A Decade of the Berkeley Math Circle: The American Experience [1] 0821846833, 9780821846834

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Mathematical Circles Library

A Decade of the Berkeley Math Circle The American Experience, Volume I Zvezdelina Stankova Tom Rike Editors

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Advisory Board for the MSRI/Mathematical Circles Library Zuming Feng Tony Gardiner Kiran Kedlaya Nikolaj N. Konstantinov Silvio Levy Walter Mientka Bjorn Poonen

Alexander Shen Tatiana Shubin (Chair) Zvezdelina Stankova Ravi Vakil Ivan Yashchenko Paul Zeitz Joshua Zucker

2000 Mathematics Subject Classification. Primary 00–01, 00A07; Secondary 00A08.

For additional information and updates on this book, visit www.ams.org/bookpages/mcl-1

Library of Congress Cataloging-in-Publication Data A decade of the Berkeley Math Circle : the American experience / Zvezdelina Stankova, Tom Rike, editors. p. cm. — (MSRI mathematical circles library ; v. 1–) Includes bibliographical references and index. ISBN 978-0-8218-4683-4 (alk. paper) 1. Mathematics—Study and teaching (Middle school)—California—San Francisco Bay Area. 2. Mathematics—Study and teaching (Secondary)—California—San Francisco Bay Area. 3. Berkeley Math Circle. I. Stankova, Zvezdelina, 1969– II. Rike, Tom, 1943– III. Mathematical Sciences Research Institute (Berkeley, Calif.) QA13.5.C22S363 2008 510.71
27946—dc22

2008030521

Copying and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy a chapter for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for such permission should be addressed to the Acquisitions Department, American Mathematical Society, 201 Charles Street, Providence, Rhode Island 02904-2294, USA. Requests can also be made by e-mail to [email protected]. c 2008 by Zvezdelina Stankova and Thomas Rike
Printed in the United States of America. ∞ The paper used in this book is acid-free and falls within the guidelines


established to ensure permanence and durability. Visit the AMS home page at http://www.ams.org/ Visit the MSRI home page at http://www.msri.org/ 10 9 8 7 6 5 4 3 2

18 17 16 15 14 13

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Contents Foreword

vii

Introduction 1. Top-Tier Math Circles 2. Why, What, and for Whom? 3. Notation and Technicalities 4. The Art of Being a Mathematician and Problem Solving 5. Acknowledgments

ix ix xi xv xvi xvii

Session 1. Inversion in the Plane. Part I 1. Why Inversion? Motivation 2. Inversion as a Transformation 3. Definition of Inversion 4. Basic Properties of Inversion 5. Problem Solving Techniques with Inversion 6. Solving Our First Problems with Inversion 7. How Does Inversion Affect Distances? 8. Proof of Ptolemy’s Theorem 9. How Are Inversion Problems Created? 10. Hints and Solutions to Selected Problems

1 1 3 4 5 9 10 15 17 19 22

Session 2. Combinatorics. Part I 1. Two Counting Conundrums 2. Multiplication, Menus, and Encoding 3. Addition and Partition 4. Division: A Cure for Uniform Overcounting 5. Balls in Urns and Other Applications 6. Sororities of Numbers: A Promise Fulfilled

25 25 26 31 35 40 44

Session 3. Rubik’s Cube. Part I 1. Getting Started and Some Notation 2. Encoding the Rubik’s Cube Mathematically 3. Some Basic Features of Rubik’s Cube Moves 4. Visualizing Permutations 5. Cycle Structure of Permutations

47 47 49 51 55 57

iii

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iv

CONTENTS 6. Applications of Cycle Structure to the Cube 7. Conclusions

59 62

Session 4. Number Theory. Part I 1. Wearing the Crown of Mathematics 2. Remainders: Where It All Began 3. Congruences in Z 4. Properties of Congruences 5. Remainders Learn to Ride a Bike 6. Twists in the Brute-Force Approach 7. Pairs and Divisibility 8. Hints and Solutions to Selected Problems

63 63 67 70 72 75 78 81 84

Session 5. A Few Words About Proofs. Part I 1. Why Prove Things? 2. Proofs versus Non-proofs 3. Proof by Contradiction 4. Proofs of Possibility and Impossibility 5. Some Problems Need Two Proofs! 6. Hints and Solutions to Selected Problems

87 87 88 91 93 97 100

Session 6. Mathematical Induction 1. Examples and Conjectures 2. Mathematical Induction and Proof 3. Mathematical Induction in Action 4. Strong Induction 5. Mathematical Induction in Other Areas 6. A Word of Caution 7. Hints and Solutions to Selected Problems

103 103 106 108 115 120 122 123

Session 7. Mass Point Geometry 1. Introduction 2. Definition and Properties of Mass Points 3. Fundamental Examples 4. Angle Bisectors and Altitudes 5. Areas, Space, and Splitting Masses 6. Ceva, Menelaus, and Associativity of Addition 7. Examples of Contest Problems 8. History and Sources 9. Hints and Solutions to Selected Problems

127 127 130 134 136 139 142 145 151 151

Session 8. More on Proofs. Part II 1. Proof by Induction Again 2. Extremes Are Naturally Demanding 3. The Pigeonhole Principle 4. Hints and Solutions to Selected Problems

155 155 162 165 175

Session 9. Complex Numbers. Part I 1. A Problem from Geometry 2. Some History

179 179 180

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CONTENTS 3. 4. 5. 6. 7. 8.

v

Complex Numbers via Geometry Basic Operations on Complex Numbers Complex Multiplication Another Form of Complex Numbers Summary: What Have We Learned? Hints and Solutions to Selected Problems

183 183 187 192 194 196

Session 10. Stomp. Games with Invariants 1. Warm-up Classics 2. Invariants with Numbers 3. Stomp 4. Tilings and More Invariants 5. Escape of the Clones 6. Hints and Solutions to Selected Problems

203 203 206 211 215 216 219

Session 11. Favorite Problems at BMC. Part I 1. The Search for the Missing Circle 2. Inscribed and Central Angles 3. “Going-in-Circles” 4. Going “off on a Tangent” Leads Straight to the Point! 5. When Phantom Circles Team up with Tangents 6. Constructing Cyclic Trapezoids “out of Thin Air” 7. Hints and Solutions to Selected Problems

225 225 227 229 233 235 239 243

Session 12. Monovariants. Part I 1. China Shops and Chocolate Bars 2. Walking around a Mansion 3. Finite versus Infinite 4. Monovariant Teamwork 5. Women and Men Walking around the Mansion 6. Non-numerical Monovariants 7. Mansion Appendix for the Advanced Reader 8. Hints and Solutions to Selected Problems

249 249 251 254 257 260 263 268 279

Epilogue 1. What Comes from Within 2. The Culture of Circles 3. Eastern European vs. USA Math Circles 4. History and Power 5. Does the U.S. Need Top-Tier Math Circles?

285 285 286 287 290 294

Symbols and Notation

299

Abbreviations

301

Biographical Data

303

Bibliography

309

Credits

313

Index

315

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To Zvezda’s husband, Dmitri, and to Tom’s wife, Peggy, for making this book possible, for their infinite patience and love over the course of eighteen months of hard work, and . . . to the instructors of the Berkeley Math Circle for donating their time and effort over the last decade, for leading inspiring sessions full of mathematical challenges and wonders, and for sharing their passion for mathematics with the circlers. Zvezda Stankova and Tom Rike

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Foreword The Mathematical Sciences Research Institute (MSRI) came into being in Berkeley, California, in 1982, the realization of the vision of Shiing-Shen Chern, Calvin Moore, and Isadore Singer. These founders sought to establish a research institute with a constantly renewed faculty that would pursue research in the mathematical sciences in a manner forever fresh and young. Academic institutions dedicate themselves to research, teaching, and service in varying degrees, and MSRI’s main thrust is research. But from the beginning, there has been a desire and a commitment to take beautiful mathematics to the street and serve a hungry public. In 1998 a confluence of people and experiences at MSRI brought Math Circles to the San Francisco Bay Area: David Eisenbud and Hugo Rossi, the Institute’s Director and Deputy Director, were deeply committed to sharing the beauty of mathematics with the outside world; Zvezdelina Stankova was a postdoctoral scholar at MSRI that year and brought years of experience in and passion for math circles from her native Bulgaria; Paul Zeitz, Professor of Mathematics at the University of San Francisco and coach of the winning American team at a recent International Mathematics Olympiad, added his skills to the team; Tatiana Shubin, Professor of Mathematics at San Jose State University, brought her passion and experience of math circles from the former Soviet Union; and Tom Davis, an applied mathematician at Silicon Graphics, joined forces to offer after-school math programs that were advanced, challenging, fun, and beautiful. We have come to know many others since, including Mark Saul in New York and the Kaplans in Boston, who share our passion for math circles. And the students came, first in Berkeley and San Jose, then in San Francisco, Stanford, Oakland, and Davis, with open minds and willing hearts to learn about mathematics of which they previously could not have dreamt. They worked on problem solving with faculty who had a depth of understanding and a love of mathematics beyond anything they had ever encountered. There were no attendance lists, no tests, no being forced to do anything. The beauty of the mathematics attracted them to want more and more. We vii

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FOREWORD

have come to learn that many students, girls and boys, of different socioeconomic and racial backgrounds thrive in math circles and come to love the experience. Both professors and students find in math circles a situation that is rare in classroom mathematics instruction. Math circles are voluntary, extracurricular, after-school programs. The students who are there are much less likely to be motivated by the need to satisfy an academic requirement, prepare for a career, or enhance a resumé. They are, for the most part, there because they love mathematics. The teachers encounter students who are willing and hungry to learn, while the students encounter teachers with expertise and enthusiasm far beyond the usual classroom experience. Teachers and students look forward with anticipation to the next meeting. The MSRI Mathematical Circles Library contains books that share the experiences we have gathered from around the world and make them available to you. If you are interested in starting a circle, or persuading someone else to lead a circle, or adding to the circle you are in, these books and the MSRI website http://www.mathcircles.org will be excellent resources. We are deeply indebted to the Templeton Foundation, the Moscow Center for Continuous Mathematics Education, the American Mathematical Society for their support, and to John Ewing and Edward Dunne for shepherding this project. Robert L. Bryant Director Mathematical Sciences Research Institute

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Introduction “The Berkeley Math Circle was really critical in my development. It was the best method available not only to get a flow of mathematical ideas and problems to think about each week but also to meet other interested students and professional mathematicians from all over the Bay Area. You get stimulation from exchanging ideas with other people that you don’t get from reading books at home. I can also testify to the usefulness of studying mathematics even for students who don’t plan on doing it as a career. For someone who wants to go into, say, law, policy analysis, philosophy, economics, or computer science, the kind of logical, abstract thinking that mathematics develops is really the best preparation. I realize that the Circle is most interested in attracting students whose lifelong passion is for mathematics, but it also helps others along the way.” Gabriel Carroll, BMC alumnus Perfect IMO ’01 score Four-time Putnam Fellow Ph.D. student in economics, MIT

1. Top-Tier Math Circles1 This book is based on material from a dozen of the 320 sessions of the Berkeley Math Circle (BMC), held over the past 10 years. In recent discussions, parents have described BMC as a top-tier math circle, calling for the following two definitions. 1.1. Math circles are weekly math programs that attract middle and high school students to mathematics by exposing them to intriguing and intellectually stimulating topics, rarely encountered in classrooms. Math circles vary in their organization, styles of sessions, and goals. But they all have one thing in common: to inspire in students an understanding of and a lifelong love for mathematics. 1 Excerpts from [98] by Marc Whitlow and Mike Breen (BMC Parents), Zvezdelina Stankova (BMC Director), and Tatiana Shubin (SJMC Director) ix

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INTRODUCTION

1.2. Top-tier math circles prepare our best young minds for their future roles as mathematics leaders. Sessions are taught by accomplished mathematicians and explore advanced mathematical areas. They provide an educational opportunity for top pre-college mathematics students, not offered in any other setting in the U.S. education system. In addition to learning advanced mathematics topics, students are taught the technical writing skills needed to convey the solutions of complex problems. As an example of a top-tier math circle, the Berkeley Math Circle is fashioned after the leading models in Eastern Europe, where math circles originated over a century ago. A typical weekly session has 20 to 30 students and lasts 2 hours. Like top-tier universities, BMC • challenges students with beautiful, difficult mathematical theories, • introduces them to powerful problem solving techniques, • constantly provokes deep thought, and • inspires the creation of original ideas. Topics covered at BMC include combinatorics, graph theory, linear algebra, geometric transformations, recursive sequences, series, set theory, group theory, number theory, elliptic curves, algebraic geometry, applications to computer science, natural sciences, economics, and many more. Each topic is taught by an expert in the field who has the ability to challenge the students and support them as they attempt to meet these challenges. All problems require students to come up with mathematical proofs. Proofs put forward by the students are not always the most eloquent. Only an accomplished mathematician can understand where a student might be heading in his/her proof and offer assistance through this challenge.2 The sessions are fast paced and intellectually demanding. It is hard to convey just how advanced this subject matter is without actually attending a session, but comparable levels can be found in advanced undergraduate and beginning graduate courses. The Monthly Contests (MC)3 at BMC can also convey the depth of the material. These are take-home exams of five hard, thought-provoking problems, requiring independent research. They develop not only advanced understanding, but also technical writing skills: the students must describe on paper, convincingly and without gaps, how they solved a problem. This is a fundamental skill and key to making intellectual property contributions; it is a unique feature of the top-tier math circles, not found in middle or high schools, where students are taught to meet state standards on questions that take less than a minute to answer. In contrast, monthly contest problems may take the best students hours or days of concentrated thought. Only a few participants are capable of solving all the problems; yet, through the attempt everyone learns about the real world of mathematical research. 2 For examples of noteworthy past and present instructors who have brought their world expertise to BMC, see the Epilogue. 3 Session 11 on circle geometry is based on the monthly contests.

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2. WHY, WHAT, AND FOR WHOM?

xi

1.3. The next generation of math leaders. The students of BMC come from a variety of socio-economic and ethnic backgrounds. The proportion of female to male students is approximately 2:3. This is an amazingly high fraction considering the trend of other high-level math programs, which are “male-dominated” or “male-only”. Excellent role models for the female students are provided by the three female directors of the top-tier math circles in Berkeley [8], San Jose [80], and Los Angeles [57]; but perhaps even more important to the students are the outstanding lectures given by nine female professors and graduate students. Currently, BMC does not actively recruit participants. Students and their parents find out about the circles in a variety of ways: from the circle’s web site, http://mathcircle.berkeley.edu/, by word of mouth, through a local university, and in publications. There is no formal selection process, and the math circle doors are open to any student. Needless to say, the BMC students are usually years ahead of their peers: they often complete most of high school mathematics by age 13 (8th grade), some take many college math major courses by the time they graduate from high school, and a few of the top circlers venture into graduate courses and serious mathematical research even before entering a university. The accomplishments of students who have benefited from BMC can be measured in many ways. For example, a number of these students have gone on to win International Math Olympiad medals and Putnam awards, and the majority have been admitted to top-tier universities. BMC and the other top-tier math circles not only produce highly accomplished students – they produce and train the next generation of leaders in mathematics.4

2. Why, What, and for Whom? Running the Berkeley Math Circle for ten years has taught us many lessons about math education in the U.S. and has even helped us understand better our own childhood education and origins of our passion for mathematics. To share this experience with you, the reader, is the purpose of this book : • to present you with beautiful theories, problem solving techniques, and mathematical insights; • to provide you with an abundance of exercises and problems to work on and with ready materials for math circle sessions. 2.1. The middle or high school student who is interested in expanding his/her math horizon and going well beyond anything that the regular math classroom can offer, who is brave enough to tackle non-trivial math ideas and work on hard problems for hours, who loves challenges and is motivated to overcome them: this is the ideal reader of the book. 4 To learn about the need for top-tier math circles, we direct the reader to the Epilogue.

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xii

INTRODUCTION

Don’t confuse the above description with “top” or “brilliant” student: you will never know if you are talented in math unless you give it a try. And you may be pleasantly surprised by what you find out: that mathematics is a whole lot more than “adding fractions,” “algebraic manipulations,” or “endless quadratic equations” in homework assignments. You will discover that calculus is not the “pinnacle” of mathematical knowledge (as thought by the general public): it is only one of many beginnings, part of the subject of real analysis. Indeed, other wonderful topics are awaiting you here: • inversion in geometry and circle geometry, • abstract algebra via Rubik’s cube and number theory, • a mass point “hybrid” between algebra and geometry, • combinatorics and complex numbers, • game theory via invariants and monovariants, and, of course, • plenty of proof methods and problem solving techniques. 2.2. Prerequisites. To read the book comfortably, you do not need to have calculus under your belt. However, familiarity with basic geometry and algebra concepts and theorems will definitely be helpful, e.g., lines, circles, triangles, rectangles, trapezoids, and quadrilaterals in general; similarity criteria for triangles and the Pythagorean Theorem; equal alternate interior angles for parallel lines and bisecting diagonals in a parallelogram; integers, divisibility and remainders; operations on fractions and real numbers, intervals and sets of numbers; and manipulations of algebraic expressions written with letters. Occasionally, functions will play an important role; hence having studied some basic (pre-calculus) examples will not hurt, e.g., linear and quadratic functions, polynomials, exponential and trigonometric functions, as well as their graphs. The above concepts will be re-introduced via examples in the book. But if you feel that you need more solid background, we direct you to several wonderful books that should be part of any budding mathematician’s library: • Geometry, Book 1 by Kiselev (cf. [51]), • Algebra by Gelfand and Shen (cf. [35]), • Functions and Graphs, The Method of Coordinates, and Trigonometry by Gelfand et al. (cf. [33, 32, 34]), • for the older reader, 103 Trigonometry Problems from the Training of the USA IMO Team by Andreescu and Feng (cf. [3]). 2.3. The middle or high school teacher who wishes to start a math circle in his/her school or teach a specially designed problem solving class will find this book invaluable. Most of the topics are introductory and independent of each other, and hence they can be pursued in just about any order during the first year of a new math circle. Yet, as you read through the sessions, you will notice how some of them also contain harder material suitable for intermediate level and the second year of a math circle. Other than the several open problems sprinkled throughout the text, the only truly advanced Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

2. WHY, WHAT, AND FOR WHOM?

xiii

part is the Mansion Appendix in the last session: it is original research by BMCer Evan O’Dorney and can be left for the die-hards. Running a math circle, especially for a teacher, is a hard task. But it is possible. In the 1960’s, Tom Rike (an editor for this book and a veteran high school math teacher) was working on his master’s degree. While browsing in the library one day, he ran across The USSR Olympiad Problem Book (cf. [82]). It contained problems written for talented 7th –10th graders; yet, he could not solve any of these “elementary” problems. In his own words: “My abstract algebra had been too abstract, and I did not have the concrete examples that I needed. I never took a class in number theory because it sounded too elementary. I had developed the real number system starting from the Peano axioms, but I didn’t really understand the fundamentals of the natural numbers, prime numbers. This was an epiphany for me. I felt as though I had been challenged by some force outside me and did not know how to respond.”

