively, such that $\frac{\overrightarrow{AM}}{\overrightarrow{MD}}\cdot \frac{\overrightarrow{CN}}{\overrightarrow{NB}}=1$. Prove that: $$|MN|\leq\max\{|AB|, |CD|\}.$$ %Težišče %_____________________________________ \item Points $T$ and $T'$ are the centroids of $n$-sided polygons $A_1A_2...A_n$ and $A'_1A'_2...A'_n$. Calculate: $$\overrightarrow{A_1A'_1}+\overrightarrow{A_2A'_2}+\cdots+\overrightarrow{A_nA'_n}.$$ \item Prove that quadrilaterals $ABCD$ and $A'B'C'D'$ have a common centroid exactly when: $$\overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}+ \overrightarrow{DD'}=\overrightarrow{0}.$$ \item Let $P$, $Q$, $R$ and $S$ be the centroids of triangles $ABD$, $BCA$, $CDB$ and $DAC$. Prove that quadrilaterals $PQRS$ and $ABCD$ have a common centroid. \item Let $A_1A_2A_3A_4A_5A_6$ be an arbitrary hexagon and $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ and $B_6$ are the centroids of triangles $A_1A_2A_3$, $A_2A_3A_4$, $A_3A_4A_5$, $A_4A_5A_6$, $A_5A_6A_1$ and $A_6A_1A_2$, respectively. Prove that these centroids determine a hexagon with three pairs of parallel sides. \item Let $A$, $B$, $C$ and $D$ be four different points. Points $T_A$, $T_B$, $T_C$ and $T_D$ are the centroids of triangles $BCD$, $ACD$, $ABD$ and $ABC$, respectively. Prove that the lines $AT_A$, $BT_B$, $CT_C$ and $DT_D$ intersect at one point $T$. In what ratio does point $T$ divide these lines? \item Let $CC_1$ be the centroid of the triangle $ABC$ and $P$ any point on the side $AB$ of this triangle. The parallel $l$ of the line $CC_1$ through the point $P$ intersects the line $AC$ and $BC$ in points $M$ and $N$. Prove that: $$\overrightarrow{PM} + \overrightarrow{PN}= \overrightarrow{AC} + \overrightarrow{BC}.$$ %Hamilton and Euler %______________________________________________________ \item Let $A$, $B$, $C$ be points of a plane that lie on the same side of the line $p$, and $O$ a point on the line $p$, for which $|\overrightarrow{OA}| = |\overrightarrow{OB}| = |\overrightarrow{OC}| =1$. Prove that then it also holds: $$|\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}| \geq 1.$$ \item Calculate the angles determined by the vectors $\overrightarrow{OA}$, $\overrightarrow{OB}$ and $\overrightarrow{OC}$, if the points $A$, $B$ and $C$ lie on a circle with center $O$ and in addition it holds: $$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = \overrightarrow{0}.$$ \item Let $A$, $B$, $C$ and $D$ be points that lie on a circle with center $O$, and it holds $$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = \overrightarrow{0}.$$ Prove that $ABCD$ is a rectangle. \item Let $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ be vectors of a plane, for which $|\overrightarrow{a}| = |\overrightarrow{b}| = |\overrightarrow{c}| =x$. Investigate, in which case it also holds $|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}| = x$. %Tales %______________________________________________________ \item Divide the distance $AB$: (\textit{a}) into five equal distances,\\ (\textit{b}) in the ratio $2:5$,\\ (\textit{c}) into three distances, which are in the ratio $2:\frac{1}{2}:1$. \item The distance $AB$ is given. Only by using the rule with the possibility of drawing parallels (constructions in affine geometry) plot the point $C$, if it is: (\textit{a}) $\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}$ \hspace*{3mm} (\textit{b}) $\overrightarrow{AC}=\frac{3}{5}\overrightarrow{AB}$ \hspace*{3mm} (\textit{c}) $\overrightarrow{AC}=-\frac{4}{7}\overrightarrow{AB}$ \item The distances $a$, $b$ and $c$ are given. Plot the distance $x$, so that it will: (\textit{a}) $a:b=c:x$ \hspace*{3mm} (\textit{b}) $x=\frac{a\cdot b}{c}$ \hspace*{3mm} (\textit{c}) $x=\frac{a^2}{c}$\hspace*{3mm}\\ (\textit{č}) $x=\frac{2ab}{3c}$\hspace*{3mm} (\textit{č}) $(x+c):(x-c)=7:2$ \item Let $M$ and $N$ be points on the arm $OX$, $P$ point on the arm $OY$ angle $XOY$ and $NQ\parallel MP$ and $PN\parallel QS$ ($Q\in OY$, $S\in OX$). Prove that $|ON|^2=|OM|\cdot |OS|$ (for the distance $ON$ in this case we say that it is \index{geometrijska sredina daljic}\pojem{geometrijska sredina} of the distances $OM$ and $OS$). \item Let $ABC$ be a triangle and $Q$, $K$, $L$, $M$, $N$ and $P$ such points on the sides $AB$, $AC$, $BC$, $BA$, $CA$ and $CB$, that $AQ\cong CP\cong AC$, $AK\cong BL\cong AB$ and $BM\cong CN\cong BC$. Prove that $MN$, $PQ$ and $LK$ are three parallel. \item Let $P$ be the center of the centroid $AA_1$ of the triangle $ABC$. The point $Q$ is the intersection of the line $BP$ with the side $AC$. Calculate the ratios $AQ:QC$ and $BP:PQ$. \item The points $P$ and $Q$ lie on the sides $AB$ and $AC$ of the triangle $ABC$, and $\frac{|\overrightarrow{PB}|}{|\overrightarrow{AP}|} +\frac{|\overrightarrow{QC}|}{|\overrightarrow{AQ}|}=1$. Prove that the centroid of this triangle lies on the line $PQ$. \item Let $a$, $b$ and $c$ be three sides with a common starting point $S$ and $M$ point on the side $a$. If the point $M$ "moves" along the side $a$, the ratio of the distance of this point from the lines $b$ and $c$ is constant. Prove it. \item Let $D$ be a point that lies on the side $BC$ of the triangle $ABC$ and $F$ and $G$ points in which the line passing through the point $D$ and parallel to the centroid $AA_1$, intersects the lines $AB$ and $AC$. Prove that the sum $|DF|+|DG|$ is constant if the point $D$ "moves" along the side $BC$. \item Draw a triangle with the following data: (\textit{a}) $v_a$, $r$, $b-c$ \hspace*{3mm} (\textit{b}) $\beta$, $r$, $b-c$ \end{enumerate} % DEL 6 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - %________________________________________________________________________________ % IZOMETRIJE %________________________________________________________________________________ \del{Isometries} \label{pogIZO} %________________________________________________________________________________ \poglavje{Isometries. Identity Map} \label{odd6Ident} We have formally defined isometries or isometric transformations of a plane $\mathcal{I}:\mathbb{E}^2\rightarrow \mathbb{E}^2$ in section \ref{odd2AKSSKL} as transformations of a plane that preserve the relation of congruence of pairs of points. We later used them to define the relation of congruence of figures. We intuitively identify them as movements of a plane. Some of them we already know from before (we have not formally introduced them in this book) - rotation and translation (Figure \ref{sl.izo.6.1.1.pic}). Even reflection over a line is an isometry. But this differs from the aforementioned two isometries because it is not a true movement (in a plane). In order to see it as a movement, we need to go to a space where we can rotate the plane by $180^0$. This changes the orientation of the plane. \begin{figure}[!htb] \centering \input{sl.izo.6.1.1.pic} \caption{} \label{sl.izo.6.1.1.pic} \end{figure} We call such isometries, which change the orientation of the plane, \index{izometrija!indirektna} \pojem{indirect}. Reflection over a line is therefore an indirect isometry. For those isometries that preserve the orientation of the plane, we say that they are \index{izometrija!direktna}\pojem{direct}. Rotation and translation are examples of direct isometries. At this point, we will not prove the fact that every isometry of a plane is either direct or indirect, or that if one isometry preserves the orientation of one figure, it also preserves the orientation of all other figures. It is clear that the composition of two direct or two indirect isometries represents a direct isometry. Similarly, the composition of one direct and one indirect isometry is an indirect isometry. Intuitively, with direct isometries we can bring the figure and its image with movement in the plane to the point of overlap. For indirect isometries this is not possible with free movement in the plane - it is necessary to use movement in three-dimensional space. Except for the conditions of directness and indirectness, another very important property of isometries is the number of fixed points. We recall that we call a point fixed under an isometry if it is mapped to itself by that isometry (section \ref{odd2AKSSKL}). Intuitively, a rotation has exactly one fixed point – its center. A translation has no fixed points. A reflection over a line has infinitely many, but they are all fixed points on the axis of that reflection. Is it possible for an isometry to have three non-collinear fixed points? Intuitively, it is clear (and we have also formally proven it in Theorem \ref{IizrekABCident}) that all other points in the plane are also fixed under such an isometry. We have called such an isometry an \index{isometry!identical} identical isometry or the identity, and we have denoted it with $\mathcal{E}$. Obviously, the identity is also a direct isometry, since it preserves all figures. Because of its importance, we will write Theorem \ref{IizrekABCident} once again, but in a slightly different form. \bizrek \label{IizrekABC2} The identity map $\mathcal{E}$ is the only isometry of a plane having three non-collinear fixed points. \eizrek With the following theorem, we will provide another criterion for the identity. \bizrek \label{izo2ftIdent} A direct isometry of a plane that has at least two fixed points is the identity map. \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.1.2.pic} \caption{} \label{sl.izo.6.1.2.pic} \end{figure} \textbf{\textit{Proof.}} We assume that the direct isometry plane $\mathcal{I}$ has at least two fixed points $A$ and $B$ or $A'=\mathcal{I}(A)=A$ and $B'=\mathcal{I}(B)=B$ (Figure \ref{sl.izo.6.1.2.pic}). Let $X$ be an arbitrary point that does not lie on the line $AB$. Let $X'=\mathcal{I}(X)$. Because $\mathcal{I}$ is a direct isometry, the triangles $AXB$ and $AX'B$ are oriented the same, so the point $X'$ lies in the plane with the edge $AB$ and with the point $X$. From $\mathcal{I}:A,B,X\mapsto A,B,X'$ it follows $XA\cong X'A$, $XB\cong X'B$ and $AB\cong AB$. Because also $XA\cong XA$ and $XB\cong XB$, by \ref{izomEnaC'} $X=X'$. Therefore, $X$ is also a fixed point. Because $A$, $B$ and $X$ are three non-linear points of the isometry $\mathcal{I}$, $\mathcal{I}=\mathcal{E}$ (statement \ref{IizrekABC2}). \kdokaz In the following we will formally introduce and consider different types of isometries. %________________________________________________________________________________ \poglavje{Reflections} \label{odd6OsnZrc} Although the basic reflection is an intuitively already known transformation, we will first give a formal definition. Let $s$ be a line of some plane. The transformation of this plane, in which each point of the line $s$ is fixed and in which each point $X$, which does not lie on the line $s$, is mapped to such a point $X'$, that $s$ is the perpendicular bisector of the segment $XX'$, is called \index{reflection!over a line} \pojem{basic reflection} or \index{reflection!basic}\pojem{reflection over a line} $s$ and is denoted by $\mathcal{S}_s$ (Figure \ref{sl.izo.6.2.1.pic}). The line $s$ is \index{axis!of reflections} \pojem{the axis of reflections}. Because from the notation $\mathcal{S}_s$ it is already clear that it is a reflection over the line $s$, we will call it shorter: reflection $\mathcal{S}_s$. \begin{figure}[!htb] \centering \input{sl.izo.6.2.1.pic} \caption{} \label{sl.izo.6.2.1.pic} \end{figure} If the figure $\phi$ is mapped onto the figure $\phi'$ by the reflection $\mathcal{S}_s$, or $\mathcal{S}_s: \phi \rightarrow \phi'$, we will say that the figures $\phi$ and $\phi'$ are \pojem{symmetric} with respect to the axis $s$ (Figure \ref{sl.izo.6.2.2.pic}). The axis $s$ is the \pojem{axis of symmetry} of the figures $\phi$ and $\phi'$. If $\phi=\phi'$ or $\mathcal{S}_s: \phi \rightarrow \phi$, we will say that the figure $\phi$ is \index{figure!axis-symmetric} \pojem{axis-symmetric} or \pojem{axis-centered}. The line $s$ is the \index{axis!symmetry of a figure}\pojem{axis of symmetry} or \index{centered line} \pojem{centered line} of this figure (Figure \ref{sl.izo.6.2.2.pic}). \begin{figure}[!htb] \centering \input{sl.izo.6.2.2.pic} \caption{} \label{sl.izo.6.2.2.pic} \end{figure} From the definition it is already clear that all the fixed points of the reflection $\mathcal{S}_s$ lie on the axis $s$ of this reflection. Although it is intuitively clear, we need to prove the following property of the defined mapping. \bizrek \label{izozrIndIzo} A reflection is an opposite isometry. \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.2.3.pic} \caption{} \label{sl.izo.6.2.3.pic} \end{figure} \textbf{\textit{Proof.}} Let $\mathcal{S}_s$ be a reflection over the line $s$ (Figure \ref{sl.izo.6.2.3.pic}). From the definition it is clear that it represents a bijective mapping. It remains to prove that for any points $X$ and $Y$, which are mapped by $\mathcal{S}_s$ into the points $X'$ and $Y'$, it holds that $XY\cong X'Y'$. We will consider several cases. If $X=X'$ and $Y=Y'$, this relation is automatically fulfilled (statement \ref{sklRelEkv}). Let $X\neq X'$. We will mark with $X_s$ the center of the line segment $XX'$. Because $s$ is the symmetry of the line segment $XX'$, by definition $X_s\in s$ and $XX'\perp s$. If $Y=Y'$, the triangles $XX_sY$ and $X'X_sY$ (or $X'X_sY'$) are congruent (statement \textit{SAS} \ref{SKS}), therefore $XY\cong X'Y'$. We will now look at an example of $X\neq X'$ and $Y\neq Y'$. Similarly, for the center $Y_s$ of the line $YY'$ it holds that $Y_s\in s$ and $YY'\perp s$. The triangles $XX_sY_s$ and $X'X_sY_s$ are congruent (by the theorem \textit{SAS}, so there exists an isometry $\mathcal{I}$, such that $\mathcal{I}: X, X_s,Y_s\mapsto X', X_s,Y_s$. This maps the line segment $X_sY_sX$ onto the line segment $X_sY_sX'$ (by axiom \ref{aksIII2}). Because the isometry $\mathcal{I}$ preserves angles, the line segment $Y_sY$ is mapped onto the line segment $Y_sY'$, the point $Y$ is mapped onto the point $Y'$. From $\mathcal{I}: X, Y\mapsto X', Y'$ it follows at the end that $XY\cong X'Y'$. Let $A,B\in s$ and $C\notin s$. Then $\mathcal{S}_s(A)=A$, $\mathcal{S}_s(B)=A$ and $\mathcal{S}_s(C)=C'\neq C$. This means that the triangle $ABC$ is mapped onto the triangle $ABC'$ by the basic reflection $\mathcal{S}_s$. Because $C,C' \div s$ or $C,C' \div AB$, the two triangles are differently oriented, so the basic reflection $\mathcal{S}_s$ is an indirect isometry. \kdokaz It is clear that in the proof of the previous theorem it also holds that $\mathcal{I}=\mathcal{S}_s$. We can only determine this at the end, when we prove that $\mathcal{S}_s$ is an isometry and use theorem \ref{IizrekABC}. We will now prove some simple properties of reflections across a line. \bizrek \label{izoZrcPrInvol} For an arbitrary line $p$ it holds that $$\mathcal{S}_p^2=\mathcal{E}\hspace*{1mm}\textrm{ i.e. } \hspace*{1mm}\mathcal{S}_p^{-1}=\mathcal{S}_p.$$ \eizrek \textbf{\textit{Proof.}} It is enough to prove that $\mathcal{S}_p^2(X)=X$ for every point $X$ of the plane. Let $\mathcal{S}_p(X)=X'$. If $X\in p$ or $X=X'$, the relation $\mathcal{S}_p^2(X)=X$ is automatically fulfilled. If $X\notin p$, by definition the line $p$ is the perpendicular bisector of the line segments $XX'$ (and also $X'X$), so $\mathcal{S}_p(X')=X$ or $\mathcal{S}_p^2(X)=X$. \kdokaz Every isometry (as well as every mapping) $f:\mathbb{E}^2\rightarrow \mathbb{E}^2$, for which $f^2=\mathcal{E}$, is called \pojem{involution}\index{involution}. The basic reflection is therefore an involution. \bizrek \label{zrcFiksKroz} Let $l$ be an arbitrary circle with the centre $S$ and $p$ a line in the plane. If $S\in p$, then $\mathcal{S}_p(l)=l$. \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.2.4.pic} \caption{} \label{sl.izo.6.2.4.pic} \end{figure} \textbf{\textit{Proof.}} Let $X\in l$ be an arbitrary point of the circle $l$ and $\mathcal{S}_p(X)=X'$ (Figure \ref{sl.izo.6.2.4.pic}). Because $\mathcal{S}_p:S,X\mapsto S, X'$, we have $SX\cong SX'$ or $X'\in l$. Therefore $\mathcal{S}_p(l)\subseteq l$. Similarly, an arbitrary point $Y$ of the circle $l$ is the image of the point $Y'=\mathcal{S}_p^{-1}(Y)=\mathcal{S}_p(Y)$, which lies on this circle. Therefore $\mathcal{S}_p(l)\supseteq l$. Therefore $\mathcal{S}_p(l)=l$. \kdokaz \bzgled \label{izoSimVekt} If $\mathcal{S}_s:A, B\mapsto A', B'$, the vector $\overrightarrow{v}=\overrightarrow{AB}+\overrightarrow{A'B'}$ is parallel to the line $s$. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.2.5.pic} \caption{} \label{sl.izo.6.2.5.pic} \end{figure} \textbf{\textit{Proof.}} Let $A_s$ and $B_s$ be the centres of the lines $AA'$ and $BB'$ (Figure \ref{sl.izo.6.2.5.pic}). Because $p$ is the symmetry of these lines, $A_s, B_s\in p$. Therefore the vector \begin{eqnarray*} \overrightarrow{v}&=&\overrightarrow{AB}+\overrightarrow{A'B'}=\\ &=&(\overrightarrow{AA_s}+\overrightarrow{A_sB_s}+\overrightarrow{B_sB})+ (\overrightarrow{A'A_s}+\overrightarrow{A_sB_s}+\overrightarrow{B_sB'})=\\ &=& 2\overrightarrow{A_sB_s} \end{eqnarray*} is parallel to the line $s$. \kdokaz We will continue with the properties of the basic reflection in the next section, but now let's look at the use of this isometry. \bzgled \label{HeronProbl} (Heron's\footnote{This problem was posed by \index{Heron}\textit{Heron of Alexandria} (20--100). In his work 'Catoprica' he formulated a law that says that a beam that goes from point $A$ and is reflected from line $p$ through point $B$, passes the shortest possible path.} problem) \index{problem!Heron's} Two points $A$ and $B$ are given on the same side of a line $p$. Find a point $X$ on the line $p$ such that the sum $|AX|+|XB|$ is minimal. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.2.6.pic} \caption{} \label{sl.izo.6.2.6.pic} \end{figure} \textbf{\textit{Solution.}} Let $A'$ be the image of point $A$ under reflection $\mathcal{S}_p$ (Figure \ref{sl.izo.6.2.6.pic}). With $X$ we mark the intersection of line $p$ and $A'B$ (points $A'$ and $B$ are on different sides of line $p$). We will prove that $X$ is the desired point. Let $Y\neq X$ be an arbitrary point on line $p$. Because reflection $\mathcal{S}_p$ is an isometry that maps line segment $AX$ to line segment $A'X$ and line segment $AY$ to line segment $A'Y$, we have $AX\cong A'X$ and $AY\cong A'Y$. If we also use the triangle inequality - \ref{neenaktrik} (for triangle $A'YB$), we get: $$|AX| + |XB| = |A'X| + |XB| = |A'B| < |A'Y| + |YB| = |AY| + |YB|,$$ which was to be proven. \kdokaz \bzgled Let $k$ and $l$ be circles on the same side of a line $p$ in the same plane. Construct a point $S$ on the line $p$ such that the tangents from this point to the circles $k$ and $l$ determine the congruent angles with the line $p$. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.2.7.pic} \caption{} \label{sl.izo.6.2.7.pic} \end{figure} \textbf{\textit{Solution.}} Let $q$ and $r$ be the tangents in that order of the circles $k$ and $l$, which intersect on the line $p$ in the point $S$ and with it determine the corresponding angle (Figure \ref{sl.izo.6.2.7.pic}). The line $p$ is the angle bisector of the angle determined by the tangents $q$ and $r$, therefore $\mathcal{S}_p(r)=q$. The line $q$ is therefore also a tangent of the circle $l'=\mathcal{S}_p(l)$. This means that we can plan the line $q$ as the common tangent of the circles $k$ and $l'$ (see example \ref{tang2ehkroz}). Then $S=q\cap p$ and $r=\mathcal{S}_p(q)$. \kdokaz \bzgled \label{FagnanLema} Let $AP$, $BQ$ and $CR$ be the altitudes of a triangle $ABC$. If $P'=\mathcal{S}_{AB}(P)$ and $P''=\mathcal{S}_{AC}(P)$, prove that $P'$, $R$, $Q$ and $P''$ are collinear points. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.2.9a.pic} \caption{} \label{sl.izo.6.2.9a.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.2.9a.pic}) First, from $\angle BRC=90^0$ and $\angle BQC=90^0$ by Tales' theorem \ref{TalesovIzrKroz} it follows that points $R$ and $Q$ lie on a circle with diameter $BC$. Therefore $BRQC$ is a chordal quadrilateral and by theorem \ref{TetivniPogojZunanji} it holds that $\angle ARQ\cong \angle BCQ = \gamma$. Similarly (from the chordality of $CARP$ quadrilateral) it also holds that $\angle BRP\cong\gamma$, so $\angle ARQ\cong \angle BRP$. Because $\mathcal{S}_{AB}:P, R, B\mapsto P', R, B$, it holds that $\angle BRP\cong \angle BRP'$. From the previous relations it follows that $\angle ARQ\cong \angle BRP'$. Triangle $ABC$ is acute, which means that points $P$ and $Q$ are on the same side of line $AB$. From this it follows that points $Q$ and $P'$ are on different sides of line $AB$. From the proven relation $\angle ARQ\cong \angle BRP'$ it now follows that $P'$, $R$ and $Q$ are collinear points. Similarly, points $P''$, $R$ and $Q$ are collinear, which means that all four points $P'$, $R$, $Q$ and $P''$ lie on the same line. \kdokaz \bizrek \label{FagnanLema1} Let $A$ be a point not lying on lines $p$ and $q$ in the same plane. Construct points $B\in p$ and $C\in q$ such that the perimeter of the triangle $ABC$ is minimal. \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.2.8.pic} \caption{} \label{sl.izo.6.2.8.pic} \end{figure} \textbf{\textit{Solution.}} Let $A'=\mathcal{S}_p(A)$, $A''=\mathcal{S}_q(A)$ and $B$ and $C$ be the intersection points of line $A'A''$ with lines $p$ and $q$ (Figure \ref{sl.izo.6.2.8.pic}). We will prove that $ABC$ is the desired triangle. If $B_1$ and $C_1$ are any points (where $B_1\neq B$ or $C_1\neq C$) that lie on the lines $p$ and $q$, the perimeter of the triangle $AB_1C_1$ is equal to the length of the broken line $A'B_1C_1A''$ ($AB_1\cong A'B_1$ and $AC_1\cong A'' C_1$), and the perimeter of the triangle $ABC$ is equal to the length of the line segment $A'A''$ ($AB\cong A'B$ and $AC\cong A'' C$). From Theorem \ref{neenakIzlLin} it follows that the first length is greater than the second, so the perimeter of the triangle $ABC$ is less than the perimeter of the triangle $AB_1C_1$. \kdokaz \bzgled For a given acute triangle $ABC$ determine the inscribed triangle of minimal perimeter \index{problem!Fagnano}(Fagnano's problem\footnote{\textit{G. F. Fagnano} \index{Fagnano, G. F.} (1717--1797), Italian mathematician, posed this problem in 1775 and solved it with the methods of differential calculus. A more elementary solution to this problem was later given by the Hungarian mathematician \index{Fejer, L.} \textit{L. Fejer} (1880--1959).}). \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.2.9.pic} \caption{} \label{sl.izo.6.2.9.pic} \end{figure} \textbf{\textit{Solution.}} We mark for $AP$, $BQ$ and $CR$ the altitudes of the triangle $ABC$ (Figure \ref{sl.izo.6.2.9.pic}). If $X$ is any point on the side $BC$, then the minimal perimeter of the triangle $XYZ$, where the vertices $Y$ and $Z$ lie on the sides $AC$ and $AB$, is obtained in the way described in the previous Theorem \ref{FagnanLema1}. The points $Y$ and $Z$ are therefore the intersections of the sides $AC$ and $AB$ with the line $X'X''$ ($X'=\mathcal{S}_{AB}(X)$, $X''=\mathcal{S}_{AC}(X)$). In this case, the perimeter of the triangle $XYZ$ is equal to the length of the line segment $X'X''$. So it remains for us to determine for which point $X$ of side $BC$ is the perimeter of triangle $XYZ$ the smallest or the distance $X'X''$ the shortest. From the properties of reflection it follows that $AX'\cong AX\cong AX''$, $\angle X'AB\cong \angle BAX$ and $\angle X''AC\cong \angle CAX$ or $\angle X'AX''=2\angle BAC$. As $X'AX''$ is an isosceles triangle $X'AX''$ is constant (depending on the point $X$), therefore its base $X'X''$ is the shortest when its shortest leg $AX'$ (by \ref{SkladTrikLema}) or the distance $AX\cong AX'\cong AX''$. This is fulfilled when $X$ is the foot of the altitude of triangle $ABC$ from point $A$ or $X=P$. From \ref{FagnanLema} it follows that $Y=Q$ and $Z=R$. So the inscribed triangle with the smallest perimeter is actually the pedal triangle of triangle $ABC$. \kdokaz \bzgled Let $P$ and $Q$ be interior points of an angle $aOb$. Construct a point $X$ on the side $a$ of this angle such that the rays $XP$ and $XQ$ intersect the side $b$ at points $Y$ and $Z$ such that $XY\cong XZ$. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.2.10.pic} \caption{} \label{sl.izo.6.2.10.pic} \end{figure} \textbf{\textit{Solution.}} Let $X$, $Y$ and $Z$ be the sought points (Figure \ref{sl.izo.6.2.10.pic}). Let $S$ be the center of the segment $YZ$ and $P'=\mathcal{S}_a(P)$. The measure of the angle $aOb$ we denote by $\omega$. The triangle $XYZ$ is equilateral, therefore the circumcenter $XS$ is at the same time the altitude and the symmetry of the internal angle $XYZ$ of this triangle. So it holds that $\angle OSX=90^0$ and $\angle YXS\cong \angle ZXS$. From $\mathcal{S}_a:P, X, O\mapsto P', X, O$ it follows that $\angle PXO\cong \angle P'XO$. Now we can calculate the measure of the angle $P'XQ$: \begin{eqnarray*} \angle P' XQ &=& \angle P' XP + \angle PXQ = 2\angle OXP + 2\angle PXS\\ &=& 2\angle OXS = 2(90° - \omega) =180° - 2\omega. \end{eqnarray*} Since the points $P'$ and $Q$ are known to us, the point $X$ is the intersection of the segment $a$ and the arc with the chord $P'Q$ and the central angle $180° - 2\omega$ (the theorem \ref{ObodKotGMT}). \kdokaz \bzgled Let $ABCD$ be a rectangle such that $|AB|=3|BC|$. Suppose that $E$ and $F$ are points on the side $AB$ such that $AE\cong EF\cong FB$. Prove $$\angle AED+\angle AFD+\angle ABD=90^0.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.2.11.pic} \caption{} \label{sl.izo.6.2.11.pic} \end{figure} \textbf{\textit{Solution.}} Let $F'=\mathcal{S}_{CD}(F)$ and $B'=\mathcal{S}_{CD}(B)$ (Figure \ref{sl.izo.6.2.11.pic}). Then $DF'\cong DF$ and $\angle DF'F\cong \angle DFF'$. From the similarity of triangles $DAF$ and $F'B'B$ (the \textit{SAS} theorem \ref{SKS}) it follows that $DF\cong F'B$ and $\angle DFA\cong \angle B'BF'$. It is also true that $\angle FF'B\cong \angle B'BF'$ (the \ref{KotiTransverzala} theorem). Therefore: $$\angle DF'B=\angle DF'F+\angle FF'B=\angle DFF'+\angle DFA=\angle AFF'=90^0.$$ Because $DF'\cong F'B$, $DF'B$ is an isosceles right triangle with the hypotenuse $BD$, therefore (the \ref{enakokraki} theorem) $\angle DBF'=\angle BDF'=45^0$. Because $DAE$ is also an isosceles right triangle with the hypotenuse $DE$, $\angle AED=45^0$ or $\angle AED=\angle DBF'$. Therefore: $$\angle AED+\angle AFD+\angle ABD=\angle DBF'+\angle F'BB'+ \angle ABD=\angle ABC=90^0,$$ which was to be proven. \kdokaz At the next example we will describe the standard procedure of using isometries in design tasks. We assume that it is necessary to construct points $X\in k$ and $Y\in l$, where $k$ and $l$ are given circles (it can also be lines or a circle and a line), but we know that for some isometry $\mathcal{I}$ it is true that $\mathcal{I}(X)=Y$. In this case from $X\in k$ it follows that $Y=\mathcal{I}(X)\in \mathcal{I}(k)$. Because $Y$ is also in $l$, we can construct the point $Y$ from the condition $Y\in l\cap \mathcal{I}(k)$. \bzgled Two circles $k$ and $l$ on different sides of a line $p$ are given. Construct a square $ABCD$ such that $A\in k$, $C\in l$ and $B,D\in p$. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.2.12.pic} \caption{} \label{sl.izo.6.2.12.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.2.12.pic}) Let $ABCD$ be the sought square. Then $\mathcal{S}_p(C)=A$. The point $C$ lies on the circle $l$, therefore its image -- point $A$ -- after reflection $\mathcal{S}_p$ lies on the image of the circle $l$ -- circle $l'$, which we can plot. But the point $A$ also lies on the circle $k$, therefore $A\in l'\cap k$. In this way, we first find the vertex $A$, and then $C=\mathcal{S}_p(A)$. The center of the square $S$ is found as the center of the diagonal $AC$. In the end, the vertices $B$ and $D$ are the intersections of the line $p$ and the circle with center $S$ and radius $SA$. The number of solutions depends on the number of intersections of the circles $l'$ and $k$. \kdokaz \bnaloga\footnote{38. IMO Argentina - 1997, Problem 2.} The angle at $A$ is the smallest angle of triangle $ABC$. The points $B$ and $C$ divide the circumcircle of the triangle into two arcs. Let $U$ be an interior point of the arc between $B$ and $C$ which does not contain $A$. The perpendicular bisectors of $AB$ and $AC$ meet the line $AU$ at $V$ and $W$, respectively. The lines $BV$ and $CW$ meet at $T$. Show that $$|AU| =|TB| + |TC|.$$ \enaloga \begin{figure}[!htb] \centering \input{sl.izo.6.2.IMO1.pic} \caption{} \label{sl.izo.6.2.IMO1.pic} \end{figure} \textbf{\textit{Solution.}} Let's mark with $q$ and $p$ the sides $AB$ and $AC$ of the triangle $ABC$ (Figure \ref{sl.izo.6.2.IMO1.pic}). Their intersection -- point $O$ -- is the center of the circumscribed circle $l$ of this triangle. According to the assumption, $V=AU\cap q$ and $W=AU\cap p$. Let $D$ be the other intersection of the line segment $CW$ with the circle $l$. With $\mathcal{S}_p$ we mark the reflection over the line $p$. $\mathcal{S}_p$ maps the points $A$, $C$, $W$ and $O$ in order, into the points $C$, $A$, $W$ and $O$ and the line segment $AW$ into the line segment $CW$. Because $O\in p$, by the theorem \ref{zrcFiksKroz} $\mathcal{S}_p(l) =l$. From this follows: $$\mathcal{S}_p(U) =\mathcal{S}_p(AW\cap l)= \mathcal{S}_p(AW)\cap \mathcal{S}_p(l)= CW\cap l=D.$$ Therefore $\mathcal{S}_p:\hspace*{1mm} A,U\mapsto C,D$, so $AU\cong CD$ and $AD\cong CU$. From the similarity of the line segments $AD$ and $CU$ follows the similarity of the corresponding angles (theorem \ref{SklTetSklObKot}) or equivalently $\angle ABD\cong\angle UAC$. But the angles $BDC$ and $BAC$ are also similar over the line segment $BC$ (theorem\ref{ObodObodKot}). Because the point $V$ lies on the line $q$, which is the symmetry of the line segment $AB$, $\mathcal{S}_q:\hspace*{1mm} A,B,V\mapsto B,A,V$. This means that $\angle BAV\cong\angle ABV$. From the similarity of the corresponding angles it follows: \begin{eqnarray*} \angle BDT&=&\angle BDC\cong\angle BAC=\\ &=&\angle BAU+ \angle UAC=\\ &=&\angle ABV+\angle ABD=\\ &=&\angle DBT. \end{eqnarray*} Therefore the triangle $DTB$ is isosceles and by the theorem \ref{enakokraki} it holds that $TD\cong TB$. In the end we get: $$|AU|= |CD|=|CT|+|TD|=|CT|+|TB|,$$ which was to be proven. \kdokaz \bnaloga \footnote{50. IMO Germany - 2009, Problem 4.} Let $ABC$ be a triangle with $|AB|=|AC|$. The angle bisectors of $\angle CAB$ and $\angle ABC$ meet the sides $BC$ and $CA$ at $D$ and $E$, respectively. Let $K$ be the incentre of triangle $ADC$. Suppose that $\angle BEK=45^0$. Find all possible values of $\angle CAB$. \enaloga \textbf{\textit{Solution.}} We denote by $S$ the centre of the inscribed circle of triangle $ABC$. This means that $S$ is the intersection of the altitudes $AD$ and $BE$ of angles $\alpha=\angle CAB$ and $\beta=\angle ABC$. The point $S$ also lies on the altitude $CF$ ($F\in AB$) of angle $\gamma=\angle ACB$. From $|AB|=|AC|$ it follows $\beta\cong\gamma$, therefore $\angle SBC\cong\angle SCB$. By assumption, the point $K$ is the centre of the inscribed circle of triangle $ADC$, therefore the point $K$ lies on the altitude $CS$ of angle $\angle ACD=\angle ACB$. The segment $DK$ is the altitude of angle $ADC$. From the similarity of triangles $ADB$ and $ADC$ (the \textit{SAS} theorem \ref{SKS}) it follows that $\angle ADC\cong\angle ADB=90^0$. From this it follows that $\angle SDK=45^0$. Let $E'=\mathcal{S}_{SC}(E)$. Because $CS$ is the altitude of angle $ACB$, the segment $CA$ is mapped onto the segment $CB$ by the reflection $\mathcal{S}_{SC}$. Therefore $E'\in CB$. Because $S,K,C\in SC$, we have $\mathcal{S}_{SC}:\hspace*{1mm}S,K,C\mapsto S,K,C$. \begin{figure}[!htb] \centering \input{sl.izo.6.2.IMO2a.pic} \caption{} \label{sl.izo.6.2.IMO2a.pic} \end{figure} We will consider two cases. If $E'=D$ (Figure \ref{sl.izo.6.2.IMO2a.pic}), then $\angle SEC\cong \angle SDC=90^0$. Triangles $ABE$ and $CBE$ are similar in this case (the \textit{ASA} theorem \ref{KSK}), therefore $AB\cong BC$. Triangle $ABC$ is isosceles or $\angle BAC=60^0$. \begin{figure}[!htb] \centering \input{sl.izo.6.2.IMO2.pic} \caption{} \label{sl.izo.6.2.IMO2.pic} \end{figure} Let $E'\neq D$ (Figure \ref{sl.izo.6.2.IMO2.pic}). From $\mathcal{S}_{SC}:\hspace*{1mm}S,K\mapsto S,K$ it follows that $\angle SE'K \cong \angle SEK=45^0$. Because $\angle SDK \cong \angle SE'K=45^0$, by izreku \ref{ObodKotGMT} the points $S$, $K$, $D$ and $E'$ are concircle. From izreku \ref{TalesovIzrKroz} it then follows that $\angle SKE'\cong \angle SDE'=90^0$. Because the reflection $\mathcal{S}_{SC}$ maps the angle $SKE$ into the angle $SKE'$, $\angle SKE\cong \angle SKE'=90^0$ (the points $E$, $K$ and $E'$ are therefore collinear). From this and izreku \ref{VsotKotTrik} for the triangle $SKE$ it follows that $\angle KSE=45^0$. It is the outer angle of the triangle $SBC$, so by izreku \ref{zunanjiNotrNotr} $\angle SBC +\angle SCB=45^0$ or $\beta+\gamma=90^0$. From this relation it follows that $\angle BAC=90^0$. The possible values of the angle $\angle CAB$ are therefore $60^0$ and $90^0$. \kdokaz %________________________________________________________________________________ \poglavje{More on Reflections} \label{odd6Sopi} In the initial research on isometries, we found that every direct isometry with at least two fixed points represents the identity. It is intuitively clear that an opposite isometry of the plane with at least one fixed point is a reflection. We will prove an even stronger statement. \bizrek \label{izo1ftIndZrc} An opposite isometry of the plane with at least one fixed point is a reflection. The axis of this reflection passes through this point. \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.3.1.pic} \caption{} \label{sl.izo.6.3.1.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.3.1.pic}) Let $A$ be a fixed point of an indirect isometry $\mathcal{I}$ of some plane. Because it is indirect, the isometry $\mathcal{I}$ is not the identity, so there is such a point $B$ of this plane that $\mathcal{I}(B)=B'\neq B$. Let $p$ be the perpendicular bisector of the segment $BB'$. Therefore, $\mathcal{I}$ is an isometry that maps the point $A$ and $B$ to the point $A$ and $B'$, so $AB\cong AB'$. This means that the point $A$ lies on the perpendicular bisector $p$ of the segment $BB'$. We prove that the isometry $\mathcal{I}$ represents the reflection over the line $p$ or $\mathcal{I} = \mathcal{S}_p$. The composition $\mathcal{S}_p \circ \mathcal{I}$ (of two indirect isometries) is a direct isometry with two fixed points $A$ and $B$, so based on the aforementioned formula \ref{izo2ftIdent} it represents the identity. Therefore, $\mathcal{S}_p \circ \mathcal{I}=\mathcal{E}$. From this it follows that $ \mathcal{I}=\mathcal{S}_p^{-1} \circ\mathcal{E}$ or $\mathcal{I} = \mathcal{S}_p$. The following formula is very useful. \bizrek \label{izoTransmutacija} Let $\mathcal{I}$ be an isometry and $p$ an arbitrary line in the plane. Suppose that this isometry maps the line $p$ to a line $p'$. If $\mathcal{S}_p$ is a reflection with axis $p$, then $$\mathcal{I}\circ \mathcal{S}_p\circ \mathcal{I}^{-1} = \mathcal{S}_{p'}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.3.2.pic} \caption{} \label{sl.izo.6.3.2.pic} \end{figure} \textbf{\textit{Proof.}} Let $Y$ be an arbitrary point of the line $p'$ and $X = \mathcal{I}^{-1}(Y)$ (Figure \ref{sl.izo.6.3.2.pic}). Since $\mathcal{I}(p)=p'$ or $\mathcal{I}^{-1}(p')=p$, the point $X$ lies on the line $p$ and is $\mathcal{S}_p(X)=X$. Therefore, it holds: $$\mathcal{I}\circ\mathcal{S}_p\circ\mathcal{I}^{-1}(Y)= \mathcal{I}\circ\mathcal{S}_p(X)= \mathcal{I}(X)=Y.$$ Thus, the composition $\mathcal{I}\circ\mathcal{S}_p\circ\mathcal{I}^{-1}$ is an indirect isometry with a fixed point $Y$, therefore, according to the previous statement \ref{izo1ftIndZrc}, it represents the central reflection. But $Y$ is an arbitrary point of the line $p'$, which means that the axis of this reflection is just the line $p'$ or $\mathcal{I}\circ \mathcal{S}_p\circ \mathcal{I}^{-1} = \mathcal{S}_{p'}$. \kdokaz The transformation $\mathcal{I}\circ\mathcal{S}_p\circ\mathcal{I}^{-1}$ from the previous statement is called the \index{transmutation!reflection across a line} \pojem{transmutation} of the reflection $\mathcal{S}_p$ with the isometry $\mathcal{I}$. We prove a simple consequence of the previous statement. \bzgled \label{izoZrcKomut} Two reflections in the plane commute if and only if their axis are perpendicular or coincident, i.e. $$\mathcal{S}_p \circ \mathcal{S}_q = \mathcal{S}_q \circ \mathcal{S}_p \Leftrightarrow (p\perp q \vee p=q).$$ \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.3.3.pic} \caption{} \label{sl.izo.6.3.3.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.3.3.pic}) First, $\mathcal{S}_p \circ \mathcal{S}_q = \mathcal{S}_q \circ \mathcal{S}_p \Leftrightarrow \mathcal{S}_p= \mathcal{S}_q \circ \mathcal{S}_p \circ \mathcal{S}_q^{-1}$. If we use the previous statement \ref{izoTransmutacija} for $\mathcal{I} = \mathcal{S}_q$, we get that the last equality is equivalent to $\mathcal{S}_p = \mathcal{S}_{p'}$, where $p' = \mathcal{S}_q(p)$. This is exactly the case when $p=p'=\mathcal{S}_q(p)$ i.e. when $p\perp q$ or $p=q$. \kdokaz The composite of two basic reflections $\mathcal{I}=\mathcal{S}_p \circ \mathcal{S}_q$ will be very important in further research isometries. We immediately notice that the composite of two indirect isometries represents a direct isometry. If the axes of these reflections coincide, we have already determined that it is the identity (statement \ref{izoZrcPrInvol}). It is interesting to ask what $\mathcal{I}$ represents in the general case, when the axes of $p$ and $q$ of these reflections are different and coplanar; especially in the case when they intersect or are parallel. We will find the answer to this question in the following sections. In the next statement we will find out something about the fixed points of the composite of two basic reflections. \bizrek \label{izoKomppqX} Let $p$ and $q$ be two distinct coplanar lines. Then $$\mathcal{S}_q\circ\mathcal{S}_p(X)=X \Leftrightarrow p\cap q=\{X\}.$$ \eizrek \textbf{\textit{Proof.}} ($\Leftarrow$) Trivial, because from $X\in p$ and $X\in q$ it follows $\mathcal{S}_p(X)= \mathcal{S}_q(X)=X$ i.e. $\mathcal{S}_q\circ\mathcal{S}_p(X)=X$. ($\Rightarrow$) Let $\mathcal{S}_q\circ\mathcal{S}_p(X)=X$ in $\mathcal{S}_p(X)=X'$. From this it follows that $\mathcal{S}_q(X')=X$. Assume that $X\neq X'$. In this case, according to the definition of the axis of reflection $p$ and $q$ are the symmetrals of the line segment $XX'$. But this is not possible, because the line segment has one symmetral, while $p$ and $q$ are different by assumption. Therefore, $X=X'$. From this it follows that $\mathcal{S}_p(X)=\mathcal{S}_q(X)=X$, which means that $X\in p$ and $X\in q$ or $X\in p\cap q$. Because $p$ and $q$ are different lines, by Axiom \ref{AksI1} $p\cap q=\{X\}$. \kdokaz If the lines $p$ and $q$ intersect, from the previous statement it follows that the composite $\mathcal{S}_q\circ\mathcal{S}_p$ has only one fixed point, which is their intersection point. If $p$ and $q$ are parallel, the composite $\mathcal{S}_q\circ\mathcal{S}_p$ has no fixed points. For further research of isometries, the following concept is very characteristic. The set of all lines of a plane that pass through one point $S$ of this plane, is called the \index{šop!konkurentnih premic} \pojem{set of concurrent lines} (or \index{šop!eliptični}\pojem{elliptic set}) with center $S$ and is denoted by $\mathcal{X}_S$. The set of all lines of a plane that are parallel to one line $s$ of this plane, is called the \index{šop!vzporednic} \pojem{set of parallel lines} (or \index{šop!parabolični}\pojem{parabolic set} or \index{snop premic}\pojem{set of lines}) and is denoted by $\mathcal{X}_s$. \begin{figure}[!htb] \centering \input{sl.izo.6.3.4.pic} \caption{} \label{sl.izo.6.3.4.pic} \end{figure} In both cases we will talk about the \pojem{set of lines} (Figure \ref{sl.izo.6.3.4.pic}). If we will not specifically emphasize whether it is a set of concurrent lines or a set of parallel lines, we will denote this set only by $\mathcal{X}$. In connection with the introduced concepts, the following statement is valid. \bizrek \label{izoSop} Let $p$, $q$ and $r$ be lines in the plane. The product $\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p$ is a reflection if and only if the lines $p$, $q$ and $r$ belong to the same family of lines. The axis of the new reflection also belongs to this family. \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.3.5.pic} \caption{} \label{sl.izo.6.3.5.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.3.5.pic}) ($\Leftarrow$) First, let us assume that the lines $p$, $q$ and $r$ belong to the same family of lines $\mathcal{X}$. We will consider two cases. \textit{1)} Let $\mathcal{X}$ be a family of concurrent lines with a center $S$ or $p, q, r\in \mathcal{X}_S$. In this case, the composition $\mathcal{I}=\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p$ is an indirect isometry with a fixed point $S$, so according to the izorek \ref{izo1ftIndZrc} it represents a reflection $\mathcal{S}_t$ with an axis that goes through the point $S$ or $t\in \mathcal{X}_S$. \textit{2)} Let $\mathcal{X}$ be a set of parallel lines and $n$ any common rectangle of lines $p$, $q$ and $r$. In this case, the composite $\mathcal{I}=\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p$ is an indirect isometry and it holds that $\mathcal{I}(n)=n$. We will now prove that there exists a fixed point on line $n$ of this isometry. Let $A$ be any point on line $n$ and $\mathcal{I}(A)=A'$. If $A$ is not a fixed point of isometry $\mathcal{I}$, with $O$ we denote the center of line $AA'$ and with $O'=\mathcal{I}(O)$. $\mathcal{I}$ is an indirect isometry, therefore vectors $AO$ and $A'O'$ are congruent and have different orientation on line $n$. From this and from the fact that $O$ is the center of line $AA'$, it follows that $\overrightarrow{A'O}=\overrightarrow{OA}=\overrightarrow{A'O'}$. Therefore $O=O'$ or $O$ is a fixed point of isometry $\mathcal{I}$, which represents some basic reflection $\mathcal{S}_t$ (statement \ref{izo1ftIndZrc}). In this case $\mathcal{S}_t(n)=\mathcal{I}(n)=n$. This means that $t=n$ or $t\perp n$. The first option is not possible, because from $\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p=\mathcal{S}_n$ it follows that $\mathcal{S}_q \circ \mathcal{S}_p=\mathcal{S}_r\circ \mathcal{S}_n$. The last relation is not possible, because lines $p$ and $q$ are parallel, lines $r$ and $n$ are perpendicular and they intersect (statement \ref{izoKomppqX}). Therefore $t\perp n$, which means that $t$ is also a line from the set of parallel lines $\mathcal{X}$. ($\Rightarrow$) We will now assume that $\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p=\mathcal{S}_t$ for some line $t$. In this case it holds that $\mathcal{S}_q \circ \mathcal{S}_p=\mathcal{S}_r \circ\mathcal{S}_t$. We will again consider two cases. \textit{1)} We will assume that $p$ and $q$ intersect in some point $S$. In this case, it holds that $\mathcal{S}_r \circ\mathcal{S}_t(S) = \mathcal{S}_q \circ \mathcal{S}_p(S)=S$. According to statement \ref{izoKomppqX}, lines $r$ and $t$ also intersect in point $S$, therefore $p, q, r, t\in \mathcal{X}_S$. \textit{2)} Let's assume that the lines $p$ and $q$ are parallel. From the aforementioned equation \ref{izoKomppqX} it follows that the composite $\mathcal{S}_q \circ \mathcal{S}_p$ has no fixed points. Then neither does the composite $\mathcal{S}_r \circ \mathcal{S}_t$, so $r\parallel t$. Let $n$ be the common perpendicular of the lines $p$ and $q$. Then $\mathcal{S}_r \circ\mathcal{S}_t(n) = \mathcal{S}_q \circ \mathcal{S}_p(n)=n$. From this it follows that $\mathcal{S}_t(n)=\mathcal{S}_r(n)=n'$. If $n\neq n'$, then both parallels $p$ and $q$ are axes of symmetry of the lines $n$ and $n'$, which is not possible. Therefore $n=n'$ and $\mathcal{S}_t(n)=\mathcal{S}_r(n)=n$. So $r,t\perp n$, which means that the lines $p$, $q$, $r$ and $t$ belong to the family of lines that are all perpendicular to the line $n$. \kdokaz The consequence of the proven statement is the following equation. \bizrek \label{izoSop2n+1} The product of an odd number of reflections with the axes from the same family of lines is also a reflection. \eizrek \textbf{\textit{Proof.}} We will carry out the proof by induction on the number $n\geq 1$, where $m=2n+1$ (odd) is the number of lines. For $n=1$ or $m=3$, the statement is a direct consequence of the previous equation \ref{izoSop}. We assume that for every $k\in \mathbb{N}$ ($k\geq 1$) and every $(2 k+1)$-tuple of lines $p_1,p_2,\ldots ,p_{2k+1}$, which are all from the same family, the composite $\mathcal{S}_{p_{2k+1}}\circ \cdots \circ\mathcal{S}_{p_2}\circ\mathcal{S}_{p_1}$ represents some basic reflection $\mathcal{S}_p$ (inductive assumption). It is necessary to prove that then for $k+1$ and every $(2(k+1)+1)$-terica of lines $p_1,p_2,\ldots ,p_{2(k+1)+1}$, which are all from the same family of lines, the composition $\mathcal{I}=\mathcal{S}_{p_{2(k+1)+1}}\circ \cdots \circ\mathcal{S}_{p_2}\circ\mathcal{S}_{p_1}$ represents a certain basic reflection $\mathcal{S}_{p'}$. But: \begin{eqnarray*} \mathcal{I}&=&\mathcal{S}_{p_{2(k+1)+1}}\circ \cdots \circ\mathcal{S}_{p_2}\circ\mathcal{S}_{p_1}=\\ &=&\mathcal{S}_{p_{2k+3}}\circ\mathcal{S}_{p_{2k+2}}\circ \mathcal{S}_{p_{2k+1}}\circ \cdots \circ\mathcal{S}_{p_2}\circ\mathcal{S}_{p_1}=\\ &=& \mathcal{S}_{p_{2k+3}}\circ\mathcal{S}_{p_{2k+2}}\circ \mathcal{S}_p= \hspace*{14mm} \textrm{ (by the induction hypothesis)}\\ &=& \mathcal{S}_{p'} \hspace*{48mm} \textrm{ (by \ref{izoSop}),} \end{eqnarray*} which had to be proven. \kdokaz We shall also prove some direct consequences of \ref{izoSop}. \bzgled \label{iropqrrqp} If $p$, $q$ and $r$ are lines from the same family of lines $\mathcal{X}$, then $$\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p = \mathcal{S}_p \circ \mathcal{S}_q \circ \mathcal{S}_r.$$ \ezgled \textbf{\textit{Proof.}} Because $p$, $q$ and $r$ are lines from the same family of lines $\mathcal{X}$, by the previous \ref{izoSop} it follows that $\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p =\mathcal{S}_t$ for some line $t\in\mathcal{X}$. Therefore: $$(\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p)^2 =\mathcal{S}_t^2=\mathcal{E},\hspace*{3mm}\textrm{or,}$$ $$\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p \circ\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p=\mathcal{E}.$$ If we multiply the last relation from the left by $\mathcal{S}_r$, $\mathcal{S}_q$ and $\mathcal{S}_p$, we get the desired equality. \kdokaz \bzgled \label{izoSopabc} If $a$, $b$ and $c$ are lines such that $\mathcal{S}_a(b)=c$, then those lines are from the same family of lines. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.3.6a.pic} \caption{} \label{sl.izo.6.3.6a.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.3.6a.pic}) From the transmutation theorem \ref{izoTransmutacija} it follows that $\mathcal{S}_a\circ\mathcal{S}_b\circ\mathcal{S}_a= \mathcal{S}_{\mathcal{S}_a(b)}= \mathcal{S}_c$. By multiplying the obtained relation $\mathcal{S}_a\circ\mathcal{S}_b\circ\mathcal{S}_a= \mathcal{S}_c$ first from the right with $\mathcal{S}_a$, and then from the left with $\mathcal{S}_c$, we get $\mathcal{S}_c\circ\mathcal{S}_a\circ\mathcal{S}_b=\mathcal{S}_a$. By theorem \ref{izoSop} the lines $a$, $b$ and $c$ belong to the same family of lines. \kdokaz It is clear that the line $a$ from the previous theorem is the line of symmetry of the lines $b$ and $c$. With the following theorem we will determine the line $t$ from the relation $\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p =\mathcal{S}_t$ for lines $p$, $q$ and $r$ from the same family of lines. \bizrek \label{izoSoppqrt} If $p$, $q$, $r$ and $t$ are lines such that $\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p =\mathcal{S}_t$, then the axis of symmetry of the lines $p$ and $r$ is also the axis of symmetry of the lines $q$ and $t$. \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.3.6.pic} \caption{} \label{sl.izo.6.3.6.pic} \end{figure} \textbf{\textit{Proof.}} From $\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p =\mathcal{S}_t$ it follows that the lines $p$, $q$, $r$ and $t$ belong to the same set $\mathcal{X}$ (statement \ref{izoSop}). Let $s$ be the line of symmetry of the lines $p$ and $r$ (Figure \ref{sl.izo.6.3.6.pic}). This means that $\mathcal{S}_s(p)=r$, so according to the transmutation statement \ref{izoTransmutacija} also $\mathcal{S}_s\circ\mathcal{S}_p\circ\mathcal{S}_s =\mathcal{S}_{\mathcal{S}_s(p)} =\mathcal{S}_r$. From this it further follows: $$\mathcal{S}_t = \mathcal{S}_r \circ\mathcal{S}_q \circ \mathcal{S}_p = (\mathcal{S}_s\circ\mathcal{S}_p\circ\mathcal{S}_s)\circ \mathcal{S}_q\circ\mathcal{S}_p = \mathcal{S}_s\circ(\mathcal{S}_p\circ \mathcal{S}_s\circ \mathcal{S}_q)\circ \mathcal{S}_p.$$ Because $\mathcal{S}_s(p)=r$, from the previous example \ref{izoSopabc} it follows that the lines $s$, $p$ and $r$ belong to the same set, so $s\in \mathcal{X}$. From statement \ref{iropqrrqp} it follows $\mathcal{S}_p\circ \mathcal{S}_s\circ \mathcal{S}_q = \mathcal{S}_q\circ \mathcal{S}_s\circ \mathcal{S}_p$. If we use the transmutation statement \ref{izoTransmutacija} again, we get: $$\mathcal{S}_t = \mathcal{S}_s\circ\mathcal{S}_p\circ \mathcal{S}_s\circ \mathcal{S}_q\circ \mathcal{S}_p= \mathcal{S}_s\circ\mathcal{S}_q\circ \mathcal{S}_s\circ \mathcal{S}_p\circ \mathcal{S}_p= \mathcal{S}_s\circ\mathcal{S}_q\circ \mathcal{S}_s= \mathcal{S}_{\mathcal{S}_s(q)}.$$ From $\mathcal{S}_t =\mathcal{S}_{\mathcal{S}_s(q)}$ it follows $\mathcal{S}_s(q)=t$, which means that the line $s$ is the line of symmetry of the lines $q$ and $t$. \kdokaz In the next planning task we will illustrate two methods of solving. \bizrek Let $p$, $q$ and $r$ be lines in the plane. Construct a triangle $ABC$ such that its vertices $B$ and $C$ lie on the line $p$ and the lines $q$ and $r$ are perpendicular bisectors of the sides $AB$ and $AC$. \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.3.7.pic} \caption{} \label{sl.izo.6.3.7.pic} \end{figure} \textbf{\textit{Solution.}} (first way) Let $q$ and $r$ be the perpendicular bisectors of the sides $AB$ and $AC$ (Figure \ref{sl.izo.6.3.7.pic}\textit{a}). Then $\mathcal{S}_q(B)=A$ and $\mathcal{S}_r(C)=A$. The points $B$ and $C$ lie on the line $p$, so the point $A$ lies on the images of this line with respect to the reflections $\mathcal{S}_q$ and $\mathcal{S}_r$. Therefore, we can construct the point $A$ as the intersection of the lines $p'_q=\mathcal{S}_q(p)$ and $p'_r=\mathcal{S}_r(p)$. \textbf{\textit{Solution.}} (second way) The intersection of the lines $q$ and $r$ is the center of the circumscribed circle of this triangle (see \ref{SredOcrtaneKrozn}); we denote it by $O$ (Figure \ref{sl.izo.6.3.7.pic}\textit{b}). Therefore, we can also construct the third perpendicular bisector $s$ of the side $BC$, which is perpendicular to the line $p$ at the point $O$. The composition $\mathcal{S}_q \circ \mathcal{S}_r \circ \mathcal{S}_s$ by \ref{izoSop} is a certain basic reflection $\mathcal{S}_l$ (because $q, r, s\in \mathcal{X}_O$). The fixed points of this composition are $O$ and $B$, so $\mathcal{S}_q \circ \mathcal{S}_r \circ \mathcal{S}_s = \mathcal{S}_{OB}$. Therefore, we can construct the line $OB=l$ as the perpendicular bisector of the segment $XX'$, where $X$ is an arbitrary point and $X'=\mathcal{S}_q \circ \mathcal{S}_r \circ \mathcal{S}_s(X)$. In the case when $X=X'$, we have $l=OX$. Finally, we obtain the point $B$ as the intersection of the lines $p$ and $l$. \kdokaz \bzgled Let $ABCDE$ be a pentagon with a right angle at the vertex $A$. The perpendicular bisectors of the sides $AE$, $BC$, and $CD$ (lines $p$, $q$, and $r$) intersect at a point $O$ and the perpendicular bisectors of the sides $AB$ and $DE$ (lines $x$ and $y$) intersect at a point $S$ ($S\neq O$). Prove that one of the vertices of a triangle $KLM$, such that $p$, $q$ and $r$ are the perpendicular bisectors of its sides, belongs to the line $OS$. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.3.8.pic} \caption{} \label{sl.izo.6.3.8.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.3.8.pic}) Let $\mathcal{I} = \mathcal{S}_y \circ\mathcal{S}_r \circ\mathcal{S}_q \circ\mathcal{S}_x \circ\mathcal{S}_p$. $\mathcal{I}$ is an indirect isometry, with a fixed point $E$, so by \ref{izo1ftIndZrc} it represents some reflection $S_l$. Because $\angle EAB$ is a right angle, simetrali $p$ and $x$ are perpendicular, so the reflections $\mathcal{S}_p$ and $\mathcal{S}_x$ commute (\ref{izoZrcKomut}). So we have: $$\mathcal{I} =\mathcal{S}_l= \mathcal{S}_y \circ\mathcal{S}_r \circ\mathcal{S}_q \circ\mathcal{S}_x \circ\mathcal{S}_p= \mathcal{S}_y \circ (\mathcal{S}_r \circ\mathcal{S}_q \circ\mathcal{S}_p) \circ\mathcal{S}_x.$$ Lines $p$, $q$ and $r$ belong to the same $\mathcal{X}_O$, so the composite $\mathcal{S}_r \circ\mathcal{S}_q \circ\mathcal{S}_p$ represents some reflection $\mathcal{S}_t$, where it also holds that $t\in\mathcal{X}_O$ or the $t$-axis goes through the point $O$ (\ref{izoSop}). So we have $\mathcal{S}_l= \mathcal{S}_y \circ\mathcal{S}_t \circ\mathcal{S}_x$, so lines $x$, $t$ and $y$ belong to the same, or the $t$-line goes through the point $S=x\cap y$. Because the $t$-line goes through different points $O$ and $S$, $t=OS$. So $\mathcal{S}_{OS}=\mathcal{S}_t=\mathcal{S}_r \circ\mathcal{S}_q \circ\mathcal{S}_p$. Let $p$, $q$ and $r$ be the simetrali of sides $KL$, $LM$ and $MK$ of triangle $KLM$. In this case $\mathcal{S}_{OS}(K)=\mathcal{S}_t(K)=\mathcal{S}_r \circ\mathcal{S}_q \circ\mathcal{S}_p(K)=K$, so point $K$ lies on the line $OS$. \kdokaz %________________________________________________________________________________ \poglavje{Rotations} \label{odd6Rotac} So far we have learned about isometries that have more than one fixed point. We have proven that there are two such types of isometries, namely identity as a direct isometry with at least two fixed points (Theorem \ref{izo2ftIdent}) and reflection as an indirect isometry with at least one fixed point (Theorem \ref{izo1ftIndZrc}). In this section we will learn about a type of direct isometries that have exactly one fixed point. Let $S$ be an arbitrary point and $\omega\neq 0$ be an oriented plane. The transformation of this plane, in which the point $S$ is fixed, each other point $X\neq S$ of this plane is mapped to such a point $X'$, that $\measuredangle XSX'\cong \omega$ and $SX'\cong SX$, is called a rotation with center $S$ for angle $\omega$; we denote it with $\mathcal{R}_{S,\omega}$ (Figure \ref{sl.izo.6.4.1.pic}). \begin{figure}[!htb] \centering \input{sl.izo.6.4.1.pic} \caption{} \label{sl.izo.6.4.1.pic} \end{figure} From the definition itself, the following theorems directly follow. \bizrek \label{RotacFiksT} The only fixed point of a rotation $\mathcal{R}_{S,\omega}$ is its center $S$ i.e. $$\mathcal{R}_{S,\omega}(X)=X \hspace*{1mm} \Leftrightarrow \hspace*{1mm} X=S.$$ \eizrek \bizrek \label{rotacEnaki} Rotations $\mathcal{R}_{S,\alpha}$ and $\mathcal{R}_{B,\beta}$ are equal if and only if $A=B$ and $\alpha=\beta$ i.e. $$\mathcal{R}_{S,\alpha}=\mathcal{R}_{B,\beta} \hspace*{1mm} \Leftrightarrow \hspace*{1mm} A=B \hspace*{1mm} \wedge\hspace*{1mm} \alpha=\beta.$$ \eizrek \bizrek The inverse transformation of a rotation is a rotation with the same center and congruent angle with opposite orientation, i.e. $$\mathcal{R}_{S,\omega}^{-1}=\mathcal{R}_{S,-\omega}.$$ \eizrek Although it is intuitively clear, we need to prove that rotation is an isometry. \bizrek Rotations are isometries of the plane. \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.4.2.pic} \caption{} \label{sl.izo.6.4.2.pic} \end{figure} \textbf{\textit{Proof.}} Let $\mathcal{R}_{S,\omega}$ be an arbitrary rotation. From the definition it is clear that it represents a bijective mapping. It remains to be proven that for any points $X$ and $Y$ and their images $X'$ and $Y'$ ($\mathcal{R}_{S,\omega}:X,Y\mapsto X', Y'$) it holds $X'Y'\cong XY$ (Figure \ref{sl.izo.6.4.2.pic}). If one of the points $X$ or $Y$ is equal to the center of the rotation $S$ (for example $X=S$), the relation $X'Y'\cong XY$ is automatically fulfilled, since in this case, according to the definition of the rotation, it holds $SY'\cong SY$. Assume that none of the points $X$ and $Y$ is the center of the rotation $S$. First, according to the definition of the rotation $SX'\cong SX$ and $SY'\cong SY$. Then (from relation \ref{orientKotVsota}): \begin{eqnarray*} \measuredangle Y'SX'&=& \measuredangle Y'SX+\measuredangle XSX'= \measuredangle Y'SX + \omega=\\ &=& \measuredangle Y'SX + \measuredangle YSY'= \measuredangle YSY'+ \measuredangle Y'SX=\\ &=&\measuredangle YSX \end{eqnarray*} This means that the triangle $X'SY'$ and $XSY$ are congruent (\textit{SAS} theorem \ref{SKS}), therefore $X'Y'\cong XY$. \kdokaz As every isometry, also the rotation maps a line into a line. In the next theorem we will see the relation between a line and its image under the rotation. \bizrek \label{rotacPremPremKot} A line and its rotated image determine an angle congruent to the angle of this rotation. If the angle of the rotation measures $180^0$, the line and its image are parallel. \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.4.3.pic} \caption{} \label{sl.izo.6.4.3.pic} \end{figure} \textbf{\textit{Proof.}} Let $p'$ be the image of the line $p$ under the rotation $\mathcal{R}_{S,\omega}$ (Figure \ref{sl.izo.6.4.3.pic}). If $S\in p$, the proof is trivial. Assume that $S\notin p$. We denote with $P$ the orthogonal projection of the center $S$ onto the line $p$ and its image $P'=\mathcal{R}_{S,\omega}(P)$. Because $S\notin p$, it follows that $P\neq S$ or $P'\neq P$. From $P\in p$ it follows that $P'\in p'$. Because the isometry preserves the angles, from $SP\perp p$ it follows that $SP'\perp p'$. If $\omega=180^0$, the points $P$, $S$ and $P'$ are collinear and the lines $p$ and $p'$ have a common perpendicular $PP'$, which means that $p\parallel p'$. In the case $\omega\neq 180^0$, the lines intersect (in the opposite case, from $p\parallel p'$ it would follow that $SP\parallel SP'$, which is not possible). Let $V$ be their intersection point. The angle, determined by the lines $p$ and $p'$, is equal to the angle determined by the lines $SP$ and $SP'$ (the angle with perpendicular legs - statement \ref{KotaPravokKraki}), which is just the angle of rotation $\omega$. \kdokaz In the proof of the previous statement, the process of designing the image $p'$ of the line $p$ under the rotation $\mathcal{R}_{S,\omega}$ is also described. First, we draw the orthogonal projection $P$ of the center $S$ onto the line $p$, and then its image $P=\mathcal{R}_{S,\omega}(P')$. The line $p'$ is obtained as the perpendicular of the line $SP'$ in the point $P'$. Of course, another way would be to rotate two arbitrary points of the line $p$. In the next important statement, we will see how to express the rotation using the basic reflections. \bizrek \label{rotacKom2Zrc} Any rotation $\mathcal{R}_{S,\omega}$ can be expressed as the product of two reflections $\mathcal{S}_p$ and $\mathcal{S}_q$ where $p$ and $q$ are arbitrary lines, such that $S=p\cap q$ and $\measuredangle pq=\frac{1}{2}\omega$.\\ The reverse is also true - the product of two reflections $\mathcal{S}_p$ and $\mathcal{S}_q$ ($S=p\cap q$) is a rotation with the centre $S$ and the angle $\omega=2\cdot\measuredangle pq$, i.e.\\ $$\mathcal{R}_{S,\omega}=\mathcal{S}_q\circ\mathcal{S}_p \hspace*{1mm} \Leftrightarrow \hspace*{1mm} S=p\cap q \hspace*{1mm} \wedge\hspace*{1mm} \measuredangle pq=\frac{1}{2}\omega.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.4.4.pic} \caption{} \label{sl.izo.6.4.4.pic} \end{figure} \textbf{\textit{Proof.}} ($\Leftarrow$) Assume that $p$ and $q$ are lines that intersect in the point $S$. Let $\omega =2\cdot\measuredangle pq$ (Figure \ref{sl.izo.6.4.4.pic}). We will prove that $\mathcal{R}_{S,\omega}=\mathcal{S}_q\circ\mathcal{S}_p$, or $\mathcal{R}_{S,\omega}(X)=\mathcal{S}_q\circ\mathcal{S}_p(X)$ for every point $X$ of the plane. We will consider two cases. \textit{1)} If $X=S$, then $\mathcal{S}_p(X)=\mathcal{S}_q(X)=X$ (because the lines $p$ and $q$ intersect in the point $S$). So $\mathcal{R}_{S,\omega}(S)=S=\mathcal{S}_q\circ\mathcal{S}_p(S)$. \textit{2)} Let $X\neq S$ and $\mathcal{S}_q\circ\mathcal{S}_p(X)=X'$. Prove that also $\mathcal{R}_{S,\omega}(X)=X'$. Let $\mathcal{S}_p(X)=X_1$. Then $\mathcal{S}_q(X_1)=X'$. Because of this, first $SX\cong SX_1\cong SX'$, and then: \begin{eqnarray*} \measuredangle XSX'&=& \measuredangle XSX_1+ \measuredangle X_1SX'=\\ &=& 2\cdot\measuredangle p,SX_1+2\cdot\measuredangle SX_1,q=\\ &=& 2\cdot(\measuredangle p,SX_1+ \measuredangle SX_1,q) =\\ &=& 2\cdot\measuredangle pq=\omega. \end{eqnarray*} By the definition of rotation, $\mathcal{R}_{S,\omega}(X)=X'$. ($\Rightarrow$) Let $\mathcal{R}_{S,\omega}=\mathcal{S}_q\circ\mathcal{S}_p$. Then it is $\mathcal{S}_q\circ\mathcal{S}_p(S)=\mathcal{R}_{S,\omega}(S)=S$. By izrek \ref{izoKomppqX} $S=p\cap q$. It remains to prove $\measuredangle pq=\frac{1}{2}\omega$ or $\omega =2\cdot\measuredangle pq$. We denote $\widehat{\omega} =2\cdot\measuredangle pq$. By the first part of the proof ($\Leftarrow$) it is $\mathcal{S}_q\circ\mathcal{S}_p=\mathcal{R}_{S,\widehat{\omega}}$. Therefore $\mathcal{R}_{S,\omega}=\mathcal{S}_q\circ\mathcal{S}_p= \mathcal{R}_{S,\widehat{\omega}}$. From izrek \ref{rotacEnaki} it follows $\omega =\widehat{\omega} =2\cdot\measuredangle pq$. \kdokaz \bizrek \label{RotacDirekt} A rotation is a direct isometry. \eizrek \textbf{\textit{Proof.}} A direct consequence of the previous izrek \ref{rotacKom2Zrc}, because the composition of two basic reflections is an indirect isometry. \kdokaz Similarly to the basic reflection, also for the rotation, there is an izrek about transmutation\index{transmutation!rotations}. \bizrek \label{izoTransmRotac} For an arbitrary rotation $\mathcal{R}_{O,\alpha}$ and an arbitrary isometry $\mathcal{I}$ is $$\mathcal{I}\circ \mathcal{R}_{O,\alpha}\circ\mathcal{I}^{-1}= \mathcal{R}_{\mathcal{I}(O),\alpha'},$$ where: $\alpha'=\alpha$, if $\mathcal{I}$ a direct isometry, or $\alpha'=-\alpha$, if $\mathcal{I}$ is an opposite isometry. \eizrek \textbf{\textit{Proof.}} By izrek \ref{rotacKom2Zrc} we can write the rotation $\mathcal{R}_{O,\alpha}$ as a composite of two basic reflections, namely $\mathcal{R}_{O,\omega}=\mathcal{S}_q\circ\mathcal{S}_p$, where $O=p\cap q$ and $\measuredangle pq=\frac{1}{2}\alpha$. If we use the transmutation theorem for basic reflections, we get: \begin{eqnarray*} \mathcal{I}\circ \mathcal{R}_{O,\alpha}\circ\mathcal{I}^{-1}&=& \mathcal{I}\circ\mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{I}=\\ &=& \mathcal{I}\circ\mathcal{S}_q\circ\mathcal{I}^{-1} \circ\mathcal{I}\circ\mathcal{S}_p\circ\mathcal{I}^{-1}=\\ &=&\mathcal{S}_{q'}\circ\mathcal{S}_{p'}=\\ &=&\mathcal{R}_{O',\alpha'}, \end{eqnarray*} where $p'$ and $q'$ are the images of lines $p$ and $q$ under the isometry $\mathcal{I}$, $O'=p'\cap q'$ and $\alpha'=2\cdot\measuredangle p'q'$. Because $O=p\cap q$, we have $\mathcal{I}(O)= \mathcal{I}(p\cap q)=\mathcal{I}(p)\cap \mathcal{I}(q)=p'\cap q'=O'$. Because isometries preserve the relation of congruence of angles, we have $\alpha'=2\cdot\measuredangle p'q'=2\cdot\measuredangle pq=\alpha$, if $\mathcal{I}$ is direct, or $\alpha'=2\cdot\measuredangle p'q'=-2\cdot\measuredangle pq=-\alpha$, if $\mathcal{I}$ is indirect. \kdokaz In the next theorem we will prove the important fact that for two non-parallel lines there exists a rotation that transforms the first line into the second. \bizrek Let $AB$ and $A'B'$ be congruent line segments that are not parallel. There is exactly one rotation that maps the line segment $AB$ to the line segment $A'B'$, such that the point $A$ maps to the point $A'$ and the point $B$ to the point $B'$. \eizrek \begin{figure}[!htb] \centering \input{sl.skl.3.1.10Rotac.pic} \caption{} \label{sl.skl.3.1.10Rotac.pic} \end{figure} \textbf{\textit{Proof.}} Let $S$ be the intersection of the line segments $AA'$ and $BB'$, and $\omega=\measuredangle AB, A'B'$. We will prove that $\mathcal{R}_{S,\omega}$ is the desired rotation. Because the point $S$ lies on the line segments $AA'$ and $BB'$, we have $SA\cong SA'$ and $SB\cong SB'$ (Figure \ref{sl.skl.3.1.10Rotac.pic}). Then from $AB\cong A'B'$ by the \textit{SSS} \ref{SSS} it follows that $\triangle SAB\cong \triangle SA'B'$ (see also example \ref{načrt1odd3}). Therefore, we have $\measuredangle ASB\cong \measuredangle A'SB'$, or: $$\measuredangle ASA'=\measuredangle ASB+\measuredangle BSA'= \measuredangle A'SB'+\measuredangle BSA'= \measuredangle BSB'.$$ By the definition of rotation, we have $\mathcal{R}_{S,\measuredangle ASA'}:A,B\mapsto A',B'$. By izreku \ref{rotacPremPremKot} we have $\measuredangle ASA'=\measuredangle AB, A'B'=\omega$, so $\mathcal{R}_{S,\omega}:A,B\mapsto A',B'$. Because rotation as an isometry preserves the relation $\mathcal{B}$, the line segment $AB$ is mapped by this rotation to the line segment $A'B'$. Let $\mathcal{R}_{\widehat{S},\widehat{\omega}}$ be another rotation, for which we have $\mathcal{R}_{\widehat{S},\widehat{\omega}}:A,B\mapsto A',B'$. Then $\mathcal{R}^{-1}_{S,\omega} \circ \mathcal{R}_{\widehat{S},\widehat{\omega}}$ is a direct isometry with two fixed points $A$ and $B$, so by izreku \ref{izo2ftIdent} it represents the identity $\mathcal{E}$, or $\mathcal{R}_{\widehat{S},\widehat{\omega}} = \mathcal{R}_{S,\omega}$. \kdokaz In the following examples we will use the fact that a triangle $ABC$ is equilateral if and only if $\mathcal{R}_{A,60^0}(B)=C$. \bzgled Let $P$ be an interior point of an equilateral triangle $ABC$.\\ a) Prove that $|PA|+|PB|\geq |PC|.$\\ b) Suppose that $\angle BPA=\mu$, $\angle CPA= \nu$ and $\angle BPC= \xi$. Calculate the interior angles of a triangle, with sides that are congruent to the line segments $PA$, $PB$ in $PC$. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.4.5.pic} \caption{} \label{sl.izo.6.4.5.pic} \end{figure} \textbf{\textit{Proof.}} If $P'=\mathcal{R}_{A,60^0}(P)$, the triangle $APP'$ is equilateral (Figure \ref{sl.izo.6.4.5.pic}). Because $\mathcal{R}_{A,60^0}(B)=C$, the line segment $BP$ is mapped to the corresponding line segment $CP'$ by this rotation. Therefore, the triangle $PP'C$ has sides that are congruent to the line segments $PA$, $PB$ and $PC$. By the triangle inequality - statement \ref{neenaktrik} - we have $|PA|+|PB|=|PP'|+|P'C|\geq PC.$ Let us calculate the angles of the triangle $PP'C$ as well. The rotation $\mathcal{R}_{A,60^0}$ maps the triangle $ABP$ to the triangle $ACP'$, so the triangles are congruent and $\angle BPA=\angle CP'A$. From this and from statement \ref{VsotKotTrik} it follows that: \begin{eqnarray*} \angle CP'P &=& \angle CP'A-\angle PP'A=\angle BPA-60^0=\mu-60^0,\\ \angle CPP' &=& \angle CPA-60^0=\nu-60^0,\\ \angle PCP' &=& 180^0 -(\mu-60^0)-(\nu-60^0)=300^0-(\mu+\nu)=\\ &=& 300^0-(360^0-\xi)=\xi-60^0, \end{eqnarray*} which had to be calculated. \kdokaz \bzgled Let $ABCD$ be a rhombus, such that the interior angle at the vertex $A$ is equal to $60^0$. If a line $l$ intersects the sides $AB$ and $BC$ of this rhombus at points $P$ and $Q$ such that $|BP|+|BQ|=|AB|$, then $PQD$ is a regular triangle. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.4.6.pic} \caption{} \label{sl.izo.6.4.6.pic} \end{figure} \textbf{\textit{Proof.}} From $AB\cong AD$ and $\angle DAB=60^0$ it follows that the triangle $ABD$ is regular (Figure \ref{sl.izo.6.4.6.pic}). Similarly, the triangle $BCD$ is also regular. This means that $\mathcal{R}_{D,60^0}:A,B\mapsto B,C$. Because $|BP|+|BQ|=|AB|$ and $|BP|+|AP|=|AB|$, we have $AP\cong BQ$. But the rotation $\mathcal{R}_{D,60^0}$ maps the segment $AB$ to the segment $BC$, so because of the condition $AP\cong BQ$ it also maps the point $P$ to the point $Q$. Therefore, $\mathcal{R}_{D,60^0}(P)=Q$, which means that the triangle $DPQ$ is regular. \kdokaz At the next example we will illustrate the use of rotation in design tasks. \bzgled Let $A$ be an interior point of an angle $pOq$. Construct points $B$ and $C$ on the sides $p$ and $q$ such that $ABC$ is a regular triangle. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.4.7.pic} \caption{} \label{sl.izo.6.4.7.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.4.7.pic}). Let $ABC$ be such a regular triangle that its vertices $B$ and $C$ lie on the sides $p$ and $q$. Then $\mathcal{R}_{A,60^0}(B)=C$. The point $B$ lies on the segment $p$, so its image - the point $C$ - lies on the segment $p'=\mathcal{R}_{A,60^0}(p)$. The vertices $C$ can be obtained as the intersection of the segments $p'$ and $q$, and then the vertex $B$ as $B=\mathcal{R}_{A,-60^0}(C)$. If the segments $p'$ and $q$ do not intersect, there is no solution. But if they lie on the same line, the task has infinitely many solutions. \kdokaz If in the previous task the condition was that the triangle $ABC$ was isosceles and right-angled at the vertex $A$, we would use the rotation with center $A$ for the angle $45^0$. Similarly, if we needed to plan the square $PQRS$ with center at the point $A$ under the condition $P\in p$, $Q\in q$ (Figure \ref{sl.izo.6.4.8.pic}). \bnaloga\footnote{1. IMO Romania - 1959, Problem 5.} An arbitrary point $M$ is selected in the interior of the segment $AB$. The squares $AMCD$ and $MBEF$ are constructed on the same side of $AB$, with the segments $AM$ and $MB$ as their respective bases. The circles circumscribed about these squares, with centres $P$ and $Q$, intersect at $M$ and also at another point $N$. Let $N'$ denote the point of intersection of the straight lines $AF$ and $BC$. (a) Prove that the points $N$ and $N'$ coincide. (b) Prove that the straight lines $MN$ pass through a fixed point $S$ independent of the choice of $M$. (c) Find the locus of the midpoints of the segments $PQ$ as $M$ varies between $A$ and $B$ \enaloga \begin{figure}[!htb] \centering \input{sl.izo.6.4.IMO1.pic} \caption{} \label{sl.izo.6.4.IMO1.pic} \end{figure} \textbf{\textit{Solution.}} Let $k$ and $l$ denote the circumscribed circles of the squares $AMCD$ and $MBEF$ (Figure \ref{sl.izo.6.4.IMO1.pic}). (\textit{i}) The rotation $\mathcal{R}_{M,-90^0}$ maps the points $A$ and $F$ in order to the points $C$ and $B$ or the line $AF$ to the line $CB$. By the statement of \ref{rotacPremPremKot} these two lines determine the angle of rotation, so $AF\perp CB$. Therefore $\angle AN'C\cong\angle BN'F =90^0$, which means (by the statement \ref{TalesovIzrKroz}) that the point $N'$ lies on both circles $k$ and $l$ with diameters $AC$ and $BF$. Because of this, $N'=N$. (\textit{ii}) Let $j$ be a circle with diameter $AB$ and $S$ the center of that arc (semicircle) $AB$, which is on the opposite side of the line $AB$ with respect to the squares $AMCD$ and $MBEF$. The point $S$ is not dependent on the choice of point $M$. We prove that all lines $MN$ go through the point $S$. Because $\angle ANB=90^0$, by the statement \ref{TalesovIzrKroz} also the point $N$ lies on the circle $j$. From the statement \ref{ObodObodKot} it follows that $\angle ANM =\angle ADM=45^0$ and $\angle MNB =\angle MEB=45^0$, so $NM$ is the perpendicular bisector of the angle $ANB$. By the statement \ref{TockaN} the perpendicular bisector $NM$ goes through the point $S$. \begin{figure}[!htb] \centering \input{sl.izo.6.4.IMO1a.pic} \caption{} \label{sl.izo.6.4.IMO1a.pic} \end{figure} (\textit{iii}) Let $O$ be the center of the line segment $PQ$. We denote with $P'$, $Q'$ and $O'$ the orthogonal projections of points $P$, $Q$ and $O$ on the line $AB$ (Figure \ref{sl.izo.6.4.IMO1a.pic}). The line $OO'$ is the median of the rectangle $P'Q'QP$, so by Theorem \ref{srednjTrapez}: $$|OO'|=\frac{1}{2}\left(|PP'|+|QQ'| \right)= \frac{1}{2}\left(\frac{1}{2}|AM|+\frac{1}{2}|MB| \right)=\frac{1}{4}|AB|.$$ Therefore, the distance of the point $O$ from the line $AB$ is constant or independent of the choice of the point $M$. Let $L$ be the center of the square $ABGH$, which is on the same side of the line $AB$ as the square $AMCD$ and $MBEF$. With $O_A$ and $O_B$ we denote the centers of the line segments $LA$ and $LB$. Because $O_AO_B$ is the median of the triangle $ALB$ with the base $AB$, $O_AO_B \parallel AB$. From $|AO_A|=\frac{1}{2}|AL|=\frac{1}{4}|AG|$ it follows that the distance of the parallels $O_AO_B$ and $AB$ is equal to $\frac{1}{4}|AB|$. From the already proven fact $d(O,AB)=\frac{1}{4}|AB|$ it follows that the point $O$ lies on the line $O_AO_B$. But, when the point $M$ moves inside the line segment $AB$, the points $P$ and $Q$ move inside the line segments $AL$ and $BL$. This means that the point $O$ is inside the triangle $ALB$ or $O$ lies on the open line segment $O_AO_B$. We will also prove that any point $O$ of the line segment $O_AO_B$ is the center of a line segment $PQ$ for a certain choice of the squares $AMCD$ and $MBEF$ or the points $M$. In this case, we get the point $M$ from the condition $|AO'|=\frac{1}{2}|AM|+\frac{1}{4}|AB|$ or $|AM|=2|AO'|-\frac{1}{2}|AB|$. Such a point $M$ always exists, if $\frac{1}{4}|AB|<|AO'|<\frac{3}{4}|AB|$, or when the point $O$ lies on the open line segment $O_AO_B$. The desired geometric location of points $M$ is therefore the line $O_AO_B$. \kdokaz %________________________________________________________________________________ \poglavje{Half-Turn} \label{odd6SredZrc} We will consider the rotation by the angle $180^0$ as a special type of isometry. The rotation of the plane $\mathcal{R}_{S,\omega}$ by the angle $\omega=180^0$ we will call the \index{zrcaljenje!središčno}\pojem{central reflection} or \index{zrcaljenje!čez točko}\pojem{reflection over a point} with \index{središče!središčnega zrcaljenja}\pojem{center} $S$ and denote it by $\mathcal{S}_S$ (Figure \ref{sl.izo.6.5.1.pic}). So $\mathcal{S}_S=\mathcal{R}_{S,180^0}$. \begin{figure}[!htb] \centering \input{sl.izo.6.5.1.pic} \caption{} \label{sl.izo.6.5.1.pic} \end{figure} From the definition it is clear that for any point $X\neq S$ it holds $\mathcal{S}_S(X)=X'$ exactly when $S$ is the center of the line $XX'$. Directly from the definition it also follows the next statement. \bizrek \label{izoSredZrcInv} A half-turn is an involution, i.e. $\mathcal{S}^2_S=\mathcal{E}$. \eizrek Because the central reflection is a type of rotation, it also has all the properties of rotation. The only fixed point of the central reflection is therefore the center of this reflection. Just like the rotation, the central reflection $\mathcal{S}_S$ can be represented as the composition of two basic reflections. The angle, which is determined by the axes, is equal to half the angle of the rotation (statement \ref{rotacKom2Zrc}) - in our case this is $\frac{180^0}{2}=90^0$. This means that the axes of the two reflections are perpendicular in the case of central reflection. In this case, according to \ref{izoZrcKomut}, the basic reflections commute (Figure \ref{sl.izo.6.5.2.pic}). So the following statement holds. \begin{figure}[!htb] \centering \input{sl.izo.6.5.2.pic} \caption{} \label{sl.izo.6.5.2.pic} \end{figure} \bizrek \label{izoSrZrcKom2Zrc} Any half-turn around a point $S$ can be expressed as the product of two reflections $\mathcal{S}_p$ and $\mathcal{S}_q$ where $p$ and $q$ are arbitrary perpendicular lines intersecting at the point $S$.\\ The reverse is also true - the product of two reflections $\mathcal{S}_p$ and $\mathcal{S}_q$ ($S=p\cap q$ and $p\perp q$) is the half-turn around the point $S$, i.e.\\ $$\mathcal{S}_S=\mathcal{S}_q\circ\mathcal{S}_p= \mathcal{S}_p\circ\mathcal{S}_q \hspace*{1mm} \Leftrightarrow \hspace*{1mm} S=p\cap q \hspace*{1mm} \wedge\hspace*{1mm} p\perp q.$$ \eizrek \bizrek \label{izoKomp3SredZrc} The product of three half-turns is a half-turn. If the centres of these half-turns are three non-collinear points, then the centre of the new half-turn is the fourth vertex of a parallelogram. \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.5.3.pic} \caption{} \label{sl.izo.6.5.3.pic} \end{figure} \textbf{\textit{Proof.}} Let $\mathcal{S}_A$, $\mathcal{S}_B$ and $\mathcal{S}_C$ be three central symmetries with centers $A$, $B$ and $C$ (Figure \ref{sl.izo.6.5.3.pic}). We denote by $p$ the line $AB$, by $a$, $b$ and $c$ the rectangles of the line $p$ through the points $A$, $B$ and $C$, and by $c'$ the rectangle of the line $c$ through the point $C$. Then by izreku \ref{izoSrZrcKom2Zrc}: $$\mathcal{S}_C \circ \mathcal{S}_B \circ \mathcal{S}_A= \mathcal{S}_{c'} \circ \mathcal{S}_c \circ \mathcal{S}_b \circ \mathcal{S}_p \circ \mathcal{S}_p \circ \mathcal{S}_a = \mathcal{S}_{c'} \circ \mathcal{S}_c \circ \mathcal{S}_b \circ \mathcal{S}_a.$$ The axes $a$, $b$ and $c$ belong to the same set of parallels $\mathcal{X}_a$, because they are all perpendicular to the line $p$. The composite $\mathcal{S}_c \circ \mathcal{S}_b \circ \mathcal{S}_a$ therefore represents the basic reflection $S_d$, where the axis $d$ belongs to the same set $\mathcal{X}_a$ (izrek \ref{izoSop}). Then both lines $c'$ and $d$ are perpendicular, so: $$\mathcal{S}_C \circ \mathcal{S}_B \circ \mathcal{S}_A= \mathcal{S}_{c'} \circ \mathcal{S}_d=\mathcal{S}_D,$$ where $D=d\cap c'$. By izreku \ref{izoSoppqrt} the pairs of lines $a$, $c$ and $b$, $d$ have a common somernica. This means that if $A$, $B$ and $C$ are three non-collinear points, the quadrilateral $ABCD$ is a parallelogram. \kdokaz The consequence of the proven statement is the following izrek. \bizrek \label{izoKomp2n+1SredZrc} The product of an odd number of half-turns is a half-turn. \eizrek \textbf{\textit{Proof.}} We will prove the statement by induction on the number $n\geq 1$, where $m=2n+1$ (odd) is the number of points. For $n=1$, i.e. $m=3$, the statement is a direct consequence of the previous izrek \ref{izoKomp3SredZrc}. Let's assume that for every $k\in \mathbb{N}$ ($k\geq 1$) and every $(2k+1)$-tuple of coplanar points $A_1,A_2,\ldots ,A_{2k+1}$ the composite $\mathcal{S}_{A_{2k+1}}\circ \cdots \circ\mathcal{S}_{A_2}\circ\mathcal{S}_{A_1}$ represents some central reflection $\mathcal{S}_A$ (inductive assumption). It is necessary to prove that then also for $k+1$ and every $(2(k+1)+1)$-tuple of coplanar points $A_1,A_2,\ldots ,A_{(2(k+1)+1)}$ the composite $\mathcal{I}=\mathcal{S}_{A_{2(k+1)+1}}\circ \cdots \circ\mathcal{S}_{A_2}\circ\mathcal{S}_{A_1}$ represents some central reflection $\mathcal{S}_{A'}$. So: \begin{eqnarray*} \mathcal{I}&=&\mathcal{S}_{A_{2(k+1)+1}}\circ \cdots \circ\mathcal{S}_{A_2}\circ\mathcal{S}_{A_1}=\\ &=&\mathcal{S}_{A_{2k+3}}\circ\mathcal{S}_{A_{2k+2}}\circ \mathcal{S}_{A_{2k+1}}\circ \cdots \circ\mathcal{S}_{A_2}\circ\mathcal{S}_{A_1}=\\ &=& \mathcal{S}_{A_{2k+3}}\circ\mathcal{S}_{A_{2k+2}}\circ \mathcal{S}_A= \hspace*{14mm} \textrm{ (by the inductive assumption)}\\ &=& \mathcal{S}_{A'} \hspace*{48mm} \textrm{ (by \ref{izoKomp3SredZrc}),} \end{eqnarray*} which was to be proven. \kdokaz A direct consequence of \ref{izoTransmRotac}, which concerns rotation, is the following theorem or izrek about transmutation of central reflection\index{transmutation!of central reflection}. \bizrek \label{izoTransmSredZrc} For an arbitrary half-turn $\mathcal{S}_{O}$ and an arbitrary isometry $\mathcal{I}$ is $$\mathcal{I}\circ \mathcal{S}_{O}\circ\mathcal{I}^{-1}= \mathcal{R}_{\mathcal{I}(O)}.$$ \eizrek Just like the basic reflection, also the central reflection is often used in planning tasks. \bzgled Let $S$ be an interior point of an angle $pOq$. Construct a square $ABCD$ with the centre $S$ such that the vertices $A$ and $C$ lie on the sides $p$ and $q$. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.5.4.pic} \caption{} \label{sl.izo.6.5.4.pic} \end{figure} \textbf{\textit{Proof.}} Let $ABCD$ be a square that satisfies the given conditions - $S$ is its center, $A\in p$ and $C\in q$ (Figure \ref{sl.izo.6.5.4.pic}). Because the point $S$ is the center of the diagonal $AC$, $\mathcal{S}_S(A)=C$. From $A\in p$ it follows that $C\in \mathcal{S}_S(p)$. We therefore obtain the point $C$ from the condition $C\in q\cap\mathcal{S}_S(p)$. Finally, $A=\mathcal{S}_S(C)$, $D=\mathcal{R}_{S,90^0}(C)$ and $B=\mathcal{R}_{S,-90^0}(C)$. \kdokaz \bzgled Let $A$ be one of the two intersections of circles $k$ and $l$. Construct a common secant of these two circles passing through the point $A$ and determine a two congruent chord with these circles. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.5.5.pic} \caption{} \label{sl.izo.6.5.5.pic} \end{figure} \textbf{\textit{Solution.}} Let $B$ be the other intersection of circles $k$ and $l$ (Figure \ref{sl.izo.6.5.5.pic}). Then one solution to the task is the line $AB$, since the distance $AB$ is the common secant of both circles. Let $s\ni A$ be a common secant of circles $k$ and $l$, which also intersects in points $K$ and $L$, so that $AK\cong AL$. In the case $K=L$ we have $K=L=B$, which is the already mentioned solution. If $K\neq L$, from the condition $A, K, L\in s$ it follows that $A$ is the center of the line $KL$. Therefore $\mathcal{S}_A(K)=L$, which means that we obtain the point $L$ from the condition $L\in l\cap \mathcal{S}_A(k)$. Finally, $K=\mathcal{S}_A(L)$. \kdokaz \bzgled Construct a quadrilateral $ABCD$ with the sides that are congruent to the four given line segments $a$, $b$, $c$, and $d$ and the line segment defined by the midpoints of the two opposite sides $AD$ and $BC$ that is congruent to the given line segment $l$. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.5.6.pic} \caption{} \label{sl.izo.6.5.6.pic} \end{figure} \textbf{\textit{Solution.}} Let $P$ and $Q$ be the midpoints of sides $AD\cong d$ and $BC\cong b$ of quadrilateral $ABCD$, so that $PQ\cong l$, $AB\cong a$ and $CD\cong c$ still hold (Figure \ref{sl.izo.6.5.6.pic}). Let $S$ be the midpoint of diagonal $AC$. We can construct triangle $PQS$ since $PQ\cong l$, $QS=\frac{1}{2}a$ and $PS =\frac{1}{2}c$ (by Theorem \ref{srednjicaTrik}). Points $C$ and $A$ lie on circles $k_d=k(P,\frac{1}{2}d)$ and $k_b=k(Q,\frac{1}{2}b)$, respectively. Since $C=\mathcal{S}_S(A)$, point $C$ also lies on the image $k'_d$ of circle $k_d$ under central reflection $\mathcal{S}_S$. Point $C$ is therefore one of the intersections of circles $k'_d$ and $k_b$. Then $A=\mathcal{S}_S(C)$, $D=\mathcal{S}_P(A)$ and $B=\mathcal{S}_Q(C)$. \kdokaz \bzgled Let $P$, $Q$, $R$, $S$ and $T$ be points in the plane. Construct a pentagon $ABCDE$ such that the points $P$, $Q$, $R$, $S$ and $T$ are the midpoints of its sides $AB$, $BC$, $CD$, $DE$ and $EA$, respectively. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.5.7.pic} \caption{} \label{sl.izo.6.5.7.pic} \end{figure} \textbf{\textit{Solution.}} Let $ABCDE$ be the desired pentagon (Figure \ref{sl.izo.6.5.7.pic}) and: $$\mathcal{I} = \mathcal{S}_T \circ \mathcal{S}_S \circ \mathcal{S}_R \circ \mathcal{S}_Q \circ \mathcal{S}_P.$$ From \ref{izoKomp2n+1SredZrc} it follows that the isometry $\mathcal{I}$ is a central reflection $\mathcal{S}_O$. Since $\mathcal{S}_O(A)=\mathcal{I}(A)=A$, we have $O=A$ or $\mathcal{I}=\mathcal{S}_A$. The point $A$ can therefore be constructed as the center of the line $XX'$, where $X$ is any point on the given plane and $X'= \mathcal{I}(X)$. \kdokaz \bzgled Let $A_iB_i$ ($i\in \{1,2,3\}$) be parallel chords of a circle $k$. Suppose that $S_1$, $S_2$ and $S_3$ are the midpoints of the line segments $A_2A_3$, $A_1A_3$ and $A_1A_2$, respectively. Prove that $C_i=\mathcal{S}_{S_i}(B_i)$ are three collinear points. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.5.4a.pic} \caption{} \label{sl.izo.6.5.4a.pic} \end{figure} \textbf{\textit{Proof.}} Let $S$ be the center of the circle $k$ (Figure \ref{sl.izo.6.5.4a.pic}). Because $SA_i\cong SB_i$ ($i\in \{1,2,3\}$), all the simetrals $s_i$ of the chords $A_iB_i$ ($i\in \{1,2,3\}$) pass through the point $S$. From $s_i\perp A_iB_i$ and the fact that the chords are parallel to each other, it follows that all the simetrals $s_i$ are equal to each other - we denote them with $s$. Therefore, $s$ is the common simetral of the chords $A_iB_i$, so $\mathcal{S}_s(A_i)=B_i$ ($i\in \{1,2,3\}$). Since $S_1$, $S_2$, and $S_3$ are the centers of the lines $C_1B_1$, $C_2B_2$, and $C_3B_3$, and the sides $A_2A_3$, $A_1A_3$, and $A_1A_2$ of the triangle $A_1A_2A_3$, according to the theorem \ref{vektSestSplosno}, $\overrightarrow{S_1S_2}= \frac{1}{2}(\overrightarrow{C_1C_2}+\overrightarrow{B_1B_2})$ or $\overrightarrow{C_1C_2}= 2\overrightarrow{S_1S_2}-\overrightarrow{B_1B_2} =\overrightarrow{A_2A_1}-\overrightarrow{B_1B_2}=\overrightarrow{A_2A_1} +\overrightarrow{B_2B_1}$. Since the points $A_2$ and $A_1$ are reflected across the line $s$ at the points $B_2$ and $B_1$, the vector $\overrightarrow{C_1C_2}=\overrightarrow{A_2A_1} +\overrightarrow{B_2B_1}$ is collinear with the line $s$ (see \ref{izoSimVekt}). Similarly, the vector $\overrightarrow{C_2C_3}$ is collinear with the line $s$, so the points $C_1$, $C_2$, and $C_3$ lie on the same line parallel to the line $s$. \kdokaz %________________________________________________________________________________ \poglavje{Translations} \label{odd6Transl} So far we have considered isometries that have at least one fixed point. Of the direct isometries we had the identity and rotation, and of the indirect ones, the basic reflection. Now we will introduce a new type of isometries that have no fixed points. Let $\overrightarrow{v}$ be an arbitrary vector of the Euclidean plane ($\overrightarrow{v}\neq \overrightarrow{0}$). The transformation of this plane, in which the point $X$ is mapped to such a point $X'$, that $\overrightarrow{XX'}=\overrightarrow{v}$, is called the \index{translacija} \pojem{translation} or the \index{vzporedni premik} \pojem{parallel shift} for the vector $\overrightarrow{v}$, and is denoted by $\mathcal{T}_{\overrightarrow{v}}$ (Figure \ref{sl.izo.6.6.1.pic}). The vector $\overrightarrow{v}$ is\index{vektor!translacije} \pojem{the vector of translation}. \begin{figure}[!htb] \centering \input{sl.izo.6.6.1.pic} \caption{} \label{sl.izo.6.6.1.pic} \end{figure} We will now prove the first basic properties of translation. \bizrek \label{translEnaki} Two translations are equal if and only the vectors of these translations are equal, i.e. $$\mathcal{T}_{\overrightarrow{v}}= \mathcal{T}_{\overrightarrow{u}}\Leftrightarrow \overrightarrow{v}=\overrightarrow{u}.$$ \eizrek \textbf{\textit{Proof.}} The part ($\Leftarrow$) is trivial. We prove part ($\Rightarrow$). For an arbitrary point $X$ we have $\mathcal{T}_{\overrightarrow{v}}(X)= \mathcal{T}_{\overrightarrow{u}}(X)=X'$. In this case $\overrightarrow{v}=\overrightarrow{u}=\overrightarrow{XX'}$. \kdokaz \bizrek A translation has no fixed points. \eizrek \textbf{\textit{Proof.}} If $X$ is a fixed point of the translation $\mathcal{T}_{\overrightarrow{v}}$ or $\mathcal{T}_{\overrightarrow{v}}(X)=X$, then $\overrightarrow{v}=\overrightarrow{XX}=\overrightarrow{0}$, which according to the definition of translation is not possible. \kdokaz \bizrek The inverse transformation of a translation is a translation with the opposite vector, i.e. $\mathcal{T}^{-1}_{\overrightarrow{v}}= \mathcal{T}_{-\overrightarrow{v}}$ \eizrek \textbf{\textit{Proof.}} Let $Y$ be an arbitrary point and $X$ such a point that $\overrightarrow{XY}=\overrightarrow{v}$. From this it follows that $\mathcal{T}_{\overrightarrow{v}}(X)=Y$. From $\overrightarrow{YX}=-\overrightarrow{v}$ it follows that $\mathcal{T}_{-\overrightarrow{v}}(Y)=X$. Because $\mathcal{T}^{-1}_{\overrightarrow{v}}(Y)=X$, it follows that $\mathcal{T}^{-1}_{\overrightarrow{v}}(Y)= \mathcal{T}_{-\overrightarrow{v}}(Y)$. Because this is true for an arbitrary point $Y$, we have $\mathcal{T}^{-1}_{\overrightarrow{v}}= \mathcal{T}_{-\overrightarrow{v}}$. \kdokaz \bizrek Translations are isometries of the plane. \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.6.2.pic} \caption{} \label{sl.izo.6.6.2.pic} \end{figure} \textbf{\textit{Proof.}} Let $\mathcal{T}_{\overrightarrow{v}}$ be an arbitrary translation. From the definition it is clear that it represents a bijective mapping. It remains to be proven that for any points $X$ and $Y$ and their images $\mathcal{T}_{\overrightarrow{v}}:X,Y\mapsto X', Y'$ it holds that $X'Y'\cong XY$ (Figure \ref{sl.izo.6.6.2.pic}). From $\overrightarrow{XX'}=\overrightarrow{v}$ and $\overrightarrow{YY'}=\overrightarrow{v}$ it follows that $\overrightarrow{XX'}=\overrightarrow{YY'}$. By \ref{vektABCD_ACBD} it is then also true that $\overrightarrow{XY}=\overrightarrow{X'Y'}$ or $X'Y'\cong XY$. \kdokaz We will now show that, just like a rotation, any translation can be expressed as the composition of two basic reflections. \bizrek \label{translKom2Zrc} Any translation $\mathcal{T}_{\overrightarrow{v}}$ can be expressed as the product of two reflections $\mathcal{S}_p$ and $\mathcal{S}_q$ where $p$ and $q$ are arbitrary parallel lines with a common perpendicular line ($P\in p$ and $Q\in q$) such that $\overrightarrow{v} =2\overrightarrow{PQ}$, i.e. $$\mathcal{T}_{\overrightarrow{v}}=\mathcal{S}_q\circ\mathcal{S}_p \hspace*{1mm} \Leftrightarrow \hspace*{1mm} p\parallel q \hspace*{1mm} \wedge\hspace*{1mm} \overrightarrow{v} =2\overrightarrow{PQ}\hspace*{1mm} (PQ\perp p,\hspace*{1mm} P\in p,\hspace*{1mm} Q\in q).$$ \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.6.3.pic} \caption{} \label{sl.izo.6.6.3.