% Version of MOP 1999 % last revised 19 May 1999 % %% Geometry project \documentclass[12pt]{report} \usepackage{amsthm, amsmath, amsfonts} %% Page formatting \textwidth=6.5in \textheight=8.0in \headheight=0.0in \headsep=0.0in \parindent=0.2in \topmargin=0.0in \oddsidemargin=0.0in \evensidemargin=0.0in %% Environments \newcounter{exc} \numberwithin{exc}{section} \newenvironment{exer}{\vspace{0.1in} \noindent \textbf{Problems for Section~\thesection} \vspace{0.1in} \begin{list}{\arabic{exc}.}{\usecounter{exc}}}{\end{list}} %% Don't need proof environment if using package amsthm %% Theorem-like environments \newtheorem{theorem}{Theorem}[chapter] \newtheorem{fact}[theorem]{Fact} \newtheorem{corr}[theorem]{Corollary} %% Abbreviations \def\beqn{\begin{equation}} \def\eeqn{\end{equation}} \def\beqa{\begin{eqnarray*}} \def\eeqa{\end{eqnarray*}} \def\bthm{\begin{theorem}} \def\ethm{\end{theorem}} \def\bpf{\begin{proof}} \def\epf{\end{proof}} \def\ii{\item} \def\lpar{\left(} \def\rpar{\right)} \def\ang{\angle} \def\sang{\sin \angle} \def\half{\frac{1}{2}} \def\vA{\vec{A}} \def\vB{\vec{B}} \def\vC{\vec{C}} \def\CC{\mathbb{C}} \def\norm#1{\lVert #1 \rVert} %% I need a new symbol for directed angle. \def\dang{\angle} \def\ray#1{\vec{#1}} %% Fix this! Maybe use AMS? \def\arc#1{\stackrel{\frown}{#1}} %% Ditto! \begin{document} %% Begin with Roman numerals \pagenumbering{roman} %% Title page \begin{titlepage} \begin{center} Notes on Euclidean Geometry \vspace{1in} Kiran Kedlaya based on notes for the Math Olympiad Program (MOP) Version 1.0, last revised August 3, 1999 \vspace{1in} \copyright Kiran S. Kedlaya. This is an unfinished manuscript distributed for personal use only. In particular, any publication of all or part of this manuscript without prior consent of the author is strictly prohibited. \bigskip Please send all comments and corrections to the author at \texttt{kedlaya@math.mit.edu}. Thank you! \end{center} \end{titlepage} %% Contents \tableofcontents \pagebreak %% Preface \section*{Introduction} This book is a compilation and distillation of my notes, as participant and later as instructor, from the Math Olympiad Program (MOP), the annual summer program to prepare U.S. high school students for the International Mathematical Olympiad (IMO). As such, it has an overt and a covert mission. The overt mission is to assist students in preparing for the USA Mathematical Olympiad (USAMO) and the IMO, as American students have historically fared poorly on problems in Euclidean geometry. The covert, and perhaps more important, mission is to introduce said students (and anyone else who happens to read this) to a lineage of mathematics stretching from ancient times to the present; hence the inclusion of advanced topics in inversive and projective geometry which may segue into the study of complex analysis, algebraic geometry, or the like. The model for this book has been the slender classic \emph{Geometry Revisited} by H.S.M. Coxeter and S. Greitzer, with which many American IMO participants, myself included, have supplemented their education in Euclidean geometry. We have gone further by including some topics neglected there (the circle of Apollonius, directed angles, concurrent perpendiculars) and providing numerous problems. Think of this book as \emph{``Geometry Revisited'' Revisited}, if you will. Some words about terminology are in order at this point. For the purposes of this book, a \emph{theorem} is an important result which I either prove or tell you where a proof can be found (usually the former, but a few of the proofs would take me too far afield). A \emph{corollary} is a result which is important in its own right, but is easily deduced from a nearby theorem. A \emph{fact} is a result which is important and/or useful, but whose proof is simple enough to be left as an \emph{exercise}. On the other hand, while an exercise is usually a routine application of standard techniques, a \emph{problem} requires some additional insight. I have attributed my source for each problem to the best of my knowledge. USAMO, IMO and Putnam problems are listed by year and number; problems from other national contests (as well as the K\"ursch\'ak contest of Hungary, the Balkan Mathematical Olympiad and others) are listed by country and year. ``Arbelos'' refers to Samuel Greitzer's student publication from 1982--1987, and ``Monthly'' to the \textit{American Mathematical Monthly}. Problems listed as ``Original'' are my own problems which have not before appeared in print. The reader stuck on a problem can consult the hints section in the back, but we strongly advise not doing so before making a considerable effort to solve the problem unaided. Note that many of the problems presented have solutions other than those we describe, so it is probably worth reading the hints even for problems which you solve on your own, in case you've found another approach. No list of acknowledgments could possibly include everyone who has made helpful suggestions, but at least I should mention those from whom I learned the subject via MOP: Titu Andreescu, R\u{a}zvan Gelca, Anne Hudson, Gregg Patruno, and Dan Ullman. Thanks also to the participants of the 1997, 1998 and 1999 MOPs for suffering through preliminary versions of this book. \chapter{Tricks of the trade} We begin with a chapter that highlights a small core of basic techniques that prove useful in a large number of problems. The point is to show how much one can accomplish even with very little advanced knowledge. %% Work backwards? \pagenumbering{arabic} \section{Slicing and dicing} One of the most elegant ways of establishing a geometric result is to dissect the figure into pieces, then rearrange the pieces so that the result becomes obvious. The quintessential example of this technique is the ancient Indian proof of the Pythagorean theorem. \begin{theorem}[Pythagoras] If $ABC$ is a right triangle with hypotenuse $BC$, then $AB^{2} + BC^{2} = AC^{2}$. \end{theorem} \begin{proof} Behold! DIAGRAM. \end{proof} Other useful dissections include a proof of the fact that the area of a triangle is half its base times its height DIAGRAM, a proof that the median to the hypotenuse of a right triangle divides the triangle into two isosceles triangles DIAGRAM, and in three dimensions, an embedding of a tetrahedron in a box (rectangular parallelepiped) DIAGRAM. \begin{exer} \ii (MOP 1997) Let $Q$ be a quadrilateral whose side lengths are $a,b,c,d$, in that order. Show that the area of $Q$ does not exceed $(ac+bd)/2$. \ii Let $ABC$ be a triangle and $M_{A}, M_{B}, M_{C}$ the midpoints of the sides $BC, CA, AB$, respectively. Show that the triangle with side lengths $AM_{A}, BM_{B}, CM_{C}$ has area $3/4$ that of the triangle $ABC$. \ii \label{ex:seventh} In triangle $ABC$, points $D, E, F$ are marked on sides $BC, CA, AB$, respectively, so that \[ \frac{BD}{DC} = \frac{CE}{EA} = \frac{AF}{FB} = 2. \] Show that the triangle formed by the lines $AD, BE, CF$ has area $1/7$ that of the triangle $ABC$. \ii \label{ex:samearea} In the hexagon $ABCDEF$, opposite sides are equal and parallel. Prove that triangles $ACE$ and $BDF$ have the same area. \ii On the Scholastic Aptitude Test (an American college entrance exam), students were once asked to determine the number of faces of the polyhedron obtained by gluing a regular tetrahedron to a square pyramid along one of the triangular faces. The answer expected by the test authors was 7, since the two polyhedra have 9 faces, 2 of which are removed by gluing, but a student taking the exam pointed out that this is incorrect. What is the correct answer? \ii \label{ex:tetocta} A regular tetrahedron and a regular octahedron have edges of the same length. What is the ratio between their volumes? \end{exer} \section{Angle chasing} A number of problems in Euclidean geometry can be solved by careful bookkeeping of angles, which allows one to detect similar triangles, cyclic quadrilaterals, and the like. We offer as an example of such ``angle chasing'' a theorem attributed to Miquel. DIAGRAM. \begin{theorem} Let $ABC$ be a triangle and let $P,Q,R$ be any points on the sides $BC, CA, AB$, respectively. Then the circumcircles of $ARQ, BPR, CQP$ pass through a common point. \end{theorem} \begin{proof} Let $T$ be the intersection of the circumcircles of $ARQ$ and $BPR$. By collinearity of points, \[ \angle TQA = \pi - \angle CQT, \quad \angle TRB = \pi - \angle ART, \quad \angle TPC = \pi - \angle BPT. \] In a cyclic quadrilateral, opposite angles are supplementary. Therefore \[ \angle TQA = \pi - \angle ART, \angle TRB = \pi - \angle BPT. \] We conclude $\angle TPC = \pi - \angle CQT$. Now conversely, a quadrilateral whose opposite angles are supplementary is cyclic. Therefore $T$ also lies on the circumcircle of $CQP$, as desired. \end{proof} A defect of the technique is that the relevant theorems depend on the configuration of the points involved, particularly on whether certain points fall between certain other points. For example, one might ask whether the above theorem still holds if $P,Q,R$ are allowed to lie on the extensions of the sides of $ABC$. It does hold, but the above proof breaks down because some of the angles claimed to be equal become supplementary, and vice versa. DIAGRAM. Euclid did not worry much about configuration issues, but later observers realized that carelessness in these matters could be disastrous. For example, consider the following pseudotheorem from \cite{bib:max}. \newtheorem*{pseudo}{Pseudotheorem} \begin{pseudo} All triangles are isosceles. \end{pseudo} \begin{proof}[Pseudoproof] Let $ABC$ be a triangle, and let $O$ be the intersection of the internal angle bisector of $A$ with the perpendicular bisector of $BC$. DIAGRAM (showing $O$ INSIDE!) Let $D, Q, R$ be the feet of perpendiculars from $O$ to $BC, CA, AB$, respectively. By symmetry across $OD$, $OB = OC$, while by symmetry across $AO$, $AQ = AR$ and $OQ = OR$. Now the right triangles $ORB$ and $OQC$ have equal legs $OR = OQ$ and equal hypotenuses $OB = OC$, so they are congruent, giving $RB = QC$. Finally, we conclude \[ AB = AR + RB = AQ + QC = AC, \] and hence the triangle $ABC$ is isosceles. \end{proof} \begin{exer} \ii Where is the error in the proof of the Pseudotheorem? \ii (USAMO 1994/3) A convex hexagon $ABCDEF$ is inscribed in a circle such that $AB=CD=EF$ and diagonals $AD,BE,CF$ are concurrent. Let $P$ be the intersection of $AD$ and $CE$. Prove that $CP/PE = (AC/CE)^{2}$. \ii (IMO 1990/1) Chords $AB$ and $CD$ of a circle intersect at a point $E$ inside the circle. Let $M$ be an interior point of the segment $EB$. The tangent line of $E$ to the circle through $D,E,M$ intersects the lines $BC$ and $AC$ at $F$ and $G$, respectively. If $AM/AB = t$, find $EG/EF$ in terms of $t$. \end{exer} \section{Sign conventions} Euclid's angles, lengths and areas were always nonnegative quantities, but often it is more convenient to allow negative quantities in a systematic fashion, so that equations like \[ AB + BC = AC \] hold without conditions on the relative positions of the objects involved. In this section, we describe standard sign conventions, their advantages (in avoiding configuration dependencies and errors as in the Pseudoproof) and limitations. When measuring segments on or parallel to a given line, one may speak of \emph{directed lengths} by fixing a choice of the positive direction, and regarding the segment $AB$ as having positive length when the ray from $A$ through $B$ points in the positive direction, and negative otherwise. Of course this depends on the choice of the positive direction, but often lengths come in ratios such as $AB/BC$, or in pairwise products such as $AB \cdot BC$; in those cases, the choice does not matter and the directed ratio or product is well-defined. \emph{Directed areas} are less complicated than directed lengths because all areas can be directed at once. Conventionally, the area of a triangle or other polygon whose vertices are denoted in counterclockwise order is taken to be positive. This convention is generally compatible with directed lengths, though in any case you should pause for a moment to decide whether your intended manipulations make sense. For example, if $A, B, C$ are collinear, the ratio of directed areas $[PAB]/[PBC]$ equals the ratio of directed lengths $AB/BC$. Directed angles are more complicated than directed lengths or areas, and the bulk of this section will be devoted to explaining their uses and drawbacks. The \emph{directed angle} $\dang(\ell_1, \ell_2)$ between lines $\ell_{1}$ and $\ell_{2}$, modulo $\pi$, is the angle of rotation required to bring $\ell_{1}$ parallel to $\ell_{2}$. The directed angle $\dang ABC$ made by the points $A,B,C$ is then defined as the angle between the lines $AB$ and $BC$. Pay close attention to the fact that angles are regarded modulo $\pi$ and not $2\pi$; this difference will help in many cases but has its own limitations. One can now verify the following ``axioms of directed angle arithmetic,'' all of which are independent of configuration. \begin{fact} The following relations hold among directed angles: \begin{enumerate} \ii $\dang ABC = -\dang CBA$. (Definition) \ii $\dang APB + \dang BPC = \dang APC$. \ii $\dang ABC = \dang ABD$ if and only if $B,C,D$ are collinear. In particular, $\dang ABC = 0$ if and only if $A,B,C$ are collinear. \ii $\dang ABD = \dang ACD$ if and only if $A,B,C,D$ are concyclic. \ii $\dang ABC = \dang ACD$ if and only if $CD$ is tangent to the circle passing through $A,B,C$. \ii $\dang ABC + \dang BCA + \dang CAB = 0$. (Angles of a triangle) \ii $2\dang ABC = \dang AOC$ if $A,B,C$ lie on a circle centered at $O$. \ii $\dang ABC$ equals $\half$ of the measure of the arc $\arc{AC}$ of the circumcircle of $ABC$. \end{enumerate} \end{fact} For example, if $A,B,C,D$ lie on a circle in that order, then we have $\ang ABD = \ang ACD$ as undirected angles. On the other hand, if theye have $\ang ABD = \pi - \ang CDA$, so in terms of directed angles \[ \dang ABD = \pi - \dang CDA = - \dang CDA = \dang ADC. \] It should be noted that this coincidence is a principal reason why one works modulo $\pi$ and not $2\pi$! (The other principal reason is of course so that collinear points always make an angle of 0.) The last two axioms ought to raise some eyebrows, because division by 2 is a dangerous thing when working modulo $\pi$. To be precise, the equation $2\angle A = 2\angle B$ of directed angles does not imply that $\angle A = \angle B$, for the possibility also exists that $\angle A = \angle B + \pi/2$. (Those familiar with elementary number theory will recognize an analogous situation: one cannot divide by 2 in the congruence $2a \equiv 2b \pmod{c}$ when $c$ is even.) This explains why we do not write $\dang ABC = \half \dang AOC$: the latter expression is not well-defined. On the other hand, directed arcs can be unambiguously measured mod $2\pi$, so dividing an arc by 2 gives an angle mod $\pi$. If all of this seems too much to worry about, do not lose hope; the conventions are easily learned with a little practice. We start with an easy example as a warmup. The result is one we will invoke in the proof of Pascal's theorem (Theorem~\ref{thm:pascal}). DIAGRAM \begin{theorem} \label{thm:parchord} Let $A$ and $B$ be the intersections of circles $\omega_{1}$ and $\omega_{2}$. Let $CD$ be any chord on $\omega_{1}$, and let $E$ and $F$ be the second intersections of the lines $CA$ and $BD$, respectively, with $\omega_{2}$. Then $EF$ is parallel to $CD$. (If $E=F$, we take $EF$ to mean the tangent to $\omega_{2}$ at $E$.) \end{theorem} \begin{proof} We chase angles as follows: \beqa \dang CDF &=& \dang CDB \qquad \mbox{(collinearity of $B,D,F$)} \\ &=& \dang CAB \qquad \mbox{(cyclic quadrilateral $ABCD$)} \\ &=& \dang EAB \qquad \mbox{(collinearity of $A,C,E$)} \\ &=& \dang EFB \qquad \mbox{(cyclic quadrilateral $ABEF$)}. \eeqa Hence the lines $CD$ and $EF$ make the same angle with $BF$, and so are parallel. \end{proof} Don't forget that directed angles can be expressed in terms of lines as well as in terms of points. This alternate perspective simplifies some proofs, as in the following example. (For a situation where this diagram occurs, see Problem~\ref{ex:postsim}.) \begin{theorem} \label{thm:presim} Let $ABC$ be a triangle. Suppose that the lines $\ell_{1}$ and $\ell_{2}$ are perpendicular, and meet each side (or its extension) in a pair of points symmetric across the midpoint of the side. Then the intersection of $\ell_{1}$ and $\ell_{2}$ is concyclic with the midpoints of the three sides. DIAGRAM. \end{theorem} \begin{proof} Let $M_{A}, M_{B}, M_{C}$ be the midpoints of the sides $BC, CA, AB$, respectively, and let $P = \ell_{1} \cap \ell_{2}$. Since the lines $\ell_{1}, \ell_{2}, BC$ form a right triangle and $M_{B}$ is the midpoint of the hypotenuse, the triangle formed by the points $P, M_{B}$ and $\ell_{2} \cap BC$ is isosceles with DIAGRAM \[ \dang(M_{B}P, \ell_{2}) = \dang(\ell_{2}, BC). \] By a similar argument, \[ \dang(\ell_{2}, M_{A}P) = \dang(CA, \ell_{2}), \] and adding these gives \[ \dang M_{B}P M_{A} = \dang BCA = \dang M_{B}M_{C}M_{A} \] since the sides of the triangle $M_{A}M_{B}M_{C}$ are parallel to those of $ABC$. We conclude that $M_{A}, M_{B}, M_{C}, P$ are concyclic, as desired. \end{proof} \begin{exer} \ii Let $A_{0}B_{0}C_{0}$ be a triangle and $P$ a point. Define a new triangle whose vertices $A_{1}, B_{1}, C_{1}$ as the feet of the perpendiculars from $P$ to $B_{0}C_{0}, C_{0}A_{0}, A_{0}B_{0}$, respectively. Similarly, define the triangles $A_{2}B_{2}C_{2}$ and $A_{3}B_{3}C_{3}$. Show that $A_{3}B_{3}C_{3}$ is similar to $A_{0}B_{0}C_{0}$. \ii %% MOP 1991/1995? Two circles intersect at points $A$ and $B$. An arbitrary line through $B$ intersects the first circle again at $C$ and the second circle again at $D$. The tangents to the first circle at $C$ and the second at $D$ intersect at $M$. Through the intersection of $AM$ and $CD$, there passes a line parallel to $CM$ and intersecting $AC$ at $K$. Prove that $BK$ is tangent to the second circle. \ii \label{ex:invcirc} Let $C_1,\,C_2,\,C_3,\,C_4$ be four circles in the plane. Suppose that $C_1$ and $C_2$ intersect at $P_1$ and $Q_1$, $C_2$ and $C_3$ intersect at $P_2$ and $Q_2$, $C_3$ and $C_4$ intersect at $P_3$ and $Q_3$, and $C_4$ and $C_1$ intersect at $P_4$ and $Q_4$. Show that if $P_1,\,P_2,\,P_3$, and $P_4$ lie on a line or circle, then $Q_1,\,Q_2,\,Q_3$, and $Q_4$ also lie on a line or circle. (This is tricky; see the proof of Theorem~\ref{thm:invcirc}.) \end{exer} \section{Working backward} A common stratagem, when trying to prove that a given point has a desired property, is to construct a phantom point with the desired property, then reason backwards to show that it coincides with the original point. We illustrate this point with an example. DIAGRAM \begin{theorem} \label{thm:back1} Suppose the triangles $ABC$ and $AB'C'$ are directly similar. Then the points $A, B, C$ and $BB' \cap CC'$ lie on a circle. \end{theorem} \begin{proof} Since we want to show that $BB' \cap CC'$ lies on the circle through $A, B, C$, and analogously on the circle through $A, B', C'$, we define the point $P$ to be the intersection of these two circles. Then \[ \dang APB = \dang ACB = \dang AC'B' = \dang APB' \] and so $P$ lies on the line $BB'$, and similarly on the line $CC'$. \end{proof} \begin{exer} \ii (IMO 1994/2) \label{ex:imo94} Let $ABC$ be an isosceles triangle with $AB = AC$. Suppose that \begin{enumerate} \item $M$ is the midpoint of $BC$ and $O$ is the point on the line $AM$ such that $OB$ is perpendicular to $AB$; \item $Q$ is an arbitrary point on the segment $BC$ different from $B$ and $C$; \item $E$ lies on the line $AB$ and $F$ lies on the line $AC$ such that $E$, $Q$, $F$ are distinct and collinear. \end{enumerate} Prove that $OQ$ is perpendicular to $EF$ if and only if $QE = QF$. \ii (Morley's theorem) \label{ex:morley} Let $ABC$ be a triangle, and for each side, draw the intersection of the two angle trisectors closer to that side. (That is, draw the intersection of the trisectors of $A$ and $B$ closer to $AB$, and so on.) Prove that these three intersections determine an equilateral triangle. \end{exer} %% Concurrence \chapter{Concurrence and Collinearity} This chapter is devoted to the study of two fundamental and reciprocal questions: when do three given points lie on a single line, and when do three given lines pass through a single point? The techniques we describe in this chapter will later be augmented by more sophisticated approaches, such as the radical axis theorem and the Pascal-Brianchon theorems. We say that three points are \emph{collinear} if they lie on a single line, and that three lines are \emph{concurrent} if they pass through a single point, or if all three are parallel. The latter convention addresses situations where a point of concurrence has moved ``to infinity''; we will formalize this in our discussion of projective geometry later. \section{Concurrent lines: Ceva's theorem} We begin with a simple but useful result, published in 1678 by the Italian engineer Giovanni Ceva (1647-1734). In his honor, a segment joining a vertex of a triangle to a point on the opposite side is called a \emph{cevian}. DIAGRAM. \begin{theorem}[Ceva] Let $ABC$ be a triangle, and let $P, Q, R$ be points on the lines $BC, CA, AB$, respectively. Then the lines $AP, BQ,$ $CR$ are concurrent if and only if (using directed lengths) \begin{equation} \label{eq:ceva} \frac{BP}{PC} \frac{CQ}{QA} \frac{AR}{RB} = 1. \end{equation} \end{theorem} \begin{proof} First suppose that the lines $AP, BQ, CR$ concur at a point $T$. Then the ratio of lengths $BP/PC$ is equal, by similar triangles, to the ratio of the distances from $B$ and $C$ to the line $AP$. On the