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होम Euclidean Geometry in Mathematical Olympiads

Euclidean Geometry in Mathematical Olympiads

Evan Chen

यह पुस्तक आपको कितनी अच्छी लगी?

फ़ाइल की गुणवत्ता क्या है?

पुस्तक की गुणवत्ता का मूल्यांकन करने के लिए यह पुस्तक डाउनलोड करें

डाउनलोड की गई फ़ाइलों की गुणवत्ता क्या है?

This is a challenging problem-solving book in Euclidean geometry, assuming nothing of the reader other than a good deal of courage.

Topics covered included cyclic quadrilaterals, power of a point, homothety, triangle centers; along the way the reader will meet such classical gems as the nine-point circle, the Simson line, the symmedian and the mixtilinear incircle, as well as the theorems of Euler, Ceva, Menelaus, and Pascal. Another part is dedicated to the use of complex numbers and barycentric coordinates, granting the reader both a traditional and computational viewpoint of the material. The final part consists of some more advanced topics, such as inversion in the plane, the cross ratio and projective transformations, and the theory of the complete quadrilateral. The exposition is friendly and relaxed, and accompanied by over 300 beautifully drawn figures.

The emphasis of this book is placed squarely on the problems. Each chapter contains carefully chosen worked examples, which explain not only the solutions to the problems but also describe in close detail how one would invent the solution to begin with. The text contains as selection of 300 practice problems of varying difficulty from contests around the world, with extensive hints and selected solutions.

This book is especially suitable for students preparing for national or international mathematical olympiads, or for teachers looking for a text for an honor class.

प्रकाशन:

Mathematical Association of America;athematical Association of America

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                Euclidean Geometry in Mathematical Olympiads With 248 Illustrations  c 2016 by  The Mathematical Association of America (Incorporated) Library of Congress Control Number: 2016933605 Print ISBN: 978-0-88385-839-4 Electronic ISBN: 978-1-61444-411-4 Printed in the United States of America Current Printing (last digit): 10 9 8 7 6 5 4 3 2 1  Euclidean Geometry in Mathematical Olympiads With 248 Illustrations  Evan Chen  Published and Distributed by The Mathematical Association of America  Council on Publications and Communications Jennifer J. Quinn, Chair Committee on Books Jennifer J. Quinn, Chair MAA Problem Books Editorial Board Gail S Nelson, Editor Claudi Alsina Scott Annin Adam H. Berliner Jennifer Roche Bowen Douglas B. Meade John H. Rickert Zsuzsanna Szaniszlo Eric R. Westlund  MAA PROBLEM BOOKS SERIES Problem Books is a series of the Mathematical Association of America consisting of collections of problems and solutions from annual mathematical competitions; compilations of problems (including unsolved problems) specific to particular branches of mathematics; books on the art and practice of problem solving, etc. Aha! Solutions, Martin Erickson The Alberta High School Math Competitions 1957–2006: A Canadian Problem Book, compiled and edited by Andy Liu The Contest Problem Book VII: American Mathematics Competitions, 1995–2000 Contests, compiled and augmented by Harold B. Reiter The Contest Problem Book VIII: American Mathematics Competitions (AMC 10), 2000– 2007, compiled and edited by J. Douglas Faires & David Wells The Contest Problem Book IX: American Mathematics Competitions (AMC 12), 2000–2007, compiled and edited by David Wells & J. Douglas Faires Euclidean Geometry in Mathematical Olympiads, by Evan Chen First Steps for Math Olympians: Using the American Mathematics Competitions, by J. Douglas Faires A Friendly Mathematics Competition: 35 Years of Teamwork in Indiana, edited by Rick Gillman A Gentle Introduction to the American Invitational Mathematics Exam, by Scott A. Annin Hungar; ian Problem Book IV, translated and edited by Robert Barrington Leigh and Andy Liu  The Inquisitive Problem Solver, Paul Vaderlind, Richard K. Guy, and Loren C. Larson International Mathematical Olympiads 1986–1999, Marcin E. Kuczma Mathematical Olympiads 1998–1999: Problems and Solutions From Around the World, edited by Titu Andreescu and Zuming Feng Mathematical Olympiads 1999–2000: Problems and Solutions From Around the World, edited by Titu Andreescu and Zuming Feng Mathematical Olympiads 2000–2001: Problems and Solutions From Around the World, edited by Titu Andreescu, Zuming Feng, and George Lee, Jr. A Mathematical Orchard: Problems and Solutions, by Mark I. Krusemeyer, George T. Gilbert, and Loren C. Larson Problems from Murray Klamkin: The Canadian Collection, edited by Andy Liu and Bruce Shawyer Trigonometry: A Clever Study Guide, by James Tanton The William Lowell Putnam Mathematical Competition Problems and Solutions: 1938– 1964, A. M. Gleason, R. E. Greenwood, L. M. Kelly The William Lowell Putnam Mathematical Competition Problems and Solutions: 1965– 1984, Gerald L. Alexanderson, Leonard F. Klosinski, and Loren C. Larson The William Lowell Putnam Mathematical Competition 1985–2000: Problems, Solutions, and Commentary, Kiran S. Kedlaya, Bjorn Poonen, Ravi Vakil USA and International Mathematical Olympiads 2000, edited by Titu Andreescu and Zuming Feng USA and International Mathematical Olympiads 2001, edited by Titu Andreescu and Zuming Feng USA and International Mathematical Olympiads 2002, edited by Titu Andreescu and Zuming Feng USA and International Mathematical Olympiads 2003, edited by Titu Andreescu and Zuming Feng USA and International Mathematical Olympiads 2004, edited by Titu Andreescu, Zuming Feng, and Po-Shen Loh  MAA Service Center P.O. Box 91112 Washington, DC 20090-1112 1-800-331-1MAA FAX: 1-240-396-5647  Dedicated to the Mathematical Olympiad Summer Program  Contents  Preface xi Preliminaries xiii 0.1 The Structure of This Book . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiii 0.2 Centers of a Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiv 0.3 Other Notations and Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xv  I  Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  1  1 Angle Chasing 1.1 Triangles and Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Cyclic Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 The Orthic Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 The Incenter/Excenter Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Directed Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Tangents to Circles and Phantom Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Solving a Problem from the IMO Shortlist . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  3 3 6 7 9 11 15 16 18  2 Circles 2.1 Orientations of Similar Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Power of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 The Radical Axis and Radical Center . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Coaxial Circles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Revisiting Tangents: The Incenter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 The Excircles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Example Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  23 23 24 26 30 31 32 34 39  3 Lengths and Ratios 3.1 The Extended Law of Sines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Ceva's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Directed Lengths and Menelaus's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 The Centroid and the Medial Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  43 43 44 46 48 vii  viii  Contents  3.5 3.6 3.7  Homothety and the Nine-Point Circle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Example Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56  4 Assorted Configurations 4.1 Simson Lines Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Incircles and Excircles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Midpoints of Altitudes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Even More Incircle and Incenter Configurations . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Isogonal and Isotomic Conjugates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Symmedians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Circles Inscribed in Segments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 Mixtilinear Incircles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  II  59 59 60 62 63 63 64 66 68 70  Analytic Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73  5 Computational Geometry 5.1 Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Ptolemy's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Example Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  75 75 77 79 81 84 91  6 Complex Numbers 95 6.1 What is a Complex Number? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 6.2 Adding and Multiplying Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 6.3 Collinearity and Perpendicularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 6.4 The Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 6.5 Useful Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 6.6 Complex Incenter and Circumcenter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 6.7 Example Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 6.8 When (Not) to use Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 6.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 7 Barycentric Coordinates 119 7.1 Definitions and First Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 7.2 Centers of the Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 7.3 Collinearity, Concurrence, and Points at Infinity . . . . . . . . . . . . . . . . . . . . . . . . . 123 7.4 Displacement Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 7.5 A Demonstration from the IMO Shortlist . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 7.6 Conway's Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 7.7 Displacement Vectors, Continued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 7.8 More Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 7.9 When (Not) to Use Barycentric Coordinates. . . . . . . . . . . . . . . . . . . . . . . . . . . . .142 7.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143  Contents  ix  III Farther from Kansas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 8 Inversion 149 8.1 Circles are Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 8.2 Where Do Clines Go?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .151 8.3 An Example from the USAMO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 8.4 Overlays and Orthogonal Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 8.5 More Overlays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 8.6 The Inversion Distance Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 8.7 More Example Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 8.8 When to Invert . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 8.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 9 Projective Geometry 169 9.1 Completing the Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 9.2 Cross Ratios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 9.3 Harmonic Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 9.4 Apollonian Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 9.5 Poles/Polars and Brocard's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 9.6 Pascal's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 9.7 Projective Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 9.8 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 9.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 10 Complete Quadrilaterals 195 10.1 Spiral Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 10.2 Miquel's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 10.3 The Gauss-Bodenmiller Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 10.4 More Properties of General Miquel Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 10.5 Miquel Points of Cyclic Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 10.6 Example Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 10.