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.6.3.pic}) ($\Leftarrow$) Let us assume that $p\parallel q$, $\overrightarrow{v}=2\overrightarrow{PQ}$, $PQ\perp p$, $P\in p$ and $Q\in q$. We need to prove that $\mathcal{T}_{\overrightarrow{v}}(X)= \mathcal{S}_q\circ\mathcal{S}_p(X)$ for any point $X$ of this plane. Let $\mathcal{S}_q\circ\mathcal{S}_p(X)=X'$. We will prove that also $\mathcal{T}_{\overrightarrow{v}}(X)=X'$ or $\overrightarrow{XX'}= \overrightarrow{v}$. We will mark $\mathcal{S}_p(X)=X_1$. In this case it is clear that $\mathcal{S}_q(X_1)=X'$. Because of this, first the line $XX'$ is the perpendicular bisector of the parallels $p$ and $q$, which means that the vector $\overrightarrow{XX'}$ is parallel to the vector $\overrightarrow{v}$. No matter the position of the point $X$, in every case the length of the line $XX'$ is twice as long as the distance between the parallels $p$ and $q$, which is equal to the length of the line $PQ$. So $\overrightarrow{XX'} =2\overrightarrow{PQ}= \overrightarrow{v}$. With this we have proven that $\mathcal{T}_{\overrightarrow{v}}(X)=X'$. ($\Rightarrow$) Let now $\mathcal{T}_{\overrightarrow{v}}=\mathcal{S}_q\circ\mathcal{S}_p$. The parallels $p$ and $q$ are in this case parallel, because otherwise the intersection point of these two lines would be a fixed point of the composition $\mathcal{S}_q\circ\mathcal{S}_p$, but the translation $\mathcal{T}_{\overrightarrow{v}}$ has no fixed points. From the first part of the proof ($\Leftarrow$) it follows that the composition $\mathcal{S}_q\circ\mathcal{S}_p$ is equal to the translation $\mathcal{T}_{\overrightarrow{v_1}}$ for the vector $\overrightarrow{v_1}$, where $\overrightarrow{v_1}=2\overrightarrow{PQ}$, $PQ\perp p$, $P\in p$ and $Q\in q$. But from $\mathcal{T}_{\overrightarrow{v}}= \mathcal{S}_q\circ\mathcal{S}_p=\mathcal{T}_{\overrightarrow{v_1}}$ it follows that $\overrightarrow{v}=\overrightarrow{v_1}$ (from statement \ref{translEnaki}) or $\overrightarrow{v}=2\overrightarrow{PQ}$. \kdokaz From the previous statement it follows that the translation as the composition of two indirect isometries is a direct isometry. \bizrek A translation is a direct isometry. \eizrek We will prove that we can represent a translation as the composition of two central reflections. \bizrek \label{transl2sred} The product of two half-turns is a translation, such that $$\mathcal{S}_Q\circ\mathcal{S}_P= \mathcal{T}_{2\overrightarrow{PQ}}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.6.4.pic} \caption{} \label{sl.izo.6.6.4.pic} \end{figure} \textbf{\textit{Proof.}} Let $X$ be an arbitrary point, $\mathcal{S}_Q\circ\mathcal{S}_P(X)=X'$ and $\mathcal{S}_P(X)=X_1$ (Figure \ref{sl.izo.6.6.4.pic}). Then $\mathcal{S}_Q(X_1)=X'$. By the definition of central reflections, points $P$ and $Q$ are the centers of the lines $XX_1$ and $X_1X'$, which means that the line $PQ$ is the median of the triangle $XX_1X'$ for the base $XX'$. By Theorem \ref{srednjicaTrikVekt}, $\overrightarrow{XX'}=2\overrightarrow{PQ}$, so $\mathcal{T}_{2\overrightarrow{PQ}}(X)=X'= \mathcal{S}_Q\circ\mathcal{S}_P(X)$. Since this is true for every point $X$, $\mathcal{S}_Q\circ\mathcal{S}_P= \mathcal{T}_{2\overrightarrow{PQ}}$. \kdokaz \bizrek \label{translKomp} The product of two translations is a translation for the vector that is the sum of the vectors of these two translations, i.e. $$\mathcal{T}_{\overrightarrow{u}}\circ \mathcal{T}_{\overrightarrow{v}}= \mathcal{T}_{\overrightarrow{v}+\overrightarrow{u}}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.6.5.pic} \caption{} \label{sl.izo.6.6.5.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.6.5.pic}). Let $P$ be an arbitrary point, $Q$ such a point that $\overrightarrow{PQ} = \frac{1}{2}\overrightarrow{v}$, and $R$ a point for which $\overrightarrow{QR} = \frac{1}{2}\overrightarrow{u}$. Because $\overrightarrow{v} + \overrightarrow{u} = 2\overrightarrow{PQ} + 2\overrightarrow{QR} = 2\overrightarrow{PR}$, by the previous statement \ref{transl2sred} it follows: $$\mathcal{T}_{\overrightarrow{u}}\circ \mathcal{T}_{\overrightarrow{v}}= \mathcal{T}_{2\overrightarrow{QR}}\circ \mathcal{T}_{2\overrightarrow{PQ}}= \mathcal{S}_R\circ\mathcal{S}_Q\circ \mathcal{S}_Q\circ\mathcal{S}_P= \mathcal{S}_R\circ\mathcal{S}_P= \mathcal{T}_{2\overrightarrow{PR}}= \mathcal{T}_{\overrightarrow{v}+\overrightarrow{u}},$$ which was to be proven. \kdokaz The consequence of the proven is the following statement. \bizrek The product of translations is commutative, i.e. $$\mathcal{T}_{\overrightarrow{u}}\circ \mathcal{T}_{\overrightarrow{v}}= \mathcal{T}_{\overrightarrow{v}}\circ \mathcal{T}_{\overrightarrow{u}}$$ \eizrek \textbf{\textit{Proof.}} By the previous statement \ref{translKomp} it is: $$\mathcal{T}_{\overrightarrow{u}}\circ \mathcal{T}_{\overrightarrow{v}}= \mathcal{T}_{\overrightarrow{v}+\overrightarrow{u}}= \mathcal{T}_{\overrightarrow{u}+\overrightarrow{v}}= \mathcal{T}_{\overrightarrow{v}}\circ \mathcal{T}_{\overrightarrow{u}},$$ which was to be proven. \kdokaz Similarly to the basic reflection, rotation and central reflection, also for translation the statement about transmutation\index{transmutation!of translation} holds. \bizrek \label{izoTransmTrans} For an arbitrary translation $\mathcal{T}_{2\overrightarrow{PQ}}$ and an arbitrary isometry $\mathcal{I}$ is $$\mathcal{I}\circ \mathcal{T}_{2\overrightarrow{PQ}}\circ\mathcal{I}^{-1}= \mathcal{T}_{2\overrightarrow{\mathcal{I}(P)\mathcal{I}(Q)}}.$$ \eizrek \textbf{\textit{Proof.}} By \ref{transl2sred} we can write the translation $\mathcal{T}_{2\overrightarrow{PQ}}$ as the composite of two central reflections, namely $\mathcal{T}_{2\overrightarrow{PQ}}=\mathcal{S}_Q\circ\mathcal{S}_P$. If we use the transmutation theorem for central reflections \ref{izoTransmSredZrc}, we get: \begin{eqnarray*} \mathcal{I}\circ \mathcal{T}_{2\overrightarrow{PQ}}\circ\mathcal{I}^{-1}&=& \mathcal{I}\circ\mathcal{S}_Q\circ\mathcal{S}_P\circ\mathcal{I}=\\ &=& \mathcal{I}\circ\mathcal{S}_Q\circ\mathcal{I}^{-1} \circ\mathcal{I}\circ\mathcal{S}_P\circ\mathcal{I}^{-1}=\\ &=&\mathcal{S}_{\mathcal{I}(Q)}\circ\mathcal{S}_{\mathcal{I}(P)}=\\ &=&\mathcal{T}_{2\overrightarrow{\mathcal{I}(P)\mathcal{I}(Q)}}, \end{eqnarray*} which is what needed to be proven. \kdokaz \bizrek \label{izoKompTranslRot} The product of a translation and a rotation (also a rotation and a translation) is a rotation with the same angle. \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.6.5a.pic} \caption{} \label{sl.izo.6.6.5a.pic} \end{figure} \textbf{\textit{Proof.}} Let $\mathcal{I}=\mathcal{R}_{S,\omega}\circ \mathcal{T}_{\overrightarrow{v}}$ be the composition of the translation $\mathcal{T}_{\overrightarrow{v}}$ and the rotation $\mathcal{R}_{S,\omega}$. Let $b$ be a line passing through the point $S$ and perpendicular to the vector $\overrightarrow{v}$ (Figure \ref{sl.izo.6.6.5a.pic}). According to \ref{translKom2Zrc} for a given line $a$, $a\parallel b$, it holds that $\mathcal{T}_{\overrightarrow{v}}=\mathcal{S}_b\circ \mathcal{S}_a$. Let $c$ be a line passing through the point $S$, for which $\measuredangle b,c=\frac{1}{2}\omega$. According to \ref{rotacKom2Zrc} is $\mathcal{R}_{S,\omega}=\mathcal{S}_c\circ \mathcal{S}_b$. Therefore: $$\mathcal{I}=\mathcal{R}_{S,\omega}\circ \mathcal{T}_{\overrightarrow{v}}=\mathcal{S}_c\circ \mathcal{S}_b\circ\mathcal{S}_b\circ \mathcal{S}_a=\mathcal{S}_c\circ \mathcal{S}_a.$$ By Playfair's axiom, the lines $a$ and $c$ intersect in some point $S_1$, according to \ref{KotiTransverzala} is $\measuredangle a,c=\measuredangle b,c=\frac{1}{2}\omega$. If we use \ref{rotacKom2Zrc} once again, we get: $$\mathcal{R}_{S,\omega}\circ \mathcal{T}_{\overrightarrow{v}}=\mathcal{S}_c\circ \mathcal{S}_a=\mathcal{R}_{S_1,\omega}.$$ In a similar way it can be proven that the composition of the rotation and translation $\mathcal{T}_{\overrightarrow{v}} \circ\mathcal{R}_{S,\omega}$ is a rotation for the same angle. \kdokaz In the following we will consider the use of translation. First, we will see how we can use translation in planning tasks. \bzgled Let $A$ and $B$ be interior points of an angle $pOq$. Construct points $C$ and $D$ on the sides $p$ and $q$ such that the quadrilateral $ABCD$ is a parallelogram. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.6.6.pic} \caption{} \label{sl.izo.6.6.6.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.6.6.pic}) Since $ABCD$ is a parallelogram, $\overrightarrow{AB}= \overrightarrow{DC} = \overrightarrow{v}$. Therefore, $\mathcal{T}_{\overrightarrow{v}}(D)= C$. From the given conditions, the point $D$ lies on the line segment $p$, so its image $C$ under the translation $\mathcal{T}_{\overrightarrow{v}}$ lies on the image $p'$ of the line segment $p$ under this translation (the image $p'$ can be obtained by drawing the image $O'$ of the point $O$, and then the parallel line segment of the line segment $p$ from the point $O'$). Since the point $C$ also lies on the line segment $q$, we can draw it from the condition $C\in p'\cap q$. The number of solutions to the task depends on whether the line segments $p'$ and $q$ have any common points. Since the line segments $p$ and $q$ are not parallel (it is the angle between them), the line segments $p'$ and $q$ do not have more than one common point. Therefore, the task has one or no solution. \kdokaz \bzgled Let $AB$ be a chord of a circle $k$, $P$ and $Q$ points of this circle lying on the same side of the line $AB$ and $d$ a line segment in the plane. Construct a point $L$ on the circle $k$ such that $XY\cong d$, where $X$ and $Y$ are intersections of the lines $LP$ and $LQ$ with the chord $AB$. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.6.7.pic} \caption{} \label{sl.izo.6.6.7.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.6.7.pic}) Although the points $X$ and $Y$ are unknown, the vector $\overrightarrow{v}=\overrightarrow{XY}$ is known, which has the same length as the distance $d$ and is parallel and has the same direction as the vector $\overrightarrow{AB}$. Also, as $\omega=\angle PLQ$ is known, because it is the central angle for the chord $PQ$ (statement \ref{ObodKotGMT}). Let $P'=\mathcal{T}_{\overrightarrow{v}}(P)$. Because $\overrightarrow{PP'}=\overrightarrow{v}=\overrightarrow{XY}$, the quadrilateral $PP'YX$ is a parallelogram, so the angles $P'YQ$ and $PLQ$ are complementary (statement \ref{KotaVzporKraki}). We can therefore construct the point $Y$ as the intersection of the chord $AB$ with the appropriate chord $P'Q$ and central angle $\omega$. Then $X=\mathcal{T}^{-1}_{\overrightarrow{v}}(Y)$. \kdokaz \bzgled \footnote{Predlog za MMO 1996. (SL 10.)} Let $H$ be the orthocentre of a triangle $ABC$ and $P$ the point lying on the circumcircle of this triangle different from its vertices. $E$ is the foot of the altitude from the vertex $B$ of the triangle $ABC$. Suppose that the quadrilaterals $PAQB$ and $PARC$ are parallelograms. The lines $AQ$ and $HR$ intersect at a point $X$. Prove that $EX\parallel AP$. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.6.8.pic} \caption{} \label{sl.izo.6.6.8.pic} \end{figure} \textbf{\textit{Solution.}} We mark with $O$ the center of the circumscribed circle of the triangle $ABC$ and with $\mathcal{T}_{\overrightarrow{PA}}$ the translation for the vector $\overrightarrow{PA}$ (Figure \ref{sl.izo.6.6.8.pic}). Since $PAQB$ and $PARC$ are parallelograms, $\overrightarrow{BQ}=\overrightarrow{PA}=\overrightarrow{CR}$. This means that the translation $\mathcal{T}_{\overrightarrow{PA}}$ maps the triangle $BPC$ into the triangle $QAR$, the altitude point $H_1$ of the triangle $BPC$ into the altitude point $H'_1$ of the triangle $QAR$. We prove that $H'_1=H$ or $H=\mathcal{T}_{\overrightarrow{PA}}(H_1)$ or $\overrightarrow{H_1H}=\overrightarrow{PA}$. If we use Hamilton's theorem \ref{Hamilton} for the triangle $ABC$ and $PBC$, which have a common center of the inscribed circle, we get $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC} =\overrightarrow{OH}$ and $\overrightarrow{OP}+\overrightarrow{OB}+\overrightarrow{OC} =\overrightarrow{OH_1}$. From this it follows first $\overrightarrow{OH}-\overrightarrow{OA}= \overrightarrow{OH_1}-\overrightarrow{OP}$, then $\overrightarrow{OH}-\overrightarrow{OH_1}= \overrightarrow{OA}-\overrightarrow{OP}$ or $\overrightarrow{H_1H}=\overrightarrow{PA}$. So $H$ is the altitude point of the triangle $ARQ$, which means that $RH\perp AQ$ or $\angle AXH=90^0$. Since $\angle AEH=90^0$ as well, the points $X$ and $E$ lie on the circle with diameter $AH$ (theorem \ref{TalesovIzrKrozObrat}), so $AXEH$ is a trapezoid. So: \begin{eqnarray*} \angle AXE&=&180^0-\angle AHE \hspace*{4mm} \textrm{(theorem \ref{TetivniPogoj})}\\ &=&180^0-\angle ACB \hspace*{4mm} \textrm{(theorem \ref{KotaPravokKraki})}\\ &=&180^0-\angle APB \hspace*{4mm} \textrm{(theorem \ref{ObodObodKot})}\\ &=&\angle QAP \hspace*{4mm} \textrm{(by theorem \ref{paralelogram}, because }PAQB\textrm{ is a parallelogram)} \end{eqnarray*} By theorem \ref{KotiTransverzala}, $EX\parallel AP$. \kdokaz %________________________________________________________________________________ \poglavje{Composition of Two Rotations} \label{odd6KompRotac} In the previous sections, we have already considered the composition of some mappings. We have found that the composition of two basic reflections is a direct isometry, namely the identity or a rotation or a translation, depending on whether the lines are equal or intersect or are parallel. In the previous section, we proved that the composition of two central reflections and also the composition of two translations is a translation. In this section, we will investigate the composition of two rotations. We start with the following statement. \bizrek \label{rotacKomp2rotac} Let $R_{A,\alpha}$ and $R_{B,\beta}$ be rotations of the plane and $\mathcal{I}=R_{B,\beta}\circ R_{A,\alpha}$. Then: \begin{enumerate} \item if $\alpha+\beta\notin\{k\cdot 360^0;k\in \mathbb{Z}\}$ and $A=B$, the product $\mathcal{I}$ is a rotation for the angle $\alpha+\beta$ with the centre $A$, \item if $\alpha+\beta\notin\{k\cdot 360^0;k\in \mathbb{Z}\}$ and $A\neq B$, the product $\mathcal{I}$ is a rotation for the angle $\alpha+\beta$ with the centre $C$,\\ where $\measuredangle CAB=\frac{1}{2}\alpha$ and $\measuredangle ABC=\frac{1}{2}\beta$, \item if $\alpha+\beta\in\{k\cdot 360^0;k\in \mathbb{Z}\}$ and $A=B$, the product $\mathcal{I}$ is the identity map, \item if $\alpha+\beta\in\{k\cdot 360^0;k\in \mathbb{Z}\}$ and $A\neq B$, the product $\mathcal{I}$ is a translation. \end{enumerate} \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.7.1.pic} \caption{} \label{sl.izo.6.7.1.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.7.1.pic}) Let us assume that the centres $A$ and $B$ are different (the case $A=B$ is trivial). Let $p$ and $q$ be such lines and planes, that $\measuredangle p,AB=\frac{1}{2}\alpha$ and $\measuredangle AB,q=\frac{1}{2}\beta$. If we decompose each of the rotations with the help of reflections on the axes (statement \ref{rotacKom2Zrc}), we get: $$\mathcal{I}=R_{B,\beta}\circ R_{A,\alpha}= \mathcal{S}_q\circ\mathcal{S}_{AB}\circ\mathcal{S}_{AB}\circ \mathcal{S}_p=\mathcal{S}_q\circ \mathcal{S}_p.$$ If $\alpha+\beta\in\{0^0,360^0,-360^0\}$, then $\frac{1}{2}\left(\alpha+\beta\right)\in\{0^0,180^0,-180^0\}$, which means that the lines $p$ and $q$ are parallel (statement \ref{KotiTransverzala}). In this case, the composition $\mathcal{I}=\mathcal{S}_q\circ\mathcal{S}_p$ is a translation (statement \ref{translKom2Zrc}). If $\alpha+\beta\notin\{0^0,360^0,-360^0\}$, then $\frac{1}{2}\left(\alpha+\beta\right)\notin\{0^0,180^0,-180^0\}$ and the lines $p$ and $q$ intersect in some point $C$ (statement \ref{KotiTransverzala}). As $\measuredangle pCq$ is the exterior angle of the triangle $ABC$, it is equal to the sum of the angles $\measuredangle p,AB$ and $\measuredangle AB,q$ (statement \ref{zunanjiNotrNotr}). Therefore $\measuredangle pCq=\frac{1}{2}\left(\alpha+\beta\right)$. The composition $\mathcal{I}=\mathcal{S}_q\circ\mathcal{S}_p$ in this case represents a rotation (statement \ref{rotacKom2Zrc}) $\mathcal{R}_{C,2\measuredangle pCq}=\mathcal{R}_{C,\alpha+\beta}$. \kdokaz In the following we will see the use of the previous statement in various tasks. \bzgled Let $P$, $Q$ and $R$ be non-collinear points. Construct a triangle $ABC$ such that $ARB$, $BPC$ and $CQA$ are regular triangles with the same orientation. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.7.2.pic} \caption{} \label{sl.izo.6.7.2.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.7.2.pic}) If $A$, $B$ and $C$ are points that satisfy the given conditions, it holds: $$\mathcal{R}_{Q,60^0} \circ \mathcal{R}_{R,60^0} \circ \mathcal{R}_{P,60^0}(C)=C.$$ But according to the previous statement \ref{rotacKomp2rotac} the composite $\mathcal{R}_{Q,60^0} \circ \mathcal{R}_{R,60^0} \circ \mathcal{R}_{P,60^0}$ is a central symmetry (because $60^0+60^0+60^0=180^0$) with a fixed point $C$, so: $$\mathcal{R}_{Q,60^0} \circ \mathcal{R}_{R,60^0} \circ \mathcal{R}_{P,60^0}=\mathcal{S}_C.$$ The vertex $C$ can therefore be constructed as the center of the line segment $XX'$, where $X'$ is the image of an arbitrary point $X$ under the composite $\mathcal{R}_{Q,60^0} \circ \mathcal{R}_{R,60^0} \circ \mathcal{R}_{P,60^0}$. Then $B=\mathcal{R}_{P,60^0}(C)$ and $A=\mathcal{R}_{R,60^0}(B)$. \kdokaz \bzgled \label{rotKompZlato} Let $BALK$ and $ACPQ$ be squares with the same orientation and $Z$ the midpoint of the line segment $PK$. Prove that $BZC$ is an isosceles right triangle with the hypotenuse $BC$. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.7.3.pic} \caption{} \label{sl.izo.6.7.3.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.7.3.pic}) Let $\mathcal{I}=\mathcal{R}_{B,90^0}\circ \mathcal{R}_{C,90^0}$. Because $90°+90°= 180^0$, according to statement \ref{rotacKomp2rotac} the isometry $\mathcal{I}$ represents the central symmetry $\mathcal{S}_Y$, where $\measuredangle YCB=\frac{1}{2}\cdot 90^0=45^0$ and $\measuredangle CBY=\frac{1}{2}\cdot 90^0=45^0$. Therefore, the point $Y$ is the vertex of the isosceles right triangle $BYC$ with the hypotenuse $BC$. But since also $\mathcal{S}_Y(P)=\mathcal{I}(P)=K$, it follows that $Y=Z$. \kdokaz \bzgled \label{rotZgl1} On each side of a quadrilateral $ABCD$ squares $BALK$, $CBMN$, $DCSR$ and $ADPQ$ are externally erected. Let $E$, $F$, $G$ and $H$ be the midpoints of the line segments $PK$, $MR$, $LN$ and $SQ$. Prove that the quadrilaterals $BFDE$ and $AGCH$ are also squares. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.7.4.pic} \caption{} \label{sl.izo.6.7.4.pic} \end{figure} \textbf{\textit{Solution.}} By the previous statement \ref{rotKompZlato} the triangles $DEB$ and $BFD$ are isosceles and right with a common hypotenuse $BD$ (Figure \ref{sl.izo.6.7.4.pic}). From this it follows that the quadrilateral $BFDE$ is a square. In the same way we get that the quadrilateral $AGCH$ is a square. \kdokaz \bzgled \label{rotZgl2} Let $A_1$, $B_1$ and $C_1$ be the centres of squares externally erected on the sides $BC$, $AC$ and $AB$ of an arbitrary triangle $ABC$ and $P$ the midpoint of the side $AC$. Prove that: \begin{enumerate} \item $C_1PA_1$ is an isosceles right triangle, \item $C_1B_1$ and $A_1A$ are perpendicular and congruent line segments, \item the line segments $AA_1$, $BB_1$ and $CC_1$ intersect at a single point\footnote{To je drugi od treh planimetričnih izrekov, ki jih je objavil italijanski matematik \index{Bellavitis, G.} \textit{G. Bellavitis} (1803--1880) na kongresu v Milanu 1844.}. \end{enumerate} \ezgled \textbf{\textit{Solution.}} \textit{1)} Let $\mathcal{I}= \mathcal{R}_{C_1,90^0} \mathcal{R}_{A_1,90^0}$ (Figure \ref{sl.izo.6.7.5.pic}). Because $90^0+90^0= 180^0$, the isometry $\mathcal{I}$ represents the central reflection $\mathcal{S}_Y$, where the point $Y$ is the vertex of an isosceles right triangle $C_1YA_1$ with the hypotenuse $C_1A_1$ (statement \ref{rotacKomp2rotac}). Because $\mathcal{S}_Y(C)=\mathcal{I}(C)=A$, it follows that $Y=P$. \begin{figure}[!htb] \centering \input{sl.izo.6.7.5.pic} \caption{} \label{sl.izo.6.7.5.pic} \end{figure} \textit{2)} With the rotation $\mathcal{R}_{P,90^0}$, the line segment $B_1C_1$ is mapped to the line segment $AA_1$ (Figure \ref{sl.izo.6.7.5a.pic}). Therefore, the line segments are congruent and perpendicular (statement \ref{rotacPremPremKot}). \begin{figure}[!htb] \centering \input{sl.izo.6.7.5a.pic} \caption{} \label{sl.izo.6.7.5a.pic} \end{figure} \textit{3)} From the proven part \textit{(2)} it follows that the lines $AA_1$, $BB_1$, $CC_1$ are the altitudes of the triangle $A_1B_1C_1$ (Figure \ref{sl.izo.6.7.5a.pic}), so they intersect in its altitude point (statement \ref{VisinskaTocka}). \kdokaz \bzgled \label{rotZgl3} Let $ABCD$ and $AB_1C_1D_1$ be squares with the same orientation, $P$ and $Q$ the midpoints of the line segments $BD_1$ and $B_1D$. Suppose that $O$ and $S$ are the centres of these squares. Prove that the quadrilateral $POQS$ is also a square. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.7.6.pic} \caption{} \label{sl.izo.6.7.6.pic} \end{figure} \textbf{\textit{Solution.}} The statement is obvious if we use part \textit{(1)} from the previous example \ref{rotZgl2} twice (Figure \ref{sl.izo.6.7.6.pic}). \kdokaz If we connect the facts from examples \ref{rotZgl1} and \ref{rotZgl3}, we get the following statement (Figure \ref{sl.izo.6.7.4a.pic}). \begin{figure}[!htb] \centering \input{sl.izo.6.7.4a.pic} \caption{} \label{sl.izo.6.7.4a.pic} \end{figure} \bzgled \label{rotZgl4} Let $BALK$, $CBMN$, $DCSR$ and $ADPQ$ be the squares externally erected on the four sides of an arbitrary quadrilateral $ABCD$. Then all quadrilaterals defined by the following vertices are also squares: \begin{enumerate} \item the point $B$, the midpoint of the line segment $MR$, the point $D$ and the midpoint of the line segment $PK$, \item the point $A$, the midpoint of the line segment $LN$, the point $C$ and the midpoint of the line segment $SQ$, \item the midpoints of the line segments $QL$, $LB$, $BD$ and $DQ$, \item the midpoints of the line segments $KM$, $MC$, $CA$ and $AK$, \item the midpoints of the line segments $NS$, $SD$, $DB$ and $BN$, \item the midpoints of the line segments $RP$, $PA$, $AC$ and $CR$. \end{enumerate} \ezgled In the next example we will use the composition of rotations in a situation where the sum of the angles of these rotations is equal to $360^0$. \bzgled \label{rotZgl5} Let $ABC$ and $A'B'C'$ be isosceles triangles of the same orientation with the bases $BC$ and $B'C'$ and $\angle BAC\cong\angle B'A'C'=\alpha$. Suppose that $A_0$, $B_0$ and $C_0$ are the midpoints of the line segments $AA'$, $BB'$ and $CC'$. Prove that $A_0B_0C_0$ is also an isosceles triangle with the base $B_0C_0$ and $\angle B_0A_0C_0\cong\alpha$. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.7.7.pic} \caption{} \label{sl.izo.6.7.7.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.7.7.pic}). Let $\mathcal{I}= \mathcal{S}_{C_0} \circ \mathcal{R}_{A',\alpha} \circ \mathcal{S}_{B_0} \circ \mathcal{R}_{A,-\alpha}$. The composite $\mathcal{I}$ is a direct isometry. Because the sum of the corresponding angles of rotation is equal to $360^0$ and $\mathcal{I}(C)=C$, it must be $\mathcal{I}=\mathcal{E}$ (statement \ref{rotacKomp2rotac}). If we use the statement about the transmutation of rotation \ref{izoTransmRotac}, we get $\mathcal{S}_{A_0} \circ\mathcal{R}_{A,\alpha} \circ \mathcal{S}_{A_0} =\mathcal{R}_{A',\alpha}$. From what has been proven and from the statement \ref{transl2sred} it follows: \begin{eqnarray*} \mathcal{E}&=&\mathcal{I}= \mathcal{S}_{C_0} \circ \mathcal{R}_{A',\alpha} \circ \mathcal{S}_{B_0} \circ \mathcal{R}_{A,-\alpha}=\\ &=& \mathcal{S}_{C_0} \circ (\mathcal{S}_{A_0} \circ\mathcal{R}_{A,\alpha} \circ \mathcal{S}_{A_0}) \circ \mathcal{S}_{B_0} \circ \mathcal{R}_{A,-\alpha}=\\ &=& (\mathcal{S}_{C_0} \circ \mathcal{S}_{A_0}) \circ\mathcal{R}_{A,\alpha} \circ (\mathcal{S}_{A_0} \circ \mathcal{S}_{B_0}) \circ \mathcal{R}_{A,-\alpha}=\\ &=& \mathcal{T}_{2\overrightarrow{A_0C_0}} \circ\mathcal{R}_{A,\alpha} \circ \mathcal{T}_{2\overrightarrow{B_0A_0}} \circ \mathcal{R}_{A,-\alpha} \end{eqnarray*} Therefore $\mathcal{E}=\mathcal{T}_{2\overrightarrow{A_0C_0}} \circ\mathcal{R}_{A,\alpha} \circ \mathcal{T}_{2\overrightarrow{B_0A_0}} \circ \mathcal{R}_{A,-\alpha}$, or $\mathcal{T}_{2\overrightarrow{C_0A_0}}= \mathcal{R}_{A,\alpha} \circ \mathcal{T}_{2\overrightarrow{B_0A_0}} \circ \mathcal{R}_{A,-\alpha}$. If we use the statement about the transmutation of translation \ref{izoTransmTrans}, we get: $$\mathcal{T}_{2\overrightarrow{C_0A_0}}= \mathcal{R}_{A,\alpha} \circ \mathcal{T}_{2\overrightarrow{B_0A_0}} \circ \mathcal{R}_{A,-\alpha}=\mathcal{T}_{2\overrightarrow{B'_0A'_0}},$$ where $\mathcal{R}_{A,\alpha}:A_0, B_0\mapsto A'_0, B'_0$. Therefore $\overrightarrow{C_0A_0}=\overrightarrow{B'_0A_0'}$ or $\overrightarrow{A_0C_0}=\overrightarrow{A'_0B'_0}$. The vector $\overrightarrow{A_0B_0}$ is transformed into the vector $\overrightarrow{A'_0B'_0}=\overrightarrow{A_0C_0}$ by the rotation $\mathcal{R}_{A,\alpha}$, so $|A_0B_0|=|A_0C_0|$ and $\measuredangle B_0A_0C_0=\measuredangle\overrightarrow{A_0B_0}, \overrightarrow{A_0C_0}=\alpha$. \kdokaz \bzgled \label{RotacZglVeck} Let $A_1A_2...A_n$ and $B_1B_2...B_n$ be regular $n$-gons with the same orientation. Suppose that $S_1$, $S_2$, ..., $S_n$ are the midpoints of the line segments $A_1B_1$, $A_2B_2$, ..., $A_nB_n$. Prove that $S_1S_2...S_n$ is also a regular $n$-gon. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.7.8.pic} \caption{} \label{sl.izo.6.7.8.pic} \end{figure} \textbf{\textit{Solution.}} Since $n$-gons $A_1A_2...A_n$ and $B_1B_2...B_n$ are regular and have the same orientation, $A_1A_2A_3$ and $B_1B_2B_3$ are equilateral and have the same orientation as the triangle with the bases $A_1A_3$ and $B_1B_3$ (Figure \ref{sl.izo.6.7.8.pic}). In addition, $\angle A_1A_2A_3\cong \angle B_1B_2B_3=\frac{(n - 2)\cdot 180^0}{n}$ (\ref{pravVeckNotrKot}). From the previous example \ref{rotZgl5} it follows that also $S_1S_2S_3$ is an equilateral triangle and $\angle S_1S_2S_3=\frac{(n - 2)\cdot 180^0}{n}$. Therefore, the sides $S_1S_2$ and $S_2S_3$ are congruent, the interior angle at the vertex $S_2$ is congruent to the angle of the regular $n$-gon. We prove analogously that all sides of the polygon $S_1S_2...S_n$ are congruent and all its interior angles are congruent, which means that this polygon is regular. \kdokaz We will consider one interesting consequence of the composition of rotations in section \ref{odd7Napoleon}. %________________________________________________________________________________ \poglavje{Glide Reflections} \label{odd6ZrcDrs} So far we have learned about some types of isometries - three direct isometries (identity, rotation and translation) and one indirect isometry (reflection in a point). Now we will define another indirect isometry that has no fixed points. Let $\mathcal{T}_{\overrightarrow{v}}$ be a translation for vector $\overrightarrow{v} = 2\overrightarrow{PQ}$ and $\mathcal{S}_{PQ}$ be a reflection over the line $PQ$. The composite $\mathcal{S}_{PQ}\circ \mathcal{T}_{\overrightarrow{v}}$ is called \index{glide reflection}\pojem{glide reflection} with the axis $PQ$ for vector $\overrightarrow{v} = 2\overrightarrow{PQ}$ (Figure \ref{sl.izo.6.8.1.pic}). We denote it by $\mathcal{G}_{2\overrightarrow{PQ}}$. From the definition it is clear that a glide reflection is determined by its axis and vector. We prove the basic properties of glide reflection. \bizrek \label{IzoZrcDrs1} Any glide reflection can be expressed as the product of three reflections.\\ The reflection and translation in the product, as the presentation of glide reflection, commute, i.e. $$\mathcal{G}_{2\overrightarrow{PQ}}= \mathcal{S}_{PQ}\circ \mathcal{T}_{2\overrightarrow{PQ}}= \mathcal{T}_{2\overrightarrow{PQ}}\circ \mathcal{S}_{PQ}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.8.2.pic} \caption{} \label{sl.izo.6.8.2.pic} \end{figure} \textbf{\textit{Proof.}} Let $\mathcal{G}_{2\overrightarrow{PQ}}= \mathcal{S}_{PQ}\circ \mathcal{T}_{2\overrightarrow{PQ}}$ with the axis $PQ$ and vector $2\overrightarrow{PQ}$. Let $p$ and $q$ be the perpendiculars of the line $PQ$ at points $P$ and $Q$ (Figure \ref{sl.izo.6.8.2.pic}). Because $p,q\perp PQ$ and $p\parallel q$, we can represent the translation $\mathcal{T}_{2\overrightarrow{PQ}}$ as the composite $\mathcal{S}_q\circ\mathcal{S}_p$. In this case, the reflection $\mathcal{S}_{PQ}$ commutes with reflections $\mathcal{S}_p$ and $\mathcal{S}_q$ (statement \ref{izoZrcKomut}). Therefore: $$\mathcal{G}_{2\overrightarrow{PQ}}= \mathcal{S}_{PQ}\circ \mathcal{T}_{2\overrightarrow{PQ}}= \mathcal{S}_{PQ}\circ\mathcal{S}_q\circ\mathcal{S}_p= \mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{S}_{PQ}= \mathcal{T}_{2\overrightarrow{PQ}}\circ\mathcal{S}_{PQ},$$ which had to be proven. \kdokaz \bizrek \label{izoZrcdrsZrcdrs} The product of a glide reflection with itself is a translation (Figure \ref{sl.izo.6.8.3.pic}). \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.8.3.pic} \caption{} \label{sl.izo.6.8.3.pic} \end{figure} \textbf{\textit{Proof.}} We use the previous statement \ref{IzoZrcDrs1} and the statement \ref{translKomp}: $$\mathcal{G}^2_{2\overrightarrow{PQ}}= \mathcal{T}_{2\overrightarrow{PQ}}\circ \mathcal{S}_{PQ}\circ \mathcal{S}_{PQ}\circ \mathcal{T}_{2\overrightarrow{PQ}}= \mathcal{T}^2_{2\overrightarrow{PQ}}= \mathcal{T}_{4\overrightarrow{PQ}},$$ which had to be proven. \kdokaz From the proof of statement \ref{IzoZrcDrs1} it follows that every glide reflection can be represented as the composition of three basic reflections, where the axis of one reflection is perpendicular to the axes of the other two. A glide reflection is therefore an indirect isometry as the composition of three basic reflections - indirect isometries. The same fact also follows from the definition of glide reflection, because it is the composition of one direct and one indirect isometry. \bizrek A glide reflection has no fixed points. \eizrek \textbf{\textit{Proof.}} By statement \ref{IzoZrcDrs1} a glide reflection can be represented as the composition of three basic reflections: $$\mathcal{G}_{2\overrightarrow{PQ}}= \mathcal{S}_{PQ}\circ\mathcal{S}_q\circ\mathcal{S}_p,$$ where $p$ and $q$ are perpendicular to the line $PQ$ in points $P$ and $Q$. If the glide reflection $\mathcal{G}_{2\overrightarrow{PQ}}$ or the composition $\mathcal{S}_{PQ}\circ\mathcal{S}_q\circ\mathcal{S}_p$ had a fixed point, $\mathcal{S}_{PQ}\circ\mathcal{S}_q\circ\mathcal{S}_p$ as an indirect isometry would represent the basic reflection (statement \ref{izo1ftIndZrc}). By statement \ref{izoSop} in this case the lines $p$, $q$ and $PQ$ would belong to the same bundle, which is not possible. The glide reflection $\mathcal{G}_{2\overrightarrow{PQ}}$ therefore has no fixed points. \kdokaz We have already used the fact from the statement \ref{IzoZrcDrs1}, that we can always represent a glide reflection as a composite of three basic reflections, where the axes of these reflections are not in the same family. But does the converse hold - that the composite of three basic reflections, whose axes are not from the same family, is always a glide reflection? In connection with this, the following statement, which is also very important for the classification of isometries, which we will carry out in the next section, will hold. \bizrek \label{izoZrcdrsprq} If lines $p$, $q$ and $r$ in the plane are not from the same family of lines, then product $\mathcal{S}_p \circ \mathcal{S}_q\circ \mathcal{S}_r$ is a glide reflection. \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.8.4.pic} \caption{} \label{sl.izo.6.8.4.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.8.4.pic}) The line $q$ intersects either the line $p$ or the line $r$, because otherwise all three lines would belong to one set of parallel lines. Without loss of generality, let $q\cap r=\{A\}$. The lines $p$, $q$ and $r$ are not from the same set, so $A\notin p$. Let $s$ be a line that goes through the point $A$ and is perpendicular to the line $p$ at the point $B$. The lines $s$, $q$ and $r$ belong to the same set $\mathcal{X}_A$, so by izrek \ref{izoSop} $\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_ s = \mathcal{S}_t$, where $t$ is also $t\in \mathcal{X}_A$. From this it follows that $\mathcal{S}_r \circ \mathcal{S}_q = \mathcal{S}_t \circ \mathcal{S}_s$ or (if we use izrek \ref{izoSrZrcKom2Zrc}): $$\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p = \mathcal{S}_t \circ \mathcal{S}_s \circ \mathcal{S}_p = \mathcal{S}_t \circ \mathcal{S}_B.$$ Let $C$ be the perpendicular projection of the point $B$ onto the line $t$ (in this case $B\neq C$, because otherwise the lines $s$ and $t$, and consequently $r$ and $q$, would overlap) and $b$ the line that is perpendicular to the line $BC$ at the point $B$. Then, by izrek \ref{izoSrZrcKom2Zrc} and \ref{translKom2Zrc}: $$\mathcal{S}_r \circ \mathcal{S}_q \circ \mathcal{S}_p = \mathcal{S}_t \circ \mathcal{S}_B= \mathcal{S}_t \circ \mathcal{S}_b \circ \mathcal{S}_{BC}= \mathcal{T}_{2\overrightarrow{BC}} \circ \mathcal{S}_{BC}= \mathcal{G}_{2\overrightarrow{BC}},$$ which was to be proven. \kdokaz \bzgled \label{izoZrcDrsKompSrOsn} Any glide reflection can be expressed as the product of a reflection and a half-turn, specifically $$\mathcal{G}_{2\overrightarrow{PQ}}=\mathcal{S}_q\circ \mathcal{S}_P,$$ where $q$ is perpendicular to the line $PQ$ in the point $Q$. \ezgled By definition, $\mathcal{G}_{2\overrightarrow{PQ}}=\mathcal{S}_{PQ} \circ \mathcal{T}_{2\overrightarrow{PQ}}$. If we use the expression \ref{transl2sred} and \ref{izoSrZrcKom2Zrc}, it is: $$\mathcal{G}_{2\overrightarrow{PQ}}=\mathcal{S}_{PQ}\circ \mathcal{T}_{2\overrightarrow{PQ}}= \mathcal{S}_{PQ}\circ\mathcal{S}_Q\circ\mathcal{S}_P= \mathcal{S}_{PQ}\circ\mathcal{S}_{PQ}\circ \mathcal{S}_q\circ\mathcal{S}_P= \mathcal{S}_q\circ\mathcal{S}_P,$$ which had to be proven. \kdokaz %________________________________________________________________________________ \poglavje{Classification of Plane Isometries. Chasles' Theorem} \label{odd6KlasifIzo} In the previous sections we investigated certain types of isometries. The question arises as to whether these are the only isometries or whether there is another type of isometry that we have not encountered yet. In this section we will prove that the answer to this question is negative and make a final classification of all isometries of the plane. First, we will consider direct isometries. So far we have mentioned the identity, rotation and translation. We will prove that these are the only types of direct isometries. First, let us recall that each of these isometries can be represented as the composition of two reflections over a line. Depending on whether the axes overlap, intersect or are parallel, we got the identity, rotation and translation. These are also the only possibilities for the mutual position of two lines in the plane. Can all direct isometries be represented as the composition of two reflections over a line? In that case, the three aforementioned direct isometries would really be the only ones. We will use this idea in the following theorem. \bizrek \label{Chaslesov+} Any direct isometry can be expressed as the product of two reflections. The only direct isometries are identity map, rotations and translations. \eizrek \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.10.1.pic}) Let $\mathcal{I} : \mathbb{E}^2\rightarrow \mathbb{E}^2$ be a direct isometry of the plane. We will carry out the proof according to the number of fixed points. \textit{1)} If a direct isometry $\mathcal{I}$ has at least two fixed points, it is, according to \ref{izo2ftIdent}, the identity. We can represent it as the composite $\mathcal{I} = \mathcal{S}_p \circ \mathcal{S}_p$ (\ref{izoZrcPrInvol}) for an arbitrary line $p$. \begin{figure}[!htb] \centering \input{sl.izo.6.10.1.pic} \caption{} \label{sl.izo.6.10.1.pic} \end{figure} \textit{2)} Let's assume that the isometry $\mathcal{I}$ has exactly one fixed point $S$. Let $p$ be an arbitrary line passing through the point $S$. The composite $\mathcal{S}_p\circ \mathcal{I}$ is an indirect isometry with a fixed point $S$, therefore, according to \ref{izo1ftIndZrc}, it represents a central reflection - for example $\mathcal{S}_q$, where the axis $q$ passes through the point $S$. So we have $\mathcal{S}_p\circ \mathcal{I} = \mathcal{S}_q$ or $\mathcal{I} =\mathcal{S}_p^{-1}\circ\mathcal{S}_q=\mathcal{S}_p\circ\mathcal{S}_q$ (\ref{izoZrcPrInvol}). The lines $p$ and $q$ intersect in the point $S$ (from $p=q$ we get $\mathcal{I}=\mathcal{S}_p \circ \mathcal{S}_p=\mathcal{E}$), therefore, according to \ref{rotacKom2Zrc}, $\mathcal{I}$ represents a rotation with the center in the point $S$. \textit{3)} Let $\mathcal{I}$ be an isometry without fixed points. Then for any point $A$ of this plane $\mathcal{I}(A) = A'\neq A$. Let $p$ be the perpendicular bisector of the line segment $AA'$ and $S_p$ the reflection over the line $p$. In this case $\mathcal{S}_p\circ \mathcal{I}(A)= \mathcal{S}_p(A')=A$. The composition $\mathcal{S} \circ \mathcal{I}_p$ is therefore an indirect isometry with a fixed point $A$, so according to theorem \ref{izo1ftIndZrc} it represents the reflection over some line $q$ - $\mathcal{S}_q$, where the axis of $q$ goes through the point $A$. As in the previous example, $\mathcal{I}=\mathcal{S}_p\circ\mathcal{S}_q$. The lines $p$ and $q$ do not intersect and are not equal, because otherwise from the already proven $\mathcal{I}$ it would represent the identity or a rotation and would have at least one fixed point. So the only possibility left is that the lines $p$ and $q$ are parallel. In this case $\mathcal{I}$ is a translation (theorem \ref{translKom2Zrc}). \kdokaz We are left with indirect isometries. The only ones mentioned so far are the basic reflection and the glide reflection. Are they also the only ones? We will answer this in the next theorem. \bizrek \label{Chaslesov-} Any opposite isometry is either a reflection either it can be represented as the product of three reflections. The only opposite isometries are reflections and glide reflections. \eizrek \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.10.2.pic}) Let $\mathcal{I} : \mathbb{E}^2\rightarrow \mathbb{E}^2$ be an indirect isometry of the plane. We will again carry out the proof depending on the number of fixed points. \textit{1)} If the isometry $\mathcal{I}$ has at least one fixed point, $\mathcal{I}$ according to theorem \ref{izo1ftIndZrc} is the reflection over a line. \textit{2)} Let's assume that $\mathcal{I}$ is an isometry without any fixed points. In this case, for any point $A$ on this plane, we have $\mathcal{I}(A) = A'\neq A$. Let $p$ be the line of symmetry of the line segment $AA'$ and $\mathcal{S}_p$ be the reflection over the line $p$. Then $\mathcal{S}_p\circ \mathcal{I}(A)= \mathcal{S}_p(A')=A$. Therefore, the composition $\mathcal{S}_p \circ \mathcal{I}$ is a direct isometry with a fixed point at $A$, which, according to the previous statement, represents a rotation or the identity (a translation has no fixed points). The other case is not possible, because otherwise $\mathcal{I} = \mathcal{S}_p$ and the isometry $\mathcal{I}$ would have fixed points, which contradicts the initial assumption. So the composition $\mathcal{S} \circ \mathcal{I}_p$ is a rotation with center at $A$, which we can represent as the composition of two reflections over the lines $q$ and $r$, which intersect at the point $A$. Therefore, $\mathcal{S}_p\circ \mathcal{I}=\mathcal{S}_q\circ \mathcal{S}_r$ or $\mathcal{I}=\mathcal{S}_p\circ \mathcal{S}_q\circ \mathcal{S}_r$. Since $A'\neq A$ and $p$ is the line of symmetry of the line segment $AA'$, the point $A$ does not lie on the line $p$. This means that the lines $p$, $q$ and $r$ are not in the same plane. Therefore, according to \ref{izoZrcdrsprq}, $\mathcal{I}=\mathcal{S}_p\circ \mathcal{S}_q\circ \mathcal{S}_r$ is a glide reflection. \kdokaz Because of the previous two statements, we can say that these isometries are the only type of isometries. We can also say that we can represent any isometry of the plane as the composition of the basic isometries, where we can choose the axes so that there are no more than three in the composition. From the proof of the aforementioned statements, it is clear that we can determine any isometry only by the number of fixed points and whether the isometry is direct or indirect. We will formulate these facts in the next two statements. \bizrek \label{IzoKompZrc} Any isometry of the plane can be expressed as the product of one, two or three reflections. \eizrek \bizrek \label{Chaslesov} \index{izrek!Chaslesov} (Chasles’\footnote{\index{Chasles, M.} \textit{M. Chasles} (1793--1880), French geometer, who derived this classification in 1831.}) The only isometries of the plane $\mathcal{I} : E^2 \rightarrow E^2$ are: identity map, reflections, rotations, translations and glide reflections. Specifically:\\ \hspace*{3mm}(i) if $\mathcal{I}$ is a direct isometry and has at least two fixed points, then $\mathcal{I}$ is the identity map,\\ \hspace*{3mm}(ii) if $\mathcal{I}$ is a direct isometry and has exactly one fixed point, then $\mathcal{I}$ is a rotation (or specially a half-turn),\\ \hspace*{3mm}(iii) if $\mathcal{I}$ is a direct isometry and has no fixed points, then $\mathcal{I}$ is a translation,\\ \hspace*{3mm}(iv) if $\mathcal{I}$ is an opposite isometry and has at least one fixed point, then $\mathcal{I}$ is a reflection,\\ \hspace*{3mm}(v) if $\mathcal{I}$ is an opposite isometry and has no fixed points, then $\mathcal{I}$ is a glide reflection. \eizrek All that we have said in this section about the classification of isometries is illustrated in the following table (Figure \ref{IzoKlas.eps}): \vspace*{-2mm} %\begin{figure}[!htb] %\centering %\input{sl.izo.6.10.3.pic} %\caption{} \label{sl.izo.6.10.3.pic} %\end{figure} \begin{figure}[!htb] \centering \includegraphics[width=0.85\textwidth]{IzoKlas.eps} \caption{} \label{IzoKlas.eps} \end{figure} %________________________________________________________________________________ \poglavje{Hjelmslev's Theorem} \label{odd6Hjelmslev} The following theorem, which refers to opposite isometries, is very useful. (Hjelmslev's\footnote{ \index{Hjelmslev, J. T.} \textit{J. T. Hjelmslev} (1873--1950), Danish mathematician.}) \index{theorem!Hjelmslev's} The midpoints of all line segments defined by corresponding pairs of points of an arbitrary indirect isometry lie on the same line. \begin{figure}[!htb] \centering \input{sl.izo.6.11.1.pic} \caption{} \label{sl.izo.6.11.1.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.11.1.pic}) Let $\mathcal{I}$ be an indirect isometry and $X'=\mathcal{I}(X)$ for an arbitrary point $X$ and $X_s$ the center of the line segment $XX'$. From Chasles' theorem \ref{Chaslesov} it follows that the only types of indirect isometries of a plane are reflection and glide reflection. We prove that in both cases the sought line is exactly the line of reflection or glide reflection. In the first case, where $\mathcal{I}=\mathcal{S}_s$, it is trivial that $X_s\in s$. Let $\mathcal{I}=\mathcal{G}_{2\overrightarrow{PQ}}= \mathcal{S}_s\circ \mathcal{T}_{2\overrightarrow{PQ}}$, where $s=PQ$. If we denote by $X_1=\mathcal{T}_{2\overrightarrow{PQ}}(X)$ and $X_2$ the center of the line segment $X_1X'$, then the line segment $X_sX_2$ is the median of the triangle $XX_1X'$ with the base $XX_1$. Therefore, (from theorem \ref{srednjicaTrik}) $X_sX_2\parallel XX_1\parallel s$ or $X_s\in s$ (Playfair's axiom \ref{Playfair}). \kdokaz The proven theorem can be used if we have congruent figures or if we find an indirect isometry that maps one set of points to another. We will illustrate this with the following examples. \bzgled Let $ABC$ and $A'B'C'$ be congruent triangles with the opposite orientation. Prove that the midpoints of the line segments $AA$', $BB'$ and $CC'$ lie on the same line. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.11.2.pic} \caption{} \label{sl.izo.6.11.2.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.11.2.pic}) Since $ABC$ and $A'B'C'$ are congruent triangles with different orientations, there exists an indirect isometry $\mathcal{I}$, which maps triangle $ABC$ to triangle $A'B'C'$. In this case, $A$ and $A'$, $B$ and $B'$, and $C$ and $C'$ are pairs of this isometry, so by \ref{Chasles-Hjelmsleva} the midpoints of the line segments $AA'$, $BB'$, and $CC'$ lie on the same line. \kdokaz \bzgled Let $A$ be a point and $p$ a line in the plane. Suppose that points $X_i$ lie on the line $p$ and $AX_iY_i$ are regular triangles with the same orientation. Prove that the midpoints of the line segments $X_iY_i$ lie on the same line. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.11.3.pic} \caption{} \label{sl.izo.6.11.3.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.11.3.pic}) The isometry $\mathcal{R}_{A,60^0}\circ\mathcal{S}_p$ is an indirect isometry, which maps points $X_i$ to points $Y_i$, so by \ref{Chasles-Hjelmsleva} the midpoints of the line segments $X_iY_i$ lie on the same line. \kdokaz %________________________________________________________________________________ \poglavje{Isometry Groups. Symmetries of Figures} \label{odd6Grupe} In section \ref{odd2AKSSKL} we found that the set $\mathfrak{I}$ of all isometries of a plane together with the operation of the composition of mappings represents the so-called structure \index{group}\pojem{group}\footnote{The theory of groups was discovered by the brilliant young French mathematician \index{Galois, E.} \textit{E. Galois} (1811--1832).}. This means that the following properties are fulfilled: \begin{enumerate} \item $(\forall f\in \mathfrak{I})(\forall g\in \mathfrak{I}) \hspace*{1mm}f\circ g\in \mathfrak{I}$, \item $(\forall f\in \mathfrak{I})(\forall g\in \mathfrak{I}) (\forall h\in \mathfrak{I}) \hspace*{1mm}(f\circ g)\circ h=f\circ (g\circ h)$, \item $(\exists e\in \mathfrak{I})(\forall f\in \mathfrak{I}) \hspace*{1mm}f\circ e=e\circ f=f$, \item $(\forall f\in \mathfrak{I})(\exists g\in \mathfrak{I}) \hspace*{1mm}f\circ g=g\circ f=e$. \end{enumerate} Property (2) is valid in general for the composition of mappings. Properties (1), (3) and (4) were introduced with the axiom \ref{aksIII4}. Property (1) means that the composition of two isometries is again an isometry, (3) and (4) refer to the identity and the inverse isometry. The group mentioned, which is determined by the set $\mathfrak{I}$ of all isometries of a plane with respect to the operation of the composition of mappings, is called the \index{group!isometries} \pojem{group of all isometries of the plane}. We will also denote it by $\mathfrak{I}$. There are also other groups of isometries, which we obtain if we take an appropriate subset of all isometries of the plane. Property (1) tells us that this set cannot be arbitrary. For example, the set of all rotations of the plane is not a group, because the composition of two rotations is not always a rotation (it can also be a translation)." From the properties of translation it follows that the set of all translations together with the identity $\mathcal{E}$ forms a group - i.e. \index{group!translation}\pojem{translation group} with the designation $\mathfrak{T}$. We call it a \index{subgroup} \pojem{subgroup} of the group $\mathfrak{I}$ of all isometries of the plane (this fact is denoted by $\mathfrak{T}<\mathfrak{I}$). In fact, every group of isometries of the plane is a subgroup of the group $\mathfrak{I}$ of all isometries of this plane. However, the group of translations is not the only subgroup of the group $\mathfrak{I}$. All direct isometries of the plane in fact form one such subgroup; we denote it by $\mathfrak{I}^+$. This means that $\mathfrak{I^+}<\mathfrak{I}$. Since translations are direct isometries, $\mathfrak{T}<\mathfrak{I^+}$ is also true. It is clear that the set of all indirect isometries does not determine a group, since the composition of two indirect isometries is a direct isometry (even the identity is not an indirect isometry). There are also finite subgroups of the group $\mathfrak{I}$ (which have a finite number of isometries). An example of such a subgroup is the so-called Klein group $\mathfrak{K}$ ($\mathfrak{K}<\mathfrak{I}$) (or \textit{Klein's quadrilateral}), which is determined by the set of isometries $\{\mathcal{E}, \mathcal{S}_p, \mathcal{S}_q, \mathcal{S}_O\}$, where $p$ and $q$ are rectangles that intersect at the point $O$. This group can also be represented by the following table: \vspace*{5mm} \hspace*{22mm}\begin{tabular}{|c||c|c|c|c|} \hline % after \\ : \hline or \cline{col1-col2} \cline{col3-col4} ... $\circ$ & $\mathcal{E}$ & $\mathcal{S}_p$ & $\mathcal{S}_q$& $\mathcal{S}_O$ \\ \hline \hline $\mathcal{E}$ & $\mathcal{E}$& $\mathcal{S}_p$& $\mathcal{S}_q$& $\mathcal{S}_O$ \\ \hline $\mathcal{S}_p$ & $\mathcal{S}_p$ & $\mathcal{E}$ & $\mathcal{S}_O$& $\mathcal{S}_q$ \\ \hline $\mathcal{S}_q$ & $\mathcal{S}_q$ & $\mathcal{S}_O$ & $\mathcal{E}$& $\mathcal{S}_p$ \\ \hline $\mathcal{S}_O$ & $\mathcal{S}_O$ & $\mathcal{S}_q$ &$\mathcal{S}_p$ & $\mathcal{E}$ \\ \hline \end{tabular} \vspace*{5mm} Since Klein's group $\mathfrak{K}$ contains the glide reflection, which is an indirect isometry, $\mathfrak{K}$ is not a subgroup of the group $\mathfrak{I}^+$. We will call the group of isometries consisting only of the identity $\{\mathcal{E}\}$ the \index{group!trivial} \pojem{trivial group} with the notation $\mathfrak{E}$. This group is obviously a subgroup of every group of isometries; e.g. $\mathfrak{E}<\mathfrak{T}<\mathfrak{I^+}<\mathfrak{I}$ or $\mathfrak{E}<\mathfrak{K}<\mathfrak{I}$. If we take only the identity $\{\mathcal{E}$ and one glide reflection $\mathcal{S}_p\}$ we get another finite group of isometries. We get the same structure if instead of the glide reflection over the line we take the glide reflection over the point, i.e. $\{\mathcal{E}$ and $\mathcal{S}_O\}$. Although the sets are different, the structure of the group is the same, as we illustrate with tables. We say that in this case the groups \index{group!isomorphic}\pojem{isomorphic}. \vspace*{5mm} \hspace*{12mm}\begin{tabular}{|c||c|c|} \hline % after \\ : \hline or \cline{col1-col2} \cline{col3-col4} ... $\circ$ & $\mathcal{E}$ & $\mathcal{S}_p$ \\ \hline \hline $\mathcal{E}$ & $\mathcal{E}$& $\mathcal{S}_p$ \\ \hline $\mathcal{S}_p$ & $\mathcal{S}_p$ & $\mathcal{E}$ \\ \hline \end{tabular} \hspace*{22mm} \begin{tabular}{|c||c|c|} \hline % after \\ : \hline or \cline{col1-col2} \cline{col3-col4} ... $\circ$ & $\mathcal{E}$ & $\mathcal{S}_O$ \\ \hline \hline $\mathcal{E}$ & $\mathcal{E}$& $\mathcal{S}_O$ \\ \hline $\mathcal{S}_O$ & $\mathcal{S}_O$ & $\mathcal{E}$ \\ \hline \end{tabular} \vspace*{5mm} We have already shown that every isometry of the plane can be represented as a composite of a finite number of reflections over a line (we can always choose at most three reflections) - the statement \ref{IzoKompZrc}. We say that reflections over a line \pojem{generate} the group $\mathfrak{I}$ of all isometries of that plane or that they are \pojem{generators} of the group $\mathcal{I}$. We recall that the composite of an even number of reflections over a line represents a direct isometry, while in the case of an indirect isometry, the number of reflections over a line is odd. There is another important type of groups of isometries that we will present in the following statement. \bizrek Let $\phi$ be a figure in the plane. The set of all isometries of that plane that map the figure $\phi$ to itself forms a group. \eizrek \textbf{\textit{Proof.}} Let $\mathfrak{G}$ be the set of all isometries of the plane that map the figure $\phi$ to itself. We will prove that this set determines a group. For any isometries $f,g\in \mathfrak{G}$ we have $f(\phi)=\phi$ and $g(\phi)=\phi$. But in this case we also have $g \circ f (\phi) = \phi$ and $f^{-1}(\phi) = \phi$. So conditions (1) and (4) are satisfied. Regarding condition (2), we have already said that the operation of composition of mappings is always satisfied. Condition (3) is also satisfied in our case, since we have $\mathcal{E}(\phi)=\phi$. \kdokaz The group from the previous statement - the set of all isometries of the plane that map the figure $\phi$ onto itself - is called the \index{group!symmetry}\emph{symmetry group} of the figure $\phi$ and is denoted by $\mathfrak{G}(\phi)$. It is clear that for every figure $\phi$ it holds that $\mathfrak{G}(\phi)<\mathfrak{I}$. We mention that in the proof that a subset of a group (in our case the group $\mathfrak{I}$) represents a group or its subgroup, it is enough to check only the conditions (\textit{i}) and (\textit{iv}). The condition (\textit{ii}) is always satisfied, while the condition (\textit{iii}) directly follows from the conditions (\textit{i}) and (\textit{iv}). There is another group that is a subgroup of the symmetry group $\mathfrak{G}(\phi)$. This group consists of the set of all direct isometries from $\mathfrak{G}(\phi)$. We call it the \index{group!rotation}\emph{rotation group} of the figure $\phi$ and denote it by $\mathfrak{G}^+(\phi)$. It is clear that for every figure its rotation group is a subgroup of its symmetry group, i.e. $\mathfrak{G}^+(\phi)<\mathfrak{G}(\phi)$. In the next example we will determine the symmetry groups of different figures. \bzgled \label{grupeSimPrimeri} Determine the symmetry group and the rotation group of: (i) a square, (ii) a rectangle, (iii) a trapezium, (iv) a line, (v) a ray, (vi) a circle. \ezgled \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.12.1.pic}) (\textit{i}) A square with the center $O$ has four sides that intersect in the point $O$ and four rotations (including the identity), therefore: $\mathfrak{G}(\phi)=\{\mathcal{S}_p, \mathcal{S}_q, \mathcal{S}_r, \mathcal{S}_s, \mathcal{E}, \mathcal{R}_{O,90^0} , \mathcal{R}_{O,180^0} , \mathcal{R}_{O,270^0} \}$ and $\mathfrak{G}^+(\phi)=\{\mathcal{E}, \mathcal{R}_{O,90^0} , \mathcal{R}_{O,180^0} , \mathcal{R}_{O,270^0} \}$. \begin{figure}[!htb] \centering \input{sl.izo.6.12.1.pic} \caption{} \label{sl.izo.6.12.1.pic} \end{figure} (\textit{ii}) A rectangle with center $O$ has only two diagonals, so $\mathfrak{G}(\phi)=\{\mathcal{S}_p, \mathcal{S}_q, \mathcal{E}, \mathcal{S}_O \}$ and $\mathfrak{G}^+(\phi)=\{\mathcal{E}, \mathcal{S}_O \}$. It is interesting to note that the first group of rectangles $\mathfrak{G}(\phi)$ is actually Klein's group $\mathfrak{K}$. (\textit{iii}) If $\phi$ is an arbitrary trapezoid, the identity is the only one that maps it onto itself. This means that $\mathfrak{G}(\phi)=\mathfrak{G}^+(\phi) =\{\mathcal{E}\}$ is a group that we have already called the trivial group. (\textit{iv}) The groups of symmetry and rotation of the line $p$ are infinite groups and specifically: \begin{eqnarray*} \mathfrak{G}(p)&=&\{\mathcal{S}_l;l\perp p\}\cup \{\mathcal{S}_p\}\cup \{\mathcal{S}_O;O\in p\}\cup \{ \mathcal{T}_{\overrightarrow{v}};\overrightarrow{v} \parallel p\}\cup \{\mathcal{E}\}\\ \mathfrak{G}^+(p)&=& \{\mathcal{S}_O;O\in p\}\cup \{ \mathcal{T}_{\overrightarrow{v}};\overrightarrow{v} \parallel p\}\cup \{\mathcal{E}\} \end{eqnarray*} (\textit{v}) For the line segment $h=OA$ we have $\mathfrak{G}(h)=\{\mathcal{E}, \mathcal{S}_{h}\}$ and $\mathfrak{G}^+(h)=\{\mathcal{E}\}$. (\textit{vi}) For the circle $k(O,r)$ all reflections through the point $O$ and all rotations with center $O$ (including the identity) belong to the group of symmetry. The group of rotations is composed only of the aforementioned rotations. So: \begin{eqnarray*} \mathfrak{G}(k)&=&\{\mathcal{S}_l;l\ni O\}\cup \{\mathcal{R}_{O,\alpha};\alpha \textrm{ any angle}\}\cup \{\mathcal{E}\}\\ \mathfrak{G}^+(k)&=&\{\mathcal{R}_{O,\alpha};\alpha \textrm{ any angle}\}\cup \{\mathcal{E}\}, \end{eqnarray*} which was to be proven. \kdokaz We have seen that a rectangle has exactly two diagonals and is also centrally symmetric. Now we will prove a general statement. \bzgled If a figure in the plane has exactly two axes of symmetry, then it also has a centre of symmetry. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.12.2.pic} \caption{} \label{sl.izo.6.12.2.pic} \end{figure} \textbf{\textit{Proof.}} Let $p$ and $q$ be the only axes of symmetry of the figure $\phi$. We shall prove that $p\perp q$ (Figure \ref{sl.izo.6.12.2.pic}). Since $\mathcal{S}_p(\phi)=\mathcal{S}_q(\phi)=\phi$, by the transmutability theorem \ref{izoTransmutacija} it is also $$\mathcal{S}_{\mathcal{S}_p(q)}(\phi)= \mathcal{S}_p\circ\mathcal{S}_q\circ\mathcal{S}_p^{-1}(\phi)=\phi.$$ The lines $p$ and $q$ are the only axes of symmetry of the figure $\phi$, therefore either $\mathcal{S}_p(q)=p$ or $\mathcal{S}_p(q)=q$. In both cases it follows that $p=q$ or $p\perp q$. However, the first possibility is excluded, since $p$ and $q$ are different by assumption. Therefore the lines $p$ and $q$ are perpendicular and intersect in a point $O$, so: $$\mathcal{S}_O(\phi)= \mathcal{S}_q\circ\mathcal{S}_p(\phi)=\mathcal{S}_q(\mathcal{S}_p(\phi))= \phi,$$ which means that the figure is centrally symmetric. \kdokaz In group terms we can write the previous theorem in the following way. \bzgled If the symmetry group of some figure contains exactly two reflections then it also contains a half-turn. \ezgled First of all, we notice that different figures can have the same symmetry group. For example, the symmetry group of a rectangle and a line is the Klein group $\mathfrak{K}$. Also, an isosceles triangle and an isosceles trapezoid have the same symmetry group $\{\mathcal{E}, \mathcal{S}_{p}\}$. Also, the first three and the fifth symmetry groups from the example \ref{grupeSimPrimeri} are finite groups, while the fourth and the sixth are infinite. From this example we can also see that the infinity of the symmetry group of a figure is not related to the boundedness of this figure. The symmetry group of a bounded figure can be finite, but it can also be infinite. The same goes for unbounded figures. For further use, we shall define the concept of boundedness more precisely. The figure $\phi$ is \index{figure!bounded} \pojem{bounded}, if there exists such a distance $AB$, that for any points $X$ and $Y$ of this figure it holds $XY|XX_n|=|n\overrightarrow{v}|$. This should hold for every $n\in \mathbb{N}$, which of course is not the case. The last relation therefore leads us to a contradiction, which means that in the group $\mathfrak{G}(\phi)$ there are no translations. \kdokaz From the previous izrek it follows, that the group of symmetry of a bounded figure also does not contain a glide reflection, since its square is a translation (example \ref{izoZrcdrsZrcdrs}). But how is it with reflections and rotations? \bizrek \label{GrupaSomer1} If a bounded figure has more axes of symmetry, then they all intersect at one point. \eizrek \textbf{\textit{Proof.}} Let $p$ and $q$ be the axes of symmetry of a bounded figure $\phi$. The lines $p$ and $q$ are not parallel, because otherwise the composition of the appropriate reflections $\mathcal{S}_p \circ \mathcal{S}_q$ would be a translation. So $p$ and $q$ intersect in some point $S$. If $r$ is a third arbitrary axe, it contains the point $S$, because otherwise (from \ref{izoZrcdrsprq}) the composition $\mathcal{S}_p \circ \mathcal{S}_q\circ \mathcal{S}_r$ would be a mirror glide. \kdokaz \bizrek \label{GrupaSomerRot} If a bounded figure has at least one axe of symmetry and at least one center of rotation, then this center lies on the axe of symmetry. \eizrek \textbf{\textit{Proof.}} We assume the opposite - let $p$ be the axe of symmetry of a bounded figure $\phi$ and $S$ the center of rotation $\mathcal{R}_{S,\alpha}$, which maps this figure into itself, but $S\notin p$. The rotation $\mathcal{R}_{S,\alpha}$ can be represented as the composition of two reflections over the lines with the points, which go through the point $S$. So $\mathcal{S}_p \circ \mathcal{R}_{S,\alpha} = \mathcal{S}_p \circ \mathcal{S}_q\circ \mathcal{S}_r$. The lines $p$, $q$ and $r$ are not in the same bundle, so (from \ref{izoZrcdrsprq}) the composition $\mathcal{S}_p \circ \mathcal{R}_{S,\alpha}$ is a mirror glide, which cannot be in the group $\mathfrak{G}(\phi)$, which means that the center $S$ lies on the axe $p$. \kdokaz \bizrek \label{GrupaRot} All rotations of a bounded figure have the same center. \eizrek \begin{figure}[!htb] \centering \input{sl.izo.6.12.4.pic} \caption{} \label{sl.izo.6.12.4.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.izo.6.12.4.pic}). Let's assume the opposite - that $\mathcal{R}_{O,\alpha}$ and $\mathcal{R}_{S,\beta}$ are rotations that map the bounded figure $\phi$ into itself and $O\neq S$. Then $O'=\mathcal{R}_{S,\beta}(O)\neq S$. By \ref{izoTransmRotac} $\mathcal{R}_{O',\alpha}=\mathcal{R}_{S,\beta}\circ \mathcal{R}_{O,\alpha}\circ\mathcal{R}_{S,\beta}^{-1}$, so $\mathcal{R}_{O',\alpha}\in \mathfrak{G}(\phi)$ and then also $\mathcal{I}=\mathcal{R}_{O',\alpha}^{-1}\circ \mathcal{R}_{O,\alpha}\in \mathfrak{G}(\phi)$. Let $p$ be a line that with the line $OO'$ in the point $O$ determines the angle $\frac{1}{2}\alpha$, and $q$ its parallel through the point $O'$. By \ref{rotacKom2Zrc} $\mathcal{I} = \mathcal{S}_q \circ \mathcal{S}_{OO'} \circ \mathcal{S}_{OO'} \circ \mathcal{S}_p = \mathcal{S}_q \circ \mathcal{S}_p$, so $\mathcal{I}$ is a translation, which is not possible. This means that $O\neq S$ is not true, so all rotations of a bounded figure have the same centre. \kdokaz From the previous equations we get the following statement. \bizrek \label{GrupaOmejenLik} If symmetry group of a bounded figure $\phi$ is not trivial, then there is a point $S$ such that the only possible isometries in this group are: \\ (i) the identity map,\\ (ii) rotations with the centre $S$,\\ (iii) reflections with axes, which all contains the point $S$. \eizrek In this way we have determined all possible groups of bounded figures. These groups are not necessarily finite. What are then the finite groups? Let's explore some examples of symmetry groups. \bzgled \label{GrupaDiederska} Determine the symmetry group and the rotation group of a regular $n$-gon. \ezgled \begin{figure}[!htb] \centering \input{sl.izo.6.12.5.pic} \caption{} \label{sl.izo.6.12.5.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.izo.6.12.5.pic}) No matter if $n$ is an even or odd number, a regular $n$-gon has exactly $n$ diagonals (see section \ref{odd3PravilniVeck}). The basic angle of rotation that transforms a $n$-gon into itself is $\theta = \frac{360^0}{n}$. Therefore, if we denote the group of symmetries and the group of rotations of a regular $n$-gon with $\mathcal{D}_n$ and $\mathcal{C}_n$, and its center with $O$, we get: \begin{eqnarray*} \mathfrak{D}_n&=&\{ \mathcal{S}_{p_1}, \mathcal{S}_{p_2},\ldots, \mathcal{S}_{p_n}, \mathcal{E}, \mathcal{R}_{O,\theta}, \mathcal{R}_{O,2\theta},\ldots \mathcal{R}_{O,(n-1)\theta}\}\\ \mathfrak{C}_n&=&\{ \mathcal{E}, \mathcal{R}_{O,\theta}, \mathcal{R}_{O,2\theta},\ldots \mathcal{R}_{O,(n-1)\theta}\}, \end{eqnarray*} which is what needed to be determined. \kdokaz The group of symmetries $\mathfrak{D}_n$ of a regular $n$-gon from the previous example is called \index{group!diedral} \pojem{diedral group}, the group of rotations of a regular $n$-gon $\mathfrak{C}_n$ is \index{group!cyclic} \pojem{cyclic group}. It is clear that $\mathfrak{C}_n<\mathfrak{D}_n$. Groups $\mathfrak{D}_n$ and $\mathfrak{C}_n$ ($n\geq 3$) can also be generalized for the cases $n < 3$, even though in this case we are no longer talking about an $n$-gon. $\mathfrak{D}_2$ is thus actually Klein's group $\mathfrak{K}$ (the group of symmetries of a line or some sort of a "2-gon"). The group $\mathfrak{C}_2$ consists of the identity and one central symmetry. This group can be represented as the group of rotations of a line. The group $\mathfrak{D}_1$ contains the identity and one reflection, but it is isomorphic to the group $\mathfrak{C}_2$ (an example we have already mentioned at the beginning of this section). The group $\mathfrak{C}_1$ is the trivial group $\mathfrak{E}$. \begin{figure}[!htb] \centering \input{sl.izo.6.12.5a.pic} \caption{} \label{sl.izo.6.12.5a.pic} \end{figure} We mention that the groups $\mathfrak{D}_n$ and $\mathfrak{C}_n$ ($n\in \mathbb{N}$) are not symmetry groups only for regular $n$-sided polygons (Figure \ref{sl.izo.6.12.5a.pic}). They can also be symmetry groups of unlimited shapes. For example, Klein's group $\mathfrak{D}_2$ is also a symmetry group of the hyperbola with equation $x\cdot y= 1$ in the Cartesian coordinate system $O_{xy}$. The asymptotes are then determined by the equations $y = x$ and $y = -x$, and the center of symmetry is the origin $O$. $\mathfrak{D}_2$ is also a symmetry group of the ellipse, given by the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. It is clear that all groups $\mathfrak{D}_n$ and $\mathfrak{C}_n$ ($n\in \mathbb{N}$) are finite groups of isometries. We will show that these are the only finite groups of isometries! \bizrek \label{GrupaLeonardo} \index{izrek!Leonarda da Vincija} (Leonardo da Vinci\footnote{Famous Italian painter, architect and inventor \index{Leonardo da Vinci} \textit{Leonardo da Vinci} (1452--1519) studied all possible symmetries of the central building and possible arrangements of chapels around it, which preserve the basic symmetry.}) The only finite groups of isometries are $\mathfrak{D}_n$ and $\mathfrak{C}_n$. \eizrek \textbf{\textit{Proof.}} Let $\mathfrak{G}$ be a finite group of isometries. If it contained a translation $\mathcal{T}_{\overrightarrow{v}}$, it would have an infinite subgroup $\{ \mathcal{E}, \mathcal{T}_{\overrightarrow{v}}, \mathcal{T}_{\overrightarrow{v}}^2, \mathcal{T}_{\overrightarrow{v}}^3,\ldots \}$, which is not possible. Therefore, $\mathfrak{G}$ is a group without translations. In a similar way as before (statement \ref{GrupaOmejenLik}), we can prove that the only possible isometries in this group (except for the identical mapping) are rotations with the same center $S$ and reflections over a line with eight, which go through the point $S$. If the group has no rotations, at most one basic reflection is possible (already two would generate a rotation). In this case, the only possible groups are $\mathfrak{D}_1$ and $\mathfrak{C}_1$. Without loss of generality, we assume that all angles of rotation are positive. Because the group $\mathfrak{G}$ is finite, there exists a rotation $\mathcal{R}_{O,\theta}$ with the smallest angle $\theta$. For the same reason, there exists a natural number $n$, for which $\mathcal{R}_{O,\theta}^n = \mathcal{E}$ (otherwise, the finite group $\mathfrak{G}$ would have an infinite subgroup $\{ \mathcal{E}, \mathcal{R}_{O,\theta}, \mathcal{R}_{O,\theta}^2, \mathcal{R}_{O,\theta}^3,\ldots \}$). Therefore, $\theta = \frac{360^0}{n}$. Let $\mathcal{R}_{O,\delta}$ be an arbitrary rotation from $\mathfrak{G}$. We prove that the angle $\delta$ is a multiple of the angle $\theta$ or $\delta=k\cdot \theta$ for some $k\in \mathbb{N}$. We assume the opposite, that such $k$ does not exist. Because $\delta>\theta$, for some $l\in \mathbb{N}$ we have $l\theta<\delta<(l+1)\theta$. If we rearrange, we get $0<\delta-l\theta<\theta$. But $\mathcal{R}_{O,\delta-l\theta}=\mathcal{R}_{O,\delta}\circ \left(\mathcal{R}_{O,\theta}^{-1}\right)^l\in \mathfrak{G}$, which is contradictory to how we defined the angle $\theta$. Therefore, all rotations from $\mathfrak{G}$ are of the form $\mathcal{R}_{O,\theta}^k$, $k\in \{ 1,2,\ldots, n\}$. If there are no basic reflections in the group, then $\mathfrak{G}$ is a cyclic group or $\mathfrak{G}=\mathfrak{C}_n$ for some $n\in \mathbb{N}$. If there are also reflections over a line in the group, their composites (of each two reflections) are rotations, so each pair of axes determines an angle, which is half of some angle of rotation. In this case, $\mathfrak{G}$ is a dihedral group or $\mathfrak{G}=\mathfrak{D}_n$ for some $n\in \mathbb{N}$. \kdokaz We emphasize once again that the final group of symmetries and the group of symmetries of a limited figure are not the same concept. The group of symmetries of limited figures is not necessarily final (for example, the group of symmetries of a circle). Similarly, unlimited figures can have a finite group of symmetries (for example, an equilateral triangle) and unlimited figures with an infinite group of symmetries (for example, a line). We will illustrate these facts with the following diagram (Figure \ref{sl.izo.6.12.6.pic}). \begin{figure}[!htb] \centering \input{sl.izo.6.12.6.pic} \caption{} \label{sl.izo.6.12.6.pic} \end{figure} We will also mention another type of infinite groups of isometries that preserve certain tessellations of the plane (see section \ref{odd3Tlakovanja}). So it's not just about the groups of symmetries of these tessellations, but also about all subgroups of these groups. For example, if we choose tiling a plane with an arbitrary parallelogram $ABCD$ (Figure \ref{sl.izo.6.12.7.pic}), the infinite group of symmetries of this tiling is generated by reflections in the vertex $A$ and the centers of the sides $AB$ and $AD$. In this group are then also reflections in the vertices, centers of sides and intersections of diagonals of all parallelograms of this tiling and all translations that are composites of two of these reflections. But one of its subgroups, which also preserves the same tiling, is the group of all mentioned translations. It is generated by two translations $\mathcal{T}_{\overrightarrow{AB}}$ and $\mathcal{T}_{\overrightarrow{AD}}$. \begin{figure}[!htb] \centering \input{sl.izo.6.12.7aa.pic} \input{sl.izo.6.12.7bb.pic} \caption{} \label{sl.izo.6.12.7.pic} \end{figure} The group of symmetries in the case of tiling with a regular triangle $ABC$ (Figure \ref{sl.izo.6.12.7.pic}) or tiling $(3,6)$ contains rotations in the vertices of the grid at angles $k\cdot 60^0$, reflections over lines, which are determined by the sides and heights of these triangles, and translations, generated by translations $\mathcal{T}_{\overrightarrow{AB}}$ and $\mathcal{T}_{\overrightarrow{AC}}$. This group has several different subgroups, all preserving tiling $(3,6)$. All such groups that preserve certain tilings are called \index{group!discrete} \pojem{discrete groups of isometries}. We formally define them as follows: A group of isometries $\mathfrak{G}$ is a discrete group of isometries, if there exists such $\varepsilon>0$, that the lengths of vectors of all translations and measures of angles of all rotations from this group are greater than $\varepsilon$. There are two types of discrete groups of isometries: \index{group!frieze} \pojem{frieze groups}\footnote{The term \textit{frieze}, which was used by the Ancient Greeks, meant a repeating border pattern.} and \index{group!wallpaper}\pojem{wallpaper groups}. In frieze groups, the subgroup of all translations is generated by one single translation, in wallpaper groups by two translations, determined by two non-linear vectors. It has been proven that there are exactly 7 frieze groups and 17 wallpaper groups\footnote{All 17 groups or the corresponding types of ornamentation were known to the Egyptians, and were often used by Muslim artists. In the Alhambra palace (Spain), the Moors painted all 17 types of ornamentation in the 14th century. The formal proof that there are exactly 17 wallpaper groups was first given by the Russian mathematician, crystallographer and mineralogist \index{Fedorov, E.} \textit{E. Fedorov} (1853–-1919) in 1891, and then completed and continued by the Hungarian mathematician \index{Pólya, G.}\textit{G. Pólya} (1887–-1985) in 1924, who is more famous for his famous book ‘‘How to Solve Mathematical Problems’’.}. Frieze groups determine 7 different types of \index{border} \pojem{borders} - strips with repeating patterns (Figure \ref{sl.izo.6.12.7}a\footnote{http://mathworld.wolfram.com/WallpaperGroups.html}), wallpaper groups 17 different types of \pojem{ornamentation} - covering a plane with identical patterns (Figure \ref{sl.izo.6.12.7}b\footnote{http://www.quadibloc.com/math/tilint.htm}). \begin{figure}[!h] \centering \includegraphics[bb=0 0 7cm 7cm]{bands.eps}\\ \vspace*{15mm} \includegraphics[width=1\textwidth]{wall17_phpbtUJbf.eps} \caption{} \label{sl.izo.6.12.7} \end{figure} %v bmp %\begin{figure}[!h] %\centering %\includegraphics[bb=0 0 7cm 7cm]{bands.bmp}\\ %\vspace*{8mm} %\includegraphics[bb=0 0 12cm 9.85cm]{wall17_phpbtUJbf.bmp} %\caption{} \label{sl.izo.6.12.7} %\end{figure} \bnaloga\footnote{40. IMO Romania - 1999, Problem 1.} Determine all finite sets $\mathbf {S}$ of at least three points in the plane which satisfy the following condition:\\ for any two distinct points $A$ and $B$ in $\mathbf {S}$, the perpendicular bisector of the line segment $AB$ is an axis of symmetry for $\mathbf {S}$. \enaloga \textbf{\textit{Solution.}} Let $\mathbf {S}=\{A_1,A_2,\ldots A_n \}$ in $\mathcal{M}=\{s_{A_iA_j};\hspace*{1mm}A_i,A_j\in \mathcal{S}\}$ be a set of all symmetries. Because $\mathbf {S}$ is a finite set (or figure), by Theorem \ref{GrupaSomer1} all perpendiculars of this set or symmetries from the set $\mathcal{M}$ go through one point - we denote it with $O$. It is clear that $O$ is the intersection of any two symmetries from $\mathcal{M}$, so $O=s_{A_1A_2}\cap s_{A_2A_3}$. Let $k$ be a circle with center $O$ that goes through the point $A_1$. We first prove that all points of the set $\mathbf {S}$ lie on this circle. Let $A_i\in \mathbf {S}$ be an arbitrary point. From the proven $O\in s_{A_1A_i}$ and $A_1 \in k$ it follows $A_i=\mathcal{S}_{s_{A_1A_i}}(A_1)\in \mathcal{S}_{s_{A_1A_i}}(k)=k$ (Figure \ref{sl.izo.6.12.IMO1.pic}). Without loss of generality, we can assume that the points $\mathbf {S}$ are ordered in sequence on the circle $k$, so that $\angle A_1OA_2<\angle A_1OA_3<\cdots \angle A_1OA_n$ (otherwise we can make a new labeling of these points). This means that $A_1A_2\ldots A_n$ represents a polygon, which is inscribed in the circle $k$. We also prove that this polygon is regular. Because this polygon is concyclic, it is enough to prove that all sides are congruent. Because $s_{A_1A_3}$ is the altitude of the set $\mathbf {S}$ (and therefore also of the polygon $A_1A_2\ldots A_n$), $A_1A_2\cong A_2A_3$. Similarly, $s_{A_iA_{i+2}}$ ($i\in \{2,3,\ldots n-1\}$) is the altitude of the polygon $A_1A_2\ldots A_n$ and $A_iA_{i+1}\cong A_{i+1}A_{i+2}$, which means that $A_1A_2\ldots A_n$ is a regular polygon. The set $\mathbf {S}$ therefore represents the vertices of a regular polygon. \kdokaz %\vspace*{31mm} \newpage \naloge{Exercises} \begin{enumerate} \item Given a line $p$ and points $A$ and $B$, which lie on opposite sides of the line $p$. Construct a point $X$, which lies on the line $p$, so that the difference $|AX|-|XB|$ is maximal. \item In the plane are given lines $p$, $q$ and $r$. Construct an equilateral triangle $ABC$, so that the vertex $B$ lies on the line $p$, $C$ on $q$, and the altitude from the vertex $A$ lies on the line $r$. \item Given a quadrilateral $ABCD$ and a point $S$. Draw a parallelogram with the center in the point $S$, so that its vertices lie on the lines of the sides of the given quadrilateral. \item Let $\mathcal{I}$ be an indirect isometry of the plane, which maps the point $A$ to the point $B$, $B$ to $A$. Prove that $\mathcal{I}$ is a central symmetry. \item Let $K$ and $L$ be points, which are symmetric to the vertex $A$ of the triangle $ABC$ with respect to the lines of symmetry of the internal angles at the vertices $B$ and $C$. Let $P$ be the point of tangency of the inscribed circle of this triangle and the sides $BC$. Prove that $P$ is the center of the segment $KL$. \item Let $k$ and $l$ be two circles on different sides of the line $p$. Draw an isosceles triangle $ABC$, so that its altitude $AA'$ lies on the line $p$, the vertex $B$ lies on the circle $k$, and the vertex $C$ lies on the circle $l$. \item Let $k$ be a circle and $a$, $b$, and $c$ be lines in the same plane. Draw a triangle $ABC$ inscribed in the circle $k$, so that its sides $BC$, $AC$, and $AB$ are parallel to the lines $a$, $b$, and $c$, respectively. \item Let $ABCDE$ be a quadrilateral with $BC\parallel DE$ and $CD\parallel EA$. Prove that the vertex $D$ lies on the perpendicular bisector of the line segment $AB$. \item The lines $p$, $q$, and $r$ lie in the same plane. Prove that the composition $\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p =\mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_r$ is equivalent to the condition that the lines $p$, $q$, and $r$ belong to the same bundle. \item Let $O$, $P$, and $Q$ be three non-collinear points. Construct a square $ABCD$ (in the plane $OPQ$) with the center at the point $O$, so that the points $P$ and $Q$ lie on the lines $AB$ and $BC$, respectively. \item Let $\mathcal{R}_{S,\alpha}$ be a rotation and $\mathcal{S}_p$ be a reflection in the same plane and $S\in p$. Prove that the composition $\mathcal{R}_{S,\alpha}\circ\mathcal{S}_p$ and $\mathcal{S}_p\circ\mathcal{R}_{S,\alpha}$ represent a reflection. \item Given a point $A$ and a circle $k$ in the same plane. Draw a square $ABCD$, so that the midpoints of the diagonals $BD$ lie on the circle $k$. \item Let $ABC$ be an arbitrary triangle. Prove: $$\mathcal{R}_{C,2\measuredangle BCA}\circ \mathcal{R}_{B,2\measuredangle ABC}\circ \mathcal{R}_{A,2\measuredangle CAB}=\mathcal{E}.$$ \item Prove that the composition of a reflection $\mathcal{S}_p$ and a central reflection $\mathcal{S}_S$ ($S\in p$) represents a reflection. \item Let $O$, $P$, and $Q$ be three non-collinear points. Construct a square $ABCD$ (in the plane $OPQ$) with the center at the point $O$, so that the points $P$ and $Q$ lie on the lines $AB$ and $CD$, respectively. \item What does the composition of a translation and a central reflection represent? \item Given are a line $p$ and circles $k$ and $l$, which lie in the same plane. Draw a line that is parallel to the line $p$, so that it determines the congruent tangents on the circles $k$ and $l$. \item Let $c$ be a line that intersects the parallels $a$ and $b$, and $l$ be a distance. Draw an equilateral triangle $ABC$, so that $A\in a$, $B\in b$, $C\in c$ and $AB\cong l$. \item Prove that the composition of a rotation and an axial reflection of a plane represents a glide reflection exactly when the center of rotation does not lie on the axis of the axial reflection. \item Let $ABC$ be an equilateral triangle. Prove that the composition $\mathcal{S}_{AB} \circ\mathcal{S}_{CA} \circ\mathcal{S}_{BC}$ represents a glide reflection. Also determine the vector and the axis of this glide reflection. \item Given are points $A$ and $B$ on the same side of a line $p$. Draw a line $XY$, which lies on the line $p$ and is congruent to the given distance $l$, so that the sum $|AX|+|XY|+|YB|$ is minimal. \item Let $ABC$ be an isosceles right triangle with the right angle at the vertex $A$. What does the composition: $\mathcal{G}_{\overrightarrow{AB}}\circ \mathcal{G}_{\overrightarrow{CA}}$ represent? \item In the same plane are given lines $a$, $b$ and $c$. Draw points $A\in a$ and $B\in b$ so that $\mathcal{S}_c(A)=B$. \item Given are lines $p$ and $q$ and a point $A$ in the same plane. Draw points $B$ and $C$ so that the lines $p$ and $q$ will be the internal angle bisectors at the vertices $B$ and $C$ of the triangle $ABC$. \item Let $s$ be the angle bisector of one of the angles determined by the lines $p$ and $q$. Prove that $\mathcal{S}_s\circ\mathcal{S}_p = \mathcal{S}_q\circ\mathcal{S}_s$. \item Let $S$ be the center of the triangle $ABC$ inscribed in the circle and $P$ be the point where this circle touches the side $BC$. Prove: $$\mathcal{S}_{SC} \circ\mathcal{S}_{SA}\circ\mathcal{S}_{SB} =\mathcal{S}_{SP}.$$ \item The lines $p$, $q$ and $r$ of a plane go through the center $S$ of the circle $k$. Draw a triangle $ABC$, which is inscribed in this circle, so that the lines $p$, $q$ and $r$ are the altitudes of the vertices $A$, $B$ and $C$ of this triangle. \item The lines $p$, $q$, $r$, $s$ and $t$ of a plane intersect in the point $O$, and the point $M$ lies on the line $p$. Draw a pentagon so that $M$ is the center of one of its sides, and the lines $p$, $q$, $r$, $s$ and $t$ are the altitudes. \item The point $P$ lies in the plane of the triangle $ABC$. Prove that the lines which are symmetric to the lines $AP$, $BP$ and $CP$ with respect to the altitudes at the vertices $A$, $B$ and $C$ of this triangle, belong to the same family. \item Calculate the angle determined by the lines $p$ and $q$, if it holds that $\mathcal{S}_p\circ\mathcal{S}_q\circ\mathcal{S}_p = \mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{S}_q$. \item Let $\mathcal{R}_{A,\alpha}$ and $\mathcal{R}_{B,\beta}$ be rotations in the same plane. Determine all points $X$ in this plane, for which it holds that $\mathcal{R}_{A,\alpha}(X)=\mathcal{R}_{B,\beta}(X)$. \item The lines $p$ and $q$ intersect at an angle of $60^0$ in the center $O$ of the equilateral triangle $ABC$. Prove that the segments determined by the sides of the triangle $ABC$ on the lines $p$ and $q$ are congruent. \item The point $S$ is the center of the regular pentagon $ABCDE$. Prove that it holds: $$\overrightarrow{SA} + \overrightarrow{SB} + \overrightarrow{SC} + \overrightarrow{SD} + \overrightarrow{SE} = \overrightarrow{0}.$$ \item Prove that the diagonals of the regular pentagon intersect in the points which are also the vertices of the regular pentagon. \item Let $ABP$ and $BCQ$ be two triangles with the same orientation and $\mathcal{B}(A,B,C)$. The points $K$ and $L$ are the midpoints of the lines $AQ$ and $PC$. Prove that $BLK$ is a triangle. \item There are three concentric circles and a line in the same plane. Draw a triangle so that its vertices are on these circles in order, and one side is parallel to the given line. \item The point $P$ is an internal point of the triangle $ABC$, so that $\angle APB=113^0$ and $\angle BPC=123^0$. Calculate the size of the angles of the triangle whose sides are consistent with the distances $PA$, $PB$ and $PC$. \item The points $P$, $Q$ and $R$ are given. Draw a triangle $ABC$ so that $P$, $Q$ and $R$ are the centers of the squares constructed over the sides $BC$, $CA$ and $AB$ of this triangle. \item Let $A$ and $B$ be points and $p$ a line in the same plane. Prove that the composite $\mathcal{S}_B\circ\mathcal{S}_p\circ\mathcal{S}_A$ is a reflection exactly when $AB\perp p$. \item Let $p$, $q$ and $r$ be tangents to the triangle $ABC$ of the inscribed circles that are parallel to its sides $BC$, $AC$ and $AB$. Prove that the lines $p$, $q$, $r$, $BC$, $AC$ and $AB$ determine such a hexagon, in which the pairs of opposite sides are consistent with the distances. \item Draw a triangle with the data: $\alpha$, $t_b$, $t_c$. \item Let $ALKB$ and $ACPQ$ be two squares outside the triangle $ABC$ drawn over the sides $AB$ and $AC$, and $X$ the center of the side $BC$. Prove that $AX\perp LQ$ and $|AX|=\frac{1}{2}|QL|$. \item Let $O$ be the center of the triangle $ABC$ and $D$ and $E$ the points of the sides $CA$ and $CB$, so that $CD\cong CE$. The point $F$ is the fourth vertex of the parallelogram $BODF$. Prove that the triangle $OEF$ is a right triangle. \item Let $L$ be the point in which the inscribed circle of the triangle $ABC$ touches its side $BC$. Prove: $$\mathcal{R}_{C,\measuredangle ACB}\circ\mathcal{R}_{A,\measuredangle BAC} \circ\mathcal{R}_{B,\measuredangle CBA} =\mathcal{S}_L.$$ \item Points $P$ and $Q$ as well as $M$ and $N$ are the centers of two squares, which are drawn outside of the opposite sides of any quadrilateral. Prove that $PQ\perp MN$ and $PQ\cong MN$. \item Let $APB$ and $ACQ$ be two right triangles, which are drawn outside of the triangle $ABC$ on the sides $AB$ and $AC$. Point $S$ is the center of the side $BC$ and $O$ is the center of the triangle $ACQ$. Prove that $|OP|=2|OS|$. \item Prove that the reflection in an axis and the translation of a plane commute only when the axis of this reflection is parallel to the vector of the translation. \item In the same plane, given are a line $p$, circles $k$ and $l$ and a distance $d$. Draw a rhombus $ABCD$ with a side, which is congruent to the distance $d$, the side $AB$ lies on the line $p$, points $C$ and $D$ in turn lie on the circles $k$ and $l$. \item Let $p$ be a line, $A$ and $B$ points which lie on the same side of the line $p$, and $d$ a distance in the same plane. Draw points $X$ and $Y$ on the line $p$ so that $AX\cong BY$ and $XY\cong d$. \item Let $H$ be the altitude of the triangle $ABC$ and $R$ the radius of the circumscribed circle of this triangle. Prove that $|AB|^2+|CH|^2=4R^2$. \item Let $EAB$ be a triangle, which is drawn on the side $AB$ of the square $ABCD$. Let also $M=pr_{\perp AE}(C)$ and $N=pr_{\perp BE}(D)$ and point $P$ be the intersection of the lines $CM$ and $DN$. Prove that $PE\perp AB$. \item Draw an equilateral triangle $ABC$ so that its vertices in turn lie on three parallel lines $a$, $b$ and $c$ in the same plane, the center of this triangle lies on the line $s$, which intersects the lines $a$, $b$ and $c$. \item If a pentagon has at least two axes of symmetry, it is regular. Prove. \item Let $A$, $B$ and $C$ be three collinear points. What does the composite $\mathcal{G}_{\overrightarrow{BC}}\circ \mathcal{S}_A$ represent? \item Let $p$, $q$ and $r$ be lines that are not in the same plane, and let $A$ be a point in the same plane. Draw a line $s$ through the point $A$ such that $\mathcal{S}_r\circ \mathcal{S}_q\circ \mathcal{S}_p(s)=s'$ and $s\parallel s'$. %new tasks %___________________________________ \item Let $Z$ and $K$ be inner points of the rectangle $ABCD$. Draw points $A_1$, $B_1$, $C_1$ and $D_1$, which in turn lie on the sides $AB$, $BC$, $CD$ and $DA$ of this rectangle, so that $\angle ZA_1A\cong\angle B_1A_1B$, $\angle A_1B_1B\cong\angle C_1B_1C$, $\angle B_1C_1C\cong\angle D_1C_1D$ and $\angle C_1D_1D\cong\angle KD_1A$. \item The point $A$ lies on the line $a$, and the point $B$ lies on the line $b$. Determine the rotation that maps the line $a$ to the line $b$ and the point $A$ to the point $B$. \item In the center of the square, two rectangles intersect. Prove that these rectangles intersect the sides of the square at the vertices of a new square. \item Given a circle $k$ and lines $a$, $b$, $c$, $d$ and $e$, which lie in the same plane. Draw a pentagon on the circle $k$ with sides that are in turn parallel to the lines $a$, $b$, $c$, $d$ and $e$. \item The point $P$ lies inside the angle $aOb$. Draw a line $p$ through the point $P$, which with the sides $a$ and $b$ determines a triangle with the smallest area. \item The parallelogram $PQKL$ is drawn in the parallelogram $ABCD$ (the vertices of the first lie on the sides of the second). Prove that the parallelograms have a common center. \item The arcs $l_1, l_2,\cdots , l_n$ lie on the circle $k$ and the sum of their lengths is less than the radius of this circle. Prove that there exists such a diameter $PQ$ of the circle $k$, that none of its endpoints lies on any of the arcs $l_1, l_2,\cdots , l_n$. \item Given is a circle $k(S,20)$. Players $\mathcal{A}$ and $\mathcal{B}$ take turns drawing circles with radii $x_i$ ($1AB$, or $b>c$. Since $\angle AMB =\omega$, by the statement \ref{ObodKotGMT} the point $M$ lies on the arc $l$ with the chord $AB$ and the central angle $\omega$. The extension $h_{B,2}$ maps the points $B$ and $M$ in a row to the points $B$ and $C$, and the arc $l$ to the arc $l'$. From $M\in l$ it follows that $C\in l'$. Since $AC=b$, it follows that the point $C$ lies on the circle $k(A,b)$ as well. So $C\in l'\cap k(A,b)$. The proven facts allow the construction. First of all, we draw the segment $AB=c$ and the arc $l$ with this chord and the central angle $\omega$, then the arc $l'=h_{B,2}(l)$, and finally the point $C$ as one of the intersections of the arc $l'$ with the circle $k(A,b)$. We will prove that the constructed triangle $ABC$ satisfies the conditions of the task. By construction, $AB=c$ immediately. From $C\in k(A,b)$ it follows that $AC=b$. Let $M=h^{-1}_{B,2}(C)$. Because $C\in l'=h_{B,2}(l)$, $M \in l$. Because $l$ is a line segment with this string and the obtuse angle $\omega$, $\angle AMB=\omega$. It remains to be proven that the point $M$ is the center of the line segment $BC$, which follows directly from the relation $M=h^{-1}_{B,2}(C)$. We will find the conditions for the number of solutions to the task. We have already mentioned that, due to the condition $\omega<90^0$, $b>c$ must be true. In the case of $b\leq c$, there is no solution. The number of solutions to the task is further dependent on the number of intersections of the line segment $l'$ with the circle $k(A,b)$. \kdokaz %________________________________________________________________________________ \poglavje{Classification of Similarity Transformations} \label{odd7PrezentTransPod} In section \ref{odd6KlasifIzo} we classified isometries, and here we will do a similar classification of similarity transformations. Also in this classification, the number of fixed points and the fact that the similarity transformation is direct or indirect will be important. In the previous two sections we have found that all isometries and central stretches represent similarity transformations. Also, their composition is a similarity transformation (statement \ref{TransPodGrupa}). Now we will prove that the converse is also true. \bizrek \label{TransPodKompHomIzo} Each similarity transformations $f$ with coefficient $k$ can be expressed as the product of one isometry and one homothety with an arbitrary centre: $$f=h_{S,k}\circ\mathcal{I}_1=\mathcal{I}_2\circ h_{S,k}.$$ \eizrek \textbf{\textit{Proof.}} Let $f$ be an arbitrary similarity transformation with coefficient $k$. If we denote the central stretch with arbitrary center $S$ and coefficient $\frac{1}{k}$ with $h_{S,\frac{1}{k}}$, then the composite $h_{S,\frac{1}{k}}\circ f$ represents a similarity transformation with coefficient $k\cdot \frac{1}{k}=1$ (statement \ref{TransPodGrupa}) or isometry. So $h_{S,\frac{1}{k}}\circ f=\mathcal{I}_1$, where $\mathcal{I}_1$ is some isometry. According to statement \ref{homotGrupa}, $f=h_{S,\frac{1}{k}}^{-1}\circ\mathcal{I}_1=h_{S,k}\circ\mathcal{I}_1$. Similarly, $f\circ h_{S,\frac{1}{k}}=\mathcal{I}_2$, where $\mathcal{I}_2$ is some isometry, so $f=\mathcal{I}_2\circ h_{S,k}$. \bizrek \label{TransPodOhranjaKote} Similarity transformations preserve the measure of angles, i.e. there map an angle to the congruent angle. \eizrek \textbf{\textit{Proof.}} The statement is a direct consequence of statements \ref{TransPodKompHomIzo} and \ref{homotOhranjaKote}. \kdokaz \bizrek \label{homotTransm} For each isometry $\mathcal{I}$ and each homothety $h_{S,k}$ it is: $$\mathcal{I}\circ h_{S,k}\circ \mathcal{I}^{-1}=h_{\mathcal{I}(S),k}$$ \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.1p.2.pic} \caption{} \label{sl.pod.7.1p.2.