other hand, that ratio is also equal to the ratio of areas $[ATB]/[CTA]$, by calculating these areas as half of base times height, with $AT$ as the base. DIAGRAM. By this argument, we get \[ \frac{BP}{PC} \frac{CQ}{QA} \frac{AR}{RB} = \frac{[ATB]}{[CTA]} \frac{[BTC]}{[ATB]} \frac{[CTA]}{[BTC]} = 1. \] Conversely, suppose that (\ref{eq:ceva}) holds; we will apply the trick of working backward. The lines $AP$ and $BQ$ meet at some point $T$, and the line $CT$ meets $AB$ at some point, which we call $R'$. By construction, $AP, BQ, CR'$ are concurrent. However, using Ceva in the other direction (which we just proved), we find that \[ \frac{BP}{PC} \frac{CQ}{QA} \frac{AR'}{R'B} = 1. \] Combining this equation with (\ref{eq:ceva}) yields \[ \frac{AR}{RB} = \frac{AR'}{R'B}. \] Since $AR + RB = AR' + R'B = AB$, adding 1 to both sides gives \[ \frac{AB}{RB} = \frac{AB}{R'B}, \] from which we conclude that $RB = R'B$, and hence $R = R'$. \end{proof} In certain cases, Ceva's Theorem is more easily applied in the following trigonometric form. \begin{fact}[``Trig Ceva''] Let $ABC$ be a triangle, and let $P$, $Q$, $R$ be any points in the plane distinct from $A, B, C,$ respectively. Then $AP, BQ, CR$ are concurrent if and only if \[ \frac{\sang CAP}{\sang APB} \frac{\sang ABQ}{\sang QBC} \frac{\sang BCR}{\sang RCA} = 1. \] \end{fact} One can either deduce this from Ceva's theorem or prove it directly; we leave both approaches as exercises. Be careful when using Trig Ceva with directed angles, as signs matter; the ratio $\sang CAP/\sang APB$ must be defined in terms of angles modulo $2\pi$, but the sign of the ratio itself only depends on the line $AP$, not on the choice of $P$ on one side or the other of $A$. \begin{exer} \ii Suppose the cevians $AP, BQ, CR$ meet at $T$. Prove that \[ \frac{TP}{AP} + \frac{TQ}{BQ} + \frac{TR}{CR} = 1. \] \ii Let $ABC$ be a triangle and $D, E, F$ points on sides $BC, CA, AB$, respectively, such that the cevians $AD, BE, CF$ are concurrent. Show that if $M, N, P$ are points on $EF, FD, DE$, respectively, then the lines $AM, BN, CP$ concur if and only if the lines $DM, EN, FP$ concur. (Many special cases of this question occur in the problem literature.) \ii (Hungary-Israel, 1997) The three squares $ACC_1A'', ABB_1'A', BCDE$ are constructed externally on the sides of a triangle $ABC$. Let $P$ be the center of $BCDE$. Prove that the lines $A'C, A''B, PA$ are concurrent. \ii (R\u{a}zvan Gelca) Let $ABC$ be a triangle and $D, E, F$ the points where the incircle touches the sides $BC, CA, AB$, respectively. Let $M, N, P$ be points on the segments $EF, FD, DE$, respectively. Show that the lines $AM, BN, CP$ intersect if and only if the lines $DM, EN, FP$ intersect. \ii (USAMO 1995/3) Given a nonisosceles, nonright triangle $\, ABC \,$ inscribed in a circle with center $O$, and let $\, A_1, \, B_1, \,$ and $\, C_1 \,$ be the midpoints of sides $\, BC, \, CA, \,$ and $\, AB, \,$ respectively. Point $\, A_2 \,$ is located on the ray $\, OA_1 \,$ so that $\, \Delta OAA_1 \,$ is similar to $\, \Delta OA_2A$. Points $\, B_2 \,$ and $\, C_2 \,$ on rays $\, OB_1 \,$ and $\, OC_1, \,$ respectively, are defined similarly. Prove that lines $\, AA_2, \, BB_2, \,$ and $\, CC_2 \,$ are concurrent. %% Need Titu's permission to release %\ii %Let $\triangle ABC$ be an equilateral triangle and $P$ a point in its %interior. Let the lines $AP, BP, CP$ meet the sides $BC, CA, AB$ at %the points $A_{1}, B_{1}, C_{1}$, respectively. Prove that %\[ %A_{1}B_{1} \cdot B_{1}C_{1} \cdot C_{1}A_{1} \geq A_{1}B \cdot B_{1}C %\cdot C_{1}A. %\] \ii Given triangle $ABC$ and points $X, Y, Z$ such that $\ang ABZ = \ang XBC$, $\ang BCX = \ang YCA$, $\ang CAY = \ang ZAB$, prove that $AX$, $BY$, $CZ$ are concurrent. (Again, many special cases of this problem can be found in the literature.) \ii Let $A, B, C, D, E, F,P$ be seven points on a circle. Show that $AD, BE, CF$ are concurrent if and only if \[ \frac{\sang APB \sang CPD \sang EPF}{\sang BPC \sang DPE \sang FPA} = -1, \] using directed angles modulo $2\pi$. (The only tricky part is the sign.) \end{exer} \section{Collinear points: Menelaos' theorem} When he published his theorem, Ceva also revived interest in an ancient theorem attributed to Menelaos (a mathematician from the first century A.D.\footnote{Which stands for ``Arbitrary Delineation.''}, not the brother of Agamemnon in Homer's \textit{Iliad}). DIAGRAM. \begin{theorem}[Menelaos] Let $ABC$ be a triangle, and let $P, Q, R$ be points on the lines $BC, CA, AB,$ respectively. Then $P, Q, R$ are collinear if and only if \[ \frac{BP}{PC} \frac{CQ}{QA} \frac{AR}{RB} = -1. \] \end{theorem} \begin{proof} Assume that $P,Q,R$ are collinear. Let $x,y,z$ be the directed distances from $A,B,C$, respectively, to the line $PQR$. DIAGRAM. Then $BP/PC = -y/z$ and so forth, so \[ \frac{BP}{PC} \frac{CQ}{QA} \frac{AR}{RB} = (-1)(-1)(-1) \frac{y}{z} \frac{z}{x} \frac{x}{y} = -1. \] The converse follows by the same argument as for Ceva's Theorem. \end{proof} An important consequence of Menelaos' theorem is the following result of Desargues, which is most easily stated by introducing two pieces of terminology. Two triangles $ABC$ and $DEF$ are said to be \emph{perspective from a point} if the lines $AD, BE, CF$ are concurrent. They are said to be \emph{perspective from a line} if the points $AB \cap DE, BC \cap EF, CD \cap FA$ are collinear. \begin{theorem}[Desargues] \label{thm:des} Two triangles $ABC$ and $DEF$ are perspective from a point if and only if they are perspective from a line. \end{theorem} \begin{proof} We only prove that if $ABC$ and $DEF$ are perpective from a point, they are perspective from a line. We leave it as an exercise to deduce the reverse implication from this (stare at the diagram); we will do this again later, using duality. Suppose that $AD, BE, CF$ concur at $O$, and let $P = BC \cap EF$, $Q = CA \cap FD$, $R = AB \cap DE$. To show that $P, Q, R$ are collinear, we want to show that \[ \frac{AR}{RB} \frac{BP}{PC} \frac{CQ}{QA} = -1 \] and invoke Menelaos' theorem. To get hold of the first term, we apply Menelaos to the points $R, D, E$ on the sides of the triangle $OAB$, giving \[ \frac{AR}{RB} \frac{BD}{DO} \frac{OE}{EA} = -1. \] Analogously, \[ \frac{BP}{PC} \frac{CE}{EO} \frac{OF}{FB} = \frac{CQ}{QA} \frac{AF}{FO} \frac{OD}{DC} = -1. \] When we multiply these three expressions together and cancel equal terms, we get precisely the condition of Menelaos' theorem. \end{proof} Another important consequence of Menelaos' theorem is the following result of Pappus. \begin{theorem}[Pappus] \label{thm:pappus} Let $A,C,E$ be three collinear points and $B,D,F$ three other collinear points. Then the points $AB \cap DE$, $BC \cap EF$, $CD \cap FA$ are collinear. \end{theorem} The proof is similar, but more complicated; we omit it, save to say that one applies Menelaos repeatedly using the triangle formed by the lines $AB, CD, EF$. If you can't make the cancellation work, see \cite{bib:cg}. Note that Desargues' and Pappus' theorems only involve points and lines, with no mention of distances or angles. This makes them ``theorems of projective geometry,'' and we will see later how projective transformations often yield simple proofs of such theorems. \begin{exer} \ii \label{ex:des2} Prove the reverse implication of Desargues' theorem. \ii \label{ex:harmcon} Let $A,B,C$ be three points on a line. Pick a point $D$ in the plane, and a point $E$ on $BD$. Then draw the line through $AE \cap CD$ and $CE \cap AD$. Show that this line meets the line $AC$ in a point $P$ that depends only on $A,B,C$. (We will study the relationship among the points $A,B,C,P$ further in the chapter on projective geometry.) \ii (MOP 1990) Let $A, B, C$ be three collinear points and $D, E, F$ three other collinear points. Let $G = BE \cap CF$, $H = AD \cap CF$, $I = AD \cap CE$. If $AI = HD$ and $CH = GF$, prove that $BI = GE$. \ii (Original) \label{ex:apcon1} Let $ABC$ be a triangle and $P$ a point in its interior, not lying on any of the medians of $ABC$. Let $A_1, B_1, C_1$ be the intersections of $PA$ with $BC$, $PB$ with $CA$, $PC$ with $AB$, respectively, and let $A_2, B_2, C_2$ be the intersections of $B_1C_1$ with $BC$, $C_1A_1$ with $CA$, $A_1B_1$ with $AB$, respectively. Prove that the midpoints of $A_1A_2, B_1B_2, C_1C_2$ are collinear. (See also Problem~\ref{ex:apcon}.) \end{exer} \section{Concurrent perpendiculars} Some of the special points of a triangle are constructed by drawing perpendiculars to the sides of a triangle. For example, the circumcenter can be constructed by drawing the perpendicular bisectors. It is convenient that a result analogous to Ceva's Theorem holds for perpendiculars; the analogy is so strong that we can safely leave the proof to the reader (see Problem~1). DIAGRAM. \begin{fact} \label{thm:concperp} Let $ABC$ be a triangle, and let $P, Q, R$ be three points in the plane. Then the lines through $P, Q, R$ perpendicular to $BC, CA, AB,$ respectively, are concurrent if and only if \[ BP^{2} - PC^{2} + CQ^{2} - QA^{2} + AR^{2} - RB^{2} = 0. \] \end{fact} A surprising consequence is that the lines through $P,Q,R$ perpendicular to $BC$, $CA$, $AB$, respectively, are concurrent if and only if the lines through $A,B,C$ perpendicular to $QR$, $RP$, $PQ$, respectively, are concurrent! \begin{exer} \ii Prove that the lines $AB$ and $CD$ are perpendicular if and only if $AC^{2} - AD^{2} = BC^{2} - BD^{2}$. (Use vectors, coordinates or Pythagoras.) Then prove Fact~\ref{thm:concperp}. \ii (Germany, 1996) Let $ABC$ be a triangcle, and construct squares $ABB_1A_2, BCC_1B_2, CAA_1C_2$ externally on its sides. Prove that the perpendicular bisectors of the segments $A_1A_2, B_1B_2, C_1C_2$ are concurrent. \ii Let $ABC$ be a triangle, $\ell$ a line and $L,M,N$ the feet of the perpendiculars to $\ell$ from $A,B,C$, respectively. Prove that the perpendiculars to $BC,CA,AB$ through $L,M,N$, respectively, are concurrent. Their intersection is called the \emph{orthopole} of the line $\ell$ and the triangle $ABC$. \end{exer} \section{Additional problems} \begin{exer} \ii (USAMO 1997/2) \label{ex:usa97} Let $ABC$ be a triangle, and draw isosceles triangles $DBC, AEC, ABF$ external to $ABC$ (with $BC, CA, AB$ as their respective bases). Prove that the lines through $A,B,C$ perpendicular to $EF,FD,DE$, respectively, are concurrent. (Several solutions are possible.) \ii (MOP 1997) Let $ABC$ be a triangle, and $D, E, F$ the points where the incircle touches sides $BC, CA, AB$, respectively. The parallel to $AB$ through $E$ meets $DF$ at $Q$, and the parallel to $AB$ through $D$ meets $EF$ at $T$. Prove that the lines $CF, DE, QT$ are concurrent. \ii (Stanley Rabinowitz) The incircle of triangle $ABC$ touches sides $BC, CA, AB$ at $D, E, F$, respectively. Let $P$ be any point inside triangle $ABC$, and let $X, Y, Z$ be the points where the segments $PA, PB, PC$, respectively, meet the incircle. Prove that the lines $DX, EY, FZ$ are concurrent. (The diagram for this problem serves as the logo for Mathpro Press, founded by Rabinowitz.) \end{exer} \chapter{Transformations} In geometry, it is often useful to study transformations of the plane (i.e.\ functions mapping the plane to itself) preserving certain properties. In fact, Felix Klein DATES went so far as to define ``geometry'' as the study of properties invariant under a particular set of transformations! \section{Rigid motions} A \emph{rigid motion} of the Euclidean plane is a map which preserves distances; that is, if $P$ maps to $P'$ and $Q$ to $Q'$, then we have $PQ = P'Q'$. The rigid motions are the following: \begin{itemize} \item Translation: each point moves a fixed distance in a fixed direction, so that $PQQ'P'$ is always a parallelogram. \item Rotation with center $O$ and angle $\theta$: each point $P$ maps to the point $P'$ such that $OP = OP'$ and $\ang POP' = \theta$, where the angle is directed and measured modulo $2\pi$ (not the usual $\pi$!). \item Reflection through the line $\ell$: each point $P$ maps to the point $P'$ such that $\ell$ is the perpendicular bisector of $PP'$. \item Glide reflection along the line $\ell$: reflection through $\ell$ followed by a translation along $\ell$. \end{itemize} \begin{theorem} Given two congruent figures, each not contained in a line, one can be mapped onto the other by a unique rigid motion. If the figures are directly congruent, the rigid motion is a translation or a rotation; if oppositely congruent, a reflection or glide reflection. \end{theorem} The use throughout this chapter of the term ``figure'' includes the assumption that all points of the figure are distinguishable, to rule out the possibility of nontrivial maps carrying the figure to itself (e.g. the symmetries of a regular polygon, or a circle). \begin{proof} We first address the uniqueness. If there were two rigid motions carrying the first figure to the second, then composing one with the inverse of the other would yield a nontrivial rigid motion leaving one entire figure in place. By assumption, however, this figure contains three noncollinear points $A, B, C$, and a point $P$ is uniquely determined by its distances to these three points (see Problem~1), so every point is fixed by the rigid motion, a contradiction. Thus the motion is unique if it exists. Now we address existence. let $A, B, C$ be three noncollinear points of the first figure, and $A', B', C'$ the corresponding points of the second figure. There exists a translation mapping $A$ to $A'$; following that with a suitable rotation (since $AB = A'B'$), we can ensure that $B$ also maps to $B'$. Now we claim $C$ maps either to $C'$ or to its reflection across $A'B'$; in other words, given two points $A,B$ and a point $C$ not on $AB$, $C$ is determined up to reflection across $AB$ by the distances $AC$ and $BC$. Of course this holds because this data fixes $C$ to lie on two distinct circles. Now if $P$ is any point of the first figure, then $P$ is uniquely determined by the distances $AP, BP, CP$, and so it must map to the corresponding point of the second figure. This completes the proof of existence. For the final assertion, first assume triangles $ABC$ and $A'B'C'$ are directly congruent. If the perpendicular bisectors of $AA'$ and $BB'$ are parallel, then $ABB'A'$ is a parallelogram, so there is a translation taking $A$ to $A'$ and $B$ to $B'$. Otherwise, let these perpendicular bisectors meet at $O$. Draw the circle through $B$ and $B'$ centered at $O$; there are (at most) two points on this circle whose distance to $A'$ is the length $AB$. One point is the reflection of $B$ across the perpendicular bisector of $AA'$; by our assumption, this cannot equal $B'$. Thus $B'$ is the other point, which is the image of $B$ under the rotation about $O$ taking $B$ to $B'$. DIAGRAM. In either case, we have a direct rigid motion taking $A$ to $A'$ and $B$ to $B'$; this rigid motion must also take $C$ to $C'$, completing the proof in this case. Next, assume $ABC$ and $A'B'C'$ are oppositely congruent. The lines through which $AB$ reflects to a line parallel to $A'B'$ form two perpendicular families of parallel lines. In each family there is one line passing through the midpoint of $AA'$; the glide reflection through this line taking $A$ to $A'$ takes $B$ either to $B'$ or to its half-turn about $A'$. In the latter case, switching to the other family gives a glide reflection taking $B$ to $B'$. As in the first case, $C$ automatically goes to $C'$, and we are done. \end{proof} In particular, the composition of two rotations is either a rotation or translation. In fact, one can say more. \begin{fact} The composition of a rotation of angle $\theta_{1}$ with a rotation of angle $\theta_{2}$ is a rotation of angle $\theta_{1} + \theta_{2}$ if this is not a multiple of $2\pi$, and a translation otherwise. \end{fact} On the other hand, given two rotations, it is not obvious where the center of their composition is; in particular, it generally depends on the order of the rotations. (In fancy language, the group of rigid motions is not commutative.) \begin{exer} \ii Prove that a point not lying on $AB$ is uniquely determined up to reflection across $AB$ by its distances to two points $A$ and $B$, and that any point at all is uniquely determined by its distances to three points $A, B, C$. (The latter is the ``triangulation'' principle used once upon a time for navigation at sea, and nowadays in the Global Positioning System.) \ii (MOP 1997) Consider a triangle $ABC$ with $AB = AC$, and points $M$ and $N$ on $AB$ and $AC$, respectively. The lines $BN$ and $CM$ intersect at $P$. Prove that $MN$ and $BC$ are parallel if and only if $\ang APM = \ang APN$. \ii (IMO 1986/2) A triangle $A_1A_2A_3$ and a point $P_0$ are given in the plane. We define $A_s = A_{s-3}$ for all $s \geq 4$. We construct a sequence of points $P_1, P_2, P_3, \dots$ such that $P_{k+1}$ is the image of $P_k$ under rotation with center $A_{k+1}$ through angle $120^\circ$ clockwise (for $k=0,1,2,\dots$). Prove that if $P_{1986} = P_0$, then the triangle $A_1A_2A_3$ is equilateral. \ii (MOP 1996) \label{ex:threeeq} %%(IMO 96 Test 1, problem 3(b)) Let $AB_1C_1, AB_2C_2, AB_3C_3$ be directly congruent equilateral triangles. Prove that the pairwise intersections of the circumcircles of triangles $AB_1C_2, AB_2C_3, AB_3C_1$ form an equilateral triangle congruent to the first three. \end{exer} \section{Homothety} Having classified the rigid motions, we move now to the \emph{similarities}, those transformations which preserve lengths up to a scalar factor. That is, $P'Q'/PQ$ does not depend on $P$ or $Q$. An important class of similarities not yet encountered are the homotheties, which are sufficiently useful for problems that we shall devote an entire section to them. Given a point $P$ and a real number $r$, the \emph{homothety} with center $P$ and ratio $r$ maps each point $P$ maps to the point $P'$ on the ray $OP$ such that $OP' = r OP$. We use the usual directed lengths convention, so $r$ may be positive or negative. DIAGRAM. Homotheties have the property that they map every side of a figure to a parallel segment. Aside from translations (which might be thought of as degenerate homotheties with center ``at infinity''), this property characterizes homotheties; the following theorem is often useful as a concurrence criterion. \begin{theorem} Two directly similar but not congruent figures with corresponding sides parallel are homothetic. In particular, the lines $AA'$, where $A$ and $A'$ are corresponding points, all pass through a common point. \end{theorem} As for rotations, we conclude that the composition of two homotheties is a homothety, though again it is less than obvious where the center is! \begin{exer} \ii Given a triangle $ABC$, construct (with straightedge and compass) a square with one vertex on $AB$, one vertex on $AC$, and two (adjacent) vertices on $BC$. \ii (USAMO 1992/4) Chords $\, \overline{AA'}, \, \overline{BB'}, \, \overline{CC'} \,$ of a sphere meet at an interior point $\, P \,$ but are not contained in a plane. The sphere through $\, A,B,C,P \,$ is tangent to the sphere through $\, A', B', C', P$. Prove that $\, AA' = BB' = CC'$. \ii (Putnam 1996/A-2) Let $C_1$ and $C_2$ be circles whose centers are 10 units apart, and whose radii are 1 and 3. Find, with proof, the locus of all points $M$ for which there exists points $X$ on $C_1$ and $Y$ on $C_2$ such that $M$ is the midpoint of the line segment $XY$. \ii Given three nonintersecting circles, draw the intersection of the external tangents to each pair of the circles. Show that these three points are collinear. \end{exer} \section{Spiral similarity} \label{sec:spir} A \emph{similarity} is any transformation that takes any figure to a figure similar to it. In other words, it preserves angles up to sign (and hence collinearity) and multiplies lengths by a constant factor. We distinguish between \emph{direct} and \emph{opposite} similarities, which respectively preserve and reverse the sign of directed angles. A \emph{spiral similarity} of angle $\theta$ and ratio $r$ centered at a point $P$ consists of a homothety of ratio $r$ centered at $P$ followed by a rotation of angle $\theta$ centered at $P$. (The order of these operations do not matter; one easy way to see this is to express both operations in terms of complex numbers.) One might imagine there are more complicated similarities, e.g. a homothety followed by a rotation about different centers. As it turns out, there are no more similarities left to describe! (This proof uses complex numbers; we will sketch a synthetic proof later.) \begin{theorem} \label{thm:sim} Every direct similarity is either a translation or a spiral similarity. \end{theorem} \begin{proof} First we show that every direct similarity can be expressed in terms of homothety, translation, and rotation. Let $A$ and $B$ be two points with images $C$ and $D$. If we perform a homothety about $A$ of ratio $CD/AB$, then a translation mapping $A$ to $C$, then a suitable rotation, we get another similarity mapping $A$ and $B$ to $C$ and $D$. On the other hand, if $P$ is any point not on the line $AB$ and $Q$ and $Q'$ are its images under the original similarity and the new similarity, then the triangles $ABP$, $CDQ$ and $CDQ'$ are all similar, implying that $C,Q,Q'$ and $D,Q,Q'$ are collinear, which means $Q = Q'$. In other words, the original similarity coincides with the new one. The basic transformations can be expressed in terms of complex numbers as follows: \begin{center} \begin{tabular}{ll} Translation by vector $v$ & $z \mapsto z + v$ \\ Homothety of ratio $r$, center $x$ & $z \mapsto r(z-x) + x$ \\ Rotation by angle $\theta$, center $x$ & $z \mapsto e^{i\theta}(z-x) + x $ \end{tabular} \end{center} The point is that each of these