For the next 30 years Tom studied olympiad problem solving, first on his own, then through workshops and math circles in the SF Bay Area. He ran his own math circle at Oakland High School and gave talks at just about all other circles around. Even though at times he was only “a few pages” ahead of the students, he kept on learning and teaching problem solving because working on math circles had come to be a large part of his life: “Although I have not attained my goal of becoming a true olympiad problem solver, the journey I have made in pursuit of this goal has been one of the most rewarding endeavors in my life.”

Hence, a word to the middle and high school teachers: keep on reading the book, despite moments of difficulty or confusion. For the motivated, persevering, and caring teacher, there will come a time when he/she will look back at the material here, smile, and effortlessly deliver it to the students at his/her own math circle. Truly gratifying. 2.4. Proofs in particular. The only topic that has been deliberately postponed till mid-book is Proofs in Sessions 5, 6, and 8. That proofs are important goes without question in Galileo’s mind: “It appears to me that those who rely simply on the weight of authority to prove any assertion, without searching out the arguments to support it, act absurdly. I wish to question freely and to answer freely without any sort of adulation. That well becomes any who are sincere in the search for truth.”

We shall learn a variety of proof methods: by contradiction, Pigeonhole Principle, and induction; by counterexample, example, or general argument; using invariants or monovariants, and others. The first session already calls for the method of inversion coupled with direct proofs, extra constructions, and contradiction. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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INTRODUCTION

The delay in formally introducing proofs is justified by our belief that humans and machines learn in essentially different ways. If we teach a machine how to do math problems, the only way for it to be perfectly “happy” is to receive the information in a linear fashion, i.e., to be fed first several chapters strictly on proofs and then to proceed with examples and more sophisticated problems from various math disciplines. On the other hand, human beings are curious and somewhat contradictory creatures: even though eventually we would like everything to be molded into an elegant, logically stable “tower” of knowledge, we also insist on seeing justification before we plunge into big enterprises. A young student, especially, needs to know why a proof is important and hence be motivated to search for proofs, instead of being turned off (as so often happens in proof-classes in school) by the stricture of long two-column proofs of relatively easy but “boring” statements. And thus, we opted to expose the reader first to four wonderful and intellectually stimulating topics, inversion in the plane, combinatorics, Rubik’s cube, and number theory. These sessions don’t avoid proof techniques; yet, they mainly concentrate on a specific mathematical topic. Hopefully, this way the beginner will be drawn to the beauty of each subject, will accumulate interest, ideas, see examples, and be convinced that without proofs we can’t really claim something is true. After such an initial level of mathematical maturity is acquired and the reader is “won over” to the math problem solving “front”, we will formally welcome proofs in Session 5. Teachers who plan on using the book in circles or classes should consider delivering the material in this order, postponing formal proofs until students have been exposed to at least two or three intriguing mathematical topics. 2.5. The parent of a middle or high school student is also among our intended audience; in fact, parents are probably the most important readers because without their support and enthusiasm, without them bringing and encouraging their children, there would hardly be any top-tier math circles in the U.S. Hence, if you are among those parents or if you are a parent new to the math circle movement, this book will provide a very strong beginning for your child. And for you as well. The back row of the BMC classroom is always packed with adults: mostly parents and teachers. A strict rule reigns in this last row: you can never speak or interfere during sessions. Yet, as you can imagine, any rule is there to be broken: it is parents that become at times exulted by the material presented and cannot keep quiet, shouting involuntarily an answer to one or another (unintentionally) provocative question. As a parent, you can do three things with this book: give it to your child (but make sure that he/she has the necessary background – see the recommended basic books); learn from it and teach your child; or give it to his/her math teacher and encourage the founding of a school-based math circle. Whatever path you choose to follow, it will eventually benefit your child and possibly a larger group of classmates. In any case, enjoy the book! Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

3. NOTATION AND TECHNICALITIES

xv

3. Notation and Technicalities “Philosophy is written in this grand book, the universe, which stands continually open to our gaze. But the book cannot be understood unless one first learns to comprehend the language and read the characters in which it is written. It is written in the language of mathematics, and its characters are triangles, circles, and other geometric figures without which it is humanly impossible to understand a single word of it; without these one is wandering in a dark labyrinth.” Galileo

3.1. Marginalia. In addition to geometric “characters”, we will also use a number of other symbols from algebra and logic. Let us examine first the non-standard margin icons which appear throughout the book. Warm-up or brute force




Exercise

Basic Pigeonhole Principle Generalized Pigeonhole Principle

Problem Basis Step Open question or one that requires extra knowledge

Inductive Step

Problem solving technique Warning Contradiction

Strong Basis Step Strong Inductive Step

The first four margin pictures refer to increasing difficulty of exercises and problems. Assigning such symbols is somewhat arbitrary since the same exercise could be easy for one person and could be a really hard problem for another; something may be beyond the knowledge of the reader early in the book, while later it may turn out to be a piece of cake. Thus, treat these symbols as a general guide to the difficulty of the material and make your own judgment after having attempted each problem. The problem solving techniques, indicated by an eye, are ubiquitous throughout the book and will be discussed in the next section. The warning road sign, the high-voltage symbol, and the pigeons will appear prominently in Session 5 on proofs. The last four margin pictures refer to the steps of basic and strong mathematical induction, to be introduced in Session 6. 3.2. Logic. Mathematical statements that are proven are referred to by standard names such as theorem, lemma, proposition, property, or corollary. Conjectures are statements that are believed to be true, but no proof for them has been supplied yet. We will avoid the formal definition environment whenever possible, but theorems and such will be phrased formally. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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INTRODUCTION

Almost all sessions have a section on Hints and Solutions to Selected Problems. There and throughout the text, you will see two symbols indicating the end of a solution. The standard square
indicates the end of a complete solution or a proof with minor gaps, which are usually mentioned and the reader is expected to easily fill them in. The diamond ♦ is at the end of an incomplete solution, partial proof, sketch of a proof, hint, or any discussion requiring more work by the reader to reach a complete proof. The text uses standard mathematical words and expressions, such as “implies,” “therefore,” “if then,” “only if,” “if and only if,” letter notations for various sets of numbers, e.g., Z for the integers, and many others. Even though some are explained and illustrated via examples, the reader is expected to be familiar with basic logic notions and notation (cf. list of Symbols and Notation on page 299). If you need to review or learn this material in depth, we refer you to the first chapter of Jacobs’ Geometry [47] on deductive reasoning. A complete list of Abbreviations can be found on page 301. 3.3. Labeling and future volumes. Subfigures within the same figure are implicitly labeled in alphabetical order. For example, Figure 7 on page 7 contains subfigures Figure 7a, 7b, and 7c, reading from left to right. Finally, more than half of the sessions are Parts I of series of sessions, to be continued in Volumes II and III of the book.

4. The Art of Being a Mathematician and Problem Solving “Perhaps I can best describe my experience of doing mathematics in terms of a journey through a dark unexplored mansion. You enter the first room of the mansion and it’s completely dark. You stumble around bumping into the furniture, but gradually you learn where each piece of furniture is. Finally, after six months or so, you find the light switch, you turn it on, and suddenly it’s all illuminated. You can see exactly where you were.” Sir Andrew John Wiles

There are no manuals on how to become a mathematician. This book will give you tips and point to possible paths, but the “art of being a mathematician” can be mastered only through personal experience. With every problem solved and every new definition or theorem learned, you will move closer to this goal. The two most important skills that you will acquire along the way are • to think creatively while still “obeying the rules” and • to make connections between problems, ideas, and even theories. 4.1. Problem solving techniques. Although all sessions in this book are based on basic knowledge from middle and high school and are, therefore, accessible to a wide range of ages and mathematical backgrounds, to do the Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

5. ACKNOWLEDGMENTS

xvii

exercises, you need to develop problem solving techniques (PST’s). Session 1 on inversion will introduce PST’s as part of a trilogy of mathematical knowledge: Concepts, Theorems, and PST’s; and throughout the book you will encounter about 100 PST’s. You will also need to learn how to fit together various mathematical parts in order to move forward in the solutions. 4.2. Muddying your hands. Do not expect each session to be a collection of clearly spelled out recipes leading to instantaneous solutions . . . . Nope! The book will encourage you to apply the newly acquired knowledge to problems and will guide you along the way but will rarely give you ready answers. “The best way to learn is to learn from your own mistakes,” said my advisor Joe Harris. Sessions 5, 6, and 8, for example, will formally discuss non-proofs, and a number of other sessions will present common problem solving pitfalls. And so, it will be you, the reader, who has to commit to mastering the new math theories and techniques by • “muddying your hands” in the problems, • going back and reviewing necessary PST’s and theory, and • persistently moving forward in the book. Nothing good comes “for free”: you will have to work hard, always with a pencil and paper in hand. Keep in mind that the math world is huge: you’ll never know everything, but you’ll learn where to find things, how to connect and use them. The rewards will be substantial.

5. Acknowledgments 5.1. Institutional support and sponsors. The Berkeley Math Circle was made possible through the years with the unwavering support of: • University of California at Berkeley Math Department, which hosts the Circle and its web site and has provided student assistants and secretarial support every year since 1999. Through faculty grants, Ivan Matić has been able to act as an associate director. The department chairs Cal Moore, Hugh Woodin, Ted Slaman, and Alan Weinstein have always been encouraging and supportive, and 16 UCB professors have delivered Circle sessions. • Mathematical Sciences Research Institute, which has overseen the project from its inception, provided funds through various sponsors, and hosted Circle meetings and events. Special thanks to Deputy Directors Hugo Rossi, Joe Buhler, Michael Singer, and Bob Megginson, Directors David Eisenbud and Robert Bryant, and Associate Director Kathy O’Hara for their leadership, understanding, and help. A number of sponsors have financially supported BMC over the years: Packard Foundation, Toyota Foundation, Clay Mathematics Institute, Mosse Foundation for Art and Education, Merriam-Webster Foundation for the Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

xviii

INTRODUCTION

Scripps National Spelling Bee; National Science Foundation and other grants from Professors Ravi Vakil (Stanford), Bjorn Poonen, Alexander Givental, and Martin Olsson (UC Berkeley); and generous private donors. 5.2. Parents and students. The BMC parents have encouraged and driven their kids to the Circle for years, brought snacks during the breaks, organized Circle parties, attended meetings, and donated time, effort, and personal funds to the Circle. We are especially grateful to Marc Whitlow, Mike Breen, Jennifer O’Dorney, Yuki Ishikawa, and Tony DeRose for their enthusiasm, leadership, and professional services provided so selflessly to BMC. A sequence of UC Berkeley student assistants have contributed to the smooth operation of the Circle by communicating with circlers, parents, instructors, and administrators and by re-designing and maintaining the web site. Joyce Yeung, Maksim Maydanskiy, Wycee de Vera, William Chen, and David Wertheimer have been exceptionally professional and caring. Many thanks go to our monthly contest coordinators: Professors Alexander Givental and Bjorn Poonen, circlers Gabriel Carroll, Andrew Dudzik, Inna Zakharevich, Neil Herriot, Maksim Maydanskiy, and Evan O’Dorney, and associate director Ivan Matić. 5.3. Professional support with the web site has been rendered on numerous occasions by Paulo de Souza, Dmitri Mironov, and Steve Sizemore. Marsha Snow, Barbara Peavy, and Tom Brown have offered valuable secretarial support over the years. BMC owes its logo design to Archer Design Inc. As one can see, dozens of people have been involved in running the Berkeley Math Circle: it is a joint operation born of the love and care for our young generation of mathematicians. The most important people in this operation are undoubtedly the BMC instructors (over 50), who have delivered the 320 sessions during the last 10 years. We would like to thank all of them! Twelve joined BMC a decade ago and have stayed with us throughout the years: Ted Alper, Tom Davis, Dmitry Fuchs, Alexander Givental, Quan Lam, Bjorn Poonen, Tom Rike, Vera Serganova, Tatiana Shubin, Zvezdelina Stankova, Paul Zeitz, and Joshua Zucker. 5.4. Book support. Edward Dunne, our AMS editor, and his staff have been very helpful in resolving technical and other issues. Gabriel Carroll is responsible for drawing half a dozen cartoons, inspired by the earlier BMC sessions. All AHSME, AIME, and USAMO problems are used with permission from the American Mathematics Competitions (AMC), Lincoln, Nebraska [1]. A number of pictures and references have been taken from Wikipedia at http://www.wikipedia.org/. With gratitude, Zvezdelina Stankova Berkeley Math Circle Director Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

Session 1 Inversion in the Plane. Part I Zvezdelina Stankova Sneak Preview. Ten years ago the BMC commenced with this very session, which taught the circlers how to transform complicated and hard geometry problems into surprisingly simpler ones by the so-called method of Inversion in the Plane. In the current Part I, we shall motivate the method of inversion, explain how it works, introduce its fundamental properties, and answer basic questions such as what happens to lines and circles and how distances change under inversion. Later in Parts II–III, we shall study more advanced properties of inversion such as preservation of angle measure, view the transformation via complex numbers and explain its origins. The problems and problem solving techniques that will be introduced along the way will gradually increase in difficulty and depth.

1. Why Inversion? Motivation The impatient reader may wish to skip directly to Section 3. With the rest, let us consider the following problem, well known in the world of mathematics as Ptolemy’s Theorem:




Problem 1. Let quadrilateral ABCD be inscribed in a circle k (cf. Fig. 1a). Prove that the sum of the products of its opposite sides equals the product of its diagonals: AB · DC + AD · BC = AC · BD.

(1)

D

C C1

k A

B1

B A1

Figure 1. Ptolemy’s Theorem and its inversion equivalent 1

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2

1. INVERSION IN THE PLANE. PART I

There are various proofs of Ptolemy’s Theorem; for instance, proofs using trigonometry, extra geometric constructions and similar triangles, via complex numbers, and others. Yet, none of these solutions is as strikingly simple and elegant as the one Figure 1b suggests. After applying our “black box” of inversion, Ptolemy’s Theorem becomes equivalent to a statement that just about any middle schooler would readily supply: “If point B1 lies on segment A1 C1 , then the two smaller distances add up to the bigger one: A1 B1 + B1 C1 = A1 C1 .” (For the complete solution of Ptolemy’s Theorem, look further into this session.) Impressive! But why introduce a whole new method just to do this one famous problem? Point taken. Let’s try a “slightly” harder problem. Problem 2. Circles k1 , k2 , and k3 are tangent pairwise, and each is tangent to a line l. A fourth circle k is tangent to k1 , k2 , and k3 , so that k and l do not intersect. Find the distance d from the center of k to l if the radius of k equals 1 (cf. Fig. 2a). k2

k1

R k3

+

1

l

R−1

k

R

Figure 2. Problem 2 and its inversion equivalent The given information here seems awfully meager: knowing only the radius of k, one must be able to find the length of the dashed segment in Figure 2a. Shouldn’t changing the radii of k1 , k2 , and k3 produce infinitely many possibilities for the desired length? Could there only be one such possibility, and how can one find it? Exercise 1 (Rhetorical). If you are familiar with coordinate geometry, set up a coordinate system and create in it 9 quadratic equations expressing the 9 tangencies among the given figures. See if you can solve your equations. Solving such a system of 9 equations will most likely be painful. Even if successful, the reader will not have a clear idea of why this long calculation works or why the answer is what it is. Instead, our method of inversion turns Problem 2 into an equivalent but simple problem involving only one right triangle with sides R, R − 1, and R + 1 (cf. Fig. 2b). Calculating the unknown R now follows easily from one application of the Pythagorean Theorem and explains completely why the original problem has the unique answer. . . d = 7. However, understanding the complete solution requires the property of preservation of angle measure, which will be discussed in Part II. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

2. INVERSION AS A TRANSFORMATION

3

We hope that the above two problems and their promised solutions have intrigued the reader enough to continue learning about inversion in the plane. The latter is a wonderful and deep tool, which can solve hard problems when used wisely and can also create new, interesting, and hard problems practically from scratch.

2. Inversion as a Transformation 2.1. Is it like anything we know? Let us recall some transformations in the plane that we know very well, for example, rotations, reflections, translations, and rescalings. Would these transformations do the job that inversion claims to do? Consider Figure 3. An elephant will look just like an elephant, whether rotated, reflected, translated, or rescaled. Similarly, any figure will preserve its shape under any of these transformations.

Figure 3. Elephant transformations in the plane Yet, the suggested inversion proof of Ptolemy’s Theorem turns the original circle in Figure 1a into the line in Figure 1b. Inversion, therefore, must be a transformation that radically changes the shapes of figures. For comparison and for fun, one possible inverted elephant is depicted in Figure 4. If the inverted elephant E  resembles anything at all, it probably resembles Aladdin’s lamp with the Genie (or whatever the reader’s imagination supplies), but it certainly doesn’t look anything like the original elephant E.1 k

E

E

O

Figure 4. Inverted elephant: E → E  1After we discuss the basic properties of inversion, the diligent reader can come back to

Figure 4 to check that the elephant inversion with respect to the drawn circle k is correct, up to translation and rescaling of the inverted elephant E
so that it fits on this page.