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 11 Personal Favorites  209  IV Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 Appendix A: An Ounce of Linear Algebra 215 A.1 Matrices and Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 A.2 Cramer's Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 A.3 Vectors and the Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 Appendix B: Hints  221  Appendix C: Selected Solutions 241 C.1 Solutions to Chapters 1–4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 C.2 Solutions to Chapters 5–7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251  x  Contents  C.3 Solutions to Chapters 8–10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 C.4 Solutions to Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 Appendix D: List of Contests and Abbreviations Bibliography Index About the Author  303 305 307 311  Preface  Give him threepence, since he must make gain out of what he learns. Euclid of Alexandria This book is an outgrowth of five years of participating in mathematical olympiads, where geometry flourishes in great vigor. The ideas, techniques, and proofs come from countless resources—lectures at MOP∗ resources found online, discussions on the Art of Problem Solving site, or even just late-night chats with friends. The problems are taken from contests around the world, many of which I personally solved during the contest, and even a couple of which are my own creations. As I have learned from these olympiads, mathematical learning is not passive—the only way to learn mathematics is by doing. Hence this book is centered heavily around solving problems, making it especially suitable for students preparing for national or international olympiads. Each chapter contains both examples and practice problems, ranging from easy exercises to true challenges. Indeed, I was inspired to write this book because as a contestant I did not find any resources I particularly liked. Some books were rich in theory but contained few challenging problems for me to practice on. Other resources I found consisted of hundreds of problems, loosely sorted in topics as broad as "collinearity and concurrence", and lacking any exposition on how a reader should come up with the solutions in the first place. I have thus written this book keeping these issues in mind, and I hope that the structure of the book reflects this. I am indebted to many people for the materialization of this text. First and foremost, I thank Paul Zeitz for the careful advice he provided that led me to eventually publish this book. I am also deeply indebted to Chris Jeuell and Sam Korsky whose careful readings of the manuscript led to hundreds of revisions and caught errors. Thanks guys! I also warmly thank the many other individuals who made suggestions and comments on early drafts. In particular, I would like to thank Ray Li, Qing Huang, and Girish Venkat for their substantial contributions, as well as Jingyi Zhao, Cindy Zhang, and Tyler Zhu, among many others. Of course any remaining errors were produced by me and I accept sole responsibility for them. Another special thanks also to the Art of Problem Solving fora, ∗ The Mathematical Olympiad Summer Program, which is a training program for the USA team at the International Mathematical Olympiad.  xi  xii  Preface  from which countless problems in this text were discovered and shared. I would also like to acknowledge Aaron Lin, who I collaborated with on early drafts of the book. Finally, I of course need to thank everyone who makes the mathematical olympiads possible—the students, the teachers, the problem writers, the coaches, the parents. Math contests not only gave me access to the best peer group in the world but also pushed me to limits that I never could have dreamed were possible. Without them, this book certainly could not have been written. Evan Chen Fremont, CA  Preliminaries  0.1 The Structure of This Book Loosely, each of the chapters is divided into the following parts. r A theoretical portion, describing a set of related theorems and tools, r One or more examples demonstrating the application of these tools, and r A set of several practice problems. The theoretical portion consists of theorems and techniques, as well as particular geometric configurations. The configurations typically reappear later on, either in the proof of another statement or in the solutions to exercises. Consequently, recognizing a given configuration is often key to solving a particular problem. We present the configurations from the same perspective as many of the problems. The example problems demonstrate how the techniques in the chapter can be used to solve problems. I have endeavored to not merely provide the solution, but to explain how it comes from, and how a reader would think of it. Often a long commentary precedes the actual formal solution, and almost always this commentary is longer than the solution itself. The hope is to help the reader gain intuition and motivation, which are indispensable for problem solving. Finally, I have provided roughly a dozen practice problems at the end of each chapter. The hints are numbered and appear in random order in Appendix B, and several of the solutions in Appendix C. I have also tried to include the sources of the problems, so that a diligent reader can find solutions online (for example on the Art of Problem Solving forums, www.aops.com). A full listing of contest acronyms appears in Appendix D. The book is organized so that earlier chapters never require material from later chapters. However, many of the later chapters approximately commute. In particular, Part III does not rely on Part II. Also, Chapters 6 and 7 can be read in either order. Readers are encouraged to not be bureaucratic in their learning and move around as they see fit, e.g., skipping complicated sections and returning to them later, or moving quickly through familiar material.  xiii  xiv  Preliminaries  0.2 Centers of a Triangle Throughout the text we refer to several centers of a triangle. For your reference, we define them here. It is not obvious that these centers exist based on these definitions; we prove this in Chapter 3. For now, you should take their existence for granted. A  A  G  H  B  C  B  A  A  O  B  C  I  C  B  C  Figure 0.2A. Meet the family! Clockwise from top left: the orthocenter H , centroid G, incenter I , and circumcenter O.  r The orthocenter of ABC, usually denoted by H , is the intersection of the perpendiculars (or altitudes) from A to BC, B to CA, and C to AB. The triangle formed by the feet of these altitudes is called the orthic triangle. r The centroid, usually denoted by G, is the intersection the medians, which are the lines joining each vertex to the midpoint of the opposite side. The triangle formed by the midpoints is called the medial triangle. r Next, the incenter, usually denoted by I , is the intersection of the angle bisectors of the angles of ABC. It is also the center of a circle (the incircle) tangent to all three sides. The radius of the incircle is called the inradius. r Finally, the circumcenter, usually denoted by O, is the center of the unique circle (the circumcircle) passing through the vertices of ABC. The radius of this circumcircle is called the circumradius.  xv  0.3. Other Notations and Conventions  These four centers are shown in Figure 0.2A; we will encounter these remarkable points again and again throughout the book.  0.3 Other Notations and Conventions Consider a triangle ABC. Throughout this text, let a = BC, b = CA, c = AB, and abbreviate A = ∠BAC, B = ∠CBA, C = ∠ACB (for example, we may write sin 12 A for sin 12 ∠BAC). We let s=  1 (a + b + c) 2  denote the semiperimeter of ABC. Next, define [P1 P2 . . . Pn ] to be the area of the polygon P1 P2 . . . Pn . In particular, [ABC] is area of ABC. Finally, given a sequence of points P1 , P2 , . . . , Pn all lying on one circle, let (P1 P2 . . . Pn ) denote this circle. We use  to distinguish a directed angle from a standard angle ∠. (Directed angles are defined in Chapter 1.) Angles are measured in degrees. Finally, we often use the notation AB to denote either the segment AB or the line AB; the use should be clear from context. In the rare case we need to make a distinction we explicitly write out "line AB" or "segment AB". Beginning in Chapter 9, we also use the shorthand AB ∩ CD for the intersection of the two lines AB and CD. In long algebraic computations which have some amount of symmetry, we may use cyclic sum notation as follows: the notation  f (a, b, c) cyc  is shorthand for the cyclic sum f (a, b, c) + f (b, c, a) + f (c, a, b). For example,   cyc  a 2 b = a 2 b + b2 c + c2 a.  Part I Fundamentals  1  CHAPTER  1  Angle Chasing  This is your last chance. After this, there is no turning back. You take the blue pill—the story ends, you wake up in your bed and believe whatever you want to believe. You take the red pill—you stay in Wonderland and I show you how deep the rabbit-hole goes. Morpheus in The Matrix Angle chasing is one of the most fundamental skills in olympiad geometry. For that reason, we dedicate the entire first chapter to fully developing the technique.  1.1 Triangles and Circles Consider the following example problem, illustrated in Figure 1.1A.  Example 1.1. In quadrilateral W XY Z with perpendicular diagonals (as in Figure 1.1A), we are given ∠W ZX = 30◦ , ∠XW Y = 40◦ , and ∠W Y Z = 50◦ . (a) Compute ∠Z. (b) Compute ∠X. W  40◦  30◦  X  50◦  Y  Z Figure 1.1A. Given these angles, which other angles can you compute?  You probably already know the following fact: Proposition 1.2 (Triangle Sum). The sum of the angles in a triangle is 180◦ . 3  4  1. Angle Chasing  As it turns out, this is not sufficient to solve the entire problem, only the first half. The next section develops the tools necessary for the second half. Nevertheless, it is perhaps surprising what results we can derive from Proposition 1.2 alone. Here is one of the more surprising theorems. Theorem 1.3 (Inscribed Angle Theorem). subtends an arc with measure 2∠ACB.  If ∠ACB is inscribed in a circle, then it  Proof. Draw in OC. Set α = ∠ACO and β = ∠BCO, and let θ = α + β. C θ O 2θ A  B  Figure 1.1B. The inscribed angle theorem.  We need some way to use the condition AO = BO = CO. How do we do so? Using isosceles triangles, roughly the only way we know how to convert lengths into angles. Because AO = CO, we know that ∠OAC = ∠OCA = α. How does this help? Using Proposition 1.2 gives ∠AOC = 180◦ − (∠OAC + ∠OCA) = 180◦ − 2α. Now we do exactly the same thing with B. We can derive ∠BOC = 180◦ − 2β. Therefore, ∠AOB = 360◦ − (∠AOC + ∠BOC) = 360◦ − (360◦ − 2α − 2β) = 2θ and we are done.  We can also get information about the centers defined in Section 0.2. For example, recall the incenter is the intersection of the angle bisectors.  Example 1.4. If I is the incenter of ABC then 1 ∠BI C = 90◦ + A. 2  5  1.1. Triangles and Circles  Proof. We have ∠BI C = 180◦ − (∠I BC + ∠I CB) 1 = 180◦ − (B + C) 2 1 = 180◦ − (180◦ − A) 2 1 = 90◦ + A. 2 A  I  B  C  Figure 1.1C. The incenter of a triangle.  Problems for this Section Problem 1.5. Solve the first part of Example 1.1. Hint: 185 Problem 1.6. Let ABC be a triangle inscribed in a circle ω. Show that AC ⊥ CB if and only if AB is a diameter of ω. Problem 1.7. Let O and H denote the circumcenter and orthocenter of an acute ABC, respectively, as in Figure 1.1D. Show that ∠BAH = ∠CAO. Hints: 540 373 A  H  B  O  C  Figure 1.1D. The orthocenter and circumcenter. See Section 0.2 if you are not familiar with these.  6  1. Angle Chasing  1.2 Cyclic Quadrilaterals The heart of this section is the following proposition, which follows directly from the inscribed angle theorem. Proposition 1.8. Let ABCD be a convex cyclic quadrilateral. Then ∠ABC + ∠CDA = 180◦ and ∠ABD = ∠ACD. Here a cyclic quadrilateral refers to a quadrilateral that can be inscribed in a circle. See Figure 1.2A. More generally, points are concyclic if they all lie on some circle. B  B  A  A  C  D  C  D  Figure 1.2A. Cyclic quadrilaterals with angles marked.  