maps has the form $z \mapsto az + b$ for some complex numbers $a,b$, and hence all direct similarities have this form. (It is easily seen that any such map is indeed a similarity; in fact, it's the composition of a spiral similarity about $0$ and a translation.) Now let $z \mapsto az + b$ be a direct similarity. If $a=1$, clearly we have a translation by $b$. Otherwise, let $t = b/(1-a)$ be the unique solution of $t = at + b$. Then our map can be written $z \mapsto a(z-t) + t$, which is clearly a spiral similarity about $t$. \end{proof} For practical applications of the relationship between complex numbers and transformations, see Appendix~\ref{nonsynth}. \begin{exer} %\ii %Fill in the missing steps in the above argument. For example, why %must a similarity multiply any two lengths by the same factor? %\ii %Give another solution to Problem~% from angles %using a spiral similarity. \ii (USAMO 1978/2) $ABCD$ and $A'B'C'D'$ are square maps of the same region, drawn to different scales and superimposed. Prove that there is only one point $O$ on the small map which lies directly over point $O'$ of the large map such that $O$ and $O'$ represent the same point of the country. Also, give a Euclidean construction (straightedge and compass) for $O$. \ii (MOP 1998) Let $ABCDEF$ be a cyclic hexagon with $AB=CD=EF$. Prove that the intersections of $AC$ with $BD$, of $CE$ with $DF$, and of $EA$ with $FB$ form a triangle similar to $BDF$. \ii \label{ex:sevenpt} Let $C_1, C_2, C_3$ be circles such that $C_1$ and $C_2$ meet at distinct points $A$ and $B$, $C_2$ and $C_3$ meet at distinct points $C$ and $D$, and $C_3$ and $C_1$ meet at distinct points $E$ and $F$. Let $P_1$ be an arbitrary point on $C_1$, and define points $P_2, \dots, P_7$ as follows: \begin{center} \begin{tabular}{ll} $P_2$ & is the second intersection of line $P_1A$ with $C_2$; \\ $P_3$ & is the second intersection of line $P_2C$ with $C_3$; \\ $P_4$ & is the second intersection of line $P_3E$ with $C_1$; \\ $P_5$ & is the second intersection of line $P_1B$ with $C_2$; \\ $P_6$ & is the second intersection of line $P_2D$ with $C_3$; \\ $P_7$ & is the second intersection of line $P_3F$ with $C_1$. \\ \end{tabular} \end{center} \noindent Prove that $P_7 = P_1$. \end{exer} \section{Affine transformations} The last type of transformation we introduce in this chapter is the most general, at the price of preserving the least structure. However, for sheer strangeness it does not rival either inversion or projective transformations, which we shall introduce much later. %%Define as preserving collinearity and length ratios? An affine transformation is most easily defined in terms of coordinates, which is the approach we shall take. An \emph{affine transformation} is a map from the plane to itself which, in coordinates, has the form \[ (x, y) \mapsto (ax + by + c, dx + ey + f) \] for some constants $a,b,c,d,e,f$. By our calculations involving complex numbers, every similarity is an affine transformations. Some more exotic affine transformations are the stretch $(x,y) \mapsto (x,cy)$ and the shear $(x,y) \mapsto (x+y,y)$. \begin{fact} \label{ex:aff} Any three noncollinear points can be mapped to any three other noncollinear points by a unique affine transformation. \end{fact} As an example of the use of the affine transformation, we offer the following theorem. \begin{theorem} Let $ABCDE$ be a convex pentagon and let $F = BC \cap DE$, $G = CD \cap EA$, $H = DE \cap AB$, $I = EA \cap BC$, $J = AB \cap DE$. Suppose that the areas of the triangles $AHI, BIJ, CJF, DFG, EGH$ are all equal. Then the lines $AF, BG, CH, DI, EJ$ are all concurrent. DIAGRAM \end{theorem} \begin{proof} Everything in the theorem is affine-invariant, so we may place three of the points anywhere we want. Let us assume $A, C, D$ form an isosceles triangle with $AC = AD$ and $\angle CDA = \pi/5$, which is to say that $A, C, D$ are three vertices of a regular pentagon. Our first observation is that since $CJF$ and $DFG$ have equal areas, so do $CJD$ and $JGD$, by adding the area of $CDH$ to both sides. By the base-height formula, this means $GJ$ is parallel to $CD$, and similarly for the other sides. In particular, since $ACD$ was assumed to be isosceles, $F$ lies on its angle bisector at $A$, and $J$ and $C$ are the reflections of $G$ and $D$ across $AF$. Next we want to show that $B$ and $E$ are mirror images across $AF$, so let $E'$ and $H'$ be the reflections of $E$ and $H$, respectively. Since the lines $FC$ and $FD$ are mirror images, we know $E'$ lies on $BD$, and similarly $H'$ lies on $AC$. Suppose that $E'D < BD$, or equivalently that $E$ is closer than $B$ to the line $CD$. Then we also have $DH' < CI'$; since $CJ = DG$, we deduce $JH' < JI$. Now it is evident that the triangle $E'H'J$, being contained in $BJI$, has smaller area; on the other hand, it has the same area as $EHG$, which by assumption has the same area as $BJI$, a contradiction. So we cannot have $E'D < BD$, or $E'D > BD$ by a similar argument. We conclude $E'D = BD$, i.e.\ $B$ and $E$ are mirror images. DIAGRAM. In particular, this implies that $BE$ is parallel to $CD$. Since we could just as well have put $B, D, E$ at the vertices of an isosceles triangle, we also may conclude $AC \parallel DE$ and so forth. Now let $\ell$ be the line through $C$ parallel to $AD$; by the above argument, we know $B$ is the intersection of $\ell$ with $DF$. On the other hand, $B$ is also the intersection of $\ell$ with the line through $A$ parallel to $CF$. If we move $F$ towards $A$ along the angle bisector of $ACD$ at $A$, the intersection of $DF$ with $\ell$ moves away from $C$, but the intersection of the parallel to $CF$ through $A$ with $\ell$ moves closer to $C$. Hence these can only coincide for at most one choice of $F$, and of course they do coincide when $ABCDE$ is a regular pentagon. We conclude that $ABCDE$ is the image of a regular pentagon under an affine transformation, which in particular implies that $AF, BG, CH, DI, EJ$ are concurrent. DIAGRAM \end{proof} \begin{exer} \ii Let $AX, BY, CZ$ be three cevians of the triangle $ABC$ satisfying \[ \frac{BX}{XC} = \frac{CY}{YA} = \frac{AZ}{ZB} = k, \] where $k$ is a given constant greater than 1. Find the ratio of the area of the triangle formed by the three cevians to the area of $ABC$. (Compare your answer with Problem~\ref{ex:seventh}, which is the case $k= 2$.) \ii (France, 1996) Let $ABC$ be a triangle. For any line $\ell$ not parallel to any side of the triangle, let $G_{\ell}$ be the centroid of the degenerate triangle whose vertices are the intersection of $\ell$ with $BC, CA, AB$ (i.e.\ the vector average of these three points). Determine the union of the $G_{\ell}$ over all $\ell$. \end{exer} % \section{Summary} % % For quick reference, we provide a summary of what properties are % preserved by which types of transformations. CHART. % % We also provide a diagram (from COX GREi) indicating which % transformations are special cases of others. DIAGRAM. %% Geometry project %% Circles \chapter{Circular reasoning} In this chapter, we study an ostensibly simple, yet intriguing object: the circle. This chapter consists mainly of ``classical'' results; the modern technique of inversion, which builds on these results and yields surprising new results and proofs of its own, occupies a subsequent chapter. \section{Power of a point} The following is truly a theorem of Euclidean geometry: it appears in the \textit{Elements}. \begin{theorem} Given a fixed circle and a fixed point $P$, draw a line through $P$ intersecting the circle at $A$ and $B$. Then the product $PA \cdot PB$ depends only on $P$ and the circle, not on the line. \end{theorem} \begin{proof} Draw another line through $P$ meeting the circle at $C$ and $D$, labeled as in one of the diagrams. Then \[ \dang PAC = \dang BAC = \dang BDC = -\dang PDB \] as directed angles, so the triangles $PAC$ and $PDB$ are (oppositely) similar, giving $PA/PD = PC/PB$, or equivalently $PA\cdot PB = PC\cdot PB$. \end{proof} The quantity $PA \cdot PB$ is called the \emph{power} of $P$ with respect to the circle. If $O$ is the center of the circle and $r$ its radius, we may choose $OP$ as our line and so express the power as \[ (OP + r)(OP - r) = OP^{2} - r^{2}. \] Note that the usual sign convention makes the power positive if $P$ lies outside the circle and negative if $P$ lies inside the circle. Also note that for $P$ outside the circle, the limiting case $A=B$ means that $PA$ is tangent to the circle at $A$. The power-of-a-point theorem has an occasionally useful converse. \begin{fact} \label{fact:power} If the lines $AB$ and $CD$ meet at $P$, and $PA \cdot PB = PC \cdot PD$ (as signed lengths), then $A,B,C,D$ are concyclic. \end{fact} \begin{exer} \ii If $A,B,C,D$ are concyclic and $AB \cap CD = E$, prove that $(AC/BC)(AD/BD) = AE/BE$. \ii (\textit{Mathematics Magazine}, Dec. 1992) Let $ABC$ be an acute triangle, let $H$ be the foot of the altitude from $A$, and let $D, E, Q$ be the feet of the perpendiculars from an arbitrary point $P$ in the triangle onto $AB, AC, AH$, respectively. Prove that \[ |AB \cdot AD - AC \cdot AE| = BC \cdot PQ. \] \ii Draw tangents $OA$ and $OB$ from a point $O$ to a given circle. Through $A$ is drawn a chord $AC$ parallel to $OB$; let $E$ be the second intersection of $OC$ with the circle. Prove that the line $AE$ bisects the segment $OB$. \ii (MOP 1995) Given triangle $ABC$, let $D, E$ be any points on $BC$. A circle through $A$ cuts the lines $AB, AC, AD, AE$ at the points $P, Q, R, S$, respectively. Prove that \[ \frac{AP \cdot AB - AR \cdot AD}{AS \cdot AE - AQ \cdot AC} = \frac{BD}{CE}. \] \ii (IMO 1995/1) Let $A,B,C, D$ be four distinct points on a line, in that order. 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. \ii (USAMO 1998/2) Let $\omega_1$ and $\omega_2$ be concentric circles, with $\omega_2$ in the interior of $\omega_1$. From a point $A$ on $\omega_1$ one draws the tangent $AB$ to $\omega_2$ ($B\in \omega_2$). Let $C$ be the second point of intersection of $AB$ and $\omega_1$, and let $D$ be the midpoint of $AB$. A line passing through $A$ intersects $\omega_2$ at $E$ and $F$ in such a way that the perpendicular bisectors of $DE$ and $CF$ intersect at a point $M$ on $AB$. Find, with proof, the ratio $AM/MC$. \end{exer} \section{Radical axis} Given two circles, one with center $O_{1}$ and radius $r_{1}$, the other with center $O_{2}$ and radius $r_{2}$, what is the set of points with equal power with respect to the two circles? By our explicit formula for the power of a point, this is simply the set of points $P$ such that $PO_{1}^{2} - r_{1}^{2} = PO_{2}^{2} - r_{2}^{2}$, or equivalently such that $PO_{1}^{2} - PO_{2}^{2} = r_{1}^{2} - r_{2}^{2}$. By Problem~2.3.1?, this set is a straight line perpendicular to $O_{1}O_{2}$; we call this line the \emph{radical axis} of the two circles. \begin{theorem}[Radical axis theorem] Let $\omega_{1}, \omega_{2}, \omega_{3}$ be three circles. Then the radical axes of $\omega_{1}$ and $\omega_{2}$, of $\omega_{2}$ and $\omega_{3}$, and of $\omega_{3}$ and $\omega_{1}$ either all coincide, or are concurrent (or parallel). \end{theorem} \begin{proof} A point on two of the radical axes has equal power with respect to all three circles. Hence if two of the axes coincide, so does the third, and otherwise if any two of the axes have a common point, this point lies on the third axis as well. \end{proof} \begin{corr} The common chords of three mutually intersecting circles lie on concurrent lines. \end{corr} If the radical axes coincide, the three circles are said to be \emph{coaxal}; otherwise, the intersection of the three radical axes is called the \emph{radical center} of the circles. (As usual, this intersection could be ``at infinity'', if the three lines are parallel.) There are three types of coaxal families, depending on whether the circles have zero, one, or two intersections with the common radical axis; these three cases are illustrated in the diagram below. DIAGRAM. A useful criterion for recognizing and applying the coaxal property is the following simple observation and partial converse. \begin{fact} If three circles are coaxal, their centers are collinear. Conversely, if three circles pass through a common point and have collinear centers, they are coaxal. \end{fact} Like the power-of-a-point theorem, the radical axis theorem has an occasionally useful converse. \begin{fact} Suppose that $ABCD$ and $CDEF$ are cyclic quadrilaterals, and the lines $AB, CD, EF$ are concurrent. Then $EFAB$ is also cyclic. \end{fact} \begin{exer} \ii Use the radical axis theorem to give another solution for Problem~\ref{ex:usa97}. \ii (MOP 1995) Let $BB', CC'$ be altitudes of triangle $ABC$, and assume $AB \neq AC$. Let $M$ be the midpoint of $BC$, $H$ the orthocenter of $ABC$, and $D$ the intersection of $BC$ and $B'C'$. Show that $DH$ is perpendicular to $AM$. \ii (IMO 1994 proposal) %% Also inversion A circle $\omega$ is tangent to two parallel lines $\ell_{1}$ and $\ell_{2}$. A second circle $\omega_{1}$ is tangent to $\ell_{1}$ at $A$ and to $\omega$ externally at $C$. A third circle $\omega_{2}$ is tangent to $\ell_{2}$ at $B$, to $\omega$ externally at $D$ and to $\omega_{1}$ externally at $E$. Let $Q$ be the intersection of $AD$ and $BC$. Prove that $QC = QD = QE$. \ii (India, 1995) Let $ABC$ be a triangle. A line parallel to $BC$ meets sides $AB$ and $AC$ at $D$ and $E$, respectively. Let $P$ be a point inside triangle $ADE$, and let $F$ and $G$ be the intersection of $DE$ with $BP$ and $CP$, respectively. Show that $A$ lies on the radical axis of the circumcircles of $PDG$ and $PFE$. \ii \label{ex:imo85} (IMO 1985/5) A circle with center $O$ passes through the vertices $A$ and $C$ of triangle $ABC$, and intersects the segments $AB$ and $BC$ again at distinct points $K$ and $N$, respectively. The circumscribed circles of the triangle $ABC$ and $KBN$ intersect at exactly two distinct points $B$ and $M$. Prove that angle $OMB$ is a right angle. \end{exer} \section{The Pascal-Brianchon theorems} \label{sec:pb} An amazing theorem about a hexagon inscribed in a circle was published by Blaise Pascal (1623-1662). His original proof, which was favorably described by Leibniz, has unfortunately been lost; we present here an ingenious proof essentially due to Jan van Yzeren (A simple proof of Pascal's hexagon theorem, textit{Monthly}, December 1993). %% AMM 100 (Dec 1993) 930-931 \begin{theorem}[Pascal] \label{thm:pascal} Let $ABCDEF$ be a hexagon inscribed in a circle. Then the intersections of $AB$ and $DE$, of $BC$ and $EF$, and of $CD$ and $FA$ are collinear. \end{theorem} \begin{proof} Let $P = AB \cap DE, Q = BC \cap EF, R = CD \cap FA$. Draw the circumcircle of $CFR$, and extend the lines $BC$ and $DE$ to meet this circle again at $G$ and $H$, respectively. By Theorem~\ref{thm:parchord}, we have $BE || GH$, $DE || HR$ and $AB || GR$. Now notice that the triangles $RGH$ and $PBE$ have parallel sides, which means they are homothetic. In other words, the lines $BG, EH, PR$ are concurrent, which means $BG \cap EH = Q$ is collinear with $P$ and $R$, as desired. \end{proof} Some time later, C.J. Brianchon (1760-1854) discovered a counterpart to Pascal's theorem for a hexagon circumscribed about a circle. We will give Brianchon's proof of his theorem, which uses the polar map to reduce it to Pascal's theorem, in Section~\ref{sec:polar}; a direct but somewhat complicated proof can be found in \cite{bib:cg}. \begin{theorem}[Brianchon] Let $ABCDEF$ be a hexagon circumscribed about a circle. Then the lines $AD, BE, CF$ are concurrent. \end{theorem} In practice, one often applies Pascal's and Brianchon's theorem in certain degenerate cases, where some of the vertices are the same. In Pascal's theorem, if we move two adjacent vertices of the hexagon very close together, the line through them approaches a tangent to the circle. (In calculus, this would be called taking a derivative!) Thus for example, in the diagram below DIAGRAM, we may conclude that $AA \cap CD, AC \cap DE, CD \cap FA$ are collinear, where $AA$ denotes the tangent at $A$. As for Brianchon's theorem, the analogous argument shows that the ``vertex'' between two collinear sides belongs at the point of tangency. DIAGRAM. \begin{exer} \ii What do we get if we apply Brianchon's theorem with three degenerate vertices? (We will encounter this fact again later.) %\ii (MOP 1997) %Let $ABC$ and $BCD$ be equilateral triangles. A line through $D$ meets %$AB$ at $M$ and $AC$ at $N$. What is the acute angle between the lines %$BN$ and $CM$? \ii Let $ABCD$ be a circumscribed quadrilateral, whose incircle touches $AB,BC,CD,DA$ at $M, N, P, Q$, respectively. Prove that the lines $AC, BD, MP, NQ$ are concurrent. \ii (MOP 1995) With the same notation, let lines $BQ$ and $BP$ intersect the inscribed circle at $E$ and $F$, respectively. Prove that $ME, NF$ and $BD$ are concurrent. \ii (Poland, 1997) \label{ex:pol1997} Let $ABCDE$ be a convex quadrilateral with $CD = DE$ and $\angle BCD = \angle DEA = \pi/2$. Let $F$ be the point on side $AB$ such that $AF/FB = AE/BC$. Show that \[ \ang FCE = \ang FDE \quad \mbox{and} \quad \ang FEC = \ang BDC. \] \end{exer} \section{Simson line} The following theorem is often called Simson's theorem, but \cite{bib:cg} attributes it to William Wallace. \begin{theorem} Let $A,B,C$ be three points on a circle. Then the feet of the perpendiculars from $P$ to the lines $AB, BC, CA$ are collinear if and only if $P$ also lies on the circle. \end{theorem} \begin{proof} The proof is by (directed) angle-chasing. Let $X,Y,Z$ be the feet of the respective perpendiculars from $P$ to $BC, CA, AB$; then the quadrilaterals $PXCY, PYAZ, PZBX$ each have two right angles, and are thus cyclic. Therefore \beqa \dang PXY &=& \dang PCY \qquad \mbox{(cyclic quadrilateral $PXCY$)} \\ &=& \dang PCA \qquad \mbox{(collinearity of $A, C, Y$)} \eeqa and analogously $\dang PXZ = \dang PBA$. Now $X, Y, Z$ are collinear if and only if $\dang PXY = \dang PXZ$, which by the above equations occurs if and only if $\dang PCA = \dang PBA$; in other words, if and only if $A, B, C, P$ are concyclic. \end{proof} For $P$ on the circle, the line described in the theorem is called the \emph{Simson line} of $P$ with respect to the triangle $ABC$. We note in passing that an alternate proof of the collinearity in this case can be given using Menelaos. \begin{exer} \ii Let $A,B,C,P,Q$ be points on a circle. Show that the (directed) angle between the Simson lines of $P$ and $Q$ with respect to the triangle $ABC$ equals half of the (directed) arc $PQ$. \ii \label{ex:simcon} Let $A,B,C,D$ be four points on a circle. Prove that the intersection of the Simson line of $A$ with respect to $BCD$ with the Simson line of $B$ with respect to $ACD$ lies on the line through $C$ and the orthocenter of $ABD$. \ii \label{ex:postsim} If $A, B, C, P, Q$ are five points on a circle such that $PQ$ is a diameter, show that the Simson lines of $P$ and $Q$ with respect to $ABC$ intersect at a point concyclic with the midpoints of the sides of $ABC$. \ii Let $I$ be the incenter of triangle $ABC$, and $D, E, F$ the projections of $I$ onto $BC, CA, AB$, respectively. The incircle of $ABC$ meets the segments $AI, BI, CI$ at $M, N, P$, respectively. Show that the Simson lines of any point on the incircle with respect to the triangles $DEF$ and $MNP$ are perpendicular. \end{exer} \section{Circle of Apollonius} The ancient geometer Apollonius DATES is most famous for his early work on conic sections. However, his name has come to be attached to another pretty geometrical construction. \begin{theorem} \label{thm:ap} Let $A,B$ be any two points and $k \neq 1$ a positive real number. Then the locus of points $P$ such that $PA/PB = k$ is a circle whose center lies on $AB$. \end{theorem} \begin{proof} One can show this synthetically, but the shortest proof involves introducing Cartesian coordinates such that $A = (a, 0)$ and $B = (b, 0)$. The condition $PA/PB = k$ is equivalent to $PA^{2} = k^{2} PB^{2}$, which in coordinates can be written \[ (x-a)^{2} + y^{2} = k^{2}[(x-b)^{2} + y^{2}]. \] Combining terms and dividing through by $1-k^{2}$, we get \[ x^{2} + \frac{2k^{2}b-2a}{1-k^{2}}x + y^{2} = \frac{k^{2}b^{2}-a^{2}}{1-k^{2}}, \] which is easily recognized as the equation of a circle whose center lies on the $x$-axis. \end{proof} This circle is called the \emph{circle of Apollonius} corresponding to the points $A,B$ and the ratio $k$. (This term usually also includes the degenerate case $k=1$, where the ``circle'' becomes the perpendicular bisector of $AB$.) \begin{exer} \ii \label{ex:simsynth} Use circles of Apollonius to give a synthetic proof of the classification of similarities (Theorem~\ref{thm:sim}). \ii (Original) \label{ex:apcon} Let $ABC$ be a triangle and $P$ a point in its interior, not lying on any of the medians of $ABC$. Let $A_1, B_1, C_1$ be the intersections of $PA$ with $BC$, $PB$ with $CA$, $PC$ with $AB$, respectively, and let $A_2, B_2, C_2$ be the intersections of $B_1C_1$ with $BC$, $C_1A_1$ with $CA$, $A_1B_1$ with $AB$, respectively. Prove that if some two of the circles with diameters $A_1A_2, B_1B_2, C_1C_2$ intersect, then they are coaxal. (The case where the circles do not meet is trickier, unless you work in the complex projective plane as described in Section~\ref{sec:alggeo}.) Note that Problem~\ref{ex:apcon1} follows from this. \end{exer} \section{Additional problems} \begin{exer} \ii %% Broken chord theorem (Archimedes' ``broken-chord'' theorem) Point $D$ is the midpoint of arc $AC$ of a circle; point $B$ is on minor arc $CD$; and $E$ is the point on $AB$ such that $DE$ is perpendicular to $AB$. Prove that $AE = BE + BC$. \ii \label{ex:convex} The convex hexagon $ABCDEF$ is such that \[ \angle BCA = \ang DEC = \ang FAE = \ang AFB = \ang CBD = \ang EDF. \] Prove that $AB = CD = EF$. \ii (Descartes' circle theorem) Let $r_1,r_2,r_3,r_4$ be the radii of four mutually externally tangent circles. Prove that \[ \sum_{i=1}^4 \frac{2}{r_1^2} = \left( \sum_{i=1}^4 \frac{1}{r_i} \right)^2. \] \end{exer} %% Geometry project %% Triangle trivia \chapter{Triangle trivia} To a triangle are associated literally hundreds of special points; in this chapter, we study but a few of the more important ones. \section{Centroid} The following is one of the few nontrivial facts proved in standard American geometry courses. The \emph{median} from $A$ in the triangle $ABC$ is the segment joining $A$ and the midpoint of $BC$. \begin{fact} The medians of a triangle are concurrent. Moreover, the point of concurrency trisects each median. DIAGRAM. \end{fact} One can easily show this using Ceva and Menelaos, or by performing an affine transformation making the triangle equilateral. There is also a physical interpretation: if equal masses are placed at each of the vertices of a triangle, the center of mass will lie at the centroid. We leave as an exercise the task of finding physical interpretations of the other special points (or see Section \ref{sec:vec}). \section{Incenter and excenters} If $P$ is a point inside triangle $ABC$, then the distances from $P$ to the sides $AB$ and $AC$ are \[ PA \sin \angle PAB \quad \mbox{and} \quad PA \sin \angle PAC \] and these are equal if and only if $\angle PAB = \angle PAC$, in other words, if $P$ lies on the internal angle bisector of $\angle A$. From this it follows that the intersection of two internal angle bisectors is equidistant from all three sides, and consequently lies on the third bisector. This intersection is the \emph{incenter} of $ABC$, and its distance to any side is the \emph{inradius}, usually denoted $r$. The terminology comes from the fact that the circle of radius $r$ centered at the incenter is tangent to all three sides of $ABC$, and thus is called the \emph{inscribed circle}, or \emph{incircle}, of $ABC$. Do not forget, though, that the angle $A$ in triangle $ABC$ has \emph{two} angle bisectors, one internal and one external. The locus of points equidistant to the two lines $AB$ and $AC$ is the union of both lines, and so one might expect to find other circles tangent to all three sides. Indeed, the internal angle bisector at $A$ concurs with the external bisectors of the other two angles (by the same argument as above); the point of concurrence is the \emph{excenter} opposite $A$, and the circle centered there tangent to all three sides is the \emph{escribed circle}, or \emph{excircle}, opposite $A$. In studying the incircle and excircles, a fundamental tool is the fact that the two tangents to a circle from an external point have the same length. This fact is equally useful is its own right, and we have included some exercises that take advantage of equal tangents. In any case, the key observation we need is that if $D,E,F$ are the points where the incircle touches $BC, CA, AB$, respectively, then $AE = AF$ and so on, so a little algebra gives \[ AE = \frac{1}{2}(AE + EB + AF + FC - BD - DC). \] This establishes the first half of the following result; the second half is no harder. \begin{fact} Let $s = (a+b+c)/2$. Then the distance from $A$ to the point where the incircle touches $AB$ is $s-a$, and the distance from $A$ to the point where the excircle opposite $C$ touches $AB$ is $s-b$. DIAGRAM \end{fact} The quantity $s$ is often called the \emph{semiperimeter} of the triangle $ABC$. It will often be helpful to know in what ratio an angle bisector divides the opposite side. The answer can be used to give another proof of the concurrence of the angle bisectors. \begin{fact}[Angle bisector theorem] If $D$ is the foot of either angle bisector of $A$ in triangle $ABC$, then (as unsigned lengths) \[ \frac{DB}{DC} = \frac{AB}{AC}. \] \end{fact} Another useful construction for studying incenters is the \begin{fact} \label{midarc} Let $ABC$ be a triangle inscribed in a circle with center $O$, and let $M$ be the second intersection of the internal angle bisector of $A$ with the circle. The following facts are true: \begin{enumerate} \ii $MO$ is perpendicular to $BC$ (that is, $M$ is the midpoint of arc $BC$); \ii The circle centered at $M$ passing through $B$ and $C$ also passes through the incenter $I$ and the excenter $I_A$ opposite $A$; that is, $MB = MI = MC = MI_A$. \ii $OI^2 = R^2 - 2Rr$, where $R$ is the circumradius and $r$ the inradius of $ABC$. \end{enumerate} \end{fact} \begin{exer} \ii Use the angle bisector theorem to give a synthetic proof of Theorem~\ref{thm:ap}. % \ii % Prove that the feet of the external angle bisectors are collinear. % \ii (Arbelos) The two common external tangent segments between two nonintersecting circle cut off a segment along one of the common internal tangents. Show that all three segments have the same length. \ii \label{ex:usamo91} (USAMO 1991/5) Let $D$ be an arbitrary point on side $AB$ of a given triangle $ABC$, and let $E$ be the interior point where $CD$ intersects the external common tangent to the incircles of triangles $ACD$ and $BCD$. As $D$ assumes all positions between $A$ and $B$, show that $E$ traces an arc of a circle. \ii (Iran, 1997) \label{ex:iran97} Let $ABC$ be a triangle and $P$ a varying point on the arc $BC$ of the circumcircle of $ABC$. Prove that the circle through $P$ and the incenters of $PAB$ and $PAC$ passes through a fixed point independent of $P$. \ii (USAMO 1999/6) \label{ex:usamo1999} Let $ABCD$ be an isosceles trapezoid with $AB \parallel CD$. The inscribed circle $\omega$ of triangle $BCD$ meets $CD$ at $E$. Let $F$ be a point on the (internal) angle bisector of $\angle DAC$ such that $EF \perp CD$. Let the circumscribed circle of triangle $ACF$ meet line $CD$ at $C$ and $G$. Prove that the triangle $AFG$ is isosceles. \ii (IMO 1992/4) \label{ex:imo1992} In the plane let $\, C \,$ be a circle, $\, L \,$ a line tangent to the circle $\, C, \,$ and $\, M \,$ a point on $\, L$. Find the locus of all points $\, P \,$ with the following property: there exists two points $\, Q, R \,$ on $\, L \,$ such that $\, M \,$ is the midpoint of $\, QR \,$ and $\, C \,$ is the inscribed circle of triangle $\, PQR$. \end{exer} \section{Circumcenter and orthocenter} The locus of points equidistant from two points $A$ and $B$ is the perpendicular bisector of the segment $AB$. Hence given a triangle $ABC$, the perpendicular bisectors of any two sides meet at a point equidistant from all three vertices, and this point must then lie on the third bisector. In short, the perpendicular bisectors of a triangle concur at a point called the \emph{circumcenter} of $ABC$. As you may have guessed, this point is the center of the unique circle passing through the three points $A,B,C$, and this is known as the \emph{circumcircle} of $ABC$. A closely related point is the \emph{orthocenter}, defined as the intersection of the altitudes of a triangle. One can apply Fact~\ref{thm:concperp} to show that these actually concur, or one can modify the proof of the following theorem to include this concurrence as a consequence. \begin{theorem} Let $ABC$ be a triangle and $O,G,H$ its circumcenter, centroid and orthocenter, respectively. Then $O,G,H$ lie on a line in that order, and $2OG = GH$. \end{theorem} The line $OGH$ is called the \emph{Euler line} of triangle $ABC$. \begin{proof} The homothety with center $G$ and ratio $-1/2$ carries $ABC$ to the medial triangle $A'B'C'$, where $A'$ is the midpoint of $BC$ and so forth. Moreover, the altitude from $A'$ in the medial triangle coincides with the perpendicular bisector of $BC$ (since both are perpendicular to $BC$ and pass through $A'$). Hence $H$ maps to $O$ under the homothety, and the claim follows. \end{proof} Some of the exercises will use the following facts about the orthocenter, which we leave as exercises in angle-chasing. DIAGRAM. \begin{fact} Let $H_{A}, H_{B}, H_{C}$ be the feet of the altitudes from $A,B,C$, respectively, and let $H$ be the orthocenter. Then the following statements hold: \begin{enumerate} \ii The triangles $AH_{B}H_{C}, H_{A}BH_{C}, H_{A}H_{B}C$ are (oppositely) similar to $ABC$. \ii The altitudes bisect the angles of the triangle $H_{A}H_{B}H_{C}$ (so $H$ is its incenter). \ii The reflections of $H$ across $BC, CA, AB$ lie on the circumcircle of $ABC$. \end{enumerate} \end{fact} The triangle formed by the feet of the altitudes is called the \emph{orthic triangle}. \begin{exer} \ii Let $ABC$ be a triangle. A circle passing through $B$ and $C$ intersects the sides $AB$ and $AC$ again at $C'$ and $B'$, respectively. Prove that $BB', CC', HH'$ are concurrent, where $H$ and $H'$ are the orthocentes of triangles $ABC$ and $A'B'C'$, respectively. \ii (USAMO 1990/5) An acute-angled triangle $ABC$ is given in the plane. The circle with diameter $\, AB \,$ intersects altitude $\, CC' \,$ and its extension at points $\, M \,$ and $\, N \,$, and the circle with diameter $\, AC \,$ intersects altitude $\, BB' \,$ and its extensions at $\, P \,$ and $\, Q \,$. Prove that the points $\, M, N, P, Q \,$ lie on a common circle. \ii Let $\ell$ be a line through the orthocenter %? $H$ of a triangle $ABC$. Prove that the reflections of $\ell$ across $AB, BC,$ and $CA$ all pass through a common point; show also that this point lies on the circumcircle of $ABC$. \ii (Bulgaria, 1997) Let $ABC$ be a triangle with orthocenter $H$, and let $M$ and $K$ denote the midpoints of $AB$ and $CH$. Prove that the angle bisectors of $\angle CAH$ and $\angle CBH$ meet at a point on the line $MK$. \ii \label{ex:ninept} Let $ABC$ be a triangle with orthocenter $H$. Define the following points: \begin{tabular}{l} let $M_A, M_B, M_C$ be the midpoints of the sides $BC, CA, AB$; \\ let $H_A, H_B, H_C$ be the feet of the altitudes from $A,B,C$; \\ let $A', B', C'$ be the midpoints of the segments $HA, HB, HC$. \end{tabular} \begin{enumerate} \ii Show that the triangle $A'B'C'$ is the half-turn of the triangle $M_AM_BM_C$ about its circumcenter. \ii Conclude that all nine points lie on a single circle, called the \emph{nine-point circle} of $ABC$. \ii Show that the center of the nine-point circle is the midpoint of $OH$. \end{enumerate} \end{exer} \section{Gergonne and Nagel points} These points are less famous than some of the others, but they make for a few interesting problems, so let's get straight to work. \begin{exer} \ii Prove that the cevians joining each vertex of $ABC$ to the point where the incircle touches the opposite sign are concurrent; this point is the \emph{Gergonne point}. Also show that the cevians joining each vertex to the point where the excircle opposite that vertex touches the opposite side are concurrent; this point is the \emph{Nagel point}. \ii \label{ex:nagel} In triangle $ABC$, let $G,I,N$ be the centroid, incenter, and Nagel point, respectively. Show that $G,I,N$ lie on a line in that order, and that $NG = 2\cdot IG$. (The proof is analogous to that for the Euler line, but somewhat trickier.) \ii Let $P,Q,R$ be the midpoints of sides $BC, CA, AB$, respectively. Show that the incenter of $PQR$ is the midpoint of $IN$. (The analogy continues!) \end{exer} \section{Isogonal conjugates} Two points $P$ and $Q$ inside triangle $ABC$ are said to be \emph{isogonal conjugates} if $\angle PAB = \angle QAC$ and so on. In other words, $Q$ is the reflection of $P$ across each of the internal angle bisectors of $ABC$. \begin{exer} \ii Prove that every point inside triangle $ABC$ has an isogonal conjugate. What happens if we allow points outside the triangle? \ii Prove that the orthocenter and the circumcenter are isogonal conjugates. \ii The isogonal conjugate of the centroid is called the \emph{Lemoine point}. Draw through the Lemoine point a line parallel to each side of the triangle, and consider its intersections with the other two sides. Show that these six points are concyclic. \ii A cevian through the Lemoine point is called a \emph{symmedian}. Show that the tangents to the circumcircle of a triangle at two vertices intersect on the symmedian of the third vertex. (Don't forget circles of Apollonius!) % Communicated by Dan Moraseski \ii Let $D, E, F$ be the feet of the symmedians of triangle $ABC$. Prove that the Lemoine point of $ABC$ is the centroid of $DEF$. \end{exer} \section{Brocard points} The exercises in this section establish the existence and several properties of the Brocard points. \begin{exer} \ii \label{ex:broc1} Show that inside any triangle $ABC$, there exists a point $P$ such that \[ \angle PAB = \angle PBC = \angle PCA. \] \ii Show that the point $P$ of the previous exercise is unique. It is one of the \emph{Brocard points} of $ABC$; the other is the isogonal conjugate of $P$, which satisfies similar relations with the vertices in reverse order. \ii Let $\omega$ be the angle such that \[ \cot \omega = \cot A + \cot B + \cot C. \] Show that the \emph{Brocard angle} $\angle PAB$ is equal to $\omega$. \ii Show that the maximum Brocard angle is $30^{\circ}$, achieved only by an equilateral triangle. (We will have more to say about this in the section on geometric inequalities.) \ii %(IMO ???) If $K$ is the area of triangle $ABC$, show that \[ a^{2} + b^{2} + c^{2} \geq 4\sqrt{3}K \] by expressing the Brocard angle in terms of $a,b,c,K$. \ii \label{ex:imo91} (IMO 1991/5) Prove that inside any triangle $ABC$, there exists a point $P$ such that one of the angles $\ang PAB, \ang PBC, \ang PCA$ measures at most $30^\circ$. \end{exer} \section{Miscellaneous} Here are a few additional problems concerning triangle trivia. Before proceeding to the problems, we state as facts a few standard formulae for the area of a triangle. \begin{fact} Let $ABC$ be a triangle with side lengths $a = BC, b = CA, c = AB$, inradius $r$, circumradius $R$, exradius opposite $A$ $r_{A}$, semiperimeter $s$, and area $K$. Then \begin{eqnarray*} K &=& \frac{1}{2} ab \sin C \quad \mbox{(Law of Sines)}\\ &=& \frac{abc}{4R} \quad \mbox{(by Extended Law of Sines)} \\ &=& rs = r_{A} (s-a) \\ &=& \sqrt{s(s-a)(s-b)(s-c)}. \quad \mbox{(Heron's formula)} \end{eqnarray*} \end{fact} \begin{exer} \ii \label{ex:stewart} Let $D$ be a point on side $BC$, and let $m = BD, n = CD$ and $t = AD$. Prove \emph{Stewart's formula}: \[ m^{2} b + n^{2} c = a (t^{2} + mn). \] DIAGRAM. \ii Use Stewart's formula to prove the Steiner-Lehmus theorem: a triangle with two equal angle bisectors must be isosceles. (A synthetic proof is possible but not easy to find.) \ii (United Kingdom, 1997) In acute triangle $ABC$, $CF$ is an altitude, with $F$ on $AB$, and $BM$ is a median, with $M$ on $CA$. Given that $BM=CF$ and $\angle MBC = \angle FCA$, prove that the triangle $ABC$ is equilateral. Also, what happens if $ABC$ is not acute? \ii The point $D$ lies inside the acute triangle $ABC$. Three of the circumscribed circles of the triangles $ABC, BCD, CDA, DAB$ have equal radii. Prove that the fourth circle has the same radius, and characterize all such sets of four points. (What happens if $ABC$ need not be acute, or $D$ need not lie in its interior?) \ii (Bulgaria, 1997) Let $ABC$ be a triangle and let $M,N$ be the feet of the angle bisectors of $B,C$, respectively. Let $D$ be the intersection of the ray $MN$ with the circumcircle of $ABC$. Prove that \[ \frac{1}{BD} = \frac{1}{AD} + \frac{1}{CD}. \] \ii %% Don't have solution Let $ABCDE$ be a cyclic pentagon such that $r_{ABC} = r_{AED}$ and $r_{ABD} = r_{ACE}$, where $r_{XYZ}$ denotes the inradius of triangle $XYZ$. Prove that $AB = AE$ and $BC = DE$. \ii %% MOP 1990--Russia IMO proposal? (MOP 1990) Let $AA_{1}, BB_{1}, CC_{1}$ be the altitudes in an acute-angled triangle $ABC$, and let $K$ and $M$ be points on the line segments $A_{1}C_{1}$ and $B_{1}C_{1}$, respectively. Prove that if the angles $MAK$ and $CAA_{1}$ are equal, then the angle $C_{1}KM$ is bisected by $AK$. \end{exer} %% Quadrilaterals \chapter{Quadrilaterals} \section{General quadrilaterals} There's not a great deal that can be said about an arbitrary quadrilateral---the extra freedom in placing an additional vertex disrupts much of the structure we found in triangles. What little there is to say we offer in the form of a few exercises. \begin{exer} \ii Prove that the midpoints of the sides of any quadrilateral form a parallelogram (known as the \emph{Varignon parallelogram}). \ii Let $ABCD$ be a convex quadrilateral, and let $\theta$ be the angle between the diagonals $AC$ and $BD$. Prove that \[ K_{ABCD} = \half AC \cdot BD \sin \theta. \] \ii Derive a formula for the area of a convex quadrilateral in terms of its four sides and a pair of opposite angles. \end{exer} \section{Cyclic quadrilaterals} The condition that the four vertices of a quadrilateral lie on a circle gives rise to a wealth of interesting structures, which we investigate in this section. We start with a classical result of Ptolemy. \begin{theorem}[Ptolemy] \label{thm:pt} Let $ABCD$ be a convex cyclic quadrilateral. Then \[ AB \cdot CD + BC \cdot DA = AC \cdot BD. \] \end{theorem} \begin{proof} Mark the point $P$ on $BD$ such that $BP = (AB \cdot CD)/AC$, or equivalently $BP/AB = CD/AC$. Since $\ang ABP = \ang ACD$, the triangles $ABP$ and $ACD$ are similar. DIAGRAM. On the other hand, we now have \[ \ang DPA = \pi - \ang APB = \pi - \ang ADC = \ang CBA. \] Thus the triangles $APD$ and $ABC$ are also similar, yielding $DP/BC = AD/AC$. Consequently \[ BD = BP + PD = \frac{AB \cdot CD}{AC} + \frac{AD \cdot BC}{AC} \] and the theorem follows. \end{proof} This proof is elegant, but one cannot help wondering, ``How could anyone think of that?'' (I wonder that myself; the proof appears in an issue of Samuel Greitzer's \emph{Arbelos}, but he gives no attribution.) The reader might enjoy attempting a proof using trigonometry or complex numbers. Another important result about cyclic quadrilateral is an area formula attributed to the ancient Indian mathematician Brahmagupta DATES. \begin{fact}[Brahmagupta] If a cyclic quadrilateral has sides $a,b,c,d$ and area $K$, then \[ K = \sqrt{(s-a)(s-b)(s-c)(s-d)}, \] where $s = (a+b+c+d)/2$ is the semiperimeter. \end{fact} Heron's formula for the area of a triangle follows from Brahmagupta's formula by regarding a triangle as a cyclic quadrilateral with one side of length 0. (Brahmagupta's formula is a rare case where Western terminology attributes a result to an Eastern mathematician; the ancient Chinese and Indian discoverers of Pascal's triangle, the solution of the Pell equation, and the arctangent series were not so fortunate.) % Let $\alpha, \beta, \gamma, % \delta$ be half of the measures of the arcs $AB,BC,CD,DA$, % respectively. By the Extended Law of Sines, what we must show is % \[ % \sin \alpha \sin \gamma + \sin \beta \sin \delta % = \sin (\alpha + \beta) \sin (\beta + \gamma). % \] % Since $\alpha + \beta + \gamma + \delta = \pi$, we can replace $\sin % \delta$ with $\sin (\alpha + \beta + \gamma)$; by the addition % formula, the previous equation is equivalent to % \[ % \sin \alpha \sin \gamma + \sin \beta \sin (\alpha + \beta) \cos % \gamma + \sin \beta \cos (\alpha + \beta) \sin \gamma % = \sin (\alpha + \beta) \sin (\beta + \gamma). % \] % Rearranging this gives % \begin{equation} \label{eq:pt1} % \sin \gamma [\sin \alpha + \sin \beta \cos (\alpha + \beta)] % = \sin (\alpha + \beta) [\sin (\beta + \gamma) - \sin \beta \cos % \gamma]. % \end{equation} % But we know that % \beqa % \cos \beta \sin (\alpha + \beta) - \sin \beta \cos (\alpha + \beta) % &=& \sin [(\alpha + \beta) - \alpha] = \sin \alpha \\ % \sin \beta \cos \gamma + \cos \beta \sin \gamma &=& \sin (\beta + % \gamma). % \eeqa % Hence both sides of (\ref{eq:pt1}) equal $\sin (\alpha+\beta) \cos % \beta \sin \gamma$, proving the result. % \end{proof} % % We will eventually give two other proofs of Ptolemy's theorem: one % with inversion, and one with complex numbers. % %\beqa %\sin \alpha \sin \gamma + \sin \beta \sin (\alpha + \beta + \gamma) %&=& \sin [(\alpha+ \beta) - \alpha] \sin \gamma + \sin \beta [\sin (\alpha + \beta) \cos %\gamma + \cos (\alpha + \beta) \sin \gamma] \\ %&=& \sin (\alpha + \beta) \cos \alpha \sin \gamma %- cos (\alpha + \beta) \sin \alpha \sin \gamma %+ \sin \beta \sin (\alpha + \beta) \cos \gamma %+ \sin \beta \cos (\alpha + \beta) \sin \gamma %\end{proof} \begin{exer} \ii (Brahmagupta) Let $ABCD$ be a cyclic quadrilateral with perpendicular diagonals. Then the line through the intersection of the diagonals and the midpoint of any side is perpendicular to the opposite side. \ii Brahmagupta's formula implies that the area of a cyclic quadrilateral depends only on the lengths of the sides and not the order in which they occur. Can you demonstrate this fact by ``slicing and dicing''? \ii Use Ptolemy's theorem and the previous problem to give a formula for the lengths of the diagonals of a cyclic quadrilateral in terms of the lengths of the sides. \ii \label{ex:inrect} Let $ABCD$ be a cyclic quadrilateral. Prove that the incenters of triangles $ABC, BCD, CDA, DAB$ form a rectangle. \ii With the same notation, prove that the sum of the inradii of $ABC$ and $CDA$ equals the sum of the inradii of $BCD$ and $DAB$. \end{exer} \section{Circumscribed quadrilaterals} The following theorem characterizes circumscribed quadrilaterals; while it can be proved directly using the equal tangents rule, it proves easier to exploit what we already know about incircles and excircles of triangles. DIAGRAM. %diagram \begin{theorem} A convex quadrilateral $ABCD$ admits an inscribed circle if and only if $AB + CD = BC + DA$. \end{theorem} \begin{proof} Let the sides $AB$ and $CD$ meet at $P$; without loss of generality, assume $A$ lies between $P$ and $B$. (We skip the limiting