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4

1. INVERSION IN THE PLANE. PART I

3. Definition of Inversion 3.1. Demystifying inversion. It is time to start doing rigorous mathematics by defining what inversion really is. To this end, take your favorite circle k in the plane and label its center with O and its radius with r: these will be the center and the radius of our inversion I(O, r). As with any transformation in the plane, inversion moves the points in the plane according to some rules. 3.2. What happens to points outside k? If X is any point outside circle k, then X will go under our inversion into point X1 defined as follows (cf. Fig. 5a): −−→ (1) Connect X to O via the ray OX. (2) Draw a tangent from X to the circle k and denote the point of tangency by T . −−→ (3) Drop a perpendicular from T to OX and denote the foot of this perpendicular by X1 . I

Thus, point X goes to point X1 , and we denote this by X → X1 , or I(X) = X1 . We also say that X1 is the image of X under I. T

Y1

T k

k Y O X1

O

X Z I

I

I

Figure 5. Define inversion: X → X1 , Y → Y1 , and Z → Z 3.3. What happens to points inside k? To find where a point Y inside circle k goes, we simply reverse the above procedure (cf. Fig. 5b): −−→ (1 ) Connect Y to O via the ray OY . −−→ (2 ) Erect a perpendicular from Y to OY until it intersects circle k in point T  . (3 ) Draw the tangent at T  to the circle k and denote its intersection −−→ with OY by Y1 . I

Thus, Y → Y1 , or I(Y ) = Y1 . Exercise 2 (Warm-up). To become comfortable with the definition of inversion, draw several points inside and outside circle k, and then follow the correct procedures to find their images under inversion. In each case, is there a single way to proceed, or can you obtain the images by following two different routes of construction? Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

4. BASIC PROPERTIES OF INVERSION

5

If we look more carefully at the two procedures above, we will notice that they involve choices: there are two tangents through point X to the circle −−→ k (see the dashed line in Figure 5a), and the perpendicular at Y to OY (in Figure 5b) intersects k in two points. Using the symmetry of the situation, the reader can easily verify that the end points X1 and Y1 are independent of these choices. 3.4. Have we missed any points? We haven’t said what happens to the points on the circle k or to point O itself. Exercise 3 (Warm-up). Pick a point Z on the circle k and attempt to find its image under inversion by following either of the two procedures above. Repeat with point O. What do you notice? Try to explain what goes on in each case. Do not look below for an answer until you have tried your best! If the points outside of k are mapped to points inside of k and the points inside k are mapped to points outside of k, it is reasonable to expect that the points on k will stay on k. Indeed, starting with point Z ∈ k and following either of the two procedures above, we will inevitably end up again with Z I after inversion (why?), i.e., Z → Z (cf. Fig. 5b). Finally, where does the center O go under I? Neither of our two procedures makes sense: we get stuck in the very beginning trying to draw a −−→ “ray” OO through the same point O and then trying to draw a perpendicular to that “ray.” However, this does not mean that we cannot send O to some other special point in the plane. But where to? The basic properties of inversion below will convince us that there is no reasonable choice in the plane for the image of O, and hence we shall say that the inversion I(O, r) is undefined at its center O.2

4. Basic Properties of Inversion 4.1. The composition I 2 . Having heard of a new mathematical “toy” like a transformation, true mathematicians get a good feeling for it by “playing” with it. For example, one of the first things they would do is to apply it several times, one after another. Exercise 4 (Warm-up). What happens to points in the plane if we apply inversion I(O, r) twice in a row? Consider separately each of the three possible cases X, Y , and Z in Figure 5. After you make your own picture, compare it with Figure 6. Applying inversion twice in a row lands us right back where we started: (2)

I

I

I

I

I

I

X → X1 → X, Y → Y1 → Y, and Z → Z → Z. 2 In Part III, we will compactify the plane to a sphere by adding one extra point O . ∞

This will allow us to extend inversion to O by defining I(O) = O∞ and I(O∞ ) = O.

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6

1. INVERSION IN THE PLANE. PART I X

X Z

Z Y

X1 Y Y1

Figure 6. Composition I 2 and swapping points: Prop. 1–3 True mathematicians also love naming any new stuff that appears. For example, the application of the same transformation I twice in a row is commonly referred to as the composition I 2 (also denoted by I ◦ I), and the transformation which doesn’t move anything around is known as the identity map id, i.e., id(X) = X for any point X. With this new terminology, we can conclude formally:3 Property 1. The composition I 2 is the identity transformation on the plane. In plain language, I swaps points in the plane in pairs (cf. Fig. 6b): (3)

I

I

I

X ↔ X1 , Y ↔ Y1 , and Z ↔ Z.

Thus, applying I twice will get us back to where we started, e.g., I 2 (X) = X. Thus, a point P inside the circle k will swap places under inversion with a point P1 outside k. Further, any point Z on k will stay where it is – this property is formally referred to as fixing k pointwise. Property 2. Inversion I swaps the planar region outside of k with the planar region inside k and fixes k pointwise. In fact, we can apply the above reasoning to any figure in the plane. For example, the original elephant E in Figure 4a is entirely inside k, and hence the inverted elephant E  = I(E) in Figure 4b must be entirely outside k. Further, applying inversion twice to elephant E will bring back elephant E, i.e., I 2 (E) = E. Property 3. Inversion I sends any figure F1 inside k to a figure F2 outside k and vice versa: I(F1 ) = F2 and I(F2 ) = F1 . Applying I twice in a row to any figure F (regardless of whether F is inside, outside, or partially inside/outside of k) returns the original figure F : I 2 (F ) = F . 4.2. How does inversion affect lines and circles? Up until now, any two lines (or circles) in the plane have looked alike to us and have acted in the same way. Yet, under inversion, we will discover that there are essentially two types of lines and two types of circles depending on whether they pass through O or not. For short, we write F  P if figure F passes through point P , and F  P otherwise. 3 Until we reach Part III of Inversion in a later volume, we’ll ignore what happens at O.

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4. BASIC PROPERTIES OF INVERSION

7

Exercise 5 (Warm-up). Take lines l  O and m  O. Invert several points on them and notice where they lie. If you invert all points on l, what figure would they form, i.e., what is I(l)? How about I(m)? k l S

T

m

k O

U V

O k1

A1

A

k3

k

k2 O B B1

Figure 7. Lines and circles under inversion: Prop. 4–7 As long as l passes through O, we can mark 4 points S, T, U , and V on l as in Figure 7a. Property 2 implies that inversion switches segment T O and −→ −−→ ray T S, as well as segment U O and ray U V . In other words, Property 4. A line l  O goes under inversion to itself, i.e., I(l) = l. The situation changes completely when the line m does not pass through O, as shown in Figure 7b. Check out the 6 marked points on m (including A). It looks like too much of a coincidence that their inverted images (including A1 ) all lie on the same “dotted” circle k1 through O. We conjecture that Property 5. A line m  O goes under inversion to a circle k1  O. It is important to understand that, despite how convincing our Figure 7b looks, it does not prove Property 5! A formal proof with similar triangles and inscribed angles will be completed in Part II. For the rest of Part I, we shall assume that Property 5 holds true, so that we can begin to do some powerful problem solving with inversion. We can now turn our attention to circles under inversion. For starters, suppose that a circle k1 passes through O. As above, we can draw lots of points on k1 and invert them, but isn’t there a shortcut? Recall that since I(m) = k1 for a line m  O (as in Property 5, cf. Fig. 7b), then by Property 3 the reverse is also true: I(k1 ) = m. We conclude that Property 6. A circle k1  O goes under inversion to a line m  O. There is only one case left to consider: what if our circle k2 does not pass through O? Note that k2 cannot go to a line n, or else I(k2 ) = n implies I(n) = k2 , yet a line never inverts to a circle  O (compare with Properties 4–5). Furthermore, k2 cannot map to a circle k1  O, or else I(k1 ) = k2 , thereby contradicting Property 6. The only remaining plausible solution for k2 , suggested by Figure 7c, is to map to a circle k3  O. We formulate below this property and postpone its formal proof until Part II. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

8

1. INVERSION IN THE PLANE. PART I

Property 7. A circle k2  O goes under inversion to a circle k3  O. Knowing what lines and circles do under inversion empowers us to resolve certain mysteries we have come upon. Exercise 6. Why does the inverted elephant E  in Figure 4b consist only of arcs of circles (with the exception of the point representing the inversion of the elephant’s eye)? What will happen to E  if we move the center O onto the intersection of E’s ear and back? The reader can easily verify that the original elephant E consists (very wisely!) only of arcs of circles and line segments and hence so will its image E  under inversion. Because the center O in Figure 4a does not lie on any of the original circles and lines in E, Properties 4–7 guarantee that all inverted parts of E  will be parts of circles but never parts of lines (why?). If the center O is moved to lie both on the ear (arc α) and on the back (arc β) of E, these two arcs will invert to rays v and w  as in Figure 8a. The picture seems also to suggest that these two rays lie on parallel lines. Is this true? Look more carefully at the two circles k1 and k2 on which arcs α and β lie: not only do they pass through O, but they are also tangent at O! l1

v k1

k O

E 

P

α

k2

β

w 

O l2

I

Figure 8. Inverted elephant E → E  and Circles tangent at O Exercise 7. Let circles k1 and k2 be tangent to each other at the center O of inversion (cf. Fig. 8b). Why do k1 and k2 invert to parallel lines? Solution: Since k1  O and k2  O, they will invert to some lines l1  O and l2  O. Suppose for contradiction that l1 and l2 are not parallel, i.e., that they intersect in some point P . But then let’s apply inversion again: where would point P go? On the one hand, I(P ) must be on circle k1 since I(l1 ) = k1 , and on the other hand, I(P ) must be on circle k2 since I(l2 ) = k2 ; yet, k1 and k2 intersect only in O, i.e., I(P ) = O. This is impossible since any point different from O will invert to a point different from O! (Why? Check the definition of inversion.) This contradiction shows that our supposition is false. We conclude that k1 and k2 must invert to parallel lines l1 || l2 .
The reasoning in this problem is indicative of a larger phenomenon: Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

5. PROBLEM SOLVING TECHNIQUES WITH INVERSION

9

Property 8. If two figures F1 and F2 intersect only at the center O, after inversion they cannot acquire any additional intersection points. Similarly, if F1 and F2 do not intersect at all, their inverted images can “intersect” at most at the center O.4 Property 8 can be seen in various settings. For example, Exercise 8. Let line l be tangent to circle k1 . Will the inverted figures I(l) and I(k1 ) still be “tangent”? How many intersection points will they have? Draw all possible types of inverted pictures.

5. Problem Solving Techniques with Inversion 5.1. Ideas and general tips. When working in any mathematical field (e.g., plane geometry, algebra, number theory, combinatorics, real analysis, etc.), we collect the newly acquired knowledge and experience roughly in three major intellectual “bags” (cf. Fig. 9a): Concepts, Theorems and Problem Solving Techniques (PST’s). Original problem

PST’s Theorems

conclude

apply I translate

Inverted problem

Inversion PST 1

solve

Concepts apply I Solution Solution to original translate back to new problem problem

Figure 9. Three bags of math knowledge and Inversion PST 1 For example, in the topic of inversion in the plane, we can put the definition of inversion in the “Concepts bag,” while the properties of inversion and the conclusions of any exercise or problem can be placed in the “Theorems bag.” Yet, it is the “PST bag” that will be indispensable when attacking and successfully solving new problems. It is the “PST bag” that will allow us to apply the newly learned terminology and theory to problem solving, and it is the “PST bag” of experience that will most likely carry over to problem solving in any other mathematical field. The “PST bag” is often neglected in mathematical literature, where the new concepts and theory occupy almost every page. In the current book, we shall devote considerable attention to PST’s, because without enough powerful PST’s a newcomer will quickly become lost in a jungle of mathematical problem solving. 4 The more advanced reader will recognize this property as injectivity, or one-tooneness, of inversion: different points invert to different images.

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10

   

1. INVERSION IN THE PLANE. PART I

5.2. Basic PST’s of inversion. Let’s start with PST’s related specifically to inversion, yet possibly exportable to other topics and fields. PST 1. After applying appropriate inversion, translate the original problem into a new simpler problem. Solve the new problem. Apply the same inversion again and translate back the results into the original problem (cf. Fig. 9b). PST 2. Turn as many circles as possible into lines, thereby simplifying the original problem. This is one of the most basic and powerful ideas in problem solving with inversion. In applying this PST, the reader must choose wisely the center of inversion: since you are the “boss,” you possess the power to choose what is most advantageous to your solution. In general, the act of choosing the center and radius of inversion is very much like choosing an appropriate coordinate system. A good choice can make a problem much easier. PST 3. How can you turn a circle k into a line? Only by placing the center O on k. To turn simultaneously two circles into lines, you must choose the center to be one of their intersection points. The particular radius of inversion, as we will see, is often irrelevant in the solution. PST 4. Draw the inverted picture away from the original picture: this will avoid crowding too many objects in one place. Note that Figures 5–7 do not adhere to PST 4, while Figure 4 and the upcoming Figure 10 do use it. What is going on in Figure 8?



While it is essential to draw correctly the inverted objects as lines or circles, respectively, and to indicate clearly how they intersect (or not) with each other, it is often overkill to try to draw a “precise” inverted picture by following the definition of inversion word-for-word. Thus, PST 5. You are encouraged to shake off the urge to draw “exact” inverted pictures and to apply your imagination to constructing the correct general type of inverted pictures instead. In which of Figures 4–8 is PST 5 applied?

6. Solving Our First Problems with Inversion 6.1. Inversion establishes collinearity and cyclicity. We say that several points are collinear if they lie on a line. The points are concyclic if they lie on a circle, and the polygon formed by these points is called cyclic.




Problem 3. Circles k1 , k2 , k3 and k4 are positioned in such a way that k1 is tangent to k2 at point A, k2 is tangent to k3 at point B, k3 is tangent to k4 at point C, and k4 is tangent to k1 at point D. Show that A, B, C, and D are either collinear or concyclic (cf. Fig. 10a).

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6. SOLVING OUR FIRST PROBLEMS WITH INVERSION

11

Solution: We shall utilize the fundamental PST 1 and solve our problem in four steps. While doing so, we will pay attention to where and how we use other PST’s, properties of inversion, and previously shown results. Step 1 (Apply inversion). If we want to turn k1 and k2 into lines k1 and k2 , we are forced to choose point A as the center of our inversion (PST 3). Thus, we apply inversion I(A, r) with center A and some radius r. For short, we shall say that we invert through A. m

k1

D

k4 D

C

A B

I(A, r) k2

C

k3 O3

k3

B



k1

 O4 k4

k2

Figure 10. Problem 3 To draw the inverted picture (cf. Fig. 10b), we use PST’s 4–5. Because k1 and k2 are tangent at A, Exercise 7 tells us that their images k1 and k2 will be parallel lines. Further, Property 8 implies that circles I(k3 ) = k3 and I(k4 ) = k4 will be tangent to each other and tangent, respectively, to lines k2 and k1 . The 3 points of tangency are denoted by C  , B  , and D , and the centers of k3 and k4 by O3 and O4 , respectively. Step 2 (Translate into a new problem). Our original Problem 3 asks us to show that A, B, C, and D are collinear or concyclic. Since A is the center of inversion, Properties 4 and 6, respectively, would imply the same conclusion in both cases: after inversion, B  , C  , and D should be collinear (why?). Aha! Can we actually prove this, given our configuration in Figure 10b? Indeed, we will below. We have reduced our complex original problem into a new problem, solvable with ordinary high school geometry! Step 3 (Solve the new problem). In Figure 10b points O3 , C , and O4 are collinear due to the tangency of circles k3 and k4 . But radii O3 B  and O4 D are perpendicular to lines k2 and k1 (due to tangencies at B  and D  ), and, as remarked earlier, k2 || k1 . Hence O3 B  || O4 D and ∠B  O3 C  = ∠D  O4 C  as alternate interior angles. This forces the similarity B  O3 C  ∼ D  O4 C  , as these are both isosceles triangles with equal vertex angles. Reaping the effects of this similarity, we conclude that ∠O3 C  B  = ∠O4 C  D . Since O3 C  O4 is already known to be a line, this implies that B  C  D is also a line, denoted by m. Step 4 (Translate back into the original problem). We are not yet done! Applying I(A, r) to line m returns B  → B, C  → C, and D → D. The effect of I(A, r) on line m depends on whether m passes through A or not: Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

12

1. INVERSION IN THE PLANE. PART I

if m  A then m will map onto itself (Property 4), thereby making A, B, C, and D collinear; if m  A then m will map onto a circle  A (Property 5), thereby making A, B, C, and D concyclic. Now we are truly done!




The reader probably noticed that the particular radius of inversion had no bearing on our solution! Along the way, we also used another PST which may seem obvious, yet it is often bypassed by novice problem solvers. PST 6. Reap the effects of similarity of triangles. After establishing that two triangles are similar – presumably by verifying that some of their corresponding angles and side ratios are equal – conclude and use the fact that the remaining angles and ratios are equal. The inquisitive reader should be wondering if it is actually possible to have the original points A, B, C, and D collinear . . . . After all, Figure 10a depicts them as concyclic! Exercise 9. In the setting of Problem 3, draw a configuration of the four circles in such a way that the points of tangency A, B, C, and D lie on a line. What would be the corresponding inverted picture under I(A, r)? 6.2. Inversion needs help with cyclicity. Try to solve the following problem without inversion and then with inversion.




Problem 4. Circles k1 , k2 , k3 , and k4 intersect cyclicly pairwise in points {A1 , A2 }, {B1 , B2 }, {C1 , C2 }, and {D1 , D2 }; in other words, k1 ∩ k2 = {A1 , A2 }, k2 ∩ k3 = {B1 , B2 }, etc. Prove the following. (a) If A1 , B1 , C1 , D1 are collinear/concyclic, then so are A2 , B2 , C2 , D2 . (b) If A1 , A2 , C1 , C2 are concyclic/collinear, then so are B1 , B2 , D1 , D2 . Whether you attempt inversion or not, you will find yourself in need of some more sophisticated condition establishing that 4 points lie on a circle: Theorem 1 (Cyclicity). Let ABCD be a “nice” quadrilateral, i.e., its sides do not intersect except at A, B, C, and D, and all internal angles are < 180◦ . Then ABCD is cyclic if and only if one pair of opposite angles add up to 180◦ , e.g., ∠ABC + ∠ADC = 180◦ (cf. Fig. 11a).5 E D

C

Y

Z I

A

M

M k

B

K

X

k2 k1

O L K

L

Figure 11. Cyclicity, Incircle and Circumcircle 5 The Circle Geometry session will have detailed proofs of this and related theorems.

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6. SOLVING OUR FIRST PROBLEMS WITH INVERSION

13

Extending CD beyond D to point E, we observe that ∠ADE completes ∠ADC to 180◦ , so that the cyclicity condition in Theorem 1 is equivalent to the following: A, B, C, and D are concyclic iff ∠ABC = ∠ADE. Armed with this statement, we can attack Problem 4. Sketch of inversion solution to Problem 4(a): We convert two of the four circles to lines by inverting, say, through A1 . Figure 12b gives one possible inverted picture, where the circle or the line through the original A1 , B1 , C1 , and D1 maps to line B1 C1 D1 . k4 D1

D1

A1 D2 A2

k1

C2

C1 k3

B2

I(A1 , r)

k2

D2


C1

C2


B1

B1 A1

k4

k2 k3

B2


A
2

k1

Figure 12. Problem 4(a)

For A2 , B2 , C2 , and D2 to be concyclic (see dashed circle in Figure 12a), their images A2 , B2 , C2 , and D2 must be concyclic in Figure 12b. To show the latter, we apply Theorem 1 to circles k3 and k4 and to the conjectured (dashed) circle in Figure 12b; e.g., ∠A2 B2 C2 = ∠B1 C1 C2 = ∠C2 D2 D1 . The same inversion I(A1 , r) will map back the dashed circle on the right either to the dashed circle on the left or to a line l through A2 , B2 , C2 , and D2 . ♦ 6.3. Incircles, circumcircles and inversion. For the next problems, the reader will need the following notions. The incircle of ABC is a circle tangent to all three sides AB, BC, and CA. The circumcircle of ABC is a circle passing through all three vertices A, B, and C (cf. Fig. 11b–c). The centers of the newly-defined circles are called the incenter and the circumcenter of ABC, respectively.