At first, this result seems not very impressive in comparison to our original theorem. However, it turns out that the converse of the above fact is true as well. Here it is more explicitly. Theorem 1.9 (Cyclic Quadrilaterals). following are equivalent:  Let ABCD be a convex quadrilateral. Then the  (i) ABCD is cyclic. (ii) ∠ABC + ∠CDA = 180◦ . (iii) ∠ABD = ∠ACD. This turns out to be extremely useful, and several applications appear in the subsequent sections. For now, however, let us resolve the problem we proposed at the beginning. W  30◦ 40◦  40◦  X  50◦  Y  Z Figure 1.2B. Finishing Example 1.1. We discover W XY Z is cyclic.  7  1.3. The Orthic Triangle  Solution to Example 1.1, part (b). Let P be the intersection of the diagonals. Then we have ∠P ZY = 90◦ − ∠P Y Z = 40◦ . Add this to the figure to obtain Figure 1.2B. Now consider the 40◦ angles. They satisfy condition (iii) of Theorem 1.9. That means the quadrilateral W XY Z is cyclic. Then by condition (ii), we know ∠X = 180◦ − ∠Z Yet ∠Z = 30◦ + 40◦ = 70◦ , so ∠X = 110◦ , as desired. In some ways, this solution is totally unexpected. Nowhere in the problem did the problem mention a circle; nowhere in the solution does its center ever appear. And yet, using the notion of a cyclic quadrilateral reduced it to a mere calculation, whereas the problem was not tractable beforehand. This is where Theorem 1.9 draws its power. We stress the importance of Theorem 1.9. It is not an exaggeration to say that more than 50% of standard olympiad geometry problems use it as an intermediate step. We will see countless applications of this theorem throughout the text.  Problems for this Section Problem 1.10. Show that a trapezoid is cyclic if and only if it is isosceles. Problem 1.11. Quadrilateral ABCD has ∠ABC = ∠ADC = 90◦ . Show that ABCD is cyclic, and that (ABCD) (that is, the circumcircle of ABCD) has diameter AC.  1.3 The Orthic Triangle In ABC, let D, E, F denote the feet of the altitudes from A, B, and C. The DEF is called the orthic triangle of ABC. This is illustrated in Figure 1.3A. A  E F  B  H  D  C  Figure 1.3A. The orthic triangle.  It also turns out that lines AD, BE, and CF all pass through a common point H , which is called the orthocenter of H . We will show the orthocenter exists in Chapter 3.  8  1. Angle Chasing  Although there are no circles drawn in the figure, the diagram actually contains six cyclic quadrilaterals. Problem 1.12. In Figure 1.3A, there are six cyclic quadrilaterals with vertices in {A, B, C, D, E, F, H }. What are they? Hint: 91 To get you started, one of them is AF H E. This is because ∠AF H = ∠AEH = 90◦ , and so we can apply (ii) of Theorem 1.9. Now find the other five! Once the quadrilaterals are found, we are in a position of power; we can apply any part of Theorem 1.9 freely to these six quadrilaterals. (In fact, you can say even more—the right angles also tell you where the diameter of the circle is. See Problem 1.6.) Upon closer inspection, one stumbles upon the following.  Example 1.13. Prove that H is the incenter of DEF . Check that this looks reasonable in Figure 1.3A. We encourage the reader to try this problem before reading the solution below. Solution to Example 1.13. Refer to Figure 1.3A. We prove that DH is the bisector of ∠EDF . The other cases are identical, and left as an exercise. Because ∠BF H = ∠BDH = 90◦ , we see that BF H D is cyclic by Theorem 1.9. Applying the last clause of Theorem 1.9 again, we find ∠F DH = ∠F BH. Similarly, ∠H EC = ∠H DC = 90◦ , so CEH D is cyclic. Therefore, ∠H DE = ∠H CE. Because we want to prove that ∠F DH = ∠H DE, we only need to prove that ∠F BH = ∠H CE; in other words, ∠F BE = ∠F CE. This is equivalent to showing that F BCE is cyclic, which follows from ∠BF C = ∠BEC = 90◦ . (One can also simply show that both are equal to 90◦ − A by considering right triangles BEA and CF A.) Hence, DH is indeed the bisector, and therefore we conclude that H is the incenter of DEF . Combining the results of the above, we obtain our first configuration. Lemma 1.14 (The Orthic Triangle). ABC with orthocenter H . Then  Suppose DEF is the orthic triangle of acute  (a) Points A, E, F , H lie on a circle with diameter AH . (b) Points B, E, F , C lie on a circle with diameter BC. (c) H is the incenter of DEF .  Problems for this Section Problem 1.15. Work out the similar cases in the solution to Example 1.13. That is, explicitly check that EH and F H are actually bisectors as well.  9  1.4. The Incenter/Excenter Lemma  Problem 1.16. In Figure 1.3A, show that AEF , BF D, and CDE are each similar to ABC. Hint: 181 A  H  B  C  X  Y  Figure 1.3B. Reflecting the orthocenter. See Lemma 1.17.  Lemma 1.17 (Reflecting the Orthocenter). Let H be the orthocenter of ABC, as in Figure 1.3B. Let X be the reflection of H over BC and Y the reflection over the midpoint of BC. (a) Show that X lies on (ABC). (b) Show that AY is a diameter of (ABC). Hint: 674  1.4 The Incenter/Excenter Lemma We now turn our attention from the orthocenter to the incenter. Unlike before, the cyclic quadrilateral is essentially given to us. We can use it to produce some interesting results. Lemma 1.18 (The Incenter/Excenter Lemma). Let ABC be a triangle with incenter I . Ray AI meets (ABC) again at L. Let IA be the reflection of I over L. Then, (a) The points I , B, C, and IA lie on a circle with diameter I IA and center L. In particular, LI = LB = LC = LIA . (b) Rays BIA and CIA bisect the exterior angles of ABC. By "exterior angle", we mean that ray BIA bisects the angle formed by the segment BC and the extension of line AB past B. The point IA is called the A-excenter∗ of ABC; we visit it again in Section 2.6. Let us see what we can do with cyclic quadrilateral ABLC. ∗ Usually the A-excenter is defined as the intersection of exterior angle bisectors of ∠B and ∠C, rather than as the reflection of I over L. In any case, Lemma 1.18 shows these definitions are equivalent.  10  1. Angle Chasing  A I B  C  L  IA Figure 1.4A. Lemma 1.18, the incenter/excenter lemma.  Proof. Let ∠A = 2α, ∠B = 2β, and ∠C = 2γ and notice that ∠A + ∠B + ∠C = 180 ⇒ α + β + γ = 90◦ . Our first goal is to prove that LI = LB. We prove this by establishing ∠I BL = ∠LI B (this lets us convert the conclusion completely into the language of angles). To do this, we invoke (iii) of Theorem 1.9 to get ∠CBL = ∠LAC = ∠I AC = α. Therefore, ◦  ∠I BL = ∠I BC + ∠CBL = β + α. All that remains is to compute ∠BI L. But this is simple, as ∠BI L = 180◦ − ∠AI B = ∠I BA + ∠BAI = α + β Therefore triangle LBI is isosceles, with LI = LB, which is what we wanted. Similar calculations give LI = LC. Because LB = LI = LC, we see that L is indeed the center of (I BC). Because L is given to be the midpoint of I IA , it follows that I IA is a diameter of (LBC) as well. Let us now approach the second part. We wish to show that ∠IA BC = 12 (180◦ − 2β) = 90◦ − β. Recalling that I IA is a diameter of the circle, we observe that ∠I BIA = ∠I CIA = 90◦ . so ∠IA BC = ∠IA BI − ∠I BC = 90◦ − β. Similar calculations yield that ∠BCIA = 90◦ − γ , as required. This configuration shows up very often in olympiad geometry, so recognize it when it appears!  Problem for this Section Problem 1.19. Fill in the two similar calculations in the proof of Lemma 1.18.  11  1.5. Directed Angles  1.5 Directed Angles Some motivation is in order. Look again at Figure 1.3A. We assumed that ABC was acute. What happens if that is not true? For example, what if ∠A > 90◦ as in Figure 1.5A? H F E  B  A  D  C  Figure 1.5A. No one likes configuration issues.  There should be something scary in the above figure. Earlier, we proved that points B, E, A, D were concyclic using criterion (iii) of Theorem 1.9. Now, the situation is different. Has anything changed? Problem 1.20. Recall the six cyclic quadrilaterals from Problem 1.12. Check that they are still cyclic in Figure 1.5A. Problem 1.21. Prove that, in fact, A is the orthocenter of H BC. In this case, we are okay, but the dangers are clear. For example, when ABC was acute, we proved that B, H , F , D were concyclic by noticing that the opposite angles satisfied ∠BDH + ∠H F B = 180◦ . Here, however, we instead have to use the fact that ∠BDH = ∠BF H ; in other words, for the same problem we have to use different parts of Theorem 1.9. We should not need to worry about solving the same problem twice! How do we handle this? The solution is to use directed angles mod 180◦ . Such angles will be denoted with a  symbol instead of the standard ∠. (This notation is not standard; should you use it on a contest, do not neglect to say so in the opening lines of your solution.) Here is how it works. First, we consider ABC to be positive if the vertices A, B, C appear in clockwise order, and negative otherwise. In particular, ABC = CBA; they are negatives. See Figure 1.5B. Then, we are taking the angles modulo 180◦ . For example, −150◦ = 30◦ = 210◦ . Why on earth would we adopt such a strange convention? The key is that our Theorem 1.9 can now be rewritten as follows.  12  1. Angle Chasing  B 50◦ A  C Figure 1.5B. Here, ABC = 50◦ and CBA = −50◦ .  Theorem 1.22 (Cyclic Quadrilaterals with Directed Angles). Points A, B, X, Y lie on a circle if and only if AXB = AY B. This seems too good to be true, as we have dropped the convex condition—there is now only one case of the theorem. In other words, as long as we direct our angles, we no longer have to worry about configuration issues when applying Theorem 1.9. Problem 1.23. Verify that parts (ii) and (iii) of Theorem 1.9 match the description in Theorem 1.22. We present some more convenient truths in the following proposition. Proposition 1.24 (Directed Angles). have the following rules.  For any distinct points A, B, C, P in the plane, we  Oblivion. AP A = 0. Anti-Reflexivity. ABC = −CBA. Replacement. P BA = P BC if and only if A, B, C are collinear. (What happens when P = A?) Equivalently, if C lies on line BA, then the A in P BA may be replaced by C. Right Angles. If AP ⊥ BP , then AP B = BP A = 90◦ . Directed Angle Addition. AP B + BP C = AP C. Triangle Sum. ABC + BCA + CAB = 0. Isosceles Triangles. AB = AC if and only if ACB = CBA. Inscribed Angle Theorem. If (ABC) has center P , then AP B = 2ACB. Parallel Lines. If AB  CD, then ABC + BCD = 0.  One thing we have to be careful about is that 2ABC = 2XY Z does not imply ABC = XY Z, because we are taking angles modulo 180◦ . Hence it does not make sense to take half of a directed angle.† Problem 1.25. Convince yourself that all the claims in Proposition 1.24 are correct. † Because of this, it is customary to take arc measures modulo 360◦ . We may then write the inscribed angle  is taken mod 360◦ .  This is okay since ABC is taken mod 180◦ but AC theorem as ABC = 12 AC.  13  1.5. Directed Angles  Directed angles are quite counterintuitive at first, but with a little practice they become much more natural. The right way to think about them is to solve the problem for a specific configuration, but write down all statements in terms of directed angles. The solution for a specific configuration then automatically applies to all configurations. Before moving in to a less trivial example, let us finish the issue with the orthic triangle once and for all.  Example 1.26. Let H be the orthocenter of ABC, acute or not. Using directed angles, show that AEH F , BF H D, CDH E, BEF C, CF DA, and ADEB are cyclic. Solution. We know that 90◦ = ADB = ADC 90◦ = BEC = BEA 90◦ = CF A = CF B because of right angles. Then AEH = AEB = −BEA = −90◦ = 90◦ and AF H = AF C = −CF A = −90◦ = 90◦ so A, E, F , H are concyclic. Also, BF C = −CF B = −90◦ = 90◦ = BEC so B, E, F , C are concyclic. The other quadrilaterals have similar stories. We conclude with one final example. Lemma 1.27 (Miquel Point of a Triangle). Points D, E, F lie on lines BC, CA, and AB of ABC, respectively. Then there exists a point lying on all three circles (AEF ), (BF D), (CDE). This point is often called the Miquel point of the triangle. It should be clear by looking at Figure 1.5C that many, many configurations are possible. Trying to handle this with standard angles would be quite messy. Fortunately, we can get them all in one go with directed angles. Let K be the intersection of (BF D) and (CDE) other than D. The goal is to show that AF EK is cyclic as well. For the case when K is inside ABC, this is an easy angle chase. All we need to do is use the corresponding statements with directed angles for each step. We strongly encourage readers to try this themselves before reading the solution that follows. First, here is the solution for the first configuration of Figure 1.5C. Define K as above. Now we just notice that ∠F KD = 180◦ − B and ∠EKD = 180◦ − C. Consequently, ∠F KE = 360◦ − (180◦ − C) − (180◦ − B) = B + C = 180◦ − A and AF EK is cyclic. Now we just need to convert this into directed angles.  