case $AB || CD$.) The quadrilateral $ABCD$ has an inscribed circle if and only if the incircle of triangle $PBC$ coincides with the excircle of triangle $PDA$. Let $Q$ and $R$ be the points of tangency of line $PB$ with the incircle of $PBC$ and the excircle of $PDA$, respectively; since both circles are tangent to the sides of the angle $\ang CPB$, they coincide if and only if $Q = R$, or equivalently $PQ = PR$. However, by the usual formulae \beqa PQ &=& \half (PB + PC - BC) = \half (PD + DC + PA + AB - BC) \\ PR &=& \half (PA + PD + DA) \eeqa and these are equal if and only if $AB + CD = BC + DA$. \end{proof} Just as with triangles, a convex quadrilateral can have an escribed circle, a circle not inside the quadrilateral but tangent to all four sides (or rather their extensions). DIAGRAM. We trust the reader can now supply the proof of the analogous characterization of quadraterals admitting an escribed circle. \begin{fact} A convex quadrilateral $ABCD$ admits an escribed circle opposite $A$ or $C$ if and only if $AB + BC = CD + DA$. \end{fact} For more problems about circumscribed quadrilaterals, flip back to Section~\ref{sec:pb}, where we study them using Brianchon's theorem. \begin{exer} \ii (IMO 1962/5) On the circle $K$ there are given three distinct points $A,B,C$. Construct (using only straightedge and compass) a fourth point $D$ on $K$ such that a circle can be inscribed in the quadrilateral thus obtained. \ii (Dick Gibbs) Let $ABCD$ be a quadrilateral inscribed in an ellipse, and let $E = AB \cap CD$ and $F = AD \cap BC$. Show that $ACEF$ can be inscribed in a hyperbola with the same foci as the ellipse. (If you're not familiar with ellipses and hyperbolae, peek ahead to Section~\ref{sec:con}.) \ii (USAMO 1998/6) \label{ex:circquads} Let $n \geq 5$ be an integer. Find the largest integer $k$ (as a function of $n$) such that there exists a convex $n$-gon $A_1A_2\cdots A_n$ for which exactly $k$ of the quadrilaterals $A_iA_{i+1}A_{i+2}A_{i+3}$ have an inscribed circle, where $A_{n+j} = A_j$. \end{exer} \section{Complete quadrilaterals} A \emph{complete quadrilateral} is the figure formed by four lines, no two parallel and no three concurrent; the \emph{vertices} of a complete quadrilateral are the six pairwise intersections of the lines. This configuration has been widely studied; we present here as exercises a number of intriguing properties of the diagram. In the following exercises, let $ABCDEF$ be the complete quadrilateral formed by the lines $ABC, AEF, DBF, DEC$. DIAGRAM \begin{exer} \ii Show that the circles with diameters $AD, BE, CF$ are coaxal. Deduce that the midpoints of the segments $AD, BE, CF$ are collinear. (Can you show the latter directly?) \ii %% Simson lines? Show that the circumcircles of the triangles $ABF$, $ACE$, $BCD$, $DEF$ pass through a common point. (Many solutions are possible.) \ii We are given five lines in the plane, no two parallel and no three concurrent. To every four of the lines, associate the point whose existence was shown in the previous exercise. Prove these five points lie on a circle. (This assertion and the previous one belong to an infinite chain of such statements: see W.K. Clifford, \textit{Collected Papers} (1877), 38-54.) %\ii (Noam Elkies) %% Attribute--Elkies? %Let $\ell_{1}, \dots, \ell_{5}$ be five lines in the plane, no two %parallel and no three concurrent, and put $\ell_{i+5} = \ell_{i}$. %Let $\omega_{i}$ be the %circumcircle of the triangle formed by $\ell_{i}, \ell_{i+2}, %\ell_{i+4}$, and let $P_{i}$ be the second intersection of %$\omega_{i}$ and $\omega_{i+2}$ (other than $\ell_{i+2} \cap %\ell_{i+4}$). Show that $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ are %concyclic. \end{exer} %% Geometry project %% Inversion \chapter{Inversive Geometry} One of the most stunning products of the revival of Euclidean geometry in the 19th century is the method of inversion, introduced by L.J. Magnus in 1831. The power of inversion lies in its ability to convert statements about circles into statements about lines, often reducing the difficult to the trivial. \section{Inversion} Let $O$ be a point in the plane and $r$ a positive real number. The \emph{inversion} with center $O$ and radius $r$ is a transformation mapping every point $P \neq O$ in the plane to the point $P'$ on the ray $\ray{OP}$ such that $OP \cdot OP' = r^2$. Since specifying a point and a positive real number is the same as specifying a circle (the point and the positive real corresponding to the center and radius, respectively, of the circle), we can also speak of inversion through a circle using the same definition. DIAGRAM What happens to the point $O$? Points near $O$ get sent very far away, in all different directions, so there is no good place to put $O$ itself. To rectify this, we define the \emph{inversive plane} as the usual plane with one additional point, called the \emph{point at infinity}. (We will use the label $\infty$ throughout this chapter for the point at infinity.) We extend inversion to the entire inversion plane by declaring that $O$ and $\infty$ are inverses of each other. As an aside, we note a natural interpretation of the inversive plane. Under stereographic projection (used in some maps), the surface of a sphere, minus the North Pole, is mapped to a plane tangent to the sphere at the South Pole as follows: a point on the sphere maps to the point on the plane collinear with the given point and the North Pole. Then the point at infinity corresponds to the North Pole, and the inversive plane corresponds to the whole sphere. In fact, inversion through the South Pole with the appropriate radius corresponds to reflecting the sphere through the plane of the equator! DIAGRAM. Returning to Euclidean geometry, we now establish some important properties of inversion. We first make an easy but important observation. \begin{fact} \label{fac:invsim} If $O$ is the center of an inversion taking $P$ to $P'$ and $Q$ to $Q'$, then the triangles $OPQ$ and $OQ'P'$ are oppositely similar. \end{fact} In particular, we have that $\dang OP'Q' = -\dang OQP$, a fact underlying our next proof. \begin{theorem} \label{thm:invcirc} The image of a (generalized) circle under an inversion is a (generalized) circle. \end{theorem} \begin{proof} Let $A,B,C,D$ be four concyclic points and $A',B',C',D'$ their images under some inversion about $O$. We now chase directed angles, using the similar triangles of Fact~\ref{fac:invsim}: \beqa \dang A'B'C' &=& \dang A'B'O + \dang OB'C' \\ &=& \dang BAO + \dang OCB \\ &=& \dang BAD + \dang DAO + \dang OCD + \dang DCB \\ &=& \dang DAO + \dang OCD \\ &=& \dang A'D'O + \dang OD'C' \\ &=& \dang A'D'C'. \eeqa We see that $A',B',C',D'$ are concyclic as well. DIAGRAM. \end{proof} Notice the way the angles are broken up and recombined in the above proof. In some cases, inversion can turn a constraint involving two or more angles in different places into a constraint about a single angle, which then is easier to work with. Some examples can be found in the problems. Inversion also turns out to ``reverse the angles between lines''. Since lines are sent to circles in general, we will have to define the angle between two circles to make sense of this statement. Given two circles $\omega_{1}$ and $\omega_{2}$, the (directed) angle between them at one of their intersections $P$ is defined as the (directed) angle from the tangent to $\omega_{1}$ at $P$ to the tangent of $\omega_{2}$ at $P$. In particular, two circles are \emph{orthogonal} if the angle between them is a right angle. Note that the angle between the two circles is only well-defined up to sign without a choice of a point of intersection, since the angle at the other intersection is reversed. (Of course, orthogonality does not depend on this choice.) Note also that a line and a circle are orthogonal if and only if the line passes through the center of the circle. \begin{fact} \label{thm:invang} The directed angle between circles (at a chosen intersection) is reversed under inversion. \end{fact} Distances don't fare as well under inversion, but one can say something using Fact~\ref{fac:invsim}. \begin{fact}[Inversive distance formula] \label{thm:invdist} If $O$ is the center of an inversion of radius $r$ sending $P$ to $P'$ and $Q$ to $Q'$, then \[ P'Q' = PQ \cdot \frac{r^2}{OP \cdot OQ}. \] \end{fact} \begin{exer} \ii Deduce Theorem~\ref{thm:invcirc} from Problem~\ref{ex:invcirc} (or use the above proof to figure out how to do that exercise). \ii Give another proof of Theorem~\ref{thm:invcirc} using the converse of the power-of-a-point theorem (Fact~\ref{fact:power}) and Fact~\ref{thm:invdist}. \ii The angle between two lines through the origin is clearly preserved under inversion. Why doesn't this contradict the fact that inversion reverses angles? \ii (IMO 1996/2) Let $P$ be a point inside triangle $ABC$ such that \[ \angle APB - \angle ACB = \angle APC - \angle ABC. \] Let $D,E$ be the incenters of triangles $APB, APC$, respectively. Prove that $AP, BD, CE$ meet in a point. (Many other solutions are possible; over 25 were submitted by contestants at the IMO!) \ii (IMO 1998 proposal) Let $ABCDEF$ be a convex hexagon such that $\angle B + \angle D + \angle F = 360^{\circ}$ and \[ \frac{AB}{BC} \cdot \frac{CD}{DE} \cdot \frac{EF}{FA} = 1. \] Prove that \[ \frac{BC}{CA} \cdot \frac{AE}{EF} \cdot \frac{FD}{DB} = 1. \] \ii Prove that the following are equivalent: \begin{enumerate} \item The points $A$ and $B$ are inverses through the circle $\omega$. \item The line $AB$ and the circle with diameter $AB$ are both orthogonal to $\omega$. \item $\omega$ is a circle of Apollonius with respect to $A$ and $B$. \end{enumerate} In particular, conclude that a circle distinct from $\omega$ is fixed (as a whole, not pointwise) by inversion through $\omega$ if and only if it is orthogonal to $\omega$. \ii Show that a set of circles is coaxal if and only if there is a circle orthogonal to all of them. Deduce that coaxal circles remain that way under inversion. Also, try drawing a family of coaxal circles and some circles orthogonal to them; the picture is very pretty. \ii \label{ex:conc} Prove that any two nonintersecting circles can be inverted into concentric circles. \end{exer} \section{The power of inversion} Steiner was able to give nearly trivial proofs of some very hard-looking statements using inversion. In this section, we take a quick look at some of his dazzling results. We start with a classical result attributed to Pappus. The figure formed by the three semicircles is known as an \emph{arbelos} (``shoemaker's knife'' in Greek), and was a favorite of Archimedes. (It was also a favorite of one-time USA IMO coach Samuel Greitzer, who for several years authored a journal for high school students of the same name.) \begin{theorem} [Pappus] \label{thm:papinv} Let $\omega$ be a semicircle with diameter $AB$. Let $\omega_{1}$ and $\omega_{2}$ be two semicircles externally tangent to each other at $C$, and internally tangent to $\omega$ at $A$ and $B$, respectively. Let $C_{1}, C_{2}, \dots$ be a sequence of circles, each tangent to $\omega$ and $\omega_{1}$, such that $C_{i}$ is tangent to $C_{i+1}$ and $C_{1}$ is tangent to $\omega_{2}$, as in the diagram. Let $r_{n}$ be the radius of $C_{n}$ and $ d_{n}$ the distance from the center of $C_{n}$ to $AB$. Then for all $n$, \[ d_{n} = 2n r_{n}. \] \end{theorem} DIAGRAM \begin{proof} Perform an inversion with center $A$, and choose the radius of inversion so that $C_{n}$ remains fixed. Then $\omega$ and $\omega_{1}$ map to lines perpendicular to $AB$ and tangent to $C_{n}$, and $C_{n-1}, \dots, C_{1}$ to a column of circles between the lines, with $\omega_{2}'$ at the bottom of the column. The relation $d_{n} = 2n r_{n}$ is now obvious. \end{proof} The following theorem is known as \emph{Steiner's porism}. \begin{theorem} Suppose two nonintersecting circles have the property that one can fit a ``ring'' of $n$ circles between them, each tangent to the next, as in the diagram. DIAGRAM. Then one can do this starting with any circle tangent to both given circles. \end{theorem} \begin{proof} By Problem~\ref{ex:conc}, a suitable inversion takes the given circles to concentric circles, while preserving tangency of circles. The result is now obvious. \end{proof} \begin{exer} \ii Suppose that, in the hypotheses of Pappus' theorem, we assume that $C_0$ is tangent to $\omega, \omega_1$ and the line $AB$ (instead of the semicircle $\omega_2$). Show that in this case $d_n = (2n-1) r_n$. \ii (Romania, 1997) \label{ex:rom97} Let $\omega$ be a circle and $AB$ a line not intersecting $\omega$. Given a point $P_0$ on $\omega$, define the sequence $P_0, P_1, \dots$ as follows: $P_{n+1}$ is the second intersection with $\omega$ of the line through $B$ and the second intersection of the line $AP_n$ with $\omega$. Prove that for a positive integer $k$, if $P_0 = P_k$ for some choice of $P_0$, then $P_0 = P_k$ for any choice of $P_0$. \end{exer} \section{Inversion in practice} So much for the power of inversion; how is it useful for real problems? The remainder of this chapter will be devoted to several examples of how inversion can be used to solve olympiad-style problems. The paradigm will almost always be: invert the given information, invert the conclusion, and proceed to solve the new problem. Beware that in somes, though, it may be necessary to superimpose the original and inverted diagrams (as in the proof of Theorem~\ref{thm:papinv}), or to compare the original and inverted diagrams (e.g.\ using Fact~\ref{thm:invdist}). A general principle behind this method is that problems with few circles are easier than those with many circles. Hence when inverting, one should find a ``busy point,'' one with many circles and lines going through it, and invert there. \begin{exer} \ii Make up an inversion problem by reversing the paradigm: start with a result that you know, invert about some point, and see what you get. The tricky part is choosing things well enough so that the resulting problem doesn't have an obvious busy point; such a problem would be too easy! \ii Given circles $C_1, C_2, C_3, C_4$ such that $C_{i}$ and $C_{i+1}$ are externally tangent for $i=1, 2,3,4$ (where $C_5=C_1$). Prove that the four points of tangency are concyclic. \ii (Romania, 1997) Let $ABC$ be a triangle, $D$ a point on side $BC$ and $\omega$ the circumcircle of $ABC$. Show that the circles tangent to $\omega, AD, BD$ and to $\omega, AD, DC$ are tangent to each other if and only if $\angle BAD = \angle CAD$. \ii (Russia, 1995) %% (Pamphlet, Russia 33) Given a semicircle with diameter $AB$ and center $O$ and a line which intersects the semicircle at $C$ and $D$ and line $AB$ at $M$ ($MB < MA$, $MD < MC$). Let $K$ be the second point of intersection of the circumcircles of triangles $AOC$ and $DOB$. Prove that angle $MKO$ is a right angle. \ii (USAMO 1993/2) Let $ABCD$ be a convex quadrilateral with perpendicular diagonals meeting at $O$. Prove that the reflections of $O$ across $AB,BC,CD,DA$ are concyclic. (For an added challenge, find a non-inversive proof as well.) \ii \label{ex:apoll} (Apollonius' problem) Given three nonintersecting circles, how many circles are tangent to all three? And how can they be constructed with straightedge and compass? \ii %% (Test 3, problem 4) (IMO 1994 proposal) The incircle of $ABC$ touches $BC, CA, AB$ at $D,E,F$, respectively. $X$ is a point inside $ABC$ such that the incircle of $XBC$ touches $BC$ at $D$ also, and touches $CX$ and $XB$ at $Y$ and $Z$, respectively. Prove that $EFZY$ is a cyclic quadrilateral. % %\ii Moved to angle chasing chapter %Let $C_1,\,C_2,\,C_3,\,C_4$ be four circles in the plane. %Suppose that $C_1$ and $C_2$ intersect at $P_1$ and $Q_1$, %$C_2$ and $C_3$ intersect at $P_2$ and $Q_2$, %$C_3$ and $C_4$ intersect at $P_3$ and $Q_3$, %and $C_4$ and $C_1$ intersect at $P_4$ and $Q_4$. Show that if %$P_1,\,P_2,\,P_3$, and $P_4$ lie on a line or circle, then %$Q_1,\,Q_2,\,Q_3$, and $Q_4$ also lie on a line or circle. \ii %(Pamphlet, Israel 8) (Israel, 1995) Let $PQ$ be the diameter of semicircle $H$. Circle $O$ is internally tangent to $H$ and tangent to $PQ$ at $C$. Let $A$ be a point on $H$ and $B$ a point on $PQ$ such that $AB$ is perpendicular to $PQ$ and is also tangent to $O$. Prove that $AC$ bisects $\angle PAB$. \ii \label{ex:ptineq} (Ptolemy's inequality) If $ABCD$ is a convex quadrilateral, then \[ AC \cdot BD \leq AB \cdot CD + BC \cdot DA, \] with equality if and only if $ABCD$ is cyclic. (See also Theorem~\ref{thm:ptineq}.) \ii (IMO 1993/2) Let $A,B,C,D$ be four points in the plane, with $C,D$ on the same side of line $AB$, such that $AC \cdot BD = AD \cdot BC$ and $\angle ADB = \pi/2 + \angle ACB$. Find the ratio $(AB \cdot CD)/(AC \cdot BD)$ and prove that the circumcircles of triangles $ACD$ and $BCD$ are orthogonal. %\ii %(R\u{a}zvan Gelca) (What's $M$?) %Let $ABC$ be a triangle and $H$ its orthocenter. The perpendiculars %from $H$ to $MA$, $MB$, $MC$ meet $BC$, $CA$, $AB$ at $A_1$, $B_1$, %$C_1$, respectively. Show that $A_1, B_1, C_1$ are collinear, and that %the line they determine is perpendicular to $MH$. % \ii (Iran, 1995) Let $M,N,P$ be the points of intersection of the incircle of $\triangle ABC$ with sides $BC,CA,AB$, respectively. prove that the orthocenter of $\triangle MNP$, the incenter of $\triangle ABC$, and the circumcenter of $\triangle ABC$ are collinear. (The paradigm does not hold here: invert through the incircle, then superimpose the original and inverted diagrams.) % MOP 1997 Test 3 \ii (MOP 1997) \label{ex:mop97} Let $ABC$ be a triangle and $O$ its circumcenter. The lines $AB$ and $AC$ meet the circumcircle of triangle $BOC$ again at $B_1$ and $C_1$, respectively. Let $D$ be the intersection of lines $BC$ and $B_1 C_1$. Show that the circle tangent to $AD$ at $A$ and having its center on $B_1C_1$ is orthogonal to the circle with diameter $OD$. % MOP 1997 Test 4 #4 %\ii %Let $P$ be a point in the plane of a triangle $ABC$. A circle passing %through $P$ intersects the circumcircles of triangles $PBC, PCA, PAB$ \end{exer} %% Geometry project %% Projective geometry \chapter{Projective Geometry} Projective geometry arose during the Renaissance, when painters switched from a flat style of drawing to perspective drawing. The artists were interested in geometric properties of figures that were invariant under ``changes of perspective,'' such as collinearity of a set of points. The formalism of projective geometry makes a discussion of such properties possible, and exposes some remarkable facts, such as the duality of points and lines. The reader should be warned that the material here is less ``contest-oriented'' than that of the preceding chapters. While several techniques (projective transformations, cross-ratio, and to a lesser extent, the polar map) can be useful on contest problems, much of the material here is considered ``too advanced'' for contests. This is a pity, as some of the most beautiful classical geometry appears in the projective setting. We hope even the most pragmatic reader will indulge us a bit as we take a rather brief tour through projective geometry. \section{The projective plane} We begin with a lengthy description of the formalism of the projective plane. The impatient reader may wish to read only the next paragraph at first, then skip to the later sections and come back to this section as needed. The \emph{projective plane} consists of the standard Euclidean plane, together with a set of points called \emph{points at infinity}, one for each collection of parallel lines. We say that a line passes through the point at infinity corresponding to its direction (and no others), and that all of the points at infinity lie on a \emph{line at infinity}. (Note that three parallel lines now indeed have a common point at infinity, which retroactively justifies our calling such lines ``concurrent''.) An alternate description of the projective plane turns out to be quite useful, and corresponds more closely to the artists' conception. View the Euclidean plane as some plane in three-dimensional space, and fix a point $O$ not on the plane (corresponding to the eye). Then each point on the plane corresponds to a line through $O$ passing through that point, but not all lines through $O$ correspond to points on the Euclidean plane. In fact, they correspond to the points at infinity. In other words, we can identify the projective plane with the set of lines in space passing through a fixed point. This description also yields a natural coordinate system for the projective plane, using what are known as \emph{homogeneous coordinates}. Each point in the projective plane can be specified with a triple of numbers $[x:y:z]$, where $x,y,z$ are not all zero. Be careful, though: for any nonzero real number $\lambda$, $[x:y:z]$ and $[\lambda x: \lambda y: \lambda z]$ are the same point! (Hence the name ``homogeneous coordinates''.) The colons are meant to remind you that it is the ratios between the coordinates that are well-defined, not the individual coordinates themselves. How are homogeneous coordinates related to the usual Cartesian coordinates on the Euclidean plane? If we embed the Euclidean plane in space as the plane $z = 1$, then the point with Cartesian coordinates $(x,y)$ has homogeneous coordinates $[x:y:1]$, and the points at infinity are the points of the form $[x:y:0]$ for some $x,y$ not both zero. \section{Projective transformations} The original definition of a projective transformation corresponded to the process of projecting an image in the ``real world'' onto an artist's canvas. Again, fix a point $O$ in three-dimensional space, and now select two planes not passing through $O$. The mapping that takes each point $P$ on the first plane to the intersection