Problem 5 (IMO proposal). The incircle of ABC is tangent to sides BC, CA, and AB at points D, E, and F , respectively. Point X is chosen inside ABC so that the incircle of XBC is tangent to BC at D, to CX at Y , and to XB at Z. Prove the following. (a) EF ZY is a cyclic quadrilateral. (b) The circumcircles of F ZD and EY D are tangent at D.6 PST 7. For the center of inversion, try a point of tangency of several circles and/or lines: these figures will invert to parallel lines and most likely rearrange the inverted picture in a symmetric and orderly manner. 6 Part (b) is added by the author.

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14

1. INVERSION IN THE PLANE. PART I

Sketch of solution to Problem 5: With the above PST in mind, we invert through D to obtain three parallel lines: B  DC  , k  , and k1 (cf. Fig. 13). The remaining four lines in the original picture, BA, CA, BX, and CX, invert to the four shown circles. Using the symmetry of the situation and plane geometry arguments, show that lines Z  F  and Y  E  are perpendicular bisectors of chords B  D and DC  , respectively, and as such, Z  F  and Y  E  are both perpendicular to B  C  . Why does this imply that E  F  Z  Y  is a rectangle? Reason that inverting back the circle through the vertices of this rectangle lands us on the desired circle through E, F , Z, and Y . C D E

k1

F

Y

k1

F

E

k

I(A1 , r)

Y XZ

B

k A

Z

B

D

C

Figure 13. Problem 5 Moreover, since lines Z  F  and Y  E  are parallel to each other and do not pass through D, they invert back to circles tangent at D (why?). These circles are nothing but the desired circumcircles of F ZD and EY D. ♦ In fact, the circumcenters of these two triangles are B and C, respectively. Indeed, BD = BZ as tangent segments from B to circle k1 (cf. Fig. 13a), and BD = BF as tangent segments from B to circle k; thus, B is the circumcenter of F ZD, and similarly C is the circumcenter of EY D. The theme of incircles and circumcircles is expanded in the following challenging problem.




Problem 6 (IMO proposal). Circles ω, ω1 and ω2 are externally tangent to each other in points C = ω ∩ ω1 , E = ω1 ∩ ω2 , and D = ω2 ∩ ω. Lines l1 and l2 are parallel and such that l1 is tangent to ω and ω1 at points G and A, respectively, and l2 is tangent to ω and ω2 at points F and B, respectively. (a) Find (with proof) 3 collinear triples and 3 concyclic quadruples among the given 7 points of tangencies.7 (b) (Advanced) Prove that AD and BC intersect in the circumcenter of CDE. Hint: The picture already looks awfully nice and orderly, as we have our two parallel lines and a bunch of circles tangent to each other and to the lines. It seems that no inversion would make the picture nicer or simplify the problem. Yet . . . invert through D and be prepared for a pleasant surprise. Figures 11b–c give some further clues on properties of the involved incircle and circumcircle. We defer the full solution until the Hints section. ♦ 7Part (a) is added by the author.

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7. HOW DOES INVERSION AFFECT DISTANCES?

15

7. How Does Inversion Affect Distances? The fact that the inverted elephant E  in Figure 4 looks nothing like the original elephant E should make the reader question whether inversion preserves distances between points. To explore this on your own, try Exercise 10 (Warm-up). Draw your circle of inversion k with radius r = 1 inch, and pick two points A and B outside k such that the distance between them is, say, AB = 3 inches. Invert A and B to points A1 and B1 . What is the maximal possible distance between A1 and B1 ? Does inversion preserve distances between points? The sheer fact that the inverted points A1 and B1 are inside circle k (cf. Fig. 14b) forces the distance A1 B1 < 2; so the original distance AB has . . . shrunk under inversion! But how do we find the exact new distance A1 B1 ? Is there such a formula? To answer this, we proceed in two steps. 7.1. Distances from O. We first find out how inversion changes distances from points to the center O. B k

T B1

r O X1

X

O

A

A1 k

Figure 14. Changing distances under inversion: Prop. 9–10




Property 9. Let X be any point different from O and I(X) = X1 . Then r2 · (4) OX1 = OX Proof: We can rewrite the desired equation (4) as an equality of two ratios: OX1 ? r r = OX ,

which reminds us of corresponding ratios in similar triangles. To explore this idea, we locate a segment in Figure 14a of length r: OT = 1 ? OT r. This makes (4) equivalent to OX OT = OX . Thus, if similar triangles are involved, it is reasonable to expect that they are OX1 T and OT X. But are they similar indeed? They share a common angle: ∠X1 OT = ∠T OX. Further, ∠OT X is right because the radius OT and the tangent XT are perpendicular: OT ⊥ XT . By construction, OX1 ⊥ T X1 , so that ∠OX1 T is also right. We conclude that ∠OX1 T = ∠OT X. The AA–criterion for similarity of triangles now implies that OX1 T ∼ OT X. Let us reap the benefits of our hard labor (PST 6): similarity of triangles implies equal ratios of corresponding sides. In our situation, two of the three OT 1 ratios in OX1 T ∼ OT X will do: OX OT = OX , and hence the desired Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

16



1. INVERSION IN THE PLANE. PART I

equation (4) follows. Note that the above arguments work equally well when X and X1 are switched (i.e., X is inside k). When X ∈ k, the formula is
trivial, since OX = OX1 = r. Above, we unknowingly applied another powerful PST: PST 8. Algebraically manipulate equations until they acquire some geometric meaning, e.g., they point to plausible similarity or congruence of triangles. Before we move on, let us analyze formula (4). It roughly says that Property 9 . The new and old distances from the center O are inversely proportional, i.e., a point X far from O will invert to a point X1 close to O, and vice versa. It is tempting now to use formula (4) to find out where point O should go under inversion. Exercise 11 (Warm-up). Suppose we can define inversion at O, i.e., for some point O1 in the plane, I(O) = O1 . What goes wrong if O1 = O? What doesn’t work well if O1 = O? Where should O1 be located? By Property 1, inversion must return O1 back to O: I(O1 ) = O. But by definition of inversion, no point inverts to O (check Figure 5.) So, the only resolution is to define I(O) = O. But then Property 2 would fail at O since O won’t map to a point outside of k; in addition, Property 9 won’t make sense: substituting X = O = X1 yields the meaningless 0 = r2 /0. Thus, if we want to preserve all the nice properties of inversion that we have shown so far, we have only one choice: to define I(O) = O1 to be a point outside the ordinary plane. Formula (4) then would yield OO1 = r2 /0, which can be “infused” with meaning as follows. As point X moves closer and closer to O, its inverted image X1 must move farther and farther away from O. Therefore OO1 can be treated as “infinitely long”, and point O1 should be located somewhere “far away at infinity”. For a precise meaning of this and for the formal construction of I(O), the reader must wait until Part III. Still, this lack of knowledge will not inhibit us from solving a number of complicated problems via inversion in Parts I–II. 7.2. Arbitrary distances under inversion.




Property 10 (Distance formula). If A and B are distinct points different from O, with I(A) = A1 and I(B) = B1 , then OAB ∼ OB1 A1 and AB · r2 · (5) A1 B1 = OA · OB Before we plunge into the proof, note the apparent interchange of letters A and B in the similarity OAB ∼ OB1 A1 . There is no mistake here: by our brand new Property 9, if A is closer to O than B, then its inverted image A1 will be farther away from O than B1 (and vice versa).

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8. PROOF OF PTOLEMY’S THEOREM

17

Proof: Looking at Figure 14b, we immediately notice that the two triangles in question, OAB and OB1 A1 , share a common ∠O. Similarity criteria require either one more pair of congruent angles or equal side ratios. We have no further information on the angles involved, but we have plenty of information on the sides, provided by Property 9: OB1 OA = · OA1 · OA = r2 = OB1 · OB ⇒ OB OA1 Thus, by SAS–criterion for similarity, we conclude OAB ∼ OB1 A1 . How do we reap wisely the effects of this similarity? The desired formula (5) involves only side lengths, including the distances AB and A1 B1 , which haven’t appeared yet in the discussion. Thus, let’s get them into the picture by concluding, for example, OB1 A1 B1 = · (6) AB OA Which side lengths in (6) must be eliminated in order to arrive at the desired formula (5)? Only OB1 should go away, and we know how to make it disappear. Indeed, Property 9 asserts that OB1 = r2 /OB, so plugging into (6) yields r2 1 AB · r2 A1 B1 = · ⇒ A1 B1 = · AB OB OA OA · OB

 




In this solution (as well as in previous arguments) we repeatedly used the following general PST’s: PST 9. Reduce the current problem to previously solved problems. Apply any theorems, propositions, properties and other statements which have been proven before. PST 10. At each and every step of the solution, keep in perspective what needs to be proved or achieved. When proving formulas, watch out for quantities that have to be eliminated, and introduce into the discussion all quantities that do appear in the final formula.

8. Proof of Ptolemy’s Theorem We are finally ready to attack Ptolemy’s Theorem (Problem 1). Recall that we are given a cyclic quadrilateral ABCD and asked to show equation (1) involving its sides and diagonals. Now that we have successfully cracked a number of problems with inversion, it is obvious that the center O must be chosen on circle k so as to transform k into line l. Yet, if we choose a random O on k, we will most likely end up with four inverted points A , B  , C  , and D on l (cf. Fig. 15a): that’s not the solution suggested by Figure 1b. Can we do better: can we have only 3 inverted points? Yes, if we choose O = D (or any other vertex of ABCD)! Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

18

1. INVERSION IN THE PLANE. PART I O

O=D

C

D

B A m

D

B A

C

C

k



k

l

B C1

A B1 A1



Figure 15. Proof of Ptolemy’s Theorem

Proof of Ptolemy’s Theorem: After applying I(D, r) (cf. Fig. 15b), everything happens on the segment A1 C1 : A1 B1 + B1 C1 = A1 C1 . Using the distance formula of Property 10 to substitute for these segments (with O = D), we obtain BC · r2 AC · r2 AB · r2 + = · DA · DB DB · DC DA · DC Canceling the irrelevant r2 and clearing denominators, we arrive victoriously at the desired equation (1): AB · DC + AD · BC = AC · BD.






Despite the fact that we achieved our goal of proving Ptolemy’s Theorem, we shouldn’t be satisfied because PST 11. A true mathematician always questions the necessity of each condition in a theorem, in an attempt to generalize or strengthen the theorem. This not only deepens understanding of the theorem, but it can also lead to creation of new, better, and bigger theorems. Thus, we question the only condition in Ptolemy’s Theorem: that A, B, C, and D are concyclic. Do we need this fact to derive equation (1)?




Problem 7 (Ptolemy’s Inequality). Prove that for any four points A, B, C, and D (7)

AB · DC + AD · BC ≥ AC · BD.

Furthermore, show that A, B, C, and D are concyclic and arranged in this order along a circle if and only if (7) is an equality. Hint: Again, apply inversion through D. If inverted points A1 , B1 , and C1 aren’t collinear, they form A1 B1 C1 , to which the Triangle Inequality applies: A1 B1 +B1 C1 ≥ A1 C1 . Note that the Triangle Inequality (a theorem deserving a proof of its own!) becomes equality iff the three points A1 , B1 , ♦ and C1 are collinear, with B1 between the other two points. For further practice, here is a complex-looking problem, which can be solved relatively easily with the methods of inversion we’ve seen so far:

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9. HOW ARE INVERSION PROBLEMS CREATED?




19

Problem 8. The chord AB separates circle γ into two parts. Circle γ1 is inscribed in one of the parts so that it is tangent to AB at point C and to γ at point D. Circle γ2 is also inscribed in the same part of γ so that it is tangent to AB, γ1 , and γ at points R, S, and T , respectively. Let P Q be the internal tangent of γ1 and γ2 , with P, Q ∈ γ, where Q is in the same part of γ with respect to AB as D and T . Let P Q intersect AB in E. Show that (a) P, C, and D are collinear and E, C, D, and Q are concyclic. Locate as many collinear and concyclic points as you can. (b) P Q · SE = SP · SQ. Hint: Inversion I(S, r) results in three parallel lines, two congruent circles and several isosceles trapezoids due to the symmetry of the situation. ♦

9. How Are Inversion Problems Created? 9.1. Parallelograms and inversion. Here is a problem which clearly involves three circles, but, as it turns out, it has nothing to do with the circles. Instead, properties of inversion and parallelograms rule the situation.




Problem 9. Let k1 and k2 be two circles intersecting at A and B. Let t1 and t2 be the tangents to k1 and k2 at point A, and let t1 ∩ k2 = {A, C} and −−→ t2 ∩ k1 = {A, D}. If E ∈ AB such that AE = 2AB, prove that ACED is cyclic (cf. Fig. 16a). Proof: If one were to choose a center of inversion for Figure 16a, which point would that be? Well, point A stands out as the most likely choice since all given circles and lines pass through A! Thus, after I(A, r), lines t1 and t2 will stay where they are, and circles k1 and k2 will invert to lines k1 and k2 parallel correspondingly to t1 and t2 (why?; fill in all details; cf. Fig. 16). Thus, we obtain a parallelogram AC  B  D , where as usual I(B) = B  , etc. t1

t2 A

t t1 A 2

k2

k1

C B

I(A, r)

k1

E D

D

k2 C

B

E

Figure 16. Problem 9 −−→ Where is point E  ? Since E ∈ AB and A is the center of −−→ −−→ E  ∈ AB  = AB. Moreover, by Property 9, AB · AB  = AE · AE  = r2 = AB · AB  ⇒ AE  = AE

inversion, then 1 AB  , 2

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20

1. INVERSION IN THE PLANE. PART I

i.e., E  is the midpoint of AB  . This makes perfect sense: since B is closer than E to the center A, then after inversion E  becomes closer than B  to A. It is time to translate the original question in the new set-up. If A, C, E, and D are concyclic, the circle through them would invert to a line through C  , D , and E  . Aha! Can we prove that C  , D , and E  are collinear? Indeed, by a well-known geometry theorem: Theorem 2 (Parallelogram Diagonals). The diagonals in a parallelogram bisect each other.



Since E  is the midpoint of the diagonal AB  in parallelogram AC  B  D , then E  is also the midpoint of the other diagonal D  C  . In particular, D  , E  , and C  are collinear. To finish the proof, we invert back: the line D  E  C   A; hence it inverts back to a circle through A, i.e., D, E, C, and A are concyclic.
PST 12. To acquire complete understanding of a problem, it is often helpful to discover where the problem came from, i.e., how it was created. After having seen the inversion solution above, the creation of Problem 9 is no more a mystery: the author must have started with a parallelogram and inverted through one of its vertices, cleverly encoding Theorem 2 in the process. Can we create our own problems in a similar way? Exercise 12. Create a new problem by inverting Figure 16b through E  and encoding Theorem 2 in the process. 9.2. Medians and inversion. For a more challenging problem creation, consider everyone’s favorite: Theorem 3 (Centroid). The three medians in a triangle intersect in one point, which divide each median in a ratio of 2 : 1 counted from the vertex of the triangle. For those not familiar with the above concepts, a median connects a vertex with the midpoint of the opposite side. Thus, if A1 , B1 , and C1 are the midpoints, respectively, of sides BC, CA, and AB in ABC, the three medians are AA1 , BB1 , and CC1 . Theorem 3 asserts that AA1 , BB1 , and CC1 meet in one point M , usually referred to as the centroid (or center of mass 8) of ABC; and, further, that AM : M A1 = BM : M B1 = CM : M C1 = 2 : 1.




Problem 10 (Intermediate). Draw a picture of Theorem 3. By inverting your picture through the centroid M or through vertex A, create at least two new problems. Make sure that you encode Theorem 3 correctly and meaningfully in your problems. 8 We shall revisit these concepts in the Mass Point Geometry session.

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9. HOW ARE INVERSION PROBLEMS CREATED?

21

Let us demonstrate the creation of one such new problem when inverting through M . The three medians invert to lines A A1 , B  B1 , and C  C1 through M , and, due to distance formula (4), the corresponding ratios are A M : M A1 = 1 : 2, etc. (why?). The sides of ABC invert to circles k1 , k2 , and k3 through M (cf. Fig. 17b). C 1

k3

C I(M, r) B1

M C1

A

k2

M

A1

B

A

B

k1 A1

C B1

Figure 17. Alternative Centroid Theorem



But how do we encode in the inverted picture, for instance, that A1 was the midpoint of BC? The trick is to realize that PST 13. Distance formula (5) can be used backwards: we can apply the inversion I(O, r) a second time and express any old distance XY in terms r 2 ·X
Y
of new distances: XY = OX
·OY
. Thus, we start with A1 C = A1 B and apply PST 13 to each side: (8)

r 2 ·A
1 C
M A
1 ·M C


=

r 2 ·A
1 B
M A
1 ·M B


⇒ A1 C  · M B  = A1 B  · M C  .

The last equality incidentally has a nice geometric formulation: in the quadrilateral inscribed in k1 the products of the opposite sides are equal. We analogously conclude the same for the quadrilaterals inscribed in k2 and k3 . Finally, we have to decide how to turn Figure 17b into an actual theorem. One possible way is as follows:




Problem 11 (Alternative Centroid Theorem 1). Let circles k1 , k2 , and k3 intersect in point M , and let their second intersection points be A for k2 and k3 , B for k3 and k1 , and C for k1 and k2 (cf. Fig.17b with all primes deleted). Extend chord AM until it intersects k1 in A1 , and analogously locate points B1 on k2 and C1 on k3 . If the resulting quadrilaterals in k1 and k2 each have equal products of opposite sides, show that the quadrilateral in k3 has the same property, and moreover, M divides each segment AA1 , BB1 , and CC1 in a ratio of 1 : 2 counted from the first vertex. The reader is encouraged to verify that inverting through M indeed reduces Problem 11 to a version of the original Centroid Theorem: “If two medians in a triangle intersect in point M , then the third median also passes through M , and M divides each median in a ratio of 2 : 1 counted from the vertices of the triangle.” The point of constructing Problem 11 is not to marvel at it, for it really looks yucky, but to realize that the medicine for dealing with such problems is to transform them via inversion to something more elegant and more easily solvable.