14  1. Angle Chasing  A A F  E D  F B B  E C  C  D A F  E B  C  D  Figure 1.5C. The Miquel point, as in Lemma 1.27.  Proof. The first two claims are just F KD = F BD = ABC and DKE = DCE = BCA. We also know that F KD + DKE + EKF = 0 and ABC + BCA + CAB = 0. The first equation represents the fact that the sum of the angles at K is 360◦ ; the second is the fact that the sum of the angles in a triangle is 180◦ . From here we derive that CAB = EKF . But CAB = EAF ; hence EAF = EKF as desired. Having hopefully convinced you that directed angles are natural and often useful, let us provide a warning on when not to use them. Most importantly, you should not use directed angles when the problem only works for a certain configuration! An example of this is Problem 1.38; the problem statement becomes false if the quadrilateral is instead ABDC. You should also avoid using directed angles if you need to invoke trigonometry, or if you need to take half an angle (as in Problem 1.38 again). These operations do not make sense modulo 180◦ .  Problems for this Section Problem 1.28. We claimed that F KD + DKE + EKF = 0 in the above proof. Verify this using Proposition 1.24. Problem 1.29. Show that for any distinct points A, B, C, D we have ABC + BCD + CDA + DAB = 0. Hints: 114 645  15  1.6. Tangents to Circles and Phantom Points  Lemma 1.30. Points A, B, C lie on a circle with center O. Show that OAC = 90◦ − CBA. (This is not completely trivial.) Hints: 8 530 109  1.6 Tangents to Circles and Phantom Points Here we introduce one final configuration and one general technique. First, we discuss the tangents to a circle. In many ways, one can think of it as Theorem 1.22 applied to the "quadrilateral" AABC. Indeed, consider a point X on the circle and the line XA. As we move X closer to A, the line XA approaches the tangent at A. The limiting case becomes the theorem below. Proposition 1.31 (Tangent Criterion). Suppose ABC is inscribed in a circle with center O. Let P be a point in the plane. Then the following are equivalent: (i) P A is tangent to (ABC). (ii) OA ⊥ AP . (iii) P AB = ACB.  P  A  O  B C Figure 1.6A. P A is a tangent to (ABC). See Proposition 1.31.  In the following example we also introduce the technique of adding a phantom point. (This general theme is sometimes also called reverse reconstruction.)  Example 1.32. Let ABC be an acute triangle with circumcenter O, and let K be a point such that KA is tangent to (ABC) and ∠KCB = 90◦ . Point D lies on BC such that KD AB. Show that line DO passes through A. This problem is perhaps a bit trickier to solve directly, because we have not developed any tools to show that three points are collinear. (We will!) But here is a different idea. We define a phantom point D as the intersection of ray AO with BC. If we can show that KD AB, then this will prove D = D, because there is only one point on BC with KD AB. Fortunately, this can be done with merely the angle chasing that we know earlier. We leave it as Problem 1.33. As a hint, you will have to use both parts of Proposition 1.31. We have actually encountered a similar idea before, in our proof of Lemma 1.27. The idea was to let (BDF ) and (CDE) intersect at a point K, and then show that K was on the  16  1. Angle Chasing  K  K A  A  O  B  D  O  C  B  DD  C  Figure 1.6B. Example 1.32, and the phantom point.  third circle as well. This theme is common in geometry. A second example where phantom points are helpful is Lemma 1.45 on page 19. It is worth noting that solutions using phantom points can often (but not always) be rearranged to avoid them, although such solutions may be much less natural. For example, another way to solve Example 1.32 is to show that KAO = KAD. Problem 1.34 is the most common example of a problem that is not easy to rewrite without phantom points.  Problems for this Section Problem 1.33. Let ABC be a triangle and let ray AO meet BC at D . Point K is selected so that KA is tangent to (ABC) and ∠KC = 90◦ . Prove that KD AB. Problem 1.34. In scalene triangle ABC, let K be the intersection of the angle bisector of ∠A and the perpendicular bisector of BC. Prove that the points A, B, C, K are concyclic. Hints: 356 101  1.7 Solving a Problem from the IMO Shortlist To conclude the chapter, we leave the reader with one last example problem. We hope the discussion is instructive.  Example 1.35 (Shortlist 2010/G1). Let ABC be an acute triangle with D, E, F the feet of the altitudes lying on BC, CA, AB respectively. One of the intersection points of the line EF and the circumcircle is P . The lines BP and DF meet at point Q. Prove that AP = AQ. In this problem there are two possible configurations. Directed angles allows us to handle both, but let us focus on just one—say P2 and Q2 . The first thing we notice is the orthic triangle. Because of it we should keep the results of Lemma 1.14 close at heart. Additionally, we are essentially given that ACBP2 is a cyclic  17  1.7. Solving a Problem from the IMO Shortlist  Q2  A  P2 F E P1  H Q1 B  D  C  Figure 1.7A. IMO Shortlist 2010, Problem G1 (Example 1.35).  quadrilateral. Let us see what we can do with that. The conclusion AP2 = AQ2 seems better expressed in terms of angles—we want to show that AQ2 P2 = Q2 P2 A. Now we already know Q2 P2 A, because Q2 P2 A = BP2 A = BCA so it is equivalent to compute AQ2 P2 . There are two ways to realize the next step. The first is wishful thinking—the hope that a convenient cyclic quadrilateral will give us AQ2 P2 . The second way is to have a scaled diagram at hand. Either way, we stumble upon the following hope: might AQ2 P2 F be cyclic? It certainly looks like it in the diagram. How might we prove that AQ2 P2 F is cyclic? Trying to use supplementary angles seems not as hopeful, because this is what we want to use as a final step. However, inscribed arcs seems more promising. We already know AP2 Q2 = ACB. Might we be able to find AF Q2 ? Yes—we know that AF Q2 = AF D and now we are certain this will succeed, because AF D is entirely within the realm of ABC and its orthic triangle. In other words, we have eliminated P and Q. In fact, AF D = ACD = ACB  18  1. Angle Chasing  since AF DC is cyclic. This solves the problem for P2 and Q2 . Because we have been careful to direct all the angles, this automatically solves the case P1 and Q1 as well—and this is why directed angles are useful. It is important to realize that the above is not a well-written proof, but instead a description of how to arrive at the solution. Below is an example of how to write the proof in a contest—one direction only (so without working backwards like we did at first), and without the motivation. Follow along in the following proof with P1 and Q1 , checking that the directed angles work out. Solution to Example 1.35. First, because AP BC and AF DC are cyclic, QP A = BP A = BCA = DCA = DF A = QF A. Therefore, we see AF P Q is cyclic. Then AQP = AF P = AF E = AH E = DH E = DCE = BCA. We deduce that AQP = BCA = QP A which is enough to imply that AP Q is isosceles with AP = AQ. This problem is much easier if Lemma 1.14 is kept in mind. In that case, the only key observation is that AF P Q is cyclic. As we saw above, one way to make this key observation is to merely peruse the diagram for quadrilaterals that appear cyclic. That is why it is often a good idea, on any contest problem, to draw a scaled diagram using ruler and compass—in fact, preferably more than one diagram. This often gives away intermediate steps in the problem, prevents you from missing obvious facts, or gives you something to attempt to prove. It will also prevent you from wasting time trying to prove false statements.  1.8 Problems Problem 1.36. Let ABCDE be a convex pentagon such that BCDE is a square with center O and ∠A = 90◦ . Prove that AO bisects ∠BAE. Hints: 18 115 Sol: p.241 Problem 1.37 (BAMO 1999/2). Let O = (0, 0), A = (0, a), and B = (0, b), where 0 < a < b are reals. Let  be a circle with diameter AB and let P be any other point on . Line P A meets the x-axis again at Q. Prove that ∠BQP = ∠BOP . Hints: 635 100 Problem 1.38. In cyclic quadrilateral ABCD, let I1 and I2 denote the incenters of ABC and DBC, respectively. Prove that I1 I2 BC is cyclic. Hints: 684 569 Problem 1.39 (CGMO 2012/5). Let ABC be a triangle. The incircle of ABC is tangent to AB and AC at D and E respectively. Let O denote the circumcenter of BCI . Prove that ∠ODB = ∠OEC. Hints: 643 89 Sol: p.241 Problem 1.40 (Canada 1991/3). Let P be a point inside circle ω. Consider the set of chords of ω that contain P . Prove that their midpoints all lie on a circle. Hints: 455 186 169  19  1.8. Problems  Problem 1.41 (Russian Olympiad 1996). Points E and F are on side BC of convex quadrilateral ABCD (with E closer than F to B). It is known that ∠BAE = ∠CDF and ∠EAF = ∠F DE. Prove that ∠F AC = ∠EDB. Hints: 245 614 Lemma 1.42. Let ABC be an acute triangle inscribed in circle . Let X be the midpoint  not containing A and define Y , Z similarly. Show that the orthocenter of of the arc BC XY Z is the incenter I of ABC. Hints: 432 21 326 195 A Y Z I B  C X  Figure 1.8A. Lemma 1.42. I is the orthocenter of XY Z.  Problem 1.43 (JMO 2011/5). Points A, B, C, D, E lie on a circle ω and point P lies outside the circle. The given points are such that (i) lines P B and P D are tangent to ω, (ii) P , A, C are collinear, and (iii) DE AC. Prove that BE bisects AC. Hints: 401 575 Sol: p.242 Lemma 1.44 (Three Tangents). Let ABC be an acute triangle. Let BE and CF be altitudes of ABC, and denote by M the midpoint of BC. Prove that ME, MF , and the line through A parallel to BC are all tangents to (AEF ). Hints: 24 335 A  E F  B  M  C  Figure 1.8B. Lemma 1.44, involving tangents to (AEF ).  Lemma 1.45 (Right Angles on Incircle Chord). The incircle of ABC is tangent to BC, CA, AB at D, E, F , respectively. Let M and N be the midpoints of BC and AC, respectively. Ray BI meets line EF at K. Show that BK ⊥ CK. Then show K lies on line MN . Hints: 460 84  20  1. Angle Chasing  A  E F  B  D  K N  C  M  Figure 1.8C. Diagram for Lemma 1.45.  Problem 1.46 (Canada 1997/4). The point O is situated inside the parallelogram ABCD such that ∠AOB + ∠COD = 180◦ . Prove that ∠OBC = ∠ODC. Hints: 386 110 214 Sol: p.242  Problem 1.47 (IMO 2006/1). Let ABC be triangle with incenter I . A point P in the interior of the triangle satisfies ∠P BA + ∠P CA = ∠P BC + ∠P CB. Show that AP ≥ AI and that equality holds if and only if P = I . Hints: 212 453 670 Lemma 1.48 (Simson Line). Let ABC be a triangle and P be any point on (ABC). Let X, Y , Z be the feet of the perpendiculars from P onto lines BC, CA, and AB. Prove that points X, Y , Z are collinear. Hints: 278 502 Sol: p.243 Z A  P  Y  B  X  C  Figure 1.8D. Lemma 1.48; the Simson line.  Problem 1.49 (USAMO 2010/1). Let AXY ZB be a convex pentagon inscribed in a semicircle of diameter AB. Denote by P , Q, R, S the feet of the perpendiculars from Y onto lines AX, BX, AZ, BZ, respectively. Prove that the acute angle formed by lines P Q and RS is half the size of ∠XOZ, where O is the midpoint of segment AB. Hint: 661  1.8. Problems  21  Problem 1.50 (IMO 2013/4). Let ABC be an acute triangle with orthocenter H , and let W be a point on the side BC, between B and C. The points M and N are the feet of the altitudes drawn from B and C, respectively. ω1 is the circumcircle of triangle BW N and X is a point such that W X is a diameter of ω1 . Similarly, ω2 is the circumcircle of triangle CW M and Y is a point such that W Y is a diameter of ω2 . Show that the points X, Y , and H are collinear. Hints: 106 157 15 Sol: p.243 Problem 1.51 (IMO 1985/1). A circle has center on the side AB of the cyclic quadrilateral ABCD. The other three sides are tangent to the circle. Prove that AD + BC = AB. Hints: 36 201  CHAPTER  2  Circles  Construct a circle of radius zero. . . Although it is often an intermediate step, angle chasing is usually not enough to solve a problem completely. In this chapter, we develop some other fundamental tools involving circles.  