of the line $OP$ with the second plane was defined as a projective transformation. (Do you see why this map makes sense over the whole projective plane?) One can also give an algebraic description of projective transformations that accommodates degenerate cases slightly more easily. In terms of homogeneous coordinates, a projective transformation takes the form \[ [x : y : z] \mapsto [ax + by + cz : dx + ey + fz: gx + hy + iz], \] where the variables $a, \dots, i$ form an invertible $3 \times 3$ matrix. From this description it is clear that affine transformations are projective as well, since they occur when $g = h = 0$. Since we have two additional parameters (it looks like three, but by homogeneity one parameter is superfluous), the following analogue of Fact~\ref{ex:aff} is no surprise. \begin{fact} Any four points, no three collinear, can be mapped to any other four such points by a unique projective transformation. \end{fact} The most common use of a projective transformation in problem-solving is to map a particular line to the point at infinity. (As with inversion, it pays to look for a ``busy'' line for this purpose.) If the statement to be proved is well-behaved under projective transformations, this can yield drastic simplifications. The ``well-behaved'' concepts mainly consist of incidence properties between points and lines (concurrence, collinearity, and the like); as with affine transformations, angles and distances are not preserved, nor are areas or ratios of lengths along segments (unlike affine transformations). We demonstrate the power of projection by proving Desargues' theorem (Theorem~\ref{thm:des}). \begin{theorem} Suppose the triangles $ABC$ and $DEF$ are such that the points $AB \cap DE$, $BC \cap EF$, $CA \cap FD$ are collinear. Then the lines $AD, BE, CF$ are concurrent. \end{theorem} \begin{proof} Apply a projective transformation to place the points $AB \cap DE$ and $BC \cap EF$ at infinity. If triangles $ABC$ and $DEF$ are perspective from a line, they now have parallel sides and so are homothetic; thus the lines $AD, BE, CF$ concur at the center of homothety (or at a point at infinity, in case $ABC$ and $DEF$ are congruent). Conversely, if the lines $AD, BE, CF$ concur at $P$, consider the homothety centered at $A$ carrying $A$ to $D$. It preserves the line $BE$ and carries the line $AB$ to the parallel line $DE$ through $B$, so it maps $D$ to $E$. Similarly, the homothety maps $C$ to $F$, and so $CA$ and $FD$ are also parallel, implying that $ABC$ and $DEF$ are perspective from a line. \end{proof} Beware that angles, circles, and other ``metric'' objects are not preserved under projection; we will learn more about getting around this difficulty later in the chapter. \begin{exer} \ii Use a projective transformation to give an alternate proof of Pappus' theorem. \ii (Original) Let $ABCDE$ be the vertices of a pentagon, and let $F = BC \cap DE$, $G = CD \cap EA$, $H = DE \cap AB$, $I = EA \cap BC$, $J = AB \cap CD$. Show that $BD \cap CE$ lies on the line $AF$ if and only if $GH \cap IJ$ does. DIAGRAM. \end{exer} \section{A conic section} \label{sec:con} A \emph{conic section} is classically defined as a cross-section of a right circular cone by a plane not passing through a vertex, where the cone extends infinitely far in \emph{both} directions. The section is a called an \emph{ellipse}, a \emph{parabola}, or a \emph{hyperbola}, depending on whether the angle between the plane and the axis of the cone is greater than, equal to, or less than $\pi/4$. DIAGRAM. \begin{theorem} An ellipse is the locus of points whose sum of distances to two fixed points is constant. Similarly, a hyperbola is the locus of points whose (absolute) difference of distances to two fixed points is constant. \end{theorem} \begin{proof} This was already known to Apollonius, but the following clever proof is attributed to Dandelin (1794-1847). We will describe only the case of the ellipse, as the hyperbola case is similar. Inscribe spheres in the cone on either side of the plane of the ellipse, one on the side of the vertex of the cone, tangent to the plane at $A$, the other tangent to the plane at $B$. DIAGRAM. For any point on the cone between the two spheres, the sum of the lengths of the tangents to the two spheres is clearly a constant. On the other hand, for any point on the cone also lying in the plane, the segments to $A$ and $B$ are also tangent to the respective spheres, so the sum of their lengths equals this constant. The result follows. \end{proof} The two points alluded to in the above theorem are called \emph{foci} (plural of \emph{focus}). The name comes from the fact that if one has an ellipse made of a reflective material and one places a light source at one focus, all of the light rays will be ``focused'' at the opposite focus (see Problem~2). In modern times, it was noted that conic sections have a nice description in terms of Cartesian coordinates. If $z^{2} = x^{2}+y^{2}$ is the equation of the cone, it is evident that any cross-section is defined by setting some quadratic polynomial in $x$ and $y$ to 0. Hence a conic section can alternatively be defined as the zero locus of a quadratic polynomial; one must impose mild extra conditions to avoid degenerate cases, such as a pair of lines (which geometrically arise from planes through the vertex of the cone). Unless we say otherwise, our conic sections will be required to be nondegenerate. Here are some standard equations for the conic sections: \begin{tabular}{ll} Ellipse & $x^2/a^2 + y^2/b^2 = 1$ \\ Parabola & $y=ax^2+bx+c$ \\ Hyperbola & $x^2/a^2 - y^2/b^2=1$ \end{tabular} Also, the equation $xy=1$ defines a \emph{rectangular} hyperbola, one with perpendicular asymptotes. (The \emph{asymptotes} of a hyperbola are its tangent lines at its intersections with the line at infinity.) \begin{exer} \ii Given an ellipse in the plane, construct its center using only a straightedge. \ii Prove that a tangent to an ellipse makes equal (undirected) angles with the segments from the two foci to the point of tangency. DIAGRAM \ii (Erd\H{o}s) An infinite set of points in the plane has the property that the distance between any two of the points is an integer. Prove that the points are all collinear. \ii Let $P$ and $Q$ be two points on an ellipse. Prove that there exist ellipses similar to the given one, externally tangent to each other, and internally tangent to the given ellipse at $P$ and $Q$, respectively, if and only if $P$ and $Q$ are antipodes. \ii Use the previous exercise to prove that the maximum distance between two points on an ellipse is the length of the major axis \emph{without} doing any calculations. \ii (Original) Prove that the convex quadrilateral $ABCD$ contains a point $P$ such that the incircles of triangles $PAB$ and $PBC$ are tangent, as are those of $PBC$ and $PCD$, of $PCD$ and $PDA$, and of $PDA$ and $PAB$, if and only if $ABCD$ has an inscribed circle. \ii Find all points on the conic $x^{2}+y^{2}=1$ with \emph{rational} coordinates $x,y$ as follows: pick a point $(x,y)$ with rational coordinates, and project the conic from $(x,y)$ onto a fixed line (e.g.\ the line at infinity). More generally, given a single rational point on a conic, this procedure allows you to describe all such points. \end{exer} \section{Conics in the projective plane} In this section, we discuss conic sections from the point of view of projective geometry. To start, we rephrase the geometric definition of a conic section. \begin{fact} A curve is a conic section if and only if it is the image of a circle under a suitable projective transformation. \end{fact} In particular, the theorems of Pascal and Brianchon continue to hold if the circle in the statement of either theorem is replaced with an arbitrary conic. From these one can deduce converse theorems, that a hexagon is inscribed in (resp. circumscribed about) a conic if and only if it satisfies the conclusion of Pascal (resp. Brianchon); thinking of Pappus' theorem, one realizes that the conics in the previous statement must be permitted to be degenerate. We also note that the classification of conics can be restated in terms of projective geometry. \begin{fact} A conic is an ellipse (or a circle) if and only if it does not meet the line at infinity. A conic is a parabola if and only if it is tangent to the line at infinity. A conic is a hyperbola if and only if it intersects the line at infinity in two distinct points. \end{fact} \begin{exer} \ii \label{ex:pascconv} Prove that a hexagon whose opposite side meets in collinear points is inscribed in a conic (which may degenerate to a pair of lines). \ii Let $ABC$ and $BCD$ be equilateral triangles. An arbitrary line through $D$ meets $AB$ at $M$ and $AC$ at $N$. Determine the acute angle between the lines $BN$ and $CM$. \ii (Poncelet-Brianchon theorem) \label{ex:poncelet} Let $A,B,C$ be three points on a rectangular hyperbola (a hyperbola with perpendicular asymptotes). Prove that the orthocenter of the triangle $ABC$ also lies on the hyperbola. There are other special points of $ABC$ which must lie on this hyperbola; can you find any? \ii (\textit{Monthly}, Oct. 1994) Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ be a hexagon circumscribed about a conic, and form the intersections $P_{i} = A_{i}A_{i+2} \cap A_{i+1}A_{i+3}$ ($i = 1, \dots, 6$, all indices modulo 6). Show that the $P_{i}$ are the vertices of a hexagon inscribed in a conic. \ii (Arbelos) Let $A, B, C$ be three noncollinear points. Draw ellipses $E_{1}, E_{2}, E_{3}$ with foci $B$ and $C$, $C$ and $A$, $A$ and $B$, respectively. Show that: \begin{enumerate} \ii Each pair of ellipses meet in exactly two points, where a point of tangency counts twice. (In general, two ellipses can meet in as many as four points.) \ii The three lines determined by these pairs of points are concurrent. \end{enumerate} \end{exer} \section{The polar map and duality} \label{sec:polar} Fix a circle $\omega$ with center $O$. The \emph{polar map} with respect to $\omega$ interchanges points and lines in the following manner: \begin{enumerate} \ii If $P$ is a finite point other than $O$, the \emph{pole} of $P$ is the line $p$ through $P'$ perpendicular to $PP'$, where $P'$ is the inverse of $P$ through $\omega$. \ii If $p$ is a finite line not passing through $O$, the \emph{polar} of $p$ is the inverse through $\omega$ of the foot of the perpendicular from $O$ to $p$. \ii If $P$ is a point at infinity, the pole of $P$ is the line through $O$ perpendicular to any line through $P$, and vice versa. \ii If $P$ is $O$, the pole of $P$ is the line at infinity, and vice versa. \end{enumerate} The polar map is also known as \emph{reciprocation}. We keep the notational convention that points are labeled with capital letters and their poles with the corresponding lowercase letters. DIAGRAM. \begin{fact} \label{thm:poleprop} The polar map satisfies the following properties: \begin{enumerate} \ii Every point is the polar of its pole, and every line is the pole of its polar. \ii The polar of the line through the points $A$ and $B$ is the intersection of the poles $a$ and $b$. \ii Three points are collinear if and only if their poles are concurrent. \end{enumerate} \end{fact} An obvious consequence of the existence of the polar map is the duality principle. \begin{fact}[Duality principle] A theorem of projective geometry remains true if the roles of points and lines are interchanged. \end{fact} For example, the dual of one direction of Desargues' theorem is the other direction. We can now give Brianchon's original proof of his theorem, using Pascal's theorem and the polar map. There's nothing to it, really: given a hexagon circumscribed about a circle $\omega$, apply the polar map with respect to $\omega$. The result is a hexagon inscribed in $\omega$, and the collinearity of the intersections of opposite sides translates back to the original diagram as the concurrence of the lines through opposite vertices. \begin{exer} \ii Make up a problem by starting with a result that you know and applying the polar map. Beware that circles not concentric with $\omega$ do not behave well under the polar map; see below. \ii State the dual of Pappus' theorem. Can you prove this directly? (A projection may help.) \ii State and prove a dual version of problem~\ref{ex:circquads}. Since circles do not dualize to circles, you will have to come up with a new proof! \ii (China, 1996) Let $H$ be the orthocenter of acute triangle $ABC$. The tangents from $A$ to the circle with diameter $BC$ touch the circle at $P$ and $Q$. Prove that $P, Q, H$ are collinear. \ii Let $ABC$ be a triangle with incenter $I$. Fix a line $\ell$ tangent to the incircle of $ABC$ (not containing any of the sides). Let $A', B', C'$ be points on $\ell$ such that \[ \angle AIA' = \angle BIB' = \angle CIC' = \pi/2. \] Show that $AA', BB', CC'$ are concurrent. \ii \label{ex:imo852} %% (R\u{a}zvan Gelca) Let $A, B, C, D$ be four points on a circle. Show that the pole of $AC \cap BD$ with respect to this circle passes through $AB \cap CD$ and $AD \cap BC$. Use this fact to give another solution to Problem~\ref{ex:imo85} (IMO 1985/6). %% \ii %% Need Titu's permission to release %Let $\triangle ABC$ be a triangle and $P$ a point not on any of its %altitudes. Choose $A',B',C'$ on $BC,CA,AB$, %respectively, so that $\angle APA' = \angle BPB' = \angle CPC' = \pi/2$. %Show that $A', B', C'$ are collinear. \ii We know what happens to points and lines under the polar map, but what about a curve? If we view the curve as a \emph{locus}, i.e.\ a set of points, its dual is a set of lines which form an \emph{envelope}, i.e.\ they are all tangent to some curve. DIAGRAM. Show that the dual of a conic, under this definition, is again a conic. However, the dual of a circle need not be a circle. \ii Let $\omega$ be a (nondegenerate) conic. Show that there exists a unique map on the projective plane, taking points to lines and vice versa, satisfying the properties in Fact~\ref{thm:poleprop}, and taking each point on $\omega$ to the tangent to $\omega$ through that point. This map is known as the \emph{polar map with respect to $\omega$} (and coincides with the first definition if $\omega$ is a circle). \ii (IMO 1998/5) Let $I$ be the incenter of triangle $ABC$. Let the incircle of $ABC$ touch the sides $BC$, $CA$, and $AB$ at $K$, $L$, and $M$, respectively. The line through $B$ parallel to $MK$ meets the lines $LM$ and $LK$ at $R$ and $S$, respectively. Prove that angle $RIS$ is acute. \end{exer} \section{Cross-ratio} From the discussion so far, it may appear that there is no useful notion of ``distance'' in projective geometry, for projective transformations do not preserve distances, or even ratios of distances along a line (which affine transformations do preserve). There is something to be salvaged here, though; the ``ratios of ratios of distances'' are preserved. Given four collinear points $A,B,C,D$, the \emph{cross-ratio} of these points is defined as the following ratio of (directed) distances: \[ \frac{AC \cdot BD}{AD \cdot BC}. \] In case one of these points is at infinity, the definition can be extended by declaring that the ratio of two infinite distances is 1. We have left the definition where all of the points lie at infinity as an exercise. In light of duality, we ought to be able to make this definition for four concurrent lines, and in fact we can: the cross-ratio of four lines $a,b,c,d$ is defined as the cross-ratio of the intersections $A,B,C,D$ of $a,b,c,d$ with some line $\ell$ not passing through the point of concurrency. The cross-ratio is well-defined by the following observation, which follows from several applications of the Law of Sines. \begin{fact} Let $a, b, c, d$ be four concurrent lines and $\ell$ a line meeting $a, b, c, d$ at $A, B, C, D$, respectively. Then \[ \frac{AC \cdot BD}{AD \cdot BC} = \frac{\sin \dang(a, c) \sin \dang(b, d)}{\sin \dang(b, c) \sin \dang(a, d)}. \] \end{fact} \begin{fact} The cross-ratio is invariant under projective transformations and the polar map. \end{fact} In case the cross-ratio is $-1$, we say $C$ and $D$ are \emph{harmonic conjugates} with respect to $A$ and $B$ (or vice versa). If you did Problem~\ref{ex:harmcon}, you witnessed the most interesting property of harmonic conjugates: if $P$ is any point not on the line and $Q$ is any point on $PC$ other than $P$ or $C$, then $AP \cap BQ$, $AQ \cap BP$ and $D$ are collinear. (Not surprisingly, this property is projection-invariant.) One nice application of cross-ratios is the following characterization of conics. \begin{fact} Given four points $A,B,C,D$, the locus of points $E$ such that the cross-ratio of the lines $AE, BE, CE, DE$ constant is a conic. \end{fact} %We can use this characterization to give a second proof of Pascal's %theorem. DO IT! \begin{exer} \ii How should the cross-ratio be defined along the line at infinity? \ii Let $A,B,C,D$ be four points on a circle. Show that for $E$ on the circle, the cross-ratio of the lines $EA, EB, EC, ED$ remains constant. \ii Let $M$ be the midpoint of chord $XY$ of a circle, and let $AB$ and $CD$ be chords passing through $M$. Let $E = AD \cap XY$ and $F = BC \cap XY$. Prove that $EM=MF$. \ii Points $A,\,B,\,C$, and $D$, in this order, lie on a straight line. A circle $k$ passes through $B$ and $C$, and $AM$, $AN$, $DK$, and $DL$ are tangent to $k$ at $M,\,N,\,K$, and $L$. Lines $MN$ and $KL$ intersect $BC$ at $P$ and $Q$. \begin{enumerate} \item Prove that $P$ and $Q$ do not depend on $k$. \item If $AD=a$, $BC=b$, and the segment $BC$ moves along $AD$, find the minimum length of segment $PQ$. \end{enumerate} \end{exer} \section{The complex projective plane: a glimpse of algebraic geometry} \label{sec:alggeo} The homogeneous coordinates we have worked with so far also make sense for complex numbers, though visualizing the result is substantially harder. The set of points they define (i.e.\ the set of proportionality classes of ordered triples of complex numbers, not all zero) is called the \emph{complex projective plane}. We define lines and conics in this new plane simply as the zero loci of linear and quadratic polynomials, respectively. One handy feature of the complex projective plane is the following characterization of circles. \begin{fact} A nondegenerate conic is a circle if and only if it passes through the points $[1:i:0]$ and $[1:-i:0]$. \end{fact} These two points are called the \emph{circular points at infinity}, or simply the \emph{circular points} for short. The fact that complex circles always meet the line at infinity in two points, while real circles to not, is a symptom of the key fact that the complex numbers are \emph{algebraically closed}, i.e.\ every polynomial with complex coefficients has a complex root. (This is the Fundamental Theorem of Algebra, first proved by Gauss.) This means, for example, that we have the following: \begin{fact} In the complex projective plane, two conics meet in exactly four points (counting points of tangency twice). \end{fact} In fact, a more general result is true, which we will not prove. \begin{theorem}[Bezout] The zero loci of two polynomials, of degrees $m$ and $n$, contains exactly $mn$ points if the loci meet transversally everywhere (i.e.\ at each intersection, each locus has a well-defined tangent line, and the tangent lines are distinct). \end{theorem} If the loci do not meet transversally, e.g.\ if they are tangent somewhere, one must correctly assign multiplicities to the intersections to make the count work. An interesting consequence of Bezout's theorem, which we will prove independently, is due to Chasles (pronounced ``shell''). The zero locus of a polynomial of degree 3 is known as a \emph{cubic curve}. \begin{theorem}[Chasles] Let $C_{1}$ and $C_{2}$ be two cubic curves meeting in exactly nine distinct points. Then any cubic curve passing through eight of the points passes through the ninth point. \end{theorem} \begin{proof} The set of homogeneous degree 3 polynomials in $x,y,z$ is a 10-dimensional vector space (check by writing a basis of monomials); let $Q_1$ and $Q_2$ be polynomials with zero loci $C_1$ and $C_2$, respectively, and let $P_1, \dots, P_9$ be the nine intersections of $C_1$ and $C_2$. Note that no four of these points lie on a line and no seven lie on a conic, or else each of $C_1$ and $C_2$ would have this line or conic as a component, and their intersection would be infinite rather than nine points. Let $d_i$ be the dimension of the space of degree 3 polynomials vanishing at $D_1, \dots, D_i$ (and put $d_0 = 10$); then for $i \leq 8$, $d_i$ equals either $d_{i-1}-1$ or $d_i$, the latter only if every cubic curve passing through $P_1, \dots, P_{i-1}$ also passes through $P_i$. However, this turns out not to be the case; see the problems. Thus $d_8 = 2$, and we already have two linearly independent polynomials in this space, namely $Q_1$ and $Q_2$. (If they were dependent, they would define the same curve, and again the intersection would be infinite.) Thus if $C$ is a cubic curve defined by a polynomial $Q$ that passes through $P_1, \dots, P_8$, then $Q = aQ_1 + bQ_2$ for some $a,b \in \CC$, and so $Q$ also vanishes at $P_9$, as desired. \end{proof} These results are just the tip of a rather sizable iceberg. The modern subject of \emph{algebraic geometry} is concerned with the study of zero loci of sets of polynomials in spaces of any dimension. It interacts with almost every other branch of mathematics, including complex analysis, topology, number theory, combinatorics, and mathematical physics. Unfortunately, the subject as practiced today has become technically involved; the novice should start with a book written in the ``classical'' style, such as Harris \cite{bib:har} or Shafarevich \cite{bib:sha}, before proceeding to a ``modern'' text such as Hartshorne \cite{bib:hart}. (If it is not already clear from the rhapsodic tone of this section, algebraic geometry, particularly in connection with number theory, ranks among the author's main research interests.) \begin{exer} \ii \label{ex:cencirc} Prove that the center of a circle drawn in the plane cannot be constructed with straightedge alone. \ii Give another proof that there is a unique conic passing through any five points, using the circular points. \ii Make up a problem by taking a projective statement you know and projecting two of the points in the diagram to the circular points. (One of my favorites is the radical axis theorem---which becomes a projective statement if you replace the circles by conics through two fixed points!) \ii Deduce Pascal's theorem from Chasles' theorem applied to a certain degenerate cubic. \ii \label{ex:chas} Prove that given eight or fewer points in the plane, no four on a line and no seven on a conic, one of which is labeled $P$, there exists a cubic curve passing through all of the points but $P$. \ii A cubic curve which is nondegenerate, and additionally has no \emph{singular point} (a point where the partial derivatives of the defining homogeneous polynomial all vanish, like the point $[0:0:1]$ on the curve $y^2z = x^3+x^2z$) is called an \emph{elliptic curve} (the apparent misnomer occurs for historical reasons). Let $E$ be an elliptic curve, and pick a point $O$ on $E$. Define ``addition'' of points on $E$ as follows: given points $P$ and $Q$, let $R$ be the third intersection of the line $PQ$ with $E$, and let $P+Q$ be the third intersection of the line $OR$ with $E$. Prove that $(P+Q)+R = P+(Q+R)$ for any three points $P,Q,R$, i.e.