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22

1. INVERSION IN THE PLANE. PART I

10. Hints and Solutions to Selected Problems Exercise 8. There are four cases to consider, depending on whether or not O lies on l and k1 . The inverted k  and l are still tangent, except in one case where they are parallel lines. ♦ Exercise 9. One possible configuration is given in Figure 18, which is not necessarily drawn to scale. Draw another (essentially different) configuration. Do all configurations contain some internally tangent circles? ♦ k4

k4 k2

k1 A

k3

B

D

C

I(A, r) A D

k3

k2

k1

B

C

Figure 18. Four collinear points of tangency Problem 4(b). Assume that A1 , A2 , C1 , and C2 are concyclic/collinear, and apply again I(A1 , r) to Figure 12a to obtain one possible inverted picture, as in Figure 19a. The inverted configuration consists of three lines passing through A2 , two (shaded) circles and one conjectured (dashed) circle k  . k1


k


k4


D1


Z2

Y2 Z2

l C1
D2
A
2

C2
B2


k3
k2


Y1 Z1

B1


Y2 X

Y1

Z1

X

Figure 19. Cyclicity and Power of a Point The fastest way to establish that k  is indeed a circle is via Theorem 4 (Power of a Point9). Let lines Y1 Y2 and Z1 Z2 intersect in X so that X is either outside both, or inside both, segments Y1 Y2 and Z1 Z2 (cf. Fig. 19b–c). Then Y1 Y2 Z1 Z2 is cyclic iff XY1 · XY2 = XZ1 · XZ2 . Problem 4(b) is now an easy game. The power of point A2 with respect to circles k3 and k4 yields A2 B1 · A2 B2 = A2 C1 · A2 C2 = A2 D1 · A2 D2 . However, the equality of the first and the third products implies (again by Theorem 4) that B1 B2 D2 D1 is cyclic. Inverting back the (dashed) circle k  produces a line or a circle in the original problem, i.e., B1 , B2 , D2 , and D1 are collinear or concyclic. ♦ 9 Theorem 4 can be proven by using/showing
XY Z ∼
XZ Y and applying 1 1 2 2

Cyclicity Theorem 1 to two equal angles from this similarity: see the marked angles in Figures 19b–c. Figure 19c will be covered in detail in the Circle Geometry session.

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10. HINTS AND SOLUTIONS TO SELECTED PROBLEMS

23

Problem 6. We don’t need inversion to locate the triples of collinear points in part (a). Indeed, our two parallel lines l1 and l2 , along with any pair of the circles, form a configuration analogous to the one in Figure 10b. Therefore, {G, D, B}, {A, C, F }, and {A, E, B} are triples of collinear points. F ω

B O2

D O

ω2 I(D, r)

ω1

O1

A

G A

E C

G

C

l2

l1

F

ω l2

D

ω1 l1

E

B

ω2

Figure 20. Inversion creates symmetry As the text suggests, apply I(D, r) (cf. Fig. 20). The inverted configuration contains two congruent circles ω1 and l2 , each tangent to the two parallel lines ω  and ω2 , and is symmetric with respect to a vertical line through G . The configuration of lines ω  and ω2 and circles ω1 and l1 is also analogous to Figure 10b; so E  , A , and G are collinear. Inverting back line E  A G yields the circle through E, A, G, and D. For another cyclic quadruple, invert back the circle through the vertices of rectangle E  B  F  C  to the circle through E, B, F , and C. For part (b), the key is that A D is parallel to ω  and ω2 (why?), so inverting it back yields line AD tangent to ω and ω2 . Similarly, line BC is tangent to ω and ω1 via I(C, r). If O, O1 , and O2 are the centers of ω, ω1 , and ω2 , respectively, the above means that AD ⊥ OO2 and BC ⊥ OO1 (why?). Since OD = OC, O1 C = O1 E, and O2 E = O2 D, the configuration made of OO1 O2 and CDE is identical to the one in Figure 11b. In fact, Theorem 5 (Incircle). Given KLM and points X, Y and Z on sides KL, LM and M K, respectively (cf. Fig. 11b), the incircle of KLM passes through X, Y , and Z iff KZ = KX, LX = LY , and M Y = M Z. In such a case, the incenter I of KLM is the circumcenter of XY Z, and I lies on the perpendiculars from X, Y , and Z to the corresponding sides of KLM . Combining everything, we conclude that the incenter of OO1 O2 is the circumcenter of CDE, and it is the intersection of lines AD and BC. ♦ Problem 8. Inversion I(S, r) translates Figure 21a into Figure 21b. As the two inverted circles are congruent, they are translations of each other along line Q P  . Hence, Q P  = S  E  (why?). Substituting the distance formulas r 2 ·QP r2 = SE ⇒ QP · SE = SP · SQ. yields the desired equality in part (b): SQ·SP For part (a), locate four isosceles trapezoids and a rectangle in Figure 21b. Each of these is inscribed in a circle (why?). Two of these circles will invert back to lines (as they contain S), and the other three will invert back to circles. ♦

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24

1. INVERSION IN THE PLANE. PART I Q T Dγ 1 S γ2 A CE R

γ2 B γ

P

T





I(S, r)

R E

A S P γ B γ1  D C Q

Figure 21. Problem 8

Exercise 12. Applying I(E, r) produces a configuration as in Figure 22a: E is still the midpoint of A B  and C  D , and the images of the parallelogram sides are four circles grouped in pairs of congruent circles tangent at E. B1

A k1

k4

C

C  A 1

E D k3

k2 B

A

B

M

C1

Figure 22. Alternative Parallelogram and Centroid Theorems To turn this configuration into a theorem, we choose a particular version of the original Parallelogram Theorem: “If AC || DB and AD || CB, then AC = DB and AD = CB, and the two diagonals AB and CD bisect each other in point E.” Inverting the statement through E leads to: Exercise 13. Let segments AB and CD intersect at E, and let the circumcircles of AEC, CBE, BDE, and DAC be k1 , k2 , k3 , and k4 , respectively. If k1 and k3 are tangent at E, and k2 and k4 are also tangent at E, show that k1 ∼ = k3 , k2 ∼ = k4 , and E is midpoint of AB and of CD. Problem 10. Applying I(A, r) results in segments AC1 and AB1 with midpoints B  and C  , respectively; A1 divides segment AM  in a ratio of 2 : 1 counted from A, and the quadrilaterals AB  A1 C  , AB  M  B1 , and AC  M  C1 are cyclic (cf. Fig. 22b). Further, the original condition A1 B = A1 C translates into quadrilateral AB  A1 C  having equal products of opposite sides (replace M by A in (8)). Similarly, PST 13 translates the condition BM : M B1 = 2 : 1 into 2AB  · M  B1 = AB1 · M  B  . We are ready to formulate




Problem 12 (Alternative Centroid Theorem 2). Let B and C be the midpoints of AC1 and AB1 , respectively. Let the circumcircles of ACC1 and ABB1 intersect for a second time at M , and let the circumcircle of ABC intersect line AM also in A1 . Then A1 divides segment AM in a ratio of 2 : 1 counted from A, and each circle contains a special inscribed quadrilateral: ABA1 C has equal products of opposite sides, and ABM B1 and ACM C1 have one product of opposite sides twice as large as the other product: 2AB · M B1 = AB1 · M B and 2AC · M C1 = AC1 · M C.

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Session 2 Combinatorics. Part I Paul Zeitz Sneak Preview. You may have seen the bumper sticker that says “Mathematicians Count Too.” That doesn’t begin to tell the story. Combining techniques from the “stone age” (when counting was done, well, with stones) with techniques from the “computer age” (when numbers are represented by binary digits), we can boldly say that the subject of combinatorics is now in its “golden age.” Through this session you will get a glimpse of the astounding world of counting: permutations and combinations, factorials, “menu-type” encodings, matchmaking and honeymooning, partitions and complements, multinomial coefficients along the Mississippi, dogs and biscuits, balls in urns, sororities of numbers and diagonals refusing to intersect . . . will all swirl in the “magical” pool of combinatorics. Unlike magic, however, revealing the “tricks” will not leave you disappointed. You will want to make these new counting techniques your own and go forth and slay your own dragons.

1. Two Counting Conundrums “Counting” sounds like a babyish topic: its technical name is enumerative combinatorics. This branch of math starts slowly, with no prerequisites at all, but quickly moves into very subtle problems that require good imagination and clarity of thought. Take, for instance, the following classic:

Figure 1. How many regions in the circle? Problem 1. There are n points in a circle, all joined with line segments. Assume that no three (or more) segments intersect in the same point. How many regions inside the circle are formed this way? 25

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26

2. COMBINATORICS. PART I

The reader can quickly verify that the answers for 1 through 5 points are correspondingly 1, 2, 4, 8, and 16 regions, all powers of 2 (cf. Fig. 1). Our intuition is screaming: “the answer for n points is 2n regions!” Alas, this is false, as one finds out by diligently counting the 31 regions in Figure 2a.1

Figure 2. Breaking the pattern vs. breaking the rules As a high school student, I was fascinated by this problem: I couldn’t stop thinking about it until I finally solved it! That’s how strong a hold a mysterious counting puzzle can have on a curious youth. Let’s not spoil the fun for you either: I will leave this problem for you to struggle with and possibly revisit it in a later session.




Meanwhile, our goal for the present session will be the solution of a 1995 problem from the Czech and Slovak National Olympiad. Problem 2. Do there exist 10,000 10-digit numbers divisible by 7, all of which can be obtained from one another by a reordering of their digits? Before we can attack this problem (which looks like a number theory question, even though it really isn’t), we need to start at the beginning. Counting involves the four basic operations of addition, subtraction, multiplication, and division. In order to count properly, you need to know which operation to do and when to do it. It turns out that multiplication is the easiest of the four, so we will start with it.

2. Multiplication, Menus, and Encoding 2.1. Menus make you multiply. Here’s an easy question: Exercise 1. A taqueria sells burritos with the following fillings: pork, grilled chicken, chicken mole, and beef. Burritos come either small, medium, or large, with or without cheese, and with or without guacamole. How many different burritos can be ordered? Solution: The answer is, of course, 4 × 3 × 2 × 2 = 48 because each different burrito is uniquely determined by making four decisions: filling, size, cheese, and guacamole, and those decisions involve, respectively, 4, 3, 2, and 2 choices.
1 Figure 2b displays 3 diagonals intersecting in a single point, which is not allowed. But even here, you will not count the “expected” 32 regions!

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2. MULTIPLICATION, MENUS, AND ENCODING



27

The order in which we make the above four decisions does not affect the outcome. We could have chosen the burrito’s size first and the filling last, etc. Any time we can use a menu analogy, we multiply. More precisely, PST 14. If the thing we are counting is the outcome of a multi-stage process, then the number of outcomes is the product of the number of choices for each stage. We can solve many problems with this approach, as long as we can carefully formulate the outcomes as a “menu” process. For the following, recall what sets and subsets are: you can think of them as collections and subcollections of objects. Exercise 2. How many subsets does a set with 8 elements have, including the empty set and the whole set itself? Solution: Can we make choosing a subset act like ordering a burrito? We need to decide which members of the set will be in the subset, and for each element this is a simple yes/no question. If the members of the set are a, b, c, d, e, f, g, and h, we need to merely ask 8 questions: “Is a in the subset?”, “Is b in the subset?”, “Is c in the subset?”, etc. So the number of ways of performing this 8-stage process is just 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 28 = 256.




Exercise 3. Given a pool of 30 students, how many ways can we choose a 3-person government consisting of a president, vice-president, and treasurer? Solution: We must make three decisions: who is president, who is vicepresident, and who is treasurer. We do not need to decide the three positions in this order; we could instead first choose the treasurer, say. Since the order is not counted, we will fix it using alphabetical order: president, then treasurer, and finally vice-president. Since there are 30 students, there are 30 possible choices for the first decision (president), but then there are only 29 left for the second decision (treasurer), and finally 28 for vice-president. In all, there will be 30 × 29 × 28 possible governments.
Exercise 4. In Exercise 3, we tacitly assumed that no one could hold more than one office. Verify that if we allowed this, the answer would be 303 . 2.2. Permutations and factorials. Exercise 3 used a very common counting entity known as a permutation. We denote the number of permutations of n objects taken k at a time by the symbol P (n, k). It is the answer to the question: Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.




28

2. COMBINATORICS. PART I

Problem 3. How many different ways can we choose k different things from a set of n objects, where the order of choice matters? Solution: Our previous observations generalize to the formula P (n, k) = n(n − 1)(n − 2) · · · (n − k + 1), because there are n choices for the first thing, (n − 1) for the second, etc., for a total of k terms in the product.
As a reality check, compare with the answer to Exercise 3: the number of governments was simply P (30, 3) = 30(30 − 1)(30 − 2). We get an important special case when k = n: P (n, k) = P (n, n) = n(n − 1)(n − 2)(n − 3) · · · 1. You probably know the more common notation for this: it is n!, called n factorial. This simply counts the number of ways that n different objects can be arranged in a row.2 For example, the number of ways we can permute ˝ is 5!, since there are 5 possibilities for the first letter, the letters of ERDOS 4 for the second, etc. Exercise 5. If you haven’t done so already, make a table of n! for n = 1, 2, . . . , 10. You should memorize it at least up to n = 7 and passively recognize the rest. Exercise 6. 10 boys and 9 girls sit in a row of 19 seats. How many ways can this be done if (a) all boys sit next to each other and all girls sit next to each other? (b) each child has only neighbors of the opposite sex? First, we warm up by solving an easier question: how many seating arrangements with no restrictions? This is just 19!. Now for the actual problem. Solution: We need to make 3 main decisions in part (a). Either the seating will have the boys on the left or on the right; then we need to seat the boys and then the girls. The first decision has 2 possible answers; the second and third have 10! and 9!, respectively. So the answer is 2 · 10! · 9!.
Since there are more boys than girls, in part (b) the seating has to be boy-girl-boy-. . . -girl-boy. But we still need to seat the individual kids. As before, there are 10! possible seating arrangements for the boys and 9! arrangements for the girls. Since we need to seat boys and girls to get a single seating arrangement, this is a 2-stage process with 10! · 9! outcomes.
2 Such rearrangements will play a prominent role in the Rubik’s Cube sessions, where twists of the faces are represented by permutations of the facelets.

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2. MULTIPLICATION, MENUS, AND ENCODING

29

2.3. If you encode, are you a spy? I often call the menu/multiplication method “encoding”, because it is sometimes very helpful to think like a computer programmer, organizing information compactly using simple coding to represent outcomes. For example, we could represent each subset in Exercise 2 with an 8-digit binary 3 number, with a 1 in position k if and only if the kth element was in the subset. Thus, the empty subset would be encoded by 00000000, while the subset consisting of the last two elements by 00000011. Exercise 7. How many ways can you choose a team from 11 people where the team must have at least one person and must have a designated captain? Solution: First, line up the 11 people in an arbitrary way (for example, by height or in alphabetical order). We will encode the team with an 11-digit binary number where digit k is a 1 if and only if person k is on the team. Since the team has at least one person, there has to be at least one digit that is 1. Underline one of these 1’s to represent the captain. How many ways can we perform this encoding? There are just 11 choices for the underlined digit 1. The remaining 10 digits of the 11-digit binary string have no restrictions; each choice of 0’s and 1’s yields a different subset of non-captains to join the captain. For example, a team where person #4 is the captain and the only person on the team is represented by 00010000000, while 00110100000 encodes a 3-person team with the same captain as before, joined by persons #3 and #6. It should be clear that the number of encodings is 11 × 210 .
Exercise 8. In a traditional village, there are 10 young men and 10 young women. The village matchmaker arranges all the marriages. In how many ways can he pair off the 20 young people? Assume (the village is traditional) that all marriages are heterosexual, i.e., a marriage is a union of a male and a female (male-male and female-female unions are not allowed).

Figure 3. Which bride do I get? 3 Instead of the 10 decimal digits 0,...,9, the binary system has only two digits: 0 and 1.

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30

2. COMBINATORICS. PART I

Solution: The most vivid visualizations are the most helpful with encoding (cf. Fig. 3 for the case of 5 men and 5 women). Imagine lining up all 10 young men, in some arbitrary-but-fixed order, such as height, wealth, looks, math ability, sense of humor, etc. Then the 10 young women line up opposite the men, in any order. Each woman marries the man directly opposite her. Clearly each different ordering of women gives rise to a different matrimonial outcome; the problem is reduced to counting such orderings, but this is just 10!.




We learn from this and previous solutions that PST 15. To “keep the chaos down” when counting and to have some control over what is happening, it is helpful to order all of the involved objects/people (or some subset of them). This order can be random or according to some specific criterion. Exercise 9. In a not-so-traditional village, there are 10 young men and 10 young women. The village matchmaker arranges all the marriages. In how many ways can he pair off the 20 men and women, if homosexual marriages (male-male or female-female) as well as heterosexual marriages are allowed? There are a number of different approaches; we will first give a menu-style solution and return to this problem later with other methods (cf. p. 40). Solution: Once again, it helps to visualize an effective matching process. In a similar way to Exercise 8, start by lining up all 20 people in some arbitrary order. Everyone has to get married; so the matchmaker points to the first person and chooses for him or her a mate from one of the other people in line. Then these two persons leave (perhaps to go on a honeymoon). This first decision had 19 possible outcomes. Now there are 18 frightened young people remaining in line. The matchmaker repeats the process, pointing at the first person in line and choosing a mate from the 17 people behind him or her in line. This process goes on until two persons remain: they are automatically mated. There will be 19 · 17 · 15 · · · 3 · 1 possible outcomes for marrying off the entire village in this sex-blind way.




You may wonder – in fact, you should wonder – if this method counts all different marriages and counts each outcome once. This is crucial for any counting problem. Here’s a way to see why it works. We need some ordering mechanism; let’s use alphabetical order. Imagine a particular outcome. This is a collection of 10 pairs of married couples. Within each couple, order the spouses alphabetically. Thus if Pat marries Dana4, we call this couple “DanaPat,” not “Pat-Dana.” 4 Of course, we may assume that no two people have the same name; if they did, we simply change one of them.

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3. ADDITION AND PARTITION



31

Then, we list the couples in alphabetical order of the first name of each couple. For example, if we only had six people named A, B, C, D, E, and F , one possible list might be A-E, B-C, and D-F . Note that the first couple in the list will always be one starting with A, and the next couple will start with the “lowest” unused letter, etc. It should be clear that this is a one-to-one encoding of all possible marriage arrangements. Certainly, each arrangement (for the entire village) will give rise to a different list. And likewise, each list (that obeys the alphabetical order rules) will give rise to a different marriage arrangement of the village. Finally, it should be clear that this encoding method is equivalent to our original argument. PST 16. When creating and using an encoding method, you must verify that there is a one-to-one correspondence between the set of encodings and all possible outcomes in the problem, i.e., that every encoding corresponds to exactly one possible outcome, and every possible outcome corresponds to exactly one encoding.