2.1 Orientations of Similar Triangles You probably already know the similarity criterion for triangles. Similar triangles are useful because they let us convert angle information into lengths. This leads to the power of a point theorem, arguably the most common sets of similar triangles. In preparation for the upcoming section, we develop the notion of similar triangles that are similarly oriented and oppositely oriented. Here is how it works. Consider triangles ABC and XY Z. We say they are directly similar, or similar and similarly oriented, if ABC = XY Z, BCA = Y ZX, and CAB = ZXY. We say they are oppositely similar, or similar and oppositely oriented, if ABC = −XY Z, BCA = −Y ZX, and CAB = −ZXY. If they are either directly similar or oppositely similar, then they are similar. We write ABC ∼ XY Z in this case. See Figure 2.1A for an illustration. Two of the angle equalities imply the third, so this is essentially directed AA. Remember to pay attention to the order of the points.  T1  T2  T3  Figure 2.1A. T1 is directly similar to T2 and oppositely to T3 .  23  24  2. Circles  The upshot of this is that we may continue to use directed angles when proving triangles are similar; we just need to be a little more careful. In any case, as you probably already know, similar triangles also produce ratios of lengths. Proposition 2.1 (Similar Triangles). The following are equivalent for triangles ABC and XY Z. (i) (ii) (iii) (iv)  ABC ∼ XY Z. (AA) ∠A = ∠X and ∠B = ∠Y . (SAS) ∠B = ∠Y , and AB : XY = BC : Y Z. (SSS) AB : XY = BC : Y Z = CA : ZX.  Thus, lengths (particularly their ratios) can induce similar triangles and vice versa. It is important to notice that SAS similarity does not have a directed form; see Problem 2.2. In the context of angle chasing, we are interested in showing that two triangles are similar using directed AA, and then using the resulting length information to finish the problem. The power of a point theorem in the next section is perhaps the greatest demonstration. However, we remind the reader that angle chasing is only a small part of olympiad geometry, and not to overuse it.  Problem for this Section Problem 2.2. Find an example of two triangles ABC and XY Z such that AB : XY = BC : Y Z, BCA = Y ZX, but ABC and XY Z are not similar.  2.2 Power of a Point Cyclic quadrilaterals have many equal angles, so it should come as no surprise that we should be able to find some similar triangles. Let us see what length relations we can deduce. Consider four points A, B, X, Y lying on a circle. Let line AB and line XY intersect at P . See Figure 2.2A. X  B A  A P  Y  P  X  Y  Figure 2.2A. Configurations in power of a point.  A simple directed angle chase gives that P AY = BAY = BXY = BXP = −P XB  B  25  2.2. Power of a Point  and AY P = AY X = ABX = P BX = −XBP . As a result, we deduce that P AY is oppositely similar to P XB. Therefore, we derive PX PA = PY PB or P A · P B = P X · P Y. This is the heart of the theorem. Another way to think of this is that the quantity P A · P B does not depend on the choice of line AB, but instead only on the point P . In particular, if we choose line AB to pass through the center of the circle, we obtain that P A · P B = |P O − r||P O + r| where O and r are the center and radius of ω, respectively. In light of this, we define the power of P with respect to the circle ω by Powω (P ) = OP 2 − r 2 . This quantity may be negative. Actually, the sign allows us to detect whether P lies inside the circle or not. With this definition we obtain the following properties. Theorem 2.3 (Power of a Point).  Consider a circle ω and an arbitrary point P .  (a) The quantity Powω (P ) is positive, zero, or negative according to whether P is outside, on, or inside ω, respectively. (b) If is a line through P intersecting ω at two distinct points X and Y , then P X · P Y = |Powω (P )| . (c) If P is outside ω and P A is a tangent to ω at a point A on ω, then P A2 = Powω (P ). Perhaps even more important is the converse of the power of a point, which allows us to find cyclic quadrilaterals based on length. Here it is. Theorem 2.4 (Converse of the Power of a Point). Let A, B, X, Y be four distinct points in the plane and let lines AB and XY intersect at P . Suppose that either P lies in both of the segments AB and XY , or in neither segment. If P A · P B = P X · P Y , then A, B, X, Y are concyclic. Proof. The proof is by phantom points (see Example 1.32, say). Let line XP meet (ABX) at Y . Then A, B, X, Y are concyclic. Therefore, by power of a point, P A · P B = P X · P Y . Yet we are given P A · P B = P X · P Y . This implies P Y = P Y . We are not quite done! We would like that Y = Y , but P Y = P Y is not quite enough. See Figure 2.2B. It is possible that Y and Y are reflections across point P . Fortunately, the final condition now comes in. Assume for the sake of contradiction that Y = Y ; then Y and Y are reflections across P . The fact that A, B, X, Y are concyclic implies that P lies in both or neither of AB and XY . Either way, this changes if we consider AB and XY . This violates the second hypothesis of the theorem, contradiction.  26  2. Circles  B A Y  P  Y  X  Figure 2.2B. It's a trap! P A · P B = P X · P Y almost implies concyclic, but not quite.  As you might guess, the above theorem often provides a bridge between angle chasing and lengths. In fact, it can appear in even more unexpected ways. See the next section.  Problems for this Section Problem 2.5. Prove Theorem 2.3. Problem 2.6. Let ABC be a right triangle with ∠ACB = 90◦ . Give a proof of the Pythagorean theorem using Figure 2.2C. (Make sure to avoid a circular proof.)  B a C  b  A  Figure 2.2C. A proof of the Pythagorean theorem.  2.3 The Radical Axis and Radical Center We start this section with a teaser.  Example 2.7. Three circles intersect as in Figure 2.3A. Prove that the common chords are concurrent. This seems totally beyond the reach of angle chasing, and indeed it is. The key to unlocking this is the radical axis. Given two circles ω1 and ω2 with distinct centers, the radical axis of the circles is the set of points P such that Powω1 (P ) = Powω2 (P ). At first, this seems completely arbitrary. What could possibly be interesting about having equal power to two circles? Surprisingly, the situation is almost the opposite.  27  2.3. The Radical Axis and Radical Center  Figure 2.3A. The common chords are concurrent.  Theorem 2.8 (Radical Axis). Let ω1 and ω2 be circles with distinct centers O1 and O2 . The radical axis of ω1 and ω2 is a straight line perpendicular to O1 O2 . In particular, if ω1 and ω2 intersect at two points A and B, then the radical axis is line AB. An illustration is in Figure 2.3B.  A O1  O2  O2  O1  B  O1  O2  O1  O2  Figure 2.3B. Radical axes on display.  Proof. This is one of the nicer applications of Cartesian coordinates—we are motivated to do so by the squares of lengths appearing, and the perpendicularity of the lines. Suppose that O1 = (a, 0) and O2 = (b, 0) in the coordinate plane and the circles have radii r1 and r2 respectively. Then for any point P = (x, y) we have Powω1 (P ) = O1 P 2 − r12 = (x − a)2 + y 2 − r12 .  28  2. Circles  Similarly, Powω2 (P ) = O2 P 2 − r22 = (x − b)2 + y 2 − r22 . Equating the two, we find the radical axis of ω1 and ω2 is the set of points P = (x, y) satisfying 0 = Powω1 (P ) − Powω2 (P )     = (x − a)2 + y 2 − r12 − (x − b)2 + y 2 − r22   = (−2a + 2b)x + a 2 − b2 + r22 − r12 which is a straight line perpendicular to the x-axis (as −2a + 2b = 0). This implies the result. The second part is an immediately corollary. The points A and B have equal power (namely zero) to both circles; therefore, both A and B lie on the radical axis. Consequently, the radical axis must be the line AB itself. As a side remark, you might have realized in the proof that the standard equation of a circle (x − m)2 + (y − n)2 − r 2 = 0 is actually just the expansion of Powω ((x, y)) = 0. That is, the expression (x − m)2 + (y − n)2 − r 2 actually yields the power of the point (x, y) in Cartesian coordinates to the circle centered at (m, n) with radius r. The power of Theorem 2.8 (no pun intended) is the fact that it is essentially an "if and only if" statement. That is, a point has equal power to both circles if and only if it lies on the radical axis, which we know much about. Let us now return to the problem we saw at the beginning of this section. Some of you may already be able to guess the ending. Proof of Example 2.7. The common chords are radical axes. Let of ω1 and ω2 , and let 23 be the radical axis of ω2 and ω3 . Let P be the intersection of these two lines. Then P ∈  12  ⇒ Powω1 (P ) = Powω2 (P )  P ∈  23  ⇒ Powω2 (P ) = Powω3 (P )  12  be the radical axis  and  which implies Powω1 (P ) = Powω3 (P ). Hence P ∈ three lines pass through P .  31  and accordingly we discover that all  In general, consider three circles with distinct centers O1 , O2 , O3 . In light of the discussion above, there are two possibilities. 1. Usually, the pairwise radical axes concur at a single point K. In that case, we call K the radical center of the three circles. 2. Occasionally, the three radical axes will be pairwise parallel (or even the same line). Because the radical axis of two circles is perpendicular to the line joining its centers, this (annoying) case can only occur if O1 , O2 , O3 are collinear.  29  2.3. The Radical Axis and Radical Center  It is easy to see that these are the only possibilities; whenever two radical axes intersect, then the third one must pass through their intersection point. We should also recognize that the converse to Example 2.7 is also true. Here is the full configuration. Theorem 2.9 (Radical Center of Intersecting Circles). Let ω1 and ω2 be two circles with centers O1 and O2 . Select points A and B on ω1 and points C and D on ω2 . Then the following are equivalent: (a) A, B, C, D lie on a circle with center O3 not on line O1 O2 . (b) Lines AB and CD intersect on the radical axis of ω1 and ω2 .  A C  P D  B  Figure 2.3C. The converse is also true. See Theorem 2.9.  Proof. We have already shown one direction. Now suppose lines AB and CD intersect at P , and that P lies on the radical axis. Then ±P A · P B = Powω1 (P ) = Powω2 (P ) = ±P C · P D. We need one final remark: we see that Powω1 (P ) > 0 if and only if P lies strictly between A and B. Similarly, Powω2 (P ) > 0 if and only if P lies strictly between C and D. Because Powω1 (P ) = Powω2 (P ), we have the good case of Theorem 2.4. Hence, because P A · P B = P C · P D, we conclude that A, B, C, D are concyclic. Because lines AB and CD are not parallel, it must also be the case that the points O1 , O2 , O3 are not collinear. We have been very careful in our examples above to check that the power of a point holds in the right direction, and to treat the two cases "concurrent" or "all parallel". In practice, this is more rarely an issue, because the specific configuration in an olympiad problem often excludes such pathological configurations. Perhaps one notable exception is USAMO 2009/1 (Example 2.21).  30  2. Circles  To conclude this section, here is one interesting application of the radical axis that is too surprising to be excluded. Proposition 2.10. In a triangle ABC, the circumcenter exists. That is, there is a point O such that OA = OB = OC. Proof. Construct a circle of radius zero (!) centered at A, and denote it by ωA . Define ωB and ωC similarly. Because the centers are not collinear, we can find their radical center O. Now we know the powers from O to each of ωA , ωB , ωC are equal. Rephrased, the (squared) length of the "tangents" to each circle are equal: that is, OA2 = OB 2 = OC 2 . (To see that OA2 really is the power, just use PowωA (O) = OA2 − 02 = OA2 .) From here we derive that OA = OB = OC, as required. Of course, the radical axes are actually just the perpendicular bisectors of the sides. But this presentation was simply too surprising to forgo. This may be the first time you have seen a circle of radius zero; it will not be the last.  