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.2.pic}) Let $\mathcal{I}(S)=S_1$. We denote $f=\mathcal{I}\circ h_{S,k}\circ \mathcal{I}^{-1}$. It is necessary to prove that $f=h_{S_1,k}$ or $f(X_1)=h_{S_1,k}(X_1)$ for an arbitrary point $X_1$ of this plane. Let us also denote $X=\mathcal{I}^{-1}(X_1)$, $X'=h_{S,k}(X)$ and $X'_1=\mathcal{I}(X_1)$. Then we have: \begin{eqnarray*} f(X_1)&=& \mathcal{I}\circ h_{S,k}\circ \mathcal{I}^{-1}(X_1)\\ &=& \mathcal{I}\circ h_{S,k}(X)\\ &=& \mathcal{I}(X')\\ &=& X'_1 \end{eqnarray*} Therefore $f(X_1)=X'_1$. From $X'=h_{S,k}(X)$ it follows that $\overrightarrow{SX'}=k\cdot\overrightarrow{SX}$. Because $\mathcal{I}$ is an isometry and $\mathcal{I}:\hspace*{1mm}S,X,X'\mapsto S_1,X_1,X'_1$, also $\overrightarrow{S_1X'_1}=k\cdot\overrightarrow{S_1X_1}$ or $h_{S_1,k}(X_1)=X'_1$. This means that for any point $X_1$ we have $f(X_1)=X'_1=h_{S_1,k}(X_1)$, so $f=h_{S_1,k}$. \kdokaz \bizrek \label{homotIzomKom} An Isometry $\mathcal{I}$ and a homothety $h_{S,k}$ commute if and only if the centre of this homothety is a fixed point of the isometry $\mathcal{I}$, i.e.: $$\mathcal{I}\circ h_{S,k}=h_{S,k}\circ\mathcal{I}\hspace*{1mm} \Leftrightarrow\hspace*{1mm}\mathcal{I}(S)=S.$$ \eizrek \textbf{\textit{Proof.}} By the previous statement (\ref{homotTransm}) we have: \begin{eqnarray*} \mathcal{I}\circ h_{S,k}=h_{S,k}\circ\mathcal{I} \hspace*{1mm} &\Leftrightarrow& \hspace*{1mm} \mathcal{I}\circ h_{S,k}\circ\mathcal{I}^{-1}=h_{S,k}\\ \hspace*{1mm} &\Leftrightarrow& \hspace*{1mm} h_{\mathcal{I}(S),k}=h_{S,k}\\ \hspace*{1mm} &\Leftrightarrow& \hspace*{1mm} \mathcal{I}(S)=S, \end{eqnarray*} which was to be proven. \kdokaz If we choose a rotation for the isometry from statement \ref{TransPodKompHomIzo} with the same centre as the central dilation, we get a very useful type of similarity transformation. The composition of a rotation and a central stretch with the same center is called a \index{rotational stretch} \pojem{rotational stretch} (Figure \ref{sl.pod.7.1p.1.pic}): $$\rho_{S,k,\omega}=h_{S,k}\circ \mathcal{R}_{S,\omega}$$ with the \index{center!of a rotational stretch}\pojem{center} $S$, the \index{coefficient!of a rotational stretch}\pojem{coefficient} $k$, and the \index{angle!of a rotational stretch}\pojem{angle} $\omega$. \begin{figure}[!htb] \centering \input{sl.pod.7.1p.1.pic} \caption{} \label{sl.pod.7.1p.1.pic} \end{figure} By \ref{homotIzomKom} we have: $$\rho_{S,k,\omega}=h_{S,k}\circ \mathcal{R}_{S,\omega}= \mathcal{R}_{S,\omega}\circ h_{S,k}.$$ Since a central stretch and a rotation are direct transformations (\ref{homotDirekt} and \ref{RotacDirekt}), a stretch rotation is also a direct transformation of similarity. It is clear that a central stretch can also be considered as a type of stretch rotation, if we assume that the identity is a rotation by the angle $0^0$: $$h_{S,k}=\rho_{S,k,0^0}.$$ Similarly, we can also see a rotation as a type of stretch rotation: $$\mathcal{R}_{S,\omega}=\rho_{S,1,\omega}.$$ \bizrek \label{rotacRaztKot} An arbitrary line and its image under a stretch rotation determine an oriented angle which is congruent to the angle of this stretch rotation: $$\rho_{S,k,\omega}(p)=p'\hspace*{1mm} \Rightarrow \hspace*{1mm} \angle p,p'=\omega.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.1p.3.pic} \caption{} \label{sl.pod.7.1p.3.pic} \end{figure} \textbf{\textit{Proof.}} Let $p'=\rho_{S,k,\omega}(p)$ be the image of the line $p$ under the stretch rotation $\rho_{S,k,\omega}=h_{S,k}\circ \mathcal{R}_{S,\omega}$ (Figure \ref{sl.pod.7.1p.3.pic}) and $p_1=\mathcal{R}_{S,\omega}(p)$. Then $h_{S,k}(p_1)=p'$. By \ref{rotacPremPremKot} we have $\measuredangle p,p_1=\omega$. By \ref{RaztPremica} we have $p_1\parallel p'$. Therefore $\measuredangle p,p'=\angle p,p_1=\omega$ (\ref{KotiTransverzala1}). \kdokaz \bizrek \label{rotacRaztKompSredZrc} The product of a half-turn and a stretch rotation with the same centre is a stretch rotation. Furthermore: $$\rho_{S,k,\omega}\circ \mathcal{S}_S= \mathcal{S}_S\circ\rho_{S,k,\omega}=\rho_{S,-k,\omega}.$$ \eizrek \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.3a.pic}) As we have already mentioned in section \ref{odd7SredRazteg}, $\mathcal{S}_S=h_{S,-1}$. By \ref{homotGrupa} we have: \begin{eqnarray*} \rho_{S,k,\omega}\circ \mathcal{S}_S= \mathcal{R}_{S,\omega}\circ h_{S,k}\circ h_{S,-1}= \mathcal{R}_{S,\omega}\circ h_{S,-k}=\rho_{S,-k,\omega}. \end{eqnarray*} Similarly we have: \begin{eqnarray*} \mathcal{S}_S\circ\rho_{S,k,\omega}= h_{S,-1}\circ h_{S,k}\circ \mathcal{R}_{S,\omega}= h_{S,-k}\circ\mathcal{R}_{S,\omega}=\rho_{S,-k,\omega}, \end{eqnarray*} which is what needed to be proven. \kdokaz \begin{figure}[!htb] \centering \input{sl.pod.7.1p.3a.pic} \caption{} \label{sl.pod.7.1p.3a.pic} \end{figure} A direct consequence is the following theorem. \bizrek \label{rotacRaztNegKoefk} For each stretch rotation is: $$\rho_{S,-k,\omega}=\rho_{S,k,180^0+\omega}.$$ \eizrek \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.3a.pic}) By the previous statement \ref{rotacRaztKompSredZrc} and the statement \ref{rotacKomp2rotac} it is: \begin{eqnarray*} \rho_{S,-k,\omega}= \mathcal{S}_S\circ\rho_{S,k,\omega}= \mathcal{S}_S\circ\mathcal{R}_{S,\omega}\circ h_{S,k} =\mathcal{R}_{S,180^0+\omega}\circ h_{S,k}= \rho_{S,k,180^0+\omega}, \end{eqnarray*} which had to be proven. \kdokaz If for the isometry from the statement \ref{TransPodKompHomIzo} we choose the reflection over a line that goes through the center of the central stretch, we get another type of similarity transformation. The composite of the central stretch $s$ and the central stretch $h_{S,k}$ with the center $S\in s$ is called the \index{osni razteg} \pojem{osni razteg} (Figure \ref{sl.pod.7.1p.1a.pic}): $$\sigma_{S,k,s}=h_{S,k}\circ \mathcal{S}_s;\hspace*{2mm} (S\in s)$$ with the \index{središče!osnega raztega}\pojem{center}, the \index{koeficient!rotacijskega raztega}\pojem{coefficient} $k$ and the \index{os!osnega raztega}\pojem{axis} $s$. \begin{figure}[!htb] \centering \input{sl.pod.7.1p.1a.pic} \caption{} \label{sl.pod.7.1p.1a.pic} \end{figure} By the statement \ref{homotIzomKom} it is: $$\sigma_{S,k,s}=h_{S,k}\circ \mathcal{S}_s= \mathcal{S}_s\circ h_{S,k}.$$ Since the central stretch is a direct and the rotation is an indirect transformation (the statement \ref{homotDirekt} and \ref{izozrIndIzo}), the axis stretch is an indirect transformation of similarity. \bizrek \label{transPod1FixTocLema} Let $f$ be a similarity transformation that is not an isometry. If $f$ maps each line to its parallel line, then $f$ is a homothety. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.1p.1bb.pic} \caption{} \label{sl.pod.7.1p.1bb.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.1bb.pic}) Let $k$ be the coefficient of similarity of the transformation $f$. Because $f$ is not an isometry, $k \neq 1$. By assumption, $f$ maps each line into a parallel line. We first prove that there are at least two lines that intersect and are not mapped into themselves. Let $X$, $Y$ and $Z$ be any three non-collinear points of the plane and let $p$, $q$ and $r$ be lines determined by the points $X$, $Y$ and $Z$: $p = XY$, $q = YZ$ and $r = XZ$. Let $p' = f (p)$, $q' = f (q)$ and $r' = f (r)$. By assumption, $p \parallel p'$, $q \parallel q'$ and $r \parallel r'$. We prove that at least one of the three lines is not mapped into itself. Assume the contrary, that $p = p'$, $q = q'$ and $r = r'$. But in this case, $f (X) = f (p \cap q) = f (p) \cap f (q) = p \cap q = X$ and similarly, for example, $f (Y) = Y$. The distance $XY$ would be mapped into itself in this case, which is not possible, since $k \neq 1$. Without loss of generality, let $p \neq p'$. In the same way, by using a triangle in which no altitude is parallel to the line $p$, we can prove that there is another line that intersects the line $p$ and is not mapped into itself by $f$. So there are lines $b$ and $c$ that intersect at point $A$ and for $b'=f(b)$ and $c'=f(c)$ it holds that $b\parallel b'$, $c\parallel c'$, $b\neq b'$ and $c\neq c'$. Let $B\in b$ and $C\in c$ be any points that are different from point $A$. We mark $A'=b\cap c$, $B'=f(B)$ and $C'=f(C)$. First, $f(A)=f(b\cap c)=f(b)\cap f(c)=b'\cap c'=A'$, $B'\in b'$ and $C'\in c'$. Because $b\parallel b'$, also $AB\parallel A'B'$. The distance $AB$ is mapped by transformation $f$ into the distance $A'B'$, so $A'B'=k\cdot AB$. The lines $AA'$ and $BB'$ are not parallel. Otherwise the quadrilateral $ABB'A'$ would be a parallelogram or $AB\cong A'B'$ (by statement \ref{paralelogram}), which is not possible because $k\neq 1$. We mark with $S$ the intersection of the lines $AA'$ and $BB'$. Because $b\parallel b'$, by Tales' statement: $$\frac{SA'}{SA}=\frac{SB'}{SB}=\frac{A'B'}{AB}=k.$$ This means that the central stretch $h_{S,k_1}$ with center $S$ and coefficient $k_1=k$ (or $k_1=-k$) maps points $A$ and $B$ into points $A'$ and $B'$. Let $\widehat{C'}=h_{S,k_1}(C)$. Because by assumption $C'=f(C)$ or $A'C'=k\cdot AC$, by statement \ref{RaztPremica} $\widehat{C'}=C'$ or $h_{S,k_1}(C)=C'$. The mapping $$g=f^{-1}\circ h_{S,k_1}$$ is by statement \ref{TransPodGrupa} a similarity transformation with similarity coefficient $\frac{1}{k}\cdot |k_1|=\frac{1}{k}\cdot k=1$, so it represents an isometry. But $g(A)=f^{-1}\circ h_{S,k_1}(A)=f^{-1}(A')=A$, or $A$ is a fixed point of isometry $g$. In a similar way we prove that also $B$ and $C$ are fixed points of isometry $g$, which means that $g=\mathcal{E}$ is the identity (by statement \ref{IizrekABC2}). So $f^{-1}\circ h_{S,k_1}=g=\mathcal{E}$ or $f=h_{S,k_1}$. \kdokaz Now we are ready for the next important statement. \bizrek \label{transPod1FixToc} Each similarity transformation other than isometry has exactly one fixed point. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.1p.1c.pic} \caption{} \label{sl.pod.7.1p.1c.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.1c.pic}) Let $f$ be a similarity transformation with coefficient $k$. By assumption, $k\neq 1$. Similarly to the proof of the previous izrek \ref{transPod1FixTocLema}, we quickly convince ourselves that $f$ cannot have two fixed points, because $k=1$ would be. We will prove that $f$ has a fixed point. According to the previous statement (\ref{transPod1FixTocLema}), we can assume that there exists at least one line $p$ that is not mapped into its parallel. Indeed, if we assume that each plane line is mapped into its parallel, then according to the statement \ref{transPod1FixTocLema} the mapping $f$ is a central stretch with a fixed point (its center). Let $p'=f(p)$ and $ p'\nparallel p$ ($\neg p'\parallel p$). We mark the intersection of lines $p$ and $p'$ with $A$. Let $A'=f(A)$. If $A'=A$, then $A$ is a fixed point of transformation $f$ and the proof is finished. So let $A'\neq A$. From $A\in p$ it follows that $A'\in p'$. We mark the parallel to the line $p$ through the point $A'$ with $q$. Because $A'\neq A$, $q\neq p$. Let $q'=f(q)$. Because $p\parallel q$ and $f:\hspace*{1mm}p,q\rightarrow p',q'$, also $p'\parallel q'$ (if $p'$ and $q'$ would intersect in a point $T$, then also lines $p$ and $q$ would intersect in the point $f^{-1}(T)$). In the same way from $q\neq p$ it follows that $q'\neq p'$. This means that lines $p$, $q$, $p'$ and $q'$ determine the parallelogram $AA'BC$, where $B=q\cap q'$ and $C=p\cap q'$. Let $B'=f(B)$. From $B\in q$ it follows that $B'\in q'$. If $B'=B$, then $B$ is a fixed point and the proof is finished. So we can further assume that $B'\neq B$. From $p'\parallel q'$ it follows that $AA'\parallel BB'$. In the case $AB\parallel A'B'$ the quadrilateral $AA'B'B$ would be a parallelogram or $AB\cong A'B'$, which is not possible, because $f:\hspace*{1mm}A,B\rightarrow A',B'$ and $k\neq 1$. So lines $AB$ and $A'B'$ intersect in a point $S$. We mark $S'=f(S)$. We will prove that $S$ is a fixed point or $S'=S$. According to Tales' statement: \begin{eqnarray} \label{eqnTransfPod1Ft1} \frac{AS}{SB}=\frac{A'S}{SB'} \end{eqnarray} Because $f:\hspace*{1mm}A,B,S\rightarrow A',B',S'$, also (statement \ref{TransPodOhranjajoRazm}): \begin{eqnarray} \label{eqnTransfPod1Ft2} \frac{AS}{SB}=\frac{A'S'}{S'B'} \end{eqnarray} Also from $S\in AB$ it follows that $S'\in A'B'$. If we connect the last two equalities \ref{eqnTransfPod1Ft1} and \ref{eqnTransfPod1Ft2}, it is: \begin{eqnarray} \label{eqnTransfPod1Ft3} \frac{A'S}{SB'}=\frac{A'S'}{S'B'}, \end{eqnarray} where $S$ and $S'$ are points of the line $A'B'$. But similarity transformations preserve the relation $\mathcal{B}$ (statement \ref{TransPodB}), or: \begin{eqnarray*} \mathcal{B}(A,S,B)\hspace*{1mm} &\Leftrightarrow& \hspace*{1mm} \mathcal{B}(A',S',B');\\ \mathcal{B}(S,A,B)\hspace*{1mm} &\Leftrightarrow& \hspace*{1mm} \mathcal{B}(S',A',B');\\ \mathcal{B}(A,B,S)\hspace*{1mm} &\Leftrightarrow& \hspace*{1mm} \mathcal{B}(A',B',S'). \end{eqnarray*} From this and from the relation \ref{eqnTransfPod1Ft3} it follows: \begin{eqnarray*} \frac{\overrightarrow{A'S}}{\overrightarrow{SB'}}= \frac{\overrightarrow{A'S'}}{\overrightarrow{S'B'}}, \end{eqnarray*} so according to the statement \ref{izrekEnaDelitevDaljiceVekt} $S'=S$ or $S$ is a fixed point of similarity transformation $f$. \kdokaz Now we can make the predicted classification of similarity transformations. \bizrek \label{transPodKlasif} The only similarity transformations of the plane are: \begin{itemize} \item isometries, \item homotheties, \item stretch rotations, \item stretch reflections. \end{itemize} \eizrek \textbf{\textit{Proof.}} Let $f$ be an arbitrary similarity transformation with coefficient $k$. If $k=1$, $f$ is an isometry. Assume that $k\neq 1$ or $f$ is not an isometry. According to Theorem \ref{transPod1FixToc}, $f$ has exactly one fixed point - we denote it with $S$. So $f(S)=S$. According to Theorem \ref{TransPodKompHomIzo}, we can represent the transformation of similarity $f$ as the composition of a central stretch with an arbitrary center (we choose the point $S$ as the center) and a coefficient $k$ and one isometry $\mathcal{I}$: $$f=\mathcal{I}\circ h_{S,k}.$$ Because $S$ is a fixed point of the transformation of similarity $f$ and the central stretch $h_{S,k}$, it is: $$S=f(S)=\mathcal{I}\circ h_{S,k}(S)=\mathcal{I}(S).$$ So $\mathcal{I}(S)=S$ or $\mathcal{I}$ is an isometry with a fixed point $S$. According to Theorem \ref{Chaslesov}, $\mathcal{I}$ can be: identity, rotation with center $S$ or reflection over a line that goes through the point $S$: \begin{eqnarray*} \mathcal{I}=\left\{ \begin{array}{l} \mathcal{E}, \\ \mathcal{R}_{S,\omega}, \\ \mathcal{S}_s (S\in s) \end{array} \right. \end{eqnarray*} Therefore: \begin{eqnarray*} f=\mathcal{I}\circ h_{S,k}=\left\{ \begin{array}{l} h_{S,k}, \\ \rho_{S,k,\omega}, \\ \sigma_{S,k,s} \end{array} \right. \end{eqnarray*} which had to be proven. \kdokaz A direct consequence is the following theorem. \bizrek \label{transPodKlasifDirInd} The only direct similarity transformations of the plane other than isometries are: \begin{itemize} \item homotheties, \item stretch rotations. \end{itemize} The only opposite similarity transformations of the plane other than isometries are: \begin{itemize} \item stretch reflections. \end{itemize} \eizrek It is very useful to know the property of the composite of two stretch rotations. \bizrek \label{RotRazKomoz} The product of two stretch rotations $\rho_{S_1,k_1,\omega_1}$ and $\rho_{S_2,k_2,\omega_2}$ is a direct isometry, a homothety or a stretch rotation. Furthermore (Figure \ref{sl.pod.7.1p.5a.pic}): \begin{eqnarray*} \rho_{S_2,k_2,\omega_2}\circ \rho_{S_1,k_1,\omega_1}= \left\{ \begin{array}{ll} \mathcal{R}_{S,\omega}, & k_1k_2=1, \hspace*{2mm} \omega=\omega_1+\omega_2\neq n\cdot 180^0; \\ \mathcal{T}_{\overrightarrow{v}}, & k_1k_2=1, \hspace*{2mm} \omega=\omega_1+\omega_2=n\cdot 180^0; \\ h_{S,k}, & k=k_1k_2\neq 1, \hspace*{2mm} \omega=\omega_1+\omega_2=n\cdot 180^0; \\ \rho_{S,k,\omega}, & k=k_1k_2\neq 1, \hspace*{2mm} \omega=\omega_1+\omega_2\neq n\cdot 180^0. \end{array} \right. \end{eqnarray*} for $n\in \mathbb{Z}$. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.1p.5.pic} \caption{} \label{sl.pod.7.1p.5.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.1p.5.pic}) Without loss of generality, assume that $k_1>0$ and $k_2>0$. For example, if $k_1<0$, we can by \ref{rotacRaztNegKoefk} write $\rho_{S,k_1,\omega}=\rho_{S,-k_1,180^0+\omega}$, where $-k_1>0$. Let $f=\rho_{S_2,k_2,\omega_2}\circ \rho_{S_1,k_1,\omega_1}$. By the formulas \ref{RaztTransPod} and \ref{TransPodGrupa}, $f$ is a similarity transformation with coefficient $k=k_1\cdot k_2$. It is a direct transformation as the composite of two direct transformations. By the formula \ref{transPodKlasifDirInd}, $f$ can be an isometry, central dilation or rotational dilation. \begin{figure}[!htb] \centering \input{sl.pod.7.1p.5.pic} \caption{} \label{sl.pod.7.1p.5.pic} \end{figure} Let $p$ be an arbitrary line and $f(p)=p'$. Calculate the measure of the oriented angle $\measuredangle p,p'$. Let $\rho_{S_1,k_1,\omega_1}(p)=p_1$ and consequently $\rho_{S_2,k_2,\omega_2}(p_1)=p'$. By the formula \ref{rotacRaztKot}, $\angle p,p_1=\omega_1$ and $\angle p_1,p'=\omega_2$. If $\omega=\omega_1+\omega_2\neq n\cdot 180^0$ for every $n\in \mathbb{Z}$, then $\angle p,p'=\omega_1+\omega_2$ (formula \ref{zunanjiNotrNotr}). If $\omega=\omega_1+\omega_2=n\cdot 180^0$ for some $n\in \mathbb{Z}$, then $p\parallel p'$ (formula \ref{KotiTransverzala}). If $k_1\cdot k_2=1$, the composite $f$ is a similarity transformation with coefficient $k=1$ and it is a direct isometry $f=\mathcal{I}\in \mathfrak{I}^+$. It can be a translation or a rotation (in the special case of identity) (formula \ref{Chaslesov+}). If $\omega=\omega_1+\omega_2\neq n\cdot 180^0$ or $\angle p,p'=\omega_1+\omega_2$, it is a rotation by the angle $\omega=\omega_1+\omega_2$ (formula \ref{rotacPremPremKot}). If $\omega=\omega_1+\omega_2=n\cdot 180^0$ or $p\parallel p'$, it is a translation (or identity). Let us now assume that $k_1\cdot k_2\neq 1$. As we have already mentioned, in this case $f$ can be a central extension or a rotational extension with a coefficient $k=k_1\cdot k_2$. If $\omega=\omega_1+\omega_2\neq n\cdot 180^0$ or $\angle p,p'=\omega_1+\omega_2$, it is a rotational extension for an angle $\omega=\omega_1+\omega_2$ (statement \ref{rotacRaztKot}). If $\omega=\omega_1+\omega_2=n\cdot 180^0$ or $p\parallel p'$, it is a central extension according to statement \ref{RaztPremica} (or a central reflection if $k_1\cdot k_2=-1$). \kdokaz A direct consequence is the following theorem. \bizrek The product of two homotheties is a direct isometry or a homothety. Furthermore: \begin{eqnarray*} h_{S_2,k_2}\circ h_{S_1,k_1}= \left\{ \begin{array}{ll} \mathcal{T}_{\overrightarrow{v}}, & k_1k_2=1; \\ h_{S,k}, & k=k_1k_2\neq \pm 1. \end{array} \right. \end{eqnarray*} \eizrek \textbf{\textit{Proof.}} The statement follows directly from the previous statement, if we write $h_{S_1,k_1}=\rho_{S_1,k_1,0^0}$ and $h_{S_2,k_2}=\rho_{S_2,k_2,0^0}$. \kdokaz \bnaloga\footnote{17. IMO Bulgaria - 1975, Problem 3.} On the sides of an arbitrary triangle $ABC$, triangles $ABR$, $BCP$, $CAQ$ are constructed externally with $\angle CBP\cong\angle CAQ=45^0$, $\angle BCP\cong\angle ACQ=35^0$, $\angle ABR\cong\angle BAR=15^0$. Prove that $\angle QRP=90^0$ and $QR\cong RP$. \enaloga \begin{figure}[!htb] \centering \input{sl.pod.7.1.IMO2.pic} \caption{} \label{sl.pod.7.1.IMO2.pic} \end{figure} \textbf{\textit{Solution.}} Let $S$ be the third vertex of the isosceles triangle $BAS$, which is drawn over the side $BA$ of the triangle $ABC$ (Figure \ref{sl.pod.7.1.IMO2.pic}). Because $\angle RBS\cong\angle SAR=45^0$ and $\angle BSR\cong\angle ASR=30^0$, the triangles $BPC$, $AQC$, $BRS$ and $ARS$ are similar to each other. Let $f_1$ and $f_2$ be the stretching rotations: \begin{eqnarray*} &&f_1=\rho_{B,k,45^0}=h_{B,k}\circ \mathcal{R}_{B,45^0}\\ &&f_2=\rho_{A,\frac{1}{k},45^0}=h_{A,\frac{1}{k}}\circ \mathcal{R}_{A,45^0}, \end{eqnarray*} where: $$k=\frac{|BC|}{|BP|}=\frac{|AC|}{|AQ|}=\frac{|BS|}{|BR|} =\frac{|AS|}{|AR|}.$$ Let $f=f_2\circ f_1$. By the statement \ref{RotRazKomoz} it is: $$f=f_2\circ f_1=\rho_{A,\frac{1}{k},45^0}\circ \rho_{B,k,45^0}=\rho_{T,1,90^0}=\mathcal{R}_{T,90^0}.$$ Then it is valid: \begin{eqnarray*} &&\mathcal{R}_{T,90^0}(P)=f(P)=f_2\circ f_1(P)=f_2(C)=Q\\ &&\mathcal{R}_{T,90^0}(R)=f(R)=f_2\circ f_1(R)=f_2(S)=R. \end{eqnarray*} From the previous (second) relation we see that $\mathcal{R}_{T,90^0}(R)=R$, which means that $R=T$ (statement \ref{RotacFiksT}) or $\mathcal{R}_{T,90^0}=\mathcal{R}_{R,90^0}$. From this and from the first relation now follows: $$\mathcal{R}_{R,90^0}(P)=Q,$$ which means that $\angle QRP=90^0$ and $QR\cong RP$. \kdokaz %________________________________________________________________________________ \poglavje{Similar Figures. Similarity of Triangles} \label{odd7PodobTrik} As we have already announced in the introduction to this chapter, similarity transformations allow us to define the concept of similarity of figures. We say that the figure $\Phi$ \index{figure!similar}\pojem{similar} to the figure $\Phi'$ from the same plane (the notation $\Phi\sim \Phi'$), if there exists a similarity transformation $f$ of this plane, which maps the figure $\Phi$ to the figure $\Phi'$ or $f(\Phi)=\Phi'$ (Figure \ref{sl.pod.7.2.1.pic}). The coefficient of similarity transformation $f$ is at the same time \index{coefficient!of similarity of figures}\pojem{the coefficient of similarity of figures} $\Phi$ and $\Phi'$. \begin{figure}[!htb] \centering \input{sl.pod.7.2.1.pic} \caption{} \label{sl.pod.7.2.1.pic} \end{figure} It is clear that for $k=1$ we get the congruence of figures as a special case of similarity. Congruent figures are therefore also similar, but the converse is not true, that is: $$\Phi\cong \Phi' \hspace*{1mm}\Rightarrow \hspace*{1mm}\Phi\sim \Phi'.$$ We prove the most important property of the similarity of figures. \bizrek The similarity of figures is an equivalence relation. \eizrek \textbf{\textit{Proof.}} It is necessary (and sufficient) to prove that the similarity of figures is reflexive, symmetric and transitive. (\textit{R}) For every figure $\Phi$ it holds that $\Phi\sim \Phi$, because the identity $\mathcal{E}$, which maps the figure $\Phi$ to itself, is a similarity transformation with the coefficient $k=1$ (Figure \ref{sl.pod.7.2.1r.pic}). \begin{figure}[!htb] \centering \input{sl.pod.7.2.1r.pic} \caption{} \label{sl.pod.7.2.1r.pic} \end{figure} (\textit{S}) We assume that for the figures $\Phi_1$ and $\Phi_2$ it holds that $\Phi_1\sim \Phi_2$ (Figure \ref{sl.pod.7.2.1s.pic}). By definition there exists a similarity transformation $f$, which maps the figure $\Phi_1$ to the figure $\Phi_2$, or $f:\hspace*{1mm}\Phi_1\rightarrow \Phi_2$. By \ref{TransPodGrupa} the inverse mapping $f^{-1}$, for which $f^{-1}:\hspace*{1mm}\Phi_2\rightarrow \Phi_1$, is also a similarity transformation. Therefore it holds that $\Phi_2\sim \Phi_1$. \begin{figure}[!htb] \centering \input{sl.pod.7.2.1s.pic} \caption{} \label{sl.pod.7.2.1s.pic} \end{figure} (\textit{T}) Let's assume that for figures $\Phi_1$, $\Phi_2$ and $\Phi_3$ it holds that $\Phi_1\sim \Phi_2$ and $\Phi_2\sim \Phi_3$ (Figure \ref{sl.pod.7.2.1t.pic}). Prove that then also $\Phi_1\sim \Phi_3$ holds. By definition, there exist similarity transformations $f$ and $g$, such that $f:\hspace*{1mm}\Phi_1\rightarrow \Phi_2$ and $g:\hspace*{1mm}\Phi_2\rightarrow \Phi_3$ hold. But by \ref{TransPodGrupa} the composition of mapping $g\circ f:\hspace*{1mm}\Phi_1\rightarrow \Phi_3$ is also a similarity transformation, therefore $\Phi_1\sim \Phi_3$ holds. \kdokaz \begin{figure}[!htb] \centering \input{sl.pod.7.2.1t.pic} \caption{} \label{sl.pod.7.2.1t.pic} \end{figure} Because the relation of similarity between figures is symmetric, in the case of $\Phi\sim \Phi'$ we will say that figures $\Phi$ and $\Phi'$ are similar. In the following we will specifically consider the similarity between triangles. Let's assume that triangles $ABC$ and $A'B'C'$ are similar, i.e. $\triangle ABC\sim\triangle A'B'C'$. This means that there exists a similarity transformation that maps triangle $ABC$ to triangle $A'B'C'$. In the case of triangles (and also polygons) we will additionally require that the vertices are mapped in order by the similarity transformation, i.e. $f:\hspace*{1mm}A,B,C\mapsto A',B',C'$. Since a similarity transformation maps lines to lines and angles to angles (\ref{TransPodKol}), sides $AB$, $BC$ and $CA$ of triangle $ABC$ are mapped to sides $A'B'$, $B'C'$ and $C'A'$, and the internal angles $BAC$, $ABC$ and $ACB$ of triangle $ABC$ are mapped to the internal angles $B'A'C'$, $A'B'C'$ and $A'C'B'$ of triangle $A'B'C'$. For pairs of elements in this mapping we will say that they are \pojem{corresponding} or \pojem{congruent}. According to the statement \ref{TransPodOhranjajoRazm} similarity transformations preserve the ratio of distances, which means that the corresponding sides are proportional, that is\footnote{By using the similarity of the isosceles right triangles with the help of the appropriate ratio, \index{Tales}\textit{Tales from Miletus} (7th-6th century BC) calculated the height of the pyramid of Cheops.} (Figure \ref{sl.pod.7.2.2.pic}): \begin{eqnarray} \label{eqnPodTrik1} \triangle ABC\sim\triangle A'B'C'\hspace*{1mm} \Rightarrow \hspace*{1mm} \frac{A'B'}{AB}=\frac{A'C'}{AC}=\frac{B'C'}{BC}=k, \end{eqnarray} where $k$ is the coefficient of similarity. \begin{figure}[!htb] \centering \input{sl.pod.7.2.2.pic} \caption{} \label{sl.pod.7.2.2.pic} \end{figure} According to the statement \ref{TransPodOhranjaKote} similarity transformations preserve angles, that is (Figure \ref{sl.pod.7.2.2.pic}): \begin{eqnarray} \label{eqnPodTrik2} \triangle ABC\sim\triangle A'B'C'\hspace*{1mm} \Rightarrow \hspace*{1mm} \left\{ \begin{array}{l} \angle B'A'C'\cong\angle BAC; \\ \angle A'B'C'\cong\angle ABC; \\ \angle A'C'B'\cong\angle ACB. \end{array} \right. \end{eqnarray} From the fact that the similarity transformation preserves angles (statement \ref{TransPodOhranjaKote}) and the ratio of distances (statement \ref{TransPodOhranjajoRazm}), it also follows that the heights, centroids, ... of one triangle are mapped into the heights, centroids, ... of the other triangle, while the ratio of the corresponding elements is preserved. In the case of similarity of triangles, we have the same problem as with congruence - it is not always easy to prove the similarity of triangles directly by definition. So, analogously to congruence, we also get the \index{statement!about similarity of triangles}\pojem{statements about similarity of triangles}. We have already seen that from the similarity of two triangles $ABC$ and $A'B'C'$ we get the ratio of the corresponding sides \ref{eqnPodTrik1} or the congruence of the corresponding angles \ref{eqnPodTrik2}. The following statements speak about what conditions are sufficient for the triangles to be similar. \bizrek \label{PodTrikSKS} Triangles $ABC$ and $A'B'C'$ are similar if two pairs of the sides of the triangles are proportional and the pair of included angles between the sides is congruent, i.e.: \begin{eqnarray*} \left. \begin{array}{l} \angle B'A'C'\cong \angle BAC\\ \frac{A'B'}{AB}=\frac{A'C'}{AC} \end{array} \right\}\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC\sim\triangle A'B'C' \end{eqnarray*} \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.2.3.pic} \caption{} \label{sl.pod.7.2.3.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.2.3.pic}) Let it be: $$k=\frac{A'B'}{AB}=\frac{A'C'}{AC}.$$ Let $B''$ and $C''$ be such points on the line segments $A'B'$ and $A'C'$, that $A'B''\cong AB$ and $A'C''\cong AC$. Triangles $A'B''C''$ and $ABC$ are similar (\textit{SAS} \ref{SKS}), therefore there exists an isometry $\mathcal{I}$, which maps triangle $ABC$ to triangle $A'B''C''$. Because $$\frac{A'B'}{A'B''}=\frac{A'B'}{AB}=k\hspace*{1mm}\textrm{ and }\hspace*{1mm}\frac{A'C'}{A'C''}=\frac{A'C'}{AC}=k,$$ or (because points $B''$ and $C''$ lie on the line segments $A'B'$ and $A'C'$) $\overrightarrow{A'B'}=k\cdot \overrightarrow{A'B''}$ and $\overrightarrow{A'C'}=k\cdot \overrightarrow{A'C''}$, central stretch $h_{A',k}$ maps points $A'$, $B''$ and $C''$ to points $A'$, $B'$ and $C'$ or triangle $A'B''C''$ to triangle $A'B'C'$. Therefore the composition $f=h_{A',k}\circ \mathcal{I}$, which is a similarity transformation, maps triangle $ABC$ to triangle $A'B'C'$, which means that $\triangle ABC\sim\triangle A'B'C'$. \kdokaz \bizrek \label{PodTrikKKK} Triangles $ABC$ and $A'B'C'$ are similar if two pairs of the angles of the triangles are congruent, i.e.: \begin{eqnarray*} \left. \begin{array}{l} \angle B'A'C'\cong\angle BAC\\ \angle A'B'C'\cong\angle ABC \end{array} \right\}\hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC\sim\triangle A'B'C' \end{eqnarray*} \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.2.4.pic} \caption{} \label{sl.pod.7.2.4.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.2.4.pic}) Let's mark: $$k=\frac{A'B'}{AB}.$$ Let $B''$ and $C''$ be such points on the line segments $A'B'$ and $A'C'$, that $A'B''\cong AB$ and $A'C''\cong AC$ holds. Then $\frac{A'B'}{A'B''}=k$ as well. Triangles $A'B''C''$ and $ABC$ are similar (by the \textit{SAS} theorem \ref{SKS}), so there exists an isometry $\mathcal{I}$, that maps triangle $ABC$ to triangle $A'B''C''$. From this similarity follows that $\angle A'B''C''\cong\angle ABC$ as well. Because $\angle A'B'C'\cong\angle ABC$ by assumption, it is also true that $\angle A'B''C''\cong\angle A'B'C'$. By \ref{KotiTransverzala} theorem $B''C''\parallel B'C'$. By Tales' \ref{TalesovIzrek} theorem $$\frac{A'C'}{A'C''}=\frac{A'B'}{A'B''}=k.$$ Because $B''$ and $C''$ are points on the line segments $A'B'$ and $A'C'$, it is also true that $\overrightarrow{A'B'}=k\cdot \overrightarrow{A'B''}$ and $\overrightarrow{A'C'}=k\cdot \overrightarrow{A'C''}$. This means that the central stretch $h_{A',k}$ maps points $A'$, $B''$ and $C''$ to points $A'$, $B'$ and $C'$ or triangle $A'B''C''$ to triangle $A'B'C'$. The composition $f=h_{A',k}\circ \mathcal{I}$, which is a similarity transformation, therefore maps triangle $ABC$ to triangle $A'B'C'$, so $\triangle ABC\sim\triangle A'B'C'$. \kdokaz \bizrek \label{PodTrikSSS} Triangles $ABC$ and $A'B'C'$ are similar if three pairs of the sides of the triangles are proportional, i.e. \begin{eqnarray*} \frac{A'B'}{AB}=\frac{A'C'}{AC}=\frac{B'C'}{BC} \hspace*{1mm}\hspace*{1mm}\Rightarrow\triangle ABC\sim\triangle A'B'C' \end{eqnarray*} \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.2.5.pic} \caption{} \label{sl.pod.7.2.5.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.2.5.pic}) Let's mark: $$k=\frac{A'B'}{AB}=\frac{A'C'}{AC}=\frac{B'C'}{BC}.$$ Let $B''$ and $C''$ be such points of the line segments $A'B'$ and $A'C'$, that $A'B''\cong AB$ and $A'C''\cong AC$ hold. Then it is also true that $$\frac{A'B'}{A'B''}=\frac{A'C'}{A'C''}=k.$$ By the converse of Tales' theorem \ref{TalesovIzrekObr} it follows that $B''C''\parallel B'C'$ and $$\frac{B'C'}{B''C''}=\frac{A'B'}{A'B''}.$$ Therefore $$\frac{B'C'}{B''C''}=\frac{A'B'}{A'B''}=\frac{A'C'}{A'C''}=k=\frac{B'C'}{BC},$$ which means that $B''C''\cong BC$ holds. From this it follows that the triangles $A'B''C''$ and $ABC$ are congruent (\textit{SSS} theorem \ref{SSS}), so there exists an isometry $\mathcal{I}$, which maps the triangle $ABC$ to the triangle $A'B''C''$. Similarly (as in the proof of the previous two theorems) the central stretch $h_{A',k}$ maps the points $A'$, $B''$ and $C''$ to the points $A'$, $B'$ and $C'$ or the triangle $A'B''C''$ to the triangle $A'B'C'$. In the end, the composition $f=h_{A',k}\circ \mathcal{I}$, which is a similarity transformation, maps the triangle $ABC$ to the triangle $A'B'C'$, so $\triangle ABC\sim\triangle A'B'C'$. \kdokaz The next theorem will be given without a proof (Figure \ref{sl.pod.7.2.6.pic}). \bizrek \label{PodTrikSSK} Triangles $ABC$ and $A'B'C'$ are similar if two pairs of the sides of the triangles are proportional and the pair of the angles opposite to the longer sides is congruent. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.2.6.pic} \caption{} \label{sl.pod.7.2.6.pic} \end{figure} We will continue by using theorems about similarity of triangles. \bzgled Let $k$ be the circumcircle of a triangle $ABC$ and $B'$ and $C'$ the foots of the perpendiculars from the vertices $B$ and $C$ on the tangent of the circle $k$ in the vertex $A$. Prove that the altitude $AD$ of this triangle is the geometric mean of the line segments $BB'$ and $CC'$, i.e.: $$|AD|=\sqrt{|BB'|\cdot |CC'|}.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.2.7.pic} \caption{} \label{sl.pod.7.2.7.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.2.7.pic}) Since $BAB'$ is congruent to the angle subtended by the chord $AB$ at the tangent of the circle $k$ in the vertex $A$ or $\angle BAB'\cong\angle ACD$ (theorem \ref{ObodKotTang}). Since also $\angle AB'B\cong\angle CDA=90^0$, we have $\triangle AB'B\sim\triangle CDA$ (theorem\ref{PodTrikKKK}). From this it follows: \begin{eqnarray} \label{eqnPodTrikZgl1a} AD:BB'=AC:BA. \end{eqnarray} In the same way we prove $\triangle AC'C\sim\triangle BDA$ or \begin{eqnarray} \label{eqnPodTrikZgl1b} CC':AD=AC:BA. \end{eqnarray} From relations \ref{eqnPodTrikZgl1a} and \ref{eqnPodTrikZgl1b} it follows $AD:BB'=CC':AD$ or $|AD|=\sqrt{|BB'|\cdot |CC'|}$. \kdokaz \bzgled \label{izrekSinusni} Let $v_a$ be the length of the altitude $AD$, $k(O,R)$ the circumcircle of a triangle $ABC$ and $b=|AC|$ and $c=|AB|$. Prove that\footnote{The theorem from this example in trigonometry represents the so-called \index{izrek!Sinusni}\textit{sinusoidal theorem}, which was proved by the Arab mathematician \index{al-Biruni, A. R.}\textit{A. R. al-Biruni} (973--1048). Namely, if we insert $\frac{v_a}{c}=\sin \beta$ into the given equality, we get $\frac{b}{\sin\beta}=2R$ and similarly $\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=2R$.}: $$bc=2R\cdot v_a.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.2.8a.pic} \caption{} \label{sl.pod.7.2.8a.pic} \end{figure} \textbf{\textit{Proof.}} We also mark $A_0=\mathcal{S}_O(A)$ (Figure \ref{sl.pod.7.2.8a.pic}). The distance $AA_0$ is the diameter of the circle $k$, so according to Theorem \ref{TalesovIzrKroz2} $\angle ACA_0=90^0=\angle ADB$. The central angles $CBA$ and $CA_0A$ over the arc $AC$ of the circle $k$ are congruent (Theorem \ref{ObodObodKot}), i.e. $\angle DBA=\angle CBA\cong\angle CA_0A$. According to Theorem \ref{PodTrikKKK}, $\triangle ABD\sim\triangle AA_0C$, so: $$\frac{AB}{AA_0}=\frac{AD}{AC},$$ i.e. $bc=2R\cdot v_a$. \kdokaz \bzgled Suppose that the bisector of the interior angle $BA$C of a triangle $ABC$ intersects its side $BC$ at the point $E$ and the circumcircle of this triangle at the point $N$ ($N\neq A$). Let's denote $b=|AC|$, $c=|AB|$ and $l_a=|AE|$. Prove that: $$|AN|=\frac{bc}{l_a}.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.2.8.pic} \caption{} \label{sl.pod.7.2.8.pic} \end{figure} Suppose that the bisector of the interior angle $BA$C of a triangle $ABC$ intersects its side $BC$ at the point $E$ and the circumcircle of this triangle at the point $N$ ($N\neq A$). Let's denote $b=|AC|$, $c=|AB|$ and $l_a=|AE|$. Prove that: $$|AN|=\frac{bc}{l_a}.$$ \begin{figure}[!htb] \centering \input{sl.pod.7.2.8.pic} \caption{} \label{sl.pod.7.2.8.pic} \end{figure} \textbf{\textit{Proof.}} Let us also mark with $AD$ the altitude ($v_a=|AD|$), with $k(O,R)$ the circumscribed circle of the triangle $ABC$ and $M=\mathcal{S}_O(N)$ (Figure \ref{sl.pod.7.2.8.pic}). According to Theorem \ref{TockaN}, the point $N$ lies on the side $BC$'s simetrical line, which means that $NM$ is the diameter of the circle $k$ and $MN\perp BC$. According to Theorem \ref{TalesovIzrKroz2}, $\angle NAM=90^0=\angle EDA$. From $AD,MN\perp BC$ it follows that $AD\parallel MN$, therefore according to Theorem \ref{KotiTransverzala} $\angle DAE\cong\angle MNA$. The triangle $ADE$ and $NAM$ are therefore similar (Theorem \ref{PodTrikKKK}), so: $$\frac{AD}{AN}=\frac{AE}{NM}.$$ If we use the previous Theorem \ref{izrekSinusni}, we get: $$|AN|=\frac{2R\cdot v_a}{l_a}=\frac{bc}{l_a},$$ which had to be proven. \kdokaz %________________________________________________________________________________ \poglavje{The Theorems of Ceva and Menelaus} \label{odd7MenelCeva} In this section we will discuss the so-called \index{theorem!theorem of the double}\emph{theorem of the double}. \btheorem \label{izrekCeva}\index{theorem!Ceva's} (Ceva\footnote{\index{Ceva, G.} \textit{G. Ceva} (1648--1734), Italian mathematician, who proved this theorem in 1678.}) Let $P$, $Q$ and $R$ be points lying on the lines containing the sides $BC$, $CA$ and $AB$ of a triangle $ABC$. Then the lines $AP$, $BQ$ and $CR$ belong to the same family of lines if and only if: \begin{eqnarray}\label{formulaCeva} \frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot \frac{\overrightarrow{AR}}{\overrightarrow{RB}}=1. \end{eqnarray} \etheorem \begin{figure}[!htb] \centering \input{sl.pod.7.5.1.pic} \caption{} \label{sl.pod.7.5.1.pic} \end{figure} %slikaNova1-3-1 %\includegraphics[width=100mm]{slikaNova1-3-1.pdf} \textbf{\textit{Proof.}} ($\Rightarrow$) We assume that the lines $AP$, $BQ$ and $CR$ belong to the same elliptic conic, that is, they intersect in a point $S$ (we leave it to the reader to prove the statement in the case when the lines are parallel). We denote the line that contains point $A$ and is parallel to line $BC$ with $p$, and its intersections with lines $BQ$ and $CR$ with $K$ and $L$ (Figure \ref{sl.pod.7.5.1.pic}). Now from Tales' theorem \ref{TalesovIzrek} it follows that $\frac{\overrightarrow{BP}}{\overrightarrow{KA}}= \frac{\overrightarrow{SP}}{\overrightarrow{SA}}= \frac{\overrightarrow{PC}}{\overrightarrow{AL}}.$ From this we obtain: \begin{eqnarray*} \frac{\overrightarrow{BP}}{\overrightarrow{PC}}= \frac{\overrightarrow{KA}}{\overrightarrow{AL}}. \end{eqnarray*} Similarly, we also have: \begin{eqnarray*} \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}= -\frac{\overrightarrow{QC}}{\overrightarrow{QA}}= -\frac{\overrightarrow{CB}}{\overrightarrow{AK}}= \frac{\overrightarrow{CB}}{\overrightarrow{KA}},\\ \frac{\overrightarrow{AR}}{\overrightarrow{RB}}= -\frac{\overrightarrow{RA}}{\overrightarrow{RB}}= -\frac{\overrightarrow{AL}}{\overrightarrow{BC}}= \frac{\overrightarrow{AL}}{\overrightarrow{CB}}, \end{eqnarray*} which means that $$\frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot \frac{\overrightarrow{AR}}{\overrightarrow{RB}}= \frac{\overrightarrow{KA}}{\overrightarrow{AL}}\cdot \frac{\overrightarrow{CB}}{\overrightarrow{KA}}\cdot \frac{\overrightarrow{AL}}{\overrightarrow{CB}}=1.$$ ($\Leftarrow$) Now we assume that relation (\ref{formulaCeva}) holds and that $S$ is the intersection of lines $BQ$ and $CR$ (the case when $BQ$ and $CR$ are parallel, we again leave it to the reader). If we now denote with $P'$ the intersection of line $AS$ with line $BC$, from the first part of the proof it follows (Figure \ref{sl.pod.7.5.1.pic}): $$\frac{\overrightarrow{BP'}}{\overrightarrow{P'C}}\cdot \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot \frac{\overrightarrow{AR}}{\overrightarrow{RB}}=1.$$ By the assumption (\ref{formulaCeva}) we then have $\frac{\overrightarrow{BP}}{\overrightarrow{PC}}= \frac{\overrightarrow{BP'}}{\overrightarrow{P'C}}$ or $P=P'$ (statement \ref{izrekEnaDelitevDaljiceVekt}). Therefore, lines $AP$, $BQ$, $CR$ intersect in point $S$. \kdokaz We observe that the statement is true in the case when the lines $AP$, $BQ$ and $CR$ intersect in one point (elliptical cone), as well as in the case when these lines are parallel (parabolic cone). It seems that we could more easily prove the statement if we introduced points at infinity and thus generalized the concept of (elliptical) cone also to the case of parallel lines. These are the ideas that led to the development of so-called \index{geometrija!projektivna}\pojem{projective geometry}. \bizrek \index{izrek!Menelajev}\label{izrekMenelaj} (Menelaj\footnote{\index{Menelaj} \textit{The ancient Greek mathematician Menelaus of Alexandria} (1st century) proved this statement in his work \textit{Spharica}, namely in the case of spherical triangles. He mentions the statement in the plane as already known. Since the previous documents about this are not preserved, we call the statement Menelaus's statement. Because of its similarity, Menelaus's and Ceva's statements are also called \textit{theorems of the twins}. The 1500-year period that separates these two discoveries is testimony to how geometry gradually developed during this period.}) Let $P$, $Q$ and $R$ be points lying on the lines containing the sides $BC$, $CA$ and $AB$ of a triangle $ABC$. Then the points $P$, $Q$ and $R$ are collinear if and only if: \begin{eqnarray}\label{formulaMenelaj} \frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot \frac{\overrightarrow{AR}}{\overrightarrow{RB}}=-1. \end{eqnarray} \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.5.2.pic} \caption{} \label{sl.pod.7.5.2.pic} \end{figure} \textbf{\textit{Proof.}} ($\Rightarrow$) Let $P$, $Q$ and $R$ be points of a line $l$. We denote with $A'$, $B'$ and $C'$ the intersection points of the perpendiculars from the vertices $A$, $B$ and $C$ of the triangle on the line $l$ (Figure \ref{sl.pod.7.5.2.pic}). By using Tales' theorem (\ref{TalesovIzrek}) we get: \begin{eqnarray*} \frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot \frac{\overrightarrow{AR}}{\overrightarrow{RB}}= -\frac{\overrightarrow{PB}}{\overrightarrow{PC}}\cdot \frac{\overrightarrow{QC}}{\overrightarrow{QA}}\cdot \frac{\overrightarrow{RA}}{\overrightarrow{RB}}= -\frac{\overrightarrow{BB'}}{\overrightarrow{CC'}}\cdot \frac{\overrightarrow{CC'}}{\overrightarrow{AA'}}\cdot \frac{\overrightarrow{AA'}}{\overrightarrow{BB'}}= -1. \end{eqnarray*} ($\Leftarrow$) Let now the relation (\ref{formulaMenelaj}) be satisfied. With $P'$ we denote the intersection of the lines $QR$ and $BC$. If the lines $QR$ and $BC$ were parallel, from this by Tales' theorem it would follow $\frac{\overrightarrow{CQ}}{\overrightarrow{QA}}= \frac{\overrightarrow{RB}}{\overrightarrow{AR}}$ and then from (\ref{formulaMenelaj}) also $\frac{\overrightarrow{BP}}{\overrightarrow{PC}}=-1$ or $\overrightarrow{BP}=\overrightarrow{CP}$, which is not possible. Let therefore $P'=QR\cap BC$. Then the points $P'$, $Q$, and $R$ are collinear. By the proven in the first part of the theorem we have: \begin{eqnarray*} \frac{\overrightarrow{BP'}}{\overrightarrow{P'C}}\cdot \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot \frac{\overrightarrow{AR}}{\overrightarrow{RB}}=-1. \end{eqnarray*} From this and the assumed relations (\ref{formulaMenelaj}) it follows $\frac{\overrightarrow{BP}}{\overrightarrow{PC}}= \frac{\overrightarrow{BP'}}{\overrightarrow{P'C}}$ or $P=P'$ (theorem \ref{izrekEnaDelitevDaljiceVekt}), which means that the points $P$, $Q$ and $R$ are collinear. \kdokaz We continue with the use of two proven theorems. \bizrek The lines, joining the vertices of a triangle to the tangent points of the incircle, intersect at one point (so-called \index{točka!Gergonova}\pojem{Gergonne\footnote{To trditev je dokazal \index{Gergonne, J. D.}\textit{J. D. Gergonne} (1771--1859), francoski matematik. V 19. stoletju je bila posvečena posebna pozornost metričnim lastnostim trikotnika, tako so bile odkrite tudi druge karakteristične točke trikotnika (glej naslednja zgleda).} point} \color{blue} of this triangle). \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.5.3.pic} \caption{} \label{sl.pod.7.5.3.pic} \end{figure} \textbf{\textit{Proof.}} Let $P$, $Q$ and $R$ be the points in which the inscribed circle of the triangle $ABC$ (see izrek \ref{SredVcrtaneKrozn}) touches its sides $BC$, $CA$ and $AB$ (Figure \ref{sl.pod.7.5.3.pic}). We prove that the lines $AP$, $BQ$ and $CR$ intersect at one point. By izrek \ref{TangOdsek} the corresponding tangent lines are: $BP\cong BR$, $CP\cong CQ$ and $AQ\cong AR$. Because $\mathcal{B}(B,P,C)$, $\mathcal{B}(C,Q,A)$ and $\mathcal{B}(A,R,B)$, we have $\frac{\overrightarrow{BP}}{\overrightarrow{PC}} \cdot \frac{\overrightarrow{CQ}}{\overrightarrow{QA}} \cdot \frac{\overrightarrow{AR}}{\overrightarrow{RB}} >0$. Therefore: $$\frac{\overrightarrow{BP}}{\overrightarrow{PC}} \cdot \frac{\overrightarrow{CQ}}{\overrightarrow{QA}} \cdot \frac{\overrightarrow{AR}}{\overrightarrow{RB}}= \frac{|BP|}{|PC|}\cdot\frac{|CQ|}{|QA|}\cdot\frac{|AR|}{|RB|}=1.$$ By Ceva's izrek \ref{izrekCeva} the lines $AP$, $BQ$ and $CR$ belong to one family. Because of Pash's axiom \ref{PaschIzrek} the lines $AP$ and $BQ$ intersect, it is an elliptic family, which means that the lines $AP$, $BQ$ and $CR$ intersect at one point. \kdokaz \bzgled Prove that the lines, joining the vertices of a triangle to the tangent points of the excircles to the opposite sides, intersect at one point (so-called \index{točka!Nagelova}\pojem{Nagel\footnote{\index{Nagel, C. H.}\textit{C. H. Nagel} (1803--1882), German mathematician, who published this theorem in 1836. Each of the lines in this theorem divides the area of the triangle into two equal parts, so the theorem is also called \index{izrek!o polobsegu trikotnika}\pojem{the statement about the half-perimeter of the triangle}. point} \color{green1} of this triangle). \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.5.6.pic} \caption{} \label{sl.pod.7.5.6.pic} \end{figure} \textbf{\textit{Proof.}} We use the notation from the big task (\ref{velikaNaloga}). We prove that the lines $AP_a$, $BQ_b$ and $CR_c$ intersect at one point (Figure \ref{sl.pod.7.5.6.pic}). If we use the fact $\frac{\overrightarrow{BP_a}}{\overrightarrow{P_aC}} \cdot \frac{\overrightarrow{CQ_b}}{\overrightarrow{Q_bA}} \cdot \frac{\overrightarrow{AR_c}}{\overrightarrow{R_cB}} >0$ and the relations from the big task (\ref{velikaNaloga}), we get: \begin{eqnarray*} \frac{\overrightarrow{BP_a}}{\overrightarrow{P_aC}} \cdot \frac{\overrightarrow{CQ_b}}{\overrightarrow{Q_bA}} \cdot \frac{\overrightarrow{AR_c}}{\overrightarrow{R_cB}}&=& \frac{|BP_a|}{|P_aC|}\cdot\frac{|CQ_b|}{|Q_bA|}\cdot\frac{|AR_c|}{|R_cB|}=\\&=& \frac{s-c}{s-b}\cdot\frac{s-a}{s-c}\cdot\frac{s-b}{s-a}= 1 \end{eqnarray*} By Ceva's theorem \ref{izrekCeva} the lines $AP_a$, $BQ_b$ and $CR_c$ belong to one pencil. Since, as a consequence of Pash's axiom \ref{PaschIzrek}, the lines $AP_a$ and $BQ_b$ intersect, it is an elliptic pencil, which means that the lines $AP_a$, $BQ_b$ and $CR_c$ intersect at one point. \kdokaz \bzgled Prove that the lines, joining the vertices of a triangle to the points dividing opposite sides in the ratio of squares of adjacent sides, intersect at one point (so-called \index{točka!Lemoinova} \pojem{Lemoine\footnote{\index{Lemoine, E. M. H.}\textit{E. M. H. Lemoine} (1751--1816), French mathematician.} point} \color{green1} of this triangle). \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.5.5.pic} \caption{} \label{sl.pod.7.5.5.pic} \end{figure} \textbf{\textit{Proof.}} Let $X$, $Y$ and $Z$ be points on sides $BC$, $AC$ and $AB$ of triangle $ABC$, such that $\frac{|BX|}{|XC|}=\frac{|BA|^2}{|AC|^2}$, $\frac{|CY|}{|YA|}=\frac{|CB|^2}{|BA|^2}$ and $\frac{|AZ|}{|ZB|}=\frac{|AC|^2}{|CB|^2}$ (Figure \ref{sl.pod.7.5.5.pic}). Because $\frac{\overrightarrow{BX}}{\overrightarrow{XC}} \cdot \frac{\overrightarrow{CY}}{\overrightarrow{YA}} \cdot \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}} >0$, we have: \begin{eqnarray*} \frac{\overrightarrow{BX}}{\overrightarrow{XC}} \cdot \frac{\overrightarrow{CY}}{\overrightarrow{YA}} \cdot \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}}&=& \frac{|BX|}{|XC|}\cdot\frac{|CY|}{|YA|}\cdot\frac{|AZ|}{|ZB|}=\\ &=& \frac{|BA|^2}{|AC|^2}\cdot\frac{|CB|^2}{|BA|^2}\cdot\frac{|AC|^2}{|CB|^2}= 1. \end{eqnarray*} By Ceva's theorem \ref{izrekCeva} lines $AX$, $BY$ and $CZ$ belong to one pencil. Because of Pash's axiom \ref{PaschIzrek} lines $AX$ and $BY$ intersect, it is an elliptic pencil, which means that lines $AX$, $BY$ and $CZ$ intersect in one point. \kdokaz \bzgled Let $Z$ and $Y$ be points of the sides $AB$ and $AC$ of a triangle $ABC$ such that $AZ:ZB=1:3$ and $AY:YC=1:2$, and $X$ the point, in which the line $YZ$ intersects the line containing the side $BC$ of this triangle. Calculate $\overrightarrow{BX}:\overrightarrow{XC}$. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.5.4.pic} \caption{} \label{sl.pod.7.5.4.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.5.4.pic}) Since the points $X$, $Y$ and $Z$ are collinear, by Menelaus's theorem \ref{izrekMenelaj} it follows that: $$\frac{\overrightarrow{BX}}{\overrightarrow{XC}} \cdot\frac{\overrightarrow{CY}}{\overrightarrow{YA}} \cdot\frac{\overrightarrow{AZ}}{\overrightarrow{ZB}}=-1.$$ If we use the given conditions, we get: $$\frac{\overrightarrow{BX}}{\overrightarrow{XC}} \cdot\frac{2}{1} \cdot\frac{1}{3}=-1$$ or $\overrightarrow{BX}:\overrightarrow{XC}=-3:2$. \kdokaz \bzgled Prove that the tangents of the circumcircle of an arbitrary scalene triangle $ABC$ at its vertices $A$, $B$ and $C$ intersect the lines containing the opposite sides of this triangle at three collinear points\footnote{This is a special case of Pascal's theorem \ref{izrekPascalEvkl} (section \ref{odd7PappusPascal}). \index{Pascal, B.} \textit{B. Pascal} (1623--1662), French mathematician and philosopher.}. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.12.2a.pic} \caption{} \label{sl.pod.7.12.2a.pic} \end{figure} \textbf{\textit{Solution.}} Let $X$, $Y$ and $Z$ be the points of intersection of the tangent to the circle $k$ of the triangle $ABC$ in the vertices $A$, $B$ and $C$ with the sides $BC$, $AC$ and $AB$ of this triangle (Figure \ref{sl.pod.7.12.2a.pic}). We prove that $X$, $Y$ and $Z$ are collinear points. By \ref{ObodKotTang} we have $\angle XAB\cong\angle ACB$. Because the triangle $AXB$ and $CXA$ have a common internal angle at the vertex $X$, we have $\triangle AXB\sim \triangle CXA$ (\ref{PodTrikKKK}), so: $$\frac{|AX|}{|CX|}=\frac{|AB|}{|AC|}=\frac{|BX|}{|AX|}.$$ From $\frac{|AX|}{|CX|}=\frac{|BX|}{|AX|}$ it follows that $|XB|\cdot |XC|=|XA|^2$. If we divide this relation by $|XC|^2$, we get $\frac{|BX|}{|CX|}=\frac{|AX|^2}{|CX|^2}$. Since $\frac{|AX|}{|CX|}=\frac{|AB|}{|AC|}$, we finally get: \begin{eqnarray*} && \frac{|BX|}{|XC|}=\frac{|BA|^2}{|AC|^2} \end{eqnarray*} and similarly \begin{eqnarray*} && \frac{|CY|}{|YA|}=\frac{|CB|^2}{|BA|^2},\\ && \frac{|AZ|}{|ZB|}=\frac{|AC|^2}{|CB|^2}. \end{eqnarray*} Because the points $X$, $Y$ and $Z$ do not lie on the sides $BC$, $AC$ and $AB$, we have $\frac{\overrightarrow{BX}}{\overrightarrow{XC}} \cdot \frac{\overrightarrow{CY}}{\overrightarrow{YA}} \cdot \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}} <0$, so: \begin{eqnarray*} \frac{\overrightarrow{BX}}{\overrightarrow{XC}} \cdot \frac{\overrightarrow{CY}}{\overrightarrow{YA}} \cdot \frac{\overrightarrow{AZ}}{\overrightarrow{ZB}}&=& -\frac{|BX|}{|XC|}\cdot\frac{|CY|}{|YA|}\cdot\frac{|AZ|}{|ZB|}=\\ &=& -\frac{|BA|^2}{|AC|^2}\cdot\frac{|CB|^2}{|BA|^2}\cdot\frac{|AC|^2}{|CB|^2}= -1. \end{eqnarray*} By Menelaus' theorem \ref{izrekMenelaj}, the points $X$, $Y$ and $Z$ are collinear. \kdokaz The next property of quadrilaterals is a continuation of examples \ref{TetivniVcrtana} and \ref{TetivniVisinska}. Namely, we are dealing with the following problem. Given is a quadrilateral $ABCD$ with the characteristic points of the triangles $BCD$, $ACD$, $ABD$ and $ABC$ as vertices. What does the quadrilateral represent, that has the centroid of the triangle $BCD$, $ACD$, $ABD$ and $ABC$ as vertices? The answer is trivial if the centroid of the circumscribed circle is given, since in that case $ABCD$ is a quadrilateral with one point as the centroid of the circumscribed circles and the sought quadrilateral does not exist. The cases of the centroids of the circumscribed circles and the altitude points are treated in the aforementioned examples \ref{TetivniVcrtana} and \ref{TetivniVisinska}. The next example gives the answer for the case of the centroids of the triangles, in the case of an arbitrary quadrilateral. \bzgled \label{TetivniTezisce} Let $ABCD$ be an arbitrary quadrilateral and $T_A$ the centroid of the triangle $BCD$, $T_B$ the centroid of the triangle $ACD$, $T_C$ the centroid of the triangle $ABD$ and $T_D$ the centroid of the triangle $ABC$. Prove that:\\ a) a) the lines $AT_A$, $BT_B$, $CT_C$ and $DT_D$ intersect at one point which is the common centroid of the quadrilaterals $ABCD$ and $T_AT_BT_CT_D$,\\ b) the quadrilateral $T_AT_BT_CT_D$ is similar to the quadrilateral $ABCD$ with the coefficient of similarity equal to $\frac{1}{3}$. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.2.9.pic} \caption{} \label{sl.pod.7.2.9.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.2.9.pic}) The lines $AT_A$, $BT_B$, $CT_C$ and $DT_D$ intersect at one point which is the common centroid of the quadrilaterals $ABCD$ and $T_AT_BT_CT_D$. The quadrilateral $T_AT_BT_CT_D$ is similar to the quadrilateral $ABCD$ with the coefficient of similarity equal to $\frac{1}{3}$. \textit{a)} We mark with $P$, $K$, $Q$, $L$, $M$ and $N$ the centers of the lines $AB$, $BC$, $CD$, $DA$, $AC$ and $BD$. Since the point $T_A$ is the center of gravity of the triangle $BCD$, the intersection of its medians $BQ$, $CN$ and $DK$, it also holds that $QT_A:T_AB=1:2$ (\ref{tezisce}) or $\overrightarrow{QT_A}=\frac{1}{3}\overrightarrow{QB}$. Similarly, from the triangle $ACD$ we get that $\overrightarrow{QT_B}=\frac{1}{3}\overrightarrow{QA}$. From the last two relations and \ref{vektVektorskiProstor} we obtain: $$\overrightarrow{T_BT_A}=\overrightarrow{T_BQ}+\overrightarrow{QT_A} =\frac{1}{3}\overrightarrow{AQ}+\frac{1}{3}\overrightarrow{QB}= \frac{1}{3}\left(\overrightarrow{AQ}+\overrightarrow{QB} \right)=\frac{1}{3}\overrightarrow{AB}.$$ Therefore: \begin{eqnarray} \label{eqnCevaTez1} \overrightarrow{T_BT_A}=\frac{1}{3}\overrightarrow{AB}, \end{eqnarray} so $T_BT_A\parallel AB$. We mark with $T$ the intersection of the lines $AT_A$ and $BT_B$. By Tales' theorem and \ref{eqnCevaTez1} it holds that: \begin{eqnarray} \label{eqnCevaTez2} \frac{\overrightarrow{TT_A}}{\overrightarrow{TA}} =\frac{\overrightarrow{TT_B}}{\overrightarrow{TB}}= \frac{\overrightarrow{T_AT_B}}{\overrightarrow{AB}}=-\frac{1}{3}. \end{eqnarray} The lines $AT_A$ and $BT_B$ therefore intersect in the point $T$, which divides it in the ratio $2:1$. Similarly, the lines $AT_A$ and $CT_C$ or the lines $AT_A$ and $DT_D$ intersect in a point which divides it in the ratio $2:1$, and that is (because of the line $AT_A$) precisely the point $T$. This means that the lines $AT_A$, $BT_B$, $CT_C$ and $DT_D$ intersect in the point $T$. We will now prove that point $T$ is the center of mass of the quadrilaterals $ABCD$ and $T_AT_BT_CT_D$. Since $$\frac{\overrightarrow{QT_B}}{\overrightarrow{T_BA}}\cdot \frac{\overrightarrow{AP}}{\overrightarrow{PB}}\cdot \frac{\overrightarrow{BT_A}}{\overrightarrow{T_AQ}}= \frac{1}{2}\cdot\frac{1}{1}\cdot\frac{2}{1}=1, $$ by Ceva's theorem \ref{izrekCeva} for the triangle $QAB$ the lines $AT_A$, $BT_B$ and $PQ$ intersect in one point - point $T$. Therefore point $T$ lies on the line $PQ$. Similarly, point $T$ lies on the line $KL$, which means that point $T$ is the intersection of the diagonals $PQ$ and $KL$ of the quadrilateral $PKQL$. Since this is a Varignon parallelogram of the quadrilateral $ABCD$, point $T$ is, by theorem \ref{vektVarignon}, the center of mass of the quadrilateral $ABCD$. Since from \ref{eqnCevaTez2} it also follows that: $$\overrightarrow{TT_A}+\overrightarrow{TT_B}+\overrightarrow{TT_C}+\overrightarrow{TT_D}= -\frac{1}{3}\cdot\left( \overrightarrow{TA}+\overrightarrow{TB}+\overrightarrow{TC}+\overrightarrow{TD} \right)=-\frac{1}{3}\cdot \overrightarrow{0}=\overrightarrow{0},$$ point $T$ is also the center of mass of the quadrilateral $T_AT_BT_CT_D$. \textit{b)} From \ref{eqnCevaTez2} it follows that $\overrightarrow{TT_A}=-\frac{1}{3}\overrightarrow{TA}$, $\overrightarrow{TT_B}=-\frac{1}{3}\overrightarrow{TB}$, $\overrightarrow{TT_C}=-\frac{1}{3}\overrightarrow{TC}$ and $\overrightarrow{TT_D}=-\frac{1}{3}\overrightarrow{TD}$, or $h_{T,-\frac{1}{3}}:\hspace*{1mm}A,B,C,D\mapsto T_A,T_B,T_C,T_D$, which means that the quadrilateral $T_AT_BT_CT_D$ is similar to the quadrilateral $ABCD$ with the similarity coefficient $\frac{1}{3}$. \kdokaz %________________________________________________________________________________ \poglavje{Harmonic Conjugate Points. Apollonius Circle} \label{odd7Harm} Since it is very important for what follows, we will write theorem \ref{izrekEnaDelitevDaljiceVekt} from section \ref{odd5LinKombVekt} in another form. \bizrek \label{HarmCetEnaSamaDelitev} If $A$ and $B$ are different points on the line $p$ and $\lambda\neq -1$ an arbitrary real number, then there is exactly one point $L$ on the line $p$ such that: $$ \frac{\overrightarrow{AL}}{\overrightarrow{LB}}=\lambda.$$ \eizrek In section \ref{odd5TalesVekt} we found out how to divide the distance $AB$ in the ratio $m:n$ (statement \ref{izrekEnaDelitevDaljice}). We will now use this idea to construct the point $L$ from the previous statement \ref{HarmCetEnaSamaDelitev}. \bzgled The points $A$ and $B$ and the real number $\lambda\neq -1$ are given. Construct a point $L$ on the line $AB$ such that: $$\frac{\overrightarrow{AL}}{\overrightarrow{LB}}=\lambda.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.4.1.pic} \caption{} \label{sl.pod.7.4.1.pic} \end{figure} \textit{\textbf{Solution.}} (Figure \ref{sl.pod.7.4.1.pic}) In the case $\lambda=0$ the proof is direct - then $L=A$. Let $\lambda\neq 0$. Construct any parallel lines $a$ and $b$ through the points $A$ and $B$. With $P$ and $Q$ we mark such points on the lines $a$ and $b$, that $P,Q\div AB$ and $|AP|=|\lambda|$ and $|BQ|=1$. We also mark $Q'=\mathcal{S}_B(Q)$. We will consider two cases. \textit{1)} Let $\lambda>0$. It is clear that for the sought point $L$ it must hold that $\mathcal{B}(A,L,B)$. With $L_1$ we mark the intersection of the lines $AB$ and $PQ$. By Tales' statement \ref{TalesovIzrek} it holds: $$\frac{\overrightarrow{AL_1}}{\overrightarrow{L_1B}}= \frac{\overrightarrow{AP}}{\overrightarrow{QB}}=\lambda.$$ \textit{2)} If $\lambda<0$, the sought point does not lie on the line $AB$. With $L_1$ we mark the intersection of the lines $AB$ and $PQ'$. Because $\lambda\neq -1$, this intersection exists. By Tales' statement \ref{TalesovIzrek} it holds: $$\frac{\overrightarrow{AL_2}}{\overrightarrow{L_2B}}= \frac{\overrightarrow{AP}}{\overrightarrow{Q'B}} =-\frac{\overrightarrow{AP}}{\overrightarrow{BQ'}}=-\left(-\lambda\right)=\lambda,$$ which was to be proven. \kdokaz From the previous example it is therefore clear that for each of the conditions $\frac{\overrightarrow{AL}}{\overrightarrow{LB}}=\lambda$ ($\lambda>0$) or $\frac{\overrightarrow{AL}}{\overrightarrow{LB}}=\lambda$ ($\lambda<0$ and $\lambda\neq -1$) there is exactly one solution for the point $L$ on the line $AB$. In the first case, the point $L$ lies on the line $AB$, so we say that this is the \index{delitev daljice!notranja}\pojem{inner division} of the line $AB$ in the ratio $\lambda$ ($\lambda>0$). In the second case, the point $L$ does not lie on the line $AB$, so we say that this is the \index{delitev daljice!zunanja}\pojem{outer division} of the line $AB$ in the ratio $|\lambda|$ ($\lambda<0$ and $\lambda\neq -1$). We can also say that for the condition $\frac{AL}{LB}=\lambda$ ($\lambda>0$, $\lambda\neq 1$) there are two solutions for the point $L$ on the line - the inner or the outer division. If $L_1$ and $L_2$ are the inner and outer division of the line (for the same $\lambda$), it holds: $$\frac{\overrightarrow{AL_1}}{\overrightarrow{L_1B}}: \frac{\overrightarrow{AL_2}}{\overrightarrow{L_2B}}=-1.$$ If we define the expression $\frac{\overrightarrow{AL_1}}{\overrightarrow{L_1B}}: \frac{\overrightarrow{AL_2}}{\overrightarrow{L_2B}}$ as the \index{dvorazmerje parov točk}\pojem{ratio of pairs of points} $(A,B)$ and $(L_1,L_2)$ and denote $$d(A,B;L_1,L_2)=\frac{\overrightarrow{AL_1}}{\overrightarrow{L_1B}}: \frac{\overrightarrow{AL_2}}{\overrightarrow{L_2B}},$$ we see that for the inner and outer division of the line $AB$ with points $L_1$ and $L_2$ the corresponding ratio is $-1$, or $d(A,B;L_1,L_2)=-1$ This gives us the idea for a new definition. We say that different collinear points $A$, $B$, $C$ and $D$ determine \index{harmonična četverica točk} \pojem{harmonično četverico točk} oz. that the pair $(A, B)$, \pojem{harmonično konjugiran} s parom $(C, D)$, oznaka $\mathcal{H}(A,B;C,D)$, if: \begin{eqnarray}\label{formulaHarmEvkl} \frac{\overrightarrow{AC}}{\overrightarrow{CB}}= -\frac{\overrightarrow{AD}}{\overrightarrow{DB}}, \end{eqnarray} or: \begin{eqnarray*} d(A,B;C,D)=\frac{\overrightarrow{AC}}{\overrightarrow{CB}}: \frac{\overrightarrow{AD}}{\overrightarrow{DB}}=-1. \end{eqnarray*} We prove the basic properties of the defined relation. As we have already mentioned, the points $C$ and $D$, for which $\mathcal{H}(A,B;C,D)$ is true, represent the internal and external division of the distance $AB$ in some ratio $\lambda$ ($\lambda>0$ and $\lambda\neq 1$). There are therefore an infinite number of such pairs $C$, $D$. But if one of the points $C$ or $D$ is given, the other is uniquely determined. We express this in other words in the following proposition. \bizrek \label{HarmCetEnaSamaTockaD} Let $C$ be a point on a line $AB$ different from the points $A$ and $B$ and also from the midpoint of the line segment $AB$, then there is exactly one point $D$ such that $\mathcal{H}(A,B;C,D)$. \eizrek \textit{\textbf{Proof.}} Let $\frac{\overrightarrow{AC}}{\overrightarrow{CB}}=\lambda$. Because $C$ is different from $A$, $B$ and the midpoint of the line segment $AB$, $\lambda\neq 0$ and $\lambda\neq 1$. We are looking for a point $D$, for which $\mathcal{H}(A,B;C,D)$ is true, or $d(A,B;C,D)=-1$ or equivalently $\frac{\overrightarrow{AD}}{\overrightarrow{DB}}=-\lambda$. Because $-\lambda\neq -1$, by \ref{HarmCetEnaSamaDelitev} there is only one point $D$, for which this is fulfilled. \kdokaz \bizrek \label{HarmCetEF} Let $A$, $B$, $C$ and $D$ be four different collinear points on a line $p$ and $O$ a point not lying on this line. Suppose that a line that is parallel to the line $OA$ through the point $B$ intersects the lines $OC$ and $OD$ at the points $E$ and $F$. Then: $$\mathcal{H}(A,B;C,D) \hspace*{1mm} \Leftrightarrow \hspace*{1mm} \mathcal{S}_B(E)=F.$$ \eizrek \textit{\textbf{Proof.}} (Figure \ref{sl.pod.7.4.2.pic}) ($\Rightarrow$) If $\mathcal{H}(A,B;C,D)$ is true, by Tales' theorem \ref{TalesovIzrek}: $$\frac{\overrightarrow{AO}}{\overrightarrow{EB}}= \frac{\overrightarrow{AC}}{\overrightarrow{CB}}= -\frac{\overrightarrow{AD}}{\overrightarrow{DB}}= -\frac{\overrightarrow{AO}}{\overrightarrow{FB}}= \frac{\overrightarrow{AO}}{\overrightarrow{BF}},$$ it follows that $\overrightarrow{EB}=\overrightarrow{BF}$ or $\mathcal{S}_B(E)=F$. ($\Leftarrow$) Now suppose that $\mathcal{S}_B(E)=F$. From this it follows that $\overrightarrow{EB}=\overrightarrow{BF}$, so (again by Tales' theorem \ref{TalesovIzrek}): $$\frac{\overrightarrow{AC}}{\overrightarrow{CB}}= \frac{\overrightarrow{AO}}{\overrightarrow{EB}}= \frac{\overrightarrow{AO}}{\overrightarrow{BF}}= -\frac{\overrightarrow{AO}}{\overrightarrow{FB}}= -\frac{\overrightarrow{AD}}{\overrightarrow{DB}},$$ so by definition $\mathcal{H}(A,B;C,D)$. \kdokaz \begin{figure}[!htb] \centering \input{sl.pod.7.4.2.pic} \caption{} \label{sl.pod.7.4.2.pic} \end{figure} The previous theorem allows us to effectively construct the fourth point in a harmonic set of points. \bzgled \label{HarmCetEnaSamaTockaDKonstr} Let $C$ be a point that lies on a line $AB$. Suppose that $C$ is different from the points $A$, $B$ and the midpoint of the line segment $AB$. Construct a point $D$ such that $\mathcal{H}(A,B;C,D)$. \ezgled \textit{\textbf{Solution.}} According to the theorem \ref{HarmCetEnaSamaTockaD} there is only one such point $D$, for which $\mathcal{H}(A,B;C,D)$. Now we will construct it. Let $O$ be an arbitrary point, which does not lie on the line $AB$ (Figure \ref{sl.pod.7.4.2.pic}), and $l$ is a parallel to the line $AO$ in the point $B$. With $E$ we mark the intersection of the lines $OC$ and $l$ and $F=\mathcal{S}_B(E)$. The point $D$ is the intersection of the lines $OF$ and $AB$. By the previous theorem \ref{HarmCetEF} it is $\mathcal{H}(A,B;C,D)$. \kdokaz In a similar way as in the previous example, for given collinear points $A$, $B$ and $D$ (the point $D$ does not lie on the line $AB$) we can plan such a point $C$, that $\mathcal{H}(A,B;C,D)$ is valid. It is intuitively clear that from $\mathcal{H}(A,B;C,D)$ it follows $\mathcal{H}(A,B;D,C)$, because if the points $C$ and $D$ divide the line $AB$ in the same ratio, then the same is true for the points $D$ and $C$. It is interesting that from $\mathcal{H}(A,B;C,D)$ it also follows $\mathcal{H}(C,D;A,B)$, which means that if the points $C$ and $D$ divide the line $AB$ in the same ratio, then the points $A$ and $B$ divide the line $CD$ in the same ratio. We prove both properties formally. \bizrek a) $\mathcal{H}(A,B;C,D) \hspace*{1mm} \Rightarrow \hspace*{1mm} \mathcal{H}(A,B;D,C)$; \\ \hspace*{22mm}b) $\mathcal{H}(A,B;C,D) \hspace*{1mm} \Rightarrow \hspace*{1mm} \mathcal{H}(C,D;A,B)$; \eizrek \textit{\textbf{Proof.}} $$a)\hspace*{1mm}\mathcal{H}(A,B;C,D) \hspace*{1mm} \Rightarrow \hspace*{1mm} \frac{\overrightarrow{AC}}{\overrightarrow{CB}}= -\frac{\overrightarrow{AD}}{\overrightarrow{DB}} \hspace*{1mm} \Rightarrow \hspace*{1mm} \frac{\overrightarrow{AD}}{\overrightarrow{DB}}= -\frac{\overrightarrow{AC}}{\overrightarrow{CB}} \hspace*{1mm} \Rightarrow \hspace*{1mm} \mathcal{H}(A,B;D,C).$$ $$b)\hspace*{1mm}\mathcal{H}(A,B;C,D) \hspace*{1mm} \Rightarrow \hspace*{1mm} \frac{\overrightarrow{AC}}{\overrightarrow{CB}}= -\frac{\overrightarrow{AD}}{\overrightarrow{DB}} \hspace*{1mm} \Rightarrow \hspace*{1mm} \frac{\overrightarrow{CA}}{\overrightarrow{AD}}= -\frac{\overrightarrow{CB}}{\overrightarrow{BD}} \hspace*{1mm} \Rightarrow \hspace*{1mm} \mathcal{H}(C,D;A,B),$$ which had to be proven. \kdokaz \bizrek \label{izrek 1.2.1} Let $A$, $B$, $C$ and $D$ be different collinear points. Then $\mathcal{H}(A,B;C,D)$ if and only if there exists such a quadrilateral $PQRS$, that: $$A=PQ\cap RS, \hspace*{2mm}B=QR\cap PS, \hspace*{2mm} C\in PR \hspace*{2mm} \textrm{and} \hspace*{2mm} D\in QS.$$ \eizrek \textit{\textbf{Proof.}} ($\Rightarrow$) Let $PQRS$ be a quadrilateral, such that $A=PQ\cap RS$, $B=QR\cap PS$, $C\in PR$ and $D\in QS$ (Figure \ref{sl.pod.7.4.7a.pic}). By Menelaus' theorem \ref{izrekMenelaj} for the triangle $ABP$ and the line $QS$ we get: $$\frac{\overrightarrow{AD}}{\overrightarrow{DB}}\cdot \frac{\overrightarrow{BS}}{\overrightarrow{SP}}\cdot \frac{\overrightarrow{PQ}}{\overrightarrow{QA}}=-1.$$ Similarly, by Ceva's theorem \ref{izrekCeva} for the same triangle and the point $R$ we get: $$\frac{\overrightarrow{AC}}{\overrightarrow{CB}}\cdot \frac{\overrightarrow{BS}}{\overrightarrow{SP}}\cdot \frac{\overrightarrow{PQ}}{\overrightarrow{QA}}=1.$$ From these two relations it follows $\frac{\overrightarrow{AC}}{\overrightarrow{CB}}= -\frac{\overrightarrow{AD}}{\overrightarrow{DB}}$ or $\mathcal{H}(A,B;C,D)$. \begin{figure}[!htb] \centering \input{sl.pod.7.4.7a.pic} \caption{} \label{sl.pod.7.4.7a.pic} \end{figure} ($\Leftarrow$) Now, let's assume that $\mathcal{H}(A,B;C,D)$ holds. Without loss of generality, let the point $C$ be between the points $A$ and $B$; in the other two cases the proof proceeds in the same way. Let $P$ be an arbitrary point outside the line $AB$ and $Q$ an arbitrary point between the points $A$ and $P$. By Pasch's\footnote{\index{Pasch, M.} \textit{M. Pasch} (1843--1930), German mathematician, who introduced the relation of order of points in his work \textit{Predavanja o novejši geometriji} (Lectures on Modern Geometry) from 1882.} \ref{PaschIzrek} axiom, the line $PC$ and the line $QB$ intersect in some point $R$ and the line $AR$ and the line $PB$ intersect in some point $S$, therefore the line $AB$ and the line $QS$ intersect in some point $D_1$ (from $AB\parallel QS$ it follows that $C$ is the center of the line segment $AB$, which is not possible due to $\mathcal{H}(A,B;C,D)$). Now, from the first part of the proof ($\Rightarrow$) it follows $\mathcal{H}(A,B;C,D_1)$. Since by the assumption we also have $\mathcal{H}(A,B;C,D)$, by the uniqueness of the fourth point of a harmonic quadrilateral of points (theorem \ref{HarmCetEnaSamaTockaD}) it follows $D=D_1$. Therefore $PQRS$ is the desired quadrilateral. \kdokaz We observe that the previous statement is also valid in the case when $C$ is the center of the line $AB$ and $D$ is a point at infinity (Figure \ref{sl.pod.7.4.7a.pic}). The statement on the right side of the equivalence from the previous statement (the existence of the appropriate quadrilateral) could be accepted as the definition of the harmonic quadruple of points in Euclidean geometry. In projective geometry, this definition is even more natural, since in such a definition we do not use metric. From the previous statement it also follows that the relation of harmonic quadruple is preserved in the so-called \pojem{central projection} in space. Indeed, if $A$, $B$, $C$ and $D$ are points for which $\mathcal{H}(A,B;C,D)$ holds, and $A'$, $B'$, $C'$ and $D'$ are the central projections of these points, then from the existence of the appropriate quadrilateral for the quadruple $A$, $B$, $C$, $D$ it follows that the appropriate quadrilateral exists also for the quadruple $A'$, $B'$, $C'$, $D'$. The latter is the central projection of the first quadrilateral, since the central projection preserves collinearity. Of course, we should also include in the previous consideration the case when some of the central projections are points at infinity. \index{geometrija!projektivna}In this sense, projective geometry can be described as the geometry that deals with objects and properties that are preserved in the central projection. Because "being a harmonic quadruple of points" is one of such properties, it is the subject of study in projective geometry. Even more - in this geometry, this relation is one of the basic concepts (see \cite{Mitrovic})." We have already seen that on the line $AB$ there are exactly two points that divide the distance $AB$ in the ratio $\lambda>0$, $\lambda\neq 1$ (in the case $\lambda= 1$ this is only one point - the center of the line). These points represent the inner and outer division of the line in this ratio and together with points $A$ and $B$ determine a harmonic quadruple of points. The question arises: What does the set of all such points $X$ in the plane mean, so that $\frac{AX}{XB}=\lambda$ ($\lambda\neq 1$)? The answer is given by the following theorem. \bizrek \label{ApolonijevaKroznica} Suppose that $A$ and $B$ are points in the plane and $\lambda>0$, $\lambda\neq 1$ an arbitrary real number. If $C$ and $D$ are points of the line $AB$ such that: $$\frac{\overrightarrow{AC}}{\overrightarrow{CB}}= -\frac{\overrightarrow{AD}}{\overrightarrow{DB}}=\lambda,$$ i.e. $\mathcal{H}(A,B;C,D)$, then the set of all points $X$ of this plane such that: $$\frac{AX}{XB}=\lambda,$$ is a circle with the diameter $CD$ (so-called Apollonius Circle\footnote{\index{Apolonij} \textit{Apolonij iz Perge} (3.-- 2. st. pr. n. š.), starogrški matematik.}). \index{krožnica!Apolonijeva} \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.4.3.pic} \caption{} \label{sl.pod.7.4.3.pic} \end{figure} \textit{\textbf{Proof.}} Let $k$ be a circle with diameter $CD$. Let $X$ be an arbitrary point of this plane and $p$ a parallel line through point $B$ to line $AX$. With $E$ and $F$ we mark the intersections of lines $XC$ and $XD$ with line $p$ (Figure \ref{sl.pod.7.4.3.pic}). By Theorem \ref{HarmCetEF}, point $B$ is the center of the line $EF$, because $\mathcal{H}(A,B;C,D)$. It needs to be proven that: $$\frac{AX}{XB}=\lambda \hspace*{1mm} \Leftrightarrow \hspace*{1mm} X\in k.$$ ($\Leftarrow$) Let $X\in k$. Then $\angle EXF=\angle CXD=90^0$ and $BE\cong BF\cong BX$ (statement \ref{TalesovIzrKroz2}). From this and from Tales's statement \ref{TalesovIzrek} it follows: $$\frac{AX}{XB}=\frac{AX}{EB}=\frac{AC}{CB}=\lambda.$$ ($\Rightarrow$) Let $\frac{AX}{XB}=\lambda$. From this and from Tales's statement \ref{TalesovIzrek} it follows: $$\frac{AX}{XB}=\lambda=\frac{AC}{CB}=\frac{AX}{EB},$$ so $XB\cong EB\cong BF$. Therefore $EXF$ is a right angled triangle (statement \ref{TalesovIzrKroz2}). Then $\angle CXD=\angle EXF=90^0$, which means that the point $X$ lies on the circle $k$ above the diameter $CD$ (statement \ref{TalesovIzrKroz2}). \kdokaz For given points $A$ and $B$ and for different values $\lambda\in \mathbb{R}$ ($\lambda>0$, $\lambda\neq 1$) we have different Apollonius's circles (Figure \ref{sl.pod.7.4.4a.pic}). We will mark them with $\mathcal{A}_{AB,\lambda}$, but if we know which distance it is, we can also write it shorter $k_{\lambda}$. In the case $\lambda=1$ we are actually looking for a set of all such points $X$ of this plane, for which: $\frac{AX}{XB}=1$, i.e. $AX\cong BX$. Then the sought set does not represent a circle, but the symmetry $s_{AB}$ of the distance $AB$. \begin{figure}[!htb] \centering \input{sl.pod.7.4.4a.pic} \caption{} \label{sl.pod.7.4.4a.pic} \end{figure} \bzgled Points $A$ and $B$ and a line $l$ in the plane are given. Construct the point $L$ on the line $l$ such that $LA:LB=5:2$. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.4.4.pic} \caption{} \label{sl.pod.7.4.4.pic} \end{figure} \textit{\textbf{Solution.}} (Figure \ref{sl.pod.7.4.4.pic}) Let's first draw a point $C$ on the line $AB$, so that $AC:CB=5:2$ (see example \ref{izrekEnaDelitevDaljice}), and then the fourth point of the harmonic quartet $\mathcal{H}(A,B;C,D)$ (see example \ref{HarmCetEnaSamaTockaDKonstr}). Construct the Apollonius circle $k_{\frac{5}{2}}$ over the diameter $CD$. The point $L$ is then the one for which $L\in k\cap l$. By the previous theorem (\ref{ApolonijevaKroznica}) we have: $$\frac{AL}{LB}=\frac{AC}{CB}=\frac{5}{2}.$$ The task has 0, 1 or 2 solutions, depending on the number of intersections of the circle $k$ and the line $l$. \kdokaz \bizrek \label{HarmOhranjaVzporProj} A parallel projection preserves the relation of a harmonic conjugate points, i.e. if $A$, $B$, $C$, $D$ and $A'$, $B'$, $C'$, $D'$ are two quartets of collinear points such that $AA'\parallel BB'\parallel CC'\parallel DD'$, then: $$\mathcal{H}(A,B;C,D)\hspace*{1mm}\Rightarrow\hspace*{1mm} \mathcal{H}(A',B';C',D').$$ \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.4.5.pic} \caption{} \label{sl.pod.7.4.5.pic} \end{figure} \textit{\textbf{Solution.}} (Figure \ref{sl.pod.7.4.5.pic}) Let $\mathcal{H}(A,B;C,D)$ or $\frac{\overrightarrow{AC}}{\overrightarrow{CB}}: \frac{\overrightarrow{AD}}{\overrightarrow{DB}}=-1$. By Tales' theorem \ref{TalesovIzrek} we have: $$\frac{\overrightarrow{A'C'}}{\overrightarrow{C'B'}}: \frac{\overrightarrow{A'D'}}{\overrightarrow{D'B'}}= \frac{\overrightarrow{AC}}{\overrightarrow{CB}}: \frac{\overrightarrow{AD}}{\overrightarrow{DB}}=-1,$$ which means that $\mathcal{H}(A',B';C',D')$ is also true. \kdokaz It is interesting that the central projection also preserves the relation of harmonic points. Therefore, the mentioned relation is the subject of research in projective geometry (see \cite{Mitrovic}). \bizrek \label{HarmCetSimKota} Suppose that a line $BC$ intersects the bisectors of the interior and the exterior angle at the vertex $A$ of a triangle $ABC$ at points $E$ and $F$. Then: a) $BE:CE=BF:CF=AB:AC$, b) $\mathcal{H}(B,C;E,F)$. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.4.6.pic} \caption{} \label{sl.pod.7.4.6.pic} \end{figure} \textit{\textbf{Solution.}} Let $L$ be an arbitrary point for which $\mathcal{B}(C,A,L)$ holds. We mark with $M$ and $N$ the points in which the parallel to the line $AC$ through the point $B$ intersects in order the simetrals $AE$ and $AF$ of the internal and the external angle (Figure \ref{sl.pod.7.4.6.pic}). \textit{a)} By izrek \ref{KotiTransverzala} we have: \begin{eqnarray*} \angle BMA &\cong& \angle CAM=\angle CAE\cong\angle BAE=\angle BAM\\ \angle BNA &\cong& \angle LAN=\angle LAF\cong\angle BAF=\angle BAN \end{eqnarray*} Therefore, $AMB$ and $ANB$ are isosceles triangles with the bases $AM$ and $AN$ (izrek \ref{enakokraki} and it holds: $$BM\cong BA\cong BN.$$ From this and from the consequence of Tales' theorem \ref{TalesovIzrekDolzine} (because $AC\parallel NM$) we get: \begin{eqnarray*} \frac{BE}{EC}&=& \frac{BM}{AC}=\frac{BA}{AC},\\ \frac{BF}{FC}&=& \frac{BN}{AC}=\frac{BA}{AC}. \end{eqnarray*} \textit{b)} Because $\mathcal{B}(B,E,C)$ and $\neg\mathcal{B}(B,F,C)$, we have: \begin{eqnarray*} \frac{\overrightarrow{BE}}{\overrightarrow{EC}}&=& \frac{BA}{AC},\\ \frac{\overrightarrow{BF}}{\overrightarrow{FC}}&=& -\frac{BA}{AC}. \end{eqnarray*} Therefore: \begin{eqnarray*} \frac{\overrightarrow{BE}}{\overrightarrow{EC}}= -\frac{\overrightarrow{BF}}{\overrightarrow{FC}}, \end{eqnarray*} which means that $\mathcal{H}(B,C;E,F)$ holds. \kdokaz We will continue by using the previous theorem. \bzgled \label{HarmTrikZgl1} Construct a triangle $ABC$, with given: $a$, $t_a$, $b:c=2:3$. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.4.8.pic} \caption{} \label{sl.pod.7.4.8.pic} \end{figure} \textit{\textbf{Solution.}} Let $ABC$ be a triangle, in which the side $BC\cong a$, the median $AA_1\cong t_a$ and $AC:AB=2:3$ (Figure \ref{sl.pod.7.4.8.pic}). So we can construct the side $BC$ and its center $A_1$. From the given conditions, the vertex $A$ lies on the circle $k(A_1,t_a)$ and the Apollonius circle $\mathcal{A}_{BC,\frac{3}{2}}$, because $AB:AC=3:2$ (statement \ref{ApolonijevaKroznica}). So we get the vertex $A$ as one of the intersections of the circles $k$ and $\mathcal{A}_{BC,\frac{3}{2}}$. Although we do not need this fact in the construction, we mention that the Apollonius circle $\mathcal{A}_{BC,\frac{3}{2}}$ represents the circle with the diameter $EF$, where $E$ and $F$ are defined as in the statement \ref{HarmCetSimKota}. \kdokaz \bzgled Construct a parallelogram, with the sides congruent to given line segments $a$ and $b$, and the diagonals in the ratio $3:7$. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.4.9.pic} \caption{} \label{sl.pod.7.4.9.pic} \end{figure} \textit{\textbf{Solution.}} (Figure \ref{sl.pod.7.4.9.pic}) Let $S$ be the intersection of the diagonals of the sought parallelogram $ABCD$, in which $AB\cong a$, $BC\cong b$ and $AC:BD=3:7$. We also mark with $P$ the center of the side $AB$. Since the diagonals of a parallelogram are divided in half, the line segments $SA$ and $SB$ are in the ratio $2:3$. This means that we can first construct the triangle $ASB$, similar to the previous example \ref{HarmTrikZgl1}: $AB\cong a$, $SP=\frac{1}{2}b$ and $SA:SB=3:7$. \kdokaz \bizrek \label{harmVelNal} Suppose that $k(S, r)$ is the incircle and $k_a (S_a, r_a)$ the excircle of a triangle $ABC$ and $P$ and $P_a$ the touching points of these circles with the aide $BC$. Let $A'$ be the foot of the altitude $AA'$ and $E$ intersection of the bisectors of the interior angle at the vertex $A$ and the side $BC$. If $L$ and $L_a$ are foots of the perpendiculars from the points $S$ and $S_a$ on the line $AA'$, then: \hspace*{2mm} (i) $\mathcal{H}(A,E;S,Sa)$ \hspace*{2mm} (ii) $\mathcal{H}(A,A';L,La)$ \hspace*{2mm} (iii) $\mathcal{H}(A',E;P,Pa)$. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.3.7.pic} \caption{} \label{sl.pod.7.3.7.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.3.7.pic}) \textit{(i)} The lines $CS$ and $CS_a$ are the altitudes of the triangle $ACE$, therefore by \ref{HarmCetSimKota} we have $\mathcal{H}(A,E;S,S_a)$. \textit{(ii)} The points $A$, $A'$, $L$ and $L_a$ are the orthogonal projections of the points $A$, $E$, $S$ and $S_a$ on the line $AA'$. By \ref{HarmOhranjaVzporProj} we have $\mathcal{H}(A,A';L,La)$. \textit{(iii)} Similarly to the previous statement, since the points $A'$, $E$, $P$ and $P_a$ are the orthogonal projections of the points $A$, $E$, $S$ and $S_a$ on the line $BC$. \kdokaz \bzgled Construct a triangle $ABC$, with given: \hspace*{4mm} (i) $r$, $a$, $v_a$ \hspace*{5mm} (ii) $v_a$, $r$, $b-c$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.4.10.pic} \caption{} \label{sl.pod.7.4.10.pic} \end{figure} \textbf{\textit{Solution.}} We use the notation from the big exercise \ref{velikaNaloga} and the statement \ref{harmVelNal} (Figure \ref{sl.pod.7.4.10.pic}). In both cases \textit{(i)} and \textit{(ii)} we can use the fact $\mathcal{H}(A,A';L,La)$ from the statement \ref{harmVelNal}, because from $AA'\cong v_a$ and $LA'\cong r_a$ we can plot the fourth point in the harmonic quadruple $\mathcal{H}(A,A';L,La)$. So we get $r_a\cong A'L_a$. \textit{(i)} We use the relation $RR_a\cong a$ (big exercise \ref{velikaNaloga}), we plot the circles $k(S,SR)$ and $k_a(S_a,R_a)$, and then their three common tangents (example \ref{tang2ehkroz}). \textit{(ii)} We use the relation $PP_a\cong b-c$ (big exercise \ref{velikaNaloga}), we plot the circles $k(S,SP)$ and $k_a(S_a,P_a)$, and then their three common tangents. \kdokaz In the previous example we saw how, with the help of the proven relation $\mathcal{H}(A,A';L,La)$, we can get $r_a$ from the elements of the triangle $v_a$ and $r$. In a similar way, from each known pair of a triple $(v_a,r,r_a)$ we get the third element. We will write this fact in the form $\langle v_a,r,r_a\rangle$. \bzgled Construct a cyclic quadrilateral $ABCD$, with the sides congruent to given line segments $a$, $b$, $c$ in $d$. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.4.11.pic} \caption{} \label{sl.pod.7.4.11.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.4.11.pic}) Let $ABCD$ be a cyclic quadrilateral with sides congruent to given line segments $a$, $b$, $c$ and $d$, $h_{A,k}$ central stretch with center $A$ and coefficient $k=\frac{AD}{AB}=\frac{d}{a}$, $\mathcal{R}_{A,\alpha}$ rotation with center $A$ for the oriented angle $\alpha=\angle BAD$ and the composite: $$\rho_{A,k,\alpha}=\mathcal{R}_{A,\alpha}\circ h_{A,k}$$ rotational stretch with center $A$, coefficient $k$ and angle $\alpha$. Let $B'=h_{A,k}(B)$ and $C'=h_{A,k}(C)$. Because $|AB'|=k\cdot |AB|=\frac{|AD|}{|AB|}\cdot |AB|= |AD|$ and $\angle B'AD=\angle BAD= \alpha$, it follows that $R_{A,\alpha}(B')=D$, so $\rho_{A,k,\alpha}(B)=D$. Let $E=R_{A,\alpha}(C')$ or $E=\rho_{A,k,\alpha}(C)$. Then the triangle $ADE$ is the image of the triangle $ABC$ under the rotational stretch $\rho_{A,k,\alpha}$ (which is a similarity transformation), so these two triangles are similar with a similarity coefficient of $k$. From this and from the statement \ref{TetivniPogoj} it follows that: $$\angle EDC=\angle EDA+\angle ADC=\angle CBA+\angle ADC=180^0,$$ which means that the points $E$, $D$ and $C$ are collinear. The distance $ED$ can be constructed (using Tales' theorem \ref{TalesovIzrek}), because: $$ED\cong B'C'=k\cdot BC=\frac{AD}{AB}\cdot BC=\frac{b\cdot d}{a}.$$ After the construction of the points $C$, $D$ and $E$, we can also construct the point $A$, because: $$\frac{AE}{AC}=\frac{AC'}{AC}=k,$$ so the point $A$ lies on the Apollonius circle $\mathcal{A}_{EC,k}$, as well as on the circle $k(D,d)$. \kdokaz \bnaloga\footnote{44. IMO Japan - 2003, Problem 4.} $ABCD$ is cyclic. The foot of the perpendicular from $D$ to the lines $AB$, $BC$, $CA$ are $P$, $Q$, $R$, respectively. Show that the angle bisectors of $\angle ABC$ and $\angle CDA$ meet on the line $AC$ if and only if $RP \cong RQ$. \enaloga \begin{figure}[!htb] \centering \input{sl.pod.7.3.IMO1.pic} \caption{} \label{sl.pod.7.3.IMO1.pic} \end{figure} \textbf{\textit{Solution.}} Without loss of generality, we assume that $\mathcal{B}(B,C,Q)$ and $\mathcal{B}(B,P,A)$ hold. According to \ref{SimpsPrem}, the points $P$, $Q$ and $R$ are collinear and lie on \index{premica!Simsonova} Simson's line (Figure \ref{sl.pod.7.3.IMO1.pic}). From the proof of this theorem it also follows that $\angle CDQ\cong \angle ADP$, which means that $CDQ$ and $ADP$ are similar right triangles, so we have: $$CD:CQ=AD:AP \textrm{ or } CD:AD=CQ:AP.$$ Let us denote $P_1=\mathcal{S}_A(P)$, from the previous relation we get: \begin{eqnarray} \label{eqn.pod.7.3.IMO1} CD:AD=CQ:AP_1. \end{eqnarray} ($\Rightarrow$) First, let us assume that the altitudes of the angles $ABC$ and $CDA$ intersect at the point $E$, which lies on the line $AC$. According to \ref{HarmCetSimKota}, we then have: $BC:BA=CE:EA=DC:DA$. From this and from the proven relation \ref{eqn.pod.7.3.IMO1} it follows that $BC:BA=CQ:AP_1$ or $BC:CQ=BA:AP_1$. Because $\mathcal{B}(B,C,Q)$ and $\mathcal{B}(B,A,P_1)$ hold, from the previous relation we get $BC:BQ=BA:BP_1$. By the converse of Tales' theorem \ref{TalesovIzrekObr}, $CA\parallel QP_1$ or $AR\parallel QP_1$, so by Tales' theorem \ref{TalesovIzrek} $PR:RQ=PA:AP_1=1:1$ or $PR\cong RQ$. ($\Leftarrow$) Now, let $PR\cong RQ$ hold. Because $A$ is the center of the line segment $PP_1$, $RA$ is the median of the triangle $PQP_1$. From this it follows that $AR\parallel QP_1$ or $CA\parallel QP_1$. By Tales' theorem, we have $BC:CQ=BA:AP_1$ or $BC:BA=CQ:AP_1$. From the proven relation \ref{eqn.pod.7.3.IMO1} we now get $BC:BA=CD:AD$. Let $E_1$ and $E_2$ be the intersections of the altitudes of the angles $ABC$ and $CDA$ with the line $AC$. If we use the previous relation and \ref{HarmCetSimKota}, we get:\\ $CE_1:AE_1=BC:BA=CD:AD=CE_2:AE_2$. Because the points $E_1$ and $E_2$ both lie on the line segment $AC$, $E_1=E_2$ (\ref{HarmCetEnaSamaDelitev}), so the altitudes of the angles $ABC$ and $CDA$ intersect on the line $AC$. \kdokaz %________________________________________________________________________________ \poglavje{The Right Triangle Altitude Theorem. Euclid's Theorems} \label{odd7VisinEvkl} In algebra and mathematical analysis, the concepts of arithmetic or geometric mean of two (or more) numbers are known. At this point, we will first define the so-called arithmetic and geometric mean of two line segments. We say that $x$ \index{aritmetična sredina daljic} \pojem{arithmetic mean} of line segments $a$ and $b$, if it holds: \begin{eqnarray} \label{eqnVisEvkl1} |x|=\frac{1}{2}\left( |a|+|b| \right). \end{eqnarray} Similarly, $y$ \index{geometrijska sredina daljic}\pojem{geometric (also geometrical) mean} of line segments $a$ and $b$, if it holds: \begin{eqnarray} \label{eqnVisEvkl2} |y|= \sqrt{|a|\cdot |b|}. \end{eqnarray} Therefore, arithmetic or geometric mean of two line segments is a line segment with a length that is equal to the arithmetic or geometric mean of the lengths of these two line segments. Relations \ref{eqnVisEvkl1} and \ref{eqnVisEvkl2} will often be written in a shorter form: \begin{eqnarray*} x=\frac{1}{2}\left( a+b \right)\hspace*{1mm} \textrm{ or }\hspace*{1mm} y=\sqrt{ab}. \end{eqnarray*} It is clear that we can construct the arithmetic mean of two given line segments in a very simple way. In this section we will derive the construction of the geometric mean of two given line segments. We will first prove the main statement that relates to right-angled triangles. \bizrek \index{izrek!višinski}\label{izrekVisinski} The altitude on the hypotenuse of a right-angled triangle is the geometric mean of the line segments into which it divides the hypotenuse.\\ (The right triangle altitude theorem) \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.6E.1.pic} \caption{} \label{sl.pod.7.6E.1.pic} \end{figure} \textbf{\textit{Proof.}} Let $ABC$ be a right-angled triangle with hypotenuse $AB$ and altitude $CC'$. We denote $v_c=|CC'|$, $b_1=|AC'|$ and $a_1=|BC'|$ (Figure \ref{sl.pod.7.6E.1.pic}). First, $\angle ACC'=90^0-\angle CAC'=\angle CBC'$ and $\angle BCC'=90^0-\angle CBC'=\angle CAC'$. From the similarity of the right-angled triangles $CC'A$ and $BC'C$ (statement \ref{PodTrikKKK}) it follows that: $$\frac{CC'}{BC'}=\frac{C'A}{C'C},$$ hence: \begin{eqnarray} \label{eqnVisinski} v_c^2=a_1\cdot b_1. \end{eqnarray} \kdokaz \bizrek \index{izrek!Evklidov}\label{izrekEvklidov} Each leg of a right-angled triangle is the geometric mean of the hypotenuse and the line segment of the hypotenuse adjacent to the leg.\\ (Euclid's\footnote{Starogrški filozof in matematik \index{Evklid}\textit{Evklid iz Aleksandrije} (3. st. pr. n. š.).} theorems) \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.6E.2.pic} \caption{} \label{sl.pod.7.6E.2.pic} \end{figure} \textbf{\textit{Proof.}} Let $ABC$ be a right-angled triangle with hypotenuse $AB$ and altitude $CC'$. We denote $a=|BC|$, $b=|AC|$, $c=|AB|$, $b_1=|AC'|$ and $a_1=|BC'|$ (Figure \ref{sl.pod.7.6E.2.pic}). As in the proof of statement \ref{izrekVisinski}, $\angle ACC'=90^0-\angle CAC'=\angle CBA$ and $\angle CAC'=\angle CAB$. From the similarity of the right-angled triangles $CC'A$ and $BCA$ (statement \ref{PodTrikKKK}) it follows that: $$\frac{CA}{BA}=\frac{C'A}{CA},$$ or: \begin{eqnarray} \label{eqnEvklidov1} b^2=b_1\cdot c. \end{eqnarray} Similarly, we have: \begin{eqnarray} \label{eqnEvklidov2} a^2=a_1\cdot c. \end{eqnarray} \kdokaz \bzgled \label{EvklVisPosl} If $c$ is the length of the hypotenuse, $a$ and $b$ the lengths of the legs and $v_c$ the length of the altitude on the hypotenuse of a right-angled triangle $ABC$, then $$c\cdot v_c=a\cdot b.$$ \ezgled \textbf{\textit{Proof.}} By multiplying relations \ref{eqnEvklidov1} and \ref{eqnEvklidov2} from Euclid's theorem \ref{izrekEvklidov} and inserting the relation \ref{eqnVisinski} from the altitude theorem \ref{izrekVisinski} we get: $$a^2b^2=a_1b_1 c^2=v^2c^2,$$ from which it follows that $ab=cv_c$. \kdokaz Now we will derive the predicted construction. \bzgled Construct a line segment $x$ that is the geometric mean of given line segments $a$ and $b$, i.e. $x=\sqrt{ab}$. \ezgled \textbf{\textit{Proof.