3. Addition and Partition 3.1. The limitations of multiplication. In Exercise 3, each decision in a multi-stage process influenced the number of decisions for the next stage. This is a pretty common situation, and easy to handle, as long as the number of decisions required at each stage stays constant. Exercise 10. How many even 3-digit numbers have no repeating digits? Before we solve this problem, let’s look at some simpler questions. First of all, how many 3-digit numbers are there? The answer is 900, which we can see in at least two ways: • Just count the numbers from 100 to 999, inclusive: there is a total of 999 − 100 + 1 numbers. • There are 9 choices for the first digit (no zero allowed) and 10 for each of the other two, yielding 9 × 10 × 10. Next, let’s ask the slightly harder question: how many 3-digit numbers have no repeating digits? There are 9 choices for the first digit, as before, but this means that now there will be 9 (instead of 10) choices for the second digit and 8 (instead of 10) choices for the third, yielding 9 × 9 × 8 = 648 numbers. Finally, let’s tackle the original question. How about this argument: • Start at the rightmost digit; to ensure that the number is even, there are 5 choices (0, 2, 4, 6, or 8). • There will now be 9 (instead of 10) choices for the middle digit. • There will now be 7 (instead of 9) choices for the first digit. • So the answer is 5 × 9 × 7 = 315. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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2. COMBINATORICS. PART I

This answer is not correct, due to a subtle error that stems from a common beginner’s mistake: trying to convert every problem into a multiplication. Can you spot the flaw? It was clever to start with the rightmost digit, but certain decisions will affect the number of choices for future stages in different ways. If the rightmost digit is zero, then the first digit will have fewer restrictions, since it is never zero! The correct way to do this is to PST 17. Break the outcomes into several separate, mutually exclusive cases, i.e., every possible outcome must belong to exactly one of these cases. Solution to Exercise 10: All outcomes boil down to whether 0 is the last digit or not; so only two mutually exclusive cases are needed. Case 1: The rightmost digit is zero. In this case, we have only two decisions to make. There are 9 numbers left, and we are free to use any of them for the other two digits. So there are 9 choices for (say) the first digit, and then 8 choices for the second digit, a total of 8 × 9 = 72 choices.

?

?

0

? ? 2, 4, 6, 8

Figure 4. Two mutually exclusive cases Case 2: The rightmost digit is not zero. Then there are 4 choices for this digit (2, 4, 6, or 8). Again, 9 numbers are left to use (including zero). Since we cannot use zero for the first digit, let’s decide the first digit next: there will be 8 choices. Now there are 8 numbers left for the second digit (since zero was left in the pool). So there will be 4×8×8 = 256 choices in this case. The answer, then, is the sum 72 + 256 = 328 numbers.




3.2. Partition leads to addition. We had to add to get the answer in Exercise 10 above because the outcomes split into two separate cases. A decomposition of the entire set into subsets that are pairwise disjoint (as in PST 17) is called a partition: imagine partitioning an office room into smaller cubicles. Partitioned outcomes are counted by adding. More precisely: PST 18. Whenever we partition the outcomes of something into several cases, each requiring different counting methods, we add the number of outcomes in each case to get the total number of outcomes. Note that this is not the same as multiple stages. In the previous exercise each case involved multiple stages. In general, you multiply when each outcome has a menu-inspired encoding. You add if you are forced to

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3. ADDITION AND PARTITION

33

partition the outcomes into separate cases; then for each case you may use completely different strategies (for example, some may involve encoding, and others may not). Exercise 11. An n-bit string is an n-digit binary number, i.e., a string of just zeros and ones. How many 10-bit strings contain exactly 5 consecutive zeros (no more, no less)? For example, we would not count 0000000111 (too many consecutive zeros), but we would count 1110000011 and 0011000001. Solution: We will break the problem into 6 cases, depending on where the block of 5 zeros lives inside the larger 10-bit string. The leftmost position of the 5-zero block can be position 1, 2, 3, 4, 5, or 6. In case 1, we start with 5 zeros; then we must have a 1 (to prevent excess zeros), and then we can have anything we want for the remaining 4 digits: 0 0 0 0 0 1 ? ? ? ?. So we have just 24 “free” choices. Case 6 works exactly the same way, by symmetry. However, cases 2–5 are different because now the 5-digit zero block “floats” inside the larger string and needs to be surrounded by a 1 on its left and on its right. For instance, case 4 can be represented by ? ? 1 0 0 0 0 0 1 ?. The remaining 3 digits are now free, so each of these cases has 23 outcomes. Hence the total number of outcomes is the sum 24 + 23 + 23 + 23 + 23 + 24 = 64.




Note that the cases of your partition cannot overlap at all. If they do, we will need to employ subtraction to fix the overcounting. This is the hardest operation of all, which we will discuss in depth in a later session. However, in very simple situations, we can use subtraction as a simple consequence of partitioning. 3.3. Counting the complement is very different from “counting on a compliment.” Exercise 12. Three different flavors of pie are available, and seven children are each given a slice of pie in such a way that at least two children get different flavors. How many ways can this be done? Solution: We need to be clear on the meaning of “different” here. In human societies, children tend to be respected as individuals; so let’s give each child a number and each flavor a letter. Then we can encode each outcome with a 7-letter string that uses the three letters. For example, a a a b b c a represents the outcome where children #1–3 got flavor a, children #4–5 got b, etc. Thus each different 7-letter string uniquely encodes a different outcome. Let’s first not worry about the condition that at least two kids get different flavors. By the simplest menu reasoning, there are 37 different strings. This includes outcomes with different flavors, as well as outcomes without different flavors. How many of the latter are there? Well, the only way that Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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2. COMBINATORICS. PART I

no two kids have a different flavor is if all the kids get the same flavor, and there are only 3 such outcomes (namely, a a a a a a a , b b b b b b b, c c c c c c c). Thus we can break up all outcomes into two cases: all flavors the same, and not all flavors the same. This is a partition of all outcomes, so the total must be 37 . Consequently, the number of outcomes where at least two kids
get different flavors (i.e., not all flavors the same) is 37 − 3. PST 19. The method above, called counting the complement, is used when we can partition the total set of outcomes into the things we are interested in, and the rest (its “negation” or complement). If the total is easy to count and the negation is easy to count, then we count the complement and our answer is just the difference of the two. Figure 5a depicts the general situation with a subset A and its complement Ac . When you count the complement Ac , you usually get a number which is easy-to-find or small compared to the total sum. We took advantage of this possibility in Exercise 12, where |Ac | = 3 and |A| = 37 − 3. A different application of partitioning arises when we can divide the set of total outcomes into two sets that have the same number of elements. The following is a nice example, which first appeared as a problem in the Bay Area Math Meet (BAMM). Exercise 13 (BAMM). How many subsets of the set {1, 2, 3, 4, . . . , 30} have the property that the sum of their elements is greater than 232? Solution: Notice that in the set S = {1, 2, 3, 4, . . . , 30} the sum of all the elements is 1 + 2 + 3 + · · · + 30 = 30 · 31/2 = 465. Let A be a subset of S, let Ac denote the complement of A (the elements of S which are not in A), and let Σ(X) denote the sum of the elements of any set X. Then Σ(A) plus Σ(Ac ) must equal 465. Because 465 = 232+233, if Σ(A) > 232, then Σ(Ac ) ≤ 232. For instance, if A = {2, 4, 6, . . . , 30}, then Ac = {1, 3, 5, . . . , 29} (cf. Fig. 5b) and Σ(A) = 240 > 232 > 225 = Σ(Ac ). complements

?

A Ac

{2, 4, 6, . . . , 30} = A ↔ Ac = {1, 3, 5, . . . , 29}

Figure 5. Complements in general and Twin subsets In other words, there is a one-to-one correspondence between subsets whose element sum is greater than 232 and subsets whose element sum is not (namely, A ↔ Ac ). Hence the number of subsets whose element sum is greater than 232 is exactly half of the total number of subsets of S. But the
number of subsets of S is 230 ; so the answer is 229 . Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

4. DIVISION: A CURE FOR UNIFORM OVERCOUNTING

35

4. Division: A Cure for Uniform Overcounting 4.1. The Mississippi formula. The number of different permutations of ˝ is 5!, as we saw on page 28. But ERDOS Exercise 14. How many different permutations does GAUSS have? Solution: It is easy to see that 5! is too large by exactly a factor of 2, because GAUSS has two letters that are the same. If we temporarily distinguish the two S’s by using subscripts, we see that 5! counts the permutations AS1 GUS2 and AS2 GUS1 as two different things when, in fact, they are indistinguishable (since the letters do not actually have subscripts). So each of the 5! = 120 different (subscripted) permutations can be arranged into two columns where in the first column, S1 is to the left of S2 , and in the second column it is the other way around. Hence the correct answer is 5!/2 = 60.
Suppose we now wished to Exercise 15. Count the different permutations of RAMANUJAN. Solution: This 9-letter name has 3 A’s and 2 N’s, so lots of overcounting occurs if we temporarily distinguish these letters and start with a provisional answer of 9!. Focus on one particular permutation, for example A2 MA1 RJN2 UN1 A3 . If we just permute the subscripts for each letter, we get a new rearrangement of the 9 letters, but one that is indistinguishable without subscripts. For example, one such permutation is A1 MA3 RJN1 UN2 A2 . How many such rearrangements are there? All we need to do is permute the 3 subscripts of the A’s (in 3! = 6 ways) and then permute the 2 subscripts of the N’s (in 2! = 2 ways), while leaving all the other letters in place. There are 3! × 2! = 12 different subscripted rearrangements that appear the same without subscripts. 9! ·
Hence the number of distinguishable permutations is 3!2! This is called the Mississippi formula, because it can be used to count the number of distinguishable permutations of the word MISSISSIPPI. The answer is, of course, 11!/(4!4!2!). If we wanted to, we could include the M, which appears only once, and write instead 11! , 4!4!2!1! which of course leaves the value unchanged. Expressions like the fraction above are called multinomial coefficients. We’ll see why, shortly. 4.2. Binomial coefficients. If there is such as thing as a multinomial coefficient, it must have a simpler cousin. Let’s look at a simple two-letter Mississippi problem: Exercise 16. How many 10-bit strings have exactly 4 zeros? Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

36

2. COMBINATORICS. PART I

Solution: This is merely asking us to count the number of ways of arranging 4 zeros and 6 ones in a string. By the Mississippi formula, it is 10! · (1) 4!6! Alternatively, we could start with 10 blank places and then choose locations for each zero, with the understanding that the remaining places will be occupied by ones. For the first zero, there are 10 possible places. The next zero has 9 possible locations. The third and fourth zeros have, respectively, 8 and 7 possible slots. By the menu principle, this 4-stage process has 10 · 9 · 8 · 7, or P (10, 4), possible outcomes. However, the order of choice doesn’t matter. For example, we could have picked place numbers 4, 9, 2, 3, in order, resulting in zeros at those locations. But we could have also chosen 9, 2, 3, 4, which would give the same result. So the product 10 · 9 · 8 · 7 has overcounted by a factor of 4!, and the actual answer is 10 · 9 · 8 · 7 · (2) 4! It is easy to see that this expression is equal to (1) above.


Expressions such as (1) are called binomial coefficients, denoted by
 n n! = k!(n − k)! k and pronounced as “n choose k”. Exercise 16 demonstrated the following facts about binomial coefficients.   Theorem 1 (Binomial Coefficients). The quantity nk (a) counts the number of ways to permute an n-bit string consisting of k zeros and n − k ones, (b) also counts the number of different ways to choose k places out of an n-place string where the order of choice doesn’t matter, and   = n(n−1)···(n−k+1) · (c) satisfies nk = P (n,k) k! k! Theorem 1(b) introduces a very important concept. The selection of k “places” from a pool of n places where order is not important is also called a combination, often to distinguish it from a permutation (where order counts). Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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37

This is somewhat old-fashioned terminology, and we have always found it confusing. We prefer the more rigorous interpretation that   Theorem 1 . nk counts the number of different k-element subsets chosen from an n-element set. Note that for sets, order doesn’t matter. For example, the sets {a, b, c} and {b, a, c} are the same. This subset interpretation has many interesting consequences. Here are a few for you to verify. If you have trouble, read the solution below. Try to minimize, if you can, use of formulas analogous to (1) on page 36 and algebraic manipulation. Instead, try to rely on Theorem 1 , that is, describe the involved quantities as certain subsets of a bigger set.




Theorem 2 (Properties of Binomial Coefficients).    n  . (a) For any nonnegative integers n and r with r ≤ n, nr = n−r n n (b) Because n = 0 = 1, the only logical value for 0! is 1.    n  n+1 = r+1 . (c) For any positive integer n with r < n, nr + r+1 n n n n (d) 0 + 1 + 2 + · · · + n = 2n for any positive integer n.   10 Concrete “solution”: In part (a), 10 3 = 7 because picking 3 places in a 10-bit string to be 0’s is equivalent to picking 7 places to be 1’s. ♦ n 0 has to equal 1 in part (b), since it counts the number of 0-element subsets of an n-element set. There is only one, namely the empty set. Plugging r = 0 into formula (1) yields 0! = 1. Note that we are not compelled to define 0!, but if we do, we must assign it the value of 1 in order to be consistent with formula (1). ♦ Let’s keep things concrete with a specific case in part (c). Why is


 13 13 14 + = ? 4 5 5 The right-hand side (RHS) counts the number of ways of awarding an ice cream cone to each of 5 lucky children from a class of 14. Suppose one of the children is named Ramanujan. We can partition the outcomes into two cases: either Ramanujan gets ice cream or Ramanujan does not get ice cream. Ramanujan 13 4

=

=

13 5

Figure 6. Does Ramanujan get ice cream? For the first case, we need to choose 4 more lucky kids from among the 13 kids remaining. For the second case, we need to eliminate the unlucky Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

38

2. COMBINATORICS. PART I

Ramanujan from consideration and choose 5 lucky kids from the remaining 13. These cases correspond to the two terms of the left-hand side (LHS). ♦ We solve a concrete case for part (d) also. If n = 11, the RHS represents (by Exercise 2) the number of subsets of an 11-element set, including the empty set and the entire set. The LHS is a sum representing partitioning these subsets into 12 cases: the number of empty subsets, the number of 1-element subsets, the number of 2-element subsets, etc. ♦ 4.3. Pascal’s Triangle all over. Seeing the properties in Theorem 2, some readers may recognize the connection between the so-called Pascal’s Triangle and our binomial coefficients. Briefly, Pascal’s Triangle is a triangle of numbers: its top row consists of a single 1, and each consecutive row has one more number than the previous; the first and the last number in each row are 1’s, while any other number is the sum of the two numbers directly above it. 0 1

1 1 1 1 1

2 3

4 5

1 1 3

1

6 10

4 10

1 5

1

1 0 1 2 0 2 1 2 3 0 3 1 3 2 3 4 0 4 1 4 2 4 3 4 5 0 5 1 5 2 5 3 5 4 5 0

1

2

3

4

5

Figure 7. Pascal’s Triangle and Binomial coefficients Figure 7 displays the first 6 rows of Pascal’s Triangle and the corresponding binomial coefficients. As one can easily check, the two triangles have identical entries. This correspondence means that whatever we can prove about binomial coefficients can be translated in terms of Pascal’s Triangle, and vice versa. Thus, property (a) in Theorem 2 indicates the symmetry of Pascal’s Triangle across a vertical line; (b) tells us about the end 1’s on each row; (c) is the defining addition property of Pascal’s Triangle; and (d) gives the sum of numbers on each row of Pascal’s Triangle. The combinatorics of Pascal’s Triangle is very rich, and it deserves a separate session of its own. Without further detours, we shall leave this topic for now and return to our binomial coefficients. 4.4. Some mathematical “etymology.” It is time to explain why binomial coefficients have this name. Consider raising the binomial (x + y) to the second power. We start with the product (x + y)2 = (x + y)(x + y). Now let’s expand it like a beginning algebra student, writing each “raw” term as it is first spewed out without combining like terms or switching the order of multiplications. Our expansion will be x(x+y)+y(x+y) = xx+xy+yx+yy. Of course, we then clean this up to get x2 + 2xy + y 2 . Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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39

Now try it for a higher power. Consider (x + y)7 = (x + y)(x + y)(x + y)(x + y)(x + y)(x + y)(x + y). If we multiply it all out, without combining like terms, we now get 27 = 128 raw terms: all monomials with coefficients of 1, such as xxxxxxx or xxxyyyx or xxyxyyx. Notice that the last two monomials are like terms, since both simplify to x4 y 3 . Also note that each 7-term string with 4 x’s and 3 y’s represents a different raw term. For example, the string xxxxyyy arose by multiplying the x’s from the first four binomials of the product with the y’s from the last three binomials. The string xyxyxyx arose by multiplying the x from the first term with the y from the second term, etc. Thus the number of raw terms is equal to the number of 7-term strings made of x’s and y’s. For instance, the number to x4 y 3 is, by the Mississippi 7 7of raw terms that simplify formula, equal to 4 or, equivalently, 3 . Thus the coefficient of x4 y 3 in   the expansion will be 74 = 73 . Similarly, the coefficients of x2 y 5 and of y 7   will be 75 and 77 , respectively. More generally, x7−k y k will appear with   coefficient k7 . We can write out the whole expansion, then, as



 7 7 0 7 6 1 7 5 2 7 0 7 7 x y + x y + x y +··· + x y . (x + y) = 0 1 2 7 This is an example, for the exponent 7, of the so-called Binomial Theorem 5, which expands the binomial expression (x + y)n . Now you can probably imagine where the name multinomial coefficients comes from. For instance, if you expand the 4-variable expression (m + i + s + p)11 and combine the like terms, the monomial corresponding to the 11! m1 i4 s4 p2 . With this said, permutations of “Mississippi” will appear as 1!4!4!2! it’s time for some popular folklore: a student response to an exam question.

5 The Binomial Theorem can be succinctly written as (x + y)n = Pn `n´xn−k y k . k=0 k

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40

2. COMBINATORICS. PART I

5. Balls in Urns and Other Applications Armed with encoding, partitioning, and binomial coefficients, we can now investigate some really interesting problems, including Problem 2. The key is to use creative encoding. 5.1. Choosing a honeymoon location. Let’s begin by revisiting Exercise 9, the marriage problem in the non-traditional village. Suppose each couple went on a different honeymoon. We could label each honeymoon location with one of the ten letters A, B, C, . . . , J . Now line up the 20 young people in some arbitrary order. Imagine that the matchmaker has twenty stickers: two labeled “A ”, two labeled “B ”, etc. The matchmaker then goes around putting one sticker on each person. That tells that person where he or she will honeymoon. Clearly this is a way to form all possible marriages, and by the Mississippi formula there will be 20! 20! = 10 2!2! · · · 2! 2   10

such marriages. However, we overcounted, since we are distinguishing the honeymoon locations. For example, the outcome where Pat and Dana go to honeymoon C and the outcome where Pat and Dana go to honeymoon H are counted as two different outcomes. To cure this overcounting, we need to divide by 10!, since there were 10 different honeymoon locations. So the correct answer is 20! · 10!210 Here’s another approach.   Choose two young people to go on honeymoon the A. This can be done in 20 2 different ways. Now pick two people from  18 remaining group to go on honeymoon B. This can be done in 2 ways. Continuing this process, the number of ways of arranging marriages where the honeymoons are distinguishable will be




 20 18 16 4 2 ··· , 2 2 2 2 2 and once again we need to divide by 10! to get the answer to the original question. It is a simple and fun exercise to verify that, indeed, the three approaches yield the same number.