Problems for this Section Lemma 2.11. Let ABC be a triangle and consider a point P in its interior. Suppose that BC is tangent to the circumcircles of triangles ABP and ACP . Prove that ray AP bisects BC.  A  P B  C  Figure 2.3D. Diagram for Lemma 2.11.  Problem 2.12. Show that the orthocenter of a triangle exists using radical axes. That is, if AD, BE, and CF are altitudes of a triangle ABC, show that the altitudes are concurrent. Hint: 367  2.4 Coaxial Circles If a set of circles have the same radical axes, then we say they are coaxial. A collection of such circles is called a pencil of coaxial circles. In particular, if circles are coaxal, their centers are collinear. (The converse is not true.) Coaxial circles can arise naturally in the following way.  31  2.5. Revisiting Tangents: The Incenter  Figure 2.4A. Two pencils of coaxial circles.  Lemma 2.13 (Finding Coaxial Circles). Three distinct circles 1 , 2 , 3 pass through a point X. Then their centers are collinear if and only if they share a second common point. Proof. Both conditions are equivalent to being coaxial.  2.5 Revisiting Tangents: The Incenter We consider again an angle bisector. See Figure 2.5A. For any point P on the angle bisector, the distances from P to the sides are equal. Consequently, we can draw a circle centered at P tangent to the two sides. Conversely, the two tangents to any circle always have equal length, and the center of that circle lies on the corresponding angle bisector. A  P  B  C  Figure 2.5A. Two tangents to a circle.  From these remarks we can better understand the incenter. Proposition 2.14. In any triangle ABC, the angle bisectors concur at a point I , which is the center of a circle inscribed in the triangle. Proof. Essentially we are going to complete Figure 2.5A to obtain Figure 2.5B. Let the angle bisectors of ∠B and ∠C intersect at a point I . We claim that I is the desired incenter. Let D, E, F be the projections of I onto BC, CA, and AB, respectively. Because I is on the angle bisector of ∠B, we know that I F = I D. Because I is on the angle bisector of ∠C, we know that I D = I E. (If this reminds you of the proof of the radical center, it should!) Therefore, I E = I F , and we deduce that I is also on the angle bisector of ∠A. Finally, the circle centered at I with radius I D = I E = I F is evidently tangent to all sides.  32  2. Circles  A x  x  E F  I  z  y B  y  D  z  C  Figure 2.5B. Describing the incircle of a triangle.  The triangle DEF is called the contact triangle of ABC. We can say even more. In Figure 2.5B we have marked the equal lengths induced by the tangents as x, y, and z. Considering each of the sides, this gives us a system of equations of three variables y+z=a z+x =b x + y = c. Now we can solve for x, y, and z in terms of a, b, c. This is left as an exercise, but we state the result here. (Here s = 12 (a + b + c).) Lemma 2.15 (Tangents to the Incircle). If DEF is the contact triangle of ABC, then AE = AF = s − a. Similarly, BF = BD = s − b and CD = CE = s − c.  Problem for this Section Problem 2.16. Prove Lemma 2.15.  2.6 The Excircles In Lemma 1.18 we briefly alluded the excenter of a triangle. Let us consider it more completely here. The A-excircle of a triangle ABC is the circle that is tangent to BC, the extension of AB past B, and the extension of AC past C. See Figure 2.6A. The A-excenter, usually denoted IA , is the center of the A-excircle. The B-excircle and Cexcircles are defined similarly and their centers are unsurprisingly called the B-excenter and the C-excenter. We have to actually check that the A-excircle exists, as it is not entirely obvious from the definition. The proof is exactly analogous to that for the incenter, except with the angle bisector from B replaced with an external angle bisector, and similarly for C. As a simple corollary, the incenter of ABC lies on AIA . Now let us see if we can find similar length relations as in the incircle. Let X be the tangency point of the A-excircle on BC and B1 and C1 the tangency points to rays AB and  33  2.6. The Excircles  A E F B  I D  X  C C1  B1 IA Figure 2.6A. The incircle and A-excircle.  AC. We know that AB1 = AC1 and that AB1 + AC1 = (AB + BB1 ) + (AC + CC1 ) = (AB + BX) + (AC + CX) = AB + AC + BC = 2s. We have now obtained the following. Lemma 2.17 (Tangents to the Excircle). excircle, then AB1 = AC1 = s.  If AB1 and AC1 are the tangents to the A-  Let us make one last remark: in Figure 2.6A, the triangles AI F and AIA B1 are directly similar. (Why?) This lets us relate the A-exradius, or the radius of the excircle, to the other lengths in the triangle. This exradius is usually denoted ra . See Lemma 2.19.  Problems for this Section Problem 2.18. Let the external angle bisectors of B and C in a triangle ABC intersect at IA . Show that IA is the center of a circle tangent to BC, the extension of AB through B, and the extension of AC through C. Furthermore, show that IA lies on ray AI . Lemma 2.19 (Length of Exradius). Prove that the A-exradius has length ra =  s r. s−a  Hint: 302  Lemma 2.20. Let ABC be a triangle. Suppose its incircle and A-excircle are tangent to BC at X and D, respectively. Show that BX = CD and BD = CX.  34  2. Circles  2.7 Example Problems We finish this chapter with several problems, which we feel are either instructive, classical, or too surprising to not be shared.  Example 2.21 (USAMO 2009/1). Given circles ω1 and ω2 intersecting at points X and Y , let 1 be a line through the center of ω1 intersecting ω2 at points P and Q and let 2 be a line through the center of ω2 intersecting ω1 at points R and S. Prove that if P , Q, R, and S lie on a circle then the center of this circle lies on line XY .  O3  Q  X  S  O1  R  P  O2  Y Figure 2.7A. The first problem of the 2009 USAMO.  This was actually a very nasty USAMO problem, in the sense that it was easy to lose partial credit. We will see why. Let O3 and ω3 be the circumcenter and circumcircle, respectively, of the cyclic quadrilateral P QRS. After drawing the diagram, we are immediately reminded of our radical axes. In fact, we already know that that lines P Q, RS, and XY concur at a point X, by Theorem 2.9. Call this point H . Now, what else do we know? Well, glancing at the diagram∗ it appears that O1 O3 ⊥ RS. And of course this we know is true, because RS is the radical axis of ω1 an ω3 . Similarly, we notice that P Q is perpendicular to O1 O3 . Focus on O1 O2 O3 . We see that H is its orthocenter. Therefore the altitude from O3 to O1 O2 must pass through H . But line XY is precisely that altitude: it passes through H and is perpendicular to O1 O2 . Hence, O3 lies on line XY , and we are done. Or are we? Look at Theorem 2.9 again. In order to apply it, we need to know that O1 , O2 , O3 are not collinear. Unfortunately, this is not always true—see Figure 2.7B. Fortunately, noticing this case is much harder than actually doing it. We use phantom points. Let O be the midpoint of XY . (We pick this point because we know this is where O3 ∗  And you are drawing large scaled diagrams, right?  35  2.7. Example Problems  P X  O  R  O1  O2  S  Y Q  Figure 2.7B. An unnoticed special case.  must be for the problem to hold.) Now we just need to show that OP = OQ = OR = OS, from which it will follow that O = O3 . This looks much easier. It should seem like we should be able to compute everything using just repeated applications of the Pythagorean theorem (and the definition of a circle). Trying this, OP 2 = OO12 + O1 P 2 = OO12 + (O2 P 2 − O1 O22 ) = OO12 + r22 − O1 O22 . Now the point P is gone from the expression, but the r2 needs togo if we hope to get a symmetric expression. We can get rid of it by using O2 X = r2 = XO 2 + OO22 . OP 2 = OO12 + (O2 X2 + OX2 ) − O1 O22 = OX2 + OO12 + OO22 − O1 O22 =  1 XY 2  2  + OO12 + OO22 − O1 O22 .  This is symmetric; the exact same calculations  with 2 Q, R, and S yield the same results. We conclude OP 2 = OQ2 = OR 2 = OS 2 = 12 XY + OO12 + OO22 − O1 O22 as desired. Having presented the perhaps more natural solution above, here is a solution with a more analytic flavor. It carefully avoids the configuration issues in the first solution.  Solution to Example 2.21. Let r1 , r2 , r3 denote the circumradii of ω1 , ω2 , and ω3 , respectively.  36  2. Circles  We wish to show that O3 lies on the radical axis of ω1 and ω2 . Let us encode the conditions using power of a point. Because O1 is on the radical axis of ω2 and ω3 , Powω2 (O1 ) = Powω3 (O1 ) ⇒ O1 O22 − r22 = O1 O32 − r32 . Similarly, because O2 is on the radical axis of ω1 and ω3 , we have Powω1 (O2 ) = Powω3 (O2 ) ⇒ O1 O22 − r12 = O2 O32 − r32 . Subtracting the two gives (O1 O22 − r22 ) − (O1 O22 − r12 ) = (O1 O32 − r32 ) − (O2 O32 − r32 ) ⇒ r12 − r22 = O1 O32 − O2 O32 ⇒ O2 O32 − r22 = O1 O32 − r12 ⇒ Powω2 (O3 ) = Powω1 (O3 ) as desired. The main idea of this solution is to encode everything in terms of lengths using the radical axis. Effectively, we write down the givens as equations. We also write the desired conclusion as an equation, namely Powω2 (O3 ) = Powω1 (O3 ), then forget about geometry and do algebra. It is an unfortunate irony of olympiad geometry that analytic solutions are often immune to configuration issues that would otherwise plague traditional solutions. The next example is a classical result of Euler. Lemma 2.22 (Euler's Theorem). Let ABC be a triangle. Let R and r denote its circumradius and inradius, respectively. Let O and I denote its circumcenter and incenter. Then OI 2 = R(R − 2r). In particular, R ≥ 2r. The first thing we notice is that the relation is equivalent to proving R 2 − OI 2 = 2Rr. This is power of a point, clear as day. So, we let ray AI hit the circumcircle again at L. Evidently we just need to show AI · I L = 2Rr. This looks much nicer to work with—noticing the power expressions gave us a way to clean up the problem statement, and gives us some structure to work on. We work backwards for a little bit. The final condition appears like similar triangles. So perhaps we may rewrite it as AI 2R = . r IL There are not too many ways the left-hand side can show up like that. We drop the altitude from I to AB as F . Then AI F has the ratios that we want. (You can also drop the foot to  37  2.7. Example Problems  A  F  K  I O  B  C  L Figure 2.7C. Proving Euler's theorem.  AC, but this is the same thing.) All that remains is to construct a similar triangle with the lengths 2R and I L. Unfortunately, I L does not play well in this diagram. But we hope that by now you recognize I L from Lemma 1.18! Write BL = I L. Then let K be the point such that KL is a diameter of the circle. Then KBL has the dimensions we want. Could the triangles in question be similar? Yes: ∠KBL and ∠AF I are both right angles, and ∠BAL = ∠BKL by cyclic quadrilaterals. Hence this produces AI · I L = 2Rr and we are done. As usual, this is not how a solution should be written up in a contest. Solutions should be only written forwards, and without explaining where the steps come from. Solution to Lemma 2.22. Let ray AI meet the circumcircle again at L and let K be the point diametrically opposite L. Let F be the foot from I to AB. Notice that ∠F AI = ∠BAL = ∠BKL and ∠AF I = ∠KBL = 90◦ , so AI KL 2R AI = = = r IF LB LI and hence AI · I L = 2Rr. Because I lies inside ABC, we deduce the power of I with respect to (ABC) is 2Rr = R 2 − OI 2 . Consequently, OI 2 = R(R − 2r). The construction of the diameter appears again in Chapter 3, when we derive the extended law of sines, Theorem 3.1. Our last example is from the All-Russian Mathematical Olympiad, whose solution is totally unexpected. Please ponder it before reading the solution.  Example 2.23 (Russian Olympiad 2010). Triangle ABC has perimeter 4. Points X and Y lie on rays AB and AC, respectively, such that AX = AY = 1. Segments BC and XY intersect at point M. Prove that the perimeter of either ABM or ACM is 2.  38  2. Circles  A  Y M  B  C  X Figure 2.7D. A problem from the All-Russian MO 2010.  What strange conditions have been given. We are told the lengths AX = AY = 1 and the perimeter of ABC is 4, and effectively nothing else. The conclusion, which is an either-or statement, is equally puzzling. Let us reflect the point A over both X and Y to two points U and V so that AU = AV = 2. This seems slightly better, because AU = AV = 2 now, and the "two" in the perimeter is now present. But what do we do? Recalling that s = 2 in the triangle, we find that U and V are the tangency points of the excircle, call it a . Set IA the excenter, tangent to BC at T . See Figure 2.7E. A Y B  M  T  C V  X  U  IA Figure 2.7E. Adding an excircle to handle the conditions.  Looking back, we have now encoded the AX = AY = 1 condition as follows: X and Y are the midpoints of the tangents to the A-excircle. We need to show that one of ABM or ACM has perimeter equal to the length of the tangent. Now the question is: how do we use this? Let us look carefully again at the diagram. It would seem to suggest that in this case, ABM is the one with perimeter two (and not ACM). What would have to be true in order to obtain the relation AB + BM + MA = AU ? Trying to bring the lengths closer  39  2.8. Problems  to the triangle in question, we write AU = AB + BU = AB + BT . So we would need BM + MA = BT , or MA = MT . So it would appear that the points X, M, Y have the property that their distance to A equals the length of their tangents to the A-excircle. This motivates a last addition to our diagram: construct a circle of radius zero at A, say ω0 . Then X and Y lie on the radical axis of ω0 and a ; hence so does M! Now we have MA = MT , as required. Now how does the either-or condition come in? Now it is clear: it reflects whether T lies on BM or CM. (It must lie in at least one, because we are told that M lies inside the segment BC, and the tangency points of the A-excircle to BC always lie in this segment as well.) This completes the solution, which we present concisely below. Solution to Example 2.23. Let IA be the center of the A-excircle, tangent to BC at T , and to the extensions of AB and AC at U and V . We see that AU = AV = s = 2. Then XY is the radical axis of the A-excircle and the circle of radius zero at A. Therefore AM = MT . Assume without loss of generality that T lies on MC, as opposed to MB. Then AB + BM + MA = AB + BM + MT = AB + BT = AB + BU = AU = 2 as desired. While we have tried our best to present the solution in a natural way, it is no secret that this is a hard problem by any standard. It is fortunate that such pernicious problems are rare.  2.8 Problems Lemma 2.24. Let ABC be a triangle with IA , IB , and IC as excenters. Prove that triangle IA IB IC has orthocenter I and that triangle ABC is its orthic triangle. Hints: 564 103 Theorem 2.25 (The Pitot Theorem). Let ABCD be a quadrilateral. If a circle can be inscribed† in it, prove that AB + CD = BC + DA. Hint: 467 A B  D  C  Figure 2.8A. The Pitot theorem: AB + CD = BC + DA.  † The converse of the Pitot theorem is in fact also true: if AB + CD = BC + DA, then a circle can be inscribed inside ABCD. Thus, if you ever need to prove AB + CD = BC + DA, you may safely replace this with the "inscribed" condition.  40  2. Circles  Problem 2.26 (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. Hints: 260 73 409 Sol: p.244 Problem 2.27 (BAMO 2012/4). Given a segment AB in the plane, choose on it a point M different from A and B. Two equilateral triangles AMC and BMD in the plane are constructed on the same side of segment AB. The circumcircles of the two triangles intersect in point M and another point N. (a) Prove that AD and BC pass through point N . Hints: 57 77 (b) Prove that no matter where one chooses the point M along segment AB, all lines MN will pass through some fixed point K in the plane. Hints: 230 654 Problem 2.28 (JMO 2012/1). Given a triangle ABC, let P and Q be points on segments AB and AC, respectively, such that AP = AQ. Let S and R be distinct points on segment BC such that S lies between B and R, ∠BP S = ∠P RS, and ∠CQR = ∠QSR. Prove that P , Q, R, S are concyclic. Hints: 435 601 537 122 Problem 2.29 (IMO 2008/1). Let H be the orthocenter of an acute-angled triangle ABC. The circle A centered at the midpoint of BC and passing through H intersects the sideline BC at points A1 and A2 . Similarly, define the points B1 , B2 , C1 , and C2 . Prove that six points A1 , A2 , B1 , B2 , C1 , and C2 are concyclic. Hints: 82 597 Sol: p.244 Problem 2.30 (USAMO 1997/2). Let ABC be a triangle. Take points D, E, F on the perpendicular bisectors of BC, CA, AB respectively. Show that the lines through A, B, C perpendicular to EF , F D, DE respectively are concurrent. Hints: 596 2 611 Problem 2.31 (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. Hints: 49 159 134 Problem 2.32 (USAMO 1998/2). Let C1 and C2 be concentric circles, with C2 in the interior of C1 . From a point A on C1 one draws the tangent AB to C2 (B ∈ C2 ). Let C be the second point of intersection of ray AB and C1 , and let D be the midpoint of AB. A line passing through A intersects C2 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. Hints: 659 355 482  Problem 2.33 (IMO 2000/1). Two circles G1 and G2 intersect at two points M and N . Let AB be the line tangent to these circles at A and B, respectively, so that M lies closer to AB than N. Let CD be the line parallel to AB and passing through the point M, with C on G1 and D on G2 . Lines AC and BD meet at E; lines AN and CD meet at P ; lines BN and CD meet at Q. Show that EP = EQ. Hints: 17 174  2.8. Problems  41  Problem 2.34 (Canada 1990/3). Let ABCD be a cyclic quadrilateral whose diagonals meet at P . Let W , X, Y , Z be the feet of P onto AB, BC, CD, DA, respectively. Show that W X + Y Z = XY + W Z. Hints: 1 414 440 Sol: p.245 Problem 2.35 (IMO 2009/2). Let ABC be a triangle with circumcenter O. The points P and Q are interior points of the sides CA and AB, respectively. Let K, L, and M be the midpoints of the segments BP , CQ, and P Q, respectively, and let  be the circle passing through K, L, and M. Suppose that the line P Q is tangent to the circle . Prove that OP = OQ. Hints: 78 544 346 Problem 2.36. Let AD, BE, CF be the altitudes of a scalene triangle ABC with circumcenter O. Prove that (AOD), (BOE), and (COF ) intersect at point X other than O. Hints: 553 79 Sol: p.245  Problem 2.37 (Canada 2007/5). Let the incircle of triangle ABC touch sides BC, CA, and AB at D, E, and F , respectively. Let ω, ω1 , ω2 , and ω3 denote the circumcircles of triangles ABC, AEF , BDF , and CDE respectively. Let ω and ω1 intersect at A and P , ω and ω2 intersect at B and Q, ω and ω3 intersect at C and R. (a) Prove that ω1 , ω2 , and ω3 intersect in a common point. (b) Show that lines P D, QE, and RF are concurrent. Hints: 376 548 660 Problem 2.38 (Iran TST 2011/1). In acute triangle ABC, ∠B is greater than ∠C. Let M be the midpoint of BC and let E and F be the feet of the altitudes from B and C, respectively. Let K and L be the midpoints of ME and MF , respectively, and let T be on line KL such that T A BC. Prove that T A = T M. Hints: 297 495 154 Sol: p.246  CHAPTER  3  Lengths and Ratios  As one, who versed in geometric lore, would fain Measure the circle  Dante, The Divine Comedy  3.1 The Extended Law of Sines Aside from angles and similar triangles, one way to relate angles to lengths is through the law of sines. A more thorough introduction to the true power of trigonometry occurs in Section 5.3, but we see that it already proves useful here in our study of lengths. Theorem 3.1 (The Extended Law of Sines). In a triangle ABC with circumradius R, we have a b c = = = 2R. sin A sin B sin C This so-called "extended form" contains the final clause of 2R at the end. It has the advantage that it makes the symmetry more clear (if sina A = 2R is true, then the other parts follow rather immediately). The extended form also gives us a hint of a direct proof: A A X B  O B  O  C X  C Figure 3.1A. Proving the law of sines.  Proof. As discussed above we only need to prove sina A = 2R. Let BX be a diameter of the circumcircle, as in Figure 3.1A. Evidently BXC = BAC. Now consider triangle  43  44  3. Lengths and Ratios  BXC. It is a right triangle with BC = a, BX = 2R, and either ∠BXC = A or ∠BXC = 180◦ − A (depending on whether ∠A is acute). Either way, sin A = sin ∠BXC =  a 2R  and the proof ends here. The law of sines will be used later to provide a different form of the upcoming Ceva's theorem, namely Theorem 3.4.  Problem for this Section Theorem 3.2 (Angle Bisector Theorem). Let ABC be a triangle and D a point on BC so that AD is the internal angle bisector of ∠BAC. Show that AB DB = . AC DC Hint: 417  3.2 Ceva's Theorem In a triangle, a cevian is a line joining a vertex of the triangle to a point on the interior∗ of the opposite side. A natural question is when three cevians of a triangle are concurrent. This is answered by Ceva's theorem. A  A  Y  Y Z  B  Z  P  X  P  C B  X X  C  Figure 3.2A. Three cevians are concurrent as in Ceva's theorem.  Theorem 3.3 (Ceva's Theorem). concur if and only if  Let AX, BY , CZ be cevians of a triangle ABC. They BX CY AZ · · = 1. XC Y A ZB  The proof is by areas: we use the fact that if two triangles share an altitude, the ratio of the areas is the ratio of their bases. This trick is very useful in general. ∗ Some authors permit cevians to land on points on the extensions of the opposite side as well. For this chapter we assume cevians lie in the interior of the triangle unless otherwise specified.  45  3.2. Ceva's Theorem  Proof. Let us first assume the cevians concur at P , and try to show the ratios multiply to 1. Since BAX and XAC share an altitude, as do BP X and XP C, we derive [BAX] [BP X] BX = = . XC [XAC] [XP C] . For example, Now we are going to use a little algebraic trick: if ab = xy , then ab = xy = a+x b+y 4 10 4+10 14 since 6 = 15 , both are equal to 6+15 = 21 . Applying this to the area ratios yields BX [BAX] − [BP X] [BAP ] = = . XC [XAC] − [XP C] [ACP ] But now the conclusion is imminent, since CY [CBP ] AZ [ACP ] = and = YA [BAP ] ZB [CBP ] whence multiplying gives the desired BX · CY · AZ = 1. XC Y A ZB Now how do we handle the other direction? Dead simple with phantom points. Assume AX, BY , CZ are cevians with BX CY AZ · · = 1. XC Y A ZB Let BY and CZ intersect at P , and let ray AP meet BC at X (right half of Figure 3.2A). By our work already done, we know that BX CY AZ · · = 1. X C Y A ZB Thus  BX XC  =  BX , XC  which is enough to imply X = X .  The proof above illustrated two useful ideas—the use of area ratios, and the use of phantom points. As you might guess, Ceva's theorem is extremely useful for showing that three lines are concurrent. It can also be written in a trigonometric form. Theorem 3.4 (Trigonometric Form of Ceva's Theorem). of a triangle ABC. They concur if and only if  Let AX, BY , CZ be cevians  sin ∠BAX sin ∠CBY sin ∠ACZ = 1. sin ∠XAC sin ∠Y BA sin ∠ZCB The proof is a simple exercise—just use the law of sines. With this, the existence of the orthocenter, the incenter, and the centroid are all totally straightforward. For the orthocenter† , we compute sin(90◦ − B) sin(90◦ − C) sin(90◦ − A) = 1. sin(90◦ − C) sin(90◦ − A) sin(90◦ − B) † Actually we need to handle the case where ABC is obtuse separately, since in that case two of the altitudes fall outside the triangle. We develop the necessary generalization in the next section, when we discuss directed lengths in Menelaus's theorem.  46  3. Lengths and Ratios  For the incenter, we compute sin 12 A sin 12 B sin 12 C sin 12 A sin 12 B sin 12 C  = 1.  We could also have used the angle bisector theorem in the standard form of Ceva's theorem, giving cab = 1. bca Finally, for the centroid we have 111 =1 111 and we no longer have to take the existence of our centers for granted!  Problems for this Section Problem 3.5. Show the trigonometric form of Ceva holds. Problem 3.6. Let AM, BE, and CF be concurrent cevians of a triangle ABC. Show that EF BC if and only if BM = MC.  3.3 Directed Lengths and Menelaus's Theorem The analogous form of Ceva's theorem is called Menelaus's theorem, which specifies when three points on the sides of a triangle (or their extensions) are collinear. Theorem 3.7 (Menelaus's Theorem). Let X, Y , Z be points on lines BC, CA, AB in a triangle ABC, distinct from its vertices. Then X, Y , Z are collinear if and only if BX CY AZ · · = −1 XC Y A ZB where the lengths are directed. Here we have introduced ratios of directed lengths. Given collinear points A, Z, B, we AZ is positive if Z lies between A and B, and negative otherwise. (This is say that the ratio ZB much the same idea as the signs we used in defining the power of a point.) We always say explicitly when lengths are taken to be directed. Notice the similarity to Ceva's theorem. The use of −1 instead of 1 is important—for if X, Y , Z each lie in the interiors of the sides, it is impossible for the three to be collinear! Essentially the directed lengths are simply encoding two cases of Menelaus's theorem: when either one or three of {X, Y, Z} lie outside the corresponding side. It is easy to check that the sign of the directed ratio is negative precisely in these cases. There are many proofs of Menelaus's theorem that we leave to other sources. The proof we give shows one direction; if the ratios multiply to −1, then the points are collinear. (The other direction then follows using phantom points.) It is inspired by a proof to Monge's theorem (Theorem 3.22), and it is so surprising that we could not resist including it.  47  3.3. Directed Lengths and Menelaus's Theorem  A  A Z Y  B  B  C  X  C Y  X  Z  Figure 3.3A. The two cases of Menelaus's theorem.  Proof. First, suppose that the points X, Y , Z lie on the sides of the triangle in such a way that BX CY AZ · · = −1. XC Y A ZB Then it is possible to find nonzero real numbers p, q, r for which q BX =− , r XC  r CY =− , p YA  p AZ =− . q ZB  Now we go into three dimensions! Let P be the plane of triangle ABC (this page) and construct point A1 such that A1 A ⊥ P and AA1 = p; we take A1 to be above the page if p > 0 and below the page otherwise. Now define B1 and C1 analogously, so that BB1 = q and CC1 = r.  Figure 3.3B. The 3D proof of Menelaus's theorem.  One can easily check (say, by similar triangles) that the points B1 , C1 , and X are collinear. Indeed, just consider the right triangles C1 CX and B1 BX, and note the ratios of the legs. Similarly, line A1 B1 passes through Z and A1 C1 passes through Y . But now consider the plane Q of the triangle A1 B1 C1 . The intersection of planes P and Q is a line. However, this line contains the points X, Y , Z, so we are done. It also turns out that Ceva's theorem (as well as its trigonometric form) can be generalized using directed lengths. We can write this in the following manner. This should be taken as the full form of Ceva's theorem.  48  3. Lengths and Ratios  Theorem 3.8 (Ceva's Theorem with Directed Lengths). Let ABC be a triangle and X, Y , Z be points on lines BC, CA, AB distinct from its vertices. Then lines AX, BY , CZ are concurrent if and only if AZ BX CY · · =1 ZB XC Y A where the ratios are directed. The condition is equivalent to sin ∠BAX sin ∠CBY sin ∠ACZ =1 sin ∠XAC sin ∠Y BA sin ∠ZCB where either exactly one or exactly three of X, Y , Z lie strictly inside sides BC, CA, AB. Because exactly two altitudes land outside the sides in an obtuse triangle, this generalization lets us complete the proof that the orthocenter exists for obtuse triangles. (What about for right triangles?)  3.4 The Centroid and the Medial Triangle A  z  L  y G  z x B  N y x  M  C  Figure 3.4A. Area ratios on the centroid of a triangle.  We can say even more about the centroid than just its existence by again considering area ratios. Consider Figure 3.4A, where we have added the midpoints of each of the sides (the triangle they determine is called the medial triangle). Notice that 1=  [GMB] BM = MC [CMG]  as discussed before in our proof of Ceva's theorem. Consequently [GMB] = [CMG] and so we mark their areas with an x in Figure 3.4A. We can similarly define y and z. But now, by the exact same reasoning, 1=  BM [AMB] x + 2z = = . MC [CMA] x + 2y  Hence y = z. Analogous work gives x = y and x = z. So that means the six areas of the triangles are all equal.  49  3.5. Homothety and the Nine-Point Circle  In that vein, we deduce [GAB] 2z AG = = = 2. GM [MGB] x This yields an important fact concerning the centroid of the triangle. Lemma 3.9 (Centroid Division). 2 : 1 ratio.  The centroid of a triangle divides the median into a  Just how powerful can area ratios become? Answer: you can build a whole coordinate system around them. See Chapter 7.  3.5 Homothety and the Nine-Point Circle First of all, what is a homothety? A homothety or dilation is a special type of similarity, in which a figure is dilated from a center. See Figure 3.5A. h(A) A O B  C h(B) h(C)  Figure 3.5A. A homothety h with center O acting on ABC.  More formally, a homothety h is a transformation defined by a center O and a real number k. It sends a point P to another point h(P ), multiplying the distance from O by k. The number k is the scale factor. It is important to note that k can be negative, in which case we have a negative homothety. See Figure 3.5B. h(C)  A  h(B) O h(A)  B  C  Figure 3.5B. A negative homothety with center O.  In other words, all this is a fancy special case of similar triangles. Homothety preserves many things, including but not limited to tangency, angles (both vanilla and directed), circles, and so on. They do not preserve length, but they work well enough: the lengths are simply all multiplied by k. Furthermore, given noncongruent parallel segments AB and XY (what happens if AB = XY ?), we can consider the intersection point O of lines AX and BY . This is the  50  3. Lengths and Ratios  center of a homothety sending one segment to the other. (As is the intersection of lines AY and BX—one of these is negative.) As a result, parallel lines are often indicators of homotheties. A consequence of this is the following useful lemma. Lemma 3.10 (Homothetic Triangles). Let ABC and XY Z be non-congruent triangles such that AB XY , BC Y Z, and CA ZX. Then lines AX, BY , CZ concur at some point O, and O is a center of a homothety mapping ABC to XY Z. Convince yourself that this is true. The proof is to take a homothety h with X = h(A) and Y = h(B) and then check that we must have Z = h(C). One famous application of homothety is the so-called nine-point circle. Recall Lemma 1.17, which states that the reflection of the orthocenter over BC, as well as the reflection over the midpoint of BC, lies on (ABC). In Figure 3.5C, we have added in the reflections over the other sides as well. A  H  N9 O  B  C  Figure 3.5C. The nine-point circle.  We now have nine points on (ABC) with center O, the three reflections of H over the sides, the three reflections of H over the midpoints, and the vertices of the triangle themselves. Let us now take a homothety h at H (meaning with center H ) and with scale factor 12 . This brings all the reflections back onto the sides of ABC, while also giving us as an added bonus the midpoints of AH , BH , CH . In addition, O gets mapped to the midpoint of OH , say N9 .  51  3.6. Example Problems  On the other hand homothety preserves circles, so astonishingly enough, these nine points remain concyclic. We even know the center of the circle—it is the image h(O) = N9 , called the nine-point center. We even know the radius! It is just half of the circumradius (ABC). This circles is called the nine-point circle. Lemma 3.11 (Nine-Point Circle). Let ABC be a triangle with circumcenter O and orthocenter H , and denote by N9 the midpoint of OH . Then the midpoints of AB, BC, CA, AH , BH , CH , as well as the feet of the altitudes of ABC, lie on a circle centered at N9 . Moreover, the radius of this circle is half the radius of (ABC). We will see several more applications of homothety in Chapter 4, but this is one of the most memorable. A second application is the Euler line—the circumcenter, orthocenter, and centroid are collinear as well! We leave this famous result as Lemma 3.13; see Figure 3.5D.  Problems for this Section Problem 3.12. Give an alternative proof of Lemma 3.9 by taking a negative homothety. Hints: 360 165 348  A  N9 G  O  H B  C  Figure 3.5D. The Euler line of a triangle.  Lemma 3.13 (Euler Line). In triangle ABC, prove that O, G, H (with their usual meanings) are collinear and that G divides OH in a 2 : 1 ratio. Hints: 426 47 314  3.6 Example Problems Our first example is from the very first European Girl's Math Olympiad. It is a good example of how recognizing one of our configurations (in this case, the reflections of the orthocenters) can lead to an elegant solution.  Example 3.14 (EGMO 2012/7). Let ABC be an acute-angled triangle with circumcircle  and orthocenter H . Let K be a point of  on the other side of BC from A. Let L be the reflection of K across AB, and let M be the reflection of K across BC. Let E be the second point of intersection of  with the circumcircle of triangle BLM. Show that the lines KH , EM, and BC are concurrent.  52  3. Lengths and Ratios  E  L  A  M  H  B  C K  Figure 3.6A. From the first European Girl's Olympiad.  Upon first reading the problem, there are two observations we can make about it. 1. There are a lot of reflections. 2. The orthocenter does not do anything until the last sentence, when it magically appears as the endpoint of one of the concurrent lines. This is a pretty tell-tale sign. What does the orthocenter have to do with reflections and the circumcircle? We need to tie in the orthocenter somehow, otherwise it is just floating in the middle of nowhere. How do we do this? These questions motivate us to reflect H over BC and AB to points HA and HC , corresponding to the reflections of K across these segments. This move incorporates both the observations above. At this point we realize that MHA and H K concur on BC for obvious reasons. So the problem is actually asking to show that HA , M, and E are collinear. This is certainly progress.  L  HC  E A  M  H  B  C K  HA  Figure 3.6B. Adding in some reflections.  53  3.6. Example Problems  At this point we can instead let E be the intersection of HA M with  and try to show that BLE M is concyclic. We are motivated to use phantom points to handle collinearity (since "concyclic" is easier to show), and we choose E because HA and M are simpler— they are just reflections of given points. (Of course, it is probably possible to rewrite the proof without phantom points.) In any case, it suffices to prove LE M = LBM. However, we can compute LBM easily. It is just LBK + KBM = 2 (ABK + KBC) = 2ABC. So now we have reduced this to showing that LE M = 2ABC. Examining the scaled diagram closely suggests that L, HC , and E might be collinear. Is this true? It would sure seem so. To see how useful our conjecture might be, we quickly conjure HC E HA = HC BHA = 2ABC. Thus the desired conclusion is actually equivalent to showing these three points are collinear. Now we certainly want to establish this. How do we go about proving this? Angle chasing seems the most straightforward. It would suffice to prove that LHC B = E HC B; the latter is equal to E HA B, which by symmetry happens to equal BH K. So we need LHC B = BH K—which is clear by symmetry. Solution to Example 3.14. Let HA and HC be the reflections of H across BC and BA, which lie on . Let E be the second intersection of line HA M with . By construction, lines E M and H K concur on BC. First, we claim that L, HC , and E are collinear. By reflections, LHC B = −KH B = MHA B and MHA B = E HA B = E HC B as desired. Now, LE M = HC E HA = HC BHA = 2ABC and LBM = LBK + KBM = 2ABK + 2KBC = 2ABC so B, L, E , M are concyclic. Hence E = E and we are done. The second example is similar in spirit.  Example 3.15 (Shortlist 2000/G3). Let O be the circumcenter and H the orthocenter of an acute triangle ABC. Show that there exist points D, E, and F on sides BC, CA, and AB respectively such that OD + DH = OE + EH = OF + F H and the lines AD, BE, and CF are concurrent.  54  3. Lengths and Ratios  The weird part of this problem is the sum condition. Why OD + DH = OE + EH = OF + F H ? The good news is that at least we can (try to) pick the points D, E, F . So we focus on using this to get rid of the strange condition. Are there any choices of D, E, F that readily satisfy the condition, and which induce concurrent cevians? Having a ruler and compass is important h

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