}} We will derive the construction in two ways. \begin{figure}[!htb] \centering \input{sl.pod.7.6E.3.pic} \caption{} \label{sl.pod.7.6E.3.pic} \end{figure} \textit{1)} First, let's draw points $P$, $Q$ and $R$ so that $PQ\cong a$, $QR\cong b$ and $\mathcal{B}(P,Q,R)$ (Figure \ref{sl.pod.7.6E.3.pic}), then the circle $k$ over the diameter $PR$ and finally the point $X$ as the intersection of the circle $k$ with the rectangle line $PR$ at point $Q$. We prove that $x=QX$ is the desired line segment. By Theorem \ref{TalesovIzrKroz2}, $PXR$ is a right triangle with hypotenuse $PR$, $XQ$ is its altitude on this hypotenuse. By the altitude theorem \ref{izrekVisinski}, $x$ is the geometric mean of line segments $a$ and $b$. \textit{2)} Without loss of generality, let $a>b$ (if $a=b$ then also $x=a$). \begin{figure}[!htb] \centering \input{sl.pod.7.6E.4.pic} \caption{} \label{sl.pod.7.6E.4.pic} \end{figure} First, let's draw points $P$, $Q$ and $R$ so that $PQ\cong a$, $QR\cong b$ and $\mathcal{B}(P,R,Q)$ (Figure \ref{sl.pod.7.6E.4.pic}), then the circle $k$ over the diameter $PQ$ and finally the point $X$ as the intersection of the circle $k$ with the rectangle line $PQ$ at point $R$. We will now prove that $x=QX$ is the desired distance. According to the \ref{TalesovIzrKroz2} theorem, $PXQ$ is a right angled triangle with hypotenuse $PQ$, and $XR$ is its height on that hypotenuse. According to Euclid's \ref{izrekEvklidov} theorem, $x$ is the geometric mean of the distances $a$ and $b$. \kdokaz In the next example, we will consider another important construction. \bzgled \label{konstrKoren} Construct a line segment $x=e\sqrt{6}$, for a given line segment $e$. \ezgled \textbf{\textit{Proof.}} Similarly to the previous example, we will carry out the construction in two ways. \begin{figure}[!htb] \centering \input{sl.pod.7.6E.5.pic} \caption{} \label{sl.pod.7.6E.5.pic} \end{figure} \textit{1)} First, let's draw points $P$, $Q$ and $R$, such that: $PQ=3e$, $QR=2e$ and $\mathcal{B}(P,Q,R)$ (Figure \ref{sl.pod.7.6E.5.pic}), then the circle $k$ over the diameter $PR$ and finally the point $X$ as the intersection of the circle $k$ with the perpendicular line $PR$ at point $Q$. We will now prove that $x=QX$ is the desired distance. According to the \ref{TalesovIzrKroz2} theorem, $PXR$ is a right angled triangle with hypotenuse $PR$, and $XQ$ is its height on that hypotenuse. According to the \ref{izrekVisinski} theorem, $x$ is the geometric mean of the distances $PQ=3e$ and $QR=2e$, i.e. $x=\sqrt{3e\cdot 2e}=e\sqrt{6}$. \begin{figure}[!htb] \centering \input{sl.pod.7.6E.6.pic} \caption{} \label{sl.pod.7.6E.6.pic} \end{figure} \textit{2)} First, let's draw points $P$, $Q$ and $R$, such that: $PQ=3e$, $QR=2e$ and $\mathcal{B}(P,R,Q)$ (Figure \ref{sl.pod.7.6E.6.pic}), then the circle $k$ over the diameter $PQ$ and finally the point $X$ as the intersection of the circle $k$ with the perpendicular line $PQ$ at point $R$. We will prove that $x=QX$ is the desired distance. By the \ref{TalesovIzrKroz2} theorem, $PXQ$ is a right angled triangle with hypotenuse $PQ$, $XR$ is its height on that hypotenuse. By \ref{izrekEvklidov} theorem, $x$ is the geometric mean of the distances $PQ=3e$ and $QR=2e$, i.e. $x=\sqrt{3e\cdot 2e}=e\sqrt{6}$. \kdokaz In the previous example, we have described the process of constructing the distance $x=e\sqrt{n}$, where $e$ is a given distance, and $n\in \mathbb{N}$ is a given natural number. We always have two options for construction - using the altitude theorem or the Pythagorean theorem. But if $n$ is a composite number, for example $n=12$, we can choose $PQ=3e$ and $QR=4e$, or $PQ=6e$ and $QR=2e$. If $n$ is a prime number, for example $n=7$, it is most advantageous to choose $PQ=7e$ and $QR=e$. %________________________________________________________________________________ \poglavje{Pythagoras' Theorem} \label{odd7Pitagora} We will prove the famous Pythagorean theorem. \bizrek \label{PitagorovIzrek} \index{izrek!Pitagoras} The square of the length of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the lengths of both legs, i.e. for a right-angled triangle $ABC$ with the hypotenuse of length $c$ and the legs of lengths $a$ and $b$ is $$a^2+b^2=c^2$$ (Pythagoras'\footnote{Predpostavlja se, da je bil ta izrek znan že Egipčanom (pribl. 3000 let pr. n. š.) in Babiloncem (pribl. 2000 let pr. n. š.), toda starogrški filozof in matematik \index{Pitagora}\textit{Pitagora z otoka Samosa} (582--497 pr. n. š.) ga je verjetno prvi dokazal. Prvi pisni dokument dokaza Pitagorovega izreka je dal \index{Evklid}\textit{Evklid iz Aleksandrije} (3. st. pr. n. š.) v svojem delu ‘‘Elementi’’. Pri Starih Grkih se je Pitagorov izrek običajno nanašal na zvezo med ploščinami kvadratov nad stranicami pravokotnega trikotnika.} theorem) \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.6.0.pic} \caption{} \label{sl.pod.7.6.0.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.6.0.pic}) If we use the statement \ref{izrekEvklidov} and the labels from the proof of this statement, we get: $$a^2 + b^2 = c\cdot a_1 + c\cdot b_1 = c\cdot (a_1 + b_1) = c^2,$$ which was to be proven. \kdokaz In the previous form, Pythagorean Theorem applies to the squares of the lengths of the sides of a right-angled triangle. The second form, which applies to the relationship between the areas of the squares over the sides of a right-angled triangle, will be considered in section \ref{odd8PloTrik}. Pythagorean Theorem allows us to calculate the third side if the first two sides of a right-angled triangle are given. If we denote the length of the hypotenuse with $c$ and the length of the catheti with $a$ and $b$, from Pythagorean Theorem we get: \begin{eqnarray*} c&=&\sqrt{a^2+b^2}\\ a&=&\sqrt{c^2-b^2}\\ b&=&\sqrt{c^2-a^2} \end{eqnarray*} \bizrek \label{PitagorovIzrekObrat} \index{izrek!Obratni Pitagorov} Let $ABC$ be an arbitrary triangle. If $$|AC|^2+|BC|^2=|AB|^2,$$ then $ABC$ is a right-angled triangle with the right angle at the vertex $C$.\\ (Converse of Pythagorean Theorem) \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.6.0a.pic} \caption{} \label{sl.pod.7.6.0a.pic} \end{figure} \textbf{\textit{Proof.}} We mark: $c=|AB|$, $b=|AC|$ and $a=|BC|$ (Figure \ref{sl.pod.7.6.0a.pic}). So $a^2+b^2=c^2$ is true, which means: \begin{eqnarray} \label{eqnPitagObrat1} |AB|=c=\sqrt{a^2+b^2}. \end{eqnarray} Let $A'B'C'$ be such a right-angled triangle with a right angle at the vertex $C'$, that $A'C'\cong AC$ and $B'C'\cong BC$. By Pythagoras' theorem $|A'B'|^2= |A'C'|^2+|B'C'|^2=b^2+a^2$ or: \begin{eqnarray} \label{eqnPitagObrat2} |A'B'|=\sqrt{a^2+b^2}, \end{eqnarray} therefore from \ref{eqnPitagObrat1} and \ref{eqnPitagObrat2} it follows that $A'B'\cong AB$. The triangles $ABC$ and $A'B'C'$ are therefore congruent (by the \textit{SSS} \ref{SSS} theorem), from which it follows that $\angle ACB\cong \angle A'C'B'=90^0$, which means that $ABC$ is a right-angled triangle with a right angle at the vertex $C$. \kdokaz If the lengths of the sides $a$, $b$ and $c$ of the right-angled triangle $ABC$ (where $c$ is the length of the hypotenuse) are natural numbers, we say that the triple $(a,b,c)$ represents the \index{pitagorejska trojica}\pojem{Pythagorean triplet}. According to Pythagoras' \ref{PitagorovIzrek} theorem, for the Pythagorean triplet $(a, b, c)$ it is true that $a^2+b^2=c^2$. According to the converse Pythagoras' theorem, if for natural numbers $a$, $b$ and $c$ it is true that $a^2+b^2=c^2$, then $(a,b,c)$ is a Pythagorean triplet. The most famous Pythagorean triplet\footnote{This triplet was known to the Ancient Egyptians and Babylonians. It is also called the \index{trikotnik!egipčanski}\pojem{Egyptian triangle}, because the Ancient Egyptians used it to determine the right angle on the ground.} is $(3, 4, 5)$ (Figure \ref{sl.pod.7.6.0b.pic}), because $3^2+4^2=5^2$. We often encounter the Pythagorean triplets $(5, 12, 13)$ and $(7, 24, 25)$ as well. In reality, there are an infinite number of Pythagorean triples. From just one such triple $(a,b,c)$, we can get an infinite number of them: $(ka, kb, kc)$ (for any $k\in \mathbb{N}$). Obviously, in this case, we are dealing with similar right-angled triangles. We say that a \index{pitagorejska trojica!primitivna}\pojem{Pythagorean triple} is \pojem{primitive}, if $a$, $b$ and $c$ do not have a common divisor. It turns out that there are also an infinite number of primitive Pythagorean triples. We can calculate new Pythagorean triples using the following method. If $m$ and $n$ are any natural numbers ($m > n$), then: \begin{eqnarray*} a &=& m^2 - n^2,\\ b &=& 2mn,\\ c &=& m^2 + n^2. \end{eqnarray*} With a simple calculation, we can convince ourselves that $(a, b, c)$ really is a Pythagorean triple. We will use Pythagoras' theorem to calculate other shapes in the following sections. \bzgled \label{PitagorovPravokotnik} If $a$ and $b$ are the lengths of the sides and $d$ the lengths of the diagonal of a rectangle, then: $$d=\sqrt{a^2+b^2}.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.6.0c.pic} \caption{} \label{sl.pod.7.6.0c.pic} \end{figure} \textbf{\textit{Proof.}} Let $ABCD$ be any rectangle. We label: $a=|AB|$, $b=|BC|$ and $d=|AC|$ (Figure \ref{sl.pod.7.6.0c.pic}). Because $ABC$ is a right-angled triangle with hypotenuse $AC$, by Pythagoras' theorem \ref{PitagorovIzrek}, $d^2=a^2+b^2$ or $d=\sqrt{a^2+b^2}$. \kdokaz A direct consequence (for $a=b$) is the following statement. \bzgled \label{PitagorovKvadrat} If $a$ is the length of the side and $d$ the length of the diagonal of a square, then (Figure \ref{sl.pod.7.6.0d.pic}): $$d=a\sqrt{2}.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.6.0d.pic} \caption{} \label{sl.pod.7.6.0d.pic} \end{figure} The fact that $\sqrt{2}$ is an irrational number ($\sqrt{2}\notin \mathbb{Q}$) or that it cannot be written in the form of a fraction, results in the fact that the diagonal $d$ of a square and the side $a$ are not \index{proportional lines}\pojem{proportional} or \index{comparable lines}\pojem{comparable} lines\footnote{The ancient Greeks did not yet know of irrational numbers. The Pythagoreans - a philosophical school founded by the Greek mathematician \index{Pythagoras} \textit{Pythagoras of Samos} (582--497 BC) - were of the belief that everything is a number. They meant rational numbers and believed that any two lines are proportional. The first to discover the opposite was the Greek mathematician and philosopher from the Pythagorean school \index{Hipas}\textit{Hipas of Metapontum} (5th century BC). Several legends are associated with this discovery, which was therefore in complete contrast to Pythagorean philosophy - from his suicide to the fact that the Pythagoreans drowned him or that they simply excluded him from their circle.}. This means that there is no line $e$ as a unit such that for some natural numbers $n,m\in \mathbb{N}$ it would hold that $d=n\cdot e$ and $a=m\cdot e$. \bzgled \label{PitagorovEnakostr} Suppose that $a$ is the length of the side of an equilateral triangle $ABC$. If $v$ is the length of the altitude, $R$ the circumradius and $r$ the inradius of that triangle, then $$v=\frac{a\sqrt{3}}{2},\hspace{2mm} R=\frac{a\sqrt{3}}{3}, \hspace{2mm} r=\frac{a\sqrt{3}}{6}.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.6.0e.pic} \caption{} \label{sl.pod.7.6.0e.pic} \end{figure} \textbf{\textit{Proof.}} Let $AA'$ be the altitude of the triangle $ABC$ (Figure \ref{sl.pod.7.6.0e.pic}). In section \ref{odd3ZnamTock} we have proven that in an equilateral triangle all four characteristic points are equal ($O=S=T=V$). This means that the center $T$ of the triangle $ABC$ is at the same time the center of the circumscribed and inscribed circle of this triangle. Also, $T$ is at the same time the altitude point of this triangle, which means that $AA'$ is also the median, therefore the point $A'$ is the center of the side $BC$, or $|BA'|=\frac{a}{2}$. By the theorem \ref{tezisce} it follows that $AT:TA'=2:1$. Therefore: \begin{eqnarray} \label{eqnPitagEnakostr1} R=|TA|=\frac{2}{3}|AA'|=\frac{2}{3}v \hspace*{1mm} \textrm{ and } \hspace*{1mm} r=|TA'|=\frac{1}{3}|AA'|=\frac{1}{3}v. \end{eqnarray} Since $ABA'$ is a right triangle with hypotenuse $AB$, by the Pythagorean theorem \ref{PitagorovIzrek} \begin{eqnarray*} v=\sqrt{a^2-\left(\frac{a}{2} \right)^2}=\sqrt{\frac{3a^2}{4}}=\frac{a\sqrt{3}}{2}. \end{eqnarray*} If we use the relation from \ref{eqnPitagEnakostr1}, we get $R=\frac{a\sqrt{3}}{3}$ and $r=\frac{a\sqrt{3}}{6}$. \kdokaz \bzgled \label{PitagorovRomb} If $a$ is the length of the side and $e$ and $f$ the lengths of the diagonals of a rhombus, then $$\left(\frac{e}{2}\right)^2+\left(\frac{f}{2}\right)^2=a^2.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.6.0f.pic} \caption{} \label{sl.pod.7.6.0f.pic} \end{figure} \textbf{\textit{Proof.}} Let $S$ be the intersection of the diagonals of the rhombus $ABCD$ (Figure \ref{sl.pod.7.6.0f.pic}). By the theorem \ref{paralelogram} $S$ is the center of the diagonals $AC$ and $BD$. Therefore $|SA|=\frac{e}{2}$ and $|SB|=\frac{f}{2}$. By the theorem \ref{RombPravKvadr} the diagonals $AC$ and $BD$ are perpendicular, therefore $ASB$ is a right triangle with hypotenuse of length $|AB|=c$ and the cathets of lengths $|SA|=\frac{e}{2}$ and $|SB|=\frac{f}{2}$. The desired relation is now a direct consequence of the Pythagorean theorem \ref{PitagorovIzrek}. \kdokaz \bzgled If $c$ is the length of the hypotenuse, $a$ and $b$ the lengths of the legs and $v_c$ the length of the altitude on the hypotenuse of a right-angled triangle, then $$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{v_c^2}.$$ \ezgled \textbf{\textit{Proof.}} If we use the Pythagorean Theorem \ref{PitagorovIzrek} and the statement from Example \ref{EvklVisPosl}, we get: $$\frac{1}{a^2}+\frac{1}{b^2}=\frac{a^2+b^2}{a^2b^2} =\frac{c^2}{a^2b^2}=\frac{c^2}{c^2v_c^2}=\frac{1}{v_c^2},$$ which was to be proven. \kdokaz In Example \ref{konstrKoren} we considered two ways of constructing the line segment $x=e\sqrt{n}$ ($n\in \mathbb{N}$) using the altitude and Euclid's theorem. At this point we will solve this task with the help of the Pythagorean theorem - again in two ways. \bzgled \label{konstrKorenPit} Construct a line segment $e\sqrt{7}$, for a given line segment $e$. \ezgled \textbf{\textit{Solution.}} We will carry out the construction in two ways. \textit{1)} (Figure \ref{sl.pod.7.6.3.pic}). The idea is to design a right-angled triangle $ABC$ with the cathetus $a=e\sqrt{7}$, where the other cathetus $b=n\cdot e$ and the hypotenuse $c=m\cdot e$ ($n$ and $m$ are natural or at least rational numbers). By the Pythagorean Theorem \ref{PitagorovIzrek} we have $\left(e\sqrt{7}\right)^2+b^2=c^2$ or $m^2 e^2-n^2e^2=7e^2$ or equivalently $(m+n)(m-n)=7$. We get one solution of this equation by $m$ and $n$ if we solve the system: \begin{eqnarray*} && m+n=7\\ && m-n=1 \end{eqnarray*} By adding and subtracting the equations from this system, we therefore get one possibility $m=4$ and $n=3$; from this we get $c=4e$ and $b=3e$. This idea allows us to construct it. Let's first draw a right angled triangle $ABC$ with hypotenuse $AB=4e$ and cathetus $AC=3e$. According to Pythagoras' theorem \ref{PitagorovIzrek} $BC^2=\left(4e\right)^2-\left(3e\right)^2=7e^2$ or $BC=e\sqrt{7}$. \begin{figure}[!htb] \centering \input{sl.pod.7.6.3.pic} \caption{} \label{sl.pod.7.6.3.pic} \end{figure} \textit{2)} (Figure \ref{sl.pod.7.6.3.pic}). Let's first construct a right angled triangle $A_0A_1A_2$ with cathetuses $A_0A_1=A_1A_2=e$, then a right angled triangle $A_0A_2A_3$ with cathetuses $A_0A_2$ and $A_2A_3=e$. We continue the process and construct a sequence of right angled triangles $A_0A_{n-1}A_n$ with cathetuses $A_0A_{n-1}$ and $A_{n-1}A_n=e$. According to Pythagoras' theorem: \begin{eqnarray*} A_0A_2&=&\sqrt{A_0A_1^2+A_1A_2^2}=\sqrt{e^2+e^2}=e\sqrt{2}\\ A_0A_3&=&\sqrt{A_0A_2^2+A_2A_3^2}=\sqrt{2e^2+e^2}=e\sqrt{3}\\ &\vdots&\\ A_0A_n&=&\sqrt{A_0A_{n-1}^2+A_{n-1}A_n^2}=\sqrt{(n-1)e^2+e^2}=e\sqrt{n} \end{eqnarray*} We therefore get the desired distance from the sixth right angled triangle: $A_0A_7=e\sqrt{7}$. \kdokaz Some of the following examples relate to the use of Pythagoras' theorem in circles. \bzgled \label{PitagoraCofman} (Example from the book \cite{Cofman}) Let $a$ and $b$ be two circles of the same radius $r=1$, touching each other externally at the point $P$, and $t$ their common external tangent. If $c_1$, $c_2$, $\ldots$, $c_n$,... is a sequence of circles touching circles $a$ and $b$, the first of them touching the line $t$ and each circle from the sequence touches the previous one. Calculate the diameter of the circle $c_n$.\\ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.6.4.pic} \caption{} \label{sl.pod.7.6.4.pic} \end{figure} \textbf{\textit{Proof.}} Let us denote: with $A$ and $B$ the centers of the circles $a$ and $b$, with $A'$ and $B'$ the points of tangency of these circles with the tangent $t$, with $S_n$ the centers of the circles $c_n$ ($n =1,2,\ldots$), with $r_n$ their radii, with $d_n=2r_n$ their diameters, with $P_n$ the points of tangency of the circles $c_n$ and $c_{n-1}$ (or the circles $c_1$ and the line $t$) and with $S_n'$ and $X_n$ the perpendicular projections of the points $S_n$ or $P_n$ on the line $AA'$ (Figure \ref{sl.pod.7.6.4.pic}). Let $x_n=|AX_n|$. If we use the Pythagorean Theorem \ref{PitagoraovIzrek} for the triangle $AS_nS_n'$, we get $|AS_n|^2=|S_nS_n'|^2+|S_n'A|^2$ or $\left(1+r_n \right)^2=1+\left(x_n-r_n \right)^2$. From this, solving for $r_n$, we get: \begin{eqnarray} \label{eqnPitagoraCofman1} r_n=\frac{x_n^2}{2\left(1+x_n \right)}. \end{eqnarray} On the other hand, it is clear that $x_n-x_{n+1}=d_n=2r_n$. If we connect this with the relation \ref{eqnPitagoraCofman1}, we get: \begin{eqnarray*} x_{n+1}=x_n-2r_n=x_n-\frac{x_n^2}{1+x_n}=\frac{x_n}{1+x_n}. \end{eqnarray*} Because in this case $x_1=1$, by direct calculation we get $x_2=\frac{1}{2}$, $x_3=\frac{1}{3}$, $\ldots$ From this we intuitively conclude that: \begin{eqnarray} \label{eqnPitagoraCofman2} x_n=\frac{1}{n}. \end{eqnarray} We will prove the relation \ref{eqnPitagoraCofman2} formally - by using mathematical induction. The relation \ref{eqnPitagoraCofman2} is clearly true for $n=1$. We assume that the relation \ref{eqnPitagoraCofman2} is true for $n=k$ ($x_k=\frac{1}{k}$) and prove that from this it follows that it is true for $n=k+1$ ($x_{k+1}=\frac{1}{k+1}$): \begin{eqnarray*} x_{k+1}=\frac{x_k}{1+x_k}=\frac{\frac{1}{k}}{1+\frac{1}{k}}=\frac{1}{k+1}. \end{eqnarray*} With this we have proven that the relation \ref{eqnPitagoraCofman2} is true for every $n\in \mathbb{N}$. At the end, from relations \ref{eqnPitagoraCofman1} and \ref{eqnPitagoraCofman2} it follows: \begin{eqnarray*} d_n=2r_n=\frac{x_n^2}{1+x_n}= \frac{\left(\frac{1}{n}\right)^2}{1+\frac{1}{n}}=\frac{1}{n(n+1)} \end{eqnarray*} or \begin{eqnarray} \label{eqnPitagoraCofman3} d_n=\frac{1}{n(n+1)}, \end{eqnarray} which was needed to calculate. \kdokaz With the help of the statement from the previous example or relation \ref{eqnPitagoraCofman3} we can come to an interesting infinite sequence. Namely, $c_1, c_2, \ldots c_n,\ldots$ is an infinite sequence of circles and in this case it holds: \begin{eqnarray*} d_1+d_2+\cdots + d_n+\cdots=|PP_1|=1. \end{eqnarray*} Now, from relation \ref{eqnPitagoraCofman3} we get: \begin{eqnarray*} \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots +\frac{1}{n\cdot (n+1)}+\cdots=1, \end{eqnarray*} or \begin{eqnarray*} \sum_{n=1}^{\infty}\frac{1}{n (n+1)}=1. \end{eqnarray*} \bizrek \label{Jung} Let $\mathcal{P}=\{A_1,A_2,\ldots A_n\}$ be a finite set of points in the plane and $d=\max \{|A_iA_j|;\hspace*{1mm}1\leq i,j\leq n\}$. Prove that there exists a circle with radius $\frac{d}{\sqrt{3}}$, which contains all the points of this set (\index{izrek!Jungov}Jung's theorem\footnote{Nemški matematik \index{Jung, H. W. E.}\textit{H. W. E. Jung} (1876--1953) je leta 1901 dokazal splošno trditev za $n$-razsežni primer.}). \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.6.1a.pic} \caption{} \label{sl.pod.7.6.1a.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.6.1a.pic}) According to the statement \ref{lemaJung} it is enough to prove that for every three of these $n$ points there exists an appropriate circle. Let these points be $A$, $B$ and $C$. According to the assumption none of the sides of the triangle $ABC$ exceeds $d$. Without loss of generality let $BC$ be the longest side of this triangle. Then $|AC|, |AB| \leq |BC| \leq d$. We will consider two cases: \textit{1)} In the case when the triangle $ABC$ is rectangular or top-rectangular, the circle $\mathcal{K}'(S, \frac{|BC|}{2})$ (where $S$ is the center of the side $BC$) contains all three points $A$, $B$ and $C$. Because $\frac{|BC|}{2}\leq \frac{d}{2}<\frac{d}{\sqrt{3}}$, points $A$, $B$ and $C$ are also in the circle $\mathcal{K}(S, \frac{d}{\sqrt{3}})$. \textit{2)} If the triangle $ABC$ is acute, then $\angle BAC \geq 60^0$, because $BC$ is the longest side of this triangle. Let $BA'C$ be a right triangle, so that points $A$ and $A'$ are on the same side of the line $BC$, and $\mathcal{K}'(S,R)$ is a circle defined by the circumscribed circle of this triangle. Because of the condition $\angle BAC \geq 60^0$, point $A$ is either on the edge of the circle $\mathcal{K}'$ or its inner point. The radius of this circle is, according to the \ref{PitagorovEnakostr} equality, equal to $R=\frac{|BC|\sqrt{3}}{3}=\frac{|BC|}{\sqrt{3}}$. Because $|BC|\leq d$, points $A$, $B$ are also in the circle $\mathcal{K}(S, \frac{d}{\sqrt{3}})$. \kdokaz \bizrek Let $M$ and $N$ be common points of two congruent circles $k$ and $l$ with a radius $r$. Points $P$ and $Q$ are the intersections of these two circles with a line defined by their centres such that $P$ and $Q$ are on the same side of the line $MN$. Prove that $$|MN|^2 + |PQ|^2 = 4r^2.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.6.1.pic} \caption{} \label{sl.pod.7.6.1.pic} \end{figure} \textbf{\textit{Proof.}} Let $O$ and $S$ be the centers of these two circles and $\overrightarrow{v} = \overrightarrow{OS}$ (Figure \ref{sl.pod.7.6.1.pic}). The translation $\mathcal{T}_{\overrightarrow{v}}$ maps the circle $k$ to the circle $l$ and the point $M$ to some point $M'$. Since the point $M$ lies on the circle $k$, its image $M'$ lies on the circle $l$. For the same reasons, $\mathcal{T}_{\overrightarrow{v}}(P )=Q$. Therefore, $\overrightarrow{MM'} = \overrightarrow{v} = \overrightarrow{OS} = \overrightarrow{PQ}$, which means that the lines $MM'$ and $OS$ are parallel and $MM'\cong OS\cong PQ$. Because the line $MN$ is perpendicular to the line $OS$ ($OS$ is the perpendicular bisector of the segment $MN$), the line $MN$ is also perpendicular to the parallel $MM'$ of the line $OS$. Therefore, $\angle NMM'$ is a right angle, so $NM'$ is the diameter of the circle $l$. If we use the Pythagorean Theorem \ref{PitagorovIzrek}, we get: $$|MN|^2 + |PQ|^2 = |MN|^2 + |MM'|^2 = |NM'|^2 = 4r^2,$$ which was to be proven. \kdokaz \bzgled Lines $b$ and $c$ and a point $A$ in the same plane are given. Construct a square $ABCD$ such that the vertices $B$ and $C$ lie on the lines $b$ and $c$, respectively. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.6.2.pic} \caption{} \label{sl.pod.7.6.2.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.6.2.pic}) We use the rotational dilation $\rho_{A,\sqrt{2},45^0}$. Since $\rho_{A,\sqrt{2},45^0}(B)=C$, we can plan the point $C$ from the condition $C\in c\cap\rho_{A,\sqrt{2},45^0}(b)$. \kdokaz \bnaloga\footnote{1. IMO, Romania - 1959, Problem 4.} Construct a right triangle with given hypotenuse $c$ such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle. \enaloga \begin{figure}[!htb] \centering \input{sl.pod.7.6.IMO1.pic} \caption{} \label{sl.pod.7.6.IMO1.pic} \end{figure} \textbf{\textit{Solution.}} Let's assume that $ABC$ is a right angled triangle with hypotenuse $AB\cong c$ and its center of gravity is on the hypotenuse $CM=t_c$ geometrical middle of the cathetus $AC=b$ and $BC=a$ or $t_c=\sqrt{a b}$ (Figure \ref{sl.pod.7.6.IMO1.pic}). From the \ref{TalesovIzrKroz} follows $t_c=\frac{1}{2}\cdot c$. If we use the Pythagorean theorem (\ref{PitagorovIzrek}), we get: $$(a+b)^2=a^2+b^2+2ab=c^2+2t^2_c=c^2+2\left( \frac{1}{2}\cdot c\right)^2= \frac{6}{4}\cdot c^2$$ or $$a+b=\frac{\sqrt{6}}{2}\cdot c.$$ The latter relation allows the construction. First, we describe the auxiliary construction of the distance, which has the length $\frac{\sqrt{6}}{2}\cdot c$. First, we draw the line $UV=c$ and its center $W$, then the right angled triangle $TUW$ ($TU=c$ and $\angle TUW=90^0$) and the right angled triangle $TWZ$ ($WZ=\frac{1}{2}\cdot c$ and $\angle TWZ=90^0$). Now we will construct the triangle $ABC$. Draw the line $DB\cong TZ$ and $\angle BDX=45^0$. The point $A$ will be one of the intersections of the circle $k(B,c)$ and the arc $DX$. In the end, draw the point $C$ as the intersection of the line $BD$ with the line $s_{DA}$ of the line $DA$. We prove that the triangle $ABC$ satisfies all the conditions of the task. First, from the construction it follows that $A\in k(B,c)$, which means that $AB=c$. The point $C$ by construction lies on the line $DA$, so $DC\cong AC$ or (\ref{enakokraki}) $\angle DAC\cong \angle ADC =45^0$. From the triangle $ACD$ is by \ref{VsotKotTrik} $\angle ACD=90^0$. Because by construction $\mathcal{B}(D,C,B)$, is also $\angle ACB=90^0$ and $ABC$ is a right angled triangle with hypotenuse $AB=c$. We will now prove that its centroid $CM=t_c$ is the geometrical mean of its cathets $AC=b$ and $BC=a$ or $t_c=\sqrt{a b}$. By construction, the triangles $TUW$ and $TWZ$ are right angled and $UW=WZ=\frac{1}{2}\cdot c$ and $TW=c$, so from Pythagoras' theorem it follows: \begin{eqnarray*} |TZ|^2&=&|TW|^2+|WZ|^2=\\ &=& |TU|^2+|UW|^2+|WZ|^2=\\ &=& c^2+\left(\frac{c}{2}\right)^2+\left(\frac{c}{2}\right)^2=\\ &=& \frac{6}{4}\cdot c^2 \end{eqnarray*} By construction, $BD\cong TZ$, so $|BD|^2=\frac{6}{4}\cdot c^2$. Therefore: \begin{eqnarray*} \frac{6}{4}\cdot c^2&=&|BD|^2=\left(|BC|+|CD|\right)^2=\\ &=&\left(|BC|+|CA|\right)^2=(a+b)^2=\\ &=&a^2+b^2+2ab=\\ &=&c^2+2ab\\ \end{eqnarray*} From this it follows that $\left(\frac{c}{2} \right)^2=ab$. Because according to the theorem \ref{TalesovIzrKroz} $t_c=\frac{c}{2}$, in the end we get $t_c^2=ab$. The number of solutions to the task depends on the number of intersections of the circle $k(B,c)$ and the line $DX$. \kdokaz \bnaloga\footnote{23. IMO, Hungary - 1982, Problem 5.} The diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$ are divided by the inner points $M$ and $N$, respectively, so that $$\frac{AM}{AC}=\frac{CN}{CE}=r.$$ Determine $r$ if $B$, $M$ and $N$ are collinear. \enaloga \begin{figure}[!htb] \centering \input{sl.skk.4.2.IMO2.pic} \caption{} \label{sl.skk.4.2.IMO2.pic} \end{figure} \textbf{\textit{Solution.}} Let $O$ be the centre of the regular hexagon $ABCDEF$ (Figure \ref{sl.skk.4.2.IMO2.pic}) and $a$ the length of its side. From the assumption $\frac{AM}{AC}=\frac{CN}{CE}=r$ and from the fact $AC\cong CE$ it follows that $CM\cong EN$ and $AM\cong CN$. The triangles $ACB$ and $CED$ are congruent (the \textit{SSS} theorem \ref{SSS}), therefore $\angle ACB\cong\angle CED$. This means that the triangles $BCM$ and $DEN$ are also congruent (the \textit{ASA} theorem \ref{KSK} or $\angle BMC\cong \angle DNE$. The triangle $ACE$ is regular, therefore $\angle ACE=60^0$. So: \begin{eqnarray*} \angle BND&=&\angle BNC+\angle CND\\ &=&\angle BMC- \angle ECA+180^0-\angle DNE=120^0 \end{eqnarray*} Because also $\angle BOD=120^0$ and $CO\cong CD\cong CB$, the point $N$ lies on the circle $k(C,CO)$. Therefore $AM\cong CN\cong CB=a$. Let $v$ be the length of the altitude of the regular triangle $OCD$ and use the Pythagorean theorem: $$r=\frac{CN}{CE}=\frac{a}{2v}=\frac{\sqrt{3}}{3},$$ which had to be calculated. \kdokaz Let $ABC$ be an acute-angled triangle with circumcentre $O$. Let $P$ on $BC$ be the foot of the altitude from $A$. Suppose that $\angle BCA>\angle ABC+30^0$. Prove that $\angle CAB+\angle COP<90^0$. \bnaloga\footnote{42. IMO, USA - 2001, Problem 1.} Let $ABC$ be an acute-angled triangle with circumcentre $O$. Let $P$ on $BC$ be the foot of the altitude from $A$. Suppose that $\angle BCA>\angle ABC+30^0$. Prove that $\angle CAB+\angle COP<90^0$. \enaloga \begin{figure}[!htb] \centering \input{sl.pod.7.6.IMO3.pic} \caption{} \label{sl.pod.7.6.IMO3.pic} \end{figure} \textbf{\textit{Solution.}} Let $A_1$ be the centre of the side $BC$, $P'$ and $A'$ the other intersections of the lines $AP$ and $AO$ with the circumscribed circle $k(O,R)$ of the triangle $ABC$, and $Q$ the orthogonal projection of the point $A'$ on the line $BC$ (Figure \ref{sl.pod.7.6.IMO3.pic}). Let $\alpha$, $\beta$ and $\gamma$ be the internal angles of the triangle $ABC$ at the vertices $A$, $B$ and $C$. Since $AA'$ is the diameter of the circle $k$, $\angle AP'A'=90^0$ (by the statement \ref{TalesovIzrKroz}). This means that $PP'A'Q$ is a rectangle and it holds: \begin{eqnarray} PQ\cong A'P'. \label{pod.7.6.IMO31} \end{eqnarray} The point $O$ is the center of the diameter $AA'$, $A_1$ is by Tales' statement \ref{TalesovIzrek} the center of the line $PQ$. The point $A_1$ is thus the common center of the lines $BC$ and $PQ$. By the statement \ref{TockaNbetagama} it is also $\angle A'AP=\angle OAP=\gamma-\beta$. From the given condition $\angle BCA>\angle ABC+30^0$ or $\gamma-\beta>30^0$ it thus follows: \begin{eqnarray} \angle A'AP>30^0. \label{pod.7.6.IMO32} \end{eqnarray} By the statement \ref{SredObodKot} it is $\angle BOC=2\alpha$. Since $BOC$ is an isosceles triangle with the base $BC$, $\angle OCB=\frac{1}{2}\left( 180^0-2\alpha\right)=90^0-\alpha$ (by the statement \ref{enakokraki} and \ref{VsotKotTrik}). The condition $\angle CAB+\angle COP<90^0$ or $\angle COP<90^0-\alpha$, which we want to prove, is thus equivalent to the condition $\angle COP<\angle OCB=\angle OCP$ (it holds $\angle OCB=\angle OCP$, because $ABC$ is an acute-angled triangle and $\mathcal{B}(B,P,C)$ is obtuse-angled). The latter is by the statement \ref{vecstrveckot} equivalent to the condition $CP|BC|^2. \label{pod.7.6.IMO3b} \end{eqnarray} From inequality \ref{pod.7.6.IMO32} and from \ref{SredObodKot} it follows that $\angle A'OP'>60^0$, so in the equilateral triangle $OA'P'$ $\angle A'OP'$ is the largest angle (\ref{enakokraki} and \ref{VsotKotTrik}). This means that $A'P'>OA'=R$ is true. From \ref{pod.7.6.IMO31} it follows that $PQ\cong A'P'>R$. In the end: \begin{eqnarray*} \hspace*{-1.5mm} 4|OA_1|^2+2|BC|\cdot|PQ|>2|BC|\cdot|PQ|>2|BC|\cdot R=|BC|\cdot 2R>|BC|^2, \end{eqnarray*} which had to be proven. \kdokaz %_______________________________________________________________________________ \poglavje{Napoleon Triangles} \label{odd7Napoleon} We will first define two triangles, which we will adapt to any triangle $ABC$. We will construct equilateral triangles on the outside of its sides. The centers of these triangles are the vertices of the so-called \index{trikotnik!Napoleonov zunanji}\pojem{outer Napoleon triangle} (Figure \ref{sl.pod.7.7n.1.pic}). If we construct equilateral triangles on the inside in the same way, we get the so-called \index{trikotnik!Napoleonov notranji}\pojem{inner Napoleon triangle}. \begin{figure}[!htb] \centering \input{sl.pod.7.7n.1.pic} \caption{} \label{sl.pod.7.7n.1.pic} \end{figure} \bizrek The outer and inner Napoleon triangles are equilateral. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.7n.2.pic} \caption{} \label{sl.pod.7.7n.2.pic} \end{figure} \textbf{\textit{Proof.}} Let $P$,$ Q$ and $R$ be the centers of the regular triangles, which are constructed on the outside of the sides of any triangle $ABC$, so $PQR$ is the outer Napoleon triangle of triangle $ABC$ (Figure \ref{sl.pod.7.7n.2.pic}). We will prove that $PQR$ is an equilateral triangle. Let: $$\mathcal{I}=\mathcal{R}_{R,120^0}\circ \mathcal{R}_{P,120^0}\circ\mathcal{R}_{Q,120^0}.$$ Since $120^0+120^0+120^0=360^0$, the composite $\mathcal{I}$ is either the identity or a translation (statement \ref{rotacKomp2rotac}). However, the latter option is ruled out, since $\mathcal{I}(A)=A$. Therefore $\mathcal{I}=\mathcal{E}$ or: $$\mathcal{R}_{R,120^0}\circ \mathcal{R}_{P,120^0}\circ\mathcal{R}_{Q,120^0}=\mathcal{E}.$$ It follows that: $$\mathcal{R}_{R,120^0}\circ\mathcal{R}_{P,120^0}= \mathcal{R}^{-1}_{Q,120^0}=\mathcal{R}_{Q,-120^0}=\mathcal{R}_{Q,240^0}.$$ Let $Q'$ be the third vertex of the equilateral triangle $PQ'R$. By statement \ref{rotacKomp2rotac}, we have: $$\mathcal{R}_{R,120^0}\circ\mathcal{R}_{P,120^0}=\mathcal{R}_{Q',240^0}.$$ Therefore $\mathcal{R}_{Q,240^0}=\mathcal{R}_{Q',240^0}$ or $Q=Q'$, which means that $PQR$ is an equilateral triangle. In a similar way, we prove (by using rotations for an angle of $-120^0$) that the inner Napoleon triangle is also equilateral. \kdokaz We shall now prove an additional property of the two Napoleon triangles. \bizrek The outer and inner Napoleon triangles of an arbitrary triangle $ABC$ have the same centre, which is at the same time the centroid of the triangle $ABC$. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.7n.3.pic} \caption{} \label{sl.pod.7.7n.3.pic} \end{figure} \textbf{\textit{Proof.}} Let $PQR$ and $P'Q'R'$ be the outer and inner Napoleon triangles of the triangle $ABC$ (Figure \ref{sl.pod.7.7n.3.pic}). We denote by $Y$ and $A_1$ the centres of the lines $PR$ and $BC$. We shall first prove: \begin{enumerate} \item The triangles $QP'C$ and $RBP'$ are congruent and similar to the triangle $ABC$. \item The quadrilaterals $ARP'Q$ and $CP'RQ'$ are parallelograms. \item $\overrightarrow{YA_1}=\frac{1}{2}\overrightarrow{AQ}$. \end{enumerate} \textit{1}) First, from $\angle BCP'\cong \angle ACQ=30^0$ it follows that $\angle P'CQ\cong\angle BCA$ (with the rotation $\mathcal{R}_{C.30^0}$, the angle $\angle P'CQ$ is mapped into the angle $\angle BCA$). It is also true that $\frac{CQ}{CA}=\frac{CP'}{CB}=\frac{\sqrt{3}}{3}$ (by the Pythagorean equality \ref{PitagorovEnakostr}, because $CQ$ and $CP'$ are the radii of the circumscribed circles of two isosceles triangles with sides $CA$ and $CB$), so the triangles $ABC$ and $QP'C$ are similar (by the similarity theorem \ref{PodTrikSKS}) with the similarity coefficient $k=\frac{\sqrt{3}}{3}$. Similarly, the triangles $ABC$ and $RBP'$ are also similar with the same coefficient of similarity, which means that the triangles $QP'C$ and $RBP'$ are congruent. \textit{2}) From the congruence of the triangles $QP'C$ and $RBP'$ and from the fact that the triangles $RAB$ and $QAC$ are of the same size, it follows: $QP'\cong RB\cong RA$ and $P'R\cong CQ\cong QA$, so the quadrilateral $ARP'Q$ is a parallelogram. Similarly, we prove that the quadrilateral $CP'RQ'$ is a parallelogram. \textit{3}) The line $YA_1$ is the median of the triangle $RPP'$ with the base $RP'$, so by the theorems \ref{srednjicaTrikVekt} and \ref{vektParalelogram}: $$\overrightarrow{YA_1}=\frac{1}{2}\overrightarrow{RP'}=\frac{1}{2}\overrightarrow{AQ}$$ We only need to prove the initial claim. Because $\overrightarrow{YA_1}=\frac{1}{2}\overrightarrow{AQ}$, the centroid $AA_1$ of the triangle $ABC$ and the centroid $QY$ of the triangle $PQR$ intersect in a point which they divide in the ratio $2:1$. So this is the common centroid of the triangles $ABC$ and $PQR$ (by the theorem \ref{tezisce}). Similarly, it can be proven that the triangles $ABC$ and $P'Q'R'$ also have a common centroid. \kdokaz %________________________________________________________________________________ \poglavje{Ptolemy's Theorem} \label{odd7Ptolomej} We will now prove a very important theorem which is related to the cord quadrilaterals. \bizrek \label{izrekPtolomej} \index{izrek!Ptolomejev}(Ptolemy’s\footnote{\index{Ptolomej Aleksandrijski}\textit{Ptolomej Aleksandrijski} (2. st.) je dokazal ta izrek v svojemu delu ‘‘Veliki zbornik’’.} theorem) If $ABCD$ is a cyclic quadrilateral then the product of the lengths of its diagonals is equal to the sum of the products of the lengths of the pairs of opposite sides, i.e. $$|AC|\cdot |BD|=|AB|\cdot |CD|+|BC|\cdot |AD|.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.7.1.pic} \caption{} \label{sl.pod.7.7.1.pic} \end{figure} \textbf{\textit{Proof.}} Brez škode za splošnost predpostavimo, da je $\angle CBD\leq\angle ABD$. Naj bo $L$ takšna točka diagonale $AC$, da velja $\angle ABL\cong \angle CBD$ (Figure \ref{sl.pod.7.7.2.pic}). Ker je še $\angle BAL=\angle BAC\cong\angle BDC$ (izrek \ref{ObodObodKot}), sta trikotnika $ABL$ in $DBC$ podobna (izrek \ref{PodTrikKKK}), zato je $AB:DB=AL:DC$ oziroma: \begin{eqnarray} \label{eqnPtolomej1} |AB|\cdot |CD|=|AL|\cdot |BD|. \end{eqnarray} Ker je $\angle BCL=\angle BCA\cong\angle BDA$ (izrek \ref{ObodObodKot}) in $\angle LBC\cong\angle ABD$ (iz predpostavke $\angle CBD\cong\angle ABL$), velja tudi $\triangle BCL\sim \triangle BDA$ (izrek \ref{PodTrikKKK}), zato je $BC:BD=CL:DA$ oziroma: \begin{eqnarray} \label{eqnPtolomej2} |BC|\cdot |AD|=|CL|\cdot |BD|. \end{eqnarray} Po seštevanju relacij \ref{eqnPtolomej1} in \ref{eqnPtolomej2} dobimo iskano enakost. \kdokaz Trditev iz zgleda \ref{zgledTrikABCocrkrozP} bomo na tem mestu dokazali na bolj enostaven način - z uporabo Ptolomejevega izreka. \bzgled \label{zgledTrikABCocrkrozPPtol} Let $k$ be the circumcircle of a regular triangle $ABC$. If $P$ is an arbitrary point lying on the shorter arc $BC$ of the circle $k$, then $$|PA|=|PB|+|PC|.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.7.2a.pic} \caption{} \label{sl.pod.7.7.2a.pic} \end{figure} \textbf{\textit{Solution.}} The quadrilateral $ABPC$ is a cyclic quadrilateral (Figure \ref{sl.pod.7.7.2a.pic}), therefore from Ptolomey's theorem \ref{izrekPtolomej} it follows that $|PA|\cdot |BC|=|PB|\cdot |AC|+|PC|\cdot |AB|$. Because $ABC$ is an equilateral triangle, i.e. $|BC|=|AC|= |AB|$, from the previous relation it follows that $|PA|=|PB|+|PC|$. \kdokaz Another application of Ptolomey's theorem relates to an isosceles trapezium. \bzgled Let $a$ and $c$ be the bases, $b$ the leg and $d$ the diagonal isosceles trapezium. Prove that $$ac=d^2-b^2.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.7.4.pic} \caption{} \label{sl.pod.7.7.4.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.7.4.pic}) By theorem \ref{trapezTetivEnakokr} the isosceles trapezium is cyclic, therefore we can use Ptolomey's theorem \ref{izrekPtolomej} and obtain $d^2=ac+b^2$ or $ac=d^2-b^2$. \kdokaz It is interesting that if we take the special case $a=c$ in the previous theorem (Figure \ref{sl.pod.7.7.4.pic}), when the isosceles trapezium is a rectangle, we obtain the formula for the diagonal of a rectangle $d^2=a^2+b^2$, which is otherwise a direct consequence of Pythagoras' theorem \ref{PitagorovIzrek}. In this sense, we can also say that Pythagoras' theorem is a consequence of Ptolomey's theorem, if we complete the right-angled triangle to a rectangle. \begin{figure}[!htb] \centering \input{sl.pod.7.7.5.pic} \caption{} \label{sl.pod.7.7.5.pic} \end{figure} Of course, the use of Ptolemy's theorem in a square leads to the well-known relationship (theorem \ref{PitagorovKvadrat}) between the diagonal $d$ and the side $a$ of the square: $d^2=2a^2$ or $d=a\sqrt{2}$ (Figure \ref{sl.pod.7.7.5.pic}). We know that for every regular $n$-gon there is a circumscribed circle \ref{sredOcrtaneKrozVeck}. This allows us to use Ptolemy's theorem, if in a regular $n$-gon (in the case $n>4$) we choose the appropriate four vertices. Now we will use this idea in the cases $n=5$ and $n=7$. \bzgled \label{PtolomejPetkotnik} Let $a$ be the side and $d$ the diagonal of a regular pentagon ($5$-gon). Prove that $$d=\frac{1+\sqrt{5}}{2}a.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.7.3.pic} \caption{} \label{sl.pod.7.7.3.pic} \end{figure} \textbf{\textit{Solution.}} Let $A_1A_2A_3A_4A_5$ be a regular pentagon with side $a$ and diagonal $d$ (Figure \ref{sl.pod.7.7.3.pic}). If we use Ptolemy's theorem \ref{izrekPtolomej} for the quadrilateral $A_1A_2A_3A_5$, we get: $d^2=ad+a^2$ or $d^2-ad-a^2=0$. The positive solution of this quadratic equation in $d$ is precisely $d=\frac{1+\sqrt{5}}{2}a$. \kdokaz The previous example allows us to construct a regular pentagon with a side that is congruent to the line $a$. Indeed, if $ABCDE$ is the desired regular pentagon, we can first plan the triangle $ABC$, where $AB\cong BC\cong a$ and $AC =\frac{1+\sqrt{5}}{2}a$. To construct $\frac{1+\sqrt{5}}{2}a$, we use the Pythagorean Theorem (\ref{PitagorovIzrek}) for a right triangle with sides $a$ and $2a$ - its hypotenuse measures $a\sqrt{5}$. We add this hypotenuse to the line $a$ and divide the resulting line into two lines. The center of the drawn circle of the regular pentagon $ABCDE$ is also the center of the drawn circle of the triangle $ABC$ (Figure \ref{sl.pod.7.7.3a.pic}). \begin{figure}[!htb] \centering \input{sl.pod.7.7.3a.pic} \caption{} \label{sl.pod.7.7.3a.pic} \end{figure} In a similar way, we can construct a regular pentagon that is inscribed in a given circle. We do this by first constructing a regular pentagon with any side, and then using the central extension (Figure \ref{sl.pod.7.7.3a.pic}). We will talk more about the construction of regular $n$-gons in section \ref{odd9LeSestilo}. \bzgled Let $a$ be the side, $d$ the shorter and $D$ the longer diagonal of a regular heptagon ($7$-gon). Prove that $$\frac{1}{a}=\frac{1}{D}+\frac{1}{d}$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.7.2.pic} \caption{} \label{sl.pod.7.7.2.pic} \end{figure} \textbf{\textit{Solution.}} Let $A_1A_2A_3A_4A_5A_6A_7$ be a regular heptagon and $k$ the drawn circle (Figure \ref{sl.pod.7.7.2.pic}). The quadrilateral $A_1A_2A_3A_5$ is a tetragon, so from Ptolomey's theorem \ref{izrekPtolomej} it follows that $ad+aD=dD$ or $\frac{1}{a}=\frac{1}{D}+\frac{1}{d}$. \kdokaz If we use the Ptolomey Theorem multiple times, we get one generalization of task \ref{zgledTrikABCocrkrozPPtol}: \bzgled Let $P$ be an arbitrary point of the shorter arc $A_1A_{2n+1}$ of the circumcircle of a regular polygon $A_1A_2\cdots A_{2n+1}$. If we denote $d_i=PA_i$ ($i\in \{1,2,\cdots , 2n+1 \}$), prove that (Figure \ref{sl.pod.7.7.6.pic}) $$d_1+d_3+\cdots +d_{2n+1}=d_2+d_4+\cdots +d_{2n}.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.7.6.pic} \caption{} \label{sl.pod.7.7.6.pic} \end{figure} %________________________________________________________________________________ \poglavje{Stewart's Theorem} \label{odd7Stewart} The next theorem refers to a very interesting and important metric property of a triangle. \bizrek \label{StewartIzrek} \index{izrek!Stewartov} (Stewart's\footnote{\index{Stewart, M.}\textit{M. Stewart} (1717--1785), English mathematician, who in 1746 proved and published this theorem. His teacher - English mathematician \index{Simson, R.}\textit{R. Simson} (1687--1768) - introduced him to the theorem, which he published only in 1749. It is assumed that the theorem was known to \index{Arhimed} \textit{Archimedes of Syracuse} (3rd century BC) already.} theorem) If $X$ is an arbitrary point of the side $BC$ of a triangle $ABC$, then $$|AX|^2=\frac{|BX|}{|BC|}|AC|^2+\frac{|CX|}{|BC|}|AB|^2-|BX|\cdot |CX|.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.8.1.pic} \caption{} \label{sl.pod.7.8.1.pic} \end{figure} \textbf{\textit{Proof.}} Let $a$, $b$ and $c$ denote the lengths of the sides $BC$, $AC$ and $AB$, and $p$, $q$ and $x$ the lengths of the lines $BX$, $CX$ and $AX$, in order. With $A'$ we denote the intersection point of the altitude $v_a$ from the vertex $A$ of the triangle $ABC$ (Figure \ref{sl.pod.7.8.1.pic}). We assume that $\mathcal{B}(B,A',C)$. We first prove the claim for the case when $X=A'$. From Pythagoras' theorem \ref{PitagorovIzrek} it follows that $v_a^2 =b^2- q^2$ and $v_a^2 =c^2- p^2$. By multiplying the first equality by $p$ and the second by $q$ and adding the resulting relations and taking into account that $p+q=a$, we obtain $av_a^2=pb^2+qc^2-apq$ or: $$v_a^2=\frac{p}{a}b^2+\frac{q}{a}c^2-pq,$$ which means that in the case of altitudes the claim is true. Let now $X\neq A'$. The line $v_a$ is the altitude of the triangles $ABX$ and $ABC$ from the vertex $A$. Without loss of generality, we assume that $\mathcal{B}(B,A',X)$. If we denote the length of the line $BA'$ with $y$ and use the already proven part of the claim for altitudes, we obtain: \begin{eqnarray*} v_a^2&=&\frac{y}{p}x^2+\frac{p-y}{p}c^2-y(p-y),\\ v_a^2&=&\frac{y}{a}b^2+\frac{a-y}{a}c^2-y(a-y). \end{eqnarray*} If we equate the right sides of these two equalities and simplify, we obtain: $$x^2=\frac{p}{a}b^2+\frac{q}{a}c^2-pq,$$ which means that the claim is also true in the case when $X\neq A'$. The proof of the altitude $v_a$ is similar also in the case when $\mathcal{B}(B,A',C)$ is not true, only then we obtain $v_a^2=\frac{p}{a}b^2+\frac{q}{a}c^2+pq$. \kdokaz Stewart's theorem \ref{StewartIzrek} can also be written in another form: \bizrek \label{StewartIzrek2} Let $a=|BC|$, $b=|AC|$ and $c=|AB|$ be the length of the sides of a triangle $ABC$. If $X$ is the point that divides the side $BC$ of this triangle in the ratio $n:m$ and $x=|AX|$ then $$x^2=\frac{n}{n+m}b^2+\frac{m}{n+m}c^2-\frac{mn}{(m+n)^2}bc.$$ \eizrek The most well-known use of Stewart's theorem is for the centroids of a triangle. \bizrek \label{StwartTezisc} If $a$, $b$ and $c$ are the sides and $t_a$ the triangle median on the side $a$, then $$t_a^2=\frac{1}{2}b^2+\frac{1}{2}c^2-\frac{1}{4}a^2.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.8.2.pic} \caption{} \label{sl.pod.7.8.2.pic} \end{figure} \textbf{\textit{Proof.}} The theorem is a direct consequence of Stewart's theorem \ref{StewartIzrek2}, since in the case of the centroid $n=m=1$ (Figure \ref{sl.pod.7.8.2.pic}). \kdokaz A direct consequence is the following theorem. \bzgled \label{StwartTezisc2} If $a$, $b$ and $c$ are the sides and $t_a$, $t_b$ and $t_c$ the triangle medians on those sides, respectively, then $$t_a^2+t_b^2+t_c^2=\frac{3}{4}\left(a^2+ b^2+c^2\right).$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.8.3.pic} \caption{} \label{sl.pod.7.8.3.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.8.3.pic}) If we use the proven relation from the example \ref{StwartTezisc} three times, we get: \begin{eqnarray*} t_a^2&=&\frac{1}{2}b^2+\frac{1}{2}c^2-\frac{1}{4}a^2\\ t_b^2&=&\frac{1}{2}a^2+\frac{1}{2}c^2-\frac{1}{4}b^2\\ t_c^2&=&\frac{1}{2}a^2+\frac{1}{2}b^2-\frac{1}{4}c^2 \end{eqnarray*} By adding all three equations we get the desired relation. \kdokaz In the case of an isosceles triangle, the equality on the right side of Stewart's theorem is simplified. \bzgled \label{StewartEnakokraki} If $ABC$ is an isosceles triangle with the base $BC$ and $X$ an arbitrary point of this base, then $$|AX|^2=|AB|^2-|BX|\cdot |CX|.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.8.8.pic} \caption{} \label{sl.pod.7.8.8.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.8.8.pic}) Therefore, if $AB\cong AC$, directly from Stewart's theorem (\ref{StewartIzrek}) it follows: \begin{eqnarray*} |AX|^2 &=& \frac{|BX|}{|BC|}|AC|^2+\frac{|CX|}{|BC|}|AB|^2-|BX|\cdot |CX|=\\ &=&\left(\frac{|BX|}{|BC|}+\frac{|CX|}{|BC|}\right)|AB|^2-|BX|\cdot |CX|=\\ &=&\frac{|BX|+|CX|}{|BC|}|AB|^2-|BX|\cdot |CX|=\\ &=&|AB|^2-|BX|\cdot |CX|, \end{eqnarray*} which had to be proven. \kdokaz We will continue a little bit more using Stewart's theorem. \bzgled Let $E$ be the intersection of the side $BC$ with the bisector of the interior angle $BAC$ of a triangle $ABC$. If we denote $a=|BC|$, $b=|AC|$, $c=|AB|$, $l_a=|AE|$ and $s=\frac{a+b+c}{2}$, then $$l_a=\frac{2\sqrt{bc}}{b+c}\sqrt{s(s-a)}.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.8.4.pic} \caption{} \label{sl.pod.7.8.4.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.8.4.pic}) According to the statement \ref{HarmCetSimKota} it is: $BE:CE=c:b$. If we use Stewart's statement \ref{StewartIzrek2} for the triangle $ABC$ and the distance $AE$, we get: $$l_a^2=\frac{c}{b+c}b^2+\frac{b}{b+c}c^2-\frac{bc}{(b+c)^2}a^2.$$ After a simple rearrangement and simplification of the expression on the right side of the equality, we get the desired relation. \kdokaz \bizrek \label{izrekEulerStirik}\index{izrek!Eulerjev za štirikotnike} (Euler's\footnote{Švicarski matematik \index{Euler, L.}\textit{L. Euler} (1707--1783).} theorem for quadrilaterals) If $P$ and $Q$ are the midpoints of the diagonals $e$ and $f$ of an arbitrary quadrilateral with sides $a$, $b$, $c$ and $d$, then $$|PQ|^2=\frac{1}{4}\left(a^2+b^2+c^2+d^2-e^2-f^2 \right).$$ \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.8.5.pic} \caption{} \label{sl.pod.7.8.5.pic} \end{figure} \textbf{\textit{Proof.}} Let $P$ and $Q$ be the centers of the diagonals $AC$ and $BD$ of an arbitrary quadrilateral $ABCD$ (Figure \ref{sl.pod.7.8.5.pic}). We also mark $a=|AB|$, $b=|BC|$, $c=|CD|$, $d=|DA|$, $e=|AC|$ and $f=|BD|$. The distance $PQ$ is the median of the triangle $AQC$, so according to the statement \ref{StwartTezisc} it holds: \begin{eqnarray} \label{eqnEulStirik} |PQ|^2=\frac{1}{2}|AQ|^2+\frac{1}{2}|CQ|^2-\frac{1}{4}e^2. \end{eqnarray} Similarly, the distances $QA$ and $QC$ are the medians of the triangles $ABD$ and $CBD$, so (from the statement \ref{StwartTezisc}): \begin{eqnarray*} |AQ|^2&=&\frac{1}{2}d^2+\frac{1}{2}a^2-\frac{1}{4}f^2,\\ |CQ|^2&=&\frac{1}{2}c^2+\frac{1}{2}b^2-\frac{1}{4}f^2. \end{eqnarray*} If we insert the last two equalities into \ref{eqnEulStirik}, we get the desired relation. \kdokaz A direct consequence of the previous statement is obtained if we choose a parallelogram for the quadrilateral. \bizrek A quadrilateral is a parallelogram if and only if the sum of the squares of all of its sides is equal to the sum of the squares of its diagonals. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.8.6.pic} \caption{} \label{sl.pod.7.8.6.pic} \end{figure} \textbf{\textit{Proof.}} Let $P$ and $Q$ be the centers of the diagonals $AC$ and $BD$ of an arbitrary quadrilateral $ABCD$. We also mark $a=|AB|$, $b=|BC|$, $c=|CD|$, $d=|DA|$, $e=|AC|$ and $f=|BD|$. According to the statement \ref{paralelogram} the quadrilateral $ABCD$ is a parallelogram exactly when $P=Q$ (Figure \ref{sl.pod.7.8.6.pic}) or $|PQ|=0$. The latter, according to the previous statement \ref{izrekEulerStirik}, is true exactly when $\frac{1}{4}\left(a^2+b^2+c^2+d^2-e^2-f^2 \right)=0$ or $a^2+b^2+c^2+d^2=e^2+f^2$. \kdokaz \bizrek \label{GMTmnl} Let $A$ and $B$ be given points in the plane and $m, n, l\in R^+\setminus \{0\}$. Determine a set of all points of this plane such that $$m|AX|^2 + n|BX|^2 = l^2.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.8.7.pic} \caption{} \label{sl.pod.7.8.7.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.8.7.pic}) According to the statement from Example \ref{izrekEnaDelitevDaljice} there exists only one such point $S$ on the line $AB$, that satisfies $\overrightarrow{AS}:\overrightarrow{SB}=n:m$. Let $X$ be an arbitrary point. If we use Stewart's Theorem \ref{StewartIzrek2} for the triangle $AXB$ and the line $XS$, we get: \begin{eqnarray*} |XS|^2=\frac{m}{n+m}|AX|^2+\frac{n}{n+m}|BX|^2-\frac{nm}{(n+m)^2}|AB|^2, \textrm{ i.e.} \end{eqnarray*} \begin{eqnarray} \label{eqnStewMnozTock} |XS|^2=\frac{1}{n+m}\left(m|AX|^2+n|BX|^2\right)-\frac{nm}{(n+m)^2}|AB|^2. \end{eqnarray} The point $X$ lies on the desired set of points exactly when $m|AX|^2 + n|BX|^2 = l^2$. Because of \ref{eqnStewMnozTock} this is exactly when: \begin{eqnarray*} |XS|^2=\frac{1}{n+m}l^2-\frac{nm}{(n+m)^2}|AB|^2=c, \end{eqnarray*} where $c$ is a constant, that is not dependent on the point $X$. So if $c>0$, the desired set of points is a circle $k(S, c)$. If $c=0$, the set of points is only $\{S\}$, but if $c<0$, the set of points is an empty set. \kdokaz \bnaloga\footnote{50. IMO, Germany - 2009, Problem 2.} Let $ABC$ be a triangle with circumcentre $O$. The points $P$ and $Q$ are interior points of the sides $CA$ and $AB$, respectively. Let $K$, $L$ and $M$ be the midpoints of the segments $BP$, $CQ$ and $PQ$, respectively, and let $l$ be the circle passing through $K$, $L$ and $M$. Suppose that the line $PQ$ is tangent to the circle $l$. Prove that $|OP|=|OQ|$. \enaloga \begin{figure}[!htb] \centering \input{sl.pod.7.8.IMO2.pic} \caption{} \label{sl.pod.7.8.IMO2.pic} \end{figure} \textbf{\textit{Solution.}} We denote with $k(O,R)$ the circumscribed circle of the triangle $ABC$ (Figure \ref{sl.pod.7.8.IMO2.pic}). The lines $MK$ and $ML$ are the medians of the triangles $QPB$ and $PQC$, so (from the statement \ref{srednjicaTrikVekt}): \begin{eqnarray} \label{eqn72} \overrightarrow{MK}=\frac{1}{2}\overrightarrow{QB} \hspace*{3mm} \textrm{and} \hspace*{3mm} \overrightarrow{ML}=\frac{1}{2}\overrightarrow{PC} \end{eqnarray} From the statements \ref{KotiTransverzala} and \ref{KotaVzporKraki} it follows: \begin{eqnarray} \label{eqn73} \angle AQP \cong \angle QMK,\hspace*{3mm} \angle APQ \cong \angle PML\hspace*{3mm} \textrm{and} \hspace*{3mm} \angle BAC \cong \angle KML \end{eqnarray} Because, by assumption, the line $PQ$ is tangent to the circle $l$, by the statement \ref{ObodKotTang}: $$\angle MLK\cong\angle QMK \hspace*{3mm} \textrm{and} \hspace*{3mm} \angle MKL\cong\angle PML.$$ From this and \ref{eqn73} it follows: \begin{eqnarray*} \angle AQP \cong \angle MLK,\hspace*{3mm} \angle APQ \cong \angle MKL\hspace*{3mm} \textrm{and} \hspace*{3mm} \angle BAC \cong\angle KML, \end{eqnarray*} which means that the triangles $AQP$ and $MLK$ are similar (it is enough to prove the congruence of two pairs of corresponding angles, by the statement \ref{PodTrikKKK}). So, from the definition of similarity of figures it follows: $\frac{AQ}{ML}=\frac{AP}{MK}$. If we combine this relation with the relation \ref{eqn72}, we get $\frac{AQ}{AP}=\frac{ML}{MK}= \frac{\frac{1}{2}\cdot CP}{\frac{1}{2}\cdot BQ}=\frac{CP}{BQ}$. So it holds: \begin{eqnarray} \label{eqn74} |AQ|\cdot |BQ| = |AP|\cdot |CP| \end{eqnarray} From Stewart's statement \ref{StewartIzrek} for the isosceles triangle $AOB$ ($|OA|=|OB|=R$) it follows: $$|OQ|^2=|OA|^2\cdot \frac{QB}{AB}+|OB|^2\cdot \frac{QA}{AB} -|AQ|\cdot |BQ|=R^2-|AQ|\cdot |BQ|.$$ Similarly, from the triangle $AOC$ by the same statement we get: $$|OP|^2=R^2-|AP|\cdot |CP|.$$ From the proven relation \ref{eqn74} at the end it follows $|OQ|^2=|OP|^2$ or $|OQ|=|OP|$. \kdokaz %________________________________________________________________________________ \poglavje{Desargues' Theorem} \label{odd7Desargues} The next theorem is historically connected to the development of projective geometry. \bizrek \label{izrekDesarguesEvkl} \index{izrek!Desarguesov} (Desargues’\footnote{ \index{Desargues, G.} \textit{G. Desargues} (1591--1661), French architect, who was one of the founders of projective geometry.} theorem) Let $ABC$ and $A'B'C'$ be two triangles in the plane such that the lines $AA'$, $BB'$ and $CC'$ intersect at a point $S$ (i.e the triangles are \index{perspective triangles}\pojem{perspective with respect to the centre} \color{blue} $S$). If $P=BC\cap B'C'$, $Q=AC\cap A'C'$ and $R=AB\cap A'B'$, then the points $P$, $Q$ and $R$ are collinear (i.e. triangles are \pojem{perspective with respect to the axis} \color{blue} $PQ$). \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.10D.1.pic} \caption{} \label{sl.pod.7.10D.1.pic} \end{figure} \textit{\textbf{Proof.}} (Figure \ref{sl.pod.7.10D.1.pic}) If we use Menelaus' theorem \ref{izrekMenelaj} for the triangles $SA'B'$, $SA'C'$ and $SB'C'$, we get: \begin{eqnarray*} \hspace*{-2mm} \frac{\overrightarrow{SA}}{\overrightarrow{AA'}}\cdot \frac{\overrightarrow{A'R}}{\overrightarrow{RB'}}\cdot \frac{\overrightarrow{B'B}}{\overrightarrow{BS}}=-1, \hspace*{1mm} \frac{\overrightarrow{SA}}{\overrightarrow{AA'}}\cdot \frac{\overrightarrow{A'Q}}{\overrightarrow{QC'}}\cdot \frac{\overrightarrow{C'C}}{\overrightarrow{CS}}=-1, \hspace*{1mm} \frac{\overrightarrow{SC}}{\overrightarrow{CC'}}\cdot \frac{\overrightarrow{C'P}}{\overrightarrow{PB'}}\cdot \frac{\overrightarrow{B'B}}{\overrightarrow{BS}}=-1. \end{eqnarray*} From these three relations it follows: \begin{eqnarray*} \frac{\overrightarrow{A'Q}}{\overrightarrow{QC'}}\cdot \frac{\overrightarrow{C'P}}{\overrightarrow{PB'}}\cdot \frac{\overrightarrow{B'R}}{\overrightarrow{RA'}}=-1. \end{eqnarray*} Therefore, according to Menelaus' theorem \ref{izrekMenelaj} (in the opposite direction) for the triangle $A'B'C'$ the points $P$, $Q$ and $R$ are collinear. \kdokaz We prove the converse statement in a similar way. \bizrek \label{izrekDesarguesObr} \index{izrek!Desarguesov obratni} Let $ABC$ and $A'B'C'$ be two triangles in the plane such that the lines $AA'$ and $BB'$ intersect at the point $S$. If $P=BC\cap B'C'$, $Q=AC\cap A'C'$ and $R=AB\cap A'B'$ are collinear points, then also $S\in CC'$.\\ (Converse of Desargues’ theorem) \eizrek The following theorems are in a certain way similar to Desargues' theorem \ref{izrekDesarguesEvkl}. \bizrek \label{izrekDesarguesZarkVzp} Let $ABC$ and $A'B'C'$ be two triangles in the plane such that the lines $AA'$, $BB'$ and $CC'$ are parallel to each other. If $P=BC\cap B'C'$, $Q=AC\cap A'C'$ and $R=AB\cap A'B'$, then the points $P$, $Q$ and $R$ are collinear. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.10D.2.pic} \caption{} \label{sl.pod.7.10D.2.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.10D.2.pic}) After using Tales' Theorem \ref{TalesovIzrek} three times, we get: \begin{eqnarray*} \frac{\overrightarrow{BP}}{\overrightarrow{PC}}= \frac{\overrightarrow{BB'}}{\overrightarrow{C'C}},\hspace*{4mm} \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}= \frac{\overrightarrow{CC'}}{\overrightarrow{A'A}},\hspace*{4mm} \frac{\overrightarrow{AR}}{\overrightarrow{RB}}= \frac{\overrightarrow{AA'}}{\overrightarrow{B'B}}. \end{eqnarray*} After multiplying these three relations, we get: \begin{eqnarray*} \frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot \frac{\overrightarrow{CQ}}{\overrightarrow{QA}}\cdot \frac{\overrightarrow{AR}}{\overrightarrow{RB}}=-1, \end{eqnarray*} and therefore, according to Menelaus' Theorem \ref{izrekMenelaj} (in the opposite direction), for the triangle $ABC$ the points $P$, $Q$ and $R$ are collinear. \kdokaz If $P=BC\cap B'C'$, $Q=AC\cap A'C'$ and $R=AB\cap A'B'$, then the points $P$, $Q$ and $R$ are collinear. \bizrek \label{izrekDesarguesOsNesk} Let $ABC$ and $A'B'C'$ be two triangles in the plane such that the lines $AA'$, $BB'$ and $CC'$ intersect at a point $S$. If $BC\parallel B'C'$ and $AC\parallel A'C'$, then also $AB\parallel A'B'$. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.10D.3.pic} \caption{} \label{sl.pod.7.10D.3.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.10D.3.pic}) Because $BC\parallel B'C'$ and $AC\parallel A'C'$, from Tales' theorem \ref{TalesovIzrek} it follows $\frac{\overrightarrow{SB}}{\overrightarrow{SB'}}= \frac{\overrightarrow{SC}}{\overrightarrow{SC'}}$ and $\frac{\overrightarrow{SA}}{\overrightarrow{SA'}}= \frac{\overrightarrow{SC}}{\overrightarrow{SC'}}$. From the previous two relations it follows first $\frac{\overrightarrow{SB}}{\overrightarrow{SB'}}= \frac{\overrightarrow{SA}}{\overrightarrow{SA'}}$, so by Tales' theorem (in the opposite direction) \ref{TalesovIzrekObr} it is also true that $AB\parallel A'B'$. \kdokaz The formulations of the previous three theorems are very similar, even though the proofs of these theorems are essentially different. The formulations would not differ at all, if we had assumed that all parallels (in one direction) of a plane intersect at the same point at infinity, and that all points at infinity of a plane determine exactly one line at infinity. This was actually the main motivation for the development of projective geometry, in which every two lines in a plane intersect (see \cite{Mitrovic}). In the following we will see some consequences of Desargues' theorem. \bzgled \label{zgled 3.2} Let $p$, $q$ and $r$ be three lines in the plane which intersect at the same point, and points $A$, $B$ and $C$ of this plane which do not belong to these lines. Construct a triangle whose vertices belong to the given lines, and the sides contain the given points. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.10D.4.pic} \caption{} \label{sl.pod.7.10D.4.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.10D.4.pic}) Let $PQR$ be such a triangle, that its vertices $P$, $Q$ and $R$ belong to the lines $p$, $q$ and $r$, and sides $QR$, $PR$ and $PQ$ contain points $A$, $B$ and $C$. Let $S$ be the common point of lines $p$, $q$ and $r$. If $P'Q'R'$ is an arbitrary triangle that is perspective to the triangle $PQR$ with respect to the center $S$, whose sides $R'Q'$ and $R'P'$ contain the points $A$ and $B$ (the condition regarding the point $C$ is omitted), then by Desargues' Theorem \ref{izrekDesarguesEvkl} the triangle $PQR$ and $P'Q'R'$ are perspective with respect to some line $s$. Therefore, the line $s$ contains the points $A$, $B$ and $Z=PQ\cap P'Q'$. We construct the triangle $PQR$ by first constructing the auxiliary triangle $P'Q'R'$, where the point $R'\in r$ is arbitrary. Then we construct the point $Z$ as the intersection of the lines $AB$ and $P'Q'$. With the points $Z$ and $C$ the side $PQ$ is determined. In the proof that $PQR$ is the desired triangle, we use the converse of Desargues' Theorem \ref{izrekDesarguesObr}. \kdokaz %________________________________________________________________________________ \poglavje{Power of a Point} \label{odd7Potenca} One of the most interesting characteristics of a circle that highlights some of its metric properties is the power of a point\footnote{The term power was first used in this sense by the Swiss mathematician \index{Steiner, J.}\textit{J. Steiner} (1769--1863).}. Before moving on to the definition, we prove the following theorem. \bizrek \label{izrekPotenca} Suppose that $P$ is an arbitrary point in the plane of a circle $k(S,r)$. For any line of this plane containing the point $P$ and intersecting the circle $k$ at points $A$ and $B$, the expression $\overrightarrow{PA}\cdot \overrightarrow{PB}$ (Figure \ref{sl.pod.7.12.1b.pic}) is constant, furthermore: $$\overrightarrow{PA}\cdot \overrightarrow{PB} = |PS|^2 - r^2.$$ If $P$ is an exterior point of the circle $k$ and $PT$ its tangent at a point $T$, then: $$\overrightarrow{PA}\cdot \overrightarrow{PB} = |PT|^2.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.12.1b.pic} \caption{} \label{sl.pod.7.12.1b.pic} \end{figure} \textbf{\textit{Proof.}} We will consider three possible cases. \textit{1)} (Figure \ref{sl.pod.7.12.1.pic}) Let $P$ be an external point of the circle $k$. In this case, there is no $\mathcal{B}(A,P,B)$, so (the equivalence of \ref{eqnMnozVektRelacijaB} from section \ref{odd5DolzVekt}): \begin{eqnarray} \label{eqnPotenIzr1} \overrightarrow{PA}\cdot \overrightarrow{PB}>0. \end{eqnarray} Without loss of generality, assume that $\mathcal{B}(P,A,B)$ is true. Because $\angle PTA\cong\angle TBA=\angle TBP$ (the statement of \ref{ObodKotTang}) and $\angle TPA=\angle BPT$, the triangles $PAT$ and $PTB$ are similar (the statement of \ref{PodTrikKKK}), so $PA:PT=PT:PB$. If we use the Pythagorean theorem, we get: $$|PA|\cdot |PB| = |PT|^2 = |PS|^2 - r^2.$$ From this, due to relation \ref{eqnPotenIzr1}, it follows: $$\overrightarrow{PA}\cdot \overrightarrow{PB} = |PS|^2 - r^2.$$ \begin{figure}[!htb] \centering \input{sl.pod.7.12.1.pic} \caption{} \label{sl.pod.7.12.1.pic} \end{figure} \textit{2)} (Figure \ref{sl.pod.7.12.1a.pic}) If the point $P$ lies on the circle $k$, then $P=A$ or $P=B$, therefore: $$\overrightarrow{PA}\cdot \overrightarrow{PB} = 0 = |PS|^2 - r^2.$$ \begin{figure}[!htb] \centering \input{sl.pod.7.12.1a.pic} \caption{} \label{sl.pod.7.12.1a.pic} \end{figure} \textit{3)} (Figure \ref{sl.pod.7.12.1a.pic}) Let $P$ be an inner point of the circle $k$. In this case, $\mathcal{B}(A,P,B)$, so (the equivalence of \ref{eqnMnozVektRelacijaB} from section \ref{odd5DolzVekt}): \begin{eqnarray} \label{eqnPotenIzr2} \overrightarrow{PA}\cdot \overrightarrow{PB}<0. \end{eqnarray} Let $A_1$ and $B_1$ be the intersection points of the line $SP$ with the circle $k$ (without loss of generality, let $\mathcal{B}(A_1,S,P)$). Because of the compatibility of the corresponding circumferential angles (the statement of \ref{ObodObodKot}), $\triangle APA_1\sim \triangle B_1PB$ (the statement of \ref{PodTrikKKK}), so $AP:B_1P = PA_1:PB$, therefore $$|PA|\cdot |PB|=|PA_1|\cdot |PB_1|= \left(r+|PS|\right)\cdot\left(r-|PS|\right)=r^2-|PS|^2.$$ From this, due to relation \ref{eqnPotenIzr1}, it follows $$\overrightarrow{PA}\cdot \overrightarrow{PB} = |PS|^2 - r^2,$$ which had to be proven. \kdokaz The constant product $\overrightarrow{PA}\cdot \overrightarrow{PB}$ from the previous statement of \ref{izrekPotenca} is called \index{potenca točke}\pojem{potenca točke} $P$ with respect to the circle $k$ and is denoted $p(P,k)$. According to the previous statement of \ref{izrekPotenca}, the power of the point $P$ with respect to the circle $k(S,r)$ is the number $|PS|^2 - r^2$. This number is positive, negative or zero, depending on whether $P$ is an outer or inner or point on the circle $k$. So: \begin{eqnarray*} p(P,k)\hspace*{1mm}\left\{ \begin{array}{ll} >0, & \textrm{if } OP>r; \\ =0, & \textrm{if } OP=r; \\ <0, & \textrm{if } OP0$, by the previous statement (\ref{PotencOsLema}) the desired set is a rectangle of the line $S_1S_2$. In the case $r_1=r_2$ it is clear that $|S_1P|=|S_2P|$, thus the desired set is the line $S_1S_2$'s symmetry. We denote this line with $p$. In the special case when the circles intersect in points $A$ and $B$, we have $p(A,k_1)=p(A,k_2)=0$ and $p(B,k_1)=p(B,k_2)=0$. This means that points $A$ and $B$ lie on the desired line $p$, thus this is the line $AB$. If the circles touch in point $T$, we have $p(T,k_1)=p(T,k_2)=0$. This means that the point $T$ lies on the line $p$, which is the rectangle of the line $S_1S_2$. Thus the line $p$ is the common tangent of two given circles through their point of contact $T$. \kdokaz The line $p$ from the previous statement is called the \index{potent!line} \pojem{potent line} of two circles. We will denote the potent line of circles $k_1$ and $k_2$ with $p(k_1,k_2)$. It is interesting to find out how to construct the power line of two given circles $k_1$ and $k_2$ effectively. In the cases where the circles intersect or touch, we have already given the answer in Theorem \ref{PotencnaOs} (Figure \ref{sl.pod.7.12.4.pic}). It remains to construct the power line in the case where the circles have no common points. One option is to use Theorem \ref{PotencOsLema} directly. A slightly faster procedure is related to the construction of the auxiliary circle $l$, which intersects the given circle at points $A$ and $B$ or $C$ and $D$. Then the intersection of the lines $AB$ and $CD$ - point $X$ - lies on the desired power line $p(k_1,k_2)$. Indeed, from $p(k_1,l)=AB$ and $p(k_2,l)=CD$ it follows that $X\in p(k_1,l)$ and $X\in p(k_1,l)$ or $p(X,p_1)=p(X,l)=p(X,p_2)$. Theorem \ref{PotencnaOs} does not consider one case - when $k_1(S,r_1)$ and $k_2(s,r_2)$ are concentric circles. In this case, the mentioned set of points is an empty set. We get this from the condition $|S_1P|^2-|S_2P|^2=r_1^2-r_2^2$. Namely, if $S_1=S_2$ and $r_1\neq r_2$, we get the condition $0=r_1^2-r_2^2\neq 0$, which is not satisfied for any point $P$. Let us also define some concepts related to the power line of two circles. Let the set of all such circles of a plane be such that each has two power lines $p$ that are equal for each pair of circles of this set. We call this set \index{set of circles}\pojem{set of circles}. The line $p$ is the \index{power!line}\pojem{power line} of this set of circles (Figure \ref{sl.pod.7.12.5.pic}). \begin{figure}[!htb] \centering \input{sl.pod.7.12.5.pic} \caption{} \label{sl.pod.7.12.5.pic} \end{figure} Let $p$ be the power line of a set of circles. From Theorem \ref{PotencnaOs} it is clear that all the centers of the circles of this set lie on the same line $s$, which is perpendicular to the line $p$. We will consider three cases. \textit{1)} If at least two circles of a family intersect in points $A$ and $B$, then $p=AB$, so all circles of this family go through points $A$ and $B$. In this case we say that it is a \index{family of circles!elliptic}\pojem{elliptic family of circles}. \textit{2)} If at least two circles of a family touch in point $T$, then $p$ is a perpendicular of line $s$ in point $T$, so all circles have a common tangent $p$ in point $T$. In this case we say that it is a \index{family of circles!parabolic}\pojem{parabolic family of circles}. \textit{3)} If no two circles of a family have common points, we say that it is a \index{family of circles!hyperbolic}\pojem{hyperbolic family of circles}. In this case the following property is valid: if an arbitrary circle intersects circles of this family (it is not necessary that it intersects all of them) in points $A_i$ and $B_i$, then all lines $A_iB_i$ go through one point that lies on line $p$. \bizrek \label{PotencnoSr} Let $k$, $l$ and $j$ be three non-concentric circles with non-collinear centres. Then there is exactly one point that has the same power with respect to all three circles. This point is the intersection of their three radical axes $p(k,l)$, $p(l,j)$ and $p(k,j)$. \eizrek \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.6.pic}) Because the centres of circles $k$, $l$ and $j$ are three non-linear points, no two of lines $p(k,l)$, $p(l,j)$ and $p(k,j)$ are parallel. Let $P=p(k,l)\cap p(l,j)$. Then $$p(P,k)=p(P,l)=p(P,j),$$ i.e. $P\in p(k,j)$, which means that the power lines $p(k,l)$, $p(l,j)$ and $p(k,j)$ intersect in point $P$. If for another point $\widehat{P}$ of this plane it is valid that $p(\widehat{P},k)=p(\widehat{P},l)=p(\widehat{P},j)$, then $\widehat{P}\in p(k,l),\hspace*{1mm}p(l,j),\hspace*{1mm}p(k,j)$, so $\widehat{P}=P$. \kdokaz The point from the previous statement (\ref{PotencnoSr}) is called the \index{središče!potenčno} \pojem{potential center} of three circles. We will denote the potential center of circles $k$, $l$ and $j$ with $p(k,l,j)$. \begin{figure}[!htb] \centering \input{sl.pod.7.12.6.pic} \caption{} \label{sl.pod.7.12.6.pic} \end{figure} In the case when the centers of three circles are three collinear points, and the circles are not from the same pencil, all three radical axes are parallel, because they are all perpendicular to the common center of these circles (statement \ref{PotencnaOs}). From this and statement \ref{PotencnoSr} we directly get the following statement (Figure \ref{sl.pod.7.12.6.pic}). Let $k$, $l$ and $j$ be three non-concentric circles with non-collinear centres. Then there is exactly one point that has the same power with respect to all three circles. This point is the intersection of their three radical axes $p(k,l)$, $p(l,j)$ and $p(k,j)$. Radical axes of three circles in the plane that are not from the same pencil and no two of them are concentric, belong to the same family of lines. \bizrek \label{PotencnoSrSop} Radical axes of three circles in the plane that are not from the same pencil and no two of them are concentric, belong to the same family of lines. \eizrek A special case of statement \ref{PotencnoSr}, when each two lines intersect, is the following theorem. \bzgled Let $k$, $l$, and $j$ be three circles of some plane with nonlinear centres and: \begin{itemize} \item $A$ and $B$ intersections of the circles $k$ and $l$, \item $C$ and $D$ intersections of the circles $l$ and $j$, \item $E$ and $F$ intersections of the circles $j$ and $k$. \end{itemize} Prove that the lines $AB$, $CD$ and $EF$ intersect at a single point. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.12.7.pic} \caption{} \label{sl.pod.7.12.7.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.7.pic}) According to the \ref{PotencnaOs} theorem, $p(k,l)=AB$, $p(l,j)=CD$ and $p(j,k)=EF$ are the corresponding power axes. According to the \ref{PotencnoSr} theorem, they intersect at one point - the power center $P=p(k,l,j)$ of these three circles. \kdokaz The next statement is very similar. \bzgled Let $k$, $l$, and $j$ be three circles of some plane with nonlinear centres and: \begin{itemize} \item $t_1$ the common tangent of the circles $k$ in $l$, \item $t_2$ the common tangent of the circles $l$ in $j$, \item $t_3$ the common tangent of the circles $j$ in $k$. \end{itemize} Prove that the lines $t_1$, $t_2$ and $t_3$ intersect at a single point. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.12.8.pic} \caption{} \label{sl.pod.7.12.8.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.8.pic}) According to the \ref{PotencnaOs} theorem, $p(k,l)=t_1$, $p(l,j)=t_2$ and $p(j,k)=t_3$ are the corresponding power axes. According to the \ref{PotencnoSr} theorem, they intersect at one point - the power center $P=p(k,l,j)$ of these three circles. \kdokaz \bzgled a) Suppose that circles $k$ and $l$ are touching each other externally, and a line $t$ is is the common tangent of these circles at their common point. Let $AB$ be a second common tangent of these circles at touching points $A$ and $B$. Prove that the midpoint the line segment $AB$ lies on the tangent $t$.\\ b) Suppose that circles $k$ and $l$ intersect at points $P$ and $Q$. Let $AB$ be a common tangent of these circles at touching points $A$ and $B$. Prove that the midpoint of the line segment $AB$ lies on line $PQ$. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.12.9.pic} \caption{} \label{sl.pod.7.12.9.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.9.pic}) \textit{a)} Let $S$ be the midpoint of the line segment $AB$. Then we have: $p(S,k)= |SA|^2 = |SB|^2 = p(S,l)$, so the point $S$ lies on the power curve of circles $k$ and $l$ or on the line $t$ (statement \ref{PotencnaOs}). \textit{b)} Just as in the previous example, only that the power curve of circles $k$ and $l$ is the line $AB$ in this case. \kdokaz \bizrek \label{EulerjevaFormula} \index{formula!Eulerjeva} (Euler's\footnote{\index{Euler, L.} \textit{L. Euler} (1707--1783), švicarski matematik.} formula) If $k(S,r)$ is the incircle and $l(O,R)$ the circumcircle of an arbitrary triangle, then $$|OS|^2=R^2- 2Rr.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.12.10.pic} \caption{} \label{sl.pod.7.12.10.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.10.pic}) We denote by $A$, $B$ and $C$ the vertices of a triangle, $NM$ the diameter of the circumscribed circle $l$, which is perpendicular to the side $BC$ (and also $A,N\div BC$). By \ref{TockaN} the point $N$ lies on the bisector of the internal angle at the vertex $A$ or on the altitude $AS$ (\ref{SredVcrtaneKrozn}). By \ref{TockaN.NBNC} we have $NS\cong NC$. If we use the power of the point $S$ with respect to the circle $l$ (\ref{PotencnoSr}) and the fact $\mathcal{B}(A,S,N)$, we get $p(S,l)= |SO|^2 - R^2 = \overrightarrow{SA}\cdot \overrightarrow{SN} = -|SA|\cdot |SN| = -|SA|\cdot |CN|$, thus: \begin{eqnarray} \label{eqnEulFormOS} |SO|^2 - R^2 = -|SA|\cdot |CN|. \end{eqnarray} We denote by $Q$ the point of contact of the inscribed circle $k$ with the side $AC$ of the triangle $ABC$. By \ref{TangPogoj} and \ref{TalesovIzrKroz2} we have $\angle AQS\cong\angle MCN=90^0$, from \ref{ObodObodKot} we also get $\angle SAQ=\angle NAC\cong\angle NMC$. Therefore, the triangles $AQS$ and $MCN$ are similar (\ref{PodTrikKKK}), so: $$\frac{AS}{MN}=\frac{SQ}{NC}$$ or $|AS|\cdot |NC|=|MN|\cdot |SQ|=2Rr$. If we insert this into \ref{eqnEulFormOS}, we get: $$|SO|^2 = R^2-|SA|\cdot |CN|=R^2-2Rr,$$ which had to be proven. \kdokaz The next task is a special case of the previous formula and is therefore its consequence (Figure \ref{sl.pod.7.12.10a.pic}). \begin{figure}[!htb] \centering \input{sl.pod.7.12.10a.pic} \caption{} \label{sl.pod.7.12.10a.pic} \end{figure} \bnaloga\footnote{4. IMO, Czechoslovakia - 1962, Problem 6.} Consider an isosceles triangle. Let $r$ be the radius of its circumscribed circle and $\rho$ the radius of its inscribed circle. Prove that the distance $d$ between the centres of these two circles is $$d = \sqrt{r(r-2\rho)}.$$ \enaloga The next statement is also a direct consequence of \ref{EulerjevaFormula}. \bizrek If $k(S,r)$ is the incircle and $l(O,R)$ the circumcircle of an arbitrary triangle, then $$R\geq 2r.$$ Equality is achieved for an equilateral triangle. \eizrek The next design task is one of the ten Apollonius' problems on the tangency of circles, which we will investigate in more detail in section \ref{odd9ApolDotik}. \bzgled Construct a circle through two given points $A$ and $B$ and tangent to a given line $t$. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.12.11.pic} \caption{} \label{sl.pod.7.12.11.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.12.11.pic}) Let $k$ be the desired circle, which passes through points $A$ and $B$ and is tangent to the line $t$ in point $T$. If the lines $AB$ and $t$ are parallel, the third point $T$ of the circle $k$ is obtained as the intersection of the line $t$ with the perpendicular of the line $AB$. Let $P$ be the intersection of the lines $AB$ and $t$. We will use the power of point $P$ with respect to the circle $k$. With $l$ we denote any circle, which passes through points $A$ and $B$. By Theorem \ref{PotencnaOs} the line $AB$ is the power line of the circles $k$ and $l$, therefore $p(P,k)=p(P,l)$. We denote with $PT$ and $PT_1$ the tangents of the circles $k$ and $l$ in their points $P$ and $P_1$. Then it holds (Theorem \ref{izrekPotenca}): $$|PT|^2=p(P,k)=p(P,l)=|PT_1|^2$$ or $|PT|=|PT_1|$. The last relation allows us to construct the third point $T$ of the circle $k$. \kdokaz \bzgled Let $E$ be the intersection of the bisector of the interior angle at the vertex $A$ with the side $BC$ of a triangle $ABC$ and $A_1$ the midpoint of this side. Let $P$ and $Q$ be intersections of the circumcircle of the triangle $AEA_1$ with the sides $AB$ and $AC$ of the triangle $ABC$. Prove that: $$BP\cong CQ.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.12.12.pic} \caption{} \label{sl.pod.7.12.12.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.12.12.pic}) We'll mark the circumcircle of the triangle $AEA_1$ with $k$. If we use the power of points $B$ and $C$ according to the circle $k$, the relation $\mathcal{B}(B,P,A)$ and $\mathcal{B}(C,Q,A)$ (because of the assumption that points $P$ and $Q$ lie on the sides $AB$ and $AC$ of the triangle $ABC$) and the equivalence \ref{eqnMnozVektRelacijaB} from section \ref{odd5DolzVekt}, we get: \begin{eqnarray*} p(B,k)&=&|BP|\cdot |BA|=|BE|\cdot |BA_1|,\\ p(C,k)&=&|CP|\cdot |CA|=|CE|\cdot |CA_1|. \end{eqnarray*} From this and the relation $BA_1\cong CA_1$ and the theorem \ref{HarmCetSimKota} we get: \begin{eqnarray*} \frac{|BP|\cdot |BA|}{|CP|\cdot |CA|} =\frac{|BE|\cdot |BA_1|}{|CE|\cdot |CA_1|} =\frac{|BE|}{|CE|} =\frac{|BA|}{|CA|}. \end{eqnarray*} Therefore $\frac{|BP|\cdot |BA|}{|CP|\cdot |CA|}=\frac{|BA|}{|CA|}$ or $|BP|=|CP|$. \kdokaz \bzgled Construct a circle that is perpendicular to three given circles $k$, $l$ in $j$. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.12.13.pic} \caption{} \label{sl.pod.7.12.13.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.12.13.pic}) We will first assume that the centers of these circles $k$, $l$ and $j$ are nonlinear points. Let $x$ be the desired circle with center $P$, which is perpendicular to the circles $k$, $l$ and $j$, and $A\in x\cap k$, $B\in x\cap l$ and $C\in x\cap j$. Because $x\perp k,j,l$, $PA$, $PB$ and $PC$ are tangents of these circles from point $P$ (by \ref{pravokotniKroznici}). By \ref{izrekPotenca} we have $p(P,k)=|PA|^2$, $p(P,l)=|PB|^2$ and $p(P,j)=|PC|^2$. Because points $A$, $B$ and $C$ lie on the circle $x$ with center $P$, we have $|PA|^2=|PB|^2=|PC|^2$, i.e. $p(P,k)=p(P,l)= p(P,j)$. This means that $P=p(k,l,j)$ is the power center of the circles $k$, $l$ and $j$. Therefore, the desired circle $x$ can be constructed by first drawing its center $P=p(k,l,j)$, then the radius $PA$, where the line $PA$ is tangent to the circle $k$ at point $A$. It is clear that in the case where $P$ is an internal point of one of the circles $k$, $l$, $j$, the task has no solution. Even in the case where the centers of the circles are collinear points, the desired circle does not exist (it is a "degenerate circle" or a line representing their common center). \kdokaz \bzgled Circles $k(O,r)$ and $l(S,\rho)$ and a point $P$ in the same plane are given. Construct a line passing through the point $P$, which determine congruent chords on the circles $k$ and $l$. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.12.14.pic} \caption{} \label{sl.pod.7.12.14.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.pod.7.12.14.pic}) Let $p$ be a line that goes through the point $P$, the circles $k$ and $l$ intersect in such points $A$ and $B$ or $C$ and $D$, that $AB\cong CD$. Let $\overrightarrow{v}$ be a vector that is determined by the centers of the lines $AB$ and $CD$. Then $\mathcal{T}_{\overrightarrow{v}}:\hspace*{1mm}A,B\mapsto C,D$, the circle $k$ is transformed by this translation into the circle $k'$, that goes through the points $C$ and $D$. So the circles $k'$ and $l$ intersect in the points $C$ and $D$. Thus the problem of constructing the line $p$ is translated into the problem of constructing the vector $\overrightarrow{v}$ or the point $O'= \mathcal{T}_{\overrightarrow{v}}(O)$, which represents the center of the circle $k'$. The line, that is determined by the centers $S$ and $O'$ of the circles $l$ and $k'$, is perpendicular to their common chord $CD$. Because $\overrightarrow{OO'}= \overrightarrow{v} \parallel p$, it follows $$\angle OO'S=90^0.$$ By \ref{TalesovIzrKroz2} the point $O'$ lies on the circle above the diameter $OS$. The point $P$ lies on the power line $p$ of the circles $k'$ and $l$ (\ref{PotencnaOs}). From this it follows $PL\cong PK$, where $PL$ and $PK$ are tangents to the circles $l$ and $k'$ in the points $L$ and $K$. This means that we can construct a triangle that is similar to the right triangle $PKO'$ ($PK\cong PL$, $\angle PO'K=90^0$ and $KO'\cong r$), and thus also $d$, which is similar to the distance $PO'$. So the point $O'$ belongs to the intersection of the circle with center $P$ and radius $d$ and the circle above the diameter $OS$. \kdokaz \bnaloga\footnote{36. IMO Canada - 1995, Problem 1.} Let $A$, $B$, $C$, $D$ be four distinct points on a line, in that order $\mathcal{B}(A,B,C,D)$. The circles with diameters $AC$ and $BD$ intersect at $X$ and $Y$. The line $XY$ meets $BC$ at $Z$. Let $P$ be a point on the line $XY$ other than $Z$. The line $CP$ intersects the circle with diameter $AC$ at $C$ and $M$, and the line $BP$ intersects the circle with diameter $BD$ at $B$ and $N$. Prove that the lines $AM$, $DN$, $XY$ are concurrent. \enaloga \begin{figure}[!htb] \centering \input{sl.pod.7.12.IMO1.pic} \caption{} \label{sl.pod.7.12.IMO1.pic} \end{figure} \textbf{\textit{Solution.}} Let $k_1$ and $k_2$ be the circles over the segments $AC$ and $BD$ (Figure \ref{sl.pod.7.12.IMO1.pic}). By the theorem \ref{KroznPresABpravokOS} the line $XY$ is perpendicular to the central line $BC$ of these two circles and is also their power line (theorem \ref{PotencnaOs}). Let $S_1=AM\cap XY$ and $S_2=DN\cap XY$. It is enough to prove $S_1=S_2$. Because $M\in k_1$, by the theorem \ref{TalesovIzrek} $\angle AMC=90^0$ or $AS_1\perp MC$. Because of this the $\angle MCA\cong\angle AS_1Z$ (angle with perpendicular legs - theorem \ref{KotaPravokKraki}). From this similarity of angles it follows that $\frac{AZ}{PZ}=\frac{ZS_1}{ZC}$ or $|ZS_1|=\frac{|ZC|\cdot |ZA|}{|PZ|}$. But $|ZC|\cdot |ZA|=p(Z,k_1)=|ZX|\cdot |ZY|=|ZX|^2$. From this it follows: $$|ZS_1|=\frac{|ZX|^2}{|PZ|}.$$ In the same way it can be proven that: $|ZS_2|=\frac{|ZX|^2}{|PZ|}$. Because $S_1$ and $S_2$ are on the same line segment $ZP$, $S_1=S_2$. \kdokaz \bzgled Let $P$ be an arbitrary point in the plane of a triangle $ABC$ which does not lie on any of lines containing altitudes of this triangle. Suppose $A_1$ is a point, in which a perpendicular line of the line $AP$ at the point $P$ intersects the line $BC$. Analogously we can also define points $B_1$ and $C_1$. Prove that $A_1$, $B_1$ and $C_1$ are three collinear points. \ezgled \textbf{\textit{Solution.}} We mark with $A_C$ and $B_C$ the orthogonal projections of points $A$ and $B$ on the line $CP$. Similarly, $A_B$ and $C_B$ are the orthogonal projections of points $A$ and $C$ on the line $BP$, and $B_A$ and $C_A$ are the orthogonal projections of points $B$ and $C$ on the line $AP$ (Figure \ref{sl.pd.7.4.6.pic}). By Tales' theorem \ref{TalesovIzrek} it is: \begin{eqnarray} \label{4.1}\frac{AC_1}{C_1B} \cdot \frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A}= \frac{A_CP}{PB_C} \cdot \frac{B_AP}{PC_A} \cdot \frac{C_BP}{PA_B} \end{eqnarray} \begin{figure}[!htb] \centering \input{sl.pd.7.4.6.pic} \caption{} \label{sl.pd.7.4.6.pic} \end{figure} From $\angle AA_CC\cong\angle AC_AC=90^0$ it follows that points $A_C$ and $C_A$ lie on the circle with diameter $AC$. Therefore the power of point $P$ on this circle is equal (by theorem \ref{izrekPotenca}) $\overrightarrow{PC}\cdot \overrightarrow{PA_C}= \overrightarrow{PA}\cdot \overrightarrow{PC_A}$. Similarly, $\overrightarrow{PB}\cdot \overrightarrow{PC_B}= \overrightarrow{PC}\cdot \overrightarrow{PB_C}$ and $\overrightarrow{PB}\cdot \overrightarrow{PA_B}= \overrightarrow{PA}\cdot \overrightarrow{PB_A}$. From these relations we get: $\frac{PA_C}{PC_A}=\frac{PA}{PC}$, $\frac{PC_B}{PB_C}=\frac{PC}{PB}$ and $\frac{PB_A}{PA_B}=\frac{PB}{PA}$. If we insert these relations in \ref{4.1}, we get $\frac{AC_1}{C_1B} \cdot \frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A}=1$. Because $\frac{\overrightarrow{AC_1}}{\overrightarrow{C_1B}} \cdot \frac{\overrightarrow{BA_1}}{\overrightarrow{A_1C}} \cdot \frac{\overrightarrow{CB_1}}{\overrightarrow{B_1A}}<0$, it is $\frac{\overrightarrow{AC_1}}{\overrightarrow{C_1B}} \cdot \frac{\overrightarrow{BA_1}}{\overrightarrow{A_1C}} \cdot \frac{\overrightarrow{CB_1}}{\overrightarrow{B_1A}}=-1$. By Menelaus' theorem points $A_1$, $B_1$ and $C_1$ are collinear. \kdokaz \bnaloga\footnote{41. IMO, S. Korea - 2000, Problem 1.} $AB$ is tangent to the circles $CAMN$ and $NMBD$. $M$ lies between $C$ and $D$ on the line $CD$, and $CD$ is parallel to $AB$. The chords $NA$ and $CM$ meet at $P$; the chords $NB$ and $MD$ meet at $Q$. The rays $CA$ and $DB$ meet at $E$. Prove that $PE\cong QE$. \enaloga \begin{figure}[!htb] \centering \input{sl.pod.7.12.IMO3.pic} \caption{} \label{sl.pod.7.12.IMO3.pic} \end{figure} \textbf{\textit{Solution.}} We mark with $L$ the intersection of the lines $MN$ and $AB$ and with $k$ and $l$ the circumscribed circles $CAMN$ and $NMBD$ (Figure \ref{sl.pod.7.12.IMO3.pic}). By Theorem \ref{PotencnaOs} the line $MN$ is the power line of the circles $k$ and $l$, therefore for its point $L\in MN$ we have $|LA|^2=p(L,k)=p(L,l)=|LB|^2$, which means that $L$ is the center of the line segment $AB$. Because, by assumption, $AB\parallel CD$, i.e. $AB\parallel PQ$, by Tales' theorem $MP:MQ=LA:LB=1$. Therefore the point $M$ is the center of the line segment $PQ$ or $MP\cong MQ$. If we use Theorem \ref{ObodKotTang} and \ref{KotiTransverzala}, we get: \begin{eqnarray*} \angle EAB &\cong& \angle AMC \cong\angle MAB\\ \angle EBA &\cong& \angle BMQ \cong\angle MBA \end{eqnarray*} This means that the triangles $AEB$ and $AMB$ are similar (by the \textit{ASA} Theorem \ref{KSK}), namely, they are symmetric with respect to the line $AB$. This means that $EM\perp AB$. Because $AB\parallel PQ$, $EM\perp PQ$ or $\angle PME\cong\angle QME =90^0$. We have already proven that $MP\cong MQ$, therefore the triangles $PME$ and $QME$ are similar (by the \textit{SAS} Theorem \ref{SKS}), from which it follows that $PE\cong QE$. \kdokaz \bnaloga\footnote{40. IMO, Romania - 1999, Problem 5.} Two circles $k_1$ and $k_2$ are contained inside the circle $k$, and are tangent to $k$ at the distinct points $M$ and $N$, respectively. $k_1$ passes through the center of $k_2$. The line passing through the two points of intersection of $k_1$ and $k_2$ meets $k$ at $A$ and $B$. The lines $MA$ and $MB$ meet $k_1$ at $C$ and $D$, respectively. Prove that $CD$ is tangent to $k_2$. \enaloga \begin{figure}[!htb] \centering \input{sl.pod.7.12.IMO4.pic} \caption{} \label{sl.pod.7.12.IMO4.pic} \end{figure} \textbf{\textit{Solution.}} We mark with $O_1$ and $O_2$ the centers of the circles $k_1$ and $k_2$, with $r_1$ and $r_2$ their radii and with $E$ the other point of intersection of the line $AN$ with the circle $k_2$ (Figure \ref{sl.pod.7.12.IMO4.pic}). Without loss of generality we assume $r_1\geq r_2$. We will first prove that the line $CE$ is the common tangent of the circles $k_1$ and $k_2$. Let $\widehat{E}$ be the other intersection of the drawn circle $k'$ of the triangle $CMN$ and the circle $k_2$. The point $A$ lies on the power axes of the circles $k_1$ and $k_2$ or $k_1$ and $k'$, so according to the \ref{PotencnoSr} theorem, the point $A$ is the power center of the circles $k_1$, $k_2$ and $k'$. This means that the point $A$ lies on the power line of the circles $k_2$ and $k'$ - the line $N\widehat{E}$. From this it follows that $\widehat{E}\in AN\cap k_2$ or $\widehat{E}=E$. So the points $M$, $C$, $E$ and $N$ are conciliatory and according to the \ref{ObodKotTang} theorem, $\angle ACE \cong\angle ANM$. We mark with $L$ any point of the common tangent of the circles $k$ and $k_1$ in the point $M$, which lies in the plane with the edge $AC$, in which the points $B$ and $D$ are not. According to the same theorem \ref{ObodKotTang} (with respect to the circles $k$ and $k_1$), we have $\angle LMA \cong\angle MBA$ and $\angle LMC \cong\angle MDC$. From the \ref{ObodObodKot} theorem (for the circle $k$ and the cord $AM$), we also get $\angle ANM \cong\angle ABM$. If we connect the proven relations, we get: $$\angle ACE \cong \angle ANM \cong \angle ABM\cong \angle LMA \cong\ \angle CDM. $$ From $\angle ACE \cong \angle CDM$ and according to the \ref{ObodKotTang} theorem, it follows that $EC$ is the tangent of the circle $k_1$. Because in the proof we have not yet used the fact that $O_2\in k_1$, we analogously prove that $CE$ is the tangent of the circle $k_2$. Let $T$ be the intersection of line $O_2O_1$ and circle $k_2$. It is enough to prove that $T\in CD$ and $\angle CTO_2=90^0$. Let $O'_2$ be the orthogonal projection of point $O_2$ on line $O_1C$. Because $CE$ is the common tangent of circles $k_1$ and $k_2$, the radii $O_1C$ and $O_2E$ are perpendicular to this tangent. Therefore, $CEO_2O'_2$ is a rectangle and $O'_2C\cong O_2E=r_2$ holds. From this it follows that $O_1O'_2=r_1-r_2=O_1T$. This means that the triangle $O_1O'_2O_2$ is similar (by the \textit{SAS} theorem \ref{SKS}), to triangle $O_1TC$, so $\angle CTO_1\cong\angle O_2O'_2O_1=90^0$ or $\angle CTO_2=90^0$. Line $CT$ and line $AB$ are perpendicular to the center $O_1O_2$ of the two circles. Therefore, $CT\parallel AB$. Because (due to the already proven relation $\angle ABM\cong \angle CDM$), also $CD\parallel AB$. By Playfair's axiom \ref{Playfair}, line $CT$ and line $CD$ are the same line, or $T\in CD$. This means that line $CD$ is tangent to circle $k_2$ at point $T$. \kdokaz %________________________________________________________________________________ \poglavje{The Theorems of Pappus and Pascal} \label{odd7PappusPascal} Theorems in this section are historically connected to the development of \index{geometry!projective}projective geometry. \bizrek \label{izrek Pappus} \index{theorem!Pappus'}(Pappus'\footnote{\index{Pappus} \textit{Pappus} from Alexandria (3rd century), one of the last great ancient Greek geometers. He proved this theorem in the Euclidean case, using metric. But the fundamental role of Pappus' theorem in projective geometry was discovered only sixteen centuries later.} theorem) Let $A$, $B$ and $C$ be three different points of a line $p$ and $A'$, $B'$ and $C'$ three different points of another line $p'$ in the same plane. Then the points $X=BC'\cap B'C$, $Y=AC'\cap A'C$ and $Z=AB'\cap A'B$ are collinear. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.10.1.pic} \caption{} \label{sl.pod.7.10.1.pic} \end{figure} \textbf{\textit{Proof.}} Let $L=AB'\cap BC'$, $M=AB'\cap CA'$ and $N=CA'\cap BC'$ (Figure \ref{sl.pod.7.10.1.pic}). We use Menelaus' theorem (\ref{izrekMenelaj}) five times with respect to the triangle $LMN$ and the lines $BA'$, $AC'$, $CB'$, $AB$, $A'B'$: \begin{eqnarray*} & & \frac{\overrightarrow{LZ}}{\overrightarrow{ZM}}\cdot \frac{\overrightarrow{MA'}}{\overrightarrow{A'N}}\cdot \frac{\overrightarrow{NB}}{\overrightarrow{BL}}=-1,\\ & & \frac{\overrightarrow{LA}}{\overrightarrow{AM}}\cdot \frac{\overrightarrow{MY}}{\overrightarrow{YN}}\cdot \frac{\overrightarrow{NC'}}{\overrightarrow{C'L}}=-1,\\ & & \frac{\overrightarrow{LB'}}{\overrightarrow{B'M}}\cdot \frac{\overrightarrow{MC}}{\overrightarrow{CN}}\cdot \frac{\overrightarrow{NX}}{\overrightarrow{XL}}=-1,\\ & & \frac{\overrightarrow{LA}}{\overrightarrow{AM}}\cdot \frac{\overrightarrow{MC}}{\overrightarrow{CN}}\cdot \frac{\overrightarrow{NB}}{\overrightarrow{BL}}=-1,\\ & & \frac{\overrightarrow{LB'}}{\overrightarrow{B'M}}\cdot \frac{\overrightarrow{MA'}}{\overrightarrow{A'N}}\cdot \frac{\overrightarrow{NC'}}{\overrightarrow{C'L}}=-1. \end{eqnarray*} From these five relations (if we multiply the first three, then insert the fourth and fifth into the resulting relation) we get: $$\frac{\overrightarrow{LZ}}{\overrightarrow{ZM}}\cdot \frac{\overrightarrow{MY}}{\overrightarrow{YN}}\cdot \frac{\overrightarrow{NX}}{\overrightarrow{XL}}=-1.$$ By the converse of Menelaus' theorem (\ref{izrekMenelaj}) the points $X$, $Y$ and $Z$ are collinear. \kdokaz We now prove Pascal's\footnote{It is not known how the French mathematician and philosopher \index{Pascal, B.} \textit{B. Pascal} (1623--1662) proved this statement for a circle, because the original proof is lost. However, we can assume that he used the results and methods of that time, which means that he probably used Menelaus' theorem.} theorem for a circle. \bizrek \index{izrek!Pascalov} \label{izrekPascalEvkl} Let $A$, $B$, $C$, $D$, $E$ and $F$ be arbitrary points on some circle $k$. Then the points $X=AE\cap BD$, $Y=AF\cap CD$ and $Z=BF\cap CE$ are collinear. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.10.2.pic} \caption{} \label{sl.pod.7.10.2.pic} \end{figure} \textbf{\textit{Proof.}} We mark $L=AE\cap BF$, $M=AE\cap CD$ and $N=CD\cap BF$. If we use Menelaus' theorem (\ref{izrekMenelaj}) three times for the triangle $LMN$ and the lines $BD$, $AF$ and $CE$, we get (Figure \ref{sl.pod.7.10.2.pic}): \begin{eqnarray*} \hspace*{-2mm} \frac{\overrightarrow{LX}}{\overrightarrow{XM}}\cdot \frac{\overrightarrow{MD}}{\overrightarrow{DN}}\cdot \frac{\overrightarrow{NB}}{\overrightarrow{BL}}=-1,\hspace*{1mm} \frac{\overrightarrow{LA}}{\overrightarrow{AM}}\cdot \frac{\overrightarrow{MY}}{\overrightarrow{YN}}\cdot \frac{\overrightarrow{NF}}{\overrightarrow{FL}}=-1,\hspace*{1mm} \frac{\overrightarrow{LE}}{\overrightarrow{EM}}\cdot \frac{\overrightarrow{MC}}{\overrightarrow{CN}}\cdot \frac{\overrightarrow{NZ}}{\overrightarrow{ZL}}=-1. \end{eqnarray*} If we use the powers of the points $M$, $N$ and $L$ with respect to the circle $k$ (theorem \ref{izrekPotenca}) or consider the similarity of the corresponding triangles, we get: \begin{eqnarray*} \overrightarrow{MC}\cdot\overrightarrow{MD}= \overrightarrow{MA}\cdot\overrightarrow{ME},\hspace*{3mm} \overrightarrow{NC}\cdot\overrightarrow{ND}= \overrightarrow{NF}\cdot\overrightarrow{NB},\hspace*{3mm} \overrightarrow{LA}\cdot\overrightarrow{LE}= \overrightarrow{LB}\cdot\overrightarrow{LF}. \end{eqnarray*} From the previous six relations it follows: \begin{eqnarray*} \frac{\overrightarrow{LX}}{\overrightarrow{XM}}\cdot \frac{\overrightarrow{MY}}{\overrightarrow{YN}}\cdot \frac{\overrightarrow{NZ}}{\overrightarrow{ZL}}=-1. \end{eqnarray*} Therefore, according to Menelaus' theorem (\ref{izrekMenelaj}, the inverse direction), the points $X$, $Y$ and $Z$ are collinear. \kdokaz The previous statement, which refers to a circle, can be generalized to any conic section\footnote{The study of conic sections began with the Ancient Greeks. The terms ellipse, parabola, hyperbola were first used by the Greek mathematician \index{Apolonij} \textit{Apolonij} from Perga (262--200 BC) in his famous work \textit{Razprava o presekih stožca} (Discussion of the Intersections of a Cone), which consists of eight books, in which he defines a conic section as the intersection of a plane with a circular cone. A commentary on this work by Apolonij was written by the Greek philosopher and mathematician, and the last representative of ancient science, \index{Hipatija} \textit{Hipatija} from Alexandria (370--415). The interest in conic sections was revived by the German astronomer \index{Kepler, J.} \textit{J. Kepler} (1571--1630) and the French mathematician and philosopher \index{Pascal, B.} \textit{B. Pascal} (1623--1662) in the 17th century.} or a curve of the second class. In the Euclidean case, these are: ellipse, parabola and hyperbola. We can determine this by defining a conic section as the intersection of the generators of all sides of the cone and some plane. Because of this, we generally look at a conic section $\mathcal{K}$ as the central projection of some circle $k$ onto some plane. The center of projection is the apex of the cone $S$. We have already mentioned that a central projection preserves collinearity. If $A'$, $B'$, $C'$, $D'$, $E'$ and $F'$ are points on the conic section $\mathcal{K}$ and $X'$, $Y'$ and $Z'$ are the corresponding points, defined as in statement \ref{izrekPascalEvkl}, then these points are the images of some points $A$, $B$, $C$, $D$, $E$, $F$, $X$, $Y$ and $Z$. In this case, the first six points are on the circle $k$, and the last three are collinear by statement \ref{izrekPascalEvkl}. It follows that the points $X'$, $Y'$ and $Z'$ are also collinear (Figure \ref{sl.pod.7.10.3.pic}). This means that statement \ref{izrekPascalEvkl} really holds in the general case for any conic section. This general statement is known as Pascal's statement for conic sections\footnote{\index{Pascal, B.} \textit{B. Pascal} (1623--1662), French mathematician and philosopher, who as a sixteen-year-old proved this important statement about conic sections, and published it in 1640, but at that time the statement did not directly apply to projective geometry.}. \begin{figure}[!htb] \centering \input{sl.pod.7.10.3.pic} \caption{} \label{sl.pod.7.10.3.pic} \end{figure} The ideas we have discussed in this section lead us to the conclusion that we can also derive Pascal's theorem in projective geometry. Even more - in projective geometry we can define and study also the conics, but it is not possible to distinguish between ellipse, hyperbola or parabola (see \cite{Mitrovic}). %________________________________________________________________________________ \poglavje{The Golden Ratio} \label{odd7ZlatiRez} We say that point $Z$ of the line $AB$ divides this line in the ratio of the \index{golden!ratio}\pojem{golden ratio}\footnote{Such a division was already considered by \index{Pythagoras}\textit{Pythagoras from the island of Samos} (582--497 BC), an ancient Greek philosopher and mathematician. The first known records of the golden ratio were created by the ancient Greek mathematician \index{Euclid} \textit{Euclid of Alexandria} (3rd century BC). In his famous work \textit{Elements} he posed the problem: ‘‘\textit{Given a line, divide it into two unequal parts so that the area of the rectangle whose length is equal to the length of the whole line and whose height is equal to the length of the shorter part of the line is equal to the area of the square drawn on the longer part of the line.}’’ The term golden ratio, which we use today, was introduced by \index{Leonardo da Vinci}\textit{Leonardo da Vinci} (1452--1519), an Italian painter, architect and inventor. The golden ratio has been used by people for thousands of years in painting and architecture.}, if the ratio of the length of the whole line to the longer part is equal to the ratio of the longer part to the shorter part (Figure \ref{sl.pod.7.15.1.pic}) or: \begin{eqnarray} \label{eqnZlatiRez} AB:AZ=AZ:ZB. \end{eqnarray} \begin{figure}[!htb] \centering \input{sl.pod.7.15.1.pic} \caption{} \label{sl.pod.7.15.1.pic} \end{figure} If we now insert the shorter part "into" the longer part, we get the same ratio. Indeed, because $AZ:ZB=AB:AZ=(AZ+ZB):AZ$, it also holds that $$ZB:(AZ-ZB)=AZ:ZB.$$ We can continue this process (Figure \ref{sl.pod.7.15.2.pic}) \begin{figure}[!htb] \centering \input{sl.pod.7.15.2.pic} \caption{} \label{sl.pod.7.15.2.pic} \end{figure} Of course, the question arises as to how to construct such a point $Z$. We will describe this construction in the next example. \bzgled For a given line $AB$, construct a point $Z$ that divides the line segment into the golden ratio. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.15.3.pic} \caption{} \label{sl.pod.7.15.3.pic} \end{figure} \textbf{\textit{Solution.}} First, let's draw a circle $k(S,SB)$ with radius $|SB|=\frac{1}{2}|AB|$, which touches the line $AB$ at point $B$ (Figure \ref{sl.pod.7.15.3.pic}). Let's also construct the intersections of this circle with the line $AS$ - we'll mark them with $X$ and $Y$ (so that $\mathcal{B}(A,X,S)$ is true). We now get the point $Z$ as the intersection of the line $AB$ and the circle $l(A,AX)$. We will prove that $Z$ is the desired point. If we use the power of point $A$ with respect to the circle $k$ (statement \ref{izrekPotenca}), we get: $$\hspace*{-1.5mm} |AB|^2=p(A,k)=|AX|\cdot |AY|=|AX|\cdot(|AX|+|XY|)=|AZ|\cdot(|AZ|+|AB|).$$ Therefore: \begin{eqnarray} \label{eqnZlatiRez2} |AB|^2=|AZ|\cdot(|AZ|+|AB|) \end{eqnarray} Next: \begin{eqnarray*} |AB|^2=|AZ|\cdot(|AZ|+|AB|)\hspace*{1mm}&\Leftrightarrow &\hspace*{1mm} \frac{|AB|}{|AZ|}=\frac{|AZ|+|AB|}{|AB|}\\ \hspace*{1mm}&\Leftrightarrow &\hspace*{1mm} \frac{|AB|}{|AZ|}=\frac{|AZ|}{|AB|}+1\\ \hspace*{1mm}&\Leftrightarrow &\hspace*{1mm} \frac{|AB|}{|AZ|}-1=\frac{|AZ|}{|AB|}\\ \hspace*{1mm}&\Leftrightarrow &\hspace*{1mm} \frac{|AB|-|AZ|}{|AZ|}=\frac{|AZ|}{|AB|}\\ \hspace*{1mm}&\Leftrightarrow &\hspace*{1mm} \frac{|BZ|}{|AZ|}=\frac{|AZ|}{|AB|} \end{eqnarray*} which is equivalent to relation \ref{eqnZlatiRez}. This means that point $Z$ divides the line segment $AB$ in the golden ratio. \kdokaz In the next example we will calculate the value of the ratio determined by the golden ratio. \bizrek \label{zlatiRezStevilo} If a point $Z$ divides a line segment $AB$ into the golden ratio ($AZ$ is the longer part), then \begin{eqnarray*} && AZ:ZB=AB:AZ=\frac{\sqrt{5}+1}{2}, \hspace*{1mm}\textrm{ i.e.}\\ && AZ=\frac{\sqrt{5}-1}{2}AB. \end{eqnarray*} \eizrek \textbf{\textit{Proof.}} Relation \ref{eqnZlatiRez2} from the previous statement is equivalent to the relation: $$|AB|^2-|AB|\cdot |AZ|-|AZ|^2=0.$$ If we solve this quadratic equation for $|AZ|$, we get: $$|AZ|=\frac{\sqrt{5}+1}{2}|AB|,$$ from which the desired equalities follow. \kdokaz The number $$\Phi=\frac{\sqrt{5}+1}{2}$$ from the previous equation, which therefore represents the value of the ratio determined by the golden cut, is called the \index{number!golden}\pojem{golden number}. Of course, it is an irrational number ($\Phi\notin \mathbb{Q}$). Its approximate value is: $$\Phi=\frac{\sqrt{5}+1}{2}\doteq 1.62.$$ Similarly, it is: $$\frac{\sqrt{5}-1}{2}\doteq 0.62,$$ which means that the longer part $AZ$ of the golden cut is approximately $62\%$ of the entire distance $AB$. . \bzgled \label{zlatiRezKonstr} Construct a golden rectangle $ABCD$ with two sides $a$ and $b$ in the golden ratio - so-called \index{golden!rectangle}\pojem{golden rectangle}\color{green1}. \ezgled \begin{figure}[!htb] \centering \input{sl.pod.7.15.4.pic} \caption{} \label{sl.pod.7.15.4.pic} \end{figure} \textbf{\textit{Solution.}} We draw any line segment $AB$, then the point $Z$, which divides the line segment $AB$ in the golden cut so that $AZ$ is the longer part (see the previous example \ref{zlatiRezKonstr}). In the end, we get $D=\mathcal{R}_{A,90^0}(Z)$ and $C=\mathcal{T}_{\overrightarrow{AB}}(D)$ (Figure \ref{sl.pod.7.15.4.pic}). \kdokaz If we divide the golden rectangle along its longer side in the golden ratio, we get a square and another golden rectangle. We divide the new golden rectangle in the same way and continue the process. If we connect the corresponding vertices (the places of the diagonals of the squares that determine the cusp), so that we draw circular arcs with a central angle of $90^0$, we get an approximate construction of the so-called \index{golden!spiral}\pojem{golden spiral} or \index{logarithmic!spiral}\pojem{logarithmic spiral} (Figure \ref{sl.pod.7.15.5.pic}). \begin{figure}[!htb] \centering \input{sl.pod.7.15.5.pic} \caption{} \label{sl.pod.7.15.5.pic} \end{figure} Logarithmic spiral\footnote{The proposal for the name logarithmic spiral was made by the French mathematician \index{Varignon, P.}\textit{P. Varignon} (1645--1722). The logarithmic spiral has the property that each straight line from the center of the spiral intersects it at the same angle. It appears in various forms in nature: from spirals in sunflower petals, snail shells and spider webs to distant galaxies. We also call it the spiral mirabilis (miraculous spiral) - this name was proposed by the Swiss mathematician \index{Bernoulli, J.}\textit{J. Bernoulli} (1667–-1748) because of its wonderful properties.} is a curve defined by equations in parametric form: \begin{eqnarray*} x&=&a e^{bt}\cdot\cos t \\ y&=&a e^{bt}\cdot\sin t, \end{eqnarray*} where $t\in \mathbb{R}$ is a parameter, $a$ and $b$ are arbitrary real constants and $e\doteq 2,72$ is the \index{number!Euler's }\pojem{Euler's\footnote{Number $e$ is an irrational number and represents the value of the limit $\lim_{n\rightarrow \infty}\left(1+\frac{1}{n} \right)^n$ =e. It was named after the Swiss mathematician \index{Euler, L.}\textit{L. Euler} (1707--1783).} number.}. The previous construction (with circular arcs) is, as we have already mentioned, approximate, but it represents a very good approximation of this curve (Figure \ref{sl.pod.7.15.5a.pic}). \begin{figure}[!htb] \centering \input{sl.pod.7.15.5a.pic} \caption{} \label{sl.pod.7.15.5a.pic} \end{figure} We will continue with a regular pentagon. \bzgled \label{zlatiRezPravPetk1} A diagonal and a side of a regular pentagon are in the golden ratio. \ezgled \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.15.6.pic}) Let's mark with $a$ the side and with $d$ the diagonal of the regular pentagon $ABCDE$. From \ref{PtolomejPetkotnik} it follows: $$d=\frac{1+\sqrt{5}}{2}a.$$ Therefore, according to \ref{zlatiRezStevilo}, the segments $d$ and $a$ are in the golden ratio. \kdokaz \begin{figure}[!htb] \centering \input{sl.pod.7.15.6.pic} \caption{} \label{sl.pod.7.15.6.pic} \end{figure} \bzgled Two diagonals of a regular pentagon intersect at a point that divides them in the golden ratio\footnote{The Pythagoreans knew this property of the pentagon. The Pythagorean school was founded by the Greek philosopher and mathematician \index{Pitagora}\textit{Pitagora from the island of Samos} (582--497 BC) in Croton in southern Italy. His students were engaged in philosophy, mathematics and natural sciences. For their distinctive sign they chose the pentagram, which is composed of the pentagon's diagonals. The influence of the Pythagorean school on the mathematics of the Ancient Greeks lasted for several centuries after Pythagora's death.}. \ezgled \textbf{\textit{Proof.}} We denote by $P$ the intersection of the diagonals $AC$ and $BD$ of a regular pentagon $ABCDE$ and by $k$ its circumscribed circle (Figure \ref{sl.pod.7.15.6.pic}). The angles $EAD$, $DAC$, $CAB$ and $DBC$, which are adjacent to the chords $ED$, $CD$ and $CB$, are congruent (statement \ref{SklTetSklObKot}). Because according to statement \ref{pravVeckNotrKot} the interior angle $EAB$ of a regular pentagon is equal to $108^0$, the angles $EAD$, $DAC$, $CAB$ (or $PAB$) and $DBC$ are equal to $36^0$. From this we obtain $\angle PBA=\angle DBA=\angle CBA-\angle CBD=108^0-36^0= 72^0$. From the sum of the angles of triangle $ABP$ according to statement \ref{VsotKotTrik} it follows $\angle APB=180^0-\angle PBA-\angle PAB=72^0$. Therefore, the triangle $PAB$ is isosceles with the base $PB$ (statement \ref{enakokraki}), or $AP\cong AB$. From this it follows: $$AC:AP=AC:AB.$$ If we use the previous statement \ref{zlatiRezPravPetk1}, we conclude that the point $P$ divides the diagonal $AC$ in the golden ratio. \kdokaz If in a regular pentagon $ABCDE$ we choose the vertices $A$, $B$ and $D$, we get an isosceles triangle $ABD$ with a base that is equal to the side $a$ of the pentagon $ABCDE$, and the legs are equal to the diagonals $d$ of this pentagon. The leg and the base of this triangle are therefore in the golden ratio (Figure \ref{sl.pod.7.15.7.pic}). Therefore, the triangle $ABD$ is called the \index{golden!triangle}\pojem{golden triangle}. The angles of the golden triangle measure $72^0$, $72^0$ and $36^0$. \begin{figure}[!htb] \centering \input{sl.pod.7.15.7.pic} \caption{} \label{sl.pod.7.15.7.pic} \end{figure} Similarly to the golden rectangle, if we use the similitudes of the corresponding internal angles at the base of the golden triangle, we get a sequence of golden triangles, all of which are similar. With the help of the corresponding circular arcs, we can also derive another approximate construction of the golden (or logarithmic) spiral (Figure \ref{sl.pod.7.15.7.pic}). %________________________________________________________________________________ \poglavje{Morley's Theorem and Some More Theorems} \label{odd7Morly} We say that the segments $SP$ and $SQ$ \index{trisektrisa}\pojem{trisektrisi} of the angle $ASB$, if the points $P$ and $Q$ lie in this angle and it holds $$\angle ASP\cong\angle PSQ\cong\angle QSB,$$ or if it is a segment that divides the angle into three congruent angles (Figure \ref{sl.pod.7.16.0.pic}). \begin{figure}[!htb] \centering \input{sl.pod.7.16.0.pic} \caption{} \label{sl.pod.7.16.0.pic} \end{figure} We first prove an auxiliary assertion - lemma. \bizrek \label{izrekMorleyLema} Let $Y'$, $Z$, $Y$ and $Z'$ be points in the plane such that $Y'Y\cong ZY\cong ZZ'$ and $$\angle Z'ZY\cong \angle ZYY'=180^0-2\alpha>60^0.$$ If $A$ is a point in this plane that is on the different side of the line $Z'Y'$ with respect to the point $Z$ and also $Y'AZ'=3\alpha$, then the points $A$, $Y'$, $Z$, $Y$ and $Z'$ are concyclic and also $$\angle Z'AZ\cong \angle ZAY\cong \angle YAY'=\alpha.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.16.1a.pic} \caption{} \label{sl.pod.7.16.1a.pic} \end{figure} \textbf{\textit{Proof.}} The given condition $180^0-2\alpha>60^0$ is equivalent to the condition $\alpha<60^0$. Let $s$ be the perpendicular bisector of the line segment $ZY$, and $\mathcal{S}_s$ be the reflection over the line $s$ (Figure \ref{sl.pod.7.16.1a.pic}). Thus $\mathcal{S}_s(Z)=Y$. Because $\mathcal{S}_s$ is an isometry that preserves angles and lengths of line segments, $\mathcal{S}_s(Z')=Y'$ as well. From $ZY,Z'Y'\perp s$ it follows that $ZY\parallel Z'Y'$, which means that the trapezoid $ZYY'Z'$ is isosceles. By the \ref{trapezTetivEnakokr} theorem, this trapezoid is tangent - we denote its inscribed circle with $k$. By the \ref{KotiTransverzala} theorem, the angles $YZZ'$ and $Y'Z'Z$ are supplementary, so we have: \begin{eqnarray} \label{eqnMorleyLema1} \angle ZZ'Y'=180^0-\angle Z'ZY=2\alpha. \end{eqnarray} From $Z'Z\cong ZY$ by the \ref{SklTetSklObKot} theorem it follows that $\angle ZZ'Y\cong \angle YZ'Y'$, so by the relation \ref{eqnMorleyLema1} $\angle ZZ'Y\cong \angle YZ'Y'=\alpha$. The triangle $Z'ZY$ is isosceles with the base $Z'Y$, so by the \ref{enakokraki} theorem it is also $\angle ZYZ'\cong \angle ZZ'Y=\alpha$. Analogously, $\angle YY'Z\cong \angle ZY'Z'=\alpha$ and $\angle YZY'\cong \angle YY'Z=\alpha$. If we put everything together, we get $\angle ZZ'Y\cong \angle YZ'Y'\cong \angle ZYZ'\cong \angle YY'Z\cong \angle ZY'Z'\cong\angle YZY'=\alpha$, so: \begin{eqnarray} \label{eqnMorleyLema2} \angle Z'Y'Z \cong \angle Y'Z'Y= \alpha. \end{eqnarray} Because $\angle Z'ZY'=\angle Z'ZY-\angle Y'ZY=180^0-2\alpha-\alpha=180^0-3\alpha$ (this angle exists because $\alpha<60^0$) or $\angle Z'ZY'+\angle Z'AY'=180^0-3\alpha+3\alpha=180^0$, by the \ref{TetivniPogoj} theorem the quadrilateral $Z'ZY'A$ is tangent, which means that the point $A$ lies on the circle $k$. From the relation \ref{eqnMorleyLema2} and the \ref{ObodObodKot} theorem it follows: \begin{eqnarray*} &&\angle Z'AZ\cong \angle Z'Y'Z=\alpha,\\ &&\angle Y'AY\cong \angle Y'Z'Y=\alpha. \end{eqnarray*} Finally, we have: \begin{eqnarray*} \angle ZAY=\angle Z'AY'-\angle Z'AZ-\angle Y'AY =3\alpha-\alpha-\alpha=\alpha, \end{eqnarray*} which was to be proven. \kdokaz Now we are ready for the basic theorem. \bizrek \label{izrekMorley}\index{izrek!Morleyev} If $X$, $Y$ and $Z$ are three points of intersection of the adjacent angle trisectors of a triangle $ABC$, then $XYZ$ is an equilateral triangle.\\ (Morley's\footnote{\index{Morley, F.}\textit{F. Morley} (1860--1937), English mathematician, discovered this property of a triangle in 1904, but published it only 20 years later. At that time, the theorem was published as a task in the journal \textit{Educational Times}. Here we will give one of the solutions proposed at that time.} theorem) \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.16.1.pic} \caption{} \label{sl.pod.7.16.1.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.16.1.pic}) Let $X$ and $V$ be the points of intersection of the respective trisectors of the internal angles $ABC$ and $ACB$ of the triangle $ABC$, so that $X$ is the internal point of the triangle $BVC$. We also denote by $Z_1$ and $Y_1$ the points of the trisector $BV$ and $CV$, for which $\angle Z_1XV \cong \angle VXY_1=30^0$. We first prove that $XY_1Z_1$ is an equilateral triangle. Since the lines $BX$ and $CX$ are the angle bisectors of the angles $VBC$ and $VCB$, the point $X$ is the center of the inscribed circle of the triangle $BVC$, so the line $VX$ is the angle bisector of the angle $BVC$ (theorem \ref{SredVcrtaneKrozn}). From the similarity of the triangles $VXZ_1$ and $VXY_1$ (theorem \textit{ASA} \ref{KSK}) it follows that $XZ_1\cong XY_1$. Since $\angle Z_1XY_1=60^0$, $XY_1Z_1$ is an equilateral triangle (theorem \ref{enakokraki}). It is enough to prove that $Y_1=Y$ and $Z_1=Z$ or $\angle BAZ_1\cong\angle Z_1AY_1\cong\angle Z_1AC$. From the congruence of the triangles $VXZ_1$ and $VXY_1$ it follows that $Z_1V \cong Y_1V$. Therefore, $VZ_1Y_1$ is an isosceles triangle with a base $Z_1Y_1$, so according to the theorem \ref{enakokraki}: \begin{eqnarray} \label{eqnMorley1} \angle VZ_1Y_1\cong\angle VY_1Z_1. \end{eqnarray} Let $Z'$ and $Y'$ be such points on the sides $AB$ and $AC$ of the triangle $ABC$, that $BZ'\cong BX$ and $CY'\cong CX$. The triangles $BZ'Z_1$ and $BXZ_1$ are congruent (theorem \textit{SAS} \ref{SKS}), so $Z'Z_1\cong XZ_1$. Also, $\angle Z'Z_1B\cong\angle XZ_1B$ or the line $BV$ is the perpendicular bisector of the angle $Z'Z_1X$. Similarly, $Y'Y_1\cong XY_1$. From this and from the fact that $XY_1Z_1$ is an equilateral triangle, it follows: \begin{eqnarray} \label{eqnMorley2} Z'Z_1\cong Z_1Y_1\cong Y_1Y'. \end{eqnarray} We denote with $3\alpha$, $3\beta$, $3\gamma$ the measures of the internal angles of the triangle $ABC$, at the vertices $A$, $B$ and $C$. It is clear that $3\alpha+3\beta+3\gamma=180^0$, therefore: \begin{eqnarray} \label{eqnMorley3} 2\alpha+2\beta+2\gamma=120^0. \end{eqnarray} If we use the fact that the line $BV$ is the perpendicular bisector of the convex or non-convex angle $Z'Z_1X$ and the relations \ref{eqnMorley1} and \ref{eqnMorley3}, after simple calculations we get: \begin{eqnarray*} \angle Z'Z_1Y_1&=&\angle Z'Z_1V+\angle VZ_1Y_1=\\ &=&\angle VZ_1X+\angle VZ_1Y_1=\\ &=&60^0+2\angle VZ_1Y_1=\\ &=&60^0+\angle VZ_1Y_1+\angle VY_1Z_1=\\ &=&60^0+180^0-\angle Z_1VY_1=\\ &=&60^0+2\beta+2\gamma=\\ &=&60^0+120^0-2\alpha=\\ &=&180^0-2\alpha. \end{eqnarray*} Similarly, $\angle Y'Y_1Z_1=180^0-2\alpha$, so (if we take into account the relation \ref{eqnMorley2}) by theorem \ref{izrekMorleyLema} $\angle Z'AZ_1\cong\angle Z_1AY_1\cong\angle Y_1AY'= \alpha$, which was to be proven. \kdokaz \bizrek (Leibniz's\footnote{\index{Leibniz, G. W.}\textit{G. W. Leibniz} (1646--1716), German mathematician.} theorem) \label{izrekLeibniz}\index{theorem!Leibniz's} If $T$ is the centroid and $X$ an arbitrary point in the plane of a triangle $ABC$, then $$|XA|^2 + |XB|^2 + |XC|^2 = \frac{1}{3}\left(|AB|^2 +|BC|^2 +|CA|^2\right) + 3|XT|^2 ,\textrm{ i.e.}$$ $$|XA|^2 + |XB|^2 + |XC|^2 = |TA|^2 +|TB|^2 +|TC|^2 + 3|XT|^2.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.16.2.pic} \caption{} \label{sl.pod.7.16.2.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.16.2.pic}) Let $A_1$ be the center of the side $BC$ or $|AA_1|=t_a$ its centroid. Because $AT:TA_1=2:1$ (statement \ref{tezisce}), by Stewart's theorem \ref{StewartIzrek2} with respect to the triangle $AXA_1$ it follows: \begin{eqnarray} \label{eqnLeibniz1} |XT|^2=\frac{1}{3}|XA|^2+\frac{2}{3}|XA_1|^2-\frac{2}{9}t_a^2 \end{eqnarray} If we use Stewart's theorem \ref{StewartIzrek2} again (or its consequence for the centroid \ref{StwartTezisc}) with respect to the triangle $BXC$, we get: \begin{eqnarray} \label{eqnLeibniz2} |XA_1|^2=\frac{1}{2}|XB|^2+\frac{1}{2}|XC|^2-\frac{1}{4}|BC|^2 \end{eqnarray} if we insert \ref{eqnLeibniz2} in \ref{eqnLeibniz1}: \begin{eqnarray} \label{eqnLeibniz3} |XT|^2=\frac{1}{3}\left(|XA|^2+|XB|^2+|XC|^2\right)-\frac{1}{6}|BC|^2-\frac{2}{9}t_a^2 \end{eqnarray} and similarly: \begin{eqnarray} \label{eqnLeibniz4a} \hspace*{-4mm} |XT|^2&=&\frac{1}{3}\left(|XA|^2+|XB|^2+|XC|^2\right)-\frac{1}{6}|AC|^2-\frac{2}{9}t_b^2\\ \hspace*{-4mm} |XT|^2&=&\frac{1}{3}\left(|XA|^2+|XB|^2+|XC|^2\right)-\frac{1}{6}|AB|^2-\frac{2}{9}t_c^2\label{eqnLeibniz4} \end{eqnarray} By adding the equality from \ref{eqnLeibniz3} - \ref{eqnLeibniz4} we get: \begin{eqnarray*} 3|XT|^2&=&|XA|^2+|XB|^2+|XC|^2-\frac{1}{6}\left(|BC|^2+|AC|^2+|AB|^2\right)-\\ &-&\frac{2}{9}\left(t_a^2+t_b^2+t_c^2\right) \end{eqnarray*} Finally, from the statement \ref{StwartTezisc2} it follows: \begin{eqnarray*} 3|XT|^2=|XA|^2+|XB|^2+|XC|^2-\frac{1}{3}\left(|BC|^2+|AC|^2+|AB|^2\right) \end{eqnarray*} or both relations from the statement. \kdokaz A direct consequence of Leibniz's formula is the following statement. \bizrek A point in the plane of a triangle for which the squared distances from its vertices has a minimum value is its centroid. \eizrek \textbf{\textit{Proof.}} By Leibnitz's formula for any point $X$ in the plane of the triangle $ABC$ with centroid $T$ it holds: $$|XA|^2 + |XB|^2 + |XC|^2 = \frac{1}{3}\left(|AB|^2 +|BC|^2 +|CA|^2\right) + 3|XT|^2.$$ The minimum of the sum $|XA|^2 + |XB|^2 + |XC|^2$ with respect to $X$ is therefore achieved when $|XT|$ is the smallest, which is when $X=T$. \kdokaz \bizrek (Carnot's\footnote{\index{Carnot, L. N. M.}\textit{L. N. M. Carnot} (1753--1823), French mathematician.} theorem)\index{theorem!Carnot's} Let $P$, $Q$ and $R$ be points on the lines containing the sides $BC$, $CA$ and $AB$ of a triangle $ABC$. Perpendicular lines on the lines $BC$, $CA$. and $AB$ at the points $P$, $Q$ and $R$ intersect at one point if and only if \begin{eqnarray} \label{eqnCarnotIzrek1} |BP|^2 - |PC|^2 + |CQ|^2 - |QA|^2 + |AR|^2 - |RB|^2 = 0. \end{eqnarray} \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.16.3.pic} \caption{} \label{sl.pod.7.16.3.pic} \end{figure} \textbf{\textit{Proof.}} We mark with $p$, $q$ and $r$ the perpendiculars of the lines $BC$, $CA$ and $AB$ through the points $P$, $Q$ and $R$ (Figure \ref{sl.pod.7.16.3.pic}). ($\Rightarrow$) We first assume that the lines $p$, $q$ and $r$ intersect at some point $L$. If we use the Pythagorean theorem \ref{PitagorovIzrek} six times, we get: \begin{eqnarray*} \begin{array}{cc} |AR|^2+|RL|^2=|AL|^2 & \hspace{6mm} -|AQ|^2-|QL|^2=-|AL|^2 \\ |BP|^2+|PL|^2=|BL|^2 & \hspace{6mm} -|BR|^2-|RL|^2=-|BL|^2\\ |CQ|^2+|QL|^2=|CL|^2 & \hspace{6mm} -|CP|^2-|PL|^2=-|CL|^2 \end{array} \end{eqnarray*} If we add all six equalities, we get relation \ref{eqnCarnotIzrek1}. ($\Leftarrow$) Now let's assume that the relation \ref{eqnCarnotIzrek1} is true. The rectangles $q$ and $r$ of the sides $AC$ and $AB$ are not parallel (because otherwise, as a result of Playfair's axiom \ref{Playfair1}, the points $A$, $B$ and $C$ would be collinear). We denote by $\widehat{L}$ the intersection of the lines $q$ and $r$. We denote by $\widehat{P}$ the orthogonal projection of the point $\widehat{L}$ onto the line $BC$. Because the rectangles of the sides of the triangle $ABC$ intersect at the points $\widehat{P}$, $Q$ and $R$ in the point $\widehat{L}$, from the first part of the proof ($\Rightarrow$) it follows: \begin{eqnarray} \label{eqnCarnotIzrek2} |B\widehat{P}|^2 - |\widehat{P}C|^2 + |CQ|^2 - |QA|^2 + |AR|^2 - |RB|^2 = 0. \end{eqnarray} From \ref{eqnCarnotIzrek1} and \ref{eqnCarnotIzrek2} it follows: $$|B\widehat{P}|^2 - |\widehat{P}C|^2=|BP|^2 - |PC|^2.$$ In the same way as at the end of the second part of the proof of the \ref{PotencOsLema} theorem, we get $P=\widehat{P}$, i.e. the lines $p$, $q$ and $r$ intersect in one point. \kdokaz \bizrek (Butterfly theorem\footnote{An English mathematician \index{Horner, W. J.}\textit{W. J. Horner} (1786--1837) published a proof of this theorem in 1815.}) \index{theorem!butterfly} Let $S$ be the midpoint of chord $PQ$ of a circle $k$. Suppose that $AB$ and $CD$ are arbitrary chords of this circle passing through the point $S$. If $X$ and $Y$ are the points of intersection of the chords $AD$ and $BC$ with the chord $PQ$, then $S$ is the midpoint of the line segment $XY$. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.16.4.pic} \caption{} \label{sl.pod.7.16.4.pic} \end{figure} \textbf{\textit{Proof.}} Let's mark $x=|SX|$ and $y=|SY|$ (Figure \ref{sl.pod.7.16.4.pic}). Let $X_1$ and $X_2$ or $Y_1$ and $Y_2$ be the perpendicular projections of points $X$ or $Y$ on the lines $AB$ and $CD$. By Tales' theorem (\ref{TalesovIzrekDolzine}) it is: \begin{eqnarray*} \frac{x}{y}=\frac{XX_1}{YY_1}=\frac{XX_2}{YY_2} \end{eqnarray*} If we use the latter equality, then similarity of triangles $AXX_1$ and $CYY_2$ or triangles $DXX_2$ and $BYY_1$ (statement \ref{PodTrikKKK}) and finally the power of points $X$ and $Y$ with respect to the circle $k$ (statement \ref{izrekPotenca}), we get: \begin{eqnarray*} \frac{x^2}{y^2}&=& \frac{|XX_1|}{|YY_1|}\cdot\frac{|XX_2|}{|YY_2|}= \frac{|XX_1|}{|YY_2|}\cdot\frac{|XX_2|}{|YY_1|}=\\ &=& \frac{|AX|}{|CY|}\cdot\frac{|DX|}{|BY|}=\frac{|AX|\cdot |DX|}{|CY|\cdot |BY|}=\\ &=& \frac{|XP|\cdot |XQ|}{|YP|\cdot |YQ|}= \frac{\left(|PS|-x\right)\cdot\left(|PS|+x\right)} {\left(|PS|+y\right)\cdot\left(|PS|-y\right)}=\\ &=& \frac{|PS|^2-x^2}{|PS|^2-y^2}. \end{eqnarray*} From $$\frac{x^2}{y^2}=\frac{|PS|^2-x^2}{|PS|^2-y^2}$$ it finally follows that $x=y$. \kdokaz \bizrek \index{krožnica!Taylorjeva} \label{izrekTaylor} Let $A'$, $B'$ and $C'$ be the foots of the altitudes of a triangle $ABC$. The foot of the perpendiculars from the points $A'$, $B'$ and $C'$ on the lines containing the adjacent sides of that triangle lie on a circle, so-called \pojem{Taylor\footnote{\index{Taylor, B.} \textit{B. Taylor} (1685--1731), angleški matematik.} circle} \color{blue} of that triangle. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.16.5.pic} \caption{} \label{sl.pod.7.16.5.pic} \end{figure} \textbf{\textit{Proof.}} Let $A_c$ and $A_b$ be the orthogonal projections of points $C'$ and $B'$ on the line $BC$, $B_a$ and $B_c$ be the orthogonal projections of points $A'$ and $C'$ on the line $AC$ and $C_a$ and $C_b$ be the orthogonal projections of points $A'$ and $B'$ on the line $AB$ (Figure \ref{sl.pod.7.16.5.pic}). Triangle $A'B'C'$ is the pedal triangle of triangle $ABC$, so from the proof of Theorem \ref{PedalniVS} it follows $\angle B'A'C\cong\angle BAC=\alpha$ and $\angle C'B'A\cong\angle CBA=\beta$. Because $\angle B'B_aA'=\angle B'A_bA'=90^0$, by Theorem \ref{TalesovIzrKroz2} $A'A_bB_aB'$ is a trapezoid, so $\angle A_bB_aC\cong B'A'C=\alpha$ (Theorem \ref{TetivniPogojZunanji}). From this, by Theorem \ref{KotiTransverzala} it follows $A_bB_a \parallel BA$. Triangle $C'A_cB_c$ and $VA'B'$ are perspective with respect to point $C$ (this means that the lines $C'V$, $A_cA'$ and $B_cB'$ intersect at point $C$). Because $C'A_c\parallel VA'$ and $C'B_c\parallel VB'$, by the generalization of Desargues' Theorem \ref{izrekDesarguesOsNesk} it follows that $A_cB_c\parallel A'B'$. By Theorem \ref{KotiTransverzala1} it follows that $\angle B_cA_cA'\cong B'A'C=\alpha$. Because therefore $\angle B_cA_cA'=\alpha=A_bB_aC$, $A_cA_bB_aB_c$ is a trapezoid (Theorem \ref{TetivniPogojZunanji}); we denote its circumscribed circle with $k$. It remains to prove that the points $C_a$ and $C_b$ also lie on the circle $k$. Analogously to the proven $A_cB_c\parallel A'B'$, it is also $C_aB_a\parallel C'B'$ and analogously to the proven, that $\angle A_bB_aC\cong B'A'C=\alpha$, it is also $\angle C_aA_cB\cong\angle A'C'B=\gamma$. From the parallelism $C_aB_a\parallel C'B'$ it follows (from the statement \ref{KotiTransverzala1}) $\angle C_aB_aA \cong\angle C'B'A = \beta$, therefore: $$\angle C_aB_aA_b=180^0-\angle A_bB_aC-\angle C_aB_aA= 180^0-\alpha-\beta=\gamma.$$ Therefore $\angle C_aA_cB\cong\angle C_aB_aA_b=\gamma$, which means (from the statement \ref{TetivniPogojZunanji}), that $C_aA_cA_bB_a$ is a cyclic quadrilateral, therefore the point $C_a$ lies on the circle $k$. Analogously we prove that also the point $C_b$ lies on the circle $k$. \kdokaz A direct consequence is the following theorem. \bizrek Let $A'$, $B'$ and $C'$ be the foots of the altitudes of a triangle $ABC$. The foot of the perpendiculars from the points $A'$, $B'$ and $C'$ on the lines containing the adjacent sides of that triangle determine three congruent line segments. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.16.6.pic} \caption{} \label{sl.pod.7.16.6.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.16.6.pic}) In the proof of the previous statement \ref{izrekTaylor} we have determined that $A_bB_a\parallel BA$, therefore the quadrilateral $C_bC_aA_bB_a$ is a trapezoid. This trapezoid is also a cyclic (the previous statement \ref{izrekTaylor}) and therefore it is also an isosceles trapezoid (\ref{trapezTetivEnakokr}). Its diagonals are congruent (the statement \ref{trapezEnakokraki}), or. it is $A_bC_b\cong C_aB_a$. Analogously it is also $A_bC_b\cong A_cB_c$. \kdokaz \vspace*{12mm} %Tales \poglavje{The basics of Geometry} \label{osn9Geom} It remains to prove that the points $C_a$ and $C_b$ also lie on the circle $k$. Analogously to the proven $A_cB_c\parallel A'B'$, it is also $C_aB_a\parallel C'B'$ and analogously to the proven, that $\angle A_bB_aC\cong B'A'C=\alpha$, it is also $\angle C_aA_cB\cong\angle A'C'B=\gamma$. From the parallelism $C_aB_a\parallel C'B'$ it follows (from the statement \ref{KotiTransverzala1}) $\angle C_aB_aA \cong\angle C'B'A = \beta$, therefore: $$\angle C_aB_aA_b=180^0-\angle A_bB_aC-\angle C_aB_aA= 180^0-\alpha-\beta=\gamma.$$ Therefore $\angle C_aA_cB\cong\angle C_aB_aA_b=\gamma$, which means (from the statement \ref{TetivniPogojZunanji}), that $C_aA_cA_bB_a$ is a cyclic quadrilateral, therefore the point $C_a$ lies on the circle $k$. Analogously we prove that also the point $C_b$ lies on the circle $k$. \kdokaz A direct consequence is the following theorem. \bizrek Let $A'$, $B'$ and $C'$ be the foots of the altitudes of a triangle $ABC$. The foot of the perpendiculars from the points $A'$, $B'$ and $C'$ on the lines containing the adjacent sides of that triangle determine three congruent line segments. \eizrek \begin{figure}[!htb] \centering \input{sl.pod.7.16.6.pic} \caption{} \label{sl.pod.7.16.6.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.pod.7.16.6.pic}) In the proof of the previous statement \ref{izrekTaylor} we have determined that $A_bB_a\parallel BA$, therefore the quadrilateral $C_bC_aA_bB_a$ is a trapezoid. This trapezoid is also a cyclic (the previous statement \ref{izrekTaylor}) and therefore it is also an isosceles trapezoid (\ref{trapezTetivEnakokr}). Its diagonals are congruent (the statement \ref{trapezEnakokraki}), or. it is $A_bC_b\cong C_aB_a$. Analogously it is also $A_bC_b\cong A_cB_c$. \kdokaz \vspace*{12mm} %Tales \poglavje{The basics of Ge \item Let $S$ be the intersection of the diagonal $AC$ and $BD$ of the trapezoid $ABCD$. Let $P$ and $Q$ be the intersections of the parallels to the bases $AB$ and $CD$ through the point $S$ with the sides of this trapezoid. Prove that $S$ is the center of the line $PQ$. \item Let $ABCD$ be a trapezoid with the base $AB$, the point $S$ be the intersection of its diagonals and $E$ be the intersection of the altitudes of this trapezoid. Prove that the line $SE$ goes through the centers of the bases $AB$ and $CD$. \item Let $P$, $Q$ and $R$ be points in which an arbitrary line through the point $A$ intersects the altitudes of the sides $BC$ and $CD$ and the diagonal $BD$ of the parallelogram $ABCD$. Prove that $|AR|^2=|PR|\cdot |QR|$. \item The points $D$ and $K$ lie on the sides $BC$ and $AC$ of the triangle $ABC$ so that $BD:DC=2:5$ and $AK:KC=3:2$. Calculate the ratio in which the line $BK$ divides the line $AD$. \item Let $P$ be a point on the side $AD$ of the parallelogram $ABCD$ such that $\overrightarrow{AP}= \frac{1}{n}\overrightarrow{AD}$, and $Q$ be the intersection of the lines $AC$ and $BP$. Prove that: $$\overrightarrow{AQ}=\frac{1}{n + 1}\overrightarrow{AC}.$$ %Homotetija \item Given are: the point $A$, the lines $p$ and $q$ and the lines $m$ and $n$. Draw a line $s$ through the point $A$, which intersects the lines $p$ and $q$ in such points $X$ and $Y$, that $XA:AY= m:n$. \item Given are: the point $S$, the lines $p$, $q$ and $r$ and the lines $m$ and $n$. Draw a line $s$ through the point $S$, which intersects the lines $p$, $q$ and $r$ in such points $X$, $Y$ and $Z$, that $XY:YZ=m:n$. \item In the given triangle $ABC$ draw such a rectangle $PQRS$, that the side $PQ$ lies on the side $BC$, the points $R$ and $S$ lie on the sides $AB$ and $AC$, and also $PQ=2QR$. \item Draw:\\ (\textit{a}) a rhombus, if given are the side $a$ and the ratio of the diagonals $e:f$;\\ (\textit{b}) a trapezoid, if given are: the angles $\alpha$ and $\beta$ at one of the bases, the ratio of this base and the height $a:v$ and the other base $c$. \item Let $A$ be a point inside the angle $pSq$, $l$ a line and $\alpha$ an angle in some plane. Draw a triangle $APQ$, such that: $P\in p$, $Q\in q$, $\angle PAQ\cong \alpha$ and $PQ\parallel l$. \item Draw a circle $l$, that touches the given circle $k$ and the given line $p$, if the point of contact is given:\\ (\textit{a}) $l$ and $p$;\hspace*{3mm} (\textit{b}) $l$ and $k$. \item Draw a circle, that touches the arms of the angle $pSq$ and:\\ (\textit{a}) goes through the given point,\\ (\textit{b}) touches the given circle. \item Let $p$ and $q$ be lines, $S$ a point ($S\notin p$ and $S\notin q$) and the lines $m$ and $n$. Draw circles $k$ and $l$, that touch each other from the outside in the point $S$, the first one touches the line $p$, the second line $q$, and the ratio of the radii is $m:n$. \item In a plane are given: a line $p$ and points $B$ and $C$, that lie on the circle $k$. Draw a point $A$ on the circle $k$, so that the centroid of the triangle $ABC$ lies on the line $p$. \item Let $k$ be a circle with the diameter $PQ$. Draw a square $ABCD$, so that $A,B\in PQ$ and $C,D\in k$. \item In the same plane are given: lines $p$ and $q$, a point $A$ and lines $m$ and $n$. Draw a rectangle $ABCD$, so that $B\in p$, $D\in q$ and $AB:AD=m:n$. \item In the given triangle $ABC$ draw a triangle, so that its sides are parallel to the given lines $p$, $q$ and $r$. \item Let $p$, $q$ and $r$ be three lines of some plane. Draw a line $t$, that is perpendicular to the line $p$ and intersects the lines $p$, $q$ and $r$ in such points $P$, $Q$ and $R$, that $PQ\cong QR$. %Similarity of triangles \item Let $P$ be an inner point of the triangle $ABC$ and $A_1$, $B_1$ and $C_1$ be the orthogonal projections of the point $P$ on the sides of the triangle $BC$, $AC$ and $AB$. Similarly, the points $A_2$, $B_2$ and $C_2$ are determined by the point $P$ and the triangle $A_1B_1C_1$,..., the points $A_{n+1}$, $B_{n+1}$ and $C_{n+1}$ with the point $P$ and the triangle $A_nB_nC_n$... Which of the triangles $A_1B_1C_1$, $A_2B_2C_2$, ... are similar to the triangle $ABC$? \item Let $k$ be the inscribed circle of the quadrilateral $ABCD$, $E$ be the intersection of its diagonals and $CB\cong CD$. Prove that $\triangle ABC \sim\triangle BEC$. \item Let $ABCD$ be a parallelogram. In the points $E$ and $F$ of the triangle $ABC$ the inscribed circle intersects the line $AD$ and $CD$. Prove that $\triangle EBC\sim\triangle EFD$. \item Let $AA'$ and $BB'$ be the altitudes of the acute angled triangle $ABC$. Prove that $\triangle ABC\sim\triangle A'B'C$. \item Let the altitude $AD$ of the triangle $ABC$ be tangent to the inscribed circle of this triangle. Prove that $|AD|^2=|BD|\cdot |CD|$. \item In the triangle $ABC$ let the internal angle at the vertex $A$ be twice as large as the internal angle at the vertex $B$. Prove that $|BC|^2= |AC|^2+|AC|\cdot |AB|$. \item Prove that the radii of the inscribed circles of two similar triangles are proportional to the corresponding sides of these two triangles. \item The quadrilateral $ABCD$ is inscribed in a circle with center $S$. The diagonals of this quadrilateral are perpendicular and intersect at the point $E$. The line through the point $E$ and perpendicular to the side $AD$ intersects the side $BC$ at the point $M$. \\ (\textit{a}) Prove that the point $M$ is the center of the segment $BC$.\\ (\textit{b}) Determine the set of all points $M$, if the diagonal $BD$ changes and is always perpendicular to the diagonal $AC$. \item Let $t$ be the tangent to the inscribed circle $l$ of the triangle $ABC$ at the vertex $A$. Let $D$ be such a point of the line $AC$, that $BD\parallel t$. Prove that $|AB|^2=|AC|\cdot |AD|$. \item The altitude point of an acute angled triangle should divide its altitude in an equal ratio (from the vertex to the knife point of the altitude). Prove that it is an isosceles triangle. \item In the triangle $ABC$, the altitude $BD$ touches the circumscribed circle of this triangle. Prove:\\ (\textit{a}) that the difference of the angles at the base $AC$ is equal to $90^0$,\\ (\textit{b}) that $|BD|^2=|AD|\cdot |CD|$. \item The circle with the center on the base $BC$ of the isosceles triangle $ABC$ touches the sides $AB$ and $AC$. The points $P$ and $Q$ are the intersections of these sides with any tangent of this circle. Prove that $4\cdot |PB|\cdot |CQ|=|BC|^2$. \item Let $V$ be the altitude point of the acute angled triangle $ABC$, the point $V$ the center of the altitude $AD$, and the altitude $BE$ the point $V$ divides in the ratio $3:2$. Calculate the ratio in which $V$ divides the altitude $CF$. \item Let $S$ be an external point of the circle $k$. $P$ and $Q$ are the points in which the circle $k$ touches its tangent from the point $S$, $X$ and $Y$ are the intersections of these circles with any line passing through the point $S$. Prove that $XP:YP=XQ:YQ$. \item Let $D$ be a point lying on the side $BC$ of the triangle $ABC$. The points $S_1$ and $S_2$ shall be the centers of the circumscribed circles of the triangles $ABD$ and $ACD$. Prove that $ \triangle ABC\sim\triangle AS_1S_2$. \item The point $P$ lies on the hypotenuse $BC$ of the triangle $ABC$. The perpendicular of the line $BC$ at the point $P$ intersects the line $AC$ and $AB$ at the points $Q$ and $R$ and the circumscribed circle of the triangle $ABC$ at the point $S$. Prove that $|PS|^2=|PQ|\cdot |PR|$. \item The point $A$ lies on the leg $OP$ of the right angle $POQ$. Let $B$, $C$ and $D$ be such points of the leg $OQ$ that $\mathcal{B}(O,B,C)$, $\mathcal{B}(B,C,D)$ and $OA\cong OB\cong BC\cong CD$. Prove that also $\triangle ABC\sim\triangle DBA$. \item Draw a triangle if the following data is known:\\ (\textit{a}) $\alpha$, $\beta$, $R+r$, \hspace*{3mm} (\textit{b}) $a$, $b:c$, $t_c-v_c$,\hspace*{3mm} (\textit{c}) $v_a$, $v_b$, $v_c$. \item Let $AB$ and $CD$ be the bases of an isosceles trapezoid $ABCD$, and let $r$ be the radius of the inscribed circle. Prove that $|AB|\cdot |CD|=4r^2$. %Harmonic cetverica \item Given is a circle $k$ and points $A$ and $B$. Draw a point $X$ on the circle $k$ such that $AX:XB=2:5$. \item Draw a triangle with the given data:\\ (\textit{a}) $a$, $v_a$, $b:c$, \hspace*{3mm} (\textit{b}) $a$, $t_a$, $b:c$,\hspace*{3mm} (\textit{c}) $a$, $b$, $b:c$,\\ (\textit{d}) $a$, $\alpha$, $b:c$,\hspace*{3mm} (\textit{e}) $a$, $l_a$, $b:c$. \item Draw a triangle if the following data is given:\\ (\textit{a}) $v_a$, $r$, $\alpha$, \hspace*{3mm} (\textit{b}) $v_a$, $r_a$, $a$, \hspace*{3mm} (\textit{c}) $v_a$, $t_a$, $b-c$. \item Draw a parallelogram, where one side and the corresponding altitude are consistent with the given distances $a$ and $v_a$, and the diagonals are in the ratio $3:5$. \item Let $E$ be the intersection of the internal angle $BAC$ of the triangle $ABC$ with its side $BC$. Prove that $$\overrightarrow{AE}=\frac{|AC|}{|AB|+|AC|}\cdot\overrightarrow{AB}+ \frac{|AB|}{|AB|+|AC|}\cdot\overrightarrow{AC}.$$ \item Given are four collinear points for which $\mathcal{H}(A,B;C,D)$ holds. Draw a point $L$, from which the distances $AC$, $CB$ and $BD$ are seen at the same angle. \item Let $AE$ ($E\in BC$) be the internal angle of the triangle $ABC$ and let $a=|BC|$, $b=|AC|$ and $c=|AB|$. Prove that $$|BE|=\frac{ac}{b+c} \hspace*{1mm} \textrm{ and } \hspace*{1mm} |CE|=\frac{ab}{b+c}.$$ \item Let $AE$ ($E\in BC$) and $BF$ ($F\in AC$) be the internal angle bisectors and $S$ the center of the inscribed circle of triangle $ABC$. Prove that $ABC$ is an isosceles triangle (with the base $AB$) exactly when $AS:SE=BS:SF$. %Menelaus' Theorem \item Prove that the external angle bisectors of any triangle intersect the opposite sides in three collinear points. %Pythagoras' Theorem \item Given $a$, $b$ and $c$ ($a>b$), construct such a distance $x$, that:\\ (\textit{a}) $x=\sqrt{a^2+b^2}$, \hspace*{3mm} (\textit{b}) $x=\sqrt{a^2-b^2}$, \hspace*{3mm} (\textit{c}) $x=\sqrt{3ab}$,\\ (\textit{d}) $x=\sqrt{a^2+bc}$, \hspace*{3mm} (\textit{e}) $x=\sqrt{3ab-c^2}$, \hspace*{3mm} (\textit{f}) $x=\frac{a\sqrt{ab+c^2}}{b+c}$. %Stewart's Theorem \item Let $a$, $b$ and $c$ be the sides of a triangle and $a^2+b^2=5c^2$. Prove that the centroids corresponding to sides $a$ and $b$ are perpendicular to each other. \item Let $a$, $b$, $c$ and $d$ be the sides, $e$ and $f$ the diagonals and $x$ the distance determined by the midpoints of sides $b$ and $d$ of a quadrilateral. Prove: $$x^2 = \frac{1}{4} \left(a^2 +c^2 -b^2 -d^2 +e^2 +f^2 \right).$$ \item Let $a$, $b$ and $c$ be the sides of triangle $ABC$. Prove that the distance of the centroid $A_1$ of side $a$ from the foot $A'$ of the altitude on that side is equal to: $$|A_1A'|=\frac{|b^2-c^2|}{2a}.$$ %Pappus and Pascal \item Let ($A$, $B$, $C$) and ($A_1$, $B_1$, $C_1$) be two triples of collinear points of a plane that are not on the same line. If $AB_1\parallel A_1B$ and $AC_1\parallel A_1C$, then also $CB_1\parallel C_1B$. (\textit{Pappus' Theorem}\footnote{Pappus of Alexandria\index{Pappus} (3rd century BC), Greek mathematician. This is a generalization of Pappus' Theorem (see Theorem \ref{izrek Pappus}), if we choose points $X$, $Y$ and $Z$ at infinity.}) %Desargues' Theorem \item Let $P$, $Q$ and $R$ be points of sides $BC$, $AC$ and $AB$ of triangle $ABC$, such that lines $AP$, $BQ$ and $CR$ are from the same pencil. Prove: If $X=BC\cap QR$, $Y=AC\cap PR$ and $Z=AB\cap PQ$, then points $X$, $Y$ and $Z$ are collinear. \item Let $AA'$, $BB'$ and $CC'$ be altitudes of triangle $ABC$ and $X=B'C'\cap BC$, $Y=A'C'\cap AC$ and $Z=A'B'\cap AB$. Prove that points $X$, $Y$ and $Z$ are collinear points. \item Let $A$ and $B$ be points outside line $p$. Draw the intersection of lines $p$ and $AB$ without drawing line $AB$ directly. \item Let $p$ and $q$ be lines of a plane that intersect in point $S$, which is "outside the paper", and $A$ is a point of that plane. Draw the line that goes through points $A$ and $S$. \item Draw a triangle so that its vertices lie on three given parallel lines and the carriers of its sides go through three given points. %Ppotenca \item Given is a circle $k(S,r)$.\\ (\textit{a}) Which values can the power of a point have with respect to circle $k$?\\ (\textit{b}) What is the smallest value of this power and for which point is this minimal value achieved?\\ (\textit{c}) Determine the set of all points for which the power with respect to circle is equal to $\lambda\in \mathbb{R}$. \item Let $k_a(S_a,r_a)$ and $l(O,R)$ be drawn and dashed circles of some triangle. Prove the equality\footnote{The statement is a generalization of Euler's formula for a circle (see Theorem \ref{EulerjevaFormula}). \index{Euler, L.} \textit{L. Euler} (1707--1783), Swiss mathematician.}: $$S_aO^2=R^2+2r_aR.$$ \item Draw a circle that goes through given points $A$ and $B$ and touches a given circle $k$. \item Prove that the centers of lines that are determined by common tangents of two circles are collinear points. \item Draw a circle that is perpendicular to two given circles and intersects the third given circle in points that determine the diameter of that third circle. \item Let $M$ and $N$ be the intersection points of sides $AB$ and $AC$ of triangle $ABC$ with a line, which passes through the center of the inscribed circle of this triangle and is parallel to its side $BC$. Express the length of the line $MN$ as a function of the lengths of the sides of triangle $ABC$. \item Let $AA_1$ be the median of triangle $ABC$. Points $P$ and $Q$ shall be the intersection points of the altitudes $AA_1B$ and $AA_1C$ with sides $AB$ and $AC$. Prove that $PQ\parallel BC$. \item In triangle $ABC$ let the sum (or difference) of the internal angles $ABC$ and $ACB$ be equal to a right angle. Prove that $|AB|^2+|AC|^2=4r^2$, where $r$ is the radius of the circumscribed circle of this triangle. \item Let $AD$ be the altitude of triangle $ABC$. Prove that the sum (or difference) of the internal angles $ABC$ and $ACB$ is equal to a right angle exactly when: $$\frac{1}{|AB|^2}+\frac{1}{|AC|^2}=\frac{1}{|AD|^2}.$$ \item Express the distance between the centroid and the center of the circumscribed circle of a triangle as a function of the lengths of its sides and the radius of the circumscribed circle. \item Prove that in triangle $ABC$ the altitude of the external angle at the vertex $A$ and the altitudes of the internal angles at the vertices $B$ and $C$ intersect the opposite sides in three collinear points. \item Prove that in triangle $ABC$ with altitude $AD$, the center of the circumscribed circle and the point, in which the side $BC$ touches the circumscribed circle of this triangle, are three collinear points. \item Prove Simson's theorem \ref{SimpsPrem} by using Menelaus' theorem \ref{izrekMenelaj}. \item Through point $M$ of side $AB$ of triangle $ABC$ a line is constructed, which intersects the line $AC$ in point $K$. Calculate the ratio, in which the line $MK$ divides the side $BC$, if $AM:MB=1:2$ and $AK:AC=3:2$. \item Let $A_1$ be the center of side $BC$ of triangle $ABC$ and let $P$ and $Q$ be such points of sides $AB$ and $AC$, that $BP:PA=2:5$ and $AQ:QC=6:1$. Calculate the ratio, in which the line $PQ$ divides the median $AA_1$. \item Prove that in any triangle, the lines determined by the vertices and the points of contact of one of the inscribed circles of the opposite sides intersect in a common point. \item What does the set of all points representing the intersection of two lines tangent to two given circles with respect to the centers and the points of contact of the circles represent? \item Let $PP_1$ and $QQ_1$ be the external tangents of the circles $k(O,r)$ and $k_1(O_1,r_1)$ (the points $P$, $P_1$, $Q$ and $Q_1$ are the corresponding points of contact). Let $S$ be the intersection of these two tangents, $A$ one of the intersections of the circles $k$ and $k_1$, and $L$ and $L_1$ the intersections of the line $SO$ with the lines $PQ$ and $P_1Q_1$. Prove that $\angle LAO\cong\angle L_1AO_1$. \item Prove that the side of a regular pentagon is equal to the larger part of the division of the radius of the inscribed circle of the pentagon in the golden ratio. \item Let $a_5$, $a_6$ and $a_{10}$ be the sides of a regular pentagon, hexagon and decagon, which are inscribed in the same circle. Prove that: $$a_5^2=a_6^2+a_{10}^2.$$ \item Let $t_a$, $t_b$ and $t_c$ be the centroids and $s$ the semiperimeter of a triangle. Prove that: $$t_a^2+t_b^2+t_c^2\geq s^2.$$ % zvezek - dodatni MG \end{enumerate} % DEL 8 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - %________________________________________________________________________________ % PLOŠČINA %________________________________________________________________________________ \del{Area of Figures} \label{pogPLO} %________________________________________________________________________________ \poglavje{Area of Figures. Definition} \label{odd8PloLik} In this section we will define the concept of the area of a certain class of figures. The area is based on the concept of the length of a line in a certain system of measurement and on the theory of infinitesimal calculus\footnote{The first steps in this direction were made by the Greek mathematician \index{Euclid}\textit{Euclid} (408--355 BC) and \index{Archimedes}\textit{Archimedes} (287--212 BC) in the calculation of the volume of bodies.}. Let $\Omega_0$ be a square with a side length of $1$ in a given measuring system. The aforementioned square represents the \index{unit of measurement of area}\pojem{unit of measurement of area} - we also call it the \index{unit square}\pojem{unit square}. Intuitively, the area of a figure $\Phi$ represents the number of unit squares that are consistent with square $\Omega_0$ and with which we can ''cover'' figure $\Phi$. Now we will approach the definition of area more formally. Let $\Phi$ be an arbitrary figure in a plane. The unit square $\Omega_0$ defines the tiling $(4,4)$ (see section \ref{odd3Tlakovanja}) - we denote it with $\mathcal{T}_0$. With $\underline{S}_0$ we denote the number of squares of tiling $\mathcal{T}_0$, which are in figure $\Phi$, or are its subset. With $\overline{S}_0$ we denote the number of squares of tiling $\mathcal{T}_0$, which have at least one common point with figure $\Phi$ (Figure \ref{sl.plo.8.1.1.pic}). It is clear that it holds: $$0\leq\underline{S}_0\leq\overline{S}_0.$$ \begin{figure}[!htb] \centering \input{sl.plo.8.1.1.pic} \caption{} \label{sl.plo.8.1.1.pic} \end{figure} If we divide each side of the square into $10$ lines, we can divide the unit square $\Omega_0$ into $10^2$ consistent squares, which are all consistent with one of these squares $\Omega_1$, which has a side length of $\frac{1}{10}$. Square $\Omega_1$ defines a new tiling $\mathcal{T}_1$. Similarly to the previous example, let $\underline{S}_1$ be the number of squares of tiling $\mathcal{T}_1$, which are in figure $\Phi$, and $\overline{S}_1$ be the number of squares of tiling $\mathcal{T}_1$, which have at least one common point with figure $\Phi$. It is clear that $\underline{S}_1\leq\overline{S}_1$ and also: $$0\leq\underline{S}_0\leq\frac{\underline{S}_1}{10^2} \leq\frac{\overline{S}_1}{10^2}\leq\overline{S}_0.$$ If we continue the process, we get a sequence of squares $\Omega_n$, which have a side length of $\frac{1}{10^n}$, tiling $\mathcal{T}_n$ and pairs of numbers $\underline{S}_n$ and $\overline{S}_n$, for which the following is true: $$0\leq\underline{S}_0\leq\frac{\underline{S}_1}{10^2}\leq \cdots \leq \frac{\underline{S}_n}{10^{2n}}\leq\cdots\leq \frac{\overline{S}_n}{10^{2n}}\leq\cdots \leq\frac{\overline{S}_1}{10^2}\leq\overline{S}_0.$$ The sequence $\frac{\underline{S}_n}{10^{2n}}$ is increasing and bounded from above, so it is convergent by a known theorem of mathematical analysis and has the following limit: $$\underline{S}=\lim_{n\rightarrow\infty}\frac{\underline{S}_n}{10^{2n}}.$$ Similarly, the sequence $\frac{\overline{S}_n}{10^{2n}}$ is decreasing and bounded from below, so it is convergent and has the following limit: $$\overline{S}=\lim_{n\rightarrow\infty}\frac{\overline{S}_n}{10^{2n}}.$$ If $\underline{S}=\overline{S}$, we say that the figure $\Phi$ is measurable. The number $S=\underline{S}=\overline{S}$ is its \index{area of a figure}\pojem{area}. We denote it by $p(\Phi)$ or $p_{\Phi}$. It is intuitively clear that the area of a figure $\Phi$ is not dependent on the location of the unit square - that is, it is only dependent on the system of measuring lengths, in which the side of the unit square has a length of 1. We will not formally prove this fact here. We will now prove the first important property of area. \bizrek \label{ploscDaljice} The area of an arbitrary point is 0. The area of an arbitrary line segment is 0. \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.1.2.pic} \caption{} \label{sl.plo.8.1.2.pic} \end{figure} \textbf{\textit{Proof.}} \textit{1)} Let $A$ be an arbitrary point. It is clear that for every $n\in \mathbb{N}$ it holds: $\underline{S}_n=0$ and $0\leq\overline{S}_n\leq4$ (the point lies in at most four paving tiles $T_n$). So it holds: $$\underline{S}=\lim_{n\rightarrow\infty}\frac{\underline{S}_n}{10^{2n}} =\lim_{n\rightarrow\infty}\frac{0}{10^{2n}}=0$$ and $$0\leq\overline{S}=\lim_{n\rightarrow\infty}\frac{\overline{S}_n}{10^{2n}}\leq \lim_{n\rightarrow\infty}\frac{4}{10^{2n}}=0,$$ thus $p_A=S=\underline{S}=\overline{S}=0$ \textit{2)} Let $AB$ be an arbitrary distance (Figure \ref{sl.plo.8.1.2.pic}) and $|AB|=d$. With $k$ we denote the integer part of the number $d$, i.e. $k=[d]$. We choose a unit square $\Omega_0$, so that one of its vertices is the point $A$, one of its sides lies on the distance $AB$. In this case $\underline{S}_n=0$ and $\overline{S}_n\leq (k+1)\cdot 10^n$. Then it holds: $$\underline{S}=\lim_{n\rightarrow\infty}\frac{\underline{S}_n}{10^{2n}} =\lim_{n\rightarrow\infty}\frac{0}{10^{2n}}=0$$ and $$0\leq\overline{S}=\lim_{n\rightarrow\infty}\frac{\overline{S}_n}{10^{2n}}\leq \lim_{n\rightarrow\infty}\frac{(k+1)10^n}{10^{2n}}= \lim_{n\rightarrow\infty}\frac{k+1}{10^n}=0,$$ thus $p_{AB}=S=\underline{S}=\overline{S}=0.$ \kdokaz The next theorem, which refers to the basic properties of the area, we will state without a proof (see \cite{Lucic}). \bizrek \label{ploscGlavniIzrek} Let $p$ be an area, defined on a set $\mu$ of measurable figures of a plane. Then it holds: \begin{enumerate} \item $p(\Omega_0)=1$; \item $(\forall \Phi\in\mu)\hspace*{1mm}p(\Phi)\geq 0$; \item $(\forall \Phi_1,\Phi_2\in\mu)\hspace*{1mm} \left(\Phi_1\cong\Phi_2 \hspace*{1mm}\Rightarrow\hspace*{1mm}p(\Phi_1)=p(\Phi_2)\right)$; \item $(\forall \Phi_1,\Phi_2\in\mu)\hspace*{1mm} \left(p(\Phi_1\cap\Phi_2)=0 \hspace*{1mm}\Rightarrow\hspace*{1mm}p(\Phi_1\cup\Phi_2)= p(\Phi_1)+p(\Phi_2)\right)$. \end{enumerate} \eizrek The following theorem is very useful. \bizrek \label{ploscLomljenke} The area of any break is equal to 0. \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.1.3.pic} \caption{} \label{sl.plo.8.1.3.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.1.3.pic}) The claim is a direct consequence of theorems \ref{ploscDaljice} and \ref{ploscGlavniIzrek} (\textit{4}). \kdokaz %________________________________________________________________________________ \poglavje{Area of Rectangles, Parallelogram, and Trapezoids} \label{odd8PloParalel} In the following we will derive formulas for the area of certain quadrilaterals. \bizrek \label{ploscPravok} If $p$ is the area of a rectangle with sides of lengths $a$ and $b$, then it is: $$p=ab.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.2.1a.pic} \caption{} \label{sl.plo.8.2.1a.pic} \end{figure} \textbf{\textit{Proof.}} Let $ABCD$ be a rectangle with sides $AB$ and $AD$ of lengths $|AB|=a$ and $|AD|=b$ (Figure \ref{sl.plo.8.2.1a.pic}). Choose a unit square $\Omega_0$, so that one of its vertices is point $A$, one of its sides lies on the line $AB$. Let $\mathcal{T}_n$ ($n\in \mathbb{N}\cup\{0\}$) be the corresponding tiling that is determined by the unit square $\Omega_0$. We denote with $a_n=[a\cdot 10^n]$ and $b_n=[b\cdot 10^n]$ (where $[x]$ represents the integer part of number $x$). First, we have: \begin{eqnarray} \label{eqnPloscPrav1a} \frac{a_n}{10^n}\leq a\leq \frac{a_n+1}{10^n} \hspace*{2mm} \textrm{ and } \hspace*{2mm} \frac{b_n}{10^n}\leq b\leq \frac{b_n+1}{10^n}, \end{eqnarray} then: \begin{eqnarray} \label{eqnPloscPrav2a} \underline{S}_n=\frac{a_n\cdot b_n}{10^{2n}} \hspace*{2mm} \textrm{ and } \hspace*{2mm} \overline{S}_n=\frac{\left(a_n+1\right)\cdot \left(b_n+1\right)}{10^{2n}}. \end{eqnarray} From \ref{eqnPloscPrav2a} and \ref{eqnPloscPrav1a} we get: \begin{eqnarray*} 0&\leq&\lim_{n\rightarrow\infty}\left(\overline{S}_n-\underline{S}_n\right)=\\ &=&\lim_{n\rightarrow\infty}\left(\frac{\left(a_n+1\right)\cdot \left(b_n+1\right)}{10^{2n}}-\frac{a_n\cdot b_n}{10^{2n}}\right)=\\ &=& \lim_{n\rightarrow\infty}\frac{a_n+ b_n+1}{10^{2n}}\leq\\ &\leq& \lim_{n\rightarrow\infty}\frac{a+b+1}{10^n}=\\ &=& 0. \end{eqnarray*} Therefore: \begin{eqnarray*} \lim_{n\rightarrow\infty}\left(\overline{S}_n-\underline{S}_n\right)=0, \end{eqnarray*} which means that $\underline{S}=\lim_{n\rightarrow\infty}\underline{S}_n =\lim_{n\rightarrow\infty}\overline{S}_n=\overline{S}$, which means that the rectangle $ABCD$ is a measurable figure with the area $p=S=\underline{S}=\overline{S}$. We will now prove $p=ab$. From \ref{eqnPloscPrav1a} we get: \begin{eqnarray*} \frac{a_n\cdot b_n}{10^{2n}} \leq ab\leq \frac{\left(a_n+1\right)\cdot \left(b_n+1\right)}{10^{2n}}, \end{eqnarray*} Since this is true for every $n\in \mathbb{N}$, it follows: \begin{eqnarray*} \hspace*{-2mm} \underline{S}=\lim_{n\rightarrow\infty}\underline{S}_n= \lim_{n\rightarrow\infty}\frac{a_n\cdot b_n}{10^{2n}} \leq ab\leq \lim_{n\rightarrow\infty}\frac{\left(a_n+1\right)\cdot \left(b_n+1\right)}{10^{2n}}=\lim_{n\rightarrow\infty}\overline{S}_n=\overline{S}. \end{eqnarray*} From $p=S=\underline{S}=\overline{S}$ at the end it follows $p=ab$. \kdokaz A direct consequence of the previous \ref{ploscPravok} is the following claim (Figure \ref{sl.plo.8.2.2.pic}). \bizrek \label{ploscKvadr} If $p_{\square}$ is the area of a square with a side of length $a$, then: $$p_{\square}=a^2.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.2.2.pic} \caption{} \label{sl.plo.8.2.2.pic} \end{figure} We will now derive formulas for the area for some other quadrilaterals. \bizrek \label{ploscParal} If $p$ is the area of a parallelogram with sides of length $a$ and $b$ and the corresponding heights of length $v_a$ and $v_b$, then it is: $$p=av_a=bv_b.$$ \eizrek \textbf{\textit{Proof.}} Let $ABCD$ be a parallelogram with side $AB$ of length $|AB|=a$ and the corresponding height $v_a$. We mark with $E$ and $F$ the orthogonal projections of the vertices $D$ and $C$ on the line $AB$. The quadrilateral $EFCD$ is a rectangle with sides $|CD|=a$ and $|FC|=v_a$, so according to the theorem \ref{ploscPravok}: \begin{eqnarray} p_{EFCD}=av_a. \label{eqnPloscPrav2} \end{eqnarray} The right-angled triangle $AED$ and $BFC$ are congruent (theorem \textit{ASA} \ref{KSK}) so according to the theorem \ref{ploscGlavniIzrek} \textit{3)}: \begin{eqnarray} p_{AED}=p_{BFC}. \label{eqnPloscPrav1} \end{eqnarray} Without loss of generality, we assume that $\angle BAD\leq 90^0$. In this case, the points $B$ and $E$ are on the same side of the point $A$ (in the case $E=A$ it is a rectangle $ABCD$ and the statement follows directly from the theorem \ref{ploscPravok}). We will consider several possible cases: $\mathcal{B}(A,E,B)$, $E=B$ and $\mathcal{B}(A,B,E)$. \textit{1)} Let $\mathcal{B}(A,E,B)$ (Figure \ref{sl.plo.8.2.3.pic}). \begin{figure}[!htb] \centering \input{sl.plo.8.2.3.pic} \caption{} \label{sl.plo.8.2.3.pic} \end{figure} If we use the relations \ref{eqnPloscPrav2} and \ref{eqnPloscPrav1} and theorems \ref{ploscGlavniIzrek} \textit{4)} and \ref{ploscDaljice}, we get: \begin{eqnarray*} p_{ABCD}=p_{AED}+p_{EBCD}=p_{BFC}+p_{EBCD}=p_{EFCD}=av_a. \end{eqnarray*} \textit{2)} In the case $E=B$ (Figure \ref{sl.plo.8.2.3a.pic}) similarly from the relations \ref{eqnPloscPrav1} and \ref{eqnPloscPrav1} and theorems \ref{ploscGlavniIzrek} \textit{4)} and \ref{ploscDaljice} we get: \begin{eqnarray*} p_{ABCD}=p_{AED}+p_{BCD}=p_{BFC}+p_{BCD}=p_{EFCD}=av_a. \end{eqnarray*} \begin{figure}[!htb] \centering \input{sl.plo.8.2.3a.pic} \caption{} \label{sl.plo.8.2.3a.pic} \end{figure} \textit{3)} We assume that $\mathcal{B}(A,B,E)$ (Figure \ref{sl.plo.8.2.3b.pic}). \begin{figure}[!htb] \centering \input{sl.plo.8.2.3b.pic} \caption{} \label{sl.plo.8.2.3b.pic} \end{figure} The lines $BC$ and $DE$ intersect in some point $L$. By the formulas \ref{ploscGlavniIzrek} \textit{4)} and \ref{ploscDaljice} it is: $p_{AED}=p_{ABLD}+p_{BEL}$ and $p_{BFC}=p_{EFCL}+p_{BEL}$. Because by the relation \ref{eqnPloscPrav1} $p_{AED}=p_{BFC}$, it is also: \begin{eqnarray*} p_{ABLD}=p_{EFCL}. \end{eqnarray*} From this and from the formulas \ref{ploscGlavniIzrek} \textit{4)} and \ref{ploscDaljice} it follows: \begin{eqnarray*} p_{ABCD}=p_{ABLD}+p_{DLC}=p_{EFCL}+p_{DLC}=p_{EFCD}=av_a, \end{eqnarray*} which was to be proven. \kdokaz \bizrek \label{ploscTrapez} If $p$ is the area of a trapezoid with bases of lengths $a$ and $c$ and height of length $v$, then it is: $$p=\frac{a+c}{2}\cdot v.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.2.4.pic} \caption{} \label{sl.plo.8.2.4.pic} \end{figure} \textbf{\textit{Proof.}} Let $ABCD$ be a trapezoid with bases $AB$ and $CD$ of lengths $|AB|=a$ and $|CD|=c$ and a height of length $v$ (Figure \ref{sl.plo.8.2.4.pic}). With $S$ we mark the center of the leg $BC$ and $\mathcal{S}_S:\hspace*{1mm}A,D\mapsto A', D'$. Because $S$ is the common center of the lines $AA'$ and $DD'$, by \ref{paralelogram} $AD'A'D$ is a parallelogram. Because $\mathcal{S}_S(B)=C$ and $\mathcal{B}(A,B,D')$, $\mathcal{B}(A',C,D)$ is also true. The isometry $\mathcal{S}$ maps the trapezoid $ABCD$ into a similar trapezoid $A'CBD'$, therefore $|BD'|=|CD|=c$ or $|AD'|=a+c$, and by \ref{ploscGlavniIzrek} \textit{3)} also $p_{ABCD}=p_{A'CBD'}$. So the parallelogram $AD'A'D$ with a base $AD'$ of length $|AD'|=a+c$ and a height which is equal to the height of the trapezoid $ABCD$ of length $v$, is divided into two similar trapezoids $ABCD$ and $A'CBD'$ with equal areas. From this and from \ref{ploscGlavniIzrek} \textit{4)}, \ref{ploscDaljice} and \ref{ploscParal} it follows: \begin{eqnarray*} 2\cdot p_{ABCD}=p_{ABCD}+p_{A'CBD'}=p_{AD'A'D}=\left(a+c\right)v, \end{eqnarray*} or the desired relation. \kdokaz If we use \ref{srednjTrapez}, we see that the expression $\frac{a+c}{2}$ represents the length of the median of the trapezoid, so for the area of the trapezoid the formula: $$p=mv,$$ where $m$ is the length of the median of the trapezoid and $v$ is its height (Figure \ref{sl.plo.8.2.4a.pic}) is also true. \begin{figure}[!htb] \centering \input{sl.plo.8.2.4a.pic} \caption{} \label{sl.plo.8.2.4a.pic} \end{figure} \bzgled \label{ploscStirikPravok} If $p$ is the area of a quadrilateral with perpendicular diagonals of lengths $e$ and $f$, then: $$p=\frac{ef}{2}.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.plo.8.2.5.pic} \caption{} \label{sl.plo.8.2.5.pic} \end{figure} \textbf{\textit{Proof.}} Let $ABCD$ be a quadrilateral with perpendicular diagonals $AC$ and $BD$ of lengths $|AC|=e$ and $|BD|=f$ (Figure \ref{sl.plo.8.2.5.pic}). The parallels of these diagonals through the vertices of the quadrilateral $ABCD$ determine the rectangle $A'B'C'D'$ with sides of lengths $|A'B'|=f$ and $|B'C'|=e$. From the similarity of triangles $ASD$ and $DA'A$ (the \textit{SAS} theorem \ref{SKS}) it follows that $p_{ASD}=p_{DA'A}$ (the \ref{ploscGlavniIzrek} \textit{3)}). Similarly, $p_{ASB}=p_{BB'A}$, $p_{CSB}=p_{BC'C}$ and $p_{CSD}=p_{DD'C}$. From this and from theorems \ref{ploscGlavniIzrek} \textit{4)}, \ref{ploscDaljice} and \ref{ploscPravok} it follows: \begin{eqnarray*} 2\cdot p_{ABCD}&=&2\cdot\left(p_{ASD}+p_{ASB}+p_{CSB}+p_{CSD}\right)=\\ &=&2\cdot p_{ASD}+2\cdot p_{ASB}+2\cdot p_{CSB}+2\cdot p_{CSD}=\\ &=& p_{ASD}+ p_{DA'A}+ p_{ASB}+p_{BB'A}+\\ && + p_{CSB}+p_{BC'C} +p_{CSD}+p_{DD'C}=\\ &=& p_{A'B'C'D'}= ef, \end{eqnarray*} or the desired relation. \kdokaz As a consequence, we have the following theorem (Figure \ref{sl.plo.8.2.5a.pic}). \bzgled \label{ploscDeltoid} If $p$ is the area of a deltoid with diagonals of lengths $e$ and $f$, then: $$p=\frac{ef}{2}.$$ \ezgled \textbf{\textit{Proof.}} The theorem is a direct consequence of theorem \ref{ploscStirikPravok} and the definition of a deltoid. \kdokaz \begin{figure}[!htb] \centering \input{sl.plo.8.2.5a.pic} \caption{} \label{sl.plo.8.2.5a.pic} \end{figure} \bzgled \label{ploscRomb} If $p$ is the area of a rhombus with diagonals of lengths $e$ and $f$, then: $$p=\frac{ef}{2}.$$ \ezgled \textbf{\textit{Proof.}} The statement is a direct consequence of Theorems \ref{ploscStirikPravok} and \ref{RombPravKvadr}. \kdokaz \bzgled \label{ploscRomb1} If: $a$ is the length of a side, $v$ is the height, and $e$ and $f$ are the lengths of the diagonals of the rhombus, then: $$av=\frac{ef}{2}.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.plo.8.2.6.pic} \caption{} \label{sl.plo.8.2.6.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.2.5a.pic}) The statement is a direct consequence of Theorems \ref{ploscRomb1} and \ref{ploscParal}. \kdokaz %________________________________________________________________________________ \poglavje{Area of Triangles} \label{odd8PloTrik} In this section we will derive several formulas for the area of a triangle. \bizrek \label{PloscTrik} If $p_\triangle$ is the area of the triangle $ABC$, $v_a$, $v_b$ and $v_c$ are the lengths of the heights corresponding to the sides $BC$, $AC$ and $AB$ with lengths $a$, $b$ and $c$, then: $$p_\triangle=\frac{a\cdot v_a}{2}=\frac{b\cdot v_b}{2} =\frac{c\cdot v_c}{2}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.3.1a.pic} \caption{} \label{sl.plo.8.3.1a.pic} \end{figure} \textbf{\textit{Proof.}} Let $D$ be the fourth vertex of the parallelogram $ABCD$ (Figure \ref{sl.plo.8.3.1a.pic}). The triangles $ABC$ and $ADC$ are congruent (by \ref{paralelogram} and \ref{SSS}), therefore $p_{ABC}=p_{ADC}$ (by \ref{ploscGlavniIzrek} \textit{3)}). The parallelogram $ABCD$ has the side $BC$ and the corresponding altitude of lengths $a$ and $v_a$, therefore by \ref{ploscParal} $p_{ABCD}=av_a$. If we use \ref{ploscGlavniIzrek} \textit{4)} and \ref{ploscDaljice} as well, we get: \begin{eqnarray*} 2\cdot p_{ABC}=p_{ABC}+p_{ADC}=p_{ABCD}=av_a, \end{eqnarray*} or the desired relation $p_\triangle=\frac{a\cdot v_a}{2}$. Similarly, $p_\triangle=\frac{b\cdot v_b}{2}$ or $p_\triangle=\frac{c\cdot v_c}{2}$. \kdokaz \bizrek \label{PloscTrikVcrt} If $p_\triangle$ is the area of the triangle $ABC$ with the semi-perimeter $s=\frac{a+b+c}{2}$ and the radius of the inscribed circle $r$, then $$p_\triangle=sr.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.3.1.pic} \caption{} \label{sl.plo.8.3.1.pic} \end{figure} \textbf{\textit{Solution.}} Let $S$ be the center of the inscribed circle of the triangle $ABC$ (Figure \ref{sl.plo.8.3.1.pic}). By \ref{ploscGlavniIzrek} \textit{4)}, \ref{ploscDaljice} and \ref{PloscTrik} we have: \begin{eqnarray*} p_{ABC}=p_{SBC}+p_{ASC}+p_{ABS}&=& \frac{a\cdot r}{2}+\frac{b\cdot r}{2}+\frac{c\cdot r}{2}=\\ &=& \frac{a+b+c}{2}\cdot r=sr, \end{eqnarray*} which was to be proven. \kdokaz \bizrek \label{PloscTrikPricrt} If $p_\triangle$ is the area of the triangle $ABC$ with the semi-perimeter $s=\frac{a+b+c}{2}$ and the radius of the circumscribed circle $r_a$, then $$p_\triangle=(s-a)r_a.$$ \eizrek \textbf{\textit{Proof.}} We use the notation from the big task \ref{velikaNaloga} (Figure \ref{sl.plo.8.3.2.pic}). From the same task we have: $AR=s-a$ and $AR_a=s$. The right angled triangle $ARS$ and $AR_aS_a$ are similar (statement \ref{PodTrikKKK}), so $\frac{SR}{S_aR_a}=\frac{AR}{AR_a}$, i.e. $\frac{r}{r_a}=\frac{s-a}{s}$. If we use the previous statement \ref{PloscTrikVcrt}, we get: \begin{eqnarray*} p_{ABC}=sr=(s-a)r_a, \end{eqnarray*} which was to be proven. \kdokaz \begin{figure}[!htb] \centering \input{sl.plo.8.3.2.pic} \caption{} \label{sl.plo.8.3.2.pic} \end{figure} \bizrek \label{PloscTrikHeron} If $p_\triangle$ is the area of the triangle $ABC$ with the semiperimeter $s=\frac{a+b+c}{2}$, then it holds \index{formula!Heronova} (Heron's\footnote{\index{Heron} \textit{Heron from Alexandria}(20--100), ancient Greek matemetician.} formula): $$p_\triangle=\sqrt{s(s-a)(s-b)(s-c)}.$$ \eizrek \textbf{\textit{Proof.}} Again, we use the notation from the big task \ref{velikaNaloga} (Figure \ref{sl.plo.8.3.2.pic}). From the same task we have: $BP=s-b$ and $BP_a=s-c$. The angle $SBP$ and $BS_aP_a$ are complementary, because they have a pair of perpendicular sides (statement \ref{KotaPravokKraki}). Therefore, the right angled triangle $SBP$ and $BS_aP_a$ are similar (statement \ref{PodTrikKKK}), so $\frac{SP}{BP_a}=\frac{BP}{S_aP_a}$ i.e. $\frac{r}{s-c}=\frac{s-b}{r_a}$ i.e. $rr_a=(s-b)(s-c)$. If we use the statement \ref{PloscTrikVcrt} and \ref{PloscTrikPricrt}, we get: \begin{eqnarray*} p_{ABC}^2=sr(s-a)r_a=s(s-a)rr_a=s(s-a)(s-b)(s-c), \end{eqnarray*} which was to be proven. \kdokaz \bizrek \label{PloscTrikOcrt} If $p_\triangle$ is the area of the triangle $ABC$ with the side lengths $a$, $b$ and $c$ and $R$ is the radius of the circumscribed circle, then the following is true: $$p_\triangle=\frac{abc}{4R}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.3.3.pic} \caption{} \label{sl.plo.8.3.3.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.3.3.pic}) The statement is a direct consequence of the formulas \ref{PloscTrik} and \ref{izrekSinusni}: \begin{eqnarray*} p_{ABC}=\frac{av_a}{2}=\frac{a}{2}\cdot v_a=\frac{a}{2}\cdot \frac{bc}{2R}=\frac{abc}{4R}, \end{eqnarray*} which was to be proven. \kdokaz The previous formulas for the area of a triangle can be used to calculate the radius of the circumscribed, inscribed and excribed circles of a triangle as functions of its sides. \bzgled \label{PloscTrikOcrtVcrt} If $R$ and $r$ are the radius of the circumscribed and inscribed circles, $r_a$, $r_b$ and $r_c$ are the radius of the excribed circles and $s=\frac{a+b+c}{2}$ is the perimeter of the triangle $ABC$, then the following is true: (i) $R=\frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}$,\hspace*{7mm} (ii) $r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$,\\ (iii) $r_a=\sqrt{\frac{s(s-b)(s-c)}{s-a}}$,\hspace*{2.7mm} (iv) $r_b=\sqrt{\frac{(s-a)s(s-c)}{s-b}}$,\hspace*{2.7mm} (v) $r_c=\sqrt{\frac{(s-a)(s-b)s}{s-c}}$. \ezgled \textbf{\textit{Proof.}} The statement is a direct consequence of the formulas \ref{PloscTrikOcrt}, \ref{PloscTrikVcrt}, \ref{PloscTrikPricrt} and \ref{PloscTrikHeron}. \kdokaz \bzgled \label{PloscTrikVcrtPricrt} If $r$ is the radius of the inscribed circle, $r_a$, $r_b$ and $r_c$ are the radii of the circumscribed circles, and $p_{\triangle}$ is the area of the triangle $ABC$, then: $$p_{\triangle}=\sqrt{rr_ar_br_c}.$$ \ezgled \textbf{\textit{Proof.}} If we use the statements from the previous example \ref{PloscTrikOcrtVcrt}, we get: \begin{eqnarray*} rr_ar_br_c=s(s-a)(s-b)(s-c)=p_{\triangle}^2, \end{eqnarray*} which was to be proven. \kdokaz \bizrek \label{ploscTrikPedalni} Let $A'B'C'$ be the pedal triangle of the obtuse triangle $ABC$, $s'$ be the perimeter of the triangle $A'B'C'$, $R$ be the radius of the circumscribed circle, and $p_{\triangle}$ be the area of the triangle $ABC$, then: $$p_{\triangle} = s'R.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.3.3a.pic} \caption{} \label{sl.plo.8.3.3a.pic} \end{figure} \textbf{\textit{Proof.}} We mark with $O$ the center of the circumscribed circle of the triangle $ABC$ (Figure \ref{sl.plo.8.3.3a.pic}). By the statement \ref{PedalniLemaOcrtana}, $OA\perp B'C'$, so from the statement \ref{ploscStirikPravok} it follows that $p_{AC'OB'}=\frac{|AO|\cdot |B'C'|}{2}=\frac{R\cdot |B'C'|}{2}$. Analogously, $p_{BA'OC'}=\frac{R\cdot |A'C'|}{2}$ and $p_{CB'OA'}=\frac{R\cdot |A'B'|}{2}$. Because $ABC$ is an obtuse triangle, the point $O$ is in its interior (see section \ref{odd3ZnamTock}), so by the statements \ref{ploscGlavniIzrek} \textit{4)} and \ref{ploscDaljice}, it holds: \begin{eqnarray*} p_{\triangle}&=&p_{AC'OB'}+p_{BA'OC'}+p_{CB'OA'}=\\ &=&\frac{R}{2}\left(|B'C'|+|A'C'|+|A'B'| \right)=s'R, \end{eqnarray*} which was to be proven. \kdokaz \bzgled \label{ploscTrikPedalni1} Let $R$ and $r$ be the radii of the circumscribed and inscribed circles of the acute angled triangle $ABC$ and $s$ and $s'$ be the semi-perimeter of this triangle and its pedal triangle. Prove that: $$\frac{R}{r}=\frac{s}{s'}.$$ \ezgled \textbf{\textit{Proof.}} By the formulas \ref{PloscTrikVcrt} and \ref{ploscTrikPedalni} we have: \begin{eqnarray*} p_{ABC}=sr=s'R. \end{eqnarray*} From this we obtain the desired relation. \kdokaz In a special case we obtain formulas for the area of a right angled and an equilateral triangle. \bzgled \label{PloscTrikPravokotni} If $p_\triangle$ is the area of the right angled triangle $ABC$ with the sides of lengths $a$ and $b$, then: $$p_\triangle=\frac{ab}{2}.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.plo.8.3.4a.pic} \caption{} \label{sl.plo.8.3.4a.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.3.4a.pic}) The statement is a direct consequence of the formula \ref{PloscTrik}, since in this case $v_a=b$. \kdokaz \bizrek \label{PloscTrikEnakostr} If $p_\triangle$ is the area of the equilateral triangle $ABC$ with the side of length $a$, then: $$p_\triangle=\frac{a^2\sqrt{3}}{4}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.3.4.pic} \caption{} \label{sl.plo.8.3.4.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.3.4.pic}) The statement is a direct consequence of theorems \ref{PloscTrik} and \ref{PitagorovEnakostr}: \begin{eqnarray*} p_{ABC}=\frac{av_a}{2}=\frac{a}{2}\cdot v_a=\frac{a}{2}\cdot \frac{a\sqrt{3}}{2}=\frac{a^2\sqrt{3}}{4}, \end{eqnarray*} which had to be proven. \kdokaz We will continue using the formulas for the area of a triangle. \bzgled \label{CarnotOcrtLema} Let $P$ be an inner point of the triangle $ABC$ and $x$, $y$ and $z$ the distances of this point from the sides $BC$, $AC$ and $AB$ with lengths $a$, $b$ and $c$. If $r$ is the radius of the inscribed circle of this triangle, then: $$xa+yb+zc=2\cdot p_{\triangle ABC}=r(a+b+c).$$ \ezgled \begin{figure}[!htb] \centering \input{sl.plo.8.3.6.pic} \caption{} \label{sl.plo.8.3.6.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.3.6.pic}) $x$, $y$ and $z$ are the lengths of the altitudes of the triangles $PBC$, $APC$ and $ABP$, so according to theorems \ref{PloscTrik} and \ref{PloscTrikVcrt} it follows: \begin{eqnarray*} xa+yb+zc &=& 2\cdot p_{\triangle PBC}+2\cdot p_{\triangle APC}+2\cdot p_{\triangle ABP}\\ &=& 2\cdot p_{\triangle ABC}\\ &=& r(a+b+c), \end{eqnarray*} which had to be proven. \kdokaz \bzgled \index{izrek!Vivianijev} (Vivianijev\footnote{\index{Viviani, V.}\textit{V. Viviani} (1622--1703), Italian mathematician and physicist.} izrek) Let $P$ be an inner point of the equilateral triangle $ABC$ and $x$, $y$ and $z$ the distances of this point from the sides $BC$, $AC$ and $AB$. If $v$ is the height of this triangle, then: $$x+y+z=v.$$ \ezgled \begin{figure}[!htb] \centering \input{sl.plo.8.3.6a.pic} \caption{} \label{sl.plo.8.3.6a.pic} \end{figure} \textbf{\textit{Proof.}} By the statement from the previous example \ref{CarnotOcrtLema} and the formula \ref{PloscTrik}: $xa+ya+za=2\cdot p_{\triangle ABC}=v_aa$. \kdokaz \bizrek \label{CarnotOcrt}\index{izrek!Carnotov o očrtani krožnici}(Carnotov\footnote{\index{Carnot, L. N. M.}\textit{L. N. M. Carnot} (1753--1823), francoski matematik.} izrek o očrtani krožnici.) The sum of the distances from the center of the circumscribed circle to the sides of the triangle is equal to the sum of the radii of the circumscribed and inscribed circle of that triangle. So if $l(O,R)$ is the circumscribed and $k(S,r)$ the inscribed circle and $A_1$, $B_1$ and $C_1$ are the centers of the sides of the triangle $ABC$, it holds: $$|OA_1|+|OB_1|+|OC_1|=R+r.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.3.7.pic} \caption{} \label{sl.plo.8.3.7.pic} \end{figure} \textbf{\textit{Proof.}} We mark $x=|OA_1|$, $y=|OB_1|$ and $z=|OC_1|$ (Figure \ref{sl.plo.8.3.7.pic}). According to the statement from the example \ref{CarnotOcrtLema} it is: \begin{eqnarray} \label{eqnCarnotOcrt1} xa+yb+zc=r(a+b+c) \end{eqnarray} From the similarity of triangles $BA_1O$ and $CA_1O$ (\textit{SAS} \ref{SKS}) and \ref{SredObodKot} it follows that $\angle BAC=\frac{1}{2}\angle BOC=\angle BOA_1$. This means that $\triangle OA_1B\sim \triangle AB'B \sim \triangle AC'C$ (\ref{PodTrikKKK}) is valid, therefore: \begin{eqnarray*} \frac{OA_1}{AB'}=\frac{OB}{AB} \hspace*{2mm} \textrm{ and } \hspace*{2mm} \frac{OA_1}{AC'}=\frac{OB}{AC} \end{eqnarray*} or: \begin{eqnarray*} cx=R\cdot |AB'| \hspace*{2mm} \textrm{ and } \hspace*{2mm} bx=R\cdot |AC'| \end{eqnarray*} After adding the last two equations we get: \begin{eqnarray*} (b+c)x=R(|AB'|+|AC'|) \end{eqnarray*} and similarly: \begin{eqnarray*} (a+c)y=R\cdot (|BA'|+|BC'|)\\ (a+b)z=R\cdot (|CA'|+|CB'|). \end{eqnarray*} After adding the last three relations we get: \begin{eqnarray} \label{eqnCarnotOcrt2} (b+c)x+(a+c)y+(a+b)z=R(a+b+c). \end{eqnarray} If we finally add the equality \ref{eqnCarnotOcrt1} and \ref{eqnCarnotOcrt2} and divide the obtained equality by $a+b+c$, we get: \begin{eqnarray*} x+y+z=r+R, \end{eqnarray*} which had to be proven. \kdokaz \bnaloga\footnote{12. IMO, Hungary - 1970, Problem 1.} Let $M$ be a point on the side $AB$ of triangle $ABC$. Let $r_1$, $r_2$ and $r$ be the radii of the inscribed circles of triangles $AMC$, $BMC$ and $ABC$. Let $q_1$, $q_2$ and $q$ be the radii of the escribed circles of the same triangles that lie in the angle $ACB$. Prove that $$\frac{r_1}{q_1}\cdot\frac{r_2}{q_2}=\frac{r}{q}.$$ \enaloga \begin{figure}[!htb] \centering \input{sl.plo.8.3.IMO1.pic} \caption{} \label{sl.plo.8.3.IMO1.pic} \end{figure} \textbf{\textit{Solution.}} First, we mark $a=|BC|$, $b=|AC|$, $c=|AB|$, $x=|AM|$, $y=|BM|$ and $m=|CM|$ (Figure \ref{sl.plo.8.3.IMO1.pic}). If we use the expression \ref{PloscTrikVcrt} and \ref{PloscTrikPricrt}, we get: \begin{eqnarray*} && p_{\triangle ABC}=\frac{a+b+x+y}{2}\cdot r=\frac{a+b-x-y}{2}\cdot q\\ && p_{\triangle AMC}=\frac{b+m+x}{2}\cdot r_1=\frac{b+m-x}{2}\cdot q_1\\ && p_{\triangle ABC}=\frac{a+m+y}{2}\cdot r_2=\frac{a+m-y}{2}\cdot q_2 \end{eqnarray*} It follows that: \begin{eqnarray*} \frac{r}{q}=\frac{a+b+x+y}{a+b-x-y},\hspace*{6mm} \frac{r_1}{q_1}=\frac{b+m+x}{b+m-x},\hspace*{6mm} \frac{r_2}{q_2}=\frac{a+m+y}{a+m-y} \end{eqnarray*} Therefore, the relation $\frac{r_1}{q_1}\cdot\frac{r_2}{q_2}=\frac{r}{q}$, which we want to prove, is equivalent to: $$(a+b+x+y)(a+m-y)(b+m-x)=(a+b-x-y)(a+m+y)(b+m+x),$$ which, after rearranging, is equivalent to: $$(a+b)(a+m)x+(a+b)(b+m)y=c(a+m)(b+m)+(x+y)xy,$$ or, if we take into account $x+y=c$ and rearrange further, to: $$m^2=a^2\frac{x}{c}+b^2\frac{y}{c}-xy.$$ The last relation is true, because it represents Stewart's theorem \ref{StewartIzrek} for the triangle $ABC$ and the distance $CM$. \kdokaz %________________________________________________________________________________ \poglavje{Area of Polygons} \label{odd8PloVeck} The area of any polygon is obtained by dividing it with its diagonals into a union of triangles (Figure \ref{sl.plo.8.4.1.pic}). According to the expressions \ref{ploscGlavniIzrek} and \ref{ploscDaljice}, the area of this polygon is equal to the sum of the areas of these triangles. \begin{figure}[!htb] \centering \input{sl.plo.8.4.1.pic} \caption{} \label{sl.plo.8.4.1.pic} \end{figure} The following theorem applies specifically to a regular hexagon. \bizrek If $p$ is the area and $a$ is the length of a side of a regular hexagon, then: $$p=\frac{3\sqrt{3}\cdot a^2}{2}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.4.2.pic} \caption{} \label{sl.plo.8.4.2.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.4.2.pic}) The theorem is a direct consequence of formulas \ref{ploscGlavniIzrek}, \ref{ploscDaljice} \ref{PloscTrikEnakostr} and the fact that a regular hexagon can be divided into six regular triangles (section \ref{odd3PravilniVeck}). \kdokaz A similar formula to that in the case of triangles applies for the area of tangent polygons (formula \ref{PloscTrikVcrt}). \bizrek \label{ploscTetVec} If: $s$ is the circumference, $k(S,r)$ is the inscribed circle of the tangent polygon $A_1A_2\ldots A_n$ and $p$ is its area, then $$p = sr.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.4.3.pic} \caption{} \label{sl.plo.8.4.3.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.4.3.pic}) The triangles $A_1SA_2$, $A_2SA_3$, ... , $A_{n-1}SA_n$ and $A_nSA_1$ all have the same length of altitude from the vertex $S$, which is equal to $r$. If we use formulas \ref{ploscGlavniIzrek} \textit{4)}, \ref{ploscDaljice} and \ref{PloscTrik}, we get: \begin{eqnarray*} p &=& p_{A_1SA_2}+p_{A_2SA_3}+\cdots +p_{A_{n-1}SA_n} +p_{A_nSA_1}=\\ &=& \frac{|A_1A_2|\cdot r}{2}+\frac{|A_2A_3|\cdot r}{2}+\cdots+ \frac{|A_{n-1}A_n|\cdot r}{2}+\frac{|A_n A_1|\cdot r}{2}=\\ &=&sr, \end{eqnarray*} which is what needed to be proven. \kdokaz \bnaloga\footnote{30. IMO, Germany - 1989, Problem 2.} Let $S$, $S_a$, $S_b$ and $S_c$ be incentre and excentres of an acute-angled triangle $ABC$. $N_a$, $N_b$ and $N_c$ are midpoints of line segments $SS_a$, $SS_b$ and $SS_c$, respectively. Let $\mathcal{S}_{X_1X_2\ldots X_n}$ be the area of a polygon $X_1X_2\ldots X_n$. Prove that: $$\mathcal{S}_{S_aS_bS_c} = 2\mathcal{S}_{AN_cBN_aCN_b} \geq 4\mathcal{S}_{ABC}.$$ \enaloga \begin{figure}[!htb] \centering \input{sl.plo.8.3.IMO2.pic} \caption{} \label{sl.plo.8.3.IMO2.pic} \end{figure} \textbf{\textit{Solution.}} Let $l$ be the circle circumscribed about the triangle $ABC$ (Figure \ref{sl.plo.8.3.IMO2.pic}). By the great theorem (\ref{velikaNaloga}) the point $N_a$ is the centre of that arc $BC$ of the circle $l$, on which the point $A$ does not lie - we denote this arc with $l_{BC}$. Similarly, $N_b$ and $N_c$ are the centres of the corresponding arcs $AC$ and $AB$ on the circle $l$. First, from $SN_a\cong N_aS_a$ it follows that $\mathcal{S}_{SCN_a}=\mathcal{S}_{N_aCS_a}$ or $\mathcal{S}_{SCS_a}=2\cdot\mathcal{S}_{SCN_a}$. If we use analogous equalities and take into account that $S$ is an inner point of the triangle $S_aS_bS_c$, we get: \begin{eqnarray*} \mathcal{S}_{S_aS_bS_c} &=& \mathcal{S}_{SCS_a}+\mathcal{S}_{SBS_a}+ \mathcal{S}_{SCS_b}+\mathcal{S}_{SAS_b}+ \mathcal{S}_{SAS_c}+\mathcal{S}_{SBS_c}=\\ &=& 2\cdot\left(\mathcal{S}_{SCN_a}+\mathcal{S}_{SBN_a}+ \mathcal{S}_{SCN_b}+\mathcal{S}_{SAN_b}+ \mathcal{S}_{SAN_c}+\mathcal{S}_{SBN_c}\right)=\\ &=& 2\mathcal{S}_{AN_cBN_aCN_b}. \end{eqnarray*} We will now prove that $2\mathcal{S}_{AN_cBN_aCN_b} \geq 4\mathcal{S}_{ABC}$, or $\mathcal{S}_{AN_cBN_aCN_b} \geq 2\mathcal{S}_{ABC}$. Let $V$ be the altitude point of triangle $ABC$ and $V_a=S_{BC}(V)$. Triangle $ABC$ is acute by assumption, so point $V$ lies within it. By \ref{TockaV'} point $V_a$ lies on the arc $l_{BC}$. We have already mentioned that $N_a$ is the center of this arc, so the altitude of triangle $BNC$ is longer than the altitude of triangle $BVC$ by the same base $BC$. It follows that $\mathcal{S}_{BNC}\geq \mathcal{S}_{BVC}$. Similarly, we have $\mathcal{S}_{ANC}\geq \mathcal{S}_{AVC}$ and $\mathcal{S}_{ANB}\geq \mathcal{S}_{AVB}$. Since $V$ is an internal point of triangle $ABC$, we have: \begin{eqnarray*} \mathcal{S}_{AN_cBN_aCN_b}&=&\mathcal{S}_{ABC}+ \mathcal{S}_{BNC}+\mathcal{S}_{ANC}+\mathcal{S}_{ANB}\geq\\ &\geq& \mathcal{S}_{ABC}+ \mathcal{S}_{BVC}+\mathcal{S}_{AVC}+\mathcal{S}_{AVB}=\\ &=& 2\mathcal{S}_{ABC}, \end{eqnarray*} which was to be proven. \kdokaz %________________________________________________________________________________ \poglavje{Koch Snowflake} \label{odd8PloKoch} In this section we will consider a figure which is bounded (a subset of a circle), has a finite area, but its perimeter is infinite. First, we will define a special type of polygon. An equilateral triangle $ABC$ with a side length $a=a_0$ is denoted by $\mathcal{K}_0(a)$. The polygon $\mathcal{K}_1(a)$ is obtained if each side of the triangle $\mathcal{K}_0(a)$ is divided into three equal parts and an equilateral triangle with a side length $a_1=\frac{1}{3}\cdot a_0$ is drawn over the middle. The process is continued and the polygon $\mathcal{K}_n(a)$ is obtained if each side of the triangle $\mathcal{K}_{n-1}(a)$ is divided into three equal parts and an equilateral triangle with a side length $a_n=\frac{1}{3}\cdot a_{n-1}$ is drawn over the middle (Figure \ref{sl.plo.8.5.Koch1.pic}). \begin{figure}[!htb] \centering \input{sl.plo.8.5.Koch1.pic} \caption{} \label{sl.plo.8.5.Koch1.pic} \end{figure} If the described process is continued to infinity, we get the so-called \index{Kochova snežinka} \pojem{Koch's\footnote{This figure was defined by the Swedish mathematician \index{Koch, H.}\textit{Helge von Koch} (1870--1924) in 1904. Today, Koch's snowflake is classified as a \textit{fractal}.} snowflake} $\mathcal{K}(a)$. First, let's calculate the perimeter of the polygon $\mathcal{K}_n(a)$. From the process by which we obtained the polygon $\mathcal{K}_n(a)$, it is clear that the polygon $\mathcal{K}_n(a)$ has four times as many sides as $\mathcal{K}_{n-1}(a)$ (each side of the polygon $\mathcal{K}_{n-1}(a)$ is replaced by four sides of the polygon $\mathcal{K}_n(a)$). We denote by $s_n$ the number of sides of the polygon $\mathcal{K}_n(a)$. Obviously, $s_n$ is a geometric sequence with a quotient of $q=4$ and an initial term of $s_0=3$, so: \begin{eqnarray} \label{eqnKoch1} s_n=3\cdot 4^n \end{eqnarray} Each side $a_n$ of the polygon $\mathcal{K}_n(a)$ is, by construction, three times shorter than the side $a_{n-1}$ of the polygon $\mathcal{K}_{n-1}(a)$. So we have a geometric sequence $a_n$ again with a quotient of $q=\frac{1}{3}$ and an initial term of $a_0=a$, so: \begin{eqnarray} \label{eqnKoch2} a_n=\left( \frac{1}{3} \right)^n\cdot a \end{eqnarray} Because for the perimeter $o_n$ of the polygon $\mathcal{K}_n(a)$ it holds that $o_n=s_n\cdot a_n$, from \ref{eqnKoch1} and \ref{eqnKoch2} we get: \begin{eqnarray} \label{eqnKoch3} o_n=3\cdot \left( \frac{4}{3} \right)^n\cdot a \end{eqnarray} If we want to obtain the perimeter $o$ of the Koch snowflake $\mathcal{K}(a)$, we calculate it as $o=\lim_{n\rightarrow\infty}o_n$, but from \ref{eqnKoch3} we get (because $\frac{4}{3}>1$): \begin{eqnarray} \label{eqnKoch4} o=\lim_{n\rightarrow\infty}o_n=\lim_{n\rightarrow\infty}3\cdot \left( \frac{4}{3} \right)^n\cdot a=3a\cdot\lim_{n\rightarrow\infty}\left( \frac{4}{3} \right)^n=\infty \end{eqnarray} which means that the Koch snowflake $\mathcal{K}(a)$ has no finite perimeter, that is, the perimeter $o_n$ of the polygon $\mathcal{K}_n(a)$ increases without bound as the number of steps $n$ increases. Let's now calculate the area $p$ of the Koch snowflake $\mathcal{K}(a)$. We'll denote the area of the polygon $\mathcal{K}_n(a)$ with $p_n$. The area $p_n$ is obtained by adding a certain number of smaller triangle areas to the area $p_{n-1}$. How many of these triangles are there? By the construction of the polygon $\mathcal{K}_n(a)$, we get it from the polygon $\mathcal{K}_{n-1}(a)$ by drawing a smaller triangle at each side of the polygon $\mathcal{K}_{n-1}(a)$. So the number of these triangles is equal to $s_{n-1}$. Because of this and the \ref{PloscTrikEnakostr} equation, we have: \begin{eqnarray*} p_n=p_{n-1}+s_{n-1}\cdot \frac{a_n^2\cdot \sqrt{3}}{4}=p_{n-1} +\frac{3}{16}\cdot \left(\frac{4}{9} \right)^n\cdot a^2\sqrt{3}. \end{eqnarray*} The area $p$ of the Koch snowflake $\mathcal{K}(a)$ is therefore the infinite sum: \begin{eqnarray*} p&=&p_0+ \sum_{n=1}^{\infty}\frac{3}{16}\cdot \left(\frac{4}{9} \right)^n\cdot a^2\sqrt{3}=\\ &=&\frac{a^2\cdot \sqrt{3}}{4}+ \frac{3}{16} a^2\sqrt{3}\cdot\sum_{n=1}^{\infty} \left(\frac{4}{9}\right)^n=\\ &=&\frac{a^2\cdot \sqrt{3}}{4}+ \frac{3}{16} a^2\sqrt{3}\cdot\left(\frac{4}{9}+\left(\frac{4}{9}\right)^2+\cdots + \left(\frac{4}{9}\right)^n+\cdots \right). \end{eqnarray*} If we calculate the sum of an infinite geometric sequence and simplify it, we get: \begin{eqnarray} \label{eqnKoch5} p=\frac{2a^2\sqrt{3}}{5}. \end{eqnarray} %________________________________________________________________________________ \poglavje{Circumference and Area of a Circle} \label{odd8PloKrog} %OBSEG In section \ref{odd3NeenTrik} we have already defined the circumference of a polygon as the sum of all its sides. In this sense, the circumference represented a distance. In the following, we will understand the circumference also as the length of this distance, or the sum of the lengths of all the sides of the polygon. Here we will deal with the circumference of a circle. Although it is intuitively clear, this new concept needs to be defined first. Let $k(S,r)$ be an arbitrary circle. The corresponding circle is denoted by $\mathcal{K}(S,r)$. The circumference of this circle intuitively represents the length of the circle $k$. Let $A_1A_2\ldots A_n$ ($n\in \mathbb{N}$ and $n\geq 3$) be a regular polygon inscribed in the circle $k$. The tangents of this circle at the vertices of the polygon $A_1A_2\ldots A_n$ determine the sides of the regular polygon $B_1B_2\ldots B_n$, which is circumscribed to the circle $k$ (Figure \ref{sl.plo.8.5.1.pic}). \begin{figure}[!htb] \centering \input{sl.plo.8.5.1.pic} \caption{} \label{sl.plo.8.5.1.pic} \end{figure} We denote by $\underline{o}_n$ the circumference of the polygon $A_1A_2\ldots A_n$ and $\overline{o}_n$ the circumference of the polygon $B_1B_2\ldots B_n$. By the triangle inequality (statement \ref{neenaktrik}) we have $|A_1B_1|+|B_1A_2|>|A_1A_2|$, $|A_2B_2|+|B_2A_3|>|A_2A_3|$,..., $|A_nB_n|+|B_nA_1|>|A_nA_1|$ (Figure \ref{sl.plo.8.5.1a.pic}). If we add all these inequalities we get $\overline{o}_n>\underline{o}_n$. Similarly, $\underline{o}_n<\underline{o}_{n+1}$ and $\overline{o}_n>\overline{o}_{n+1}$. Therefore, for every $n\in \mathbb{N}$, $n\geq 3$ we have: \begin{eqnarray*} \underline{o}_1<\underline{o}_2<\cdots< \underline{o}_n<\overline{o}_n<\cdots<\overline{o}_2<\overline{o}_1. \end{eqnarray*} This means that $\underline{o}_n$ is an increasing sequence bounded from above by $\overline{o}_1$, so it is a known statement of mathematical analysis that it is convergent and has its limit: \begin{eqnarray*} \underline{o}=\lim_{n\rightarrow\infty}\underline{o}_n. \end{eqnarray*} Similarly, $\overline{o}_n$ is a decreasing sequence bounded from below by $\underline{o}_1$, so it has its limit: \begin{eqnarray*} \overline{o}=\lim_{n\rightarrow\infty}\overline{o}_n. \end{eqnarray*} \begin{figure}[!htb] \centering \input{sl.plo.8.5.1a.pic} \caption{} \label{sl.plo.8.5.1a.pic} \end{figure} We will not formally prove the intuitively clear fact that for a sufficiently large natural number $n$, the difference $\overline{o}_n-\underline{o}_n$ is arbitrarily small, i.e. $\underline{o}=\overline{o}$. \index{obseg!kroga}\pojem{Circumference of a circle} $\mathcal{K}$ with the designation $o$ then represents $o=\underline{o}=\overline{o}$. We prove the following important statement. \bizrek If $o$ is the circumference of a circle with radius $r$, then the ratio $o:2r$ is constant and thus independent of the choice of the circle. \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.5.1b.pic} \caption{} \label{sl.plo.8.5.1b.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.5.1b.pic}) Let $\mathcal{K}(S,r)$ and $\mathcal{K}'(S',r')$ be two arbitrary circles with circumferences $o$ and $o'$, and $k(S,r)$ and $k'(S',r')$ the corresponding circles. According to Theorem \ref{RaztKroznKrozn1}, there exists a central stretch that maps the circle $k$ to the circle $k'$. According to Theorem \ref{RaztTransPod}, this central stretch is a similarity transformation (denoted by $f$) with some coefficient $\lambda$. First, $r':r=\lambda$ or: \begin{eqnarray} r'=\lambda r. \label{eqnPloscKrogrr'oo'} \end{eqnarray} Let $A_1A_2\ldots A_n$ and $B_1B_2\ldots B_n$ ($n\in \mathbb{N}$ and $n\geq 3$) be inscribed and circumscribed regular polygons of the circle $k$ with circumferences $\underline{o}_n$ and $\overline{o}_n$. We denote: $$f:S,A_1,A_2,\ldots,A_n,B_1,B_2,\ldots,B_n\mapsto S',A_1',A_2',\ldots,A_n',B_1',B_2',\ldots,B_n'.$$ The polygons $A_1'A_2'\ldots A_n'$ and $B_1'B_2'\ldots B_n'$ ($n\in \mathbb{N}$ and $n\geq 3$) are inscribed and circumscribed regular polygons of the circle $k$ and for their circumferences $\underline{o'}_n$ and $\overline{o'}_n$ it holds that $\underline{o'}_n=\lambda\underline{o}_n$ and $\overline{o'}_n=\lambda\overline{o}_n$. From this and from the definition of the circumference of a circle it follows: \begin{eqnarray}\label{eqnPloscKrogrr'oo'1} o'=\underline{o'}=\lim_{n\rightarrow\infty}\underline{o'}_n= \lim_{n\rightarrow\infty}\lambda\underline{o}_n= \lambda\lim_{n\rightarrow\infty}\underline{o}_n=\lambda \underline{o}=\lambda o. \end{eqnarray} From \ref{eqnPloscKrogrr'oo'} and \ref{eqnPloscKrogrr'oo'1} it follows: $$\frac{o'}{2r'}=\frac{\lambda o}{2\lambda r}=\frac{o}{2r},$$ which was to be proven. \kdokaz The constant from the previous theorem is called \index{number!$\pi$}\pojem{number $\pi$} (\index{Archimedes' constant}\pojem{Archimedes' constant} or \index{number!Ludolf's}\pojem{Ludolf's number}) and we also denote it by $\pi$. From the previous theorem and from the definition of the number $\pi$ we get the following theorem. \bizrek \label{obsegKtoznice} If $o$ is the circumference of a circle with radius $r$, then: $$o=2r\pi.$$ \eizrek First, we will give a rough estimate of the number $\pi$. \bizrek \label{stevPiOcena} For the number $\pi$ it holds: $$3<\pi<4.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.5.2.pic} \caption{} \label{sl.plo.8.5.2.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.5.2.pic}) Let $\mathcal{K}$ be an arbitrary circle with radius $r$ and circumference $o$, and $k$ the corresponding circle. By \ref{obsegKtoznice}, $o=2r\pi$. We choose the regular inscribed polygon $n=6$ and the regular circumscribed polygon $n=4$. From the definition of the circumference of a circle it follows: $$\underline{o}_6r_1$) be two circles with the same center (the corresponding circles $k_1$ and $k_2$ are concentric). The set $\mathcal{K}_2\setminus \mathcal{K}_1$ is called \index{krožni!kolobar}\pojem{circular sector}. \bizrek \label{ploscKrozKolob} If $p_{kl}$ is the area of the circular sector determined by the circles $k_1(S,r_1)$ and $k_2(S,r_2)$ ($r_2>r_1$), then: $$p_{kl}=\left( r_2^2-r_1^2 \right)\cdot\pi.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.5.4.pic} \caption{} \label{sl.plo.8.5.4.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.5.4.pic}) We mark with $\mathcal{K}_1(S,r_1)$ and $\mathcal{K}_2(S,r_2)$ the corresponding circles of the circles $k_1(S,r_1)$ and $k_2(S,r_2)$. From $\mathcal{K}_2=\mathcal{K}_2\setminus \mathcal{K}_1\cup \mathcal{K}_1 $ and $\left( \mathcal{K}_2\setminus \mathcal{K}_1 \right) \cap \mathcal{K}_1=\emptyset$ according to the formulas \ref{ploscGlavniIzrek} \textit{4)} and \ref{ploscKrog} we get $p_{\mathcal{K}_2}=p_{\mathcal{K}_2\setminus \mathcal{K}_1} + p_{\mathcal{K}_1}$ i.e.: $$p_{kl}=p_{\mathcal{K}_2\setminus \mathcal{K}_1}= p_{\mathcal{K}_2}- p_{\mathcal{K}_1}=r_2^2\pi-r_1^2\pi =\left( r_2^2-r_1^2 \right)\cdot\pi,$$ which had to be proven. \kdokaz In a similar way as for the length of the circular arc (formula \ref{obsegKrozLok}) we get the following statement. \bizrek \label{ploscKrozIzsek} If $p_i$ is the area of the circular section with the corresponding central angle with a measure (in degrees) $\alpha$ and with a radius $r$, then: $$p_i=\frac{r^2\pi\cdot\alpha}{360^0}.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.5.5.pic} \caption{} \label{sl.plo.8.5.5.pic} \end{figure} \bzgled If in the formula for the area of the circle we replace the number $\pi$ with its approximate value $3$, we get the formula for the area of the inscribed regular dodecagon\footnote{This task was solved by the Chinese mathematician \index{Liu, H.}\textit{H. Liu} (3rd century)}. \ezgled \begin{figure}[!htb] \centering \input{sl.plo.8.5.6.pic} \caption{} \label{sl.plo.8.5.6.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.5.6.pic}) Let $A_1A_2\ldots A_{12}$ be a regular dodecagon, which is inscribed in the circle $k(S,r)$. Then $A_1A_3\ldots A_{11}$ is a regular hexagon, which is inscribed in the same circle. We denote by $p_{12}$ the area of the mentioned dodecagon. By \ref{ploscGlavniIzrek} \textit{4)} we have $p_{12}=12\cdot p_{SA_1A_2}$. Since the triangle $A_1SA_3$ is a right triangle, the height $A_1P$ of the triangle $SA_1A_2$ is equal to half of the side $A_1A_3$ of the hexagon $A_1A_3\ldots A_{11}$, that is: $$|A_1P|=\frac{1}{2}\cdot |A_1A_3|=\frac{r}{2}.$$ Then, by \ref{PloscTrik} we have: $$p_{12}=12\cdot p_{SA_1A_2}=12\cdot\frac{|SA_2|\cdot |A_1P|}{2}=12\cdot\frac{r^2}{4}=3\cdot r^2,$$ which had to be proven. \kdokaz \bzgled \label{HipokratoviLuni} Above the catheti of a right triangle we construct circles and from the thus enlarged triangle we cut out a circle above the hypotenuse (Figure \ref{sl.plo.8.5.7.pic}). Show that the area of the remainder is equal to the area of the original right triangle\footnote{A special claim (the next example in this section), if the given triangle is isosceles, was proven by \index{Hipokrat}\textit{Hipokrat from Kios} (5th century BC), an ancient Greek mathematician. The aforementioned figure is called \index{Hipokratovi luni}\pojem{Hipokratovi luni}. Hipokrat was the first to discover that some figures with a curved edge can be converted into a square with a straight edge and a hexagon.}. \ezgled \begin{figure}[!htb] \centering \input{sl.plo.8.5.7.pic} \caption{} \label{sl.plo.8.5.7.pic} \end{figure} \textbf{\textit{Solution.}} Let $ABC$ be the aforementioned triangle with hypotenuse $c$ and sides $a$ and $b$. Let $p$ be the desired area, $p_{\triangle}$ the area of the triangle $ABC$, and $p_a$, $p_b$, and $p_c$ the areas of the corresponding semicircles. By the formulas \ref{ploscGlavniIzrek} \textit{4)}, \ref{ploscKrozIzsek} and \ref{PloscTrik} and the Pythagorean theorem \ref{PitagorovIzrek}, we have: \begin{eqnarray*} p&=& p_a+p_b-p_c+p_{\triangle}=\\ &=& \left(\frac{a}{2} \right)^2\cdot \pi +\left(\frac{a}{2} \right)^2\cdot \pi-\left(\frac{a}{2} \right)^2\cdot \pi+p_{\triangle}=\\ &=& \frac{\pi}{4}\cdot \left(a^2+b^2-c^2 \right)+p_{\triangle}=\\ &=& \frac{\pi}{4}\cdot 0+p_{\triangle}=\\ &=& p_{\triangle}, \end{eqnarray*} which had to be proven. \kdokaz \bzgled \label{HipokratoviLuni2} The day is the square $PQRS$. Draw a square that has the same area as the figure that results when the circle drawn around the square $PQRS$ is cut by a circle with radius $PQ$ (Figure \ref{sl.plo.8.5.8.pic}). \ezgled \begin{figure}[!htb] \centering \input{sl.plo.8.5.8.pic} \caption{} \label{sl.plo.8.5.8.pic} \end{figure} \textbf{\textit{Solution.}} If we choose the isosceles right triangle $QSQ'$ (where $Q'=\mathcal{S}_p(Q)$), by the previous formula (\ref{HipokratoviLuni}) the area of the desired figure is equal to the area of the isosceles right triangle $SPQ$ (Figure \ref{sl.plo.8.5.8.pic}), and its area is equal to the area of the square $PLSN$, where $L$ is the center of the line $SQ$ and $N=\mathcal{S}_{SP}(L)$ (Figure \ref{sl.plo.8.5.8a.pic}). \begin{figure}[!htb] \centering \input{sl.plo.8.5.8a.pic} \caption{} \label{sl.plo.8.5.8a.pic} \end{figure} \kdokaz \bzgled Let $ABCD$ be a square which we increase by four semicircles above the sides, and from the resulting figure we cut out the circle circumscribed to the square (Figure \ref{sl.plo.8.5.9.pic}). Prove that the area of the remainder is equal to the area of the original square. \ezgled \begin{figure}[!htb] \centering \input{sl.plo.8.5.9.pic} \caption{} \label{sl.plo.8.5.9.pic} \end{figure} \textbf{\textit{Proof.}} Let $S$ be the center of the square $ABCD$. The statement is a direct consequence of the statement of the previous theorem (\ref{HipokratoviLuni2}) - we use the appropriate relation for the triangles $ASB$, $BSC$, $CSD$ and $DSA$ and add everything up. \kdokaz \bzgled \label{arbelos} Let $C$ be an arbitrary point on the diameter $AB$ of the semicircle $k$ and let $D$ be the point of intersection of this semicircle with the rectangle with diameter $AB$ at point $C$. Let $j$ and $l$ be semicircles with diameters $AC$ and $BC$. Prove that the area of the figure bounded by semicircles $k$, $l$ and $j$ is equal to the area of the circle with diameter $CD$\footnote{This task was published in his 'Book of Lemmas' by the Greek mathematician \index{Arhimed}\textit{Archimedes of Syracuse} (3rd century BC). He named the aforementioned figure \index{arbelos}\textit{arbelos} (which in Greek means 'shoemaker's knife').}. \ezgled \begin{figure}[!htb] \centering \input{sl.plo.8.5.10.pic} \caption{} \label{sl.plo.8.5.10.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.5.10.pic}) Let's mark $a=|AC|$ and $b=|BC|$ the appropriate diameters, $p$ the area of the aforementioned figure and $p_0$ the area of the circle with diameter $|CD|$. According to \ref{izrekVisinski} it is $|CD|^2 = ab$. If we use \ref{ploscKrozIzsek} and \ref{ploscKrog} as well, we get: \begin{eqnarray*} p&=&\frac{1}{2}\left(\frac{a+b}{2}\right)^2\cdot\pi- \frac{1}{2}\left(\frac{a}{2}\right)^2\cdot\pi- \frac{1}{2}\left(\frac{b}{2}\right)^2\cdot\pi=\\ &=& \frac{ab}{4}\cdot\pi=\\ &=& \left(\frac{|CD|}{2}\right)^2\cdot\pi=p_0, \end{eqnarray*} which was to be proven. \kdokaz \bzgled Let $k$ be a drawn circle of a right triangle and $\mathcal{K}$ the corresponding circle. Then draw three smaller circles, which touch the circle $k$ and two sides of the triangle, then three smaller circles, which touch the previous three circles and two sides of the triangle, and so on until infinity. Is the area of all the corresponding circles, except for the circle $\mathcal{K}$, smaller than half the area of the circle $\mathcal{K}$? \ezgled \begin{figure}[!htb] \centering \input{sl.plo.8.5.11.pic} \caption{} \label{sl.plo.8.5.11.pic} \end{figure} \textbf{\textit{Solution.}} (Figure \ref{sl.plo.8.5.11.pic}) Let $ABC$ be a given triangle and $k(S,r)$ the inscribed circle of this triangle. With $k_i(S_i ,r_i)$ ($i\in\mathbb{N}$) we denote the circles from the given sequence of circles at the vertex $A$ and with $p_i$ ($i \in \mathbb{N}$) the areas of the corresponding circles. We also denote the circle $k$ with $k_0(S_0 ,r_0)$ and the area of the corresponding circle $\mathcal{K}$ with $p=p_0$. Let $k_n(S_n ,r_n)$ and $k_{n+1}(S_{n+1} ,r_{n+1})$ be the consecutive circles from the mentioned sequence. We denote their tangent points with the side $AB$ of the triangle $ABC$ with $T_n$ and $T_{n+1}$, and the orthogonal projection of the center $S_{n+1}$ onto the line $S_nT_n$ with $L_n$. In the right triangle $S_{n+1}L_nS_n$ the length of the hypotenuse is $|S_nL_n| = r_n- r_{n+1}$ and the hypotenuse $|S_{n+1}S_n| = r_n+ r_{n+1}$. As $L_nS_{n+1}S_n$ is equal to half of the internal angle at the vertex $A$, namely $30°$. From this it follows that $\angle S_{n+1}S_nL_n=60^0$. Let $S_n'=\mathcal{S}_{S_{n+1}L_n}(S_n)$. The triangles $S_{n+1}L_nS_n$ and $S_{n+1}L_nS_n'$ are congruent (by the \textit{SAS} \ref{SKS}), so also $\angle S_{n+1}S_n'L_n=60^0$. Therefore, $S_{n+1}S_n'S_n$ is an isosceles triangle, which means that $|S_{n+1}S_n| =|S_n'S_n| =2\cdot|S_nL_n|$ or $$r_n+ r_{n+1}=2\cdot(r_n-r_{n+1}).$$ If we express $r_{n+1}$ from the last equality, we get: \begin{eqnarray*} r_{n+1}=\frac{1}{3}\cdot r_n. \end{eqnarray*} From this, by \ref{ploscKrog}, it follows: \begin{eqnarray*} p_{n+1}=\frac{1}{9}\cdot p_n. \end{eqnarray*} The sequence $p_n$ is therefore a geometric sequence with the coefficient $q=\frac{1}{9}$ and the initial value $p_0=p$, so: \begin{eqnarray*} p_n=\left( \frac{1}{9} \right)^n \cdot p. \end{eqnarray*} This means that the total area $p_A$ of the circles from the sequence at the vertex $A$ represents the sum of an infinite geometric sequence $p_n$ ($n\in \mathbb{N}$). Therefore: \begin{eqnarray*} p_A&=&p_1+p_2+p_3+\cdots=\\ &=&\frac{1}{9}\cdot p+\left(\frac{1}{9}\right)^2\cdot p+ \left(\frac{1}{9}\right)^3\cdot p+\cdots=\\ &=&\frac{1}{9}p\cdot \frac{1}{1-\frac{1}{9}}=\\ &=& \frac{1}{8}\cdot p. \end{eqnarray*} The sum of the areas of all circles at the vertices $A$, $B$ and $C$ is then equal to $\frac{3}{8}\cdot p$ and is not greater than half of $p$. \kdokaz \bnaloga\footnote{6. IMO, USSR - 1964, Problem 3.} A circle is inscribed in triangle $ABC$ with sides $a$, $b$, $c$. Tangents to the circle parallel to the sides of the triangle are constructed. Each of these tangents cuts off a triangle from triangle $ABC$. In each of these triangles, a circle is inscribed. Find the sum of the areas of all four inscribed circles (in terms of $a$, $b$, $c$). \enaloga \begin{figure}[!htb] \centering \input{sl.plo.8.5.IMO1.pic} \caption{} \label{sl.plo.8.5.IMO1.pic} \end{figure} \textbf{\textit{Solution.}} Let $k(S,\rho)$ be the inscribed circle of triangle $ABC$ with altitudes $v_a$, $v_b$ and $v_c$ and $p$ the area of the corresponding circle. We denote by $AB_aC_a$, $A_bBC_b$ and $A_cB_cC$ the new triangles, $k_a$, $k_b$ and $k_c$ the inscribed circles of these triangles with radii $\rho_a$, $\rho_b$ and $\rho_c$ and $p_a$, $p_b$ and $p_c$ the areas of the corresponding circles. (Figure \ref{sl.plo.8.5.IMO1.pic}). The triangles $ABC$ and $AB_aC_a$ are similar with the coefficient $\frac{v_a-2\rho}{v_a}$. Therefore: $$\rho_a=\frac{v_a-2\rho}{v_a}\cdot \rho.$$ Because according to the statement \ref{PloscTrikVcrt} for the area of the triangle $ABC$ it holds $p_\triangle=s\cdot \varrho$ (where $s=\frac{a+b+c}{2}$ is the semi-perimeter of the triangle $ABC$), it follows: $$\rho_a^2=\left(1-\frac{2\rho}{v_a}\right)^2\cdot \rho^2= \left(1-\frac{2\frac{P_\triangle}{s}}{2\frac{P_\triangle}{a}}\right)^2\cdot \rho^2= \frac{(s-a)^2}{s^2}\cdot \rho^2.$$ Similarly we obtain $\rho_b^2= \frac{(s-b)^2}{s^2}\cdot \rho^2$ and $\rho_c^2= \frac{(s-c)^2}{s^2}\cdot \rho^2$. From the statement \ref{PloscTrikOcrtVcrt} it follows $\rho^2=\frac{(s-a)(s-b)(s-c)}{s}$, therefore: \begin{eqnarray*} && p+p_a+p_b+p_c= \pi (\rho^2+\rho_a^2+\rho_b^2+\rho_c^2)=\\ &&=\pi \rho^2(1+\frac{(s-a)^2}{s^2}+\frac{(s-b)^2}{s^2}+\frac{(s-c)^2}{s^2})=\\ &&= \frac{(s-a)(s-b)(s-c)(s^2+(s-a)^2+(s-b)^2+(s-c)^2)}{s^3}\cdot \pi, \end{eqnarray*} which had to be expressed. \kdokaz %________________________________________________________________________________ \poglavje{Pythagoras' Theorem and Area} \label{odd8PloPitagora} In the section \ref{odd7Pitagora} we have proven Pythagoras' statement \ref{PitagorasIzrek} and predicted that we will express it in the form that relates to the areas. \bizrek \index{statement!Pythagoras for areas} \label{PitagorasIzrekPlosc}(Pythagoras' statement in the form of areas)\\ The area of the square over the hypotenuse of a right-angled triangle is equal to the sum of the areas of the squares over the two cathets of this triangle. \eizrek \begin{figure}[!htb] \centering \input{sl.plo.8.6.1.pic} \caption{} \label{sl.plo.8.6.1.pic} \end{figure} \textbf{\textit{Proof.}} At this point we will provide two proofs and one idea of the proof of this statement. \textit{1)} The statement is a direct consequence of the Pythagorean Theorem \ref{PitagorovIzrek} and the formula for the area of a square \ref{ploscKvadr} (Figure \ref{sl.plo.8.6.1.pic}). \textit{2)\footnote{This proof was published by \index{Euclid}\textit{Euclid of Alexandria} (3rd century BC) in his famous work 'Elements', which consists of 13 books.}} Let $AMNB$, $BPQC$ and $CKLA$ be squares constructed on the hypotenuse and the legs of the right triangle $ABC$. Let $D$ and $E$ be the projections of point $C$ on the lines $AB$ and $MN$ (Figure \ref{sl.plo.8.6.1.pic}). Since the area of the square $AMNB$ is equal to the sum of the areas of the rectangles $NBDE$ and $EDAM$, it is enough to prove that the areas of the squares $BPQC$ and $CKLA$ are equal to the areas of the rectangles $NBDE$ and $EDAM$. We will prove only the first equality $p_{BPQC}=p_{NBDE}$ (the second equality $p_{CKLA}=p_{EDAM}$ is then analogous). For this relation $p_{BPQC}=p_{NBDE}$ it is enough to prove the equality of the areas of the triangles $BPQ$ and $NBD$, since $p_{BPQC}=2\cdot p_{BPQ}$ and $p_{NBDE}=2\cdot p_{NBD}$. Then: \begin{eqnarray*} p_{BPQ}&=&p_{BPA}=\hspace*{3mm} \textrm{(the triangles have the same base and height)}\\ &=&p_{NBC}=\hspace*{3mm} \textrm{(the triangles are congruent due to } \mathcal{R}_{B,90^0+\angle CBA}\textrm{)}\\ &=&p_{NBD}\hspace*{4mm} \textrm{(the triangles have the same base and height)} \end{eqnarray*} Therefore, the triangles $BPQ$ and $NBD$ have the same area, which was enough to prove. \begin{figure}[!htb] \centering \input{sl.plo.8.6.2.pic} \caption{} \label{sl.plo.8.6.2.pic} \end{figure} \textit{3)\footnote{This proof (based on the illustration) is attributed to the Ancient Indians. We can say that it only needs the comment: "Look!" - so it is simple and elegant.}} The idea of the proof is given by the picture \ref{sl.plo.8.6.2.pic}. We will leave the formal proof to the reader. \kdokaz \bzgled The area of a circular sector, determined by the inscribed and circumscribed circle of a regular $n$-gon with a side length $a$, is not dependent on the number $n$. \ezgled \begin{figure}[!htb] \centering \input{sl.plo.8.6.3.pic} \caption{} \label{sl.plo.8.6.3.pic} \end{figure} \textbf{\textit{Proof.}} Let $AB$ be one side, $S$ the center of this side and $O$ the center of the inscribed or circumscribed circle of any regular $n$-gon $\mathcal{V}_n(a)$ (Figure \ref{sl.plo.8.6.3.pic}). According to the assumption, $|AB|=a$, therefore $|AS|=\frac{a}{2}$. We mark with $R$ and $r$ the radius of the inscribed and circumscribed circle of the $n$-gon $\mathcal{V}_n(a)$. From the congruence of triangles $OSA$ and $OSB$ (\textit{SSS} \ref{SSS}) it follows that $\angle OSA\cong \angle OSB=90^0$. Therefore $OSA$ is a right triangle with hypotenuse $OA$, so by the Pythagorean Theorem \ref{PitagorovIzrek} $|OA|^2-|OS|^2=|AS|^2$ or $R^2-r^2=\left(\frac{a}{2}\right)^2$. From this relation and the formula for the area of a circular sector \ref{ploscKrozKolob} we get: $$p_k=\left(R^2-r^2\right)\cdot \pi=\left(\frac{a}{2}\right)^2\cdot \pi=\frac{a^2\pi}{4},$$ which means that the area of the sector is only dependent on the length of the side of the regular $n$-gon $\mathcal{V}_n(a)$. \kdokaz \bzgled Using the statement from Example \ref{PitagoraCofman}, prove the inequality: \begin{eqnarray*} \frac{1}{1^2\cdot 2^2}+\frac{1}{2^2\cdot 3^2}+\cdots +\frac{1}{n^2 \left(n+1\right)^2}+\cdots<\frac{2}{3} \end{eqnarray*} or \begin{eqnarray*} \sum_{n=1}^{\infty}\frac{1}{n^2 \left(n+1\right)^2}<\frac{2}{3}. \end{eqnarray*} \ezgled \begin{figure}[!htb] \centering \input{sl.plo.8.6.4.pic} \caption{} \label{sl.plo.8.6.4.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.plo.8.6.4.pic}) We use the notation from the example \ref{PitagoraCofman}. According to the relation \ref{eqnPitagoraCofman3} from this example, $d_n=\frac{1}{n(n+1)}$, therefore $r_n=\frac{1}{2n(n+1)}$. If we denote the area of the corresponding circle of the circle $c_n$ with $p_n$, it is (formula \ref{ploscKrog}): \begin{eqnarray*} p_n=\pi r_n^2=\frac{\pi}{4}\frac{1}{n^2\left(n+1\right)^2}. \end{eqnarray*} Therefore, the sum of the areas of all corresponding circles is equal to: \begin{eqnarray} \label{eqnPitagoraCofmanPlo2} \sum_{n=1}^{\infty} p_n=\frac{\pi}{4}\sum_{n=1}^{\infty}\frac{1}{n^2\left(n+1\right)^2}. \end{eqnarray} But this sum is smaller than the area $p$ of the figure bounded by the line $A'B'$ and the circular arc $A'P$ and $B'P$ of the circles $a$ and $b$ with the central angle $90^0$. The area $p$ is equal to the difference of the area of the rectangle $A'ABB'$ and the area of two corresponding circular sectors with the central angle $90^0$. Therefore (formula \ref{ploscPravok} and \ref{ploscKrozIzsek}): \begin{eqnarray} \label{eqnPitagoraCofmanPlo3} p=2-2\cdot\frac{\pi}{4}=2-\frac{\pi}{2}. \end{eqnarray} Because $\sum_{n=1}^{\infty} p_n3$ (formula \ref{stevPiOcena}), we get: \begin{eqnarray*} \sum_{n=1}^{\infty}\frac{1}{n^2\left(n+1\right)^2}<\frac{8}{\pi}-2<\frac{8}{3}-2=\frac{2}{3}, \end{eqnarray*} which was to be proven. \kdokaz % Ttrikotniki \item Let $T$ be the centroid of the triangle $ABC$. Prove: $$p_{TBC}=p_{ATC}=p_{ABT}.$$ \item Let $c$ be the hypotenuse, $v_c$ be the corresponding altitude, and $a$ and $b$ be the catheti of a right triangle. Prove that $c+v_c>a+b$. % zvezek - dodatni MG \item Draw a square that has the same area as a rectangular triangle with catheti that are consistent with the given distances $a$ and $b$. % (Hipokratovi luni) \item Let $ABC$ be an isosceles right triangle with the length of the catheti $|CB|=|CA|=a$. We mark with $A_1$, $B_1$, $C_1$ and $C_2$ the points for which it holds: $\overrightarrow{CA_1}=\frac{n-1}{n}\cdot \overrightarrow{CA}$, $\overrightarrow{CB_1}=\frac{n-1}{n}\cdot \overrightarrow{CB}$, $\overrightarrow{AC_1}=\frac{n-1}{n}\cdot \overrightarrow{AB}$ and $\overrightarrow{AC_2}=\frac{n-2}{n}\cdot \overrightarrow{AB}$ for some $n\in \mathbb{N}$. Express the area of the quadrilateral determined by the lines $AB$, $A_1B_1$, $CC_1$ and $CC_2$, as a function of $a$ and $n$. \item Given is a right triangle $ABC$ with hypotenuse $AB$ and area $p$. Let $C'=\mathcal{S}_{AB}(C)$, $B'=\mathcal{S}_{AC}(B)$ and $A'=\mathcal{S}_{BC}(A)$. Express the area of the triangle $A'B'C'$ as a function of $p$. \item Let $R$ and $Q$ be points in which the inscribed circle of the triangle $ABC$ intersects its sides $AB$ and $AC$. Let the internal angle bisector of $ABC$ intersect the line $QR$ in the point $L$. Determine the ratio of the areas of the triangles $ABC$ and $ABL$. % pripremni zadaci - naloga 193 \item Determine a point inside the triangle $ABC$, for which the product of its distances from the sides of this triangle is maximal. % zvezek - dodatni MG \item The triangle $ABC$ with sides of lengths $a$, $b$ and $c$ has an inscribed circle. We draw the tangents of this circle that are parallel to the sides of the triangle. Each of the tangents inside the triangle determines the appropriate distances of lengths $a_1$, $b_1$ and $c_1$. Prove that: $$\frac{a_1}{a}+\frac{b_1}{b}+\frac{b_1}{b}=1.$$ % zvezek - dodatni MG \item Let $T$ and $S$ be the centroid and the center of the inscribed circle of the triangle $ABC$. Let also $|AB|+|AC|=2\cdot |BC|$. Prove that $ST\parallel BC$. % zvezek - dodatni MG \item Let $p$ be the area of the triangle $ABC$, $R$ the radius of the circumscribed circle and $s'$ the perimeter of the pedal triangle. Prove that $p=Rs'$. %Lopandic - nal 918 % Sstirikotniki \item Let $L$ be an arbitrary point inside the parallelogram $ABCD$. Prove that: $$p_{LAB}+p_{LCD}=p_{LBC}+p_{LAD}.$$ %Lopandic - nal 890 \item Let $P$ be the center of the leg $BC$ of the trapezoid $ABCD$. Prove: $$p_{APD}=\frac{1}{2}\cdot p_{ABCD}.$$ \item Let $o$ be the perimeter, $v$ the height and $p$ the area of the tangent trapezoid. Prove that: $p=\frac{o\cdot v}{4}$. \item Draw a line $p$, which goes through the vertex $D$ of the trapezoid $ABCD$ ($AB>CD$), so that it divides this trapezoid into two area-equal figures. \item Draw lines $p$ and $q$, which go through the vertex $D$ of the square $ABCD$ and divide it into area-equal figures. \item Let $ABCD$ be a square, $E$ the center of its side $BC$ and $F$ a point, for which $\overrightarrow{AF}=\frac{1}{3}\cdot \overrightarrow{AB}$. The point $G$ is the fourth vertex of the rectangle $FBEG$. What part of the area of the square $ABCD$ does the area of the triangle $BDG$ represent? \item Let $\mathcal{V}$ and $\mathcal{V}'$ be similar polygons with the similarity coefficient $k$. Then: $$p_{\mathcal{V}'}=k^2\cdot p_{\mathcal{V}}.$$ Prove. \item Let $a$, $b$, $c$ and $d$ be the lengths of the sides, $s$ the semiperimeter and $p$ the area of an arbitrary quadrilateral. Prove that: $$p=\sqrt{(s-a)(s-b)(s-c)(s-d)}.$$ %Lopandic - nal 924 \item Let $a$, $b$, $c$ and $d$ be the lengths of the sides and $p$ the area of the string-tangent quadrilateral. Prove that: $$p=\sqrt{abcd}.$$ %Lopandic - nal 925 % Veckotniki \item There is a rectangle $ABCD$ with sides of lengths $a=|AB|$ and $b=|BC|$. Calculate the area of the figure that represents the union of the rectangle $ABCD$ and its image under reflection across the line $AC$. \item Let $ABCDEF$ be a regular hexagon, and let $P$ and $Q$ be the centers of its sides $BC$ and $FA$. What part of the area of this hexagon is the area of the triangle $PQD$? % KKrog \item In a square, four congruent circles are drawn so that each circle touches two sides and two circles. Prove that the sum of the areas of these circles is equal to the area of the square inscribed circle. \item Calculate the area of the circle that is inscribed in the triangle with sides of lengths 9, 12 and 15. %resitev 54 \item Let $P$ be the center of the base $AB$ of the trapezoid $ABCD$, for which $|BC|=|CD|=|AD|=\frac{1}{2}\cdot |AB|=a$. Express the area of the figure determined by the base $CD$ and the shorter circular arcs $PD$ and $PC$ of the circles with centers $A$ and $B$, as a function of the base $a$. \item The cord $PQ$ ($|PQ|=d$) of the circle $k$ touches its conchoid $k'$. Express the area of the lune determined by the circles $k$ and $k'$, as a function of the cord $d$. \item Let $r$ be the radius of the inscribed circle of the polygon $\mathcal{V}$, which is divided into triangles $\triangle_1,\triangle_2,\ldots,\triangle_n$, so that no two triangles have common interior points. Let $r_1,r_2,\ldots , r_n$ be the radii of the inscribed circles of these triangles. Prove that: $$\sum_{i=1}^n r_i\geq r.$$ \end{enumerate} % DEL 3 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - %________________________________________________________________________________ % INVERZIJA %________________________________________________________________________________ \del{Inversion} \label{pogINV} \poglavje{Definition and Basic Properties of Inversion} \label{odd9DefInv} In this chapter we will define inversion - a new mapping, which in a certain way represents the "reflection" over a circle. We expect this new mapping to have similar properties to the reflection over a line. It is clear that inversion cannot be an isometry, but despite this, we will list the following desired common properties of the "two reflections" (Figure \ref{sl.inv.9.1.1.pic}): \begin{figure}[!htb] \centering \input{sl.inv.9.1.1.pic} \caption{} \label{sl.inv.9.1.1.pic} \end{figure} \begin{itemize} \item the reflection is a bijective mapping, \item the reflection is an involution, \item the reflection maps one of the areas of the plane, which the (circle) axis of the reflection determines, into the other area, \item all fixed points of the reflection lie on the (circle) axis of this reflection, \item if $X'$ is the image of the point $X$ under the reflection, then the line $XX'$ is perpendicular to the (circle) axis of this reflection. \end{itemize} The mapping, which we will now define, should therefore satisfy all these properties. Instead of the name "reflection over a circle" we will rather use the term "inversion"\index{inversion}\footnote{Inversion was first introduced by the German mathematician \index{Magnus, L. I.} \textit{L. I. Magnus} (1790--1861) in 1831. The first ideas of inversion appeared already in the ancient Greek mathematician \index{Apolonij} \textit{Apolonij iz Perge} (3.--2. st. pr. n. š.) and the Swiss mathematician \index{Steiner, J.} \textit{J. Steiner} (1796--1863)}. In the Euclidean plane we will define inversion in the following way. Let $k(O,r)$ be a circle in the Euclidean plane $\mathbb{E}^2$. The point $X'$ is the image of some point $X\in \mathbb{E}^2\setminus \{O\}$ in the \index{inversion} \pojem{inversion} $\psi_k$ with respect to the circle $k$ (its \index{inverse image} \pojem{inverse image}), if it belongs to the segment $OX$ and $|OX|\cdot |OX'|=r^2$, $k$ is the \index{circle!inversion}\pojem{circle of inversion}, $O$ is the \index{center!inversion} \pojem{center of inversion}. We will often denote the inversion $\psi_k$ with respect to the circle $k(O,r)$ by $\psi_{O,r}$. From the definition itself it immediately follows that the only fixed points of the inversion are precisely the points on the inversion circle. And all points of the inversion circle are fixed. Therefore, the equivalence holds: $$\psi_k(X)=X\hspace*{1mm}\Leftrightarrow\hspace*{1mm}X\in k.$$ The inversion $\psi_k$ is a bijective mapping of the set $\mathbb{E}^2\setminus \{O\}$. It also holds that $\psi_k^{-1}=\psi_k$, therefore $\psi_k^2$ is the identical mapping. Both properties follow directly from the definition. It also directly follows that the external points of the circle $k$ with the inversion $\psi_k$ are mapped to internal points and vice versa (without the point O). From the definition it follows that the line $XX'$ is perpendicular to the inversion circle, which means that the last desired property is also fulfilled. Now we will construct the image $X'=\psi_k(X)$ of an arbitrary point $X$ with the inversion. From everything said above (especially from the fact that $\psi_k^{-1}=\psi_k$) it follows that it is enough to describe the construction when $X$ is outside the circle $k$. Let in this case $OX$ be the tangent of the circle $k$ in its point $T$, then from $|OX|\cdot |OX'|=r^2=|OT|^2$ it follows that the triangles $OTX$ and $OX'T$ are similar and therefore $\angle TX'O=90^0$ (Figure \ref{sl.inv.9.1.2.pic}). Since the point $X'$ is also on the segment $OX$, we get this point as the perpendicular projection of the point $T$ on the segment $OX$. From the relation in the definition of inversion $|OX|\cdot |OX'|=r^2$ it follows that if the point $X$ approaches the center $O$, its image $X'$ "moves away to infinity". It also holds - if the point $X$ is close to the circle $k$, its image is also close to this circle, of course on the other side. We can write all these findings in a more formal way. \bizrek \label{invUrejenost} If $\psi_k:A,B \mapsto A',B'$ and $\mathcal{B}(O,A,B)$, then $\mathcal{B}(O,B',A')$. \eizrek The point $O$ has no image. Intuitively, its image is a point at infinity. We will say more about this in chapter \ref{odd9InvRavn}. We prove the following important property of inversion. \bizrek \label{invPodTrik} Let $O$, $A$ and $B$ be three non-collinear points and $\psi_k$ be the inversion with respect to the circle $k(O,r)$. If $A'$ and $B'$ are the images of points $A$ and $B$ in this inversion, then the triangles $OAB$ and $OB'A'$ are similar, that is: $$\psi_{O,r}:A,B\mapsto A',B' \Rightarrow \triangle OAB \sim \triangle OB'A'.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.inv.9.1.3.pic} \caption{} \label{sl.inv.9.1.3.pic} \end{figure} \textbf{\textit{Proof.}} Because $\psi_{O,r}:A,B\mapsto A',B'$, it is also $OA\cdot OA'=OB\cdot OB'=r^2$ (Figure \ref{sl.inv.9.1.3.pic}). The similarity of the triangles now follows from $OA:OB'=OB:OA'$ and $\angle AOB \cong \angle B'OA'$. \kdokaz The next statement is a direct consequence of the previous proposition. \bizrek Let $O$, $A$ and $B$ be three non-collinear points and $\psi_k$ be the inversion with respect to the circle $k(O,r)$. If $A'$ and $B'$ are the images of points $A$ and $B$ in this inversion, then: (i) $\angle OAB \cong \angle OB'A'$ and $\angle OBA \cong \angle OA'B'$, (ii) points $A$, $B$, $B'$ and $A'$ are concircular, (iii) the circle passing through points $A$, $B$, $B'$ and $A'$ is orthogonal to the inversion circle. \eizrek \begin{figure}[!htb] \centering \input{sl.inv.9.1.4.pic} \caption{} \label{sl.inv.9.1.4.pic} \end{figure} \textbf{\textit{Proof.}} \textit{(i)} The angle congruence follows from the similarity of triangles $OAB$ and $OB'A'$(the previous proposition \ref{invPodTrik}). \textit{(ii)} Let $l$ be a circle drawn through the triangle $ABA'$ (Figure \ref{sl.inv.9.1.4.pic}). From the definition of inversion it follows that $OA\cdot OA'=OB\cdot OB'=r^2$. This means that the power of the point $O$ on the circle $l$ is equal to $OA\cdot OA'=OB\cdot OB'$, thus the point $B'$ lies on this circle. We mention that the tautness of the quadrilateral $AA'B'B$ also follows directly from $\angle OAB \cong \angle OB'A'$, because then $\angle A'AB + \angle BB'A' = 180^0 $. \textit{(iii)} From $\psi_{O,r}(A)=A'$ it follows that one of the points $A$ and $A'$ is in the interior, the other in the exterior of the inversion circle $k$. By the consequence of Dedekind's axiom (\ref{DedPoslKrozKroz}) the circles $k$ and $l$ have two common points; one of them we denote by $T$. Because $OA\cdot OA'=OB\cdot OB'=r^2=OT^2$, it follows that $OT$ is a tangent to the circle $k$, which means that the circles $k$ and $l$ are perpendicular (by the statement \ref{pravokotniKroznici}). \kdokaz In the same way as in part (\textit{iii}) of the previous statement, we also prove the following claim. \bizrek \label{invPravKrozn} Let $X'$ be the image of the point $X$ ($X\neq X'$) under the inversion $\psi_k$ with respect to the circle $k(O,r)$. Then every circle that goes through the points $X$ and $X'$ (also the line $XX'$) is perpendicular to the inversion circle $k$. \eizrek In the next example we will see how we can translate an inversion with the help of certain stretches into an inversion with a concentric inversion circle. \bzgled \label{invRazteg} The composition of the inversion $\psi_{S,r}$ and the dilation $h_{S,k}$ with the same center $S$ and positive coefficient ($k>0$) is an inversion with center $S$ and coefficient $r\sqrt{k}$. \ezgled \textbf{\textit{Proof.}} Let $f=h_{S,k}\circ \psi_{S,r}$. For any point $X$, denote $X'=f(X)$. First, it is clear that the point $X'$ lies on the line segment $SX$. Also, denote $X_1=\psi_{S,k}(X)$. From the definition of inversion and dilation, it follows that $|SX_1| \cdot |SX| =r^2$ and $|SX'| = k\cdot |SX_1| $. Therefore, $|SX'| \cdot |SX| =k\cdot r^2=\left(r\sqrt{k}\right)^2$ or $\psi_{S,r\sqrt{k}}(X)=X' =f(X)$. Since this is true for any point $X$, we have $f=h_{S,k}\circ \psi_{S,r}=\psi_{S,r\sqrt{k}}$. \kdokaz %________________________________________________________________________________ \poglavje{Image of a Circle or Line Under an Inversion} \label{odd9SlokaKrozPrem} In this section, we will determine that an inversion is not a collinearity, which means that it does not preserve the relation of collinearity of points. We will consider the images of lines and circles under inversion. Since the domain of an inversion is an Euclidean plane without the center of inversion, it makes sense to introduce the following notation: if $\Phi$ is an arbitrary figure of the Euclidean plane and $S$ is a point of this plane, then $$\Phi^S= \Phi \setminus \{S\}.$$ \bizrek \label{InverzKroznVkrozn} Let $\psi_i$ be an inversion with respect to the circle $i(S,r)$ of the Euclidean plane. If $p$ is a line and $k$ is a circle of this plane, then (Figure \ref{sl.inv.9.2.1.pic}): (i) if $S\in p$, then $\psi_i(p^S)=p^S$, (ii) if $S\notin p$, then $\psi_i(p)=j^S$, where $j$ is a circle that passes through the point $S$, (iii) if $S\in k$, then $\psi_i(k^S)=q$, where $q$ is a line that does not pass through the point $S$; (iv) if $S\notin k$, then $\psi_i(k)=k'$, where $k'$ is a circle that does not pass through the point $S$. \eizrek \begin{figure}[!htb] \centering \input{sl.inv.9.2.1.pic} \caption{} \label{sl.inv.9.2.1.pic} \end{figure} \textbf{\textit{Proof.}} (\textit{i}) From the definition of inversion it follows that the image of any point $X \in p^S$ lies on the open ray $SX$, therefore $\psi_i(X)\in p^S$. Because $\psi_i^{-1}=\psi_i$, any point $Y\in p^S$ is the image of the point $\psi_i^{-1}(Y)=\psi_i(Y)\in p^S$. Therefore $\psi_i(p^S)=p^S$. (\textit{ii}) Let $P$ be the orthogonal projection of the point $S$ onto the line $p$ (Figure \ref{sl.inv.9.2.2.pic}). Because $S\notin p$, then $P\neq S$, therefore there exists an image of the point $P$ under the inversion $\psi_i$ -- we denote it with $P'$. Let $X$ be any point on the line $p$ ($X\neq P$) and $X'=\psi_i(X)$. By \ref{invPodTrik} the triangles $SPX$ and $SX'P'$ are similar, therefore $\angle SX'P' \cong \angle SPX=90^0$. Therefore the point $X'$ lies on the circle above the diameter $SP'$ - we denote it with $j$. Because $X'\neq S$, it follows that $X'\in j^S$ or $\psi_i(p)\subseteq j^S$. With the reverse procedure we prove that each point of the set $j^S$ is the image of some point on the line $p$, therefore $\psi_i(p)= j^S$. (\textit{iii}) A direct consequence of (\textit{ii}), because $\psi_i$ is an involution, or $\psi_i^{-1}=\psi_i$. (\textit{iv}) (Figure \ref{sl.inv.9.2.2.pic}) Let $P$ and $Q$ be the intersection points of the circle $k$ with the line that goes through the point $S$, and the center of the circle $k$ (for example, when the circles $i$ and $k$ are concentric, it is easy). Without loss of generality, assume that $\mathcal{B}(S,P,Q)$ is true. Let $P'=\psi_i(P)$, $Q'=\psi_i(Q)$ and $X$ be any point on the circle $k$, which is different from the points $P$ and $Q$ and $X'=\psi_i(X)$. By the statement \ref{invPodTrik} we have $\triangle SPX \sim \triangle SX'P'$ and $\triangle SQX \sim \triangle SX'Q'$. Therefore, $\angle SX'P' \cong \angle SPX$ and $\angle SX'Q'\cong \angle SQX$, which means: $$\angle P'X'Q'=\angle SX'P'-\angle SX'Q' =\angle SPX-\angle SQX=\angle PXQ=90^0.$$ Then the point $X'$ lies on the circle above the diameter $P'Q'$ - we denote it with $k'$. By reversing the process, we can also prove that every point on the circle $k'$ is the image of a point on the circle $k$, so $\psi_i(k)=k'$ We prove the statement similarly in the case when $\mathcal{B}(P,S,Q)$ is true. \kdokaz The previous statement also provides an effective way of constructing the image of a line or a circle in different cases. We will use the notation from this statement (\ref{InverzKroznVkrozn}). In case (\textit{ii}) - figure \ref{sl.inv.9.2.2a.pic} - it is enough to determine the image of the perpendicular projection of $P$ onto the center of inversion $S$ on the line $p$, which we map. The point $P'=\psi_i(P)$ with the center of inversion determines the diameter of the circle $j$. But if the line $p$ intersects the circle of inversion $i(S,r)$ for example in the points $M$ and $N$, both points are fixed and the image of the line $p$ is the circumscribed circle of the triangle $SMN$ (without the point $S$). The tangent of the circle of inversion $i$ is mapped into a circle (without the point $S$), which from the inside touches the circle $i$. \begin{figure}[!htb] \centering \input{sl.inv.9.2.2a.pic} \caption{} \label{sl.inv.9.2.2a.pic} \end{figure} In the case of (\textit{iii}) we do the inverse of the construction from (\textit{ii}) and we get the orthogonal projection on the line that is the image of the given circle. The other special cases (when the circle intersects or touches the circle of inversion) are the inverses of (\textit{ii}). Also in the case of (\textit{iv}), when the circle is mapped into a circle, the construction process is clear from the statement itself - the image $k'$ is determined by the points $P'$ and $Q'$. We just have to be careful that in this case the center of the circle $k$ is not mapped into the center of the new circle $k'$! Where is the image of the center of this circle located in this case? We denote with $O'$ the image of the center $O$ of the circle $k$ under the inversion $\psi_i$. Let $t_1=ST_1$ and $t_2=ST_2$ be the tangents of the circle $k$ at the points $T_1$ and $T_2$, and $T'_1=\psi_i(T_1)$ and $T'_2=\psi_i(T_2)$ (Figure \ref{sl.inv.9.2.3.pic}). It is clear that $T'_1, T'_2 \in k'$, but $t_1$ and $t_2$ are also tangents of the circle $k'$ (if, for example, $Y\in k'\cap t_1$, then also $X=\psi_i(Y)\in k\cap t_1=\{T_1\}$, i.e. $X=T'_1$). By \ref{invPodTrik} we have $\triangle ST_1O \sim \triangle SO'T'_1$ or $\angle SO'T'_1 \cong \angle ST_1O=90^0$. For the same reasons $\angle SO'T'_2 \cong \angle ST_2O=90^0$. This means that the points $T'_1$, $O'$ and $T'_2$ are collinear. So we can find the point $O'$ as the intersection of the line $T'_1T'_2$ and the segment $SO$. \begin{figure}[!htb] \centering \input{sl.inv.9.2.3.pic} \caption{} \label{sl.inv.9.2.3.pic} \end{figure} Another question arises in case of (\textit{iv}). Which circles are fixed under inversion? One such is of course the circle of inversion $i$ itself, since all of its points are fixed. The next theorem will reveal all other possibilities. \bizrek \label{InverzKroznFiks} Let $\psi_i$ be the inversion with respect to the circle $i(S,r)$. The only fixed circles of this inversion are the circle $i$ and the circles that are perpendicular to this circle, that is: $$\psi_i(k)=k \Leftrightarrow k=i \vee k\perp i.$$ \eizrek \begin{figure}[!htb] \centering \input{sl.inv.9.2.4.pic} \caption{} \label{sl.inv.9.2.4.pic} \end{figure} \textbf{\textit{Proof.}} ($\Leftarrow$) If $k=i$, the statement is trivial. Let $k$ and $i$ be perpendicular circles (Figure \ref{sl.inv.9.2.4.pic}). With $T$ we mark one of their intersections. The line $ST$ is tangent to the circle $k$ (statement \ref{TangPogoj}). Let $X$ be an arbitrary point on the circle $k$ ($X\notin i$) and $X'$ the other intersection of this circle with the segment $SX$. Then we have: $$p(S,k)= |SX|\cdot |SX'| = |OT|^2 = r^2,$$ so $\psi_i(X)=X'\in k$. Similarly, every point $Y$ on the circle $k$ is the image of some point on this circle, so $\psi_i(k)=k$ holds. ($\Rightarrow$) Let $\psi_i(k)=k$. If $k\neq i$, then there exists a point $X$ on the circle $k$, which does not lie on the circle $i$. Let $X'=\psi_i(X)$. It is clear that $X'\notin i$ also holds. From the definition of inversion it follows that the points $X$ and $X'$ are on the segment with the initial point $S$, which means that $S$ is an external point of the circle $k$ (because $\mathcal{B}(X,S,X')$ is not true). So there exists a tangent from the point $S$ to the circle $k$ -- with $T$ we mark the point of tangency. Then we have: $$p(S,k)=|ST|^2 = |SX| \cdot |SX'| = r^2,$$ so the point $T$ lies on the circle $i$. This means that the circles $k$ and $i$ are perpendicular (statement \ref{TangPogoj}). \kdokaz We mention that a similar statement also holds for fixed lines, only that the perpendicularity of a line with the inversion circle means that the line goes through the center of inversion. So the only fixed lines of inversion are those that go through the center of inversion (part (\textit{i}) of statement \ref{InverzKroznVkrozn}). The previous statement also holds if instead of inversion we talk about reflection over a line. The only fixed lines (circles) of this reflection are the reflection axis and those lines (circles) that are perpendicular to that axis. This will be the motivation to equalize lines and circles in a certain way, thus also the reflection axis and inversion. We will learn more about this in the next section. \bzgled Let $\gamma$ be an arc and $c$ and $p$ be lines. Draw a triangle $ABC$, if $AB\cong c$, $\angle BCA\cong \gamma$ and $|BA'|\cdot|BC|=p^2$ (the line $AA'$ is the altitude of the triangle $ABC$). \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.2.5.pic} \caption{} \label{sl.inv.9.2.5.pic} \end{figure} \textbf{\textit{Solution.}} Let $\triangle ABC$ be the desired triangle, which satisfies the conditions of the task (Figure \ref{sl.inv.9.2.5.pic}). First, from the condition $\angle BCA\cong \gamma$ it follows that the point $C$ lies on the locus $l$, which is determined by the string $AB$ and the angular dimension $\gamma$. Since $A'$ is the vertex of the altitude, the point $A'$ lies on the circle $k$ above the diameter $AB$. Let $\psi_i$ be the inversion with respect to the circle $i(B,p)$. From the condition $|BA'|\cdot|BC|=p^2$ it follows that $\psi_i(A')=C$. Since $A'$ is also in $k$, it follows that $C$ is in $k'$, where $k'=\psi_i(k)$ is a line (statement \ref{InverzKroznVkrozn}). So if we first draw the line $AB\cong c$, we get the point $C$ as the intersection of the line $k'=\psi_i(k)$ with the locus $l$. \kdokaz \bzgled \label{MiquelKroznice}(Miquel's\footnote{\index{Miquel, A.}\textit{A. Miquel} (1816–-1851), French mathematician, who published this statement in 1840.} statement about six circles) Let $k_1$, $k_2$, $k_3$ and $k_4$ be such circles that the circles $k_1$ and $k_2$ intersect in points $A$ and $A_0$, the circles $k_2$ and $k_3$ in points $B$ and $B_0$, the circles $k_3$ and $k_4$ in points $C$ and $C_0$, the circles $k_4$ and $k_1$ in points $D$ and $D_0$. If the points $A$, $B$, $C$ and $D$ are concircular, the points $A_0$, $B_0$, $C_0$ and $D_0$ are concircular or collinear. \index{statement!Miquel's.} \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.3.7a.pic} \caption{} \label{sl.inv.9.3.7a.pic} \end{figure} \textbf{\textit{Proof.}} Let $k$ be the circle determined by the points $A$, $B$, $C$ and $D$. Let $\psi_i$ be the inversion with respect to any circle $i$ with center at point $A$ (Figure \ref{sl.inv.9.3.7a.pic}) and $$\psi_i:\hspace*{1mm}B,C,D,A_0,B_0,C_0,D_0 \mapsto B',C',D',A'_0,B'_0,C'_0,D'_0.$$ Because the circles $k$, $k_1$ and $k_2$ go through point $A$, their images $k'$, $k'_1$ and $k'_2$ are lines under inversion $\psi_i$ that determine the vertices of the triangle $A_0'B'D'$. Circles $k_3$ and $k_4$, however, do not go through point $A$, so their images $k'_3$ and $k'_4$ are circles that intersect at points $C'\in B'D'$ and $C'_0$. Let $k'_0$ be the circle drawn through the triangle $A'_0D'_0B'_0$. From Example \ref{Miquelova točka} it follows that the circles $k'_3$, $k'_4$ and $k'_0$ intersect at one point - $C'_0$, i.e. $A'_0,B'_0,C'_0,D'_0\in k'_0$. Therefore, $A_0,B_0,C_0,D_0\in k_0=\psi_i(k'_0)$. By Theorem \ref{InverzKroznVkrozn} $k_0$ is a circle or a line. \kdokaz %________________________________________________________________________________ \poglavje{The Inversive Plane} \label{odd9InvRavn} The motivation for the following discussion is the fact that the domain of definition is not the whole Euclidean plane under inversion - the center of inversion does not have its image. This often makes our work more difficult, as, for example, in the formulation of Theorem \ref{InverzKroznVkrozn}, where we always have to be careful when a line or a circle goes through the center of inversion. Also, the images of a line or a circle are not always a line or a circle, because we have to exclude the center of inversion in certain cases. Because of everything mentioned above, another solution is offered; instead of excluding the center of inversion from the definition domain, we can formally add one point to the Euclidean plane that will be the image of the center of inversion at this inversion. From the definition of inversion for the image $X'$ of the point $X$ at the inversion $\psi_{S,k}$ it holds: $|SX'|\cdot |SX|=r^2$. If $S=X$, then we formally get $|SX'|\cdot 0=r^2$, which means that $|SX'|=\infty$. So, the new point is intuitively seen as a "point at infinity", so we will formally mark it with the symbol $\infty$ and call it the \index{point!at infinity}\pojem{point at infinity}. The set that arises from adding this new point to the Euclidean plane is called the \index{inversive plane} \pojem{inversive plane}, which we will mark with $\widehat{E}^2$. So: $$\widehat{E}^2=E^2 \cup \{\infty\}.$$ But we must be careful - we must not mix the inversive plane with the \index{extended Euclidean plane} \pojem{extended Euclidean plane}, which we get if we add one (infinitely distant) line to the Euclidean plane and define that the parallel lines intersect on it. The extended Euclidean plane is a model of so-called projective geometry, which we mentioned in the introduction. In the case of the inversive plane, we have added only one point $\infty$ and required that for every inversion $\psi_{S,k}$ it holds that $\psi_{S,k}(S)=\infty$. Since $\psi_{S,k}$ is an involution (it holds that $\psi_{S,k}^{-1}=\psi_{S,k}$), by agreement it also holds that $\psi_{S,k}(\infty)=S$. In section (\textit{iii}) we saw the statement \ref{InverzKroznVkrozn} that the image of a circle $k^S$ (a circle without point $S$) under an inversion $\psi_{S,k}$ is a line $q$, which does not go through point $S$. But if we add point $S$ to the circle $k^S$, we will get the set $q \cup \{\infty\}$ as the image. Therefore, we will naturally assign point $\infty$ to lines in the inversive plane. If we call such lines with added point $\infty$ $i$-circles, we can translate the statement \ref{InverzKroznVkrozn} and say that an $i$-circle is mapped to an $i$-circle under an inversion. But we also saw certain analogies with reflection over a line and inversion. Therefore, we will call both mappings in the inversive plane $i$-inversions. It is not hard to see that we can generalize the statement \ref{InverzKroznVkrozn} and write it in a simpler form. \bizrek \label{InverzInvRavKvK} An $i$-circle is mapped to an $i$-circle under an $i$-inversion. \eizrek The fact that every line in the inversive plane contains point $\infty$ means that two parallels intersect in this point. Lines, which intersect (in a regular point of the Euclidean plane) have two common points in the inversive plane. Since parallels have only one common point, we also say that they are tangent in point $\infty$. The image of two parallels under an $i$-inversion will either be parallels (if it is the axis of reflection) or two circles, which are tangent in the center of inversion (if it is an inversion). Two circles, which are tangent (because they have only one common point), are mapped either to parallels or to circles, which are tangent. The same goes for the case of a line and a circle being tangent. In all these cases in the inversive plane we can formulate them much shorter in the following way. \bizrek \label{InverzDotik} Let $\psi_i$ be an arbitrary inversion. If $i$-circles $k$ and $l$ are tangent, then $i$-circles $\psi_i(k)$ and $\psi_i(l)$ are tangent. \eizrek With the help of this expression, we will first prove the following important expression. \bizrek \label{InverzKonf} Inversion is a \index{conformal mapping}conformal mapping, which means: The angle under which the lines $p$ and $q$ intersect in the point $A$, is equal to the angle under which their inverse images $p'$ and $q'$ intersect in the corresponding point $A'$. \eizrek \begin{figure}[!htb] \centering \input{sl.inv.9.3.1.pic} \caption{} \label{sl.inv.9.3.1.pic} \end{figure} \textbf{\textit{Proof.}} Let $\psi_i$ be the inversion with respect to the circle $i(S,r)$, in which the lines $p$ and $q$ and their intersection $A$ are mapped to $p'$, $q'$ and $A'$ (Figure \ref{sl.inv.9.3.1.pic}). The case when $A=S$, is trivial, because then the lines $p$ and $q$ are stationary. We first assume that $A\notin i$. In this case it is clear that $A\neq A'$. Let $P$ be the intersection of the rectangle of the line $p$ in the point $A$ and the simetral of the line segment $AA'$ and $k_p$ the circle with the center $P$, which goes through the point $A$. From the construction of the circle $k_p$ it is clear that also the point $A'$ lies on this circle and that the line $p$ is its tangent in the point $A$. If in a similar way we mark with $Q$ the intersection of the rectangle of the line $q$ in the point $A$ and the simetral of the line segment $AA'$ and $k_q$ the circle with the center $Q$, which goes through the point $A$, it follows that the point $A'$ lies on the circle $k_q$ and that the line $q$ is its tangent in the point $A$. Because both circles $k_p$ and $k_q$ go through the pair of points $A$, $A'$ of the inversion $\psi_i$, $k_p$ and $k_q$ are perpendicular to the circle of inversion $i$ (expression \ref{invPravKrozn}). According to expression \ref{InverzKroznFiks}, then $\psi_i(k_p)=k_p$ and $\psi_i(k_q)=k_q$. The lines $p$ and $q$ intersect the circles $k_p$ and $k_q$ in that order. By \ref{InverzDotik} the intersections $i$ of the circles $p'$ and $\psi_i(k_p)=k_p$, or $q'$ and $\psi_i(k_q)=k_q$, are in correspondence. Therefore we have: $$\angle p,q \cong \angle k_p,k_q \cong \angle p',q'.$$ If $A\in i$, we apply the previous proof (with an inversion with respect to the concentric circle) using \ref{invRazteg}, because a dilation preserves the size of angles. \kdokaz It is clear that the proof would be practically the same if $p$ and $q$ were circles or a circle and a line. Therefore we can write the previous statement in a more general form. \bizrek Every $i$-inversion preserves the angle between two $i$-circles. \eizrek A special case of the previous statement concerns right angles. From this it follows that perpendicularity is an invariant of inversion and the following statement holds. \bizrek \label{InverzKonfPrav} Every $i$-inversion maps two perpendicular $i$-circles to a perpendicular $i$-circle. \eizrek Now we will use the proven statements that hold in the Euclidean plane. The fact that the circle passing through the center of inversion is mapped to a line by this inversion allows us to solve various design tasks where instead of the sought circle we will first construct its image - a line. This also applies to other tasks where we will use inversion to translate a statement that concerns a circle into an equivalent statement that concerns a line. In both cases, it is desirable to have at least one point of the circle, which we will then choose as the center of inversion. We should also mention that because of all this, we will often say that the image of a line $p$ ($S\notin p$) under inversion $\psi_{S,r}$ is just the circle $j$ ($j\ni S$) instead of $j^S$. And vice versa - in the same case we will write $\psi_{S,r}(j)=p$. \bzgled Given are the circle $k$ and the points $A$ and $B$. Draw the circle (line) $x$, which is perpendicular to the circle $k$ and passes through the points $A$ and $B$. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.3.2.pic} \caption{} \label{sl.inv.9.3.2.pic} \end{figure} \textbf{\textit{Solution.}} Let $x$ be the circle, which is perpendicular to the circle $k$ and passes through the points $A$ and $B$, and $\psi_i$ inversion with respect to any circle $i$ with center $A$ (Figure \ref{sl.inv.9.3.2.pic}). First, we assume that $A\notin k$. In this case, $k'=\psi_i(k)$ is the circle and $x'=\psi_i(x)$ is the line that goes through the point $B'=\psi_i(B)$ (statement \ref{InverzKroznVkrozn}). By statement \ref{InverzKonfPrav} from $x \perp k$ it follows that $x' \perp k'$. Therefore, $x'$ is the line that goes through the point $B'$ and is perpendicular to the circle $k'$ or goes through its center. The line $x'$ can therefore be drawn as the line that goes through the points $B'=\psi_i(B)$ and the center of the circle $k'=\psi_i(k)$. In the end, $x=\psi_i(x')$. But if $A\in k$, then $k'$ is the line (statement \ref{InverzKroznVkrozn}) and the line $x'$ is drawn as its perpendicular through the point $B'$. Since we can always draw one single perpendicular $x'$, there is exactly one solution also for the $i$-circle $x$. But $x$ is the circle exactly when $S\notin x'$ or when the points $A$, $B$ and the center of the circle $k$ are not collinear. In the case of collinearity of these points, the solution $x$ is the line. \kdokaz \bzgled \label{TriKroznInv} If among three circles $k_1$, $k_2$ and $k_3$ of a plane two touch each other from the outside, the circle $k$, which is determined by the points of contact, is perpendicular to each of the circles $k_1$, $k_2$ and $k_3$. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.3.3.pic} \caption{} \label{sl.inv.9.3.3.pic} \end{figure} \textbf{\textit{Proof.}} Let $A$ be the point of intersection of the circles $k_2$ and $k_3$, $B$ the point of intersection of the circles $k_1$ and $k_3$, and $C$ the point of intersection of the circles $k_2$ and $k_1$ (Figure \ref{sl.inv.9.3.3.pic}). Then the circle $k$ is the circumscribed circle of the triangle $ABC$. Let $\psi_{A,r}$ be the inversion with center $A$ and arbitrary radius $r$. The circles $k_2$ and $k_3$ touch at the center of the inversion $A$, so they are mapped by this inversion into the lines $k'_2$ and $k'_3$ (statement \ref{InverzKroznVkrozn}), which have no common points, i.e. $k'_2\parallel k'_3$. The circle $k_1$, which does not pass through the point $A$ and touches the circles $k_2$ and $k_3$ at the points $C$ and $B$, is mapped by the inversion $\psi_{A,r}$ into the circle $k'_1$ (statement \ref{InverzKroznVkrozn}), which touches the lines $k'_2$ and $k'_3$ (statement \ref{InverzDotik}) at the points $C'=\psi_{A,r}(C)$ and $B'=\psi_{A,r}(B)$. Therefore, the lines $k'_2$ and $k'_3$ are parallel tangents of the circle $k'_1$ at the points $C'$ and $B'$, and the line segment $B'C'$ is their common perpendicular, the distance $B'C'$ being the diameter of the circle $k'_1$. The line $B'C'$ is the image of the circle $k$, which passes through the center of the inversion $A$ and through the points $B$ and $C$ (statement \ref{InverzKroznVkrozn}), i.e. $\psi_{A,r}(k)=k'=B'C'$. Since the line $k'=B'C'$ is perpendicular to the line $k'_2$ and $k'_3$ and to the circle $k'_1$, by statement \ref{InverzKonfPrav} the circle $k$ is perpendicular to each of the circles $k_1$, $k_2$ and $k_3$. \kdokaz \bzgled Two of the four circles touch each other from the outside at the points $A$, $B$, $C$ and $D$. Prove that the points $A$, $B$, $C$ and $D$ are either concilic or collinear. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.3.3c.pic} \caption{} \label{sl.inv.9.3.3c.pic} \end{figure} \textbf{\textit{Proof.}} We mark with $k$ the circle (or line), which goes through the points $B$, $C$ and $D$. Let $k_1$, $k_2$, $k_3$ and $k_4$ be such circles, that $k_1$ and $k_2$ touch in the point $A$, $k_1$ and $k_4$ in the point $B$, $k_3$ and $k_4$ in the point $C$ and $k_3$ and $k_2$ in the point $D$. We mark with $\psi_i$ the inversion with respect to any circle $i$ with the center in $A$ (Figure \ref{sl.inv.9.3.3c.pic}). Let $B'$, $C'$ and $D'$ be the images of the points $B$, $C$ and $D$ and $k'_1$, $k'_2$, $k'_3$ and $k'_4$ be the images of the circles $k_1$, $k_2$, $k_3$ and $k_4$ in this inversion. From \ref{InverzKroznVkrozn} and \ref{InverzDotik} it follows, that $k'_3$ and $k'_4$ are the circles, which touch in the point $C'$, $k'_1$ and $k'_2$ are the parallel lines, which are tangent to the circles $k'_4$ and $k'_3$ in the points $B'$ and $D'$. From the parallelity of the lines $k'_1$ and $k'_2$ it follows $\angle k'_2,D'C' \cong \angle k'_1,B'C'$, so according to \ref{ObodKotTang} they are also consistent with the central angle $D'S_2C'$ and $B'S_1C'$. Because of this is $\angle D'C'S_2 \cong \angle B'C'S_1$, which means, that the points $B'$, $C'$ and $D'$ are collinear or $k'$ is a line. According to \ref{InverzKroznVkrozn} its image with respect to the inversion $k=\psi_i(k')$ goes through the center of the inversion, so $A,B,C,D\in k$. \kdokaz \bzgled Let $k$, $l$ and $j$ be three mutually perpendicular circles with the common tangents $AB$, $CD$ and $EF$. Prove that the circumscribed circles of the triangles $ACE$ and $ADF$ touch in the point $A$. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.3.3a.pic} \caption{} \label{sl.inv.9.3.3a.pic} \end{figure} \textbf{\textit{Proof.}} Let $AB$ be the common chord of the circles $k$ and $l$, $CD$ be the common chord of the circles $l$ and $j$ and $EF$ be the common chord of the circles $j$ and $k$ (Figure \ref{sl.inv.9.3.3a.pic}). We denote by $x$ and $y$ the ocrtani circles of the triangles $ACE$ and $ADF$ and by $\psi_i$ the inversion with respect to any circle $i$ with center $A$. Let: \begin{eqnarray*} && \psi_i:\hspace*{1mm}B,C,D,E,F \mapsto B',C',D',E',F' \hspace*{2mm} \textrm{ in }\\ && \psi_i:\hspace*{1mm}k,l,j,x,y \rightarrow k',l',j',x',y'. \end{eqnarray*} Because $A\in k, l, x, y$ and $A\notin j$, by izreku \ref{InverzKroznVkrozn} $k'=E'F'$, $l'=C'D'$, $x'=E'C'$ and $y'=D'F'$ are lines, $j'\ni C',D',E',F'$ is a circle. From the mutual perpendicularity of the circles $k$, $l$ and $j$ and the fact that $B\in k\cap l$ it follows that the lines $k'$ and $l'$ are perpendicular in the point $B'$, and the circle $j'$ is perpendicular to both lines $k'$ and $l'$ (izrek \ref{InverzKonfPrav}). This means that $C'D'$ and $E'F'$ are perpendicular diameters of the circle $j'$, so the quadrilateral $E'C'F'D'$ is a square and $E'C' \parallel D'F'$. From the parallelism of the lines $x'$ and $y'$ it follows that their inverse images (circles $x$ and $y$) touch at the center of inversion $A$. \kdokaz \bzgled Let $k$ and $l$ be the circles of some plane with centers $O$ and $S$. Let $t_i$ be the tangents of the circle $k$, which intersect the circle $l$ in the points $A_i$ and $B_i$. Prove that there exists a circle that touches all the ocrtani circles of the triangles $SA_iB_i$. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.3.4.pic} \caption{} \label{sl.inv.9.3.4.pic} \end{figure} \textbf{\textit{Proof.}} Let $\psi_l$ be the inversion with respect to the circle $l$ (Figure \ref{sl.inv.9.3.4.pic}). The lines $t_i$ are mapped by this inversion to the circles $t'_i$, which are the inscribed circles of the triangles $SA_iB_i$ (statement \ref{InverzKroznVkrozn}). Because the lines $t_i$ touch the circle $k$, all the circles $t'_i$ touch the circle $k'=\psi_l(k)$ (statement \ref{InverzDotik}). \kdokaz \bzgled Let $ST$ be the diameter of the circle $k$, $t$ the tangent of this circle at the point $T$, and $PQ$ and $PR$ its tangents at the points $Q$ and $R$. Prove that if $L$, $Q'$ and $R'$ are the intersections of the altitudes $SP$, $SQ$ and $SR$ with the line $t$, then the point $L$ is the midpoint of the segment $Q'R'$. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.3.5.pic} \caption{} \label{sl.inv.9.3.5.pic} \end{figure} \textbf{\textit{Proof.}} Let $\psi_i$ be the inversion with respect to the circle $i(S,ST)$ (Figure \ref{sl.inv.9.3.5.pic}). The point $T$ is the orthogonal projection of the center of inversion $S$ onto the line $t$. Because $\psi_i(T)=T$, it follows from statement \ref{InverzKroznVkrozn} that $\psi_i(t)=k$ or $k'=\psi_i(k)=t$. Therefore, $\psi_i(Q)=Q'$ and $\psi_i(R)=R'$. We denote $P'=\psi_i(P)$. From $L\in SP$ it follows that the points $S$, $P$, $L$ and $P'$ are collinear. The tangents $q=PQ$ and $r=PR$ of the circle $k$ are mapped to the circles $q'$ and $r'$, which pass through the point $S$ and touch the line $k'$ at the points $Q'$ and $R'$ (statements \ref{InverzKroznVkrozn} and \ref{InverzDotik}). Therefore, the line $t=k'$ is the common tangent of the circles $q'$ and $r'$, which intersect at the points $S$ and $P'$. The point $L$ lies on their power axis $SP'$, so $|LQ'|^2=|LR'|^2$, that is, the point $L$ is the midpoint of the segment $Q'R'$. \kdokaz Let $a$, $b$ and $c_0$ be circles with diameters $PQ$, $PR$ and $RQ$, where $\mathcal{B}(P,R,Q)$. Let $c_0$, $c_1$, $c_2$, ... $c_n$, ... be a sequence of circles on the same side of the line $PQ$, which each circle in the sequence touches the previous one. Prove that the distance from the center of the circle $c_n$ to the line $PQ$ is $n$ times greater than the diameter of this circle\footnote{This problem was considered by \index{Pappus} \textit{Pappus of Alexandria} (4th century). The pattern that is formed by the semicircles $a$, $b$, $c_0$, was investigated by \index{Archimedes} \textit{Archimedes of Syracuse} (3rd century BC) - see Example \ref{arbelos}.}\\ (an example from the book \cite{Cofman}). \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.3.6.pic} \caption{} \label{sl.inv.9.3.6.pic} \end{figure} \textbf{\textit{Proof.}} (Figure \ref{sl.inv.9.3.6.pic}). Let $i$ be a circle with its center in point $P$, which is perpendicular to the circle $c_n$ (circle $i$ goes through the points of tangency from point $P$ to circle $c_n$). With the inversion $\psi_i$ with respect to this circle, the line $l=PQ$ and the circles $a$ and $b$ are mapped into lines $l$, $a'$ and $b'$ (statement \ref{InverzKroznVkrozn}); in this case, the lines $a'$ and $b'$ are perpendicular to the line $l$ (statement \ref{InverzKonfPrav}). The circles $c_0$, $c_1$, $c_2$, ..., $c_n$, ... are mapped into congruent circles $c'_0$, $c'_1$, $c'_2$, ..., $c'_n$, ... with the same inversion, which are all tangent to the parallels $a'$, $b'$ (statement \ref{InverzDotik}) and are all congruent circles $c_n$, because $c'_n=\psi_i(c_n)=c_n$ (statement \ref{InverzKroznFiks}). Since $c_0'\perp l$, the center of the circle $c'_0$ lies on the line $l$, therefore the distance from the center of the circle $c_n=c'_n$ to this line is $n$ times larger than the diameter of this circle. \kdokaz \bzgled Let $t$ be a common external tangent circle of the circles $k$ and $l$, which are externally tangent in point $A$, and $c_0$, $c_1$, $c_2$, ... $c_n$, ... a sequence of circles, each circle in the sequence is tangent to the previous one, circle $c_0$ is also tangent to the circle $t$. Prove that there exists a circle (or a line), which is perpendicular to each circle from the given sequence. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.3.8.pic} \caption{} \label{sl.inv.9.3.8.pic} \end{figure} \textbf{\textit{Proof.}} Let $i$ be an arbitrary circle with center $A$ and $\psi_i$ the inversion with respect to that circle (Figure \ref{sl.inv.9.3.8.pic}). In this inversion, the circles $k$ and $l$ are mapped to the lines $k'$ and $l'$ (by \ref{InverzKroznVkrozn}), which are parallel because the circles $k$ and $l$ touch at the center of inversion $A$. The circles of the sequence $c_0$, $c_1$, $c_2$, ..., $c_n$, ... are mapped to the circles $c'_0$, $c'_1$, $c'_2$, ..., $c'_n$, ..., where each two consecutive ones touch and all the circles touch the parallels $k'$ and $l'$ (by \ref{InverzDotik}). Therefore, all the circles of this sequence are mutually tangent. The line $n'$, determined by their centers, is the perpendicular of the lines $k'$ and $l'$. The image $n=\psi_i(n')$ of $n'$ is perpendicular to the circles $c'_0$, $c'_1$, $c'_2$, ..., $c'_n$, ... (by \ref{InverzKonf}). In the end, we can conclude that $n$ is a circle exactly when the line $n'$ does not go through the point $A$, i.e. when the circles $k$ and $l$ are not tangent, otherwise $n$ is a line. \kdokaz \bzgled Let $A$, $B$, $C$ and $D$ be four arbitrary coplanar points. Prove that the angle, under which the circumscribed circles of the triangles $ABC$ and $ABD$ intersect, is equal to the angle, under which the circumscribed circles of the triangles $ACD$ and $BCD$ intersect. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.3.9.pic} \caption{} \label{sl.inv.9.3.9pic} \end{figure} \textbf{\textit{Proof.}} We mark with $k_A$, $k_B$, $k_C$ and $k_D$ the circumscribed circles of the triangles $BCD$, $ACD$, $ABD$ and $ABC$ (Figure \ref{sl.inv.9.3.9pic}). By an inversion $\psi_i$ with respect to an arbitrary circle $i$ with center $A$, our statement is transformed to the equivalent \ref{ObodKotTang}. \kdokaz \bzgled Let $A$, $B$ and $C$ be points that lie on the line $l$, and $P$ be a point outside of this line. Prove that the point $P$ and the centers of the circumscribed circles of the triangles $APB$, $BPC$ and $APC$ are four concircular points. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.3.10.pic} \caption{} \label{sl.inv.9.3.10pic} \end{figure} \textbf{\textit{Proof.}} We mark with $K_A$, $K_B$, $K_C$ the centers of the circumscribed circles of the triangles $BPC$, $APC$ and $APB$ (Figure \ref{sl.inv.9.3.10pic}). If $X=\mathcal{S}_{K_A}(P)$, $Y=\mathcal{S}_{K_B}(P)$ and $Z=\mathcal{S}_{K_C}(P)$ or $h_{P,2}(K_A)=X$, $h_{P,2}(K_B)=Y$ and $h_{P,2}(K_C)=Z$, from the concircularity of the points $X$, $Y$, $Z$ and $P$ it follows that the points $K_A$, $K_B$, $K_C$ and $P$ are concircular (from $X,Y,Z,P\in k$ it follows that $K_A,K_B,K_C,P\in h^{-1}_{P,2}$). We thus prove that the points $X$, $Y$, $Z$ and $P$ are concircular. Let $\psi_{P,r}$ be an inversion with the center $P$ and an arbitrary radius $r$ and $$\psi_{P,r}:\hspace*{1mm} A,B,C,X,Y,Z\mapsto A',B',C',X',Y',Z'.$$ The inversion $\psi_{P,r}$ maps the circumscribed circle of the triangle $BPC$, $APC$ and $APB$ into the lines $B'C'$, $A'C'$ and $A'B'$ (the statement \ref{InverzKroznVkrozn}). Because $PX$, $PY$ and $PZ$ are the diameters of these circles, the points $X'$, $Y'$ and $Z'$ are the orthogonal projections of the center of inversion $P$ onto the lines $B'C'$, $A'C'$ and $A'B'$. The line $l$ is mapped by the statement \ref{InverzKroznVkrozn} into the circumscribed circle of the triangle $A'B'C'$, which also goes through the point $P$. By the statement \ref{SimpsPrem} the points $X'$, $Y'$ and $Z'$ lie on the \index{line!Simson's} Simson's line $s$, which means (the statement \ref{InverzKroznVkrozn}), that the points $X$, $Y$ and $Z$ lie on the circle $\psi_{P,r}(s)$, which goes through the point $P$, therefore the points $X$, $Y$, $Z$ and $P$ are concurrent. \kdokaz %________________________________________________________________________________ \poglavje{Metric Properties of Inversion} \label{odd9MetrInv} It is clear from the definition itself that inversion is not an isometry. In the previous chapter we have found that inversion does not preserve the relation of collinearity of points, which means that it is not even a transformation of similarity. However, we are interested in how the distance between points changes, although in general the distance $AB$ is mapped by inversion into the segment $A'B'$. The answer will be given by the following important statement. \bizrek \label{invMetr} If $A'$ and $B'$ ($A',B'\neq S$) are the images of the points $A$ and $B$ by the inversion $\psi_{S,r}$, then it holds: $$|A'B'|=\frac{r^2\cdot |AB|}{|SA|\cdot |SB|}$$ \eizrek \begin{figure}[!htb] \centering \input{sl.inv.9.4.1.pic} \caption{} \label{sl.inv.9.4.1.pic} \end{figure} \textbf{\textit{Proof.}} We will consider two possible cases. \textit{(i)} Let the points $S$, $A$ and $B$ be nonlinear (Figure \ref{sl.inv.9.4.1.pic}). According to Theorem \ref{invPodTrik}, the triangles $ASB$ and $B'SA'$ are similar and $A'B':AB=SB':SA$. Since $B'$ is the image of point $B$ under the inversion $\psi_{S,r}$, it also holds that $|SB|\cdot |SB'|=r^2$ or $|SB'|=\frac{r^2}{|SB|}$. From these two relations it follows: $$|A'B'|=\frac{|SB'|\cdot |AB|}{|SA|}= \frac{r^2\cdot |AB|}{|SA|\cdot |SB|}.$$ \textit{(ii)} Let $S$, $A$ and $B$ be collinear points. Without loss of generality, we assume that $\mathcal{B}(S,A,B)$ holds. Then $\mathcal{B}(S,B',A')$ (Theorem \ref{invUrejenost}) holds, so $$|A'B'|=|SA'|-|SB'|= \frac{r^2}{|SA|}-\frac{r^2}{|SB|}= \frac{(|SB|- |SA|)\cdot r^2}{|SA|\cdot |SB|} = \frac{r^2\cdot |AB|}{|SA|\cdot |SB|} ,$$ which had to be proven. \kdokaz From the previous theorem we see that the distance between the images of points $A'$ and $B'$ increases if at least one of the originals $A$ and $B$ approaches the center of inversion $S$, which is logical, because we saw that the image of this center is a point at infinity. In section \ref{odd7Ptolomej} we have already proven Ptolomej's \index{theorem!Ptolomej's general} theorem (\ref{izrekPtolomej}) which relates to the chordal quadrilaterals. Now we will generalize this statement. \bizrek If $ABCD$ is an arbitrary convex quadrilateral, then: $$|AB|\cdot |CD|+|BC|\cdot |AD|\geq |AC|\cdot |BD|.$$ Equality holds if and only if $ABCD$ is a chordal quadrilateral. \eizrek \begin{figure}[!htb] \centering \input{sl.inv.9.4.2a.pic} \caption{} \label{sl.inv.9.4.2a.pic} \end{figure} \textbf{\textit{Proof.}} Let $\psi_{A,r}$ be the inversion with the center in the point $A$ (Figure \ref{sl.inv.9.4.2a.pic}). With $k$ we denote the circumscribed circle of the triangle $ABD$. Let $B'$, $C'$ and $D'$ be the images of the points $B$, $C$ and $D$ and $k'$ be the image of the circle $k$ under the inversion $\psi_{A,r}$. By the theorem \ref{InverzKroznVkrozn} the $k'$ is a line, which contains the points $B'$ and $D'$. From the previous theorem \ref{invMetr} it follows: $$|B'C'|=\frac{r^2\cdot |BC|}{|AB|\cdot |AC|}, \hspace*{2mm} |C'D'|=\frac{r^2\cdot |CD|}{|AC|\cdot |AD|} \hspace*{1mm} \textrm{ and } \hspace*{1mm} |B'D'|=\frac{r^2\cdot |BD|}{|AB|\cdot |AD|}.$$ For the points $B'$, $C'$ and $D'$ the triangle inequality \ref{neenaktrik} holds: $$B'C'+C'D'\geq B'D',$$ where the equality holds exactly when the points $B'$, $C'$ and $D'$ are collinear (and $\mathcal{B}(B',C',D')$) or when $C'\in k'$ (and $\mathcal{B}(B',C',D')$). This is exactly the case when the circle $k$ contains the point $C$ or when the quadrilateral $ABCD$ is cyclic (and convex). We can write the previous inequality also in the form: $$\frac{r^2\cdot |BC|}{|AB|\cdot |AC|}+\frac{r^2\cdot |CD|}{|AC|\cdot |AD|} \geq\frac{r^2\cdot |BD|}{|AB|\cdot |AD|}, \hspace*{2mm} \textrm{or}$$ $$|AB|\cdot |CD|+|BC|\cdot |AD|\geq |AC|\cdot |BD|,$$ which was to be proven. \kdokaz The next theorem will be a generalization of the example \ref{zgledTrikABCocrkrozP} or \ref{zgledTrikABCocrkrozPPtol}. \bzgled \label{zgledABCPinv} Let $k$ be the circumscribed circle of the regular triangle $ABC$. If $P$ is an arbitrary point in the plane of this triangle, then the following equivalence holds: the point $P$ does not lie on the circle $k$, exactly when there exists a triangle with sides, which are congruent to the distances $PA$, $PB$ and $PC$. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.4.3.pic} \caption{} \label{sl.inv.9.4.3.pic} \end{figure} \textbf{\textit{Proof.}} Let $\psi_{P,r}$ be the inversion with the center in the point $P$ (Figure \ref{sl.inv.9.4.3.pic}). Let $A'$, $B'$ and $C'$ be the images of the points $A$, $B$ and $C$ and $k'$ be the image of the circle $k$ under the inversion $\psi_{P,r}$. From the formula \ref{invMetr} it follows: $$|A'B'|=\frac{r^2\cdot |AB|}{|PA|\cdot |PB|}, \hspace*{2mm} |B'C'|=\frac{r^2\cdot |BC|}{|PB|\cdot |PC|} \hspace*{1mm} \textrm{ and } \hspace*{1mm} |A'C'|=\frac{r^2\cdot |AC|}{|PA|\cdot |PC|}.$$ Because the triangle $ABC$ is right, from the previous three relations it follows: $$A'B':B'C':A'C'=PC:PA:PB.$$ Now we can start with the proof of the desired equivalence: \begin{eqnarray*} P \notin k & \Leftrightarrow& \psi_{P,r}(k) \textrm{ represents a circle \hspace*{2mm}(formula \ref{InverzKroznVkrozn})}\\ & \Leftrightarrow& \textrm{the points }A', B', C' \textrm{ are non-collinear}\\ & \Leftrightarrow& \textrm{the distances }A'B', B'C', A'C' \textrm{ are the sides of a triangle}\\ & \Leftrightarrow& \textrm{the distances }PC, PA, PB \textrm{ are the sides of a triangle,} \end{eqnarray*} which was to be proven. \kdokaz Another generalization will be related to the Torricelli point (formula \ref{izrekTorichelijev}). \bizrek \label{izrekToricheliFerma} The point, for which the sum of the distances from the vertices of a triangle is minimal, is the \index{Torricelli, E.} \textit{E. Torricelli}'s\footnote{(1608--1647), Italian mathematician and physicist.} \index{točka!Torricellijeva}\index{točka!Fermatova}\pojem{Torricelli point} of this triangle. \eizrek \begin{figure}[!htb] \centering \input{sl.inv.9.4.4.pic} \caption{} \label{sl.inv.9.4.4.pic} \end{figure} \textbf{\textit{Proof.}} Use the same notation as in Theorem \ref{izrekTorichelijev} (Figure \ref{sl.inv.9.4.4.pic}). We have proven that the circles $k$, $l$ and $j$ intersect in Torricelli's point $P$ of the triangle $ABC$. Because $BEC$ is a right triangle, from Theorem \ref{zgledTrikABCocrkrozP} it follows that $|PB|+|PC|=|PE|$ i.e.: $$|PA|+|PB|+|PC|=|AE|.$$ We will prove that this sum is minimal only for Torricelli's point $P$. Let $P'\neq P$. Then the point $P'$ does not lie on any of the circles $k$, $l$ and $j$. Without loss of generality, let $P' \notin k$. From the previous Theorem \ref{zgledABCPinv} it follows that there exists a triangle with sides $P'E$, $P'B$ and $P'C$. Because of this and the triangle inequality (Theorem \ref{neenaktrik}) we have $|P'B| + |P'C| > |P'E|$ i.e.: $$|P'A| + |P'B| + |P'C| > |P'A| + |P'E| \geq |AE|=|PA|+|PB|+|PC|,$$ which was to be proven. \kdokaz \bnaloga\footnote{37. IMO, India - 1996, Problem 2.} Let $P$ be a point inside triangle $ABC$ such that $$\angle APB -\angle ACB = \angle APC -\angle ABC.$$ Let $D$, $E$ be the incentres of triangles $APB$, $APC$, respectively. Show that $AP$, $BD$, $CE$ meet at a point. \enaloga \begin{figure}[!htb] \centering \input{sl.inv.9.4.5.pic} \caption{} \label{sl.inv.9.4.5.pic} \end{figure} \textbf{\textit{Proof.}} Let $r>\max\{|AB|, |AC|\}$ be an arbitrary number and $\psi_{A,r}$ be the inversion with the center in the point $A$ and the radius $r$ and $B'$, $C'$ and $P'$ be the images of the points $B$, $C$ and $P$ under the inversion $\psi_{A,r}$ (Figure \ref{sl.inv.9.4.5.pic}). According to the statement \ref{invPodTrik} the triangles $ABC$ and $AC'B'$ are similar, so: $$\angle AC'B'\cong \angle ABC \textrm{ in } \angle AB'C'\cong \angle ACB.$$ From the similarity of $\triangle ABP \sim \triangle AP'B'$ and $\triangle ACP \sim \triangle AP'C'$ it follows: $$\angle AB'P'\cong \angle APB \textrm{ in } \angle AC'P'\cong \angle APC.$$ If we use the four relations of congruence of angles and the initial condition from the task ($\angle APB -\angle ACB = \angle APC -\angle ABC$), we get: \begin{eqnarray*} \angle C'B'P' &=& \angle AB'P'-\angle AB'C' = \angle APB - \angle ACB = \\ &=& \angle APC -\angle ABC = \angle AC'P' -\angle AC'B' =\\ &=& \angle B'C'P' \end{eqnarray*} Now from $\angle C'B'P'\cong \angle B'C'P'$ it follows $P'B'\cong P'C'$. If we use the relation from the statement \ref{invMetr}, we get $\frac{|PB| \cdot r^2}{|AP|\cdot |AB|}=\frac{|PC| \cdot r^2}{|AP|\cdot |AC|}$ or: $$\frac{|PB| }{ |AB|}=\frac{|PC| }{ |AC|}.$$ Let $BD \cap AP =X$ and $CE \cap AP =Y$. It is also necessary to prove that $X=Y$. Because $D$ and $E$ are the centers of inscribed circles of triangles $APB$ and $APC$, the lines $BD$ and $CE$ are the (internal) angle bisectors of $ABP$ and $ACP$. Therefore (statement \ref{HarmCetSimKota}) $$ \frac{\overrightarrow{PX}}{\overrightarrow{XA}}=\frac{PB}{BA} =\frac{PC}{CA}=\frac{\overrightarrow{PY}}{\overrightarrow{YA}}.$$ From $ \frac{\overrightarrow{PX}}{\overrightarrow{XA}} =\frac{\overrightarrow{PY}}{\overrightarrow{YA}}$ it follows (statement \ref{izrekEnaDelitevDaljice}) $X=Y$, which means that the lines $AP$, $BD$ and $CE$ intersect in the same point $X=Y$. \kdokaz %________________________________________________________________________________ \poglavje{Inversion and Harmonic Conjugate Points} \label{odd9InvHarm} In section \ref{odd7Harm} we saw that a harmonic quadruple of points is defined in two equivalent ways. For four collinear points $A$, $B$, $C$ and $D$, $\mathcal{H}(A,B;C,D)$, if one of the two (equivalent) conditions is fulfilled: \begin{itemize} \item $\frac{\overrightarrow{AC}}{\overrightarrow{CB}}=- \frac{\overrightarrow{AD}}{\overrightarrow{DB}}$, \item there exists a quadrilateral $ABCD$, such that $PQ \cap RS=A$, $PS \cap QR=B$, $PR \cap AB=C$ and $QS \cap AB=D$. \end{itemize} Now we will investigate some possibilities for equivalent definitions of the relation of harmonic quadruple of points - first with the help of inversion. \bizrek \label{invHarm} Let $AB$ be the diameter of the circle $i$ and $C$ and $D$ the points of the line $AB$, which are different from $A$ and $B$. If $\psi_i$ is the inversion with respect to the circle $i$, then it holds: $$\mathcal{H}(A,B;C,D) \hspace*{1mm} \Leftrightarrow \hspace*{1mm} \psi_i(C)=D.$$ \eizrek \textbf{\textit{Proof.}} Let $O$ be the center of the line $AB$ (Figure \ref{sl.inv.9.6.1.pic}). Then it is: \begin{eqnarray*} \mathcal{H}(A,B;C,D) &\hspace*{1mm} \Leftrightarrow \hspace*{1mm}& \frac{\overrightarrow{AC}}{\overrightarrow{CB}}=- \frac{\overrightarrow{AD}}{\overrightarrow{DB}}\\ &\hspace*{1mm} \Leftrightarrow \hspace*{1mm}& \frac{\overrightarrow{OC}-\overrightarrow{OA}} {\overrightarrow{OB}-\overrightarrow{OC}}=- \frac{\overrightarrow{OD}-\overrightarrow{OA}} {\overrightarrow{OB}-\overrightarrow{OD}}\\ &\hspace*{1mm} \Leftrightarrow \hspace*{1mm}& \frac{\overrightarrow{OC}+\overrightarrow{OB}} {\overrightarrow{OB}-\overrightarrow{OC}}=- \frac{\overrightarrow{OD}+\overrightarrow{OB}} {\overrightarrow{OB}-\overrightarrow{OD}}\\ &\hspace*{1mm} \Leftrightarrow \hspace*{1mm}& \overrightarrow{OC} \cdot \overrightarrow{OD} = OB^2\\ &\hspace*{1mm} \Leftrightarrow \hspace*{1mm}& \psi_i(C)=D, \end{eqnarray*} which had to be proven. \kdokaz \begin{figure}[!htb] \centering \input{sl.inv.9.6.1.pic} \caption{} \label{sl.inv.9.6.1.pic} \end{figure} The next statement will give the fourth possibility of an equivalent definition of the relation of harmonic quadruple of points. \bizrek \label{harmPravKrozn} Let $A$, $B$, $C$ and $D$ be different collinear points and $k$ and $l$ be circles over diameters $AB$ and $CD$. Then the equivalence holds: $$\mathcal{H}(A,B;C,D) \Leftrightarrow k\perp l.$$ \eizrek \textbf{\textit{Proof.}} Let $O$ and $S$ be the centers of the circles $k$ and $l$ (Figure \ref{sl.inv.9.6.1.pic}). From both sides of the equivalence it follows that the circles intersect. One of their intersection points is denoted by $T$. From the previous statement \ref{invHarm} it is: $$\mathcal{H}(A,B;C,D) \hspace*{1mm} \Leftrightarrow \hspace*{1mm} \psi_k(C)=D.$$ So we only need to prove: $$k\perp l \hspace*{1mm} \Leftrightarrow \hspace*{1mm} \psi_k(C)=D.$$ But now we have: \begin{eqnarray*} k\perp l \hspace*{1mm} &\Leftrightarrow& \hspace*{1mm} OT\perp TS \hspace*{2mm} \textrm{ (statement \ref{pravokotniKroznici})}\\ &\Leftrightarrow& \hspace*{1mm} OT \textrm{ is a tangent of the circle }l \hspace*{2mm}\textrm{ (statement \ref{TangPogoj})}\\ &\Leftrightarrow& \hspace*{1mm} \overrightarrow{OC}\cdot \overrightarrow{OD} = OT^2\hspace*{2mm}\textrm{ (statement \ref{izrekPotenca})}\\ &\Leftrightarrow& \hspace*{1mm} \psi_k(C)=D, \end{eqnarray*} which was to be proven. \kdokaz From the previous statement \ref{harmPravKrozn} it directly follows (already proven before) that from $\mathcal{H}(A,B;C,D)$ it follows $\mathcal{H}(C,D;A,B)$ or $\mathcal{H}(B,A;C,D)$. We already know that for three collinear points $A$, $B$ and $C$, where point $C$ is not the midpoint of line segment $AB$, there is only one point $D$, so that $\mathcal{H}(A,B;C,D)$ holds. So for three given points in a harmonic quadruple of points the fourth is uniquely determined. One of the possible constructions of this point gives us the previous statement \ref{invHarm} ($D=\psi_k(C)$). But only two points $A$ and $B$ are not enough to determine the other pair of points $C$ and $D$. There are infinitely many such pairs of points $(C,D)$, for which $\mathcal{H}(A,B;C,D)$ holds (Figure \ref{sl.inv.9.6.2.pic}). For their determination another condition is needed. We will consider such conditions in the next two examples. \begin{figure}[!htb] \centering \input{sl.inv.9.6.2.pic} \caption{} \label{sl.inv.9.6.2.pic} \end{figure} \bzgled Given are points $A$, $B$ and $S$. Construct such points $C$ and $D$, that $S$ is the midpoint of line segment $CD$ and $\mathcal{H}(A,B;C,D)$ holds. \ezgled \textbf{\textit{Solution.}} Let $k$ be a circle with diameter $AB$. If $l$ is a circle with diameter $CD$, then $S$ is the center of this circle. From \ref{harmPravKrozn} it follows that $k \perp l$ (Figure \ref{sl.inv.9.6.3.pic}). So it is enough to construct the circle $l$, because then the points $C$ and $D$ are the intersections of this circle with the line segment $AB$. But we can construct the circle $l$, if we first construct the tangent to the circle $k$ from point $S$ in point $T$. \kdokaz \begin{figure}[!htb] \centering \input{sl.inv.9.6.3.pic} \caption{} \label{sl.inv.9.6.3.pic} \end{figure} \bzgled \label{harmDaljicad} Given is a line segment $d$ and points $A$ and $B$. Construct such points $C$ and $D$, that $\mathcal{H}(A,B;C,D)$ holds and $CD\cong d$. \ezgled \textbf{\textit{Solution.}} It is enough to construct the center $S$ of the line $CD$, and then continue as in the previous example. The right triangle $OTS$ can be constructed because both sides are known $|OT| = \frac{1}{2}\cdot |AB|$ and $|ST| = \frac{1}{2}\cdot |CD|= \frac{1}{2}\cdot |d|$. From it we get the hypotenuse $OS$ (Figure \ref{sl.inv.9.6.3.pic}). \kdokaz In the next examples we will look at the use of the previous two constructions. \bzgled Draw a triangle with the data $v_a$, $l_a$ and $b-c$. \ezgled \textbf{\textit{Solution.}} We will use the labels from the big task \ref{velikaNaloga} (Figure \ref{sl.inv.9.6.4.pic}). First we can draw the right triangle $AA'E$, because $AA'\cong v_a$ and $AE\cong l_a$. By the example \ref{harmVelNal} is $\mathcal{H}(A',E;P,Pa)$. From the big task it follows $PP_a=b -c$, so we can by the previous example \ref{harmDaljicad} construct the points $P$ and $P_a$. Then we draw the center $S$ of the inscribed circle of the triangle $ABC$, at the end the inscribed circle, their tangents from the point $A$ and mark $B$ and $C$. \kdokaz \begin{figure}[!htb] \centering \input{sl.inv.9.6.4.pic} \caption{} \label{sl.inv.9.6.4.pic} \end{figure} \bzgled Draw a triangle with the data $v_a$, $a$ and $r+r_a$. \ezgled \textbf{\textit{Solution.}} Also in this case we will use the labels from the big task \ref{velikaNaloga} (Figure \ref{sl.inv.9.6.4.pic}). By the example \ref{harmVelNal} is $\mathcal{H}(A,L;A',La)$. Because $AA'\cong v_a$ and $LL_a =r+r_a$, we can by the example \ref{harmDaljicad} first draw the line $AA'\cong v_a$, then the points $L$ and $L_a$. So we get $r\cong LA'$ and $r_a\cong L_aA'$. From the great task it follows that $RR_a = a$. This means that we can draw a rectangular trapezoid $SRR_aS_a$ ($RR_a=a$, $SR=r$ and $S_aR_a=r_a$). Then we construct the inscribed circle $k(S,r)$ and the circumscribed circle $k_a(S_a,r_a)$ and finally their common tangents (two external and one internal), which are the sides of the triangle $ABC$. \kdokaz %________________________________________________________________________________ \poglavje{Feuerbach Points} \label{odd9Feuerbach} In this section we will consider some properties of the inscribed and circumscribed circles of a triangle. \bzgled Let $ABC$ be a triangle with a semiperimeter $s$ and $D$ and $E$ such points on the line $BC$ that $|AD|=|AE|=s$. Prove that the circumscribed circle of the triangle $ADE$ and the circumscribed circle of the triangle $ABC$ touch the side $BC$. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.5.0.pic} \caption{} \label{sl.inv.9.5.0.pic} \end{figure} \textbf{\textit{Proof.}} Let $l$ be the circumscribed circle of the triangle $AED$ and $P_a$, $Q_a$ and $R_a$ the points in which the circumscribed circle $k_a$ of the triangle $ABC$ touches the side $BC$ and the sides $AC$ and $AB$ (Figure \ref{sl.inv.9.5.0.pic}). With $i$ we denote the circle with center $A$ and radius $AE$ (or $AD$). From the great task \ref{velikaNaloga} it follows that $|AR_a|=|AQ_a|=s=|AE|=|AD|$, which means that the points $R_a$ and $Q_a$ lie on the circle $i$. By the theorem \ref{pravokotniKroznici} the circles $i$ and $k_a$ are perpendicular. Therefore, the circle $k_a$ is mapped to itself by the inversion $\psi_i$ (theorem \ref{InverzKroznFiks}), and the line $BC$ is mapped to the circle $l$ (theorem \ref{InverzKroznVkrozn}). Since the line $BC$ touches the circle $k_a$ at the point $P_a$, their images (the circles $l$ and $k_a$) touch at the point $T=\psi_i(P_a)$ (theorem \ref{InverzDotik}). \kdokaz \bzgled \label{InvOcrtVcrt} Let $P$, $Q$ and $R$ be points in which the incircle of triangle $ABC$ touches the sides of the triangle. Prove that the orthocenter of triangle $PQR$ and the centers of the circumscribed and incircle of triangle $ABC$ are collinear points. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.5.0a.pic} \caption{} \label{sl.inv.9.5.0a.pic} \end{figure} \textbf{\textit{Proof.}} Let $l(O,R)$ and $k(S,r)$ be the circumscribed and incircle of triangle $ABC$ and $\psi_k$ be the inversion with respect to the circle $k$ (Figure \ref{sl.inv.9.5.0a.pic}). We also denote $A'=SA \cap QR$, $B'=SB \cap PR$ and $C'=SC \cap PQ$. From the similarity of triangles $ARA'$ and $AQA'$ (the \textit{SAS} theorem \ref{SKS}) it follows that the point $A'$ is the center of the side $QR$ and that $AS \perp RQ$. Similarly, it also follows that points $B'$ and $C'$ are the centers of segments $PR$ and $PQ$ and that $BS \perp PR$ and $CS \perp PQ$. From the process of constructing the image of a point in an inversion it follows: $$\psi_k:A,B,C \mapsto A',B',C'.$$ Therefore, the circumscribed circle $l$ with inversion $\psi_k$ is mapped to the circumscribed circle of triangle $A'B'C'$ (the \ref{InverzKroznVkrozn} theorem). This is Euler's circle $e_1$ of triangle $PQR$, because it goes through the centers of its sides. The center of this circle - point $E_1$ - lies on Euler's line $e_p$ of triangle $PQR$, which is determined by its orthocenter $V_1$ and the center of the circumscribed circle $S$. We also prove that point $O$ lies on line $e_p$. Although the center of circle $l$ (point $O$) is not mapped to the center of the circle (point $E_1$) by inversion, point $E_1$ lies on line $SO$. This also means that point $O$ lies on line $SE_1=e_p$, so points $V_1$, $O$ and $S$ are collinear. \kdokaz The next simple statement is just an introduction to the so-called Feuerbach's theorem. \bizrek Let $k$ be the inscribed circle of the triangle $ABC$ and $k_a$ the circle drawn through its side $BC$. If $A_1$ is the center of the side $BC$ and $i$ is the circle with center $A_1$ and radius $\frac{1}{2}|b-c|$, then $$\psi_i:k,k_a\rightarrow k, k_a.$$ \eizrek \textbf{\textit{Proof.}} Let $P$ and $P_a$ be the points of tangency of the circles $k$ and $k_a$ with its side $BC$ (Figure \ref{sl.inv.9.5.1.pic}). From the great task \ref{velikaNaloga} it follows that the points $P$ and $P_a$ lie on the circle $i$. Therefore, by the theorem \ref{pravokotniKroznici} the circles $k$ and $k_a$ are perpendicular to the circle $i$, so $\psi_i:k,k_a\rightarrow k, k_a$ (theorem \ref{InverzKroznFiks}). \kdokaz \begin{figure}[!htb] \centering \input{sl.inv.9.5.1.pic} \caption{} \label{sl.inv.9.5.1.pic} \end{figure} \bizrek \index{krožnica!Eulerjeva} Euler's circle of the triangle intersects the inscribed circle and all three of the drawn circles of this triangle\footnote{The points of intersection are called \index{točka!Feuerbachova} \pojem{Feuerbach points} of this triangle. \index{Feuerbach, K. W.} \textit{K. W. Feuerbach} (1800--1834), German mathematician, who proved this theorem in 1822.}. \eizrek \textbf{\textit{Proof.}} We will use the notation from the great task \ref{velikaNaloga}. Let $e$ be Euler's circle of this triangle (Figure \ref{sl.inv.9.5.1.pic}). From the previous theorem the inversion $\psi_i$ with respect to the circle $i$ with center $A_1$ and radius $\frac{1}{2}|b-c|$ (or containing the points $P$ and $P_a$) maps the circles $k$ and $k_a$ to themselves. We determine the image of Euler's circle $e$ under the inversion $\psi_i$. The circle $e$ contains the center of inversion $A_1$, so it is mapped to some line $e'$ by inversion (statement \ref{InverzKroznVkrozn}). We only need to prove that the line $e'$ is tangent to the circles $k$ and $k_a$ (statement \ref{InverzDotik}). First, the line $e'$ contains the point $\psi_i(A')=E$, which is the intersection of the internal angle bisector at the vertex $A$ with the side $BC$. (statements \ref{harmVelNal} and \ref{invHarm}). But the line $e'$ also contains the points $\psi_i(B_1)=B'_1$ and $\psi_i(C_1)=C'_1$. By statement \ref{invPodTrik}, the triangles $A_1B_1C_1$ and $A_1C'_1B'_1$ are similar. It follows that the line $e'$ with the line $AB$ determines the angle $ACB$. Because the line $e'$ intersects the line $BC$ at the point $E$, which lies on the bisector of the angle $BAC$, $e'$ represents the second common tangent to the circles $k$ and $k_a$. We denote by $F'$ and $F'_a$ the points of tangency of the line $e'$ with the circles $k$ and $k_a$. Then the circle $e$ is tangent to the circles $k$ and $k_a$ at the points $F=\psi_i(F')$ and $F_a=\psi_i(F'_a)$. In the same way, we prove that $e$ is also tangent to the circles $k_b$ and $k_c$. \kdokaz %________________________________________________________________________________ \poglavje{Stainer's Theorem} \label{odd9Stainer} In this section, we will elegantly (with the help of inversion) prove Stainer's statement, which is related to the problem of the existence of a sequence of circles that are cyclically tangent to each other, and also tangent to two given, non-concentric, circles. It is clear that such a sequence does not exist in general (for any two non-concentric circles). But we will prove that if there is at least one such sequence for the two circles $k$ and $l$, then we can choose the initial element of this sequence as any circle that is tangent to the circles $k$ and $l$. We will first solve two auxiliary tasks. \bzgled \label{StainerjevLema1} Let $k_1(S_1,r_1)$ and $k_2(S_2,r_2)$ be two circles of some plane. Draw a circle with center on the line $S_1S_2$, which is perpendicular to both circles. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.7.0.pic} \caption{} \label{sl.inv.9.7.0.pic} \end{figure} \textbf{\textit{Solution.}} Let $k_1(S_1,r_1)$ and $k_2(S_2,r_2)$ be any circles. In the case that a solution exists, the center of the sought circle lies on the power line of the two circles (Figure \ref{sl.inv.9.7.0.pic}). This is the case when the intersection $S$ of the power line $l$ and the central line $S_1S_2$ is an external point of both circles or when the circles $k_1$ and $k_2$ are non-concentric (from the inside or from the outside). If we draw a tangent to one of the circles from the point $S$ (e.g. $k_1$), we also get the radius $r=ST_1$ of the circle $k(S,r)$, where the point $T_1$ is the point of tangency of the tangent from $S$ on the circle $k_1$. Let the point $T_2$ be the point of tangency of the tangent from $S$ on the circle $k_2$. Since the point $S$ lies on the power line of the two circles, we have: $$|ST_1|=p(S,k_1)=p(S,k_2)=|ST_2|$$ the circle $k(S,r)$ is perpendicular to the circle $k_1(S_1,r_1)$ and $k_2(S_2,r_2)$ by \ref{pravokotniKroznici}. \kdokaz %slika \bzgled \label{StainerjevLema2} If $k$ is a circle inside the circle $l$, there exists an inversion that maps the circles $k$ and $l$ to two concentric circles. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.7.1.pic} \caption{} \label{sl.inv.9.7.1.pic} \end{figure} \textbf{\textit{Proof.}} Let $p$ be a line, determined by the centers of the circles $k$ and $l$; it is also perpendicular to both circles. From the previous example \ref{StainerjevLema1} it follows that there exists a circle $n$ with the center on the line $p$, which is perpendicular to the circles $k$ and $l$ (Figure \ref{sl.inv.9.7.1.pic}). Because the center of the circle $n$ lies on the line $p$, the circle $n$ is also perpendicular to the line $p$. With $I$ and $J$ we denote the intersections of the circle $n$ with the line $p$. Let $i$ be an arbitrary circle with the center $I$ and $\psi_i$ the inversion with respect to this circle. Because the inversion preserves angles, the line $p$ and the circle $n$ (without the point $I$) with $\psi_i$ are mapped to the perpendicular line $p'=p$ and $n'$, the circles $k$ and $l$ to the circles $k'$ and $l'$, which are perpendicular to these two lines (the statement \ref{InverzKroznVkrozn} and \ref{InverzKonf}). Therefore, the circles $k'$ and $l'$ are concentric. \kdokaz \bzgled (Steiner's statement \index{statement!Steiner's} \footnote{\index{Steiner, J.} \textit{J. Steiner} (1769--1863), Swiss mathematician.}.) Let $l$ be a circle inside the circle $k$ and $a_1$, $a_2$, ..., $a_n$ a sequence of circles, each of which touches the circles $k$ and $l$, and each circle touches the adjacent circle in the sequence. If the circles $a_n$ and $a_1$ touch, this property is independent of the choice of the first circle $a_1$ of this sequence. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.7.2.pic} \caption{} \label{sl.inv.9.7.2.pic} \end{figure} \textbf{\textit{Proof.}} If we use the previous statement \ref{StainerjevLema2}, the proposition is translated into the case when the circles are concentric (the touch of the circles is an invariant of the inversion \ref{InverzDotik}). But in this case the new sequence of circles is simply obtained from the first one by rotation with the center in the common center of the circles $\psi_i(k)$ and $\psi_i(l)$ (Figure \ref{sl.inv.9.7.2.pic}). \kdokaz %________________________________________________________________________________ \poglavje{Problem of Apollonius} \label{odd9ApolDotik} Now we will deal with so-called \index{problem!Apolonijev} Apolonijev's problems of the intersection of circles, which can be elegantly solved by using inversion. These are problems of the following form: \textit{ \vspace*{2mm} Draw a circle $k$, which satisfies three conditions, each of which has one of the following forms: \vspace*{2mm} \begin{itemize} \item contains a given point, \item touches a given line, \item touches a given circle. \end{itemize}} It is clear that all the points, lines and circles mentioned in the above conditions lie in the same plane. We have already encountered some of these problems. For example, to construct a circle that contains a given point and touches a given line. It is not difficult to determine that there are ten Apollonius's problems. We usually list them in the following order: \begin{enumerate} \item draw a circle that contains three given points. $(A,B,C)$, \item draw a circle that contains a given point and touches a given line $(A,B,p)$, \item draw a circle that contains a given point and touches a given circle $(A,B,k)$, \item draw a circle that contains a given point and touches two given lines $(A,p,q)$, \item draw a circle that contains a given point and touches a given line and a given circle $(A,p,k)$, \item draw a circle that contains a given point and touches two given circles $(A,k_1,k_2)$, \item draw a circle that touches three given lines $(p,q,r)$, \item draw a circle that touches two given lines and a given circle $(p,q,k)$, \item draw a circle that touches a given line and two given circles $(p,k_1,k_2)$, \item draw a circle that touches three given circles $(k_1,k_2,k_3)$. \end{enumerate} We immediately see that the first and seventh problem are trivial, and also all the other problems can be solved without using inversion. However, inversion gives us a general method for solving them. This method is based on the fact that in a certain case, a circle is transformed into a line by inversion (Theorem \ref{InverzKroznVkrozn}). We will illustrate it with an example of the fifth Apollonius' problem: \bzgled Draw a circle that contains the given point $A$ and touches the given line $p$ and the given circle $k$. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.8.1.pic} \caption{} \label{sl.inv.9.8.1.pic} \end{figure} \textbf{\textit{Solution.}} Assume that $x$ is a circle that contains the point $A$ and touches the line $p$ and the circle $k$ (Figure \ref{sl.inv.9.8.1.pic}). We will consider the general case, so that the point $A$ does not lie on neither the line $p$ nor the circle $k$. With $i$ we denote the circle with the center $A$ and an arbitrary radius $r$. Let $\psi_i$ be the inversion with respect to this circle and $p'$, $k'$ and $x'$ be the images of the line $p$ and the circles $k$ and $x$ under this inversion. Because $A\notin p,k$ and $A\in x$, $p'$ and $k'$ are circles, $x'$ is a line (Theorem \ref{InverzKroznVkrozn}). The line $p$ and the circle $x$ have exactly one common point, therefore this is also true for the images $p'$ and $x'$, thus the line $x'$ is tangent to the circle $p'$. Similarly, the line $x'$ is also tangent to the circle $k'$. Therefore, the problem is reduced to constructing the line $x'$, which is the common tangent of the circles $p'$ and $k'$ (Example \ref{tang2ehkroz}). Then $x= \psi_i^{-1}(x')= \psi_i(x')$. The task has zero, one, two, three or four solutions, depending on the mutual position of the circles $p'$ and $k'$ or the number of their common tangents. In the case when $A\in p$ or $A\in k$, the line $x'$ is a line that touches one circle ($k'$ or $p'$) and is parallel to one line ($p'$ or $k'$). If $A\in p \cap k$, the task has no solution or there are infinitely many, depending on whether the circles intersect or touch. In both cases, the line $x'$ is parallel to two lines $p'$ and $k'$, but in the first case the lines $p'$ and $k'$ intersect, in the second case they are parallel. \kdokaz We immediately notice that in solving this problem, it was not important whether $p$ and $k$ were a straight line and a circle, but it was important that the images $p'$ and $k'$ were circles. But $p'$ and $k'$ are also circles in the case when, for example, $p$ and $k$ are two lines and $A\notin p,k$. So the fourth and sixth Apollonius problems are solved in the same way as the fifth problem that we just solved. We can also solve the second and third problem by using inversion with respect to any circle with center $A$. Both problems are translated into the construction of a tangent from point $B'$ to circle $p'$ (or $k'$). Problems 8, 9 and 10 are translated into problems 4, 5 and 6 in order. The idea is to first plan a circle that is concentric with the desired circle and contains the center of one of the given circles - the one with the smallest radius. In the following, we will solve a problem similar to Apollonius' problem of the intersection of circles. \bzgled Given are the point $A$, the circles $k$ and $l$, and the angles $\alpha$ and $\beta$. Draw a circle that passes through point $A$, the circles $k$ and $l$ and intersects at angles $\alpha$ and $\beta$. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.8.2.pic} \caption{} \label{sl.inv.9.8.2.pic} \end{figure} \textbf{\textit{Solution.}} Let's assume that $x$ is a circle that contains point $A$ and with circles $k$ and $l$ defines angles $\alpha$ and $\beta$ (Figure \ref{sl.inv.9.8.2.pic}). Again, we will consider the general case, when point $A$ does not lie on any of the circles $k$ and $l$. With $i$ we mark the circle with center $A$ and any radius $r$ and with $\psi_i$ the inversion with respect to that circle. Let $k'$, $l'$ and $x'$ be the images of circles $k$, $l$ and $x$ under that inversion. Because $A\notin k,l$ and $A\in x$, $k'$ and $l'$ are circles and $x'$ is a line (Theorem \ref{InverzKroznVkrozn}). Because inversion is a conformal mapping (Theorem \ref{InverzKonf}), line $x'$ intersects circles $k'$ and $l'$ under angles $\alpha$ and $\beta$. Line $x'$ can be drawn as the common tangent of two circles $k'_1$ and $l'_1$ that are concentric with circles $k'$ and $l'$. When drawing circles $k'_1$ and $l'_1$ we take into account the fact that line $x'$ with circles $k'$ and $l'$ defines the chords with central angles $2\alpha$ and $2\beta$. In the end, $x= \psi_i^{-1}(x')= \psi_i(x')$. Also in this case, the number of solutions depends on the mutual position of circles $k'_1$ and $l'_1$ i.e. the number of their common tangents. In the case when $A\in k$ or $A\in l$, $x'$ is a line that with line $k'$ (or $l'$) defines angle $\alpha$, with circle $p'$ (or $k'$) angle $\beta$. If $A\in p \cap k$, the problem in general has no solution, because it is a line $x'$, that with given lines $k'$ and $l'$ defines angles $\alpha$ and $\beta$. The problem has infinitely many solutions, when lines $k'$ and $l'$ define angles $|\beta \pm \alpha|$. \kdokaz %________________________________________________________________________________ \poglavje{Constructions With Compass Alone} \label{odd9LeSestilo} In Euclidean geometry constructions we always used a ruler and a compass, which means that we used elementary constructions\index{konstrukcije!z ravnilom in šestilom} (see section \ref{elementarneKonstrukcije}). However, by using a ruler and a compass, or the aforementioned elementary constructions, in Euclidean geometry it is not possible to derive all constructions. The most well-known are the following examples\footnote{All three problems were posed by the Ancient Greeks. Later, many famous mathematicians as well as laymen tried to solve these problems. The fact that the aforementioned constructions cannot be derived using only a ruler and a compass was only proven in the 19th century, when the French mathematician \index{Galois, E.} \textit{E. Galois} (1811--1832) developed \index{group}the theory of groups. The proof for the trisection of an angle and the doubling of a cube was first given by the French mathematician \index{Wantzel, P. L.}\textit{P. L. Wantzel} (1814--1848) in 1837. The fact that the quadrature of a circle is impossible is a consequence of the transcendence of the number $\pi$, which was proven by the German mathematician \index{Lindemann, C. L. F.}\textit{C. L. F. Lindemann} (1852–-1939) in 1882.}: \begin{itemize} \item the division of a given angle into three congruent parts (\pojem{trisection of an angle})\index{trisection of an angle}; \item the construction of a square with the same area as a given circle (\pojem{quadrature of a circle})\index{quadrature of a circle}; \item the construction of the edge of a cube that has twice the volume of a cube with a given edge $a$ (\pojem{doubling of a cube})\index{doubling of a cube}. \end{itemize} \index{konstrukcije!pravilnih $n$-kotnikov} In addition, the problem of constructing regular $n$-gon with a ruler and a compass is also known, which cannot be solved for every $n\in\{3,4,\ldots\}$\footnote{The well-known German mathematician \index{Gauss, C. F.}\textit{C. F. Gauss} (1777--1855) proved in 1796 that with a ruler and a compass, a regular $n$-gon can be constructed exactly when $n=2^k\cdot p$, where $p$ is either 1 or a prime number that can be written in the form $2^{2^l}+1$, for $l\in\{0,1,2,\ldots\}$ (i.e. \pojem{Fermatova števila}\normalcolor, which are not all prime - for example, for $n=5$, they are named after the French mathematician \index{Fermat, P.}\textit{P. Fermat} (1601--1665)). A regular $n$-gon can therefore be constructed with a ruler and a compass if $n\in\{3,4,5,6,8,10,12,16,17,\ldots\}$, but if $n\in\{7,9,11,13,14,15,18,19,\ldots\}$, this construction is not possible.}. In projective geometry, in which there is neither parallelism nor metric, we use only a ruler\index{konstrukcije!z ravnilom} or only those elementary constructions that can only be done with a ruler. It is clear that in this geometry it is not possible to design (or even define) figures such as square, parallelogram, circle, ... The question arises, what can we construct if we use only a compass or only those elementary constructions that can be done only with a compass. We consider that a line is drawn if two of its points are drawn, but we cannot use the direct construction of the intersection of two lines (with a plane). Surprisingly, in this way - only with a compass - we can carry out all the constructions that can be done with a compass and with a plane\footnote{The constructions only with the help of a compass were researched by the Italian mathematician \index{Maskeroni, L.} \textit{L. Maskeroni} (1750--1800) (after him we call them \index{konstrukcije!Maskeronijeve}\index{konstrukcije!s šestilom} \pojem{Maskeronijeve konstrukcije}) and a hundred years before him the Danish mathematician \index{Mor, G.} \textit{G. Mor} (1640--1697) in his book ‘‘Danish Euclid’’ from 1672. The theoretical basis of these constructions was given by the Austrian mathematician \index{Adler, A.} \textit{A. Adler}, who in 1890 proved that every design problem that can be solved with the help of a plane and a compass, can also be solved only with the use of a compass.}! We will continue with the following constructions, in which we will use only a compass. In the desire to show that a plane "can be replaced" with a compass, the main role will be played by inversion. In the previous use of inversion in constructions, we most often used the fact that inversion in a certain case transforms a circle into a line. So we translated the problem of constructing the desired circle with the help of inversion into the problem of constructing the desired line, which is usually easier. Now we will try the opposite - problems related to the construction of a line (and the use of a plane) we will translate with inversion into problems of constructing a circle (and the use of a compass). \bzgled \label{MaskeroniNAB} Given are points $A$ and $B$ and $n\in \mathbb{N}$. Draw such a point $P_n$, that $\overrightarrow{AP_n}=n\cdot \overrightarrow{AB}$. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.9.1.pic} \caption{} \label{sl.inv.9.9.1.pic} \end{figure} \textbf{\textit{Solution.}} We will carry out the construction inductively. For $n=1$ it is clear that $P_1=B$. First, let's draw the point $P_2$, for which $\overrightarrow{AP_2}=2\cdot \overrightarrow{AB}$ (Figure \ref{sl.inv.9.9.1.pic}). Now let's draw the circles $k_1(A,AB)$ and $k_2(B,BA)$. We mark one of their intersections with $Q_1$. Then we draw the circle $l_1(Q_1,Q_1B)$. The intersection of the circles $l_1$ and $k_2$, which is not the point $A$, we mark with $Q_2$. The point $P_2$ we get as one of the intersections of the circles $l_2(Q_2,Q_2B)$ and $k_2$ (that which is not $Q_1$). \normalcolor The relation $\overrightarrow{AP_2}=2\cdot \overrightarrow{AB}$ follows from the fact that the triangles $ABQ_1$, $Q_1BQ_2$ and $Q_2BP_2$ are all right. Assume that we have with the described procedure drawn the points $P_k$ ($k\leq n-1$), for which $\overrightarrow{AP_k}=k\cdot \overrightarrow{AB}$. We can draw the point $P_n$ in the same way. First, we draw the circle $k_n(P_{n-1},P_{n-1}P_{n-2})$. The intersection of the circles $l_{n-1}$ and $k_n$, which is not the point $P_{n-2}$, we mark with $Q_n$. The point $P_n$ we get as one of the intersections of the circles $l_n$ and $k_n$ (that which is not $Q_{n-1}$). The triangles $P_{n-2}P_{n-1}Q_{n-1}$, $Q_{n-1}P_{n-1}Q_n$ and $Q_nP_{n-1}P_n$ are all right. Therefore, $\overrightarrow{P_{n-1}P_n}=\overrightarrow{P_{n-2}P_{n-1}}$. If we use the induction assumption $\overrightarrow{AP_k}=k\cdot \overrightarrow{AB}$ for $k=n-1$, we get $\overrightarrow{AP_n}=n\cdot \overrightarrow{AB}$, which means that $P_n$ is the desired point. \kdokaz We have already mentioned the importance of inversion in our constructions. Now we are ready to prove the process of constructing the image of a point under inversion only with the help of a compass. \bzgled Given are the circle $i(S,r)$ and the point $X$. Draw the point $X'=\psi_i(X)$. \label{MaskeroniInv} \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.9.2.pic} \caption{} \label{sl.inv.9.9.2.pic} \end{figure} \textbf{\textit{Solution.}} If $X\in i$ it is trivial that $X'=X$. First, let us assume that $X$ is an external point of the inversion circle $i$ (Figure \ref{sl.inv.9.9.2.pic}). Because the center $S$ is its internal point, the circle $k(X,XS)$ intersects the inversion circle in two points (statement \ref{DedPoslKrozKroz}) - for example $P$ and $Q$, which we can draw. Then the point $X'$ is the second intersection of the circles $l_P(P,PS)$ and $l_Q(Q,QS)$ (the first one is the point $S$). We will also prove that $X'=\psi_i(X)$. Because by construction $XS\cong XP$ and $PX'\cong PS$, it follows that $\angle PX'S\cong \angle PSX'= \angle PSX\cong\angle SPX$. This means that the triangles $PSX'$ and $XPS$ are similar, therefore $\frac{|PX'|}{|XS|}=\frac{|SX'|}{|PS|}$ or $|SX|\cdot |SX'|=|PX'|\cdot |PS|=r^2$, thus $X'=\psi_i(X)$. If $X$ is an internal point of the inversion circle, then there exists a natural number $n$ such that $|n\cdot \overrightarrow{SX}|>r$. Let $X_n$ be the point for which $\overrightarrow{SX_n}=n\cdot \overrightarrow{SX}$ and $X'_n=\psi_i(X_n)$. From the previous example and the first part of the proof we can draw the points $X_n$ and $X'_n$. For the point $X'=\psi_i(X)$ it holds: $$|SX'|=\frac{r^2}{|SX|}=\frac{n\cdot r^2}{|SX_n|}=n\cdot |SX'_n|,$$ which means that we can also draw the point $X'$ using the previous example. \kdokaz \bzgled Dani sta točki $A$ in $B$. Načrtaj središče daljice $AB$. \label{MaskeroniSred} \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.9.3.pic} \caption{} \label{sl.inv.9.9.3.pic} \end{figure} \textbf{\textit{Solution.}} Let $\psi_i$ be the inversion with respect to the circle $i(A,AB)$ (Figure \ref{sl.inv.9.9.3.pic}). First, we plan such a point $X$, that $\overrightarrow{AX}=2\cdot\overrightarrow{AB}$ (example \ref{MaskeroniNAB}), then $X'=\psi_i(X)$ (example \ref{MaskeroniInv}). Because $X$ is an external point of the circle $i$, $X'$ is its internal point. Therefore $\mathcal{B}(A,X',B)$. For the point $X'$ it also holds $|AX'|=\frac{|AB|^2}{|AX|}=\frac{|AB|^2}{2\cdot |AB|}=\frac{1}{2}\cdot |AB|,$ which means that $X'$ is the center of the line segment $AB$. \kdokaz \bzgled \label{MaskeroniProj} Given are three non-collinear points $P$, $Q$ and $R$. Plan the orthogonal projection of the point $P$ onto the line $QR$. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.9.4.pic} \caption{} \label{sl.inv.9.9.4.pic} \end{figure} \textbf{\textit{Solution.}} First, we plan the centers of the line segments $PQ$ and $PR$ (example \ref{MaskeroniSred}) and label them in order with $Q_1$ and $R_1$ (Figure \ref{sl.inv.9.9.4.pic}). We plan the point $P'$ as the second intersection of the circles $k(Q_1,Q_1P)$ and $l(R_1,R_1P)$. The point $P'$ lies on the circles with diameters $PQ$ and $PR$. Therefore $\angle PP'Q=\angle PP'R=90^0$. Therefore the point $R$ lies on the line $P'Q$, which means that the points $P'$, $Q$ and $R$ are collinear. Therefore the point $P'$ is the orthogonal projection of the point $P$ onto the line $QR$. \kdokaz \bzgled \label{MaskeroniOcrt} Given are three non-collinear points $A$, $B$ and $C$. Plan the center of the circumscribed circle and the circumscribed circle of the triangle $ABC$. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.9.5.pic} \caption{} \label{sl.inv.9.9.5.pic} \end{figure} \textbf{\textit{Solution.}} Let $l$ be the circle drawn through the triangle $ABC$ with center in the point $O$ and $\psi_i$ the inversion with center $A$ and arbitrary radius (Figure \ref{sl.inv.9.9.5.pic}). Because $A\in l$, from the equation \ref{InverzKroznVkrozn} it follows, that $l'=\psi_i(l)$ is a line. If $P'$ is the orthogonal projection of the center of inversion $A$ onto the line $l'$ and $P=\psi_i(P')$, then $AP$ is the diameter of the circle $l$ (a consequence of the construction in the proof (\textit{ii}) of the equation \ref{InverzKroznVkrozn}). The proven facts allow us to construct. First, we draw an arbitrary circle $i$ with center in the point $A$ and points $B'=\psi_i(B)$ and $C'=\psi_i(C)$ (example \ref{MaskeroniInv}), then the orthogonal projection $P'$ of the point $A$ onto the line $B'C'$ (example \ref{MaskeroniProj}) and $P=\psi_i(P')$. The center of the drawn circle is obtained as the center of the segment $AP$ (example \ref{MaskeroniSred}). The circle $l(O,A)$ is the circle drawn through the triangle $ABC$. \kdokaz One of the elementary constructions with only a compass represents the construction of the intersection of two circles. With the help of inversion, we will use it for the following elementary construction, which we consider to be a construction with a ruler and a compass. \bzgled \label{MaskeroniKrPr} Given are four points $A$, $B$, $C$ and $D$. Draw the intersection of the lines $AB$ and $CD$. \ezgled \textbf{\textit{Solution.}} Let's assume that the lines $AB$ and $CD$ are not parallel. Let $S$ be any point that does not lie on either of the lines $AB$ and $CD$. It is clear that such a point exists, but an effective construction of such a point can be obtained if we use the construction of points $Q_1$ and $Q_2 $ from the example \ref{MaskeroniNAB} with respect to the distance $AB$. Let $\psi_i$ be the inversion with respect to any circle $i$ with the center at the point $S$ (Figure \ref{sl.inv.9.9.6.pic}). We draw the points $A'=\psi_i(A)$, $B'=\psi_i(B)$, $C'=\psi_i(C)$ and $D'=\psi_i(D)$ (example \ref{MaskeroniInv}) and the inscribed circles of the triangles $SA'B'$ and $SC'D'$ (example \ref{MaskeroniOcrt}) or the images of the lines $AB$ and $CD$ under the inversion $\psi_i$. The point $L'$ is the second intersection of two inscribed circles (the first intersection is the point $S$) It exists in the case when the lines $AB$ and $CD$ are not parallel. The intersection $L$ of the lines $AB$ and $CD$ is the image of the intersection $L'$ of their images - that is $L=\psi_i(L')$ (example \ref{MaskeroniInv}). \kdokaz \bzgled \label{MaskeroniKtKr}Given is a circle $k$ and points $A$ and $B$. Draw the intersection of the line $AB$ and the circle $k$. \ezgled \begin{figure}[!htb] \centering \input{sl.inv.9.9.7.pic} \caption{} \label{sl.inv.9.9.7.pic} \end{figure} \textbf{\textit{Solution.}} The construction can be carried out in the same way as in the previous case, only that $k'$ is the inscribed circle of the triangle determined by the points $C'=\psi_i(C)$, $D'=\psi_i(D)$ and $E'=\psi_i(E)$, where $C$, $D$ and $E$ are any points of the circle $k$ (Figure \ref{sl.inv.9.9.7.pic}). \kdokaz We have proven with the last two constructions, that all elementary constructions related to a line and a compass (see section \ref{elementarneKonstrukcije}), can be done only with a compass. As we have mentioned before, in doing so we count that a line is drawn if two of its points are drawn (the same goes for a distance and a segment), but we cannot use the direct construction of the intersection of two lines or a line and a circle (by using a ruler). From the last two constructions (\ref{MaskeroniKrPr} and \ref{MaskeroniKtKr}), it follows that the latter is also possible only with a compass. Since every design task that can be solved with a ruler and a compass, can be carried out with elementary constructions related to a line and a compass, from the previous it follows, that this task can only be solved with a compass. %NALOGE %________________________________________________________________________________ \naloge{Exercises} \begin{enumerate} \item Prove that the composite of two inversions $\psi_{S,r_1}$ and $\psi_{S,r_2}$ with respect to a concentric circle represents a dilation. Determine the center and coefficient of this dilation. \item Let $A$, $B$, $C$ and $D$ be four collinear points. Construct such points $E$ and $F$, that $\mathcal{H}(A,B;E,F)$ and $\mathcal{H}(C,D;E,F)$ are true. \item In a plane are given a point $A$, a line $p$ and a circle $k$. Draw a circle that goes through the point $A$ and is perpendicular to the line $p$ and the circle $k$. \item Solve the third, fourth, ninth and tenth Apollonius' problem. \item Let $A$ be a point, $p$ a line, $k$ a circle and $\omega$ an angle in some plane. Draw a circle that goes through the point $A$, touches the line $p$ and with the circle $k$ determines the angle $\omega$. \item Determine the geometric location of the points of tangency of two circles that touch the arms of a given angle in two given points $A$ and $B$. \item Draw a triangle, if the following data are known: \begin{enumerate} \item $a$, $l_a$, $v_a$ \item $v_a$, $t_a$, $b-c$ \item $b+c$, $v_a$, $r_b-r_c$ \end{enumerate} \item Let $c(S,r)$ and $l$ be a circle and a line in the same plane that do not have any common points. Let $c_1$, $c_2$, and $c_3$ also be circles in this plane that intersect each other (two at a time) and each of them intersects $c$ and $l$. Express the distance from point $S$ to line $l$ with $r$\footnote{Proposal for MMO 1982. (SL 12.)}. \item Let $ABCD$ be a regular tetrahedron. To any point $M$ on edge $CD$, we assign the point $P = f(M)$ which is the intersection of the rectangle through point $A$ on line $BM$ and the rectangle through point $B$ on line $AM$. Determine the geometric position of all points $P$, if point $M$ takes on all values on edge $CD$. \item Let $ABCD$ be a tetrahedron with perpendicular edges and $P$, $Q$, $R$, and $S$ be the points of tangency of sides $AB$, $BC$, $CD$, and $AD$ with the inscribed circle of this tetrahedron. Prove that $PR\perp QS$. \item Prove that the centers of a tetrahedron with perpendicular edges, the centers of the inscribed and circumscribed circles, and the intersection of its diagonals are collinear points (\index{izrek!Newtonov}Newton's Theorem\footnote{\index{Newton, I.}\textit{I. Newton} (1643--1727), English physicist and mathematician}). \item Let $p$ and $q$ be parallel tangents to circle $k$. Circle $c_1$ intersects line $p$ in point $P$ and circle $k$ in point $A$, circle $c_2$ intersects line $q$ and circles $k$ and $c_1$ in points $Q$, $B$, and $C$. Prove that the intersection of lines $PB$ and $AQ$ is the center of triangle $ABC$ of the circumscribed circle. \item Circles $c_1$ and $c_3$ intersect each other from the outside in point $P$. Similarly, circles $c_2$ and $c_4$ also intersect each other from the outside in the same point. Circle $c_1$ intersects circles $c_2$ and $c_4$ in points $A$ and $D$, while circle $c_3$ intersects circles $c_2$ and $c_4$ in points $B$ and $C$. Prove that the following holds\footnote{Proposal for MMO 2003. (SL 16.)}: $$\frac{|AB|\cdot|BC|}{|AD|\cdot|DC|}=\frac{|PB|^2}{|PD|^2}.$$ \item Let $A$ be a point that lies on the circle $k$. With only a compass, draw the square $ABCD$ (or its vertices), which is inscribed in the given circle. \item Given are the points $A$ and $B$. With only a compass, draw such a point $C$, that $\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}$. \item With only the help of a compass, divide the given line segment in the ratio $2:3$. \end{enumerate} % DEL 10 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - %________________________________________________________________________________ % REŠITVE IN NAPOTKI %________________________________________________________________________________ \del{Solutions and Hints} \footnotesize %REŠITVE - Aksiomi evklidske geometrije ravnine %________________________________________________________________________________ \poglavje{Introduction} There are no exercises in the first chapter. \color{viol4}$\ddot\smile$ \normalcolor %smile \poglavje{Axioms of Planar Euclidean Geometry} \begin{enumerate} \item \res{Let $P$, $Q$ and $R$ be the inner points of the sides of the triangle $ABC$. Prove that $P$, $Q$ and $R$ are non-collinear.} Assume the opposite, that points $P$, $Q$ and $R$ lie on a line $l$. This means that the line $l$ intersects all three sides of the triangle $ABC$ and by assumption does not go through any of its vertices. This is in contradiction with the consequence of \ref{PaschIzrek} Pasch's axiom \ref{AksPascheva}, therefore $P$, $Q$ and $R$ are non-collinear points. \item \res{Let $P$ and $Q$ be points of sides $BC$ and $AC$ of the triangle $ABC$ and at the same time different from its vertices. Prove that the line segments $AP$ and $BQ$ intersect in one point.} From the assumption that the point $P$ lies on the side $BC$ of the triangle $ABC$, according to the definition of the distance $BC$ it holds $\mathcal{B}(B,P,C)$. From the axiom \ref{AksII2} it follows that it does not hold $\mathcal{B}(B,C,P)$, which means that the point $B$ does not lie on the line $PC$. In the triangle $APC$ the line $BQ$ intersects the side $AC$, but does not intersect the sides $PC$, so by the consequence \ref{PaschIzrek} of Pasch's axiom \ref{AksPascheva} the line $BQ$ intersects the side $AP$ of this triangle. So the line $BQ$ intersects the line $AP$ in some point $X$ or $BQ\cap [AP]=\{X\}$. Analogously it is also $[BQ]\cap AP=\{\widehat{X}\}$. We prove that $X=\widehat{X}$. We assume the opposite, that it holds $X\neq \widehat{X}$. Because $[AP]\subset AP$ and $[BQ]\subset BQ$, it holds $X,\widehat{X}\in AP\cap BQ$. By the axiom \ref{AksI1} $AP$ and $BQ$ represent the same line or the points $A$, $B$, $P$ and $Q$ lie on the same line $l$. But on this line, which is determined by the points $B$ and $P$ (axiom \ref{AksI1}), according to the assumption also the point $C$ lies, which means that the points $A$, $B$ and $C$ are collinear. But this is in contradiction with the assumption that $ABC$ is a triangle, so $X=\widehat{X}$. This means $X\in [AP]\cap[BQ]=\{X\}$. If the line $[AP]$ and $[BQ]$ had one more common point, that point would also be the second common point of the lines $AP$ and $BQ$. We have shown that this is not possible, so $[AP]\cap[BQ]=\{X\}$. \item \res{The points $P$, $Q$ and $R$ lie in order on the sides $BC$, $AC$ and $AB$ of the triangle $ABC$ and are different from its vertices. Prove that the lines $AP$ and $QR$ intersect in one point.} We use the previous task first for the lines $AP$ and $BQ$ in the triangle $ABC$, and then once again for the line $AX$ ($\{X\}=[AP]\cap[BQ]$) and $QR$ in the triangle $ABQ$. \item \res{The line $p$, which lies in the plane of the quadrilateral, intersects its diagonal $AC$ and does not pass through any vertex of this quadrilateral. Prove that the line $p$ intersects exactly two sides of this quadrilateral.} We use the consequence \ref{PaschIzrek} of Pasch's axiom \ref{AksPascheva} twice for the line $p$ and the triangle $ABC$ and $ADC$. \item \label{nalAks3}\res{Prove that the midline is a convex figure.} Let $X$ and $Y$ be any points of the midline $pA$. It needs to be proven that the whole line segment $XY$ lies within this midline. By definition of the midline from $X,Y\in pA$ it follows that $X,A\ddot{-} p$ and $Y,A\ddot{-} p$. Because $\ddot{-} p$ is an equivalence relation, it is also true that $X,Y\ddot{-} p$. Let $Z$ be any point of the line segment $XY$. We assume that it is not the case that $Z,A\ddot{-} p$ or that $Z,A\div p$ or $Z,X\div p$ (or $Z \in p$) holds. In this case, the line segment $ZX$ intersects the line $p$ at some point $T$. The point $T$ is then the common point of the line $p$ and the line segment $XY$ (according to statement \ref{izrekAksIIDaljica}), which contradicts what has been proven about $X,Y\ddot{-} p$. Therefore, $Z,A\ddot{-} p$ holds, which means that $Z\in pA$, which in turn means that the midline $pA$ is a convex figure. \item \label{nalAks4} \res{Prove that the intersection of two convex figures is a convex figure.} Let $\Phi_1$ and $\Phi_2$ be convex figures. We prove that then $\Phi_1\cap\Phi_2$ is also a convex figure. Let $A, B\in \Phi_1\cap\Phi_2$ be any points. In this case, $A, B\in \Phi_1$ and $A,B\in\Phi_2$. Because the figures $\Phi_1$ and $\Phi_2$ are convex, it follows that $[AB]\subseteq\Phi_1$ and $[AB]\subseteq\Phi_2$ or $[AB]\subseteq\Phi_1\cap\Phi_2$, which means that $\Phi_1\cap\Phi_2$ is also a convex figure. \item \res{Prove that any triangle is a convex figure.} The triangle $ABC$ is the intersection of the midlines $ABC$, $ACB$ and $BCA$, so according to the previous two tasks \ref{nalAks3} and \ref{nalAks4}, it is a convex figure. \item \res{If $\mathcal{B}(A,B,C)$ and $\mathcal{B}(D,A,C)$, then $\mathcal{B}(B,A,D)$ is also true. Prove it.} According to the statement \ref{izrekAksIIPoltrak}, the point $A$ on the line $AB$ determines two poltraks. We denote with $p$ the poltrak $AC$ and with $p'$ its complementary (supplementary) poltrak. From $\mathcal{B}(A,B,C)$ it follows that $B,C\ddot{-} A$ from $\mathcal{B}(D,A,C)$ we have $D,C\div A$, therefore $B\in p$ and $D\in p'$. So the points $B$ and $D$ not only lie on the same poltrak with the starting point $A$, so there is no $B,D\ddot{-} A$, which means that $B,D\div A$ or $\mathcal{B}(B,A,D)$ applies. \item \res{Let $A$, $B$, $C$ and $D$ be such collinear points that $\neg\mathcal{B}(B,A,C)$ and $\neg\mathcal{B}(B,A,D)$. Prove that $\neg\mathcal{B}(C,A,D)$ applies.} From $\neg\mathcal{B}(B,A,C)$ and $\neg\mathcal{B}(B,A,D)$ it follows that $B,C\ddot{-} A$ or $B,D\ddot{-} A$. Because $\ddot{-} A$ is an equivalence relation, it is also transitive, so $C,D\ddot{-} A$ or $\neg\mathcal{B}(C,A,D)$ applies. \item \res{Let $A_1A_2\ldots,A_{2k+1}$ be an arbitrary polygon with an even number of vertices. Prove that there is no line that intersects all of its sides.} Assume the opposite, that some line $p$ intersects all sides of this polygon. Prove that in this case the points $A_1,A_3,\ldots,A_{2k+1}$ would be on the same side of the line $p$. \item \label{nalAks11}\res{If the isometry $\mathcal{I}$ maps the figure $\Phi_1$ and $\Phi_2$ into the figure $\Phi'_1$ and $\Phi'_2$, then the intersection $\Phi_1\cap\Phi_2$ with this isometry is mapped into the intersection $\Phi'_1\cap\Phi'_2$. Prove it.} If for an arbitrary point $X$ it holds that $X\in \Phi_1\cap\Phi_2$, then $X\in \Phi_1$ and $X\in\Phi_2$. From this it follows that $\mathcal{I}(X)\in \mathcal{I}(\Phi_1)$ and $\mathcal{I}(X)\in \mathcal{I}(\Phi_2)$ or $X'\in \Phi'_1$ and $X'\in \Phi'_2$ (where $X'=\mathcal{I}(X)$). So it holds that $X'\in\Phi'_1\cap\Phi'_2$. In this way we have proven $\mathcal{I}(\Phi_1\cap\Phi_2)\subseteq \Phi'_1\cap\Phi'_2$. The other inclusion $\mathcal{I}(\Phi_1\cap\Phi_2)\supseteq \Phi'_1\cap\Phi'_2$ is proven in the same way using the isometry $\mathcal{I}^{-1}$. \item \res{Prove that any two lines of a plane are parallel to each other.} We use axiom \ref{aksIII2} and theorem \ref{izrekIzoB}. \item \res{Prove that any two lines of a plane are concurrent to each other.} We use axiom \ref{aksIII2} and \ref{aksIII1}. \item \res{Let $k$ and $k'$ be two circles of a plane with centers $O$ and $O'$ and radii $AB$ and $A'B'$. Prove the equivalence: $k\cong k' \Leftrightarrow AB\cong A'B'$.} ($\Leftarrow$) Let $AB\cong A'B'$. We mark with $P$ and $P'$ any points of the circles $k$ and $k'$. Because $OP$ and $O'P'$ are radii of these circles, it holds that $OP\cong O'P'$. By theorem \ref{izrekAB} there exists an isometry $\mathcal{I}$, which maps the points $O$ and $P$ to the points $O'$ and $P'$. Let $X$ be an arbitrary point of the circle $k$ and $X'=\mathcal{I}(X)$. From $\mathcal{I}:O,X,P\rightarrow O',X',P'$ it follows that $O'X'\cong OX$. Because $X,P\in k$, it also holds that $OX\cong OP$. From the previous relations it follows that $O'X'\cong O'P'$ or $X'\in k'$. So it holds that $\mathcal{I}(k)\subseteq k'$. Similarly $\mathcal{I}(k)\supseteq k'$ (we use $\mathcal{I}^{-1}$), so $\mathcal{I}(k)= k'$. ($\Rightarrow$) Let $k\cong k'$. We assume the opposite - that $AB\not\cong A'B'$. Without loss of generality, let $AB>A'B'$. From $k\cong k'$ it follows that there exists an isometry $\mathcal{I}$, which maps the circle $k$ to the circle $k'$. Let $PQ$ be any diameter of the circle $k$. By \ref{premerInS} we have $PQ=2\cdot AB$. Let $P'=\mathcal{I}(P)$ and $Q'=\mathcal{I}(Q)$. Therefore $P'Q'$ is the diameter of the circle, for which $P'Q'\cong PQ =2\cdot AB>2\cdot A'B'$. But the relation $P'Q'>2\cdot A'B'$ is not possible (\ref{premerNajdTetiva}), so $AB\cong A'B'$. \item \res{Let $\mathcal{I}$ be a non-identical isometry of a plane with two fixed points $A$ and $B$. Let $p$ be a line of this plane, which is parallel to the line $AB$, and $A\notin p$. Prove that there are no fixed points of the isometry $\mathcal{I}$ on the line $p$.} We assume the opposite - that there exists a fixed point $C$ of the isometry $\mathcal{I}$, which lies on the line $p$. The point $C$ does not lie on the line $AB$, because in the opposite case $AB$ and $p$ would be parallel, which is in contradiction with the assumption $A\notin p$. So we have three non-collinear fixed points of the isometry $\mathcal{I}$, so by \ref{IizrekABC2} $\mathcal{I}$ is identical. This is in contradiction with the assumption, so there are no fixed points of the isometry $\mathcal{I}$ on the line $p$. \item \res{ Let $S$ be the only fixed point of the isometry $\mathcal{I}$ in some plane. Prove that if this isometry maps the line $p$ to itself, then $S\in p$.} We assume that $S\notin p$. We mark with $N$ the orthogonal projection of the point $S$ on the line $p$ and $N'=\mathcal{I}(N)$. From $S\notin p$ it follows that $N\neq S$. Because $S$ is the only fixed point of this isometry, $N'\neq N$. By the assumption $\mathcal{I}:p\rightarrow p$ we also have $N'\in p$. From $\mathcal{I}:SN,p\rightarrow SN', p$ it follows that $\angle SN',p=\angle SN,p=90^0$. In this case, there would be two different perpendiculars from the point $S$ on the line $p$, which by \ref{enaSamaPravokotnica} is not possible. Therefore the assumption $S\notin p$ is false, which means that $S\in p$. \item \label{nalAks17}\res{Prove that any two lines of a plane either intersect or are parallel. } By Axiom \ref{AksI1}, two different lines $p$ and $q$ of a plane have at most one common point. If they have one common point, they intersect by definition, if they don't have any common points, they are parallel by definition. \item \label{nalAks18}\res{If a line in a plane intersects one of two parallel lines of the same plane, then it also intersects the other parallel line. Prove.} Let $p\parallel q$ and $l\cap p =\{A\}$. If $p=q$, then it is clear that $l\cap q =\{A\}$. The other possibility is $p\cap q =\emptyset$. Then, by Playfair's Axiom, $l\cap q \neq\emptyset$. Because it can't be that $l=q$, then by the previous task \ref{nalAks17} the lines $l$ and $q$ intersect. \item \res{Prove that every isometry maps parallel lines to parallel lines.} Use task \ref{nalAks11}. \item \res{Let $p$, $q$ and $r$ be such lines of a plane that $p\parallel q$ and $r\perp p$. Prove that $r\perp q$.} The statement is a direct consequence of task \ref{nalAks18} and Theorem \ref{KotiTransverzala}. \item \res{Prove that a convex $n$-gon can't have more than three acute angles.} Use the fact that the sum of the exterior angles of a convex $n$-gon is equal to $360^0$ (Theorem \ref{VsotKotVeckZuna}). \end{enumerate} %REŠITVE - Skladnost trikotnikov %________________________________________________________________________________ \poglavje{Congruence. Triangles and Polygons} \begin{enumerate} \item \res{Let $S$ be a point that lies in the angle $pOq$, and let $A$ and $B$ be the orthogonal projections of point $S$ onto the sides $p$ and $q$ of this angle. Prove that $SA\cong SB$ if and only if the line $OS$ is the angle bisector of the angle $pOq$.} Prove the congruence of triangles $OSA$ and $OSB$ in both directions of equivalence. In the proof for ($\Rightarrow$) use Theorem \textit{SSA} \ref{SSK}, in the proof for ($\Leftarrow$) use Theorem \textit{ASA} \ref{KSK}. \item \res{Prove that the sum of the diagonals of a convex quadrilateral is greater than the sum of its opposite sides.} We use the triangle inequality for the triangles $ASB$ and $CSD$, where $S$ is the intersection of the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$. \item \res{Prove that in every triangle no more than one side is shorter than the corresponding altitude.} We mark with $a$, $b$ and $c$ the sides and with $v_a$, $v_b$, $v_c$ the corresponding altitudes of the triangle $ABC$. We assume the opposite, that for example $a \frac{3}{4}(a + b + c)$ \end{enumerate}} We use example \ref{neenTezisZgl}. \item \res{Let $p$ be a line that is parallel to the circle $k$. Prove that all points of this circle are on the same side of the line $p$.} We assume the opposite, that points $X,Y\in k$ are on different sides of the line $p$. In this case, the segment $XY$ intersects the line $p$ in some point $N$, which, according to \ref{tetivaNotrTocke}, is an interior point of the circle $k$. The line $p$, which contains the interior point $N$ of the circle $k$, is, according to \ref{DedPoslKrozPrem}, its secant, which contradicts the assumption that $p$ is parallel. \item \res{If the circle $k$ lies in some convex figure $\Phi$, then also the circle, which is determined by this circle, lies in this figure. Prove it.} It is enough to prove that any interior point $X$ of the circle $k$ lies in the figure $\Phi$. Let $AB$ be any segment that contains the point $X$ (the existence of such a segment follows from \ref{DedPoslKrozPrem}). Because $\Phi$ is a convex figure, from $A,B\in k\subset \Phi$ it follows that $X\in [AB]\subset \Phi$. \item \res{Let $p$ and $q$ be two different tangents to the circle $k$, which touch it in points $P$ and $Q$. Prove the equivalence: $p \parallel q$ if and only if $AB$ is the diameter of the circle $k$.} We use the fact $SP\perp p$ and $SQ\perp q$, where $S$ is the center of the circle $k$. \item \res{If $AB$ is a segment of the circle $k$, then the intersection of the line $AB$ and the circle, which is determined by the circle $k$, is equal to this segment. Prove it.} We denote by $\mathcal{K}$ the mentioned circle. It is necessary to prove that $[AB]=AB\cap \mathcal{K}$. The inclusion $[AB]\subseteq AB\cap \mathcal{K}$ follows directly from the axiom \ref{AksII1} and \ref{tetivaNotrTocke}. We prove the inclusion $AB\cap \mathcal{K}\subseteq [AB]$. Let $X\in AB\cap \mathcal{K}$. We assume that $X\notin [AB]$. In this case, it would be $\mathcal{B}(X,A,B)$ or $\mathcal{B}(A,B,X)$. It is not difficult to prove that in each of these two cases we would get $SX>SA$, which is not possible, because $X\in \mathcal{K}$. Therefore, $X\in[AB]$, i.e. $AB\cap \mathcal{K}\subseteq [AB]$. \item \res{Let $S'$ be the orthogonal projection of the center $S$ of the circle $k$ onto the line $p$. Prove that $S'$ is an external point of this circle exactly when the line $p$ does not intersect the circle.} We use the statement \ref{TangSekMimobKrit}. \item \res{Let $V$ be the altitude point of the triangle $ABC$, for which $CV \cong AB$. Determine the size of the angle $ACB$.} We mark with $A'$, $B'$ and $C'$ the points of the altitudes from the vertices $A$, $B$ and $C$ of the triangle $ABC$. First, the angles $C'CB$ and $A'AB$ are congruent (the angle with perpendicular legs - statement \ref{KotaPravokKraki}). From the congruence of the triangles $CVA'$ and $ABA'$ (statement \textit{ASA} \ref{KSK}) it follows that $CA'\cong AA'$. This means that $CAA'$ is an isosceles right triangle with hypotenuse $AC$, so according to statement \ref{enakokraki} $\angle ACB=\angle ACA'=45^0$. \item \res{Let $CC'$ be the altitude of the right triangle $ABC$ ($\angle ACB = 90^0$). If $O$ and $S$ are the centers of the inscribed circles of the triangles $ACC'$ and $BCC'$, the internal angle bisector of $ACB$ is perpendicular to the line $OS$. Prove it.} Let $I$ be the center of the inscribed circle of the triangle $ABC$. We prove that $I$ is the altitude point of the triangle $COS$. \item \res{Let $ABC$ be a triangle in which $\angle ABC = 15^0$ and $\angle ACB = 30^0$. Let $D$ be a point on the side $BC$ such that $\angle BAD=90^0$. Prove that $BD = 2AC$.} We mark with $S$ the center of the segment $BD$. The point $S$ is the center of the inscribed circle of the right triangle $BAD$ (statement \ref{TalesovIzrKroz2}). Therefore $SA\cong SB\cong SD$. Because $BSA$ is an isosceles triangle with the base $AB$, according to statement \ref{enakokraki} $\angle BAS\cong \angle ABS=15^0$. For the external angle $ASD$ of the triangle $BSA$ it then holds that $\angle ASD=\angle BAS + \angle ABS=30^0$ (statement \ref{zunanjiNotrNotr}). Therefore $\angle ASC=30^0=\angle ACS$, so $ASC$ is an isosceles triangle with the base $SC$ or $AS\cong AC$ (statement \ref{enakokraki}). From this it follows that $BD = 2SB=2AS=2AC$. \item \res{Prove that there exists a pentagon that can tile the plane.} We can choose a pentagon $ABCDE$, so that $ABCE$ is a square, $ECD$ is an isosceles right triangle with the base $CE$, and $A,D\div CE$. \item \res{Prove that there exists a decagon, with which it is possible to tile a plane.} We make the union of two appropriate regular hexagons - the cells of tiling $(6,3)$. \item \res{In some plane each point is painted red or black. Prove that there exists a right triangle, which has all vertices of the same color.} We use the tiling $(3,6)$. \item \res{Let $l_1,l_2,\ldots, l_n$ ($n > 3$) be arcs, which all lie on the same circle. The central angle of each arc is at most $180^0$. Prove that if each triple of arcs has at least one common point, there exists a point, which lies on each arc.} We denote with $k(S,r)$ the given circle. Let $I_i$ ($i\in \{1,2,\ldots,n\}$) be the circular segments, which are determined by the arcs $l_k$, and $J_i=I_i\setminus\{S\}$. Because for each $l_i$ the central angle is at most $180^0$, the sets $J_i$ are convex figures (prove it). By the assumption each triple of arcs $l_{i_1}$, $l_{i_2}$ and $l_{i_2}$ ($i_1,i_2,i_3\in\{1,2,\ldots,n\}$) has a common point $X_{i_1i_2i_3}$. This point also belongs to each of the figures $J_{i_1}$, $J_{i_2}$ and $J_{i_2}$. By Helly's theorem \ref{Helly} there exists a point $Y$, which belongs to each of the figures $J_1,J_2,\ldots, J_n$. If we denote with $X$ the intersection of the line segment $SY$ and the circle $k$, it follows that also the open line segment $(SX]$ belongs to each of these figures. Because for each $i$ it holds $l_i=J_i\cap k$, the point $X$ belongs to each of the arcs $l_1,l_2,\ldots, l_n$. \item \res{Let $p$ and $q$ be rectangles that intersect at point $A$. If $B, B'\in p$, $C, C'\in q$, $AB\cong AC'$, $AB'\cong AC$, $\mathcal{B}(B,A,B')$ and $\mathcal{B}(C,A,C')$, then the rectangle on the line $BC$ through point $A$ goes through the center of the line $B'C'$. Prove it.} First, by the \textit{SAS} theorem \ref{SKS}, the triangles $BAC$ and $C'AB'$ are congruent. From this it follows: $\angle AC'B'\cong\angle ABC=\beta$ and $\angle AB'C'\cong\angle ACB=90^0-\beta$. We mark with $S$ the center of the line $B'C'$, with $P$ the intersection of the lines $BC$ and $AS$. By theorem \ref{TalesovIzrKroz2}, $SA\cong SB'\cong SC'$, so $\angle CAP=\angle C'AS\cong\angle AC'S=\angle AC'B'=\beta$ (theorem \ref{enakokraki}). Because $\angle ACP=\angle ACB=90^0-\beta$, from triangle $ACP$ by theorem \ref{VsotKotTrik} $\angle APC=90^0$ or $AS\perp BC$. This means that the rectangle $AP$ of the line $BC$ goes through the center of the line $B'C'$. \item \res{Prove that the internal angle bisectors of a rectangle that is not a square intersect at points that are the vertices of a square.} We mark with $s_{\alpha}$, $s_{\beta}$, $s_{\gamma}$ and $s_{\delta}$ the internal angle bisectors at the vertices $A$, $B$, $C$ and $D$ of the rectangle $ABCD$ and $P=s_{\alpha}\cap s_{\beta}$, $Q=s_{\gamma}\cap s_{\delta}$, $L=s_{\beta}\cap s_{\gamma}$ and $K=s_{\alpha}\cap s_{\delta}$. We prove that $PKQL$ is a square. First, from $\angle PAB=45^0$ and $\angle PBA=45^0$ it follows that $ABP$ is an isosceles right triangle with the base $AB$ (theorem \ref{enakokraki} and \ref{VsotKotTrik}). Therefore $\angle APB=90^0$ and $AP\cong BP$. Analogously, all the internal angles of the quadrilateral $PKQL$ are right angles. It is enough to prove that $PK\cong PL$. From the congruence of the triangles $AKD$ and $BLC$ (theorem \ref{KSK}) it follows that $AK\cong BL$. If we connect this with the already proven $AP\cong BP$, we get $PK\cong PL$. \item \res{Prove that the internal angle bisectors of a parallelogram that is not a rhombus intersect at the vertices of a rectangle. Prove also that the diagonals of this rectangle are parallel to the sides of the parallelogram and are equal to the difference of the adjacent sides of the parallelogram.} As in the previous task, we denote by $s_{\alpha}$, $s_{\beta}$, $s_{\gamma}$ and $s_{\delta}$ the internal angle bisectors at the vertices $A$, $B$, $C$ and $D$ of the rectangle $ABCD$ and $P=s_{\alpha}\cap s_{\beta}$, $Q=s_{\gamma}\cap s_{\delta}$, $L=s_{\beta}\cap s_{\gamma}$ and $K=s_{\alpha}\cap s_{\delta}$. We prove that $PKQL$ is a rectangle. Let $E$ and $F$ be the midpoints of the sides $AD$ and $BC$ of the parallelogram $ABCD$. In the triangle $PAB$ we have $\angle APB=180^0-\angle PAB-\angle PBA=180^0-\frac{1}{2}\left(\angle DAB+\angle CBA \right)=180^0-\frac{1}{2}\cdot 180^0=90^0$ (by the \ref{VsotKotTrik} and \ref{paralelogram}). Similarly, all the other internal angles of the quadrilateral $PKQL$ are right angles. By the \ref{TalesovIzrKroz2} the point $E$ is the center of the inscribed circle of the right triangle $AKD$ with hypotenuse $AD$ and we have $EK\cong EA$. Therefore $\angle EKA\cong \angle EAK \cong\angle KAB$ (by the \ref{enakokraki}) or, by the \ref{KotiTransverzala}, $EK\parallel AB$. Similarly, $FL\parallel AB$. Since also $EF\parallel AB$ (by the \ref{srednjTrapez}) we have $KL\parallel AB$. Without loss of generality, we assume that $AB>AD$. We denote by $T$ the intersection of the bisector $s_{\alpha}$ with the side $CD$. By the \ref{KotiTransverzala} we have $\angle DTA\cong\angle TAB\cong\angle DAT$. This means that $ADT$ is an isosceles triangle, therefore $AD\cong DT$ (by the \ref{enakokraki}). Since $KL\parallel AB\parallel CT$ and $s_{\alpha}\parallel s_{\gamma}$, the quadrilateral $KLCT$ is a parallelogram. Therefore $PQ\cong KL\cong CT=CD-DT=CD-AB$. \item \res{Prove that the angle bisectors of two adjacent angles are perpendicular to each other.} The angle bisectors determine the angle that is equal to half of the corresponding extended angle. \item \res{Let $B'$ and $C'$ be the altitudes of the triangle $ABC$ from the vertices $B$ and $C$. Prove that $AB\cong AC \Leftrightarrow BB'\cong CC'$.} Prove the equivalence in both directions. In the direction ($\Rightarrow$) use the \textit{ASA} theorem \ref{KSK}, in the direction ($\Leftarrow$) use the \textit{SSA} theorem \ref{SSK}. \item \res{Prove that a triangle is right if and only if the center of the circle drawn through the triangle and its altitude point coincide. Does a similar statement hold for any two characteristic points of this triangle?} Use the fact that the sides of the triangle are perpendicular to the altitude in this case. In a similar way, one could prove that a similar statement holds for any two characteristic points of this triangle. \item \res{Prove that the obtuse triangles $ABC$ and $A'B'C'$ are congruent if and only if they have congruent altitudes $CD$ and $C'D'$, sides $AB$ and $A'B'$, and angles $ACD$ and $A'C'D'$.} From the congruence of the triangles $ABC$ and $A'B'C'$ it follows directly first that $AB\cong A'B'$, $AC\cong A'C'$, and $\angle BAC\cong \angle B'A'C'$, then that $\triangle ACD\cong\triangle A'C'D'$ (by the \textit{ASA} theorem \ref{KSK}) or $CD\cong C'D'$, and $\angle ACD\cong \angle A'C'D'$. Assume that $CD\cong C'D'$, $AB\cong A'B'$, and $\angle ACD\cong \angle A'C'D'$. From the congruence of the triangles $ACD$ and $A'C'D'$ (by the \textit{ASA} theorem \ref{KSK}) it follows that there exists an isometry $\mathcal{I}$, for which $\mathcal{I}:A,C,D \mapsto A',C',D'$ (by theorem \ref{IizrekABC}. This maps the segment $AD$ onto the segment $A'D'$. Because $B$ and $B'$ lie on the segments $AD$ and $A'D'$, and $AB\cong A'B'$, also $\mathcal{I}(B)=B'$. Therefore $\mathcal{I}:A,B,C \mapsto A',B',C'$, which means that $\triangle ABC\cong\triangle A'B'C'$. \item \res{If $ABCD$ is a rectangle and $AQB$ and $APD$ are right triangles with the same orientation, then the segment $PQ$ is congruent to the diagonal of this rectangle. Prove it.} We prove that $PAQ$ and $DAB$ are congruent triangles. \item \res{Let $BB'$ and $CC'$ be the altitudes of the triangle $ABC$ ($AC>AB$) and $D$ be such a point on the line segment $AB$, that $AD\cong AC$. The point $E$ is the intersection of the line $BB'$ with the line that goes through the point $D$ and is parallel to the line $AC$. Prove that $BE=CC'-BB'$.} Let $F$ be the intersection of the line $ED$ with the line that is in the point $B'$ parallel to the line $AB$. The quadrilateral $ADFB'$ is a parallelogram, so $FB'\cong DA$ and $\angle DFB' \cong\angle DAB'$ (by the statement \ref{paralelogram}). Because according to the assumption $AD\cong AC$, it also holds that $FB'\cong AC$. This means that the right-angled triangles $FEB'$ and $AC'C$ are similar (by the \textit{ASA} statement \ref{KSK}), so $EB'\cong C'C$ or $BE=EB'-BB'=CC'-BB'$. \item \res{Let $ABCD$ be a convex quadrilateral, for which $AB\cong BC\cong CD$ and $AC\perp BD$. Prove that $ABCD$ is a rhombus.} Because $ABCD$ is a convex quadrilateral, its diagonals intersect in some point $S$. Prove $\triangle ABS\cong\triangle CBS$ and $\triangle CBS\cong\triangle CDS$. \item \res{Let $BC$ be the base of the isosceles triangle $ABC$. If $K$ and $L$ are such points, that $\mathcal{B}(A,K,B)$, $\mathcal{B}(A,C,L)$ and $KB\cong LC$, then the center of the line segment $KL$ lies on the base $BC$. Prove.} Let $O$ be the center of the line $KL$. We mark with $M$ the fourth vertex of the parallelogram $CKBM$, the common center of their diagonals $BC$ and $KM$ (statement \ref{paralelogram}) with $S$. If we use the fact that the triangles $ABC$ and $MLC$ are of the same base with the bases $BC$ and $ML$, from the statement \ref{enakokraki} we get $\angle ABC\cong\angle ACB$ and $\angle CML\cong\angle CLM$. From the congruence of the triangles $BKC$ and $CMB$ (statement \ref{paralelogram} and \textit{SSS} \ref{SSS}) it follows that $\angle ABC=\angle KBC\cong\angle MCB$. As $ACM$ is the external angle of the triangle $MLC$, then by the statement \ref{zunanjiNotrNotr} $\angle ACM =\angle CML+\angle CLM=2\angle CML$. Therefore $\angle CML=\frac{1}{2}\angle ACM=\frac{1}{2}\left(\angle ACB+\angle BCM\right)\frac{1}{2}\left(\angle ACB+\angle ABC\right)=\angle ABC$. This means that the lines $ML$ and $BC$ are parallel (statement \ref{KotiTransverzala}). The line $OS$ is the median of the triangle $KML$ with the base $ML$, therefore by the statement \ref{srednjicaTrik} $SO\parallel MN$. By Playfair's axiom \ref{Playfair} the lines $SO$ and $BC$ are the same (coincide), therefore the point $O$ lies on the line $BC$. By Pasch's axiom \ref{AksPascheva} for the triangle $ABC$ and the line $KL$ the point $O$ lies on the side $BC$. \item \res{Let $S$ be the center of the triangle $ABC$ of the inscribed circle. The line passing through the point $S$ and parallel to the side $BC$ of this triangle intersects the sides $AB$ and $AC$ in order. Prove that $BM+NC=NM$.} We prove that the triangles $BSM$ and $SCN$ are of the same base with the bases $BS$ and $SC$. \item \res{Let $ABCDEFG$ be a convex heptagon. Calculate the sum of the convex angles determined by the broken line $ACEGBDFA$.} We use the fact that the double sum of the external angles of the heptagon determined by the intersections of the diagonals of the heptagon $ABCDEFG$ is equal to $720^0$. Result: $540^0$. \item \label{nalSkl34} \res{Prove that the centers of the sides and the vertex of any altitude of a triangle in which no two sides are congruent are the vertices of a parallelogram.} We denote with $A_1$, $B_1$ and $C_1$ the centers of sides $BC$, $CA$ and $BA$ of the triangle $ABC$ and with $A'$ the intersection point of the altitude of this triangle with the point $A$. From $|AB|\neq |AC|$ it follows that $A'\neq A_1$. We prove that the quadrilateral $A'A_1B_1C_1$ is a parallelogram. The distance $B_1C_1$ is the median of the triangle $ABC$ with the base $BC$, therefore by the statement \ref{srednjicaTrik} $B_1C_1\parallel BC$. Because $A',A_1\in BC$, also $B_1C_1\parallel A',A_1$. So the quadrilateral $A'A_1B_1C_1$ is a parallelogram. We prove also that it is a rhombus, or $C_1A'\cong B_1A_1$. But this fact follows from the statements \ref{TalesovIzrKroz2} and \ref{srednjicaTrik}, because: $C_1A'=\frac{1}{2}AB= B_1A_1$. We mention also that by using this statement we can another way prove the statement about Euler's circle \ref{EulerKroznica}. \item \res{Let $ABC$ be a right angled triangle with the right angle at the point $C$. The points $E$ and $F$ are the intersections of the internal angle bisectors at the points $A$ and $B$ with the opposite sides, $K$ and $L$ are the orthogonal projections of the points $E$ and $F$ on the hypotenuse of this triangle. Prove that $\angle LCK=45^0$.} Let $\alpha$ and $\beta$ be the inner angles at the vertices $A$ and $B$ of the triangle $ABC$. According to the theorem \ref{VsotKotTrik}, $\alpha+\beta=90^0$. According to the theorem \textit{ASA} \ref{KSK}, $\triangle ACE\cong\triangle AKE$ and $\triangle BCF\cong\triangle BLF$, therefore $EC\cong EK$ and $FC\cong FL$. So the triangles $CEK$ and $CFL$ are congruent with the bases $CK$ and $CL$, which means that $\angle ECK\cong\angle EKC$ and $\angle FCL\cong\angle FLC$. If we use the theorem \ref{zunanjiNotrNotr} for the triangle $CFL$ and the theorem \ref{VsotKotTrik} for the triangle $ALF$, we get: $\angle FCL=\frac{1}{2}\angle LFA=\frac{1}{2}\left(90^0-\alpha \right)$. Similarly, $\angle ECK=\frac{1}{2}\left(90^0-\beta \right)$. From this it follows that $\angle FCL+\angle ECK=\frac{1}{2}\left(90^0-\alpha +90^0-\beta\right)=45^0$. So $\angle LCK=90^0-(\angle FCL+\angle ECK)=90^0-45^0=45^0$. \item \res{Let $M$ be the center of the side $CD$ of the square $ABCD$ and $P$ such a point of the diagonal $AC$, that $3AP=PC$. Prove that $\angle BPM$ is a right angle.} Let $S$ be the center of the diagonal $AC$ and $V$ the center of the line $SB$. Prove that $V$ is the altitude of the triangle $PBC$ (see also the example \ref{zgledPravokotnik}). \item \res{Let $P$, $Q$ and $R$ be the centers of the sides $AB$, $BC$ and $CD$ of the parallelogram $ABCD$. The lines $DP$ and $BR$ intersect the line $AQ$ in the points $K$ and $L$. Prove that $KL= \frac{2}{5} AQ$.} Let $S$ be the center of the line $AD$ and $M$ the intersection of the lines $SC$ and $BR$. Prove first that $CM\cong AK$, the line $LQ$ is the median of the triangle $CBM$ and the line $PK$ is the median of the triangle $LAB$. \item \res{Let $D$ be the center of the hypotenuse $AB$ of the right-angled triangle $ABC$ ($AC>BC$). The points $E$ and $F$ are the intersections of the altitudes $CA$ and $CB$ with a line through $D$ and perpendicular to the line $CD$. The point $M$ is the center of the line $EF$. Prove that $CM\perp AB$.} Let $T$ be the intersection of the lines $AB$ and $CM$, and $\angle CAB=\alpha$ and $\angle CBA=\beta$. By assumption, $\alpha+\beta=180^0$. Because $D$ is the center of the hypotenuse $AB$ of the right triangle $ABC$, by Tales' theorem \ref{TalesovIzrKroz2} $DC\cong DA$. This means that $\triangle CDA$ is an isosceles triangle and $\angle DCE=\angle DCA\cong\angle DAC=\alpha$. From $FE\perp CD$ and $FC\perp CE$ it follows that $\angle CFE\cong\angle DCE=\alpha$ (by \ref{KotaPravokKraki}). The point $M$ is the center of the hypotenuse $FE$ of the right triangle $FCE$, so (similarly to the triangle $ABC$) $\angle FCM\cong\angle CFM=\angle CFE=\alpha$. In the triangle $BCT$ the sum of the two internal angles $\angle CBT+\angle BCT=\angle CBA+\angle FCM=\alpha+\beta=90^0$. By \ref{VsotKotTrik} it follows that $\angle CTB=90^0$ or $CM\perp AB$. \item \res{Let $A_1$ and $C_1$ be the centers of the sides $BC$ and $AB$ of the triangle $ABC$. The perpendicular to the internal angle at the vertex $A$ intersects the line $A_1C_1$ at the point $P$. Prove that $\angle APB$ is a right angle.} By \ref{KotiTransverzala}, $\angle C_1PA\cong\angle PAC$. Therefore $\angle C_1PA\cong\angle PAC\cong\angle PAC_1$, which means that $PAC_1$ is an isosceles triangle with the base $AP$ or $C_1A\cong C_1P$ (by \ref{enakokraki}). Because $C_1$ is the center of the side $AB$, $C_1B\cong C_1A\cong C_1P$. Therefore the point $P$ lies on the circle with the diameter $AB$, so by Tales' theorem \ref{TalesovIzrKroz2} $\angle APB=90^0$. \item \res{Let $P$ and $Q$ be such points of the sides $BC$ and $CD$ of the square $ABCD$, that the line $PA$ is the perpendicular to the angle $BPQ$. Determine the size of the angle $PAQ$.} Let $A'=pr_{\perp PQ}(A)$. Prove first that $\triangle ABP\cong\triangle AA'P$ and $\triangle ADQ\cong\triangle AA'Q$. Result: $\angle PAQ=45^0$. \item \res{Prove that the center of the circumscribed circle lies closest to the longest side of the triangle.} Let $O$ be the center of the inscribed circle of the triangle $ABC$ and $A_1$, $B_1$ and $C_1$ be the centers of its sides $BC$, $AC$ and $AB$. It is enough to prove, for example, that from $AC>AB$ it follows that $OB_1AB$. By the theorem \ref{vecstrveckot} in this case $\angle ABC>\angle ACB$. Because $B_1C_1$ is the median of the triangle $ABC$, $B_1C_1\parallel BC$ (theorem \ref{srednjicaTrik}). Therefore, by the theorem \ref{KotiTransverzala} $\angle AC_1B_1\cong \angle ABC$ and $\angle AB_1C_1\cong \angle ACB$. From $\angle ABC>\angle ACB$ it now follows: $\angle B_1C_1O=90^0-\angle AC_1B_1= 90^0-\angle ABC<90^0-\angle ACB=90^0-\angle AB_1C_1=\angle C_1B_1O$ or $\angle B_1C_1O<\angle C_1B_1O$. By the theorem \ref{vecstrveckot} for the triangle $B_1OC_1$ it is then $OB_1\gamma$ it follows that $SB\gamma$ it follows first of all that $\angle SBP=\frac{1}{2}\beta>\frac{1}{2}\gamma=\angle SCP$. Let $B'$ be the point on the segment $PC$, for which $PB'\cong PB$. From the similarity of the triangles $SPB'$ and $SPB$ (the \textit{SAS} theorem \ref{SKS}) it follows that $SB'\cong SB$ and $\angle SB'P\cong \angle SBP>\angle SCP$. We prove that $\mathcal{B}(P,B',C)$ or $PB'\angle SB'P\cong\angle SBP$ - the \ref{zunanjiNotrNotrVecji} theorem for the triangle $SCB'$). Since $\mathcal{B}(P,B',C)$, it follows that $\angle SB'C=180^0-\angle SB'P=180^0-\frac{1}{2}\beta>90^0$. In the right-angled triangle $SB'C$ it then follows that $SC\gamma$) we mark the central angles. Because according to the statement \ref{TangPogoj} $\angle ST_1P=\angle ST_2P=90^0$ we get (from the quadrilateral $PT_1ST_2$ - statement \ref{VsotKotVeck}): $\angle T_1PT_2=180^0-\gamma= \frac{\beta+\gamma}{2}-\gamma= \frac{\beta-\gamma}{2}$. \item \res{Let $L$ be the orthogonal projection of any point $K$ of the circle $k$ on its tangent $t$ through the point $T\in k$ and $X$ the point that is symmetrical to the point $L$ with respect to the line $KT$. Determine the geometric position of the points $X$.} Let's mark the center of the circle $k$ with $S$ and $\alpha=\angle LTK$. We will first prove that $S$, $X$ and $K$ are collinear points. From the congruence of the triangles $TLK$ and $TXK$ it follows that $\angle XTK\cong \angle LTK=\alpha$, $\angle TXK\cong \angle TLK=90^0$ and $\angle TKX\cong \angle TKL=90^0-\alpha$. According to the formulas \ref{ObodKotTang} and \ref{SredObodKot} $\angle TSK=2\angle LTK=2\alpha$. Therefore, from the isosceles triangle $TSK$ (formula \ref{enakokraki} and \ref{VsotKotTrik}) we get: $\angle TKS=\frac{180^0-2\alpha}{2}=90^0-\alpha=\angle TKX$ or $\angle TKS\cong\angle TKX$, which means that $X\in SK$. In the end $\angle SXT\cong\angle KXT=90^0$, which according to the formula \ref{TalesovIzrKroz2} means that the point $X$ lies on the circle $l$ with the diameter $TS$. It is not difficult to prove that the converse is also true - each point $X\in l$ can be obtained from some point $K\in k$ according to the described procedure. \item \res{Let $BB'$ and $CC'$ be the altitudes of the triangle $ABC$ and $t$ the tangent of the circumscribed circle of this triangle at point $A$. Prove that $B'C'\parallel t$.} Let's mark any point of the tangent $t$ with $P$, so that $P$ and $C$ are on different sides of this tangent. According to the formula \ref{ObodKotTang} $\angle PAC'=\angle PAB\cong\angle ACB$. Because $BCC'B'$ is a trapezoid (formula \ref{TalesovIzrKroz2}), according to the formula \ref{TetivniPogojZunanji} $\angle AC'B'\cong\angle ACB$. Therefore $\angle AC'B'\cong\angle PAC'$, which according to the formula \ref{KotiTransverzala} means that $PA\parallel B'C'$ or $B'C'\parallel t$. \item \res{In the right triangle $ABC$ above the leg $AC$ with the angle as the diameter the circle is drawn, which intersects the hypotenuse $AB$ at point $E$. The tangent of this circle at point $E$ intersects the other leg $BC$ at point $D$. Prove that $BDE$ is an isosceles triangle.} If we use the formula \ref{ObodKotTang}, we get $\angle DEC\cong\angle EAC$. Therefore: $\angle BED=90^0-\angle DEC=90^0-\angle EAC=90^0-\angle BAC=\angle ABC=\angle EBD$. From $\angle BED\cong\angle EBD$ according to the formula \ref{enakokraki} it follows that $BD\cong ED$, which means that $BDE$ is an isosceles triangle. \item \res{In a right angle with the top $A$ is inscribed a circle that touches the sides of this angle in the points $B$ and $C$. Any tangent to this circle intersects the line $AB$ and $AC$ in succession in the points $M$ and $N$ (so that $\mathcal{B}(A,M,B)$). Prove that: $$\frac{1}{3}\left(|AB|+|AC|\right) < |MB|+|NC| < \frac{1}{2}\left(|AB|+|AC|\right).$$} We denote with $T$ the point of intersection of the aforementioned circle with the line $MN$ and $s=\frac{1}{2}(|MN|+|NA|+|AM|)$ the perimeter of the triangle $MAN$. The given circle is the inscribed circle of the triangle $MAN$, so according to the great theorem it holds: \begin{eqnarray} \label{nalSkk8Eqn1} |AB|=|AC|=s. \end{eqnarray} We prove first the inequality $|MB|+|NC| <\frac{1}{2}\left(|AB|+|AC|\right)$. If we use the statement \ref{TangOdsek} and the triangle inequality \ref{neenaktrik} and the relation \ref{nalSkk8Eqn1}, we get: \begin{eqnarray*} |MB|+|NC| &=& |MT|+|NT| =|MN|=\\ &=& \frac{1}{2}\left( |MN|+|MN| \right)\\ &<& \frac{1}{2}\left( |MN|+|NA|+|AM| \right)=\\ &=& s=|AB|=\\ &=& \frac{1}{2}\left(|AB|+|AC|\right). \end{eqnarray*} We prove the second inequality: $|MB|+|NC| >\frac{1}{3}\left(|AB|+|AC|\right)$. Since $MAN$ is a right triangle with hypotenuse $MN$, according to the statement \ref{vecstrveckotHipot} it follows: \begin{eqnarray} \label{nalSkk8Eqn2} |MN|>|MA| \hspace*{1mm} \textrm{ and } \hspace*{1mm} |MN|>|NA|. \end{eqnarray} If we use the statement \ref{TangOdsek} and the relations \ref{nalSkk8Eqn2} and \ref{nalSkk8Eqn1}, we get: \begin{eqnarray*} |MB|+|NC| &=& |MT|+|NT| =|MN|=\\ &=& \frac{1}{3}\left( |MN|+|MN|+|MN| \right)\\ &>& \frac{1}{3}\left( |MN|+|NA|+|AM| \right)=\\ &=& \frac{1}{3}\cdot2s=\frac{1}{3}\cdot2\cdot|AB|=\\ &=& \frac{1}{3}\left(|AB|+|AC|\right). \end{eqnarray*} \item \res{Prove that in a right triangle the sum of the sides is equal to the sum of the diameters of the inscribed and circumscribed circle.} Let $ABC$ be a right angled triangle with a right angle at point $C$. Use the same notation as in the big task \ref{velikaNaloga}. Because $\angle PCQ=\angle BCA=90^0$, the quadrilateral $PCQS$ is a square, so by the big task (\ref{velikaNaloga}) it follows: \begin{eqnarray} \label{nalSkk9Eqn} r=|SQ|=|CP|=s-c. \end{eqnarray} From this and Tales' theorem (\ref{TalesovIzrKroz2}) it follows: $$R+r=\frac{c}{2}+s-c=b+c.$$ \item \res{The lines of symmetry of the internal angles of a convex quadrilateral intersect in six different points. Prove that four of these points are the vertices of the angle bisector quadrilateral.} Use the criterion for the angle bisector - theorem \ref{TetivniPogoj}. \item \res{Let: $c$ be the length of the hypotenuse, $a$ and $b$ be the lengths of the catheti, and $r$ be the radius of the inscribed circle of a right angled triangle. Prove that:} \begin{enumerate} \item \res{$2r + c \geq 2 \sqrt{ab}$} Use the relation \ref{nalSkk9Eqn} and the inequality between arithmetic and geometric mean. \item \res{$a + b + c > 8r$} Use the relation \ref{nalSkk9Eqn} and Pythagoras' theorem \ref{PitagorovIzrek}. \end{enumerate} \item \res{Let $P$ and $Q$ be the centers of the shorter arcs $AB$ and $AC$ of the inscribed circle of a regular triangle $ABC$. Prove that the sides $AB$ and $AC$ of this triangle divide the line $PQ$ into three congruent segments.} Let $X$ and $Y$ be the intersection points of the chord $PQ$ with the sides $AB$ and $AC$ of the triangle $ABC$. Prove that $PX\cong XY\cong YQ$ holds. By the theorem \ref{TockaN}, $BQ$ and $CP$ are the altitudes of the internal angles $ABC$ and $ACB$ of the triangle $ABC$. By the theorem \ref{ObodObodKot} we have: \begin{eqnarray*} \angle PAX &=& \angle PAB\cong\angle PCB=\frac{1}{2}\angle ACB=30^0\\ \angle APX &=& \angle APQ\cong\angle ABQ=\frac{1}{2}\angle ABC=30^0. \end{eqnarray*} Therefore: $\angle PAX = \angle APX=30^0$, which means that $APX$ is an isosceles triangle with the base $AP$ (theorem \ref{enakokraki}) or $PX\cong XA$. By the theorem \ref{zunanjiNotrNotr} we also have $\angle AXY=\angle APX+\angle PAX=60^0$. Since $\angle XAY=\angle BAC= 60^0$, $AXY$ is a right triangle, therefore $XA\cong XY\cong YA$. Similarly, $AQY$ is an isosceles triangle with the base $AQ$ or $QY\cong YA$. If we connect the proven relations of congruence of the segments, we get: $PX\cong XA\cong XY\cong AY\cong YQ$. \item \res{Let $k_1$, $k_2$, $k_3$, $k_4$ be four circles, each of which externally touches one side and two vertices of an arbitrary convex quadrilateral. Prove that the centers of these circles are the points of a concurrence.} First, prove that the two opposite angles of the quadrilateral have measures $\frac{\alpha+\beta}{2}$ and $\frac{\gamma+\delta}{2}$, where $\alpha$, $\beta$, $\gamma$ and $\delta$ are the internal angles of the initial quadrilateral, and then use theorems \ref{TetivniPogoj} and \ref{VsotKotVeck}. \item \res{The circles $k$ and $l$ externally touch in the point $A$. The points $B$ and $C$ are the points of the common external tangent of these two circles. Prove that $\angle BAC$ is a right angle.} Let $L$ be the intersection of the common tangent of the circles $k$ and $l$ in the point $A$ with the distance $BC$. Because $XA$ and $XB$ are tangents of the circle $k$ in the points $A$ and $B$, $XA\cong XB$ (statement \ref{TangOdsek}). Similarly, $XA$ and $XC$ are tangents of the circle $l$ in the points $A$ and $C$, or $XA\cong XC$. Therefore, $X$ is the center of the circle with the diameter $BC$ and the point $A$ lies on this circle. By Tales' statement \ref{TalesovIzrKroz2}, $\angle BAC=90^0$. \item \res{Let $ABCD$ be a deltoid ($AB\cong AD$ and $CB\cong CD$). Prove:} \begin{enumerate} \item \res{$ABCD$ is a tangential quadrilateral.} Use statement \ref{TangentniPogoj}. \item \res{$ABCD$ is a tautological quadrilateral exactly when $AB\perp BC$.} Use statement \ref{TetivniPogoj}. \end{enumerate} \item \res{The circles $k$ and $k_1$ touch each other from the outside in the point $T$, in which they intersect the lines $p$ and $q$. The line $p$ has also intersections with the circles $P$ and $P_1$, the line $q$ has $Q$ and $Q_1$ intersections. Prove that $PQ\parallel P_1Q_1$.} Let $t$ be the common tangent of the circles $k$ and $k_1$, $X\in t$ any point of this tangent, for which $P,X\div QQ_1$ and $Y$ the point which is symmetrical to the point $X$ with respect to $T$. By statements \ref{ObodKotTang} and \ref{sovrsnaSkladna}, we have: \begin{eqnarray*} \angle QPP_1= \angle QPT \cong\angle QTX\cong\angle Q_1TY\cong\angle Q_1P_1T =\angle Q_1P_1P \end{eqnarray*} Therefore, $\angle QPP_1 \cong\angle Q_1P_1P$, so by statement \ref{KotiTransverzala} $PQ\parallel P_1Q_1$. \item \res{Let $MN$ be the common tangent of the circles $k$ and $l$ ($M$ and $N$ are touch points), which intersect in the points $A$ and $B$. Calculate the value of the sum $\angle MAN+\angle MBN$.} Without loss of generality, assume that $d(A,MN) ||AE|-|BE||= |\left(|AC|-|EC|\right)-\left(|BD|-|ED|\right)| = ||AC|-|BD||. \end{eqnarray*} \item \res{Let $S$ be the intersection of the altitudes of the sides $AD$ and $BC$ of the trapezoid $ABCD$ with the base $AB$. Prove that the circumscribed circles of the triangles $SAB$ and $SCD$ touch each other in point $S$.} Use \ref{TangOdsek} theorem. \item \res{The lines $PB$ and $PD$ touch the circle $k(O,r)$ in points $B$ and $D$. The line $PO$ intersects the circle $k$ in points $A$ and $C$ ($\mathcal{B}(P,A,C)$). Prove that the line $BA$ is the angle bisector of the angle $PBD$.} Use \ref{enakokraki} and \ref{TangOdsek} theorems. \item \res{The quadrilateral $ABCD$ is inscribed in a circle with center $O$. The diagonals $AC$ and $BD$ are perpendicular. Let $M$ be the perpendicular projection of the center $O$ on the line $AD$. Prove that $$|OM|=\frac{1}{2}|BC|.$$} From the congruence of the triangles $OMA$ and $OMD$ (the SSA theorem \ref{SSK}) it follows that $MA\cong MD$ or that the point $M$ is the center of the line segment $AD$. We mark with $E$ the intersection of the diagonals $AC$ and $BD$. Let $D'$ be the point that is symmetrical to the point $D$ with respect to the center $O$. It is clear that $D'\in k$. By Tales' theorem \ref{TalesovIzrKroz2} we have $\angle DAD'=90^0$. The line segment $MO$ is the median of the triangle $DAD'$ for the side $AD'$, so $|OM|=\frac{1}{2}|AD'|$. It is enough to prove that $AD'\cong BC$ or (by the theorem \ref{SklTetSklObKot}) that $\angle ADD'\cong \angle BDC$. If we now use the theorem \ref{VsotKotTrik} and \ref{ObodObodKot}, we get: \begin{eqnarray*} \angle ADD' &=& 90^0-\angle AD'D =90^0- \angle ACD =\\ &=& 90^0-\angle ECD =\angle EDC =\\ &=& \angle BDC. \end{eqnarray*} \item \res{Daljici $AB$ in $BC$ sta sosednji stranici pravilnega devetkotnika, ki je včrtan krožnici $k$ s središčem $O$. Točka $M$ je središče stranice $AB$, točka $N$ pa središče polmera $OX$ krožnice $k$, ki je pravokoten na premico $BC$. Dokaži, da je $\angle OMN=30^0$.} From $\angle AOB\cong\angle BOC=\frac{360^0}{9}=40^0$ it follows that: $$\angle AOX=\angle AOB + \angle BOX=\angle AOB + \frac{1}{2}\angle BOC=60^0.$$ Since $OA\cong OB$, the triangle $AOB$ is a right triangle (theorems \ref{enakokraki} and \ref{VsotKotTrik}). Its median $AN$ is also the altitude, so $\angle ONA=90^0$. Since $\angle OMA=90^0$, by Tales' theorem the points $N$ and $M$ lie on the circle with the radius $OA$ or that the quadrilateral $ONMA$ is a square. If we now use the theorem \ref{ObodObodKot}, we get: \begin{eqnarray*} \angle OMN=\angle OAN=\frac{1}{2}\angle OAX=30^0. \end{eqnarray*} \item \res{The circles $k_1$ and $k_2$ intersect in points $A$ and $B$. Let $p$ be a line that goes through point $A$, intersects circle $k_1$ in point $C$ and circle $k_2$ in point $D$, and let $q$ be a line that goes through point $B$, intersects circle $k_1$ in point $E$ and circle $k_2$ in point $F$. Prove that $\angle CBD\cong\angle EAF$.} If we use \ref{VsotKotTrik} and \ref{ObodObodKot}, we get: \begin{eqnarray*} \angle CBD &=& 180^0-\angle BCD - \angle BDC =\\ &=& 180^0-\angle BCA - \angle BDA =\\ &=& 180^0-\angle BEA - \angle BFA =\\ &=& 180^0-\angle FEA - \angle EFA =\\ &=& \angle EAF =\\ \end{eqnarray*} \item \res{The circles $k_1$ and $k_2$ intersect in points $A$ and $B$. Draw a line $p$ that goes through point $A$, so that the length of the line segment $MN$, where $M$ and $N$ are the intersections of line $p$ with circles $k_1$ and $k_2$, is maximal.} Let $p$ be an arbitrary line that goes through point $A$ and intersects circles $k_1$ and $k_2$ in points $M$ and $N$. Let $S_1$ and $S_2$ be the centers of circles $k_1$ and $k_2$ and let $P_1$ and $P_2$ be the centers of line segments $MA$ and $NA$. Because $MN=2\cdot P_1P_2$, the problem of the maximum length of line segment $MN$ can be translated to the maximum length of line segment $P_1P_2$. From the similarity of triangles $MS_1P_1$ and $AS_1P_1$ (\textit{SSS} \ref{SSS}) it follows that $\angle S_1P_1M\cong\angle S_1P_1A=90^0$ or $\angle P_2P_1S_1=90^0$. Similarly, $\angle P_2P_1S_1=90^0$. Therefore, $S_1S_2P_2P_1$ is a right trapezoid with height $P_1P_2$, so $P_1P_2\leq S_1S_2$. Equality is achieved when $S_1S_2P_2P_1$ is a rectangle or $P_1P_2\parallel S_1S_2$. This means that line $p$ (for which $|MN|$ is maximal) is drawn as a parallel to line $S_1S_2$ through point $A$. \item \res{Let $L$ be the orthogonal projection of an arbitrary point $K$ of the circle $k$ onto its tangent at the point $T\in k$, and let $X$ be the point that is symmetric to the point $L$ with respect to the line $KT$. Determine the geometric position of the point $X$.} We denote by $S$ the center of the circle $k$ and by $k_1$ the circle with diameter $ST$. We prove that the geometric position of the point $X$ (which we denote by $\mathcal{G}$) is equal to the circle $k_1$. We need to prove the equivalence: \begin{eqnarray*} X\in \mathcal{G} \hspace*{1mm} \Leftrightarrow \hspace*{1mm} X\in k_1 \end{eqnarray*} For the points $T$ and $S$, it is clear from the definition that both $T\in \mathcal{G}$ and $T\in k_1$ hold, as well as $S\in \mathcal{G}$ and $S\in k_1$. We assume further that $X\neq T$ and $X\neq S$. We first assume $X\in \mathcal{G}$. We denote $\angle LTK=\alpha$. The triangles $XKT$ and $LKT$ are symmetric with respect to the line $KT$ (see subsection \ref{odd6OsnZrc}), so $\triangle XKT\cong \triangle LKT$ holds, i.e. $\angle XTK\cong \angle LTK=\alpha$ and $\angle KXT\cong \angle KLT=90^0$. We prove that also $\angle SXT=90^0$. It is enough to prove that the points $S$, $X$ and $K$ are collinear, i.e. that $\angle TKX=\angle TKS$ holds. First, we have (from $\triangle XKT\cong \triangle LKT$): $\angle TKX\cong\angle TKL =90^0-\alpha$. If we use the statement \ref{enakokraki} of the congruent triangles $KST$ with the base $KT$ and the statements \ref{ObodKotTang} and \ref{SredObodKot}, we get: $\angle TKS=\frac{1}{2}(180^0-\angle KST)=\frac{1}{2}(180^0-2\angle LTK)=90^0-\alpha$. Therefore $\angle TKX=\angle TKS=90^0-\alpha$, which means that the points $S$, $X$ and $K$ are collinear, i.e. $\angle SXT=90^0$. From the latter it follows that $X\in k_1$. Let's now assume that $X\in k_1$. We'll mark the intersection of the line segment $SX$ with the circle $k$ with $K$ and the point $L$ that is symmetrical to the point $X$ with respect to the line $TK$. From $X\in k_1$ follows from Tales' theorem \ref{TalesovIzrKroz2} that $\angle SXT=90^0$, so $\angle KXT=90^0$ as well. To prove that $X\in \mathcal{G}$, it is enough to prove that $X$ lies on the tangent $t$ of the circle $k$ through the point $T$. Let $L'$ be an arbitrary point on the tangent $t$, for which $L,X\ddot{-} ST$ holds. It is enough to prove $\angle LTK\cong\angle L'TK$. We'll mark $\angle LTK=\alpha$. The triangles $XKT$ and $LKT$ are symmetrical with respect to the line $KT$, so $\triangle XKT\cong \triangle LKT$ or $\angle XTK\cong \angle LTK=\alpha$ and $\angle KLT\angle KXT\cong =90^0$. If we use the formula \ref{ObodKotTang} and \ref{SredObodKot}, we get: $\angle L'TK=\frac{1}{2}\angle TSK=\frac{1}{2}(180^0-2\angle SKT)=90^0-\angle SKT=90^0-\angle XKT=\angle XTK=\alpha=\angle LTK$. Therefore $L'\in t$ or $X\in \mathcal{G}$. \item \res{Prove that a string polygon with an even number of vertices, which has all the internal angles compliant, is a regular polygon.} We'll mark the center of the circumscribed circle of this polygon with $O$ and the centers of the sides $A_iB_{i+1}$ with $B_i$ ($i\in \{1,2,\ldots , 2n-1 \}$, $A_0=A_{2n-1}$ and $A_{2n}=A_1$). We'll use the fact that $A_iOA_{i+1}$ is an isosceles triangle with the base $A_iA_{i+1}$ and we'll first prove $\triangle OB_{i-1}A_i\cong \triangle OB_{i+1}A_{i+1}$. Let $EA$ be the common tangent of the circles $k$ (larger) and $l$ (smaller) and $D$ the other intersection of the line $AB$ with the circle $l$. Let $L$ be the other intersection of the segment $AK$ with the circle $l$, $\alpha=\angle DCB$ and $\beta=\angle DBC$. By the theorem \ref{ObodKotTang} it follows: \begin{eqnarray} \label{nalSkk36Eqn1} \angle CAB=\angle CAD\cong \angle DCB=\alpha \end{eqnarray} and \begin{eqnarray} \label{nalSkk36Eqn2} \angle LDA\cong \angle EAL=\angle EAK\cong \angle ABK=\angle DBC=\beta. \end{eqnarray} Since $CDA$ is the exterior angle of the triangle $CDB$, it follows from the theorem \ref{zunanjiNotrNotr} that $\angle CDA=\angle DCB+\angle CBD=\alpha+\beta$. From this and \ref{nalSkk36Eqn2} it follows: \begin{eqnarray} \label{nalSkk36Eqn3} \angle LDC= \angle CDA-\angle LDA=\alpha+\beta-\beta=\alpha. \end{eqnarray} From the relations \ref{nalSkk36Eqn3} and \ref{nalSkk36Eqn1} and the theorem \ref{ObodObodKot}: \begin{eqnarray*} \angle KAC= \angle LAC\cong \angle LDC=\alpha=\angle CAB. \end{eqnarray*} This means that the line $AC$ is the perpendicular bisector of the angle $BAK$. The task is a special case of a more general task \ref{nalSkk47}. \item \res{Let $BC$ be the diameter of the circle $k$. Determine the geometric position of the altitude points of all triangles $ABC$, where $A$ is an arbitrary point on the circle $k$.} The desired geometric position of the points is the circle that passes through the points $B$ and $C$. Calculate the measure of the angle $BVC$, where $V$ is the altitude point of the triangle $ABC$, and use the theorem \ref{ObodKotGMT}. See also the theorem \ref{TockaV'a}. \item \res{Let $ABCDEF$ be a tetrahedral hexagon and $AB\cong DE$ and $BC\cong EF$. Prove that $CD\parallel AF$.} Let $ABCDEF$ be a tetrahedral hexagon and $AB\cong DE$ and $BC\cong EF$. Since $\angle BAD>90^0$ and $\angle BCD>90^0$, by the theorem \ref{obodKotGMTZunNotr} $A$ and $C$ are inner points of the circle $k$. This means that $AC90^0$. This means $\mathcal{B}(A',A,V)$ (where $A'$ is the altitude of this triangle) or $V$ is an external point of the circle $k$ and it holds $|\overrightarrow{OV}|>1$. We can also look at the task \ref{OlimpVekt15} (section \ref{odd5DolzVekt}). \item \res{Calculate the angles determined by the vectors $\overrightarrow{OA}$, $\overrightarrow{OB}$ and $\overrightarrow{OC}$, if the points $A$, $B$ and $C$ lie on the circle with the center $O$ and in addition it holds: $$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = \overrightarrow{0}.$$} According to Hamilton's theorem \ref{Hamilton}, $\overrightarrow{0}=\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}=\overrightarrow{OV}$, where $V$ is the altitude point of the triangle $ABC$. So $V=O$, which means that $ABC$ is an equilateral triangle and all the angles sought are equal to $120^0$. \item \res{Let $A$, $B$, $C$ and $D$ be points that lie on the circle with the center $O$, and it holds $$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = \overrightarrow{0}.$$ Prove that $ABCD$ is a rectangle.} We use Hamilton's theorem \ref{Hamilton}. \item \res{Let $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ be vectors of some plane, for which it holds $|\overrightarrow{a}| = |\overrightarrow{b}| = |\overrightarrow{c}| =x$. Investigate, in which case it also holds $|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}| = x$.} First, let's assume that $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are nonlinear vectors. Let $O$ be an arbitrary point and $A$, $B$ and $C$ be such points that $\overrightarrow{OA}=\overrightarrow{a}$, $\overrightarrow{OB}=\overrightarrow{b}$ and $\overrightarrow{OC}=\overrightarrow{c}$ (statement \ref{vektAvObst1TockaB}). From the nonlinearity of vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ it follows that points $A$, $B$ and $C$ are nonlinear. From the condition $|\overrightarrow{a}| = |\overrightarrow{b}| = |\overrightarrow{c}| =x$ it follows that $|\overrightarrow{OA}| = |\overrightarrow{OB}| = |\overrightarrow{OC}| =x$, which means that point $O$ is the center of the circumscribed circle of triangle $ABC$. By Hamilton's statement \ref{Hamilton} it is: \begin{eqnarray*} x &=& |\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}|=\\ &=& |\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}|=\\ &=& |\overrightarrow{OV}|, \end{eqnarray*} where $V$ is the altitude point, which therefore lies on the circumscribed circle of triangle $ABC$. This is possible only when $ABC$ is a right triangle. If $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are collinear vectors, the given condition is satisfied only when $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are not all of the same orientation. %Tales %______________________________________________________ \item \res{Divide the given line segment $AB$:} (\textit{a}) \res{into five equal line segments,}\\ (\textit{b}) \res{in the ratio $2:5$,}\\ (\textit{c}) \res{into three line segments, which are in the ratio $2:\frac{1}{2}:1$.} Use example \ref{izrekEnaDelitevDaljice}. \item \res{Given is the line segment $AB$. Only by using a ruler with the ability to draw parallels (constructions in affine geometry), plot point $C$, if:} (\textit{a}) \res{$\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}$,} \hspace*{3mm} (\textit{b}) \res{$\overrightarrow{AC}=\frac{3}{5}\overrightarrow{AB}$,} \hspace*{3mm} (\textit{c}) \res{$\overrightarrow{AC}=-\frac{4}{7}\overrightarrow{AB}$.} Use task \ref{nalVekt4} and example \ref{izrekEnaDelitevDaljiceNan}. \item \res{The lines $a$, $b$ and $c$ are given. Draw the line $x$, so that it will:} (\textit{a}) \res{$a:b=c:x$;} \hspace*{3mm} (\textit{b}) \res{$x=\frac{a\cdot b}{c}$,} \hspace*{3mm} (\textit{c}) \res{$x=\frac{a^2}{c}$,}\hspace*{3mm}\\ (\textit{č}) \res{$x=\frac{2ab}{3c}$,}\hspace*{3mm} (\textit{č}) \res{$(x+c):(x-c)=7:2$.} We use the Pythagorean theorem \ref{TalesovIzrek}. \item \res{Let $M$ and $N$ be points on the arm $OX$, $P$ a point on the arm $OY$ as $XOY$ and $NQ\parallel MP$ and $PN\parallel QS$ ($Q\in OY$, $S\in OX$). Prove that $|ON|^2=|OM|\cdot |OS|$ (for the line $ON$ in this case we say that it is the \index{geometrijska sredina daljic}\pojem{geometric mean} \color{green1} of the lines $OM$ and $OS$).} By the Pythagorean theorem \ref{TalesovIzrek} it is: $$\frac{|ON|}{|OM|}=\frac{|OQ|}{|OP|}=\frac{|OS|}{|ON|},$$ from which the desired relation follows. \item \res{Let $ABC$ be a triangle and $Q$, $K$, $L$, $M$, $N$ and $P$ such points of the sides $AB$, $AC$, $BC$, $BA$, $CA$ and $CB$, that $AQ\cong CP\cong AC$, $AK\cong BL\cong AB$ and $BM\cong CN\cong BC$. Prove that $MN$, $PQ$ and $LK$ are parallel.} We use the formulas \ref{vektParamPremica} and \ref{vektDelitDaljice}. \item \res{Let $P$ be the center of gravity of the centroid $AA_1$ of the triangle $ABC$. The point $Q$ is the intersection of the line $BP$ with the side $AC$. Calculate the ratios $AQ:QC$ and $BP:PQ$.} Let $R$ be the center of the line $QC$. By the formula \ref{srednjicaTrikVekt} (for the triangle $BQC$ and $AA_aR$) it is: $$\overrightarrow{PQ}=\frac{1}{2}\overrightarrow{A_1R}=\frac{1}{4}\overrightarrow{BQ},$$ so $BP:PQ=3:1$. From $\overrightarrow{PQ}=\frac{1}{2}\overrightarrow{A_1R}$ it follows that $PQ\parallel A_1R$, so by the Pythagorean theorem $AQ:QR=AP:PA_1=1:1$. Therefore $AQ\cong QR \cong RC$ or $AQ:QC=1:2$. See also the example \ref{TezisceSredisceZgled}. \item \res{The points $P$ and $Q$ lie on the sides $AB$ and $AC$ of the triangle $ABC$, and it holds that $\frac{|\overrightarrow{PB}|}{|\overrightarrow{AP}|} +\frac{|\overrightarrow{QC}|}{|\overrightarrow{AQ}|}=1$. Prove that the centroid of this triangle lies on the line $PQ$.} Let $\widehat{T}$ be the intersection of the line $PQ$ with the centroidal line $AA_1$ of this triangle. It is enough to prove that $\widehat{T}=T$. Since the points $A$, $\widehat{T}$ and $A_1$ are collinear, $\overrightarrow{A\widehat{T}}=\lambda\overrightarrow{AA_1}$ for some $\lambda\in R$. For $\widehat{T}=T$ it is enough to prove that $\lambda=\frac{2}{3}$. We denote $\alpha=\frac{|\overrightarrow{PB}|}{|\overrightarrow{AP}|}$ and $\beta=\frac{|\overrightarrow{QC}|}{|\overrightarrow{AQ}|}$. By the assumption, $\alpha+\beta=1$. From the given conditions it follows that $\overrightarrow{AB}=\left(1+\alpha\right)\cdot\overrightarrow{AP}$ and $\overrightarrow{AC}=\left(1+\beta\right)\cdot\overrightarrow{AQ}$. If we use the latter relation and the formula \ref{vektSredOSOAOB}, we get: \begin{eqnarray*} \overrightarrow{A\widehat{T}} &=& \lambda\overrightarrow{AA_1}=\\ &=& \frac{\lambda}{2}\cdot \left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\\ &=& \frac{\lambda}{2}\cdot \left(\left(1+\alpha\right)\cdot\overrightarrow{AP} +\left(1+\beta\right)\cdot\overrightarrow{AQ}\right)=\\ &=& \frac{\lambda}{2}\cdot \left(1+\alpha\right)\cdot\overrightarrow{AP} +\frac{\lambda}{2}\left(1+\beta\right)\cdot\overrightarrow{AQ}. \end{eqnarray*} Since the point $T$ lies on the line $PQ$ and since $\alpha+\beta=1$, by the formula \ref{vektParamPremica}: \begin{eqnarray*} 1&=&\frac{\lambda}{2}\cdot \left(1+\alpha\right) +\frac{\lambda}{2}\left(1+\beta\right)=\\ &=&\frac{\lambda}{2}\cdot \left(1+\alpha+1+\beta\right)=\\ &=&\frac{3\lambda}{2}\\ \end{eqnarray*} i.e. $\lambda=\frac{2}{3}$. \item \res{Let $a$, $b$ and $c$ be three segments with a common starting point $S$ and $M$ a point on the segment $a$. If the point $M$ "moves" along the segment $a$, the ratio of the distances of this point from the lines $b$ and $c$ is constant.} We use the Tales' theorem \ref{TalesovIzrek}. \item \res{Let $D$ be a point lying on the side $BC$ of the triangle $ABC$ and $F$ and $G$ points in which the line passing through the point $D$ and parallel to the median $AA_1$, intersects the lines $AB$ and $AC$. Prove that the sum $|DF|+|DG|$ is constant if the point $D$ "moves" along the side $BC$.} We prove that $|DF|+|DG|=|AA_1|$. See problem \ref{nalVekt29}. \item \res{Draw a triangle with the following data:} In both cases use the same notation as in the big problem \ref{velikaNaloga} and the statement \ref{velNalTockP'}. (\textit{a}) \res{$v_a$, $r$, $b-c$} First, we draw the rectangular triangle $SPP_a$ ($SP=r$, $\angle P=90^0$ and $PP_a=b-c$), the point $P'$ and finally use the fact that the vertex $A$ lies on the segment $P_aP'$. (\textit{b}) \res{$\beta$, $r$, $b-c$} Similarly to the previous case. \end{enumerate} %REŠITVE - Izometrije %________________________________________________________________________________ \poglavje{Isometries} \begin{enumerate} \item \res{Given is a line $p$ and points $A$ and $B$, which lie on opposite sides of the line $p$. Construct a point $X$, which lies on the line $p$, so that the difference $|AX|-|XB|$ is maximal.} Let $A'=\mathcal{S}_p(A)$. We prove that $X=A'B\cap p$ is the desired point (see also example \ref{HeronProbl}). \item \res{In the plane are given lines $p$, $q$ and $r$. Construct an equilateral triangle $ABC$, so that the vertex $B$ lies on the line $p$, $C$ on $q$, the altitude from the vertex $A$ lies on the line $r$.} Since $\mathcal{S}_r(B)=C$ and $B\in p$, the vertex $C$ we get from the condition $C\in \mathcal{S}_r(p)\cap q$. \item \res{Given is a quadrilateral $ABCD$ and a point $S$. Draw a parallelogram with the center in the point $S$, so that its vertices lie on the lines of the given quadrilateral.} We use the central reflection $\mathcal{S}_S$. \item \res{Let $\mathcal{I}$ be an indirect isometry of a plane that maps point $A$ to point $B$, $B$ to $A$. Prove that $\mathcal{I}$ is the central reflection.} Let $s$ be the line of symmetry of the segment $AB$. The composition $\mathcal{I}\circ\mathcal{S}_s$ is a direct isometry with two fixed points $A$ and $B$, therefore by izreku \ref{izo2ftIdent} it represents the identity or $\mathcal{I}\circ\mathcal{S}_s=\mathcal{E}$. If we multiply the equality by $\mathcal{S}_s$ from the right, we get $\mathcal{I}=\mathcal{S}_s$. \item \res{Let $K$ and $L$ be points that are symmetric to the vertex $A$ of the triangle $ABC$ with respect to the lines of symmetry of the internal angles at the vertices $B$ and $C$. Let $P$ be the point of tangency of the inscribed circle of this triangle and the side $BC$. Prove that $P$ is the center of the segment $KL$.} We denote by $s_\beta$ and $s_\gamma$ the mentioned lines of symmetry of the internal angles at the vertices $B$ and $C$ and by $Q$ and $R$ the points of tangency of the inscribed circle of the triangle $ABC$ with the sides $AC$ and $AB$. By izreku \ref{TangOdsek} we have $AQ\cong AR$, $BR\cong BP$ and $CP\cong CQ$. Therefore $\mathcal{S}_{s_\beta}:A,R\mapsto K,P$ and $\mathcal{S}_{s_\gamma}:A,Q\mapsto L,P$. From this it follows that $KP\cong AR\cong AQ\cong LP$, which means that $P$ is the center of the segment $KL$. \item \res{Let $k$ and $l$ be circles on different sides of a line $p$. Draw an isosceles triangle $ABC$, so that its altitude $AA'$ lies on the line $p$, the vertex $B$ on the circle $k$, and the vertex $C$ on the circle $l$.} Since $\mathcal{S}_p(B)=C$ and $B\in k$, we get the point $C$ from the condition $C\in \mathcal{S}_p(k)\cap l$. \item \res{Let $k$ be a circle and $a$, $b$ and $c$ lines in the same plane. Draw a triangle $ABC$, which is inscribed in the circle $k$, so that its sides $BC$, $AC$ and $AB$ will be parallel in order to the lines $a$, $b$ and $c$.} First, we can plan the simetrals of sides $BC$, $AC$ and $AB$ as lines that go through the center $O$ of the circle $k$ and are perpendicular in turn to the lines $a$, $b$ and $c$. We mark these simetrals with $p$, $q$ and $r$. Because the composite $\mathcal{I}=\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p$ has the fixed points $O$ and $A$, by Theorem \ref{izo1ftIndZrc} $\mathcal{I}=\mathcal{S}_{OA}$. So we can draw the line $OA$ as the simetral of the distance $XX'$, where $X$ is an arbitrary point. The point $A$ we get as the intersection of the line $OA$ and the circle $k$. Then $B=\mathcal{S}_p(A)$ and $C=\mathcal{S}_r(A)$. \item \res{Let $ABCDE$ be a rope pentagon, in which $BC\parallel DE$ and $CD\parallel EA$. Prove that the point $D$ lies on the simetral of the distance $AB$.} We mark with $O$ the center of the inscribed circle of the pentagon $ABCDE$. Because $BC\parallel DE$, the lines $BC$ and $DE$ have the same perpendicular from the center $S$, which is actually the common simetral of sides $BC$ and $DE$. We mark it with $p$. Similarly, sides $CD$ and $EA$ have the common simetral $q$. We mark with $r$ the simetral of the distance $AB$. We prove that $D\in r$. The composite $\mathcal{S}_{OD}\circ\mathcal{S}_q \circ\mathcal{S}_p\circ\mathcal{S}_r \circ\mathcal{S}_q\circ\mathcal{S}_p$ is a direct isometry, which has the fixed points $O$ and $D$, so by Theorem \ref{izo2ftIdent} it represents the identity. Therefore: $$\mathcal{S}_{OD}\circ\mathcal{S}_q \circ\mathcal{S}_p\circ\mathcal{S}_r \circ\mathcal{S}_q\circ\mathcal{S}_p=\mathcal{E}.$$ Premices $p$, $q$ and $r$ are from the same $\mathcal{X}_O$, so the composite $\mathcal{S}_q \circ\mathcal{S}_p\circ\mathcal{S}_r$ represents the basic reflection (statement \ref{izoSop}) or $$\mathcal{S}_q \circ\mathcal{S}_p\circ\mathcal{S}_r=\mathcal{S}_l =\mathcal{S}^{-1}_l=\left(\mathcal{S}_q \circ\mathcal{S}_p\circ\mathcal{S}_r\right)^{-1}=\mathcal{S}_r \circ\mathcal{S}_p\circ\mathcal{S}_q.$$ Therefore: \begin{eqnarray*} \mathcal{E}&=&\mathcal{S}_{OD}\circ\mathcal{S}_q \circ\mathcal{S}_p\circ\mathcal{S}_r \circ\mathcal{S}_q\circ\mathcal{S}_p =\\ &=&\mathcal{S}_{OD}\circ\mathcal{S}_r \circ\mathcal{S}_p\circ\mathcal{S}_q \circ\mathcal{S}_q\circ\mathcal{S}_p = \mathcal{S}_{OD}\circ\mathcal{S}_r. \end{eqnarray*} From $\mathcal{E}=\mathcal{S}_{OD}\circ\mathcal{S}_r$ follows $\mathcal{S}_{OD}=\mathcal{S}_r$ or $OD=r$ and $D\in r$. \item \res{Premice $p$, $q$ in $r$ ležijo v isti ravnini. Dokaži ekvivalenco $\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p =\mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_r$ natanko tedaj, ko premice $p$, $q$ in $r$ pripadajo istemu šopu.} Označimo $\mathcal{I}=\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p$. ($\Leftarrow$) Predpostavimo, da premice $p$, $q$ in $r$ pripadajo istemu šopu. Po izreku \ref{izoSop} izometrija $\mathcal{I}$ predstavlja osmo zrcaljenje. Iz tega sledi $\mathcal{I}=\mathcal{I}^{-1}$ oz. $\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p =\mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_r$ ($\Rightarrow$) Let us assume that $\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p =\mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_r$. From this condition it follows that $\mathcal{I}=\mathcal{I}^{-1}$ or $\mathcal{I}^2=\mathcal{E}$. Let us assume the opposite - that lines $p$, $q$ and $r$ are not from the same sheaf. According to the statement \ref{izoZrcdrsprq} it is $\mathcal{I}=\mathcal{G}_{2\overrightarrow{AB}}$ ($\overrightarrow{AB}\neq \overrightarrow{0}$). But in this case $\mathcal{E}=\mathcal{I}^2=\mathcal{G}^2_{2\overrightarrow{AB}} =\mathcal{T}_{4\overrightarrow{AB}}$ (statement \ref{izoZrcdrsZrcdrs}), which is not possible, because according to the assumption $\overrightarrow{AB}\neq \overrightarrow{0}$. Therefore lines $p$, $q$ and $r$ belong to the same sheaf. \item \res{Let $O$, $P$ and $Q$ be three non-linear points. Construct a square $ABCD$ (in the plane $OPQ$) with the center in point $O$, so that points $P$ and $Q$ in order lie on lines $AB$ and $BC$.} We use the fact that $Q,\mathcal{R}_{O,90^0}(P)\in BC$. \item \res{Let $\mathcal{R}_{S,\alpha}$ be a rotation and $\mathcal{S}_p$ be a reflection in the same plane and $S\in p$. Prove that the composition $\mathcal{R}_{S,\alpha}\circ\mathcal{S}_p$ and $\mathcal{S}_p\circ\mathcal{R}_{S,\alpha}$ represent a reflection.} We use the statement \ref{rotacKom2Zrc} and \ref{izoSop}. \item \res{Given is a point $A$ and a circle $k$ in the same plane. Draw a square $ABCD$, so that the vertices of the diagonal $BD$ lie on the circle $k$.} From $B,D\in k$ and $\mathcal{R}_{A,90^0}(B)=D$ it follows that $D\in k\cap \mathcal{R}_{A,90^0}(k)$, which allows us to construct the vertex $D$. Then $B=\mathcal{R}_{A,-90^0}(D)$ and $C=\mathcal{R}_{D,90^0}(A)$. \item \res{Let $ABC$ be an arbitrary triangle. Prove: $$\mathcal{R}_{C,2\measuredangle BCA}\circ \mathcal{R}_{B,2\measuredangle ABC}\circ \mathcal{R}_{A,2\measuredangle CAB}=\mathcal{E}.$$} According to the statement \ref{rotacKom2Zrc} it is: \begin{eqnarray*} &&\mathcal{R}_{C,2\measuredangle BCA}\circ \mathcal{R}_{B,2\measuredangle ABC}\circ \mathcal{R}_{A,2\measuredangle CAB}=\\ &&= \mathcal{S}_{CA}\circ\mathcal{S}_{CB}\circ\mathcal{S}_{BC} \circ\mathcal{S}_{BA}\circ\mathcal{S}_{AB}\circ\mathcal{S}_{AC}= \mathcal{E}. \end{eqnarray*} \item \res{Prove that the composite of the central reflection $\mathcal{S}_p$ and the central reflection $\mathcal{S}_S$ ($S\in p$) represents the central reflection.} Let $q$ be the rectangle of the line $p$ through the point $S$. According to the statement \ref{izoSrZrcKom2Zrc} it is: $$\mathcal{S}_S\circ\mathcal{S}_p= \mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{S}_p=\mathcal{S}_q.$$ \item \res{Let $O$, $P$ and $Q$ be three non-linear points. Construct the square $ABCD$ (in the plane $OPQ$) with the center in the point $O$, so that the points $P$ and $Q$ lie in succession on the lines $AB$ and $CD$.} We use the fact that $Q,\mathcal{S}_O(P)\in CD$. \item \res{What does the composite of the translation and the central reflection represent?} We use the statements \ref{transl2sred} and \ref{izoKomp3SredZrc}. \item \res{Given are the line $p$ and the circles $k$ and $l$, which lie in the same plane. Draw the line, which is parallel to the line $p$, so that it determines the corresponding tangents on the circles $k$ and $l$.} Let $K$ and $L$ be the centers of the circles $k$ and $l$. Let $q$ be the desired line that is parallel to the line $p$, and the circles $k$ and $l$ intersect in such points $A$, $B$, $C$ and $D$, so that $AB \cong CD$. Let $\mathcal{B} (A, B, C, D)$. We denote by $K_q$ and $L_q$ the orthogonal projections of the centers $K$ and $L$ onto the line $q$, and $K_p$ and $L_p$ the orthogonal projections of the centers $K$ and $L$ onto the line $p$. From the congruence of the triangles $KAK_q$ and $KBK_q$ (SSA Theorem \ref{SSK}) it follows that $K_q$ is the center of the chord $AB$. Similarly, $L_q$ is the center of the chord $CD$. From $AB \cong CD$ it then follows that $K_qB \cong L_qD$, i.e. $\overrightarrow {K_qB} = \overrightarrow {L_qD}$. Therefore: $$\overrightarrow {BD} = \overrightarrow {BD} + \overrightarrow {0} = \overrightarrow {BD} + \overrightarrow {K_qB} + \overrightarrow {DL_q} = \overrightarrow {K_qL_q} = \overrightarrow {K_pL_p}.$$ The vector $\overrightarrow {v} = \overrightarrow {K_pL_p}$ can be constructed. This means that the previous analysis allows the construction, because $\mathcal {T} _ {\overrightarrow {v}} (B) = D$, i.e. $D \in \mathcal {T} _ {\overrightarrow {v}} (k) \cap l$. \item \res {Let $c$ be a line that intersects the parallels $a$ and $b$, and $l$ a distance. Draw an isosceles triangle $ABC$ so that $A \in a$, $B \in b$, $C \in c$ and $AB \cong l$.} First, we construct an arbitrary isosceles triangle $A_1B_1C_1$ that satisfies the conditions $A_1 \in a$, $B_1 \in b$ and $A_1B_1 \cong l$, then we use the appropriate translation. \item \res {Prove that the composition of a rotation and an axis reflection of a plane represents a mirror glide exactly when the center of rotation does not lie on the axis of the axis reflection.} Use Theorems \ref {rotacKom2Zrc}, \ref {izoZrcdrsprq} and \ref {izoSop}. \item \res {Let $ABC$ be an isosceles triangle. Prove that the composition $\mathcal {S} _ {AB} \circ \mathcal {S} _ {CA} \circ \mathcal {S} _ {BC}$ represents a mirror glide. Also determine the vector and the axis of this glide.} Let $\mathcal{I}=\mathcal{S}_{AB} \circ\mathcal{S}_{CA} \circ\mathcal{S}_{BC}$. By \ref{izoZrcdrsprq} is $\mathcal{I}$ a mirror reflection. Let $A_1$, $B_1$ and $C_1$ be in order the center points of the lines $BC$, $AC$ and $AB$ of the triangle $ABC$. Because it is a right triangle is $\mathcal{I}(A_1C_1)=A_1C_1$, which means that the line $A_1C_1$ is of this mirror reflection. It is not difficult to prove that for the point $A'_1=\mathcal{I}(A_1)$ (both lie on the axis $A_1C_1$) is $\overrightarrow{A_1A'_1}=3\overrightarrow{A_1C_1}$, so $\mathcal{I}=\mathcal{G}_{3\overrightarrow{A_1C_1}}$. \item \res{Given are the points $A$ and $B$ on the same side of the line $p$. Draw the line $XY$, which lies on the line $p$ and is consistent with the given line $l$, so that the sum $|AX|+|XY|+|YB|$ is minimal.} Let $A'=\mathcal{G}_{\overrightarrow{MN}}(A)$ (where $M,N\in p$ and $MN\cong l$). The point $Y$ is obtained as the intersection of the lines $p$ and $X'Y$ (see also example \ref{HeronProbl}). \item \res{Let $ABC$ be an isosceles right triangle with a right angle at the vertex $A$. What does the composite $\mathcal{G}_{\overrightarrow{AB}}\circ \mathcal{G}_{\overrightarrow{CA}}$ represent?} Let $p$ and $q$ be the simetrali of the sides $CA$ and $AB$ of the triangle $ABC$. By \ref{izoZrcDrsKompSrOsn} is: $$\mathcal{G}_{\overrightarrow{AB}}\circ \mathcal{G}_{\overrightarrow{CA}}= \mathcal{S}_q\circ\mathcal{S}_A\circ\mathcal{S}_A\circ\mathcal{S}_p= \mathcal{S}_q\circ\mathcal{S}_p.$$ Because $ABC$ is an isosceles right triangle with a right angle at the vertex $A$, the lines $p$ and $q$ are perpendicular and intersect at the center $S$ of the hypotenuse $BC$. Therefore $\mathcal{G}_{\overrightarrow{AB}}\circ \mathcal{G}_{\overrightarrow{CA}}=\mathcal{S}_q \circ\mathcal{S}_p=\mathcal{S}_S$. \item \res{In the same plane are given the lines $a$, $b$ and $c$. Draw the points $A\in a$ and $B\in b$ so that $\mathcal{S}_c(A)=B$.} From $A\in a$ it follows that $\mathcal{S}_c(A)\in \mathcal{S}_c(a)$ or $B\in \mathcal{S}_c(a)$. Because $B\in b$ as well, we get the point $B$ from the condition $B\in \mathcal{S}_c(a)\cap b$. Then $A=\mathcal{S}_c(B)$ is also true. \item \res{Given are the lines $p$ and $q$ and the point $A$ in the same plane. Draw the points $B$ and $C$ so that the lines $p$ and $q$ will be the internal angle bisectors of the triangle $ABC$.} We use the fact that the line $BC$ is determined by the points $\mathcal{S}_p(A)$ and $\mathcal{S}_q(A)$. \item \res{Let $p$, $q$ and $r$ be lines and $K$ and $L$ points in the same plane. Draw the lines $s$ and $s'$, which go through the points $K$ and $L$ in order, so that $\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p(s)=s'$ is true.} We denote $\mathcal{I}=\mathcal{S}_r\circ\mathcal{S}_q\circ\mathcal{S}_p$. The line $s'$ is determined by the points $L$ and $\mathcal{I}(K)$, the line $s$ is determined by the points $K$ and $\mathcal{I}^{-1}(K)$. If $\mathcal{I}(K)=L$, the task has infinitely many solutions - the line $s$ is an arbitrary line, which goes through the point $K$. \item \res{Let $s$ be the line which bisects one of the angles determined by the lines $p$ and $q$. Prove that $\mathcal{S}_s\circ\mathcal{S}_p = \mathcal{S}_q\circ\mathcal{S}_s$.} We can even prove more - that the equivalence (under the assumption that the lines $p$ and $q$ intersect) holds. If we multiply the given equality by $\mathcal{S}_s$ from the right, we get the equivalent equality $\mathcal{S}_s\circ\mathcal{S}_p\circ\mathcal{S}_s = \mathcal{S}_q$. By \ref{izoTransmutacija}, this is equivalent to the equality $\mathcal{S}_{\mathcal{S}_s(p)} = \mathcal{S}_q$ or $\mathcal{S}_s(p) = q$, which is equivalent to the fact that $s$ bisects one of the angles determined by the lines $p$ and $q$. \item \label{nalIzo27} \res{Let $S$ be the center of the triangle $ABC$ inscribed in the circle and $P$ the point, in which this circle touches the side $BC$. Prove: $$\mathcal{S}_{SC} \circ\mathcal{S}_{SA}\circ\mathcal{S}_{SB} =\mathcal{S}_{SP}.$$} The equality is a direct consequence of the statement \ref{izo1ftIndZrc}, because is\\ $\mathcal{S}_{SC}\circ\mathcal{S}_{SA}\circ\mathcal{S}_{SB}:S,P\mapsto S,P$. \item \res{The lines $p$, $q$ and $r$ of a plane go through the center $S$ of the circle $k$. Draw a triangle $ABC$, which is inscribed in this circle, so that the lines $p$, $q$ and $r$ will be the altitudes of the internal angles at the vertices $A$, $B$ and $C$ of this triangle.} We use the previous task \ref{nalIzo27}. \item \res{The lines $p$, $q$, $r$, $s$ and $t$ of a plane intersect in the point $O$, the point $M$ lies on the line $p$. Draw such a pentagon that $M$ will be the center of one of its sides, the lines $p$, $q$, $r$, $s$ and $t$ will be the altitudes of the sides.} The point $M$ lies on one of the altitudes. Without loss of generality, let $M\in p$. Let $ABCDE$ be such a pentagon that the lines $p$, $q$, $r$, $s$ and $t$ are in turn the altitudes of its sides $AB$, $BC$, $CD$, $DE$ and $EA$. Let $\mathcal{I}=\mathcal{S}_t\circ\mathcal{S}_s\circ\mathcal{S}_r \circ\mathcal{S}_q\circ\mathcal{S}_p$. Because $\mathcal{I}:O,A\mapsto O,A$, by the statement \ref{izo1ftIndZrc} $\mathcal{I}=\mathcal{S}_{OA}$. The line $OA$ is obtained as the altitude of the segment $XX'$, where $X$ is an arbitrary point and $X'=\mathcal{I}(X)$, then the vertex $A$ as the intersection of the line $OA$ and the rectangle of the line $p$ through the point $M$. \item \res{The point $P$ lies in the plane of the triangle $ABC$. Prove that the lines, which are symmetrical to the lines $AP$, $BP$ and $CP$ with respect to the altitudes of at the vertices $A$, $B$ and $C$ of this triangle, belong to the same set.} Let us denote by $s_{\alpha}$, $s_{\alpha}$ and $s_{\alpha}$ the internal angle bisectors at the vertices $A$, $B$ and $C$ of the triangle $ABC$ and $a=\mathcal{S}_{s_{\alpha}}(AP)$, $b=\mathcal{S}_{s_{\beta}}(BP)$ and $c=\mathcal{S}_{s_{\gamma}}(CP)$. We will prove that the lines $a$, $b$ and $c$ belong to the same pencil. Because $\mathcal{S}_{s_{\alpha}}:AC, p\rightarrow AB, a$ is $\measuredangle CAP=\measuredangle a,AP$. Therefore $\mathcal{R}_{A,2\measuredangle CAP} =\mathcal{R}_{A,2\measuredangle a,AP}$ or $\mathcal{S}_{AC}\circ \mathcal{S}_{AP} =\mathcal{S}_a\circ \mathcal{S}_{AB}$. From this it follows that $\mathcal{S}_a=\mathcal{S}_{AC} \circ\mathcal{S}_{AP}\circ \mathcal{S}_{AB}$. Similarly: $\mathcal{S}_b=\mathcal{S}_{BA} \circ\mathcal{S}_{BP}\circ \mathcal{S}_{BC}$ and $\mathcal{S}_c=\mathcal{S}_{CB}\circ\mathcal{S}_{CP} \circ \mathcal{S}_{CA}$. Therefore it is (from \ref{izoSop} and \ref{izoTransmutacija}): \begin{eqnarray*} \mathcal{I}&=&\mathcal{S}_a\circ\mathcal{S}_b\circ\mathcal{S}_c=\\ &=& \mathcal{S}_{AC}\circ\mathcal{S}_{AP}\circ \mathcal{S}_{AB}\circ \mathcal{S}_{BA}\circ\mathcal{S}_{BP}\circ \mathcal{S}_{BC}\circ \mathcal{S}_{CB}\circ\mathcal{S}_{CP}\circ \mathcal{S}_{CA}=\\ &=& \mathcal{S}_{AC}\circ\mathcal{S}_{AP}\circ \mathcal{S}_{BP}\circ\mathcal{S}_{CP}\circ \mathcal{S}_{CA}=\\ &=& \mathcal{S}_{AC}\circ\mathcal{S}_{AX}\circ \mathcal{S}_{CA}=\\ &=& \mathcal{S}_{\mathcal{S}_{AC}(AX)}. \end{eqnarray*} By \ref{izoSop} the lines $a$, $b$ and $c$ belong to the same pencil. \item \res{Calculate the angle determined by the lines $p$ and $q$, if it holds that: $\mathcal{S}_p\circ\mathcal{S}_q\circ\mathcal{S}_p = \mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{S}_q$.} If we multiply the given equality from the left side in order by $\mathcal{S}_q$, $\mathcal{S}_p$ and $\mathcal{S}_q$, we get an equivalent equality: $\mathcal{S}_q\circ\mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_p\circ\mathcal{S}_q\circ \mathcal{S}_p = \mathcal{E}$ or $\left(\mathcal{S}_q\circ\mathcal{S}_p\right)^3 = \mathcal{E}$. If $p=q$, the last equality is clearly fulfilled. In the case $p\parallel q$, $\mathcal{S}_q\circ\mathcal{S}_p$ is a translation, so it is always the case that $\left(\mathcal{S}_q\circ\mathcal{S}_p\right)^3 =\mathcal{T}^3_{\overrightarrow{v}}=\mathcal{T}_{3\overrightarrow{v}}\neq \mathcal{E}$. If the lines $p$ and $q$ intersect, the composite $\mathcal{S}_q\circ\mathcal{S}_p$ is a rotation $\mathcal{R}_{S,\omega}$ ($p\cap q=\{S\}$ and $\omega=2\measuredangle p,q$). In this case, it is therefore the case that $\left(\mathcal{S}_q\circ\mathcal{S}_p\right)^3 =\mathcal{R}^3_{S,\omega}=\mathcal{R}_{S,3\omega}= \mathcal{E}$. This means that $3\omega=360^0$ or $\omega=120^0$, so that the lines $p$ and $q$ determine an angle of $60^0$. \item \res{Let $\mathcal{R}_{A,\alpha}$ and $\mathcal{R}_{B,\beta}$ be rotations in the same plane. Determine all points $X$ in this plane for which it is true that $\mathcal{R}_{A,\alpha}(X)=\mathcal{R}_{B,\beta}(X)$.} The condition $\mathcal{R}_{A,\alpha}(X)=\mathcal{R}_{B,\beta}(X)$ is equivalent to the condition $\mathcal{R}_{A,\alpha}\circ\mathcal{R}^{-1}_{B,\beta}(X)=X$ or $\mathcal{R}_{A,\alpha}\circ\mathcal{R}_{B,-\beta}(X)=X$. It is therefore necessary to determine the fixed points of the isometry $\mathcal{I}=\mathcal{R}_{A,\alpha}\circ\mathcal{R}_{B,-\beta}$. We will use Theorem \ref{rotacKomp2rotac}. We will consider several cases. \textit{1}) If $A=B$ and $\alpha-\beta=k\cdot360^0$ (for some $k\in \mathbb{Z}$), then $\mathcal{I}=\mathcal{E}$, so the condition is clearly true for every point in this plane. \textit{2}) If $A=B$ and $\alpha-\beta\neq k\cdot360^0$ (for every $k\in \mathbb{Z}$), then $\mathcal{I}=\mathcal{R}_{A,\alpha-\beta}$ and the condition is fulfilled only for the point $X=A$. \textit{3}) If $A\neq B$ and $\alpha-\beta=k\cdot360^0$ (for some $k\in \mathbb{Z}$), then $\mathcal{I}$ is a translation, so the condition does not apply to any point of this plane. \textit{4}) If $A\neq B$ and $\alpha-\beta\neq k\cdot360^0$ (for every $k\in \mathbb{Z}$), then $\mathcal{R}_{C,\alpha-\beta}$ (where $\measuredangle CAB=\frac{\alpha}{2}$ and $\measuredangle BAC=\frac{\beta}{2}$), which means that the condition applies only to the point $C$. \item \res{The lines $p$ and $q$ intersect at an angle of $60^0$ at the center $O$ of an isosceles triangle $ABC$. Prove that the segments determined by the sides of the triangle $ABC$ on the lines are congruent segments.} We use the rotation $\mathcal{R}_{O,120^0}$. \item \res{Let $S$ be the center of a regular pentagon $ABCDE$. Prove that: $$\overrightarrow{SA} + \overrightarrow{SB} + \overrightarrow{SC} + \overrightarrow{SD} + \overrightarrow{SE} = \overrightarrow{0}.$$} Let $\overrightarrow{SA} + \overrightarrow{SB} + \overrightarrow{SC} + \overrightarrow{SD} + \overrightarrow{SE} = \overrightarrow{SX}$. Because $ABCDE$ is a regular pentagon, we have $\mathcal{R}_{S,72^0}:A,B,C,D,E,S,X\mapsto B,C,D,E,A,S,X'$. Therefore, $ \overrightarrow{SB} + \overrightarrow{SC} + \overrightarrow{SD} + \overrightarrow{SE} +\overrightarrow{SA} = \overrightarrow{SX'}$. This means that $\overrightarrow{SX}= \overrightarrow{SX'}$ or $X=X'=\mathcal{R}_{S,72^0}(X)$. Because $S$ is the only fixed point of the rotation $\mathcal{R}_{S,72^0}$, we have $X=S$. Therefore, $\overrightarrow{SA} + \overrightarrow{SB} + \overrightarrow{SC} + \overrightarrow{SD} + \overrightarrow{SE} = \overrightarrow{SS}=\overrightarrow{0}$. \item \res{Prove that the diagonals of a regular pentagon intersect at points that are also the vertices of a regular pentagon.} We use the rotation with center at the center of a regular pentagon by an angle of $72^0$. \item \res{Let $ABP$ and $BCQ$ be two triangles with the same orientation and $\mathcal{B}(A,B,C)$. The points $K$ and $L$ are the centers of the segments $AQ$ and $PC$. Prove that $BLK$ is a triangle.} By rotating $\mathcal{R}_{B,-60^0}$, the points $A$ and $Q$ are mapped to the points $P$ and $C$. Therefore, the segment $AQ$ and its center $K$ are mapped to the segment $PC$ and its center $L$. From $\mathcal{R}_{B,-60^0}(K)=L$ it follows that $BLK$ is a triangle. \item \res{There are three concentric circles and a line in the same plane. Draw a triangle so that its vertices are on these circles in order, one side is parallel to the given line.} First, we draw an arbitrary triangle without the condition that one of its sides is parallel to the given line. We can do this by choosing an arbitrary vertex $A$ on one of the circles and using the rotation $\mathcal{R}_{A,60^0}$. Then we use the appropriate rotation with the center at the center of the concentric circles, which maps the drawn triangle into a triangle whose one side is parallel to the given line. \item \res{The point $P$ is an internal point of a regular triangle $ABC$, so that $\angle APB=113^0$ and $\angle BPC=123^0$. Calculate the size of the angles of the triangle whose sides are in correspondence with the segments $PA$, $PB$ and $PC$.} First, it is clear that $\angle APC=360^0-113^0-123^0=124^0$. We mark $P'=\mathcal{R}_{A,60^0}(P)$. $APP'$ is a regular triangle, so $PA\cong P'P$. Since $C=\mathcal{R}_{A,60^0}(B)$, we have $\triangle ABP\cong\triangle ACP'$, so $PB\cong P'C$ and $\angle APB\cong\angle AP'C$. This means that the sides $P'P$, $P'C$ and $PC$ of the triangle $PCP'$ are in correspondence with the segments $PA$, $PB$ and $PC$ in order. In this case, $\angle P'PC=124^0-60^0=64^0$, $\angle PP'C=113^0-60^0=53^0$ and $\angle PP'C=180^0-64^0-53^0=63^0$. \item \res{Given are the nonlinear points $P$, $Q$ and $R$. Draw a triangle $ABC$, so that $P$, $Q$ and $R$ are the centers of the squares that are constructed over the sides $BC$, $CA$ and $AB$ of this triangle.} We find a fixed point of the composite $\mathcal{R}_{Q,90^0}\circ\mathcal{R}_{R,90^0}\circ\mathcal{R}_{P,90^0}$. \item \res{Let $A$ and $B$ be points and $p$ a line in the same plane. Prove that the composite $\mathcal{S}_B\circ\mathcal{S}_p\circ\mathcal{S}_A$ is a central reflection exactly when $AB\perp p$.} We represent the central reflections $\mathcal{S}_A$ and $\mathcal{S}_B$ as composites $\mathcal{S}_A=\mathcal{S}_a\circ\mathcal{S}_{a_1}$ and $\mathcal{S}_B=\mathcal{S}_{b_1}\circ\mathcal{S}_b$, where $a$ and $b$ are perpendicular to the line $p$. \item \res{Let $p$, $q$ and $r$ be tangents to the triangle $ABC$ of inscribed circles, which are parallel to its sides $BC$, $AC$ and $AB$. Prove that the lines $p$, $q$, $r$, $BC$, $AC$ and $AB$ determine such a hexagon, in which the pairs of opposite sides are congruent distances.} We use the central reflection $\mathcal{S}_S$, where $S$ is the center of the inscribed circle of the triangle $ABC$. \item \res{Draw a triangle with the data: $\alpha$, $t_b$, $t_c$.} First, we can draw the altitude $BB_1\cong t_b$ and the center $T$. Because the vertex $A$ lies on the arc $l$ over the altitude $BB_1$ and the obtuse angle $\alpha$, the vertex $B$ lies on the circle $k(T,\frac{2}{3}t_c)$, and we have $C=\mathcal{S}_{B_1}(A)$, we get the vertex $C$ from the condition $C\in k(T,\frac{2}{3}t_c)\cap\mathcal{S}_{l}(A)$. \item \res{Let $ALKB$ and $ACPQ$ be squares that are drawn outside the triangle $ABC$ over the sides $AB$ and $AC$, and $X$ the center of the side $BC$. Prove that $AX\perp LQ$ and $|AX|=\frac{1}{2}|QL|$.} Let $Q'=\mathcal{S}_A(C)$. In this case it holds that $R_{A,90^0}:Q,L\mapsto Q',B$. So $QL\cong Q'B$ and $QL\perp Q'B$ (statement \ref{rotacPremPremKot}). Because $AX$ is the median of triangle $BCQ$ for the base $BQ'$, it follows that $\overrightarrow{AX}=\frac{1}{2}\overrightarrow{BQ'}$, therefore $AX\perp LQ$ and $|AX|=\frac{1}{2}|QL|$. \item \res{Let $O$ be the center of the regular triangle $ABC$ and $D$ and $E$ points on the sides $CA$ and $CB$, so that $CD\cong CE$ holds. The point $F$ is the fourth vertex of the parallelogram $BODF$. Prove that the triangle $OEF$ is regular.} Let $\mathcal{I}=\mathcal{T}_{\overrightarrow{OB}}\circ \mathcal{R}_{C,-60^0}$. By izrek \ref{izoKompTranslRot} it holds that $\mathcal{I}$ is a rotation by the same angle $-60^0$ with the center in some point $\widehat{O}$. So $\mathcal{T}_{\overrightarrow{OB}}\circ \mathcal{R}_{C,-60^0}=\mathcal{R}_{\widehat{O},-60^0}$. In this case $\mathcal{R}_{\widehat{O},-60^0}(E)=\mathcal{T}_{\overrightarrow{OB}}\circ \mathcal{R}_{C,-60^0}(E)=F$, therefore $\widehat{O}EF$ is a regular triangle. It remains to be proven that $\widehat{O}=O$ or that $O$ is a fixed point of the rotation $\mathcal{R}_{\widehat{O},-60^0}$. If $O'=\mathcal{R}_{C,-60^0}(O)$, then $O'$ is the center of the isosceles triangle $AB'C$ (where $C'=\mathcal{S}_{AC}(B)$). So $\overrightarrow{O'O}=\overrightarrow{OB}$ or $\mathcal{T}_{\overrightarrow{OB}}(O')=O$. From this it follows that $\mathcal{R}_{\widehat{O},-60^0}(O)=O$ or $\widehat{O}=O$, which means that $OEF$ is a regular triangle. \item \res{Let $L$ be the point in which the inscribed circle of triangle $ABC$ touches its side $BC$. Prove: $$\mathcal{R}_{C,\measuredangle ACB}\circ\mathcal{R}_{A,\measuredangle BAC} \circ\mathcal{R}_{B,\measuredangle CBA} =\mathcal{S}_L.$$} Let $M$ and $N$ be the points where the inscribed circle of the triangle $ABC$ touches its sides $AC$ and $AB$. By the statement in \ref{rotacKomp2rotac}, $\mathcal{R}_{C,\measuredangle ACB} \circ\mathcal{R}_{A,\measuredangle BAC}\circ \mathcal{R}_{B,\measuredangle CBA}=\mathcal{S}_{\widehat{L}}$. Since: \begin{eqnarray*} \mathcal{S}_{\widehat{L}}(L)&=& \mathcal{R}_{C,\measuredangle ACB} \circ\mathcal{R}_{A,\measuredangle BAC}\circ \mathcal{R}_{B,\measuredangle CBA}(L)=\\ &=& \mathcal{R}_{C,\measuredangle ACB} \circ\mathcal{R}_{A,\measuredangle BAC}(N)= \mathcal{R}_{C,\measuredangle ACB}(M)= L, \end{eqnarray*} it follows that $\widehat{L}=L$ or $\mathcal{R}_{C,\measuredangle ACB} \circ\mathcal{R}_{A,\measuredangle BAC}\circ \mathcal{R}_{B,\measuredangle CBA}=\mathcal{S}_L$. \item \res{The points $P$ and $Q$ as well as $M$ and $N$ are the centers of two squares that are drawn outside of the opposite sides of any quadrilateral. Prove that $PQ\perp MN$ and $PQ\cong MN$.} Let $ABCD$ be a given quadrilateral, $P$ and $Q$ the centers of the squares constructed on the sides $AB$ and $CD$, and $M$ and $N$ the centers of the squares constructed on the sides $BC$ and $AD$. The composite $\mathcal{I}=\mathcal{R}_{N,90^0}\circ \mathcal{R}_{P,90^0}$ according to the theorem \ref{rotacKomp2rotac} represents a central reflection. Because in this case $\mathcal{I}(B)=D$, the center of this reflection is actually the center of the diagonal $BD$; we denote it with $S$. So $\mathcal{R}_{N,90^0}\circ \mathcal{R}_{P,90^0}=\mathcal{S}_S$. For the point $S$ according to the same theorem \ref{rotacKomp2rotac} it holds that $\angle NPS=\frac{1}{2}90^0=45^0$ and $\angle PNS=\frac{1}{2}90^0=45^0$. This means that $PNS$ is an isosceles right-angled triangle with the base $NP$ and the right angle at the vertex $S$. Analogously, $MSQ$ is an isosceles right-angled triangle with the base $MQ$ and the right angle at the vertex $S$. From these two facts it follows that $R_{S,90^0}: M,N\mapsto Q,P$, so $MN\cong QP$ and $MN\perp QP$ (theorem \ref{rotacPremPremKot}). \item \res{Let $APB$ and $ACQ$ be two right-angled triangles outside the triangle $ABC$, constructed on the sides $AB$ and $AC$. The point $S$ is the center of the side $BC$ and $O$ is the center of the triangle $ACQ$. Prove that $|OP|=2|OS|$.} Let $\mathcal{I}=\mathcal{R}_{O,120^0}\circ \mathcal{R}_{P,60^0}$. According to the theorem \ref{rotacKomp2rotac} we have: $$\mathcal{I}=\mathcal{R}_{O,120^0}\circ \mathcal{R}_{P,60^0}=\mathcal{R}_{\widehat{S},180^0} =\mathcal{S}_{\widehat{S}},$$ where $\widehat{S}$ is the vertex of the triangle $OP\widehat{S}$ and $\angle \widehat{S}PO=\frac{1}{2}60^0=30^0$ and $\angle PO\widehat{S}=\frac{1}{2}120^0=60^0$. Because $\mathcal{S}_{\widehat{S}}(S)=\mathcal{I}(S)=S$ or $\widehat{S}=S$. If we denote $O'=\mathcal{S}_S$, then $POO'$ is a right-angled triangle, so $|OP|=|OO'|=2|OS|$. \item \res{Prove that the reflection in an axis and the translation of a plane commute if and only if the axis of this reflection is parallel to the translation vector.} If we use the statement \ref{izoTransmutacija}, we get: \begin{eqnarray*} \mathcal{T}_{\overrightarrow{v}}\circ \mathcal{S}_p= \mathcal{S}_p\circ\mathcal{T}_{\overrightarrow{v}} &\Leftrightarrow& \mathcal{T}_{\overrightarrow{v}}\circ \mathcal{T}_p\circ\mathcal{T}^{-1}_{\overrightarrow{v}}= \mathcal{S}_p\\ &\Leftrightarrow& \mathcal{S}_{\mathcal{T}_{\overrightarrow{v}}(p)}= \mathcal{S}_p \Leftrightarrow \mathcal{T}_{\overrightarrow{v}}(p)= p \Leftrightarrow \overrightarrow{v}\parallel p. \end{eqnarray*} \item \res{In the same plane are given a line $p$, circles $k$ and $l$, and a distance $d$. Draw a rhombus $ABCD$ with a side that is congruent to the distance $d$, the side $AB$ lies on the line $p$, the vertices $C$ and $D$ are in turn on the circles $k$ and $l$.} Use a translation for the vector $\overrightarrow{v}$, which is parallel to the line $p$ and $|\overrightarrow{v}|=|d|$. In this case, $D\in \mathcal{T}_{\overrightarrow{v}}(k)\cap l$. \item \res{Let $p$ be a line, $A$ and $B$ be points that lie on the same side of the line $p$, and $d$ be a distance in the same plane. Draw points $X$ and $Y$ on the line $p$ so that $AX\cong BY$ and $XY\cong d$.} Let $\overrightarrow{v}$ be a vector that is parallel to the line $p$ and $|\overrightarrow{v}|=|d|$. Let $A'=\mathcal{T}_{\overrightarrow{v}}(A)$, $Y$ be the intersection of the line $A'B$ with the line $p$, and $X=\mathcal{T}^{-1}_{\overrightarrow{v}}(Y)$. The point $Y$ lies on the line $A'B$, so $A'Y\cong YB$. Because $\mathcal{T}_{\overrightarrow{v}}:A,X\mapsto A',Y$ is a parallelogram, $AYYA'$ is a parallelogram, so $AX\cong A'Y\cong BY$. From $\mathcal{T}_{\overrightarrow{v}}(X)=Y$ it follows that $\overrightarrow{XY}=\overrightarrow{v}$ or $|XY|=|\overrightarrow{v}|=|d|$. \item \res{Let $H$ be the orthocenter of the triangle $ABC$ and $R$ be the radius of the circumscribed circle of this triangle. Prove that $|AB|^2+|CH|^2=4R^2$.} Let $O$ be the center of the circle $k(O,R)$ inscribed in the triangle $ABC$ and $A'=\mathcal{S}_O(A)$. The distance $AA'$ is the diameter of the circle $k$, so $\angle ACA'=90^0$ or $A'C\perp AC$. Because $BH\perp AC$, $A'C\parallel BH$. Similarly $A'B\parallel CH$, which means that the quadrilateral $BA'CH$ is a parallelogram. Therefore $CH\cong A'B$. Because $A'BA$ is a right triangle, by the Pythagorean Theorem \ref{PitagorovIzrek}: $|AB|^2+|CH|^2=|AB|^2+|A'B|^2=|AA'|^2=4R^2$. \item \res{Let $EAB$ be a triangle that is drawn over the side $AB$ of the square $ABCD$. Let also $M=pr_{\perp AE}(C)$ and $N=pr_{\perp BE}(D)$ and the point $P$ be the intersection of the lines $CM$ and $DN$. Prove that $PE\perp AB$.} We use a translation for the vector $\overrightarrow{CB}$. \item \res{Draw an isosceles triangle $ABC$ so that its vertices in order lie on three parallel lines $a$, $b$ and $c$ in the same plane, the center of this triangle lies on the line $s$, which intersects the lines $a$, $b$ and $c$ through the point $S_1$.} First, we draw any regular triangle $A_1B_1C_1$, where $A_1\in a$, $B_1\in b$ and $C_1\in c$, then we use a translation for the vector $\overrightarrow{S_1S}$, where $S_1$ is the center of the triangle $A_1B_1C_1$, the point $S$ is the intersection of the line $s$ with the parallel line through the point $S_1$. \item \res{If a pentagon has at least two axes of symmetry, it is regular. Prove it.} Let $\mathfrak{G}(\mathcal{V}_5)$ be the group of symmetries of our pentagon $\mathcal{V}_5$ and $p$ and $q$ its axes of symmetry. It is clear that this group is finite, so according to the theorem \ref{GrupaLeonardo} it represents either the cyclic group $\mathfrak{C}_n$ or the dihedral group $\mathfrak{D}_n$. Because it contains the axes of symmetry, $\mathfrak{G}(\mathcal{V}_5)=\mathfrak{D}_n$. In this case it is clear that $n\leq 5$. Because $\mathcal{S}_p, \mathcal{S}_q \in \mathfrak{G}(\mathcal{V}_5)=\mathfrak{D}_n$, the axes $p$ and $q$ intersect in some point $S$, the composition $\mathcal{S}_q\circ \mathcal{S}_p$ represents the rotation $\mathcal{R}_{S, \alpha}$. Because the group $\mathfrak{G}(\mathcal{V}_5)=\mathfrak{D}_n$ contains at least two axes of symmetry, $n\geq 2$. Therefore $n\in \{2,3,4,5\}$. In this case the basic rotation of this group (for the smallest angle) is $\mathcal{R}_{S, \frac{360^0}{n}}$. Because the pentagon has five vertices, the number $n$ is a divisor of the number 5, which means $n=5$. Therefore $\mathfrak{G}(\mathcal{V}_5)=\mathfrak{D}_5$, therefore $\mathcal{V}_5$ is a regular pentagon. \item \res{Let $A$, $B$ and $C$ be three collinear points. What does the composition $\mathcal{G}_{\overrightarrow{BC}}\circ \mathcal{S}_A$ represent?} We denote with $p$ the line passing through the points $A$, $B$ and $C$. Let $b$ and $a$ be the perpendiculars to the line $p$ in the points $A$ and $B$ and $s$ the symmetry of the line segment $BC$. Then: $$\mathcal{G}_{\overrightarrow{BC}}\circ \mathcal{S}_A= \mathcal{S}_s\circ \mathcal{S}_b\circ \mathcal{S}_p \circ \mathcal{S}_p\circ \mathcal{S}_a= \mathcal{S}_s\circ \mathcal{S}_b\circ \mathcal{S}_a.$$ Because the lines $a$, $b$ and $s$ are perpendicular to the line $p$, they are from the group of parallel lines, therefore according to the theorem \ref{izoSop} the composition $\mathcal{S}_s \circ \mathcal{S}_b\circ \mathcal{S}_a$ (or $\mathcal{G}_{\overrightarrow{BC}} \circ \mathcal{S}_A$) is the axis of symmetry. \item \res{Let $p$, $q$ and $r$ be lines that are not in the same plane, and let $A$ be a point in the same plane. Draw a line $s$ that goes through the point $A$ such that $\mathcal{S}_r\circ \mathcal{S}_q\circ \mathcal{S}_p(s)=s'$ and $s\parallel s'$.} According to the theorem \ref{izoZrcdrsprq}, the composite $\mathcal{S}_r\circ \mathcal{S}_q\circ \mathcal{S}_p(s)=s'$ is a reflection - we will denote it with $\mathcal{G}_{2\overrightarrow{PQ}}$. The line $s$ is parallel or perpendicular to the axis of the reflection. This is determined by the centers of the lines $XX'$ and $YY'$, where $X$ and $Y$ are any points and $\mathcal{S}_r \circ \mathcal{S}_q\circ \mathcal{S}_p: X, Y\mapsto X', Y'$. %new tasks %___________________________________ \item \res{Let $Z$ and $K$ be inner points of the rectangle $ABCD$. Draw points $A_1$, $B_1$, $C_1$ and $D_1$, which in order lie on the sides $AB$, $BC$, $CD$ and $DA$ of this rectangle, so that it holds $\angle ZA_1A\cong\angle B_1A_1B$, $\angle A_1B_1B\cong\angle C_1B_1C$, $\angle B_1C_1C\cong\angle D_1C_1D$ and $\angle C_1D_1D\cong\angle KD_1A$.} First, we draw points $Z'=S_{CB}\circ S_{AB}(Z)=S_B(Z)$ and $K'=S_{CD}\circ S_{AD}(Z)=S_D(K)$. Then we prove and use the fact that points $Z'$, $B_1$, $C_1$ and $K'$ are collinear. \item \res{The point $A$ lies on the line $a$, and the point $B$ lies on the line $b$. Determine the rotation that transforms the line $a$ into the line $b$ and the point $A$ into the point $B$.} If the lines $a$ and $b$ are parallel, the desired rotation is the central reflection with the center that is the center of the line $AB$. If the lines $a$ and $b$ intersect in the point $O$, the center of the rotation is the intersection of the simetral of the line $AB$ and the simetral of the angle $aOb$, and the rotation is equal to the angle $aOb$. \item \res{In the center of the square, two rectangles intersect. Prove that these rectangles intersect the sides of the square in the points that are the vertices of a new square.} We use the rotation $\mathcal{R}_{S,90^0}$, where $S$ is the center of the square. \item \res{Given is a circle $k$ and lines $a$, $b$, $c$, $d$ and $e$, which lie in the same plane. Draw a pentagon on the circle $k$ with sides, which in order are parallel to the lines $a$, $b$, $c$, $d$ and $e$.} First, we plan the simetrale $p$, $q$, $r$, $s$ and $t$ of the sides of the desired pentagon $ABCDE$, which are actually perpendicular to the lines $a$, $b$, $c$, $d$ and $e$ from the center $O$ of the circle $k$. The composition $\mathcal{I}= \mathcal{S}_t\circ \mathcal{S}_s\circ \mathcal{S}_r\circ \mathcal{S}_q\circ \mathcal{S}_p$ is an indirect isometry with fixed points $O$ and $A$, so according to izreku \ref{izo1ftIndZrc} $\mathcal{I}=\mathcal{S}_{OA}$. We can plan the line $OA$ as the simetral of the line $XX'$, where $X$ is an arbitrary point and $X'=\mathcal{I}(X)$. This allows us to construct the vertex $A$, and then the other vertices of the pentagon $ABCDE$. \item \res{The point $P$ lies inside the angle $aOb$. Draw a line $p$ through the point $P$, which with the sides $a$ and $b$ determines a triangle with the smallest area.} We will prove that a triangle has the smallest area, if the line $OP$ is its median. We will mark this triangle with $OAB$ ($A\in a$ and $B\in b$). Let $X$ and $Y$ be the intersections of any line (different from $AB$) through the point $P$. We will prove that the area of the triangle $XOY$ is greater than the area of the triangle $AOB$. Without loss of generality, let $\mathcal{B}(O,X,A)$ or $\mathcal{B}(O,B,Y)$. In this case, we will mark points $A$ and $B$ in the following way. Let $a'=\mathcal{S}_P(a)$, $\{B\}=b\cap a'$ and $A=\mathcal{S}_P(B)$. We will mark $X'=\mathcal{S}_P$. It is clear that $X'\in a'$. Because $\mathcal{S}_P:X,A,P\mapsto X',B,P$, the triangles $XAP$ and $X'BP$ are congruent, therefore they have the same area. If we mark $p_{V_1V_2\cdots V_n}$ the area of any polygon $V_1V_2\cdots V_n$, it is: \begin{eqnarray*} p_{AOB}&=&p_{AXPB}+p_{XAP}=\\ &=&p_{AXPB}+p_{X'BP}>p_{AXPB}+p_{X'BP}+ p_{X'BY}=\\ &=&p_{XOY}. \end{eqnarray*} \item \res{The parallelogram $PQKL$ is inscribed in the parallelogram $ABCD$ (the vertices of the first one lie on the sides of the second one). Prove that the parallelograms have a common center.} We will use the central reflection $\mathcal{S}_S$, where $S$ is the intersection of the diagonals of the parallelogram $ABCD$. \item \res{The arcs $l_1, l_2,\cdots , l_n$ lie on the circle $k$ and the sum of their lengths is less than the radius of this circle. Prove that there exists such a diameter $PQ$ of the circle $k$, that none of its endpoints lies on any of the arcs $l_1, l_2,\cdots , l_n$.} Let $S$ be the center of the circle $k$. Let $\mathcal{S}_S:l_1, l_2,\cdots , l_n\rightarrow l'_1, l'_2,\cdots , l'_n$. Assume the contrary, that for each diameter $PQ$ one of its endpoints lies on one of the arcs $l_1, l_2,\cdots , l_n$. Because $\mathcal{S}_S(P)=Q$, both endpoints lie on some arc $l_1, l_2,\cdots , l_n$, $l'_1, l'_2,\cdots , l'_n$. This means that every point of the circle $k$ lies on one of the arcs $l_1, l_2,\cdots , l_n$, $l'_1, l'_2,\cdots , l'_n$, which is not possible, since the total length of these arcs is less than the circumference of the circle $k$. \item \res{The circle $K(S,20)$ is given. The players $\mathcal{A}$ and $\mathcal{B}$ alternately draw circles with radii $x_i$ ($1b$) are given lengths, draw such a length $x$, that it is true:}\\ (\textit{a}) \res{$x=\sqrt{a^2+b^2}$,} \hspace*{3mm} (\textit{b}) \res{$x=\sqrt{a^2-b^2}$,} \hspace*{3mm} (\textit{c}) \res{$x=\sqrt{3ab}$,}\\ (\textit{d}) \res{$x=\sqrt{a^2+bc}$,} \hspace*{3mm} (\textit{e}) \res{$x=\sqrt{3ab-c^2}$,} \hspace*{3mm} (\textit{f}) \res{$x=\frac{a\sqrt{ab+c^2}}{b+c}$.} Use the Pythagorean Theorem \ref{PitagorovIzrek} and the altitude theorem \ref{izrekVisinski}. %Stewart's theorem \item \res{Let $a$, $b$ and $c$ be the sides of a triangle and let $a^2+b^2=5c^2$. Prove that the centroids $t_a$ and $t_b$ are perpendicular to each other.} Let $T$ be the centroid of the given triangle $ABC$, and $t_a$ and $t_b$ be the lengths of the corresponding centroids. We prove that $ATB$ is a right triangle. By \ref{StwartTezisc} of Stewart's theorem \ref{StewartIzrek}, we have: \begin{eqnarray*} t_a^2=\frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4}\\ t_b^2=\frac{a^2}{2}+\frac{c^2}{2}-\frac{b^2}{4}. \end{eqnarray*} From this and from the assumption $a^2+b^2=5c^2$, we get: \begin{eqnarray*} |AT|^2+|BT|^2 &=& \left(\frac{2}{3}t_a\right)^2+\left(\frac{2}{3}t_b\right)^2=\\ &=& \frac{4}{9}\left(\frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4} +\frac{a^2}{2}+\frac{c^2}{2}-\frac{b^2}{4}\right)=\\ &=& \frac{4}{9}\left(\frac{a^2+b^2}{4}+c^2\right)=\\ &=& c^2=|AB|^2. \end{eqnarray*} By the inverse Pythagorean theorem \ref{PitagorovIzrekObrat}, $ATB$ is a right triangle with hypotenuse $AB$, which means that the corresponding centroids are perpendicular. \item \res{Let $a$, $b$, $c$ and $d$ be the sides, $e$ and $f$ the diagonals, and $x$ the distance determined by the centers of sides $b$ and $d$ of a quadrilateral. Prove: $$x^2 = \frac{1}{4} \left(a^2 +c^2 -b^2 -d^2 +e^2 +f^2 \right).$$} We use the consequence \ref{StwartTezisc} of Stewart's theorem \ref{StewartIzrek}. \item \res{Let $a$, $b$ and $c$ be the sides of a triangle $ABC$. Prove that the distance from the center $A_1$ of side $a$ to the vertex $A'$ of the altitude on this side is equal to: $$|A_1A'|=\frac{|b^2-c^2|}{2a}.$$} Without loss of generality, we assume that $b\geq c$. We denote $x=|A_1A'|$, $v_a=|AA'|$ and $t_a=|AA_1|$. If we use the Pythagorean Theorem \ref{PitagorovIzrek} for the triangle $AA'A_1$ and $AA'B$, we get: \begin{eqnarray*} v_a^2 &=& t_a^2-x^2;\\ v_a^2 &=& c^2- \left( \frac{a}{2} -x \right)^2. \end{eqnarray*} After subtracting the equations and solving the obtained equation for $x$, we get: \begin{eqnarray*} x=\frac{1}{a}\left( t_a^2-c^2+ \frac{a^2}{2} \right). \end{eqnarray*} Finally, we use the relation for $t_a^2$ from the Stewart Theorem \ref{StwartTezisc}: \begin{eqnarray*} x &=& \frac{1}{a}\left( t_a^2-c^2+ \frac{a^2}{4} \right)=\\ &=& \frac{1}{a}\left( \frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4}-c^2+ \frac{a^2}{4} \right)=\\ &=& \frac{b^2-c^2}{2a}. \end{eqnarray*} %Pappus and Pascal \item \res{Let ($A$, $B$, $C$) and ($A_1$, $B_1$, $C_1$) be two collinear points of a plane that are not on the same line. If $AB_1\parallel A_1B$ and $AC_1\parallel A_1C$, then $CB_1\parallel C_1B$. (\textit{Pappus' Theorem}\footnote{Pappus of Alexandria\index{Pappus} (3rd century BC), Greek mathematician. This is a generalization of Pappus' Theorem (see Theorem \ref{izrek Pappus}), if we choose points $X$, $Y$ and $Z$ at infinity.})} We use the Tales Theorem \ref{TalesovIzrek} and the converse Tales Theorem \ref{TalesovIzrekObr}. %Desargues' Theorem \item \label{nalPodDesarg1} \res{Let $P$, $Q$ and $R$ be such points of sides $BC$, $AC$ and $AB$ of the triangle $ABC$, that the lines $AP$, $BQ$ and $CR$ are from the same pencil. Prove: If $X=BC\cap QR$, $Y=AC\cap PR$ and $Z=AB\cap PQ$, then points $X$, $Y$ and $Z$ are collinear.} We use the Desargues Theorem \ref{izrekDesarguesEvkl} for the triangle $ABC$ and $PQR$. \item \res{Let $AA'$, $BB'$ and $CC'$ be the altitudes of the triangle $ABC$ and $X=B'C'\cap BC$, $Y=A'C'\cap AC$ and $Z=A'B'\cap AB$. Prove that $X$, $Y$ and $Z$ are collinear points.} A direct consequence of the \ref{VisinskaTocka} theorem and the previous task \ref{nalPodDesarg1}. \item \res{Let $A$ and $B$ be points outside the line $p$. Draw the intersection of the lines $p$ and $AB$ without directly drawing the line $AB$.} Draw any collinear points $C$, $C'$ and $S$, then: $Y=p\cap AC$, $X=p\cap BC$, $A'=SA\cap C'Y$ and $B'=SB\cap C'X$. By the Desargues theorem \ref{izrekDesarguesEvkl} the lines $AB$ and $A'B'$ intersect at the point $Z$, which is collinear with the points $X$ and $Y$, therefore lies on the line $p$. This means that the intersection of the lines $p$ and $AB$ is obtained as the intersection of the lines $p$ and $A'B'$, that is, without directly drawing the line $AB$. \item \res{Let $p$ and $q$ be lines of a plane that intersect at the point $S$, which is "outside the paper", and $A$ a point of this plane. Draw the line that goes through the points $A$ and $S$.} Use the \ref{izrekDesarguesOsNesk} theorem. \item \res{Draw a triangle so that its vertices lie on three given parallel lines and the sides of the triangle pass through three given points.} Use the Desargues theorem \ref{izrekDesarguesEvkl}. See example \ref{zgled 3.2}. %Ppotenca \item \res{The circle $k(S,r)$ is given.} (\textit{a}) \res{What values can the power of the point have with respect to the circle $k$?} Since according to \ref{izrekPotenca} the power of a point $P$ with respect to the circle $k$ is obtained as: $p(S,k)=|PS|^2-r^2$, the value of the power is on the interval $[-r^2,\infty)$. (\textit{b}) \res{What is the smallest value of this power and for which point is this minimum value achieved?} From the previous example it is clear that the minimum value of the power is $-r^2$ and is achieved for the center $S$ of the circle $k$. (\textit{c}) \res{Determine the set of all points for which the power with respect to the circle is equal to $\lambda\in \mathbb{R}$.} The condition $p(S,k)=|PS|^2-r^2=\lambda$ and $|PS|^2=r^2+\lambda$ are equivalent. In the case when $\lambda>-r^2$, it is a circle $k(S,r^2+\lambda)$. If $\lambda=-r^2$, the desired set is equal to $\{S \}$, and in the case $\lambda<-r^2$, the empty set $\emptyset$. \item \res{Let $k_a(S_a,r_a)$ be inscribed and $l(O,R)$ be the circumscribed circle of some triangle. Prove the equality\footnote{The statement is a generalization of Euler's formula for a circle (see Theorem \ref{EulerjevaFormula}). \index{Euler, L.} \textit{L. Euler} (1707--1783), Swiss mathematician.}: $$S_aO^2=R^2+2r_aR.$$} The proof is similar to the proof of Theorem \ref{EulerjevaFormula}. \item \res{Draw a circle that passes through the given points $A$ and $B$ and touches the given circle $k$.} The desired circle is denoted by $x$. We draw any circle $j$, which passes through the points $A$ and $B$, and the circle $k$ intersects in points $C$ and $D$. We use the fact that the intersection of the lines $AB$ and $CD$ is the power center of the circles $k$, $j$ and $x$. The task is one of the ten Apollonius' problems about the touch of circles (see Section \ref{odd9ApolDotik}). \item \res{Prove that the centers of the lines that are determined by the common tangents of two circles are collinear points.} The mentioned points lie on the power line of two circles. \item \res{Draw a circle that is perpendicular to two given circles, and the third given circle intersects in points that determine the diameter of this third circle.} Let the given circles be denoted in order by $k(K, r_k)$, $j(J,r_j)$ and $l(L,r_l)$, and the desired circle by $x(X,r_x)$. Let $P\in x\cap k$, $Q\in x\cap l$ and $R, R_1\in x\cap j$. By assumption, $x\perp k, l$. By Theorem \ref{TangPogoj}, $XP$ and $XQ$ are tangents of the circles $k$ and $l$ in points $P$ and $Q$. Since $XP\cong XQ$, the center $X$ of the circle $X$ lies on the power line $p(k,l)$ of the circles $k$ and $l$. We look for another geometric location of points $X$, so that we include the condition for the circle $j$. By assumption, $RR_1$ is the diameter of the circle $j$. This means that $J$ is the center of the line $RR_1$, and from the similarity of the triangles $XJR$ and $XJR_1$ (by the \textit{SSS} theorem \ref{SSS}) it follows that $\angle XJR=90^0$. By the Pythagorean theorem \ref{PitagorovIzrek} (for $\triangle XJR$) we then have: \begin{eqnarray*} |XR|^2=|XJ|^2+r_j^2. \end{eqnarray*} From the right triangle $XQL$ we get by the same theorem: \begin{eqnarray*} |XQ|^2=r_l^2-|XL|^2. \end{eqnarray*} Since $XR\cong XQ$, from the previous two relations it follows that $|XJ|^2+r_j^2=r_l^2-|XL|^2$ or: \begin{eqnarray*} |XJ|^2+|XL|^2=r_l^2-r_j^2. \end{eqnarray*} Therefore, the point $X$ lies on some circle $g$, which we can plot (see theorem \ref{GMTmnl}). This means that we get the point $S$ as the intersection of the circle $g$ and the power line $p(k,l)$ of the circles $k$ and $l$. %Various \item \res{Let $M$ and $N$ be the intersections of the sides $AB$ and $AC$ of the triangle $ABC$ with a line that goes through the center of the inscribed circle of this triangle and is parallel to its side $BC$. Express the length of the line $MN$ as a function of the lengths of the sides of the triangle $ABC$.} We use theorem \ref{HarmCetSimKota} and prove $\frac{AS}{SE}=\frac{b+c}{a}$. Then we use \ref{TalesovIzrek} and prove $\frac{MN}{BC}=\frac{AS}{AE}$. Result: $|MN|=\frac{a(b+c)}{a+b+c}$. \item \res{Let $AA_1$ be the altitude of the triangle $ABC$. The points $P$ and $Q$ are the intersections of the altitudes of the angles $AA_1B$ and $AA_1C$ with the sides $AB$ and $AC$. Prove that $PQ\parallel BC$.} By theorem \ref{HarmCetSimKota} we have: $$\frac{AP}{PB}=\frac{AA_1}{A_1B}=\frac{AA1}{A_1C}=\frac{AQ}{QC}.$$ By the converse of theorem \ref{TalesovIzrekObr} it follows that $PQ\parallel BC$. \item \res{In the triangle $ABC$, let the sum (or difference) of the internal angles $ABC$ and $ACB$ be equal to the right angle. Prove that $|AB|^2+|AC|^2=4r^2$, where $r$ is the radius of the circumscribed circle of this triangle.} Let $\alpha=\angle BAC$, $\beta=\angle ABC$ and $\gamma=\angle ACB$ and $k(O,r)$ be the circumscribed circle of the triangle $ABC$. If $\beta+\gamma=90^0$, then $\alpha=90^0$ (by \ref{VsotKotTrik}), which means that $ABC$ is a right triangle with hypotenuse $BC$; the statement is trivial in this case as a consequence of \ref{TalesovIzrKroz2} and Pythagoras' theorem \ref{PitagorovIzrek}. Let $\beta-\gamma=90^0$ or $\gamma=\beta-90^0$. We mark $C'=\mathcal{S}_O(C)$. By \ref{TalesovIzrKroz2}, we first have $\angle CAC'=90^0$. Because $BCC'A$ is a trapezoid, by \ref{TetivniPogoj}: \begin{eqnarray*} \angle C'CA &=& 90^0-\angle AC'C=\\ &=& 90^0-(180^0-\beta)=\beta-90^0=\\ &=& \gamma =\angle BCA. \end{eqnarray*} By \ref{ObodObodKot}, $AC'\cong AB$, so by Pythagoras' theorem \ref{PitagorovIzrek} we get: \begin{eqnarray*} |AB|^2+|AC|^2=|AC'|^2+|AC|^2=|C'C|^2=4r^2. \end{eqnarray*} \item \res{Let $AD$ be the altitude of the triangle $ABC$. Prove that the sum (or difference) of the internal angles $ABC$ and $ACB$ is equal to the right angle exactly when: $$\frac{1}{|AB|^2}+\frac{1}{|AC|^2}=\frac{1}{|AD|^2}.$$} We consider the triangle $ADB$ and $CDA$. \item \res{Express the distance between the centroid and the center of the circumscribed circle of the triangle as a function of the lengths of its sides and the radius of the circumscribed circle.} We use Stewart's theorem \ref{StewartIzrek} for the triangle $OAA_1$ and \ref{StwartTezisc}; $k(O,R)$ is the circumscribed circle, $T$ is the centroid and $A_1$ is the center of the side $BC$ of the triangle $ABC$. Result: $|OT|=\sqrt{R^2-\frac{1}{9} \left(a^2+b^2+c^2 \right)}$. \item \res{Prove that in a triangle $ABC$, the external angle bisector at vertex $A$ and the internal angle bisectors at vertices $B$ and $C$ intersect the opposite sides at three collinear points.} Use Menelaus' theorem \ref{izrekMenelaj} and theorem \ref{HarmCetSimKota}. \item \res{Prove that in a triangle $ABC$, the center of the altitude $AD$, the center of the inscribed circle and the point where side $BC$ touches the circumscribed circle of this triangle are three collinear points.} Use theorem \ref{velNalTockP'}. \item \res{Prove Simson's theorem \ref{SimpsPrem} by using Menelaus' theorem \ref{izrekMenelaj}.} Let $S$ be an arbitrary point of the circumscribed circle of triangle $ABC$ and $P$, $Q$ and $R$ the orthogonal projections of this point on sides $BC$, $AC$ and $BC$. Use: $\triangle SRA\sim\triangle SPC$, $\triangle SPB\sim\triangle SQA$ and $\triangle SQC\sim\triangle SRB$. \item \res{A line through point $M$ of side $AB$ of triangle $ABC$ intersects side $AC$ at point $K$. Calculate the ratio in which line $MK$ divides side $BC$, if $AM:MB=1:2$ and $AK:AC=3:2$.} Let $P$ be the intersection of lines $MK$ and $BC$. By Menelaus' theorem $$-1=\frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot \frac{\overrightarrow{CK}}{\overrightarrow{KA}}\cdot \frac{\overrightarrow{AM}}{\overrightarrow{MB}}= \frac{\overrightarrow{BP}}{\overrightarrow{PC}}\cdot \frac{-1}{3}\cdot\frac{1}{2},$$ so $\frac{\overrightarrow{BP}}{\overrightarrow{PC}}=6:1$. \item \res{Let $A_1$ be the center of side $BC$ of triangle $ABC$ and let $P$ and $Q$ be such points of sides $AB$ and $AC$, that $BP:PA=2:5$ and $AQ:QC=6:1$. Calculate the ratio in which line $PQ$ divides the median $AA_1$.} Let $R$ be the intersection of lines $PQ$ and $BC$. Use Menelaus' theorem \ref{izrekMenelaj} first for triangle $ABC$ and line $PQ$, then for triangle $AA_1C$ and the same line. The result is $17:60$. \item \res{Prove that in an arbitrary triangle, the lines determined by the vertices and the points of tangency of one circumscribed circle with the opposite sides intersect at a common point.} We use Ceva's theorem \ref{izrekCeva} and the big task \ref{velikaNaloga}. \item \res{What does the set of all points represent, from which the tangent of the two given circles represents two compatible lines?} Because in this case the power of these points with respect to the circle is equal, the desired set is part of its power line, which is outside both circles. \item \res{Let $PP_1$ and $QQ_1$ be the external tangents of the circles $k(O,r)$ and $k_1(O_1,r_1)$ (points $P$, $P_1$, $Q$ and $Q_1$ are the corresponding touch points). Let $S$ be the intersection of these two tangents, $A$ one of the intersections of the circles $k$ and $k_1$ and $L$ and $L_1$ the intersections of the line $SO$ with the lines $PQ$ and $P_1Q_1$. Prove that $\angle LAO\cong\angle L_1AO_1$.} Without loss of generality, let $\mathcal{B}(S,O,O_1)$. Let $h_{S,\lambda}$ be the central extension with coefficient $\lambda=\frac{\overrightarrow{SO_1}}{\overrightarrow{SO}}$. Then $h_{S,\lambda}(k)=k_1$ and: $$h_{S,\lambda}:\hspace*{1mm} O,P,Q,L \mapsto O_1,P_1,Q_1,L_1.$$ Let $A_1=h_{S,\lambda}(A)$. From $h_{S,\lambda}(k)=k_1$ and $A\in k$ it follows $A_1\in k_1$. By the theorem \ref{homotOhranjaKote} we have $\angle L_1A_1O_1\cong\angle LAO$, so it is enough to prove $\angle L_1A_1O_1\cong\angle L_1AO_1$. From the similarity of the right-angled triangles $SP_1O_1$ and $SL_1P_1$ (theorem \ref{PodTrikKKK}) and the theorem \ref{izrekPotenca} we get: \begin{eqnarray*} \overrightarrow{SO_1}\cdot \overrightarrow{SL_1}=|SP_1|^2=p(S.k_1)= \overrightarrow{SA}\cdot \overrightarrow{SA_1}. \end{eqnarray*} This means that $O_1L_1AA_1$ is a string quadrilateral, so $\angle L_1A_1O_1\cong\angle L_1AO_1$. \item \res{Prove that the side of a regular pentagon is equal to the larger part of the division of the radius of the circumscribed circle of this pentagon in the golden ratio.} Let $k(S,r)$ be the center of the inscribed circle and $AB$ ($a=|AB|$) one side of the regular pentagon. In the triangle $ASB$ measure the internal angles: $\angle ASB=36^0$ and $\angle SAB=\angle SBA=72^0$. We denote by $P$ such a point that $\mathcal{B}(B,A,P)$ and $AP\cong AS$. Because $SAP$ is an isosceles triangle with the base $PS$, by \ref{enakokraki} $\angle SPA\cong\angle PSA$. From \ref{zunanjiNotrNotr} (for the triangle $SAP$) follows $\angle APS=\angle PSA=\frac{1}{2}72^0=36^0$. Therefore $\triangle SPB\sim\triangle ASB$ (\ref{PodTrikKKK}), so $\frac{PB}{SB}=\frac{SB}{AB}$ or $\frac{a+r}{r}=\frac{r}{a}$, from which directly follows our claim. \item \label{nalPod75} \res{Let $a_5$, $a_6$ and $a_{10}$ be the sides of a regular pentagon, hexagon and decagon, which are inscribed in the same circle. Prove that: $$a_5^2=a_6^2+a_{10}^2.$$} We use the similarity of triangles from the previous task \ref{nalPod75}, but we take into account the appropriate heights of these two triangles. \item \res{Let $t_a$, $t_b$ and $t_c$ be the centroids and $s$ the semi-perimeter of a triangle. Prove that: $$t_a^2+t_b^2+t_c^2\geq s^2.$$} % zvezek - dodatni MG If we use \ref{StwartTezisc2}: \begin{eqnarray*} t_a^2+t_b^2+t_c^2 &=& \frac{3}{4}\left(a^2+ b^2+c^2\right)=\\ &\geq& \frac{3}{4}\cdot 3\cdot\left(\frac{a+ b+c}{3}\right)^2=\\ &=& s^2. \end{eqnarray*} \end{enumerate} Let $A_1$ be the center of side $BC$. Then, according to the statement \ref{PloscTrik}: $p_{AA_1B}=p_{AA_1C}$ and $p_{TA_1B}=p_{TA_1C}$. If we subtract the equality, we get $p_{AA_1B}-p_{TA_1B}$ and $p_{AA_1C}-p_{TA_1C}$ or, according to the statement \ref{ploscGlavniIzrek} \textit{4)}: $p_{ABT}=p_{ATC}$. Similarly, we also get $p_{TBC}=p_{ATC}$. \item \res{Let $c$ be the hypotenuse, $v_c$ the corresponding altitude, and $a$ and $b$ the catheti of the right triangle. Prove that $c+v_c>a+b$.} % zvezek - dodatni MG According to the statement \ref{PloscTrik} for the area $p$ of this triangle, it holds that $p=\frac{1}{2}ab=\frac{1}{2}cv_c$, so $ab=cv_c$, but if we also use the Pythagorean statement \ref{PitagorovIzrek}: \begin{eqnarray*} (a+b)^2=a^2+2ab+b^2= c^2+2cv_cCD$), so that it divides this trapezoid into two figures with equal area.} Let $v$ be the height of the trapezoid and $a$ and $c$ be the lengths of its bases $AB$ and $CD$. If $X$ is a point in which the line $p$ intersects the base $AB$, from the given condition it follows: $p_{AXD}=\frac{1}{2}p_{ABCD}$ or: \begin{eqnarray*} \frac{|AX|\cdot v}{2} =\frac{1}{2}\cdot \frac{a+c}{2}\cdot v. \end{eqnarray*} Therefore, the distance $AX$ is equal to the median of the trapezoid or $|AX|=\frac{a+c}{2}$, which allows the construction of the point $X$ and the line $p$. \item \res{Draw the line $p$ and $q$, which pass through the vertex $D$ of the square $ABCD$ and divide it into areas of equal size.} If $P$ and $Q$ are points in which the line $p$ or $q$ intersects the side $AB$ or $BC$ of the square $ABCD$, we prove that $AP:PB=2:1$ and $BQ:QC=1:2$. \item \res{Let $ABCD$ be a square, $E$ be the center of its side $BC$ and $F$ be a point for which $\overrightarrow{AF}=\frac{1}{3}\cdot \overrightarrow{AB}$. The point $G$ is the fourth vertex of the rectangle $FBEG$. What part of the area of the square $ABCD$ does the triangle $BDG$ represent?} Let $a$ be the length of the side of the square $ABCD$. Let $S$ be the center of this square and $H=\mathcal{T}_{\overrightarrow{AF}}(D)$. By formulas \ref{PloscTrik}, \ref{ploscKvadr} and \ref{ploscGlavniIzrek} \textit{4)} we have: \begin{eqnarray*} p_{BDG} &=& p_{BSG}+p_{SDG}= p_{FSG}+p_{SHG}=\\ &=& p_{FSH}=\frac{1}{2}\cdot |FH| \cdot |GS|= \\ &=& \frac{1}{2}\cdot a \cdot \left( \frac{a}{2}-\frac{a}{3} \right)=\\ &=& \frac{a^2}{12} = \frac{1}{12}\cdot p_{ABCD}. \end{eqnarray*} \item \label{nalPloKoef} \res{Let $\mathcal{V}$ and $\mathcal{V}'$ be similar polygons with a similarity coefficient of $k$. Prove that: $$p_{\mathcal{V}'}=k^2\cdot p_{\mathcal{V}}.$$} First, we prove that the statement is true for triangles (we use the formula \ref{PloscTrik}), then we use the formula \ref{ploscGlavniIzrek} \textit{4)}. \item \label{nalPloHeronStirik} \res{Let $a$, $b$, $c$ and $d$ be the lengths of the sides, $s$ the semiperimeter and $p$ the area of any quadrilateral. Prove that: $$p=\sqrt{(s-a)(s-b)(s-c)(s-d)}.$$} %Lopandic - nal 924 Let $A$, $B$, $C$ and $D$ be the vertices of the given quadrilateral, so that $|AB|=a$, $|BC|=b$, $|CD|=c$ and $|DA|=d$. If $ABCD$ is a parallelogram, due to its rigidity it is a rectangle (see \ref{paralelogram} and \ref{TetivniPogoj}) and the statement is trivial. Without loss of generality, we assume that the sides $BC$ and $AD$ intersect in some point $P$ and that $a>c$. We also mark $|PC|=x$, $|PD|=y$ and $p'=p_{PCD}$. According to the theorem \ref{TetivniPogojZunanji}, $\angle PCD\cong \angle BAD$, therefore $\triangle CDP \cong \triangle ABP$ (theorem \ref{PodTrikKKK}). The similarity coefficient of these two triangles is $k=\frac{AB}{CD}=\frac{a}{c}$. Thus, according to the statement from the previous task \ref{nalPloKoef}, it holds: $$\frac{p_{ABP}}{p_{CDP}}=\frac{a^2}{c^2}$$ or, according to the theorem \ref{ploscGlavniIzrek} \textit{4)} $$\frac{p_{ABCD}+p_{CDP}}{p_{CDP}}=\frac{a^2}{c^2}.$$ From this, we get: \begin{eqnarray} \label{nalPloEqnHeronStirik1} p=\frac{a^2-c^2}{c^2}\cdot p'. \end{eqnarray} First, let us calculate $p'=p_{PCD}$. From the mentioned similarity $\triangle CDP \cong \triangle ABP$, it also follows: \begin{eqnarray*} \frac{x+b}{y}=\frac{y+d}{x}=\frac{a}{c}. \end{eqnarray*} If we rearrange the equations, we get a system of equations for $x$ and $y$: \begin{eqnarray*} && ax-cy=cd\\ && cx-ay=-bc \end{eqnarray*} and its solutions: \begin{eqnarray*} x &=& \frac{c(ad+bc)}{a^2-c^2}\\ y &=& \frac{c(ab+cd)}{a^2-c^2} \end{eqnarray*} From this, we get: \begin{eqnarray} \label{nalPloEqnHeronStirik2} x+y &=& \frac{c(b+d)}{a+c}\\ x-y &=& \frac{c(d-b)}{a-c} \end{eqnarray} Let $s'$ be the semi-perimeter of the triangle $PCD$. By using the relations \ref{nalPloEqnHeronStirik2}, we get: \begin{eqnarray*} s' &=& \frac{x+y+c}{2}=\frac{c}{a-c}\cdot (s-c)\\ s'-c &=& \frac{x+y-c}{2}=\frac{c}{a-c}\cdot (s-a)\\ s'-x &=& \frac{c+x+y}{2}=\frac{c}{a+c}\cdot (s-d)\\ s'-y &=& \frac{c+x+y}{2}=\frac{c}{a+c}\cdot (s-b) \end{eqnarray*} From the previous relations according to Heron's formula for the area of a triangle (theorem \ref{PloscTrikHeron}), it holds: \begin{eqnarray*} p' &=& p_{PCD}=\sqrt{s'(s'-c)(s'-x)(s'-y)}=\\ &=& \frac{c^2}{a^2-c^2}\sqrt{(s-a)(s-b)(s-c)(s-d)}. \end{eqnarray*} If we insert this into the relation \ref{nalPloEqnHeronStirik1}, we get: \begin{eqnarray*} p = \frac{a^2-c^2}{c^2}\cdot p'=\sqrt{(s-a)(s-b)(s-c)(s-d)}. \end{eqnarray*} \item \res{Let $a$, $b$, $c$ and $d$ be the lengths of the sides and $p$ the area of the tangent quadrilateral. Prove that: $$p=\sqrt{abcd}.$$} %Lopandic - nal 925 Let $s$ be the semiperimeter of the given quadrilateral. Since it is a tangent quadrilateral, for its sides according to \ref{TangentniPogoj} it holds that $a+c=b+d$. Therefore: \begin{eqnarray*} s-a &=& \frac{b+c+d-a}{2}=c\\ s-b &=& \frac{a+c+d-b}{2}=d\\ s-c &=& \frac{a+b+d-c}{2}=a\\ s-d &=& \frac{a+b+c-d}{2}=b. \end{eqnarray*} Since the quadrilateral is also a taut, we can use the statement from the previous task \ref{nalPloHeronStirik} and get: $$p=\sqrt{(s-a)(s-b)(s-c)(s-d)}=\sqrt{abcd}.$$ % Veckotniki \item \res{Given is a rectangle $ABCD$ with sides $a=|AB|$ and $b=|BC|$. Calculate the area of the figure which represents the union of the rectangle $ABCD$ and its image after reflection over the line $AC$.} Let $B'=\mathcal{S}_{AC}(B)$ and $D'=\mathcal{S}_{AC}(D)$, $P$ be the intersection of the lines $AB'$ and $CD$ and $S$ be the intersection of the diagonal $AC$ and $BD$ of the rectangle $ABCD$. We also mark $d=|AC|=|BD|$ and $v=|SP|$. From the similarity of the triangles $AB'C$ and $ASP$ (\ref{PodTrikKKK}) we get $v:b=\frac{d}{2}:a$ or $v=\frac{bd}{2a}$. If we use the obtained relation, \ref{PloscTrik}, \ref{ploscPravok} and \ref{ploscGlavniIzrek} and Pythagoras' theorem \ref{PitagorovIzrek}, for the area $p$ of the sought figure it holds: \begin{eqnarray*} p &=& 2\cdot \left(p_{AB'C}+p_{ADC}-p_{APC}\right)= 2\cdot \left(p_{ABC}+p_{ADC}-p_{APC}\right)=\\ &=& 2\cdot \left(p_{ABCD}-p_{APC}\right)=2\cdot \left( ab-\frac{dv}{2} \right)=\\ &=& 2\cdot \left( ab-\frac{bd^2}{4a} \right)=2ab-\frac{b(a^2+b^2)}{2a}=\\ &=& \frac{b(3a^2-b^2)}{2a}. \end{eqnarray*} \item \res{Let $ABCDEF$ be a regular hexagon, and let $P$ and $Q$ be the centers of its sides $BC$ and $FA$. What part of the area of this hexagon does the area of the triangle $PQD$ represent?} We use the formula \ref{srednjTrapez}. Result: $p_{PQD}=\frac{3}{8}p_{ABCDEF}$. % KKrog \item \res{In a square, four congruent circles are drawn so that each circle touches two sides and two other circles. Prove that the sum of the areas of these circles is equal to the area of the square of the inscribed circle.} If we denote the length of the square's side with $a$, then $r=\frac{a}{2}$ or $r_1=\frac{a}{4}$ is the radius of the inscribed circle or each of the smaller inscribed circles. Then the aforementioned sum of areas is equal to: $$4\cdot r_1^2\pi=4\left(\frac{a}{4} \right)^2\cdot \pi=\left(\frac{a}{2} \right)^2\cdot \pi=r^2\pi.$$ \item \res{Calculate the area of the circle inscribed in the triangle with sides of lengths 9, 12 and 15.} We use Heron's formula \ref{PloscTrikHeron} and the formula \ref{PloscTrikVcrt}. Result: the area of the triangle $p_{\triangle}=54$; the area of the circle $p=9\pi$. \item \res{Let $P$ be the center of the base $AB$ of the trapezoid $ABCD$, for which $|BC|=|CD|=|AD|=\frac{1}{2}\cdot |AB|=a$. Express the area of the figure determined by the base $CD$ and the shorter circular arcs $PD$ and $PC$ of the circles with centers $A$ and $B$, as a function of the base $a$.} Since, by assumption, $\overrightarrow{PB}=\overrightarrow{DC}$, the quadrilateral $PBDC$ is a parallelogram, so $BC\cong PD$. Since, by assumption, $AD\cong BC$ and $AD\cong AP$, we have $PD\cong AD\cong AP$ and $APD$ is an isosceles triangle. Therefore, $\angle PAD= 60^0$. Similarly, $\angle CBP=60^0$. The desired area $p_0$ is the difference between the area of the trapezoid $p$ and twice the area of the circular sector with central angle $60^0$: $$p_0=\frac{a^2}{12}\left(9\sqrt{3}-4\pi \right).$$ \item \res{The chord $PQ$ ($|PQ|=d$) of the circle $k$ touches its concurring circle $k'$. Express the area of the trapezoid determined by the circles $k$ and $k'$, as a function of the chord $d$.} We use the Pythagorean Theorem \ref{PitagorovIzrek}. Result: $\frac{d^2\pi}{4}$. \item \res{Let $r$ be the radius of the inscribed circle of the polygon $\mathcal{V}$, which is divided into triangles $\triangle_1,\triangle_2,\ldots,\triangle_n$, so that no two triangles have common inner points. Let $r_1,r_2,\ldots , r_n$ be the radii of the inscribed circles of these triangles. Prove that: $$\sum_{i=1}^n r_i\geq r.$$} Let $s$ be the semiperimeter and $p$ the area of the polygon $\mathcal{V}$, and let $s_i$ be the semiperimeter and $p_i$ ($i\in \{1,2,\ldots , n\}$) the area of the triangle $\triangle_i$. If we use the formula \ref{ploscTetVec} and \ref{PloscTrikVcrt} and the fact that for each $i\in \{1,2,\ldots , n\}$ we have $s\geq s_i$, we get: $$sr=p= \sum_{i=1}^n p_i = \sum_{i=1}^n s_ir_i\leq \sum_{i=1}^n sr_i=s\cdot\sum_{i=1}^n r_i.$$ \end{enumerate} %REŠITVE - Inverzija %________________________________________________________________________________ \poglavje{Inversion} \begin{enumerate} \item \res{Prove that the composition of two inversions $\psi_{S,r_1}$ and $\psi_{S,r_2}$ with respect to a concentric circle represents a dilation. Determine the center and the coefficient of this dilation.} From the definition of inversion it follows that the composition of inversions $\psi_{S,r_2}\circ\psi_{S,r_1}$ is a dilation with center $S$ and coefficient $\frac{r_2^2}{r_1^2}$. The statement is also a direct consequence of the task \ref{invRazteg}. \item \res{Let $A$, $B$, $C$ and $D$ be four collinear points. Construct such points $E$ and $F$, that $\mathcal{H}(A,B;E,F)$ and $\mathcal{H}(C,D;E,F)$ hold.} We use the formula \ref{harmPravKrozn} - first we draw the circle, which is perpendicular to the circle with diameters $AB$ and $CD$, \item \res{In a plane, given are a point $A$, a line $p$ and a circle $k$. Draw a circle that goes through point $A$ and is perpendicular to line $p$ and circle $k$.} Use inversion with center $A$. \item \res{Solve the third, fourth, ninth and tenth Apollonius problem.} For the third and fourth Apollonius problem we use inversion with center in one of the given points. For the ninth and tenth Apollonius problem we first draw a circle that is concentric with the sought circle and goes through the center of one of the circles. In this way we translate the problem to the fifth or sixth Apollonius problem. \item \res{Let: $A$ be a point, $p$ be a line, $k$ be a circle and $\omega$ be an angle in some plane. Draw a circle that goes through point $A$, touches line $p$ and with circle $k$ determines angle $\omega$.} Use inversion with center $A$. \item \res{Determine the geometric location of the points of intersection of two circles that touch at the arms of a given angle in two given points $A$ and $B$.} Use inversion with center $B$. \item \res{Draw a triangle, if the following data are known: \begin{enumerate} \item $a$, $l_a$, $v_a$ \item $v_a$, $t_a$, $b-c$ \item $b+c$, $v_a$, $r_b-r_c$ \end{enumerate}} Use the great problem (see Theorem \ref{velikaNaloga}) and the appropriate harmonic quadruples of points. \item \res{Let $c(S,r)$ and $l$ be a circle and a line in the same plane that have no common points. Let $c_1$, $c_2$ and $c_3$ be circles in this plane that touch each other (two at a time) and each of them touches also $c$ and $l$. Express the distance of point $S$ from line $l$ with $r$\footnote{Proposal for MMO 1982. (SL 12.)}.} First, we prove that there exists a circle $n$, which is perpendicular to the line $l$ and the circle $c$. Let $Y$ be one of the intersections of the circle $n$ and the rectangle on the line $l$ from the center of the circle $c$. We use the composition $f=\psi_i\circ \mathcal{R}\circ \psi_i$, where $i$ is an arbitrary circle with center $Y$, $\mathcal{R}$ is a rotation with center in the center of the circle $l'=\psi_i(l)$, which maps the tangent points of the circles $l'$ and $c'_3=\psi_i(c_3$) to the point $Y$. If: $f:\hspace*{1mm}l, c, c_1, c_2, c_3\mapsto \widehat{l}, \widehat{c}, \widehat{c_1}, \widehat{c_2}, \widehat{c_3},$ we prove that $\widehat{l}=l$, $\widehat{c}=c$, $\widehat{c_3}$ is a line parallel to the line $l$, and $\widehat{c_1}$ and $\widehat{c_2}$ are congruent circles, which touch each other and also touch the parallels $l$ and $\widehat{c_3}$. In the end, it follows that the distance from the center of the circle $c$ to the line $l$ is equal to $7r$. \item \res{Let $ABCD$ be a regular tetrahedron. To any point $M$, lying on the edge $CD$, we assign the point $P = f(M)$, which is the intersection of the rectangle through the point $A$ on the line $BM$ and the rectangle through the point $B$ on the line $AM$. Determine the geometric location of all points $P$, if the point $M$ takes all values on the edge $CD$.} The point $P$ is the altitude of the triangle $ABM$. If $S$ is the center of the edge $AB$, first we prove that $\overrightarrow{SP}\cdot \overrightarrow{SM}=\frac{a^2}{4}$, where $a$ is the edge of the regular tetrahedron. Then we use the inversion $\psi_{S,\frac{a}{2}}$ (in the plane $SCD$). The geometric location of the points is then the image of the segment $CD$ under this inversion, i.e. the corresponding circular arc with center $S$, with endpoints in the altitudes of the triangles $ACD$ and $BCD$. \item \res{Let $ABCD$ be a tangent-chord quadrilateral and $P$, $Q$, $R$ and $S$ the points of tangency of sides $AB$, $BC$, $CD$ and $AD$ with the inscribed circle of this quadrilateral. Prove that $PR\perp QS$.} We use inversion with respect to the inscribed circle. We prove that the vertices of the quadrilateral $ABCD$ are the vertices of a rectangle, and the sides of this rectangle are parallel to the distances $PR$ and $QS$. \item \res{Prove that the centers of a tangent-chord quadrilateral, the inscribed and the circumscribed circle, and the intersection of its diagonals are collinear points (\index{izrek!Newtonov}Newton's theorem\footnote{\index{Newton, I.}\textit{I. Newton} (1643--1727), English physicist and mathematician}).} We use the previous exercise and prove that the center $G$ of the rectangle from that exercise is also the center of the line determined by the center of the inscribed circle of the quadrilateral $ABCD$ (the center of inversion) and the intersection of the lines $PR$ and $QS$. The point $G$ is in fact the center of the image of the circumscribed circle under that inversion. \item \res{Let $p$ and $q$ be parallel tangents to the circle $k$. Circle $c_1$ is tangent to the line $p$ at point $P$ and to the circle $k$ at point $A$, circle $k_2$ is tangent to the line $q$ and to the circles $k$ and $k_1$ at points $Q$, $B$ and $C$. Prove that the intersection of the lines $PB$ and $AQ$ is the center of the triangle $ABC$ of the circumscribed circle.} We use inversion with center $B$ and first prove that $PB$ is a common tangent to the circles $k$ and $k_2$, and then that the intersection of the lines $PB$ and $AQ$ is the potent center of the circles $k$, $k_1$ and $k_2$. \item \res{The circles $k_1$ and $k_3$ touch each other externally in the point $P$. The circles $k_2$ and $k_2$ also touch each other externally in the same point. The circle $k_1$ intersects the circles $k_2$ and $k_4$ also in the points $A$ and $D$, the circle $k_3$ intersects the circles $k_2$ and $k_4$ also in the points $B$ and $C$. Prove that it holds\footnote{A suggestion for MMO 2003. (SL 16.)}: $$\frac{|AB|\cdot|BC|}{|AD|\cdot|DC|}=\frac{|PB|^2}{|PD|^2}.$$} Let $\psi_P$ be an inversion with an arbitrary radius $r$. This transforms the quadrilateral $A'B'C'D'$ into a parallelogram (by the statement \ref{InverzDotik}). So $A'B'\cong C'D'$ holds. If we use the statement \ref{invMetr}, we get $\frac{|AB|\cdot r^2}{|PA|\cdot|PB|}=\frac{|CD|\cdot r^2}{|PC|\cdot|PD|}$ or $\frac{|AB|}{|CD|}=\frac{|PA|\cdot|PB|}{|PC|\cdot|PD|}$. In a similar way from $C'B'\cong A'D'$ we get $\frac{|CB|}{|AD|}=\frac{|PC|\cdot|PB|}{|PA|\cdot|PD|}$. By multiplying the two relations we get: $\frac{|AB|\cdot|BC|}{|AD|\cdot|DC|}=\frac{|PB|^2}{|PD|^2}$. \item \res{Let $A$ be a point that lies on the circle $k$. With the help of a ruler only, draw a square $ABCD$ (or its vertices), that is inscribed in the given circle.} First we draw a regular hexagon $AB_1B_2CD_1D_2$, that is inscribed in the given circle. \item \res{Given are the points $A$ and $B$. With the help of a ruler only, draw such a point $C$, that $\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}$.} We use a similar procedure as in the problem \ref{MaskeroniSred}. First we draw the point $X$, for which $\overrightarrow{AX}=3\cdot \overrightarrow{AB}$, then the desired point $X'=\psi_k(X)$, where $k$ is a circle with the center $A$ and the radius $AB$. \item \res{With the help of a ruler only, divide the given segment in the ratio $2:3$.} For a given line $AB$ we first draw a point $X$, for which $\overrightarrow{AX}=2\cdot \overrightarrow{AB}$ (similarly to the previous task), then a point $Y$, for which $\overrightarrow{AY}=\frac{1}{5}\cdot \overrightarrow{AX}$ holds. $Y$ is the desired point, because $\overrightarrow{AY}=\frac{2}{5}\cdot \overrightarrow{AB}$ holds. \end{enumerate} \newpage \normalsize %________________________________________________________________________________ % LITERATURA - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - %________________________________________________________________________________ \begin{thebibliography}{1} \bibitem{Berger} Berger, M. \emph{Geometry}, Springer-Verlag, Berlin 1987. \bibitem{Cofman} Cofman, J. \emph{What to solve?}, Oxford University Press, Oxford, 1990. \bibitem{CoxeterRevisited} Coxeter, H. 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