 2 1 20 18 20! ··· = 19 · 17 · · · 3 · 1. = Exercise 17. Check that 10 10!2 10! 2 2 2 5.2. Geometry falls under the spell of combinatorics. Recall the enticing Problem 1, which asked for the number of regions in a circle. No, we shall not break our promise here – this problem is still yours to hack. But there are so many geometry problems of a similar spirit that they all Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

5. BALLS IN URNS AND OTHER APPLICATIONS

41

form a separate branch of mathematics named combinatorial geometry. This session would be amiss without at least one such example. Exercise 18. The n vertices of a polygon are arranged on the circumference of a circle so that no three diagonals intersect in the same point. (a) How many diagonals does the polygon have? (b) How many intersection points do these diagonals have? Solution: An easy application of the definition of binomial coefficients n tells us that there are 2 pairs of vertices   and hence there are that many segments joining them. Thus, there are n2 −n diagonals (we had to subtract the n sides of the polygon). Let’s call this number d.  At first you may think that the answer in part (b) is just d2 , but not all diagonals intersect. However, each intersection point in the interior is the intersection of two diagonals. The endpoints of these two diagonals are the vertices of a quadrilateral. Each choice of 4 vertices of the polygon gives rise to a different quadrilateral (cf. Fig. 8a). Once you focus on these quadrilaterals, it should be clear that there is a one-to-one correspondence between quadrilaterals formed from vertices of the polygon and interior diagonal intersection points. For instance, look at the pentagon in Figure 1: you can easily count 5 such quadrilaterals, each naccounting for one of the 5 interior
intersection points. So the answer is 4 . As a further   of a hexagon in Figure 2a:   example, check out the case we have 9 = 62 − 6 diagonals and 15 = 64 intersection points of these diagonals.

Figure 8. Quadrilaterals and Volleyball teams 5.3. Teaming up for volleyball. Exercise 19. How many ways can 10 people form two teams of 5?   Solution: You may think the answer is 10 5 , but this exactly double10 counts! To see why, note that 5 counts each 5-person choice as different from the complement, which is incorrect: once you choose a team, its complement is already automatically selected as the other team. Alternatively, one   counts the number of ways of choosing 5 people to play may realize that 10 5  
in the left half of the court (cf. Fig. 8b). The correct answer is 10 5 /2. This is really an easier version of the non-traditional marriage problem (a “marriage” is interpreted as a team of 5 people), but it may seem harder. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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2. COMBINATORICS. PART I

5.4. Dogs and biscuits are transformed into “urns” and “balls.” The next problem, on the other hand, seems quite innocent, but it is actually rather tricky. Think about it carefully before reading the solution. Problem 4. How many ways can 7 dogs consume 10 dog biscuits? The dogs are distinguishable; the biscuits are indistinguishable. Dogs do not share. Solution: Let’s arrange the dogs is some order (e.g., alphabetical). The difficulty is that we do not know how many biscuits an individual dog eats. Dog #1 could eat between 0 and 10 biscuits, and each of these 11 outcomes will affect the possibilities for the remaining dogs. Imagine a sample outcome and try to think of how we can encode generic outcomes. For example, dog biscuits

1 2 3 4 5 6 7 0 0 3 1 0 2 4

means that dog #1 ate no biscuits, while dog #7 ate 4, etc. What we want to count is the number of different tables of this sort. Instead of using numerals to indicate the number of biscuits, we can achieve some uniformity by replacing numerals with an equivalent number of symbols. For example, the following string means the same thing as the table above: |||bbb|b||bb|bbbbb| Instead of the numeral “3”, we write 3 b’s. We retain the | symbol to indicate the boundary of a table cell. This way, if two of these symbols are next to each other with nothing in between, it means that the table cell has a zero in it. For example, the first three symbols ||| translate into “zeros in the first two cells of the table.” Notice that our encoding method has more symbols than needed. The first and last symbols are always |’s. Thus we can eliminate them – they provide no information. So, for example, the string |b|b|bbb|||bbbbb is equivalent to the table dog biscuits

1 2 3 4 5 6 7 0 1 1 3 0 0 5

In this case, the first | is the rightmost boundary of the first cell, indicating a zero in the first cell, i.e., no biscuits for the first dog. If you have trouble understanding, here’s another way to think about it. Start with the following: | | | | | | These six vertical lines are the boundary lines of seven cells. Then we drop 10 b’s into the seven cells (the horizontal line is the “floor” which prevents the b’s from falling through). If n b’s lie to the left of the first vertical line, it means that the first cell has value n. If m b’s lie between the first and second lines, it means that the second cell has the value m, etc. If j b’s lie to the right of the rightmost vertical line, then the seventh cell has value j. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

5. BALLS IN URNS AND OTHER APPLICATIONS

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We don’t need the horizontal line, of course; it is just to make us think of physical cells for the b’s to drop down into. All we really are doing is creating a 16-symbol word consisting of 10 b’s and 6 |’s. For instance, the last table can be encoded on 16 slots as follows: | b | b | b b b | | | b b b b b

 16 16 By the Mississippi formula, this can be done in = ways. 10 6




This problem of course generalizes and gives rise to a useful formula that is called, among other things, the Balls in Urns or Stars-and-Bars formula. We prefer the former name because it is easier to remember: Theorem 3 (Balls in Urns). The number of ways to distribute b indistinguishable balls among u distinguishable urns is

 b+u−1 b+u−1 = · b u−1 I remember it by thinking, “It’s a hard formula; so the top is not something obvious like b + u. For the bottom, just think about dropping balls.” You’d never put u there by mistake! The Balls in Urns formula incidentally solves another equivalent, very frequently occurring problem. Exercise 20. How many ways can we choose 7 people out of 26 people, allowing for repetitions? Solution: Our intuition may suggest that the 26 people are the “balls,” being “dropped” into 7 winning slots. However, our intuition is wrong. Imagine that we are giving 7 identical awards to the 26 people, and the number of times a person is chosen corresponds to the number of awards he/she receives. Thus, the 7 awards are the “balls,” and the 26 people are the “urns.” For instance, if the people have names starting with the different 26 Latin alphabet letters, then in Figure 9, Amy received three awards, Bob – two awards, while Ivan and Lily – one award each. We can string together a name from these letters, namely, ALI BABA, but of course, the order of the letters does not matter.

 7 + 26 − 1 32 So, the answer is = = 3,365,856.
7 7

A

B

I

L

Y

Z

Figure 9. ALI BABA

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2. COMBINATORICS. PART I

6. Sororities of Numbers: A Promise Fulfilled Now we can finally tackle Problem 2. Recall that this problem asked whether there exist 10,000 10-digit numbers divisible by 7, all of which can be obtained from one another by a reordering of their digits.



6.1. Can “0” be a leader? Let us first solve a slightly simpler modified version of the problem: we shall allow “10-digit” numbers to start with the digit 0, e.g., we shall consider the number 1102 as a 10-digit number and write it as the string 0000001102. Thus, we have a total of 1010 numbers to work with. Ignore the issue of divisibility by 7 for a moment, and concentrate on understanding sets of numbers that “can be obtained from one another by a reordering of their digits.” Let us call two 10-digit numbers that can be obtained from one another in this way “sisters,” and let us call a set that is as large as possible whose elements are all sisters a “sorority.” For example, 1,111,233,999 and 9,929,313,111 are sisters who belong to a sorority with 10! 4!3!2! members, since the membership of the sorority is just the number of ways of permuting the digits. The sororities have vastly different sizes. The most “exclusive” sororities have only one member – for example, the sorority consisting entirely of 6,666,666,666; yet one sorority has 10! members – the one containing 1,234,567,890 (cf. Fig. 10). In order to solve this problem, we need to show that there is a sorority with at least 10,000 members that are divisible by 7. One approach is to look for big sororities (like the one with 10! members), but it is possible (even though it seems unlikely) that somehow most of the members will not be multiples of 7. Instead, PST 20. The crux idea is that it is not the size of the sororities that really matters, but how many sororities there are. If the number of sororities is fairly small, then even if the multiples of 7 are dispersed very evenly, “enough” of them will land in some sorority. Let’s make this more precise. Suppose it turned out that there were only 100 sororities (of course there are more than that). The number of multiples of 7 is (1010 − 1)/7 + 1 = 1,428,571,429. By the Pigeonhole Principle,6 at least one sorority will contain 1,428,571,429/100 multiples of 7, which is way more than we need.7 In any event, we have the penultimate step to work toward: compute (or at least estimate) the number of sororities. We can compute the exact number. Each sorority is uniquely determined by its collection of 10 digits where repetition is allowed . For example, one (highly exclusive) sorority can be named “ten 6’s,” while another is called 6 Part II of the Proofs session discusses the Pigeonhole Principle in detail. 7 a rounds a down to the closest integer ≤ a, while a rounds a up to the closest

integer ≥ a, e.g., 5.7 = 5 and 3.2 = 4.

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6. SORORITIES OF NUMBERS: A PROMISE FULFILLED

45

“three 4’s, a 7, two 8’s, and four 9’s.” So now the question becomes: In how many different ways can we choose 10 digits, with repetition allowed? Just as in Exercise 20, this is equivalent to putting 10 “balls” (one for each digit place) into 10 “urns” (one  each decimal digit 0, 1, 2, . . . , 9). By the Balls  for in Urns formula, this is 19 10 = 92,378. Finally, we conclude that there will be a sorority with at least 1,428,571,429/92,378 = 15,465 members divisible by 7. This is greater than 10,000.

0000000000 6666666666




1111233999 1234567890

Figure 10. Sororities of various sizes: from smallest to largest 6.2. “0” is not born to lead. Now back to our original problem. A couple of the calculations above must be adjusted, but the main idea of the proof remains the same. Since we do not allow a 10-digit number to begin with 0, we are dealing only with the numbers from 1,000,000,000 to 9,999,999,999. So there will be (1010 − 1)/7) − (109 − 1)/7 = 1,285,714,286 of them divisible by 7. Even though it has no bearing on our calculations, it is worth observing that the big sorority containing 1,234,567,890 will now have “only” 10!−9! = 9(9!) members, as we must exclude all 9! strings starting with 0. One would also expect big changes in the number of sororities, but . . . we “lose” only one sorority, namely, the sorority   corresponding to the 0-string 0000000000 (cf. Fig. 10). Thus, we have 19 10 − 1 = 92,377 sororities total. To complete the solution, we only have to divide: 1,285,714,286/92,377 = 13,919. Thus, some sorority must contain at least 13,919 members divisible by 7, so the answer is Yes.


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Session 3 Rubik’s Cube. Part I Tom Davis Sneak Preview. The 3×3×3 version of Rubik’s cube is a popular and seductive but very difficult puzzle. Its solution can be obtained using tools from a mathematical discipline called group theory. We will not discuss formal group theory in this first of three parts, but the ideas presented here will provide interesting examples of groups which we will use in later parts. Specifically, we will examine here permutations as they apply to Rubik’s cube and use them to provide crude tools that will allow you to solve a jumbled cube. (In Parts II–III, far more sophisticated tools will be developed.) Rubik’s cube provides great motivation to learn about permutations and, indirectly, about group theory. The cube itself is very interesting, so most readers will find this approach more enticing than the standard introduction to group theory in undergraduate courses, since here it has an actual, obvious, and “useful” application: to solve the cube!

1. Getting Started and Some Notation It will be far easier to follow this article if you have a physical Rubik’s cube or a computer simulation of one. It is best to have both. A computer simulation called Rubik, which runs on Macintosh or Windows machines, can be obtained at: http://www.geometer.org/rubik/ Exercise 1 (Rhetorical). If you have a jumbled physical cube and you don’t know how to solve it, you can turn the top face 45◦ , insert a screwdriver under an edge (not corner) cube, and pry out the edge cube. At this point, the cube can be completely taken apart and put together in the “solved” state. Notice that the colors on the center cubes completely determine the corner and edge cubes. When you are putting the cube back together, save an edge cube for last, turn the face missing the edge 45◦ , hook in one side of the linkage part, and press the cube in until it snaps. With Rubik, of course, you just need to press the “Reset” button. 47

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3. RUBIK’S CUBE. PART I

1.1. Basic Rubik’s cube “particles.” In all three parts of this topic, we will use the word cube to refer to the entire cube that appears to be divided into 27 smaller cubes. (In the physical cube, there is no actual small cube in the center.) We shall call these smaller cubes cubies, of which 26 are visible. There are three types of cubies: 6 lie in the center of each face of the cube, show only 1 face, and are called center cubies; 12 of them show 2 faces, called edge cubies (“edgies” ?); and 8 of them show 3 faces: the corner cubies (“cornies” ?). The entire cube has 6 faces, each of which is divided into 9 smaller faces of the individual cubies. When it is important to distinguish between the faces of the large cube and the little faces on the cubies, we’ll call the little faces facelets. We shall use this notation also when describing the twists used by the Rubik computer program. 1.2. Labeling faces and basic twists. Even though we have created color pictures as illustrations of the Rubik’s cube, it is a bad idea to use the face colors alone to describe faces or moves for two reasons. First, different cube manufacturers use different coloring schemes; and second, if you have a method, say, that flips the orientation of two adjacent edge cubies on a cube face, that method will work on any face of the cube, regardless of its color.

Figure 1. F move in befuddler notation A good way to describe moves is to imagine that the cube is held so that you are looking directly at one face, another face points at the floor and a third face at the ceiling. From your point of view, there will be a front, back, left, right, up, and down face. If the cube is held as in Figure 1a, the red face will be the front, the green face the left, the white face the top, and so on. We use the first letters of those words (conveniently all different), B, F, U, D, L, and R, to describe a quarter-turn clockwise twist about the back, front, up, down, left, and right faces, respectively.1 “Clockwise” refers to the direction to turn the face from the point of view of someone looking directly at the face. (This means, of course, that if you’re looking at the front face and want to perform a B twist, you’ll be turning the back face counterclockwise from your point of view.) An equivalent description is that a face move is accomplished by grasping that face with the right hand and giving it a one-quarter turn in the direction of the right thumb. In Figure 1b, an F move has been applied to the unjumbled cube on the left. 1 This is sometimes called the “befuddler” notation.

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2. ENCODING THE RUBIK’S CUBE MATHEMATICALLY

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We use the lower-case versions of those letters, b, f, u, etc., to refer to quarter-turn counterclockwise moves about the respective faces. We could talk about half-turns of faces, but those can just be accomplished by two quarter-turns in a row, so the moves above are certainly sufficient to solve any jumbled cube. If we need to talk about individual cubies, the same notation can be extended. The center cubies will have names like U, B, and so on. Edge cubies (with two colors) can be referred to as UR, LD, and so on. Similarly, the corner cubies have unique 3-letter names. The cubie at the corner of the left, top, and front faces is called LUF.

2. Encoding the Rubik’s Cube Mathematically 2.1. Permutations. A rearrangement of things is called a permutation.2 If the “things” are the facelets on Rubik’s cube, it is clear that Property 1. Every twist of a face is a permutation of the facelets. Obviously, in Rubik’s cube there are constraints on what rearrangements are possible, and that is what makes it so interesting. The three facelets appearing on a particular corner cubie, for example, will remain next to each other under every possible permutation. A solid understanding of permutations and how they behave will help you to solve Rubik’s cube. The cube has 54 visible facelets, so each cube movement effectively rearranges many of those 54.



Exercise 2 (Warm-up). Look closely at an F twist: how many of the 54 facelets are rearranged by this twist? How about a U twist? PST 21. The best ways to learn about any mathematical subject are to begin by looking at smaller, simpler cases and to reinforce your understanding by relating the theory to specific examples. Thus, at first, we will look at permutations of much smaller (than 54) numbers of items, where it is easy to keep track of everything. And while we discuss general properties of permutations, we encourage the reader to think about the meaning of these properties in the context of a few concrete examples. Rubik’s cube is one such concrete example, and we’ll introduce a few others as we proceed. 2.2. Inverse moves. We begin with a couple of obvious observations. If you grab the front face and give it a quarter-turn clockwise (in other words, you apply an F move), you can undo that by turning the same face a quarterturn counterclockwise (by doing an f move). 2 For more on permutations, consult the Combinatorics session.

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3. RUBIK’S CUBE. PART I

Exercise 3 (Warm-up). Perform a more complicated move, like F followed by R. How would you undo it? Will the sequence fr return to a solved cube? Why not? How about undoing FRfr? We can see what happens if we first apply FR in Figure 2a and then apply fr to that: the net result of FRfr appears in Figure 2b. This shows that fr does not undo FR.