\ that ``addition is associative''. (If you know what a group is, show that $E$ forms a group under addition, by showing that there exist inverses and an identity element.) For more on elliptic curves, and their role in number theory, see \cite{bib:st}. \ii \label{ex:rom97a} Give another solution to problem~\ref{ex:rom97} using a well-chosen projective transformation in the complex projective plane. \ii One can define addition on a curve on a singular cubic in the same fashion, as long as none of the points involved is a singular point of the cubic. Use this fact to give another solution to Problem~\ref{ex:rom97}. \ii \label{ex:flex} Let $E$ be an elliptic curve. Show that there are exactly nine points at which the tangent line at $E$ has a triple, not just a double, intersection with the curve (and so meets the curve nowhere else). These points are called \emph{flexes}. Also show that the line through any two flexes meets $E$ again at another flex. (Hence the flexes constitute a counterexample to Problem~\ref{ex:syl} in the complex projective plane!) \ii (``projective Steiner's porism'') \label{ex:projstein} Let $\omega_1$ and $\omega_2$ be two conic sections. Given a point $P_0$ on $\omega_1$, let $P_1$ be either of the points on $\omega_1$ such that the line $P_0P_1$ is tangent to $\omega_2$. then for $n \geq 2$, define $P_n$ as the point on $\omega_1$ other than $P_{n-2}$ such that $P_{n-1}P_n$ is tangent to $\omega_2$. Suppose there exists $n$ such that $P_0=P_n$ for a particular choice of $P_0$. Show that $P_0=P_n$ for any choice of $P_0$. \end{exer} \chapter{Geometric inequalities} The subject of geometric inequalities is so vast that it suffices to fill entire books, two notable examples being the volume by Bottema et al.\ \cite{bib:bott} and its sequel \cite{bib:rcige}. This chapter should thus be regarded more as a sampler of techniques than a comprehensive treatise. \section{Distance inequalities} A number of inequalities involve comparing lengths. Useful tools against such problems include: \begin{itemize} \ii Triangle inequality: in triangle $ABC$, $AB + BC > BC$. \ii Hypotenuse inequality: if $\ang ABC$ is a right angle, then $AC > BC$. \ii Ptolemy's inequality (Problem~\ref{ex:ptineq}): if $ABCD$ is a convex quadrilateral, then $AB \cdot CD + BC \cdot DA \geq AC \cdot BD$, with equality if and only if $ABCD$ is cyclic. \ii Erd\H{o}s-Mordell inequality: see Section~\ref{sec:em}. \end{itemize} Transformations can also be useful, particularly reflection. For example, to find the point $P$ on a fixed line that minimizes the sum of the distances from $P$ to two fixed points $A$ and $B$, reflect the segment $PB$ across the line and observe that the optimal position of $P$ is on the line joining $A$ to the reflection of $B$. DIAGRAM. A more dramatic example along the same lines is the following solution (by H.A. Schwarz) to \emph{Fagnano's problem}: of the triangles inscribed in a given acute triangle, which one has the least perimeter? Reflecting the triangle as shown implies that the perimeter of an inscribed triangle is at least the distance from $A$ to its eventual image, with equality when the inscribed triangle makes equal angles with each side. As noted earlier, this occurs for the orthic triangle, which is then the desired minimum. DIAGRAM. \begin{exer} \ii For what point $P$ inside a convex quadrilateral $ABCD$ is $PA+PB+PC+PD$ minimized? \ii (Euclid) Prove that the longest chord whose vertices lie on or inside a given triangle is the longest side. (This is intuitively obvious, but make sure your proof is complete.) \ii (K\"ursch\'ak, 1954) Suppose a convex quadrilateral $ABCD$ satisfies $AB+BD \leq AC+CD$. Prove that $AB < AC$. \ii (USAMO 1999/2) \label{ex:usamo992} Let $ABCD$ be a cyclic quadrilateral. Prove that \[ |AB - CD| + |AD - BC| \geq 2|AC - BD|. \] \ii (Titu Andreescu and R\u azvan Gelca) Points $A$ and $B$ are separated by two rivers. One bridge is to be built across each river so as to minimize the length of the shortest path from $A$ to $B$. Where should they be placed? (Each river is an infinite rectangular strip, and each bridge must be a straight segment perpendicular to the sides of the river. You may assume that $A$ and $B$ are separated from the intersection of the rivers by a strip wider than the two rivers combined.) \ii Prove that a quadrilateral inscribed in a parallelogram has perimeter no less than twice the length of the shorter diagonal of the parallelogram. (You may want to first consider the case where the parallelogram is a rectangle.) \ii (IMO 1993/4) For three points $P,Q,R$ in the plane, we define $m(PQR)$ as the minimum length of the three altitudes of $\triangle PQR$. (If the points are collinear, we set $m(PQR) = 0$.) Prove that for points $A,B,C,X$ in the plane, \[ m(ABC) \leq m(ABX) + m(AXC) + m(XBC). \] \ii (Sylvester's theorem) \label{ex:syl} A finite set of points in the plane has the property that the line through any two of the points passes through a third. Prove that all of the points are collinear. (As noted in exercise~\ref{ex:flex}, this result is false in the complex projective plane.) \ii (IMO 1973/4) A soldier needs to check on the presence of mines in a region having the shape of an equilateral triangle. The radius of action of his detector is equal to half the altitude of the triangle. The soldier leaves from one vertex of the triangle. What path should he follow in order to travel the least possible distance and still accomplish his mission? \ii Let $D$ be the third vertex of an equilateral triangle constructed externally on $BC$. For $P$ inside the triangle, show that $PA + PB + PC \geq AD$, and determine when equality holds. \ii Suppose the largest angle of triangle $ABC$ is not greater than $120^{\circ}$. Deduce from the previous exercise that for $P$ inside the triangle, $PA + PB + PC$ is minimized when $\ang APB = \ang BPC = \ang CPA = 120^{\circ}$. The point satisfying this condition is known variously as the \emph{Fermat point} or the \emph{Torricelli point}. \ii (IMO 1995/5) Let $ABCDEF$ be a convex hexagon with $AB=BC=CD$ and $DE=EF=FA$, such that $\angle BCD = \angle EFA = \pi/3$. Suppose $G$ and $H$ are points in the interior of the hexagon such that $\angle AGB = \angle DHE = 2\pi/3$. Prove that $AG + GB + GH + DH + HE \geq CF$. \end{exer} \section{Algebraic techniques} Another class of methods of attack for geometric inequalities involve invoking algebraic inequalities. The most commonly used is the AM-GM inequality: for $x_{1}, \dots, x_{n} > 0$, \[ \frac{x_{1} + \cdots + x_{n}}{n} \geq (x_{1}\dots x_{n})^{1/n}. \] Often all one needs is the case $n=2$, which follows from the fact that \[ (\sqrt{x_{1}} - \sqrt{x_{2}})^{2} \geq 0. \] A more sophisticated result is the Cauchy-Schwarz inequality: \[ (x_{1}^{2} + \cdots + x_{n}^{2})(y_{1}^{2} + \cdots + y_{n}^{2}) \geq (x_{1}y_{1} + \cdots + x_{n}y_{n})^{2}, \] which one proves by noting that the difference between the left side and the right is \[ \sum_{i c, b+c > a, c+a > b$ for $a,b,c$ to constitute the side lengths of a triangle translate into the more convenient conditions $x>0, y>0, z>0$. Don't forget about the possibility of ``algebraizing'' an inequality using complex numbers; see Section~\ref{sec:cplx}. \begin{exer} \ii (IMO 1988/5) $ABC$ is a triangle right-angled at $A$, and $D$ is the foot of the altitude from $A$. The straight line joining the incenters of the triangles $ABD, ACD$ intersects the sides $AB, AC$ at the points $K,L$, respectively. $S$ and $T$ denote the areas of the triangles $ABC$ and $AKL$, respectively. Show that $S \geq 2T$. \ii Given a point $P$ inside a triangle $ABC$, let $x,y,z$ be the distances from $P$ to the sides $BC, CA, AB$. Find the point $P$ which minimizes \[ \frac{a}{x} + \frac{b}{y} + \frac{c}{z}. \] \ii If $K$ is the area of a triangle with sides $a,b,c$, show that \[ ab + bc + ca \geq 4 \sqrt{3} K. \] \ii (IMO 1964/2) Suppose $a,b,c$ are the sides of a triangle. Prove that \[ a^{2}(b+c-a) + b^{2}(c+a-b) + c^{2}(a+b-c) \leq 3abc. \] \ii (IMO 1983/6) Let $a,b,c$ be the lengths of the sides of a triangle. Prove that \[ b^2c(b-c) + c^2a(c-a) + a^2b(a-b) \geq 0. \] (Beware: you may not assume that $a\geq b \geq c$ without loss of generality!) \ii (Balkan, 1996) Let $O$ and $G$ be the circumcenter and centroid of a triangle of circumradius $R$ and inradius $r$. Show that $OG^2 \leq R^2 - 2Rr$. (This proves Euler's inequality $R \geq 2r$. If you don't know how to compute $OG^2$, see problem~\ref{ex:cirort}.) \ii (Murray Klamkin) Let $n > 2$ be a positive integers, and suppose that $a_1, \dots, a_n$ are positive real numbers satisfying the inequality \[ (a_1^2 + \cdots + a_n^2)^2 > (n-1) (a_1^4 + \cdots + a_n^4). \] Show that for $1 \leq i < j < k \leq n$, the numbers $a_i, a_j, a_k$ are the lengths of the sides of a triangle. \ii \label{ex:cotident} Let $ABC$ be a triangle with inradius $r$ and circumradius $R$. Prove that \[ \frac{2r}{R} \leq \sqrt{\cos \frac{A-B}{2} \cos \frac{B-C}{2} \cos \frac{C-A}{2}}. \] \ii \label{ex:quadineq} (IMO 1995 proposal) Let $P$ be a point inside the convex quadrilateral $ABCD$. Let $E,F,G,H$ be points on sides $AB,BC,CD,DA$, respectively, such that $PE$ is parallel to $BC$, $PF$ is parallel to $AB$, $PG$ is parallel to $DA$, and $PH$ is parallel to $CD$. Let $K, K_{1}, K_{2}$ be the areas of $ABCD, AEPH, PFCG$, respectively. Prove that \[ \sqrt{K} \geq \sqrt{K_{1}} + \sqrt{K_{2}}. \] \end{exer} \section{Trigonometric inequalities and convexity} A third standard avenue of attack involves reducing a geometric inequality to an inequality involving trigonometric functions. Such inequalities can often be treated using Jensen's inequality for convex functions. A \emph{convex} function is a function $f(x)$ satisfying the rule \[ f(tx + (1-t)y) \leq tf(x) + (1-t)f(y) \] for all $x,y$ and all $t \in [0,1]$. Geometrically, this says that the area above the graph of $f$ is a convex set, i.e.\ that chords of the graph always lie above the graph. Equivalently, tangents to the graph lie below. Those of you who know calculus can check whether $f$ is convex by checking whether the second derivative of $f$ (if it exists) is always positive. (In some calculus texts, a convex function is called ``concave upward'', or occasionally is said to ``hold water''.) Also, if $f$ is continuous, it suffices to check the definition of convexity for $t=1/2$. The key fact about convex functions is \emph{Jensen's inequality}, whose proof (by induction on $n$) is not difficult. \begin{fact} Let $f(x)$ be a convex function, and let $t_{1}, \dots, t_{n}$ be nonnegative real numbers adding up to 1. Then for all $x_{1}, \dots, x_{n}$, \[ f(t_{1}x_{1} + \cdots + t_{n}x_{n}) \leq t_{1}f(x_{1}) + \cdots + t_{n}f(x_{n}). \] \end{fact} For example, the convexity of the function $(-\log x)$ implies the AM-GM inequality. As a simple example, note that in triangle $ABC$, we have $\angle A + \angle B + \angle C = \pi$, and the function $f(x) = \sin x$ is concave, so \[ \sin A + \sin B + \sin C \geq 3 \sin \pi/3 = 3\sqrt{3}/2. \] In other words, the minimum perimeter of a triangle inscribed in a fixed circle is achieved by the equilateral triangle. Also note that convexity can be used in apparently purely geometric circumstances, thanks to the following fact. (Remember, it suffices to verify this for $t=1/2$, which is easy.) \begin{fact} The distance from a fixed point $P$ is a convex function on the plane. That is, for any points $P,Q,R$, the distance from $P$ to the point (in vector notation) $tQ + (1-t)R$ is a convex function of $t$. \end{fact} \begin{exer} \ii (Bottema, 2.7) Show that in triangle $ABC$, $\sin A \sin B \sin C \leq \frac{3}{8} \sqrt{3}$. \ii Recall that the Brocard angle of triangle $ABC$ is defined by the formula \[ \cot \omega = \cot A + \cot B + \cot C. \] Prove that the Brocard angle never exceeds $\pi/6$. (Be careful: $\cot$ is only convex in the range $(0, \pi/2]$.) \ii (Bottema, 2.15) Let $\alpha, \beta, \gamma$ be the angles of a triangle. Prove that \[ \sin \frac{\beta}{2} \sin \frac{\gamma}{2} + \sin \frac{\gamma}{2} \sin \frac{\alpha}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \leq \frac 34. \] \ii Prove that of the $n$-gons inscribed in a circle, the regular $n$-gon has maximum area. \ii (Bottema, 2.59) Prove that in triangle $ABC$, \[ 1 + \cos A \cos B \cos C \geq \sqrt{3} (\sin A \sin B \sin C ). \] \ii Show that for any convex polygon $S$, the distance from $S$ to a point $P$ (the length of the shortest segment joining $P$ to a point on $S$) is a convex function of $P$. \ii (Junior Balkaniad, 1997) In triangle $ABC$, let $D,E,F$ be the points where the incircle touches the sides. Let $r, R, s$ be the inradius, circumradius, and semiperimeter, respectively, of the triangle. Prove that \[ \frac{2rs}{R} \leq DE + EF + FD \leq s \] and determine when equality occurs. \ii (MOP 1998) If $ABC$ is a acute triangle with circumcenter $O$, orthocenter $H$ and circumradius $R$, show that for any point $P$ on the segment $OH$, \[ PA + PB + PC \leq 3R. \] \end{exer} \section{The Erd\H{o}s-Mordell inequality} \label{sec:em} The following inequality is somewhat more sophisticated than the ones we have seen so far, but is nonetheless useful. It was conjectured by the Hungarian mathematician and problemist P\'al (Paul) Erd\H{o}s (1913-1996) in 1935 and first proved by Louis Mordell in the same year. \begin{theorem} For any point $P$ inside the triangle $ABC$, the sum of the distances from $P$ to $A,B,C$ is at least twice the sum of the distances from $P$ to $BC,CA,AB$. (Equality occurs only when $ABC$ is equilateral and $P$ is its center.) \end{theorem} \begin{proof} The unusually stringent equality condition should suggest that perhaps the proof proceeds in two stages, with different equality conditions. This is indeed the case. Let $X,Y,Z$ be the feet of the respective perpendiculars from $P$ to $BC,CA,AB$. We will first prove that \begin{equation} \label{eq:erdmo1} PA \geq \frac{AB}{BC} PY + \frac{AC}{BC} PZ. \end{equation} The only difference between most proofs of this theorem is in the proof of the above inequality. For example, rewrite (\ref{eq:erdmo1}) as \[ PA \sin A \geq PY \sin C + PZ \sin B, \] recognize that $PA \sin A = YZ$ by the Extended Law of Sines, and observe that the right side is the length of the projection of $YZ$ onto the line $BC$. Equality holds if and only if $YZ$ is parallel to $BC$. Putting (\ref{eq:erdmo1}) and its analogues together, we get \[ PA + PB + PC \geq PX \left( \frac{CA}{AB} + \frac{AB}{CA} \right) + PY \left( \frac{AB}{BC} + \frac{BC}{AB} \right) + PZ \left( \frac{BC}{CA} + \frac{CA}{BC} \right), \] with equality if and only if $XYZ$ is homothetic to $ABC$; this occurs if and only if $P$ is the circumcenter of $ABC$ (Problem~1). Now for the second step: we note that each of the terms in parentheses is at least 2 by the AM-GM inequality. This gives \[ PA + PB + PC \geq 2 (PX + PY + PZ), \] with equality if and only if $AB = BC = CA$. \end{proof} \begin{exer} \ii With notation as in the above proof, show that the triangles $XYZ$ and $ABC$ are homothetic if and only if $P$ is the circumcenter of $ABC$. \ii Give another proof of (\ref{eq:erdmo1}) by comparing $P$ with its reflection across the angle bisector of $A$. (Beware: the reflection may lie outside of the triangle!) \ii Solve problem~\ref{ex:imo91} using the Erd\H{o}s-Mordell inequality. \ii (IMO 1996/5) Let $ABCDEF$ be a convex hexagon such that $AB$ is parallel to $DE$, $BC$ is parallel to $EF$, and $CD$ is parallel to $FA$. Let $R_{A}, R_{C}, R_{E}$ denote the circumradii of triangles $FAB, BCD, DEF$, respectively, and let $P$ denote the perimeter of the hexagon. Prove that \[ R_{A} + R_{C} + R_{E} \geq \frac P2. \] (A certain special case of this result is equivalent to Erd\H{o}s-Mordell. Modify the proof slightly to accommodate the generalization.) \ii (Nikolai Nikolov) The incircle $k$ of the triangle $ABC$ touches its sides at the points $A_1, B_1, C_1$. For any point $K$ on $k$, let $d$ be the sum of the distances from $K$ to the sides of the triangle $A_1B_1C_1$. Prove that $KA + KB + KC > 2d$. \end{exer} \section{Additional problems} \begin{exer} \ii Prove that of all quadrilaterals with a prescribed perimeter $P$, the square has the greatest area. Can you also prove the analogous result for polygons with any number of sides? \ii What is the smallest positive real number $r$ such that a square of side length 1 can be covered by three disks of radius $r$? \ii \label{ex:incircles} Let $r$ be the inradius of triangle $ABC$. Let $r_A$ be the radius of a circle tangent to the incircle as well as to sides $AB$ and $CA$. Define $r_B$ and $r_C$ similarly. Prove that \[ r_A + r_B + r_C \geq r. \] \ii Prove that a triangle with angles $\alpha, \beta, \gamma$, circumradius $R$, and area $A$ satisfies \[ \tan \frac{\alpha}{2} + \tan \frac{\beta}{2} + \tan \frac{\gamma}{2} \leq \frac{9R^2}{4A}. \] \ii %% MOP 1995 Let $a,b,c$ be the sides of a triangle with inradius $r$ and circumradius $R$. Show that \[ \left\lvert 1 - \frac{2a}{b+c} \right\rvert \leq \sqrt{1 - \frac{2r}{R}}. \] \ii %% MOP 1995 Two concentric circles have radii $R$ and $R_{1}$ respectively, where $R_{1} > R$. $ABCD$ is inscribed in the smaller circle and $A_{1}B_{1}C_{1}D_{1}$ in the larger one, with $A_{1}$ on the extension of $CD$, $B_{1}$ on that of $DA$, $C_{1}$ on that of $AB$, and $D_{1}$ on that of $BC$. Prove that the ratio of the areas of $A_{1}B_{1}C_{1}D_{1}$ and $ABCD$ is at least $R_{1}^{2}/R^{2}$. \ii %% MOP 1995--posed for ratio 2, use Ptolemy With the same notation, prove that the ratio of the perimeters of $A_{1}B_{1}C_{1}D_{1}$ and $ABCD$ is at least $R/r$. \end{exer} \appendix \chapter{Nonsynthetic methods} \label{nonsynth} The idea of using algebra to solve geometric problems, in the guise of rectilinear coordinates, is variously attributed to Descartes or Fermat, though the trigonometric functions were known to the ancients. In any case, this section is devoted to nonsynthetic methods in Euclidean geometry, including trigonometry, vector geometry, complex numbers, and Cartesian coordinates. We have relegated this material to an appendix not to avoid offending the purists (who are probably offended already by the liberal use of nonsynthetic methods in the main text), but to avoid disrupting the logical sequence of the chapters. \section{Trigonometry} This is not a course in trigonometry; all we will do here is summarize the important facts and provide a few problems where trigonometry can or must be employed. \begin{fact}[Law of Sines] The area of triangle $ABC$ equals $\frac{1}{2} a b \sin C$. In particular, \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}. \] \end{fact} \begin{fact}[Extended Law of Sines] If $R$ is the circumradius of triangle $ABC$, then $BC = 2R \sin A$. \end{fact} \begin{fact}[Law of Cosines] In triangle $ABC$, \[ c^{2} = a^{2} + b^{2} - 2 a b \cos C. \] \end{fact} \begin{fact}[Addition formulae] The sine and cosine functions satisfy the following addition rules: \beqa \cos (A+B) &=& \cos A \cos B - \sin A \sin B \\ \cos (A-B) &=& \cos A \cos B + \sin A \sin B \\ \sin (A+B) &=& \sin A \cos B + \cos A \sin B \\ \sin (A-B) &=& \sin A \cos B - \cos A \sin B. \eeqa \end{fact} Using the addition formulae, one can convert products of sines and cosines to sums, and vice versa. \begin{fact}[Sum-to-product formulae] \beqa \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \\ \sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2} \\ \cos A + \cos B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \\ \cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}. \eeqa \end{fact} In particular, one has the double and half-angle formulas. \begin{fact}[Double-angle formula] \beqa \sin 2A &=& 2 \sin A \cos A \\ \cos 2A &=& 2 \cos^2 A - 1 = 1 - 2 \sin^2 A \\ \tan 2A &=& \frac{2 \tan A}{1 - \tan^2 A}. \eeqa \end{fact} \begin{fact}[Half-angle formula] \beqa \sin \frac A2 &=& \pm \sqrt{\frac{1 - \cos A}{2}} \\ \cos \frac A2 &=& \pm \sqrt{\frac{1 + \cos A}{2}} \\ \tan \frac A2 &=& \csc A - \cot A. \eeqa \end{fact} The half-angle formula takes a convenient form for triangles. \begin{fact} In triangle $ABC$ with sides $a,b,c$ and semiperimeter $s$, \beqa \sin \frac C2 &=& \sqrt{\frac{(s-a)(s-b)}{ab}} \\ \cos \frac C2 &=& \sqrt{\frac{s(s-c)}{ab}}. \eeqa \end{fact} It may be helpful at times to express certain other quantities associated with a triangle in terms of the angles. \begin{fact} If triangle $ABC$ has inradius $r$ and circumradius $R$, then \[ r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}. \] \end{fact} We leave the construction of other such formulae to the reader. \begin{exer} \ii For any triangle $ABC$, prove that $\tan A + \tan B + \tan C = \tan A \tan B \tan C$ and that $\cot A/2 + \cot B/2 + \cot C/2 = \cot A/2 \cot B/2 \cot C/2$. \ii Show that if none of the angles of a convex quadrilateral $ABCD$ is a right angle, then \[ \frac{\tan A + \tan B + \tan C + \tan D}{\tan A \tan B \tan C \tan D} = \cot A + \cot B + \cot C + \cot D. \] \ii Find a formula for the area of a triangle in terms of two angles and the side opposite the third angle. (More generally, given any data that uniquely determines a triangle, one can find an area formula in terms of that data. Can you come up with more examples?) %\ii %% Does this really belong here?? %Let $ABCD$ be a convex cyclic quadrilateral. Show that the sum of the %inradii of triangles $ABC$ and $CDA$ equals the sum of the inradii of %triangles $BCD$ and $DAB$. \ii Use trigonometry to give another proof of Ptolemy's theorem (Theorem~\ref{thm:pt}). \ii (USAMO 1996/5) Triangle $ABC$ has the following property: there is an interior point $P$ such that $\angle PAB = 10^{\circ}$, $\angle PBA = 20^{\circ}$, $\angle PCA = 30^{\circ}$ and $\angle PAC = 40^{\circ}$. Prove that triangle $ABC$ is isosceles. (For an added challenge, find a non-trigonometric solution!) \item (IMO 1985/1) A circle has center on the side $AB$ of a cyclic quadrilateral $ABCD$. The other three sides are tangent to the circle. Prove that $AD+DC=AB$. \end{exer} \section{Vectors} \label{sec:vec} A \emph{vector} in the place can be defined either as an arrow, where addition of arrows proceeds by the ``tip-to-tail'' rule illustrated below, or as an ordered pair $(x, y)$ recording the difference in the $x$ and $y$ coordinates between the tip and the tail. Vectors in space are defined similarly, of course using three coordinates instead of two. DIAGRAM. It is important to remember that a vector is not a point, but rather the ``difference of two points''; it encodes relative, not absolute, position. In practice, however, one chooses a point as the \emph{origin} and identifies a point with the vector from the origin to that point. (In effect, one puts the tails of all of the arrows in one place.) The standard operations on vectors include addition and subtraction, multiplication by real numbers (positive, negative or zero), and the \emph{dot product}, defined geometrically as \[ \vA \cdot \vB = \norm{\vA} \cdot \norm{\vB} \cos \angle AOB, \] where $O$ is the origin, and in coordinates as \[ (a_{x}, a_{y}) \cdot (b_{x}, b_{y}) = a_{x}b_{x} + a_{y}b_{y}. \] The key fact here is that $\vA \cdot \vB = 0$ if and only if $\vA$ and $\vB$ are perpendicular. A more exotic operation is the \emph{cross product}, which is defined for a pair of vectors in space as follows: \[ (a_{x}, a_{y}, a_{z}) \times (b_{x}, b_{y}, b_{z}) = (a_{y}b_{z} - a_{z}b_{y}, a_{z}b_{x} - a_{x}b_{z}, a_{x}b_{y} - a_{y}b_{x}). \] Geometrically speaking, $\vA \times \vB$ is perpendicular to both $\vA$ and $\vB$ and has length \[ \norm{\vA \times \vB} = \norm{\vA} \cdot \norm{\vB} \sin \angle AOB. \] This length equals the area of the parallelogram with vertices $0, \vA, \vA+\vB, \vB$, or twice the area of the triangle with vertices $0, \vA, \vB$. The sign ambiguity can be resolved by the \emph{right-hand rule}: if you point the fingers of your right hand along $\vA$, then swing them toward $\vB$, your thumb points in the direction of $\vA \times \vB$. DIAGRAM. \begin{fact} The following identities hold: \begin{enumerate} \ii Triple scalar product identity: $\vA \cdot (\vB \times \vC) = \vB \cdot (\vC \times \vA) = \vC \cdot (\vA \times \vB)$. Moreover, this quantity equals the volume of a parallelepiped with edges $\vA, \vB, \vC$. \ii Triple cross product identity: $\vA \times (\vB \times \vC) = (\vC \cdot \vA) \vB - (\vB \cdot \vA)\vC$. \end{enumerate} \end{fact} The vector equations for some of the special points of a triangle are summarized in the following table. The asterisked expressions assume the circumcenter of the triangle has been chosen as the origin; the origin-independent expressions are not nearly so pleasant to work with! \begin{center} \begin{tabular}{cc} Circumcenter* & $0$ \\ Centroid & $\frac{1}{3} (\vA + \vB + \vC)$ \\ Orthocenter* & $\vA + \vB + \vC$ \\ Incenter & $\frac{1}{a+b+c} (a\vA + b\vB + c\vC)$ \end{tabular} \end{center} \begin{exer} \ii (Romania, 1997) Let $ABCDEF$ be a convex hexagon, and let $P = AB \cap CD, Q = CD \cap EF, R = EF \cap AB$, $S = BC \cap DE, T = DE \cap FA, U = FA \cap BC$. Prove that \[ \frac{PQ}{CD} = \frac{QR}{EF} = \frac{RP}{AB} \quad \mbox{if and only if} \quad \frac{ST}{DE} = \frac{TU}{FA} = \frac{US}{BC}. \] \ii Let $P, Q, R$ be the feet of concurrent cevians in triangle $ABC$. Determine the vector expression for the point of concurrence in terms of the ratios $BP/PC$, $CQ/QA$, $AR/RB$. Use this formula to extend the above table to other special points. In particular, do so for the Nagel point and obtain an alternate solution to problem~\ref{ex:nagel}. %% Something about perpendiculars \ii (R\u{a}zvan Gelca) %% Razvan Let $ABCD$ be a convex quadrilateral and $O = AC \cap BD$. Let $M, N$ be points on $AB$ so that $AM = MN = NB$, and let $P, Q$ be points on $CD$ so that $CP = PQ = QD$. Show that triangles $MOP$ and $NOQ$ have the same area. \ii (MOP 1995) %% MOP 1995 Five points are given on a circle. A perpendicular is drawn through the centroid of the triangle formed by three of them, to the chord connecting the remaining two. Similar perpendiculars are drawn for each of the remaining nine triplets of points. Prove that the ten lines obtained in this way have a common point. \ii (MOP 1996) %% MOP 1996 Let $ABCDE$ be a convex pentagon, and let $F,G,H,I,J$ be the respective midpoints of $CD, DE, EA, AB, BC$. If $AF, BG, CH, DI$ pass through a common point, show that $EJ$ also passes through this point. \ii Let $A,B,C,D$ be four points on a circle. Use the result of problem~\ref{ex:simcon} to show that the Simson line of each point with respect to the triangle formed by the other three passes through the midpoint of the segment joining the center of the circle to the centroid of $ABCD$ (in particular, these four lines are concurrent). \ii \label{ex:cirort} %% 9 dist^{2} = a^{2} + b^{2} + c^{2} Compute the distance between the circumcenter and orthocenter of a triangle in terms of the side lengths $a,b,c$. \ii (Austria-Poland, 1979) Let $A,B,C,D$ be points in space, $M$ the midpoint of $AC$ and $N$ the midpoint of $BD$. Prove that \[ 4MN^{2} = AB^{2} + BC^{2} +CD^{2} + DA^{2} -AC^{2} - BD^{2}. \] \ii Show that the distance between the incenter and the nine-point center (see Problem~\ref{ex:ninept}) of a triangle is equal to $R/2 - r$, where $r$ and $R$ are inradius and circumradius, respectively. Deduce \emph{Feuerbach's theorem}, that the incircle and nine-point circle are tangent. (Similarly, one can show the nine-point circle is also tangent to each of the excircles.) \end{exer} \section{Complex numbers} \label{sec:cplx} In some respects, the use of complex numbers in geometry is an extension of vector geometry. The main difference is that complex numbers come equipped with a mechanism for implementing rotation, while vectors do not. They even come equipped with a mechanism for implementing reflection across the horizontal axis, namely complex conjugation. For more about the relationship between complex numbers and transformations, see Section~\ref{sec:spir}. An interesting use of complex numbers is to prove inequalities, as in the following example (compare with Problem~\ref{ex:ptineq}). \begin{theorem}[Ptolemy's inequality] \label{thm:ptineq} Let $A,B,C,D$ be four points in the plane. Then \[ AC \cdot BD \leq AB \cdot CD + BC \cdot DA, \] with equality if and only if the quadrilateral $ABCD$ is convex and cyclic. \end{theorem} \begin{proof} Regard $A,B,C,D$ as complex numbers; then we have an identity \[ (A-C)(B-D) = (A-B)(C-D) + (B-C)(A-D). \] However, the magnitude of $(A-C)(B-D)$ is precisely the product of the lengths of the segments $AC$ and $BD$, and likewise for the other terms. Thus the desired inequality is simply the triangle inequality applied to these three quantities! (The equality condition is left as an exercise.) \end{proof} \begin{exer} \ii Prove that $x,y,z$ lie at the corners of an equilateral triangle if and only if either $x + \omega y + \omega^{2} z = 0$ or $x + \omega z + \omega^{2} y = 0$, where $\omega = e^{2\pi i/3}$. \ii Let $A,C,E$ be three points on a circle. A $60^{\circ}$ rotation about the center of the circle carries $A,C,E$ to $B,D,F$, respectively. Prove that the triangle whose vertices are the midpoints of $BC, DE, FA$ is equilateral. \ii Construct equilateral triangles externally (internally) on the sides of an arbitrary triangle $ABC$. Prove that the centers of these three triangles form another equilateral triangle. This triangle is known as the \emph{inner (outer) Napoleon triangle} of $ABC$. \ii Let $P,Q,R,S$ be the centers of squares constructed externally on sides $AB,BC,CD,DA$, respectively, of a convex quadrilateral $ABCD$. Show that the segments $PR$ and $QS$ are perpendicular to each other and equal in length. \ii Let $ABCD$ be a convex quadrilateral. Construct squares $CDKL$ and $ABMN$ externally on sides $AB$ and $CD$. Show that if the midpoints of $AC, BD, KN, ML$ do not coincide, then they form a square. \ii (IMO 1977/2) Equilateral triangles $ABK$, $BCL$, $CDM$, $DAN$ are constructed inside the square $ABCD$. Prove that the midpoints of the four segments $KL, LM, MN, NK$ and the midpoints of the eight segments $AK, BK$, $BL, CL$, $CM, DM$, $DN, AN$ are the twelve vertices of a regular dodecagon. (Nowadays the IMO tends to avoid geometry problems such as this one, which have no free parameters, but they are relatively common in single-answer contests such as ARML.) \ii Use complex numbers and the circle of Apollonius (Theorem~\ref{thm:ap}) to give another proof that circles map to circles under inversion. \ii Given a point $P$ on the circumference of a unit circle and the vertices $A_1,\,A_2,\dots,A_n$ of an inscribed regular $n$-gon, prove that: \begin{enumerate} \ii $PA_1^2+PA_2^2+\cdots+PA_n^2$ is a constant (independent of $P$). \ii $PA_1^4+PA_2^4+\cdots+PA_n^4$ is a constant (independent of $P$). \end{enumerate} \ii (China, 1998) Let $P$ be an arbitrary point in the plane of triangle $ABC$ with sides $BC=a$, $CA=b$, $AB=c$, and with $PA=x$, $PB=y$, $PC=z$. Prove that \[ ayz+bzx+cxy\geq abc, \] with equality if and only if $P$ is the circumcenter of $ABC$. \end{exer} \section{Cartesian coordinates} It was first noted independently by Descartes and Fermat that any problem in Euclidean geometry could be reduced to an algebraic problem by introducing rectangular coordinates, which we now call \emph{Cartesian coordinates}. One major result of this discovery was increased interest in what we now call \emph{algebraic geometry} (see Section~\ref{sec:alggeo}). In practice, algebraizing a problem in Euclidean geometry often leads to a complicated mess whose manipulation is often more time-consuming and surely less pleasant than finding an ingenious synthetic solution. On the other hand, in some cases, the Cartesian point of view leads to unexpectedly short proofs; we have included such proofs in the text when appropriate (see, for example, Theorem~\ref{thm:ap}). In any case, if you do choose to apply coordinates to a problem, there are a few facts that it may help to know, and we summarize them in this section. The following formula for the area of a polygon is called the \emph{surveyor's formula} (or the \emph{shoelace formula}, after the mnemonic device of writing the variables in a $2\times n$ array and multiplying along the diagonals). \begin{fact} Let $(x_{1}, y_{1}), \dots, (x_{n}, y_{n})$ be the vertices of a polygon without self-intersections (but not necessarily convex). Then the signed area of the polygon is given by \[ \pm [(x_{1}y_{2} - y_{1}x_{2}) + \cdots + (x_{n-1}y_{n} - x_{n}y_{n-1}) + (x_{n}y_{1} - x_{1}y_{n})]. \] \end{fact} Note that in some cases, it is convenient to allow \emph{oblique coordinates}, in which the coordinate axes are chosen as two lines which are not necessarily parallel. (Alternatively, one may perform an affine transformation and then use normal Cartesian coordinates.) DEMONSTRATE. \begin{exer} \ii (IMO 1988/1) Consider two coplanar circles of radii $R$ and $r$ ($R > r$) with the same center. Let $P$ be a fixed point on the smaller circle and $B$ a variable point on the larger circle. The line $BP$ meets the larger circle again at $C$. The perpendicular $l$ to $BP$ at $P$ meets the smaller circle again at $A$. (If $l$ is tangent to the circle at $P$ then $A = P$.) \begin{enumerate} \item[(i)] Find the set of values of $BC^2 + CA^2 + AB^2$. \item[(ii)] Find the locus of the midpoint of $BC$. \end{enumerate} \ii Give a coordinate solution for Problem~\ref{ex:samearea}. \ii Give a coordinate proof of Pappus' theorem (Theorem~\ref{thm:pappus}), or of your favorite theorem in projective geometry. \ii \label{ex:appower} Prove that the locus of points $P$ such that the ratio of the powers of $P$ to two fixed circles $\omega_1$ and $\omega_2$ equals a constant $k\neq1$ is a circle. \ii (Greece, 1996) In a triangle $ABC$ the points $D, E, Z, H, \Theta$ are the midpoints of the segments $BC, AD, BD, ED, EZ$, respectively. If $I$ is the point of intersection of $BE$ and $AC$, and $K$ is the point of intersection of $H\Theta$ and $AC$, prove that \begin{enumerate} \ii $AK = 3 CK$; \ii $HK = 3 H\Theta$; \ii $BE = 3 EI$; \ii the area of $ABC$ is 32 times that of $E\Theta H$. \end{enumerate} \ii (Sweden, 1996) Through a point in the interior of a triangle with area $T$, draw lines parallel to the three sides, partitioning the triangle into three triangles and three parallelograms. Let $T_1, T_2, T_3$ be the areas of the three triangles. Prove that \[ \sqrt T = \sqrt{T_1} + \sqrt{T_2} + \sqrt{T_3}. \] \end{exer} %% Appendix: Hints? \chapter{Hints} \label{hints} Here are the author's suggestions on how to proceed on some of the problems. If you find another solution to a problem, so much the better---but it may not be a bad idea to try to find the suggested solution anyway! \begin{itemize} \item[\ref{ex:tetocta}] The octahedron has 4 times the volume of the tetrahedron. What happens when you glue them together at a face? \item[\ref{ex:imo94}] Prove one assertion, then work backward to prove the other. \item[\ref{ex:morley}] Construct two of the intersections of the trisectons complete the equilateral triangle, then show that its third vertex is the third intersection. If you're still stuck, see \cite{bib:cg}. \item[\ref{ex:des2}] Draw 10 points: the 6 vertices of the triangles, the three intersections of corresponding sides, and the intersection of the lines joining two pairs of corresponding vertices. If you relabel these 10 points appropriately, this diagram will turn into a case of the forward direction of Desargues! \item[\ref{ex:threeeq}] Consider the triangle $AB_1C_1$ together with the second intersection of the circumcircles of $AB_1C_2$ and $AB_3C_1$. Show that this figure is congruent to the two analogous figures formed from the other triangles. Do this by rotating $AB_1C_1$ onto $C_2AB_2$ onto $B_3C_3A$ and tracing what happens to the figure. (Or apply Theorem~\ref{thm:back1}.) \item[\ref{ex:sevenpt}] How does $P_2$ depend on $P_1$? \item[\ref{ex:imo85}] There are several solutions to this problem, but none are easy to find. In any case, before anything else, find an extra cyclic quadrilateral. \item[\ref{ex:pol1997}] Work backwards, defining $G$ as the point for which the conclusion holds. Also consider the circumcircle of $CDE$. \item[\ref{ex:simcon}] Find a cyclic hexagon. \item[\ref{ex:postsim}] Use Theorem~\ref{thm:presim}. \item[\ref{ex:convex}] Even using directed angles, the result fails for nonconvex hexagons. Figuring out why may help you determine how to use convexity here. \item[\ref{ex:simsynth}] Given segments $AB$ and $CD$, what conditions must the center $P$ of a spiral similarity carrying $AB$ to $CD$ satisfy? \item[\ref{ex:apcon}] By Ceva and Menelaos, one can show $BA_1/A_1C = BA_2/A_2C$. This means the circle with diameter $A_1A_2$ is a circle of Apollonius with respect to $B$ and $C$. \item[\ref{ex:usamo91}] The center of the circle lies at $C$. \item[\ref{ex:iran97}] The fixed point lies on the circumcircle of $ABC$. \item[\ref{ex:usamo1999}] Show that the point $F$ is the excenter of $ACD$ opposite $A$. \item[\ref{ex:imo1992}] Use homothety. \item[\ref{ex:ninept}] For (a), write the half-turn as the composition of two other homotheties and locate the fixed point. \item[\ref{ex:broc1}] What is the locus of points where one of these equalities holds? \item[\ref{ex:stewart}] Apply the Law of Cosines to the triangles $ABD$ and $ACD$. \item[\ref{ex:inrect}] Use Fact~\ref{midarc}. \item[\ref{ex:circquads}] Show that no two consecutive quadrilaterals can both have incircles. \item[\ref{ex:conc}] Which circles are orthogonal to two concentric circles? \item[\ref{ex:apoll}] Reduce to the case where two of the circles are tangent, then invert. \item[\ref{ex:mop97}] Note that $AB \cdot AB_1 = AC \cdot AC_1$. Also look at the intersection of $OA$ and $B_1 C_1$. \item[\ref{ex:pascconv}] Fix five of the points and compare the locus of sixth points making this condition hold with the conic through the five points. \item[\ref{ex:poncelet}] Apply Pascal's theorem to the hyperbola, using the intersections of the asymptotes with the line at infinity as two of the six points. \item[\ref{ex:imo852}] Draw the circle with diameter $OB$, and show that its common chord with the circle centered at $O$ is concurrent with $KN$ and $AC$. \item[\ref{ex:cencirc}] Find a projective transformation taking the circle to itself but not preserving its center. \item[\ref{ex:chas}] In fact, there exists a degenerate cubic with this property. \item[\ref{ex:rom97a}] Find a projective transformation taking the circle to a circle and the line to infinity. \item[\ref{ex:projstein}] Surprise! Reduce to the case of two concentric circles. \item[\ref{ex:usamo992}] Use the similar triangles formed by the sides and diagonals. \item[\ref{ex:cotident}] Write everything in terms of $\cot A/2$ and the like. Then turn the result into a statement about homogeneous polynomials using the identity \[ \cot \frac{A}{2} + \cot \frac B2 + \cot \frac C2 = \cot \frac{A}{2} \cot \frac B2 \cot \frac C2, \] and solve the result. \item[\ref{ex:quadineq}] Use an affine transformation to make $ABCD$ cyclic, and perform a quadrilateral analogue of the $s-a$ substitution. \item[\ref{ex:appower}] Imitate the proof of Theorem~\ref{thm:ap}. \end{itemize} %% Index %% Bibliography %% Geometry project %% Bibliography % Finally, a few suggestions for further reading. I have already % mentioned \emph{Geometry Revisited}, which is part of the New % Mathematical Library of the Mathematical Association of America; also % in this series are the three volumes of I.M. Yaglom's \emph{Geometric % Transformations}. A comprehensive treatise is \emph{A Course in % Geometry??} by Eves (publisher?). For those whose tastes run more to % the exotic, a delightful introduction to hyperbolic geometry is % \emph{Journey into Geometries} by M. Sved (MAA); for algebraic % geometry, try \emph{Algebraic Geometry} by J. Harris or \emph{Basic % Algebraic Geometry} by I. Shafarevich (both Springer-Verlag), but % beware that some abstract algebra (linear algebra and ring theory) % will become necessary along the way. %% WHat about Coxeter? \begin{thebibliography}{99} \bibitem{bib:bott} O. Bottema, R. \v{Z}. Djordjevi\'c, R. R. Jani\'c, D. S. Mitrinovi\'c, and P.M. Vasi\'c, \textit{Geometric Inequalities}, Wolters-Noordhoff, Groningen, 1968. \bibitem{bib:cg} H.S.M. Coxeter and S.L. Greitzer, \textit{Geometry Revisited}, Mathematical Association of America (NML 19), DATE. \bibitem{bib:eve} H. Eves, \textit{A Course in Geometry} (2 volumes). \bibitem{bib:gp} V. Guillemin and A. Pollock, \textit{Differential Topology}, Prentice-Hall, Englewood Cliffs, 1974. \bibitem{bib:har} J. Harris, \textit{Algebraic Geometry: A First Course}, Springer-Verlag (Graduate Texts in Mathematics 133), New York, 1992. \bibitem{bib:hart} R. Hartshorne, \emph{Algebraic Geometry}, Springer-Verlag (Graduate Texts in Mathematics 52), New York, 1977. \bibitem{bib:max} E. A. Maxwell, \textit{Fallacies in mathematics}, Cambridge University Press, 1959. \bibitem{bib:rcige} D.S. Mitrinovi\'c et al, \textit{Recent Advances in Geometric Inequalities}, Kluwer Academic Publishers, Boston, 1989. \bibitem{bib:sha} I.R. Shafarevich, \textit{Basic Algebraic Geometry} (translated by Miles Reid), Springer-Verlag, New York, 1994. \bibitem{bib:st} J.H. Silverman and J. Tate, \textit{Rational Points on Elliptic Curves}, Springer-Verlag (Undergraduate Texts in Mathematics), New York, 1992. \bibitem{bib:sve} M. Sved, \textit{Journey into Geometries}, MAA ETC. \bibitem{bib:yag} I.M. Yaglom, \emph{Geometric Transformations}, Random House (New Mathematical Library 8), New York, 1973. \end{thebibliography} \end{document}