Figure 2. fr is not FR’s inverse You can undo FR with an r followed by an f. Notice that you need to reverse the order of the moves you undo in addition to the direction of the turns. In other words, the FRfr sequence will not lead to a solved cube, while FRrf will. Becoming more ambitious, if you have just applied the sequence FRfr, to return to a solved cube, you’ll need to do an RFrf. Do you see why? Perform these moves now to return your cube to “solved.” In mathematics, if one move “undoes” another move, those moves are called inverses of each other. Thus F and f are inverses, as are U and u, etc. The inverse is often indicated with a little “−1” as an exponent. If we wanted to use this convention with our cube notation, we could write “F−1 ” in place of “f”, “U−1 ” instead of “u”, and so on. Since the standard computer keyboard does not allow you to type exponents, the lowercase versus uppercase notation is not only easier to type, but is more convenient. Keep in mind, however, that if you read almost any mathematical text that works with moves and their inverses, the use of “−1” as an exponent is the usual way to express an inverse. 2.3. Inverting sequences of moves. If we consider a series of moves as a unit, the unit is just a more complicated move that will itself have an inverse. In the example above, (FR)−1 = rf. This double-reversal idea – that rf is the inverse of FR, and RFrf is the inverse of FRfr – is very general. Property 2. If a, b, c, etc., are any moves with inverses a−1 , b−1 , c−1 , etc., then (abc · · · xyz)−1 = z −1 y −1 x−1 · · · c−1 b−1 a−1 . Because of this general principle, it is easy to write down the inverse of a sequence of cube moves: just reverse the list and then change the case of each letter to the opposite case. For example, the inverse of the sequence ffRuDlU is uLdUrFF. This will always work. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Notice also that in the case of Rubik’s cube moves another way to write the inverse of F is as FFF. In other words, if you twist the front face three more times, that’s the same as undoing the original twist: (FFF)−1 = F. We’ll look more at this idea later. 2.4. Exponential notation. Repeating moves (like applying F three times in a row)3 is such a common thing to do in math (and in life!) that we will want a notation for it. For this we use an exponential notation to indicate repetition of a move. Thus, “F3 ” is shorthand for “FFF,” where the “3” in the exponent indicates that the move is repeated three times. The exponential notation can be used to apply to a group of moves. For example, if we want to do the move FRFRFRFRFR (in other words, apply the combined FR move five times), we can indicate it as (FR)5 . Exercise 4 (Warm-up). To make sure you understand the exponential notation, why is F9 = F? Can you rewrite F6 in a simpler form? Solution: The move F applied four times in a row returns the cube to its original state so that F9 = F4 F4 F= F. Similarly, F6 = F4 F2 does exactly the
same thing as F2 : it rotates the front face 180◦ . 2.5. Is doing nothing worth anything? Since we are looking at all moves that can be performed on a cube, it is important not to forget perhaps the most important one: the move of doing nothing – of leaving the cube exactly as it was before. This is called the identity 4 move, indicated by 1. The notation 1 is not coincidental at all. Since FRB means doing “an F followed by R followed by B,” the notation FRB is designed to give the impression that we’re “multiplying” together those three moves. In the same vein, we’re also used to the idea that multiplying anything by the number 1 leaves it unchanged. Now, it is certainly true that 1F = F1 = F – doing nothing and then doing F, or doing F and then doing nothing, is the same as just doing F. Thus, the notation 1 for the Rubik’s cube identity move is reasonable both mathematically and psychologically.

3. Some Basic Features of Rubik’s Cube Moves 3.1. Commutativity and non-commutativity. We have seen above that the result of performing a sequence of moves can be thought of also as a move (but a more complex one) and that this “operation” of combining moves behaves a bit like multiplication. We also noted that the order in which we apply moves to the faces makes a difference. Figures 3a–b show the results of applying FR and RF to a solved cube; thus FR = RF. This is different 3 Recall applying inversion two times in a row in the Inversion session. 4 Compare this with the identity transformation in the Inversion session.

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from what we are used to in ordinary arithmetic, where if you multiply two numbers together, the order doesn’t matter. For example, 7 × 9 = 9 × 7, and there’s nothing special about 7 and 9.

Figure 3. FR = RF but LR = RL When the order does not matter, as in multiplication of numbers, we call the operation commutative. If it does matter, as in the application of twists to a cube or for division or subtraction of numbers (7/9 = 9/7 and 9 − 7 = 7 − 9), then we say that the operation is non-commutative. It’s easy to remember the name; you know what a commuter is: someone who commutes, or moves from one place to another, and back again. If an operation is commutative, the objects can “commute” across the operation, and the order doesn’t matter. As we saw in Figures 3a–b, the operation of applying cube moves one after the other fails to commute for some pair of operands, FR and RF, and hence it is non-commutative. Just because an operation is non-commutative, it does not mean that the result is always different when you reverse the order. In your cube, for example, FB = BF, UD = DU, and LR = RL; FF2 = F2 F, and so on. Figures 3c–d display the same result of applying RL and LR to a solved cube: LR = RL. (And in arithmetic, division is sometimes commutative: 1/(−1)=(−1)/1.) If twisting the cube faces were a commutative operation, then solving the cube would be trivial. You would just need to make sure that the total number of F turns, U turns, and so on, are multiples of 4, and you would be done. To see this on a small scale, Exercise 5 (Warm-up). Suppose your cube only allowed you to turn the front and back faces, but turns about the left, right, up, and down faces were not allowed. How do you solve the cube? Try this with your physical cube, and you’ll see that it is not a very interesting puzzle (due to the commuting FB = BF), although it might be interesting for a two-year-old. 3.2. Order. As we have observed before, if we start with a solved cube and perform the F operation on it four times, the resulting cube returns to a solved state. In other words, applying the F operation four times is the same as doing nothing, so that we can also write F4 = 1. As we do in most Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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other areas of mathematics, it is reasonable to define F0 = 1, since applying an operation zero times is the same as not applying it at all, which is our definition of 1. Similarly, F1 = F since an exponent of 1 corresponds to applying the operation once. If we look at successive powers of F: F0 = 1, F1 , F2 , F3 , and so on, then F4 is the first time that we return to the identity. For this reason, we say that the order of the operation F is 4: four moves do the job of returning to the original configuration, and no smaller number of moves will do so. What may appear at first to be somewhat amazing is that Property 3. Any cube operation has an order. In other words, if we begin with a solved cube and repeat the same sequence of moves enough times, the cube will eventually return to “solved.” Exercise 6 (Warm-up). Try to find the order of FFRR using a physical cube. Start with a solved cube and apply those four moves. You will find that the cube is a bit jumbled. Repeat the same four moves, and again, and again. Eventually (assuming you don’t make a mistake), the cube will be solved again. The total number of times you had to repeat the four-move combination is the order of that operation. Do try this. You will not have to repeat the operation too many times to get back to a solved state. Now, why is that so? In general, why will any sequence of cube moves, if repeated enough times, eventually return the cube to its starting configuration? We explore this as our first serious proof in the next subsection.

Figure 4. Order of FFRR




3.3. Existence of finite orders. Each time a Rubik’s cube operation is repeated, the facelets are rearranged. The experienced reader may know that Problem 1. There are fewer than 1 · 2 · 3 · · · 53 · 54 = 54! possible arrangements of the facelets in the Rubik’s cube.5 The less experienced reader should try to reason why there are only a finite (although very large) number of possible rearrangements. We connect this to a powerful PST, used in just about any area of mathematics. PST 22 (Pigeonhole Principle (PHP)). Given finitely many objects denoted by A1 , A2 , . . . , Ak , a list of k + 1 (or more) of these objects must contain a repetition of some objects.6 5 There are n! = 1 · 2 · 3 · · · n permutations of n objects (cf. the Combinatorics session). 6 The Pigeonhole Principle will be thoroughly discussed in the Proofs II session.

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Proof of Property 3: In our setting, applying repeatedly the same Rubik’s cube operation creates a sequence of arrangements of the facelets. Because there are only finitely many, say K, such arrangements, if we apply the operation K + 1 times, we are guaranteed by PHP eventually to repeat one of the arrangements. This does not prove yet that the cube will return to the initial configuration, but at least it shows that some arrangement will be repeated. Let us encode all of the above in mathematical notation. Call the operation P, where P stands for any combination of cube face twists. Thus, if we apply P repeatedly, eventually we will arrive at a repeat arrangement of the cubie facelets. Suppose that this first happens after m applications of P and that this repeat arrangement is the same as one that occurred at an earlier step k, with 0 ≤ k < m. Thus, {1 = P0 , P1 , P2 , . . . , Pk , Pk+1 , . . . , Pm , . . .} with Pk = Pm , and m is the smallest such number. If k = 0, we are done: 1 = P0 = Pm , so that the order of P is m. Suppose now that k > 0. Since Pk = Pm , this means that if we apply P either k times or m times to the same initial cube, we arrive at the same final cube arrangement. If we apply P−1 to that arrangement, the result will be the same, no matter whether it was arrived at after m or k steps. (Applying the same operation to the same arrangement will yield the same result.) But applying P−1 at the end of Pk and of Pm exactly undoes the final application of P: if you apply P first m times and then undo P once, that’s the same as just applying it m − 1 times, and similarly for k. Thus, we discover that well-known algebraic manipulations work on cube arrangements too: Pk P−1 = Pk−1 and Pm P−1 = Pm−1 . Therefore Pk−1 = Pm−1 , contradicting the assumption that m was the smallest value where the rearrangement repeats. Thus k must be equal to 0, so Pm = 1. This shows that every cube operation has a finite order.
3.4. Orders and solving the Rubik’s cube. Property 3 is very interesting for a couple of reasons. From a purely artistic viewpoint, if you take a solved cube and repeatedly apply the same set of operations, it will eventually return to solved again and again. But if your main goal is to solve a jumbled cube, the fact that any pattern eventually returns to solved can be actually turned into a brute-force mechanism that you could use, although your solution might be somewhat inefficient. The author solved the cube for the first time this way. 3.5. “Macros” and solving the Rubik’s cube. The usual method to solve a cube is to find combinations of moves, which, when applied as a unit (called a macro), do very specific things to the cube. For example, if you found a move (a series of twists) that would flip two edge cubies in place and the cube you were trying to solve had two edge cubies in an orientation where flipping them both would bring you closer to “solved,” you could apply the macro and bring your cube one step closer to solution. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

4. VISUALIZING PERMUTATIONS



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The question is, how do you discover such macros that do very small, specific things and leave most of the cube unaltered? It turns out (and we shall see why later) that PST 23. If you have a macro with a certain order and you apply it for half or a third of that number of steps, the result is often a useful (although sometimes very long) macro. As an example, consider the FFRR macro, with which we experimented before. You should have discovered that the order of this macro is 6 in Exercise 6. The “large” divisors of 6 are 2 and 3, so you may find interesting macros by repeating the FFRR combination twice or three times. In this case, three times is probably more useful: Exercise 7. Starting with a solved cube, demonstrate that applying FFRR three times has the net result of exchanging two pairs of edge cubies and leaving everything else exactly as it was. Check your work against the cube in Figure 4c. Thus FFRRFFRRFFRR = (FFRR)3 might be a useful macro to you as you’re solving a cube.




It’s easy to look for such macros. Simply type in to Rubik various (usually short) sets of moves and find the order k of those operations. If k is divisible by a small number like 2 or 3, or perhaps 5, divide k by that number and apply the macro a number of times equal to the quotient. Problem 2. Using the above recipe, find at least one more macro useful for solving the Rubik’s cube. To see why this strategy often produces useful macros, we will need to take a detour to learn something about the structure of permutations.

4. Visualizing Permutations 4.1. How to think of permutations? A permutation is a rearrangement of a set of objects. Keep in mind that it is the rearrangement that’s the important part; usually not the objects themselves. In some sense, the permutation that exchanges items 1 and 2 in the set {1, 2, 3} is the same as the permutation that exchanges A and B in the set {A, B, C} – the rearrangement is the same; it’s just that the names of the particular items are different in the two cases. In what follows, we’ll just consider the objects to be moved to be the numbers from 1 to N . One way to think about permutations of N objects is to visualize a set of boxes numbered 1 to N and a set of balls with the same numbers 1 to N , where each box contains a single ball. A permutation consists of taking the balls out of the boxes and putting them back, either in the same or different boxes, so that at the end, each box again contains exactly one ball. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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For example, Figure 5 illustrates a permutation of balls in the boxes labeled {1, 2, 3, 4}: the blue ball stays in box 1, the red ball moves from box 2 to box 3, the green ball moves from box 3 to box 4, and the yellow ball moves from box 4 to box 2. 1

2

3

4

1

2

3

4

Figure 5. Moving balls among boxes via permutation 1342 4.2. Two-row and one-row permutation notations. We can encode the movement of balls in Figure 5 by the two-row notation
 1234 (1) , 1342 where the top row lists the original boxes of the (blue, red, green, yellow) balls and the bottom row lists the new boxes of these balls. A similar two-row notation can encode any permutation of any number of objects. To make sure you have understood this way of thinking about permutations, do the following concrete examples: Exercise 8 (Warm-up). (a) List all permutations of {1, 2, 3} in two-row notation.
 12345 (b) Draw a diagram (as in Figure 5) for the permutation . 35124 After doing a number of examples, it becomes apparent that the top row of the two-row notation can be dropped, as long as we have a natural order on the objects we permute. Thus, the numbers 1, 2,..., N will always be listed in this order on the top row; the latter is then understood unambiguously and can be skipped altogether. In this vein, the caption of Figure 5 refers to permutation 1342: this is the same permutation as in (1), which moves 1 → 1, 2 → 3, 3 → 4, and 4 → 2. There are advantages and disadvantages to each of the two-row and onerow notations. For example, note that the permutation 1342 in Figure 5 can be described equally well by any of the following two-row notations:



 1234 2134 4321 3412 (2) or or or 1342 3142 2431 4213 or in any of 20 other forms, as long as there’s always a 1 under the 1, a 3 under the 2, a 4 under the 3, and a 2 under the 4. Exercise 9 (Warm-up). Write the permutation 213 in several two-row notations. What is the total number of such two-row notations for 213? Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

5. CYCLE STRUCTURE OF PERMUTATIONS

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4.3. Which notation is more suitable for the Rubik’s cube moves? As we observed, the one-row notation works because there is a natural order of the objects 1, 2, . . . , N . But there is no such natural order to the cubies or faces in Rubik’s cube – does a yellow face “naturally” come before a red face? Who knows?

5. Cycle Structure of Permutations 5.1. Cycle notation: why is it better? The two-row notation introduced in the previous section certainly works to describe any permutation, but there is a much better way: the so-called cycle notation. If we are looking at a particular permutation, say,
 12345678 (3) α= , 76153248 we can begin at any box and see where the contents of that box are moved by the permutation (cf. Fig. 6a). If that first ball doesn’t remain fixed, it moves to a new box (1 → 7), so the ball in that new box is moved to yet another box (7 → 4), and so on (4 → 5 → 3). Eventually, a ball has to move back to the original box since there are only a finite number of boxes (indeed, 3 → 1). This forms a cycle where each ball moves to the next position in the cycle and the moves eventually “cycle around” to the original box. We denote this cycle by (17453). (Warning: even though this looks like the one-row notation, the parentheses “( )” distinguish it as the cycle notation.) 3 1

2

3

4

5

6

7

8

5 1 7

6

2

8

4

Figure 6. Cycle notation 76153248 = (1 7 4 5 3)(2 6)(8) The cycles can have any length from 1 up to N , the total number of boxes. In our example, (17453) is a cycle of length 5; item 8 forms a cycle of length 1 (since it doesn’t move, or if you like, it moves to itself); and in addition, 2 and 6 switch places, so they form another cycle of length 2. The full cycle notation for α is (1 7 4 5 3)(2 6)(8). To interpret cycle notation, the set of items between each pair of parentheses form a cycle, with each moving to the box of the one that follows it; finally, the last one in the list moves back to the box represented by the first one in the list. Note that the cycles will be disjoint (or independent) in the sense that each item will appear in only one cycle: if an item appeared in two different cycles, then it would be attempting to follow two different paths, which is inconsistent with permutations. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Here is an exercise so standard that versions of it appear in any undergraduate abstract algebra textbook: Exercise 10. Convert (a) α = (398)(14562)(7) to two-row notation;
 12345678 (b) β = to cycle notation. 32167845 5.2. Canonical cycle notation. Note that the cycle notation is not unique, although it can be made so. All permutations in the list below are equivalent: (1)(2 3 4) (1)(3 4 2) (1)(4 2 3) (2 3 4)(1) (3 4 2)(1) (4 2 3)(1). Since the cycles are independent, we can list them in any order, and since we can begin with any element in the cycle and follow it around, a cycle of n objects can appear in any of n forms. This makes it a bit difficult to determine at a glance whether two descriptions of a permutation in cycle notation are equivalent. But if there is some sort of natural ordering to the objects, then it is possible to form a canonical cycle notation: 1. Find the smallest unused item in the list and begin a cycle with it. 2. Complete this cycle by following the movement of the objects by the permutation and close the cycle. Mark all these items as “used.” 3. If all of the objects in the permutation are used, you are done; otherwise, return to step 1. For example, the canonical form of the cycle above is (1)(2 3 4), while in Exercise 10 the two given permutations have canonical forms (14562)(398)(7) and (13)(2)(46857), respectively. Often, if the set of objects being permuted is obvious, the objects that do not move are not listed. Thus (1)(2 3 4) might be listed simply as (2 3 4). With this convention, however, there’s no reasonable way to list the identity permutation that moves nothing, so it is often listed as (1), where only one example of a non-moving object is listed, or even as 1 to indicate that it is the identity permutation. If you were listing the primitive cube operations (the face twists) in this cycle notation, the convention of leaving out 1-cycles would be a big advantage. Of the 54 facelets on a cube, a single face twist only moves 20 of them, which obviates listing 34 of the 1-cycles. 5.3. Cycle structure of a permutation. A very important feature of a permutation is captured when it is listed in cycle notation, and that is its cycle structure. For example, the cycle structure of (1)(2)(3 4 5)(6 7 8)(9 10 11 12) consists of two 1-cycles, two 3-cycles, and one 4-cycle. To see why this is important, let’s begin with a few simple examples. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

6. APPLICATIONS OF CYCLE STRUCTURE TO THE CUBE

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Consider (1 2 3). If this operation is applied three times, it is obvious that the result is the identity permutation. Each time it is applied, every element advances to the next box in the cycle; but the cycle is three steps long, so after three steps, each object will return to where it started. As another example, consider P = (17453) from Figure 6: it consists of a single 5-cycle, so that P 5 = P 10 = P 15 = P 20 = · · · = 1. In fact, Property 4. If P is a permutation consisting of a single n-cycle, i.e., P = (i1 i2 i3 · · · in ), then P n = 1. Furthermore, if you apply P repeatedly, it will return to the identity after every n applications. Next, let’s consider the permutation P = (1 2 3)(4 5 6 7): it consists of both a 3-cycle and a 4-cycle. Since the two cycles have no elements in common, if we apply P repeatedly and don’t pay any attention to the elements in the 4-cycle, we see the elements in the 3-cycle returning to their initial locations every three applications. Similarly, if we ignore the elements in the 3-cycle and pay attention only to those in the 4-cycle, then after every 4 applications of P those four elements return to their starting places. In other words, the elements in the 3-cycle return to their original locations for P 3 , P 6 , P 9 , P 12 , P 15 , and so on. Similarly, the elements in the 4-cycle return to their original locations for P 4 , P 8 , P 12 , P 16 , and so on. Notice that P 12 appears in both lists and that this is the first exponent of P that is in both lists. This means that after 12 applications of P , the elements in both the 3-cycle and the 4-cycle are returned to their starting locations, and furthermore, this is the first time that that happens. Thus P 12 = 1, and the order of P is 12. The number 12 is the least common multiple of 3 and 4, usually written as lcm(3, 4) = 12. In other words, 12 is the smallest (positive) integer that is a multiple of both 3 and 4. It should be evident from the discussion above that Property 5. If a permutation consists of two cycles of lengths m and n, then the order of that permutation is simply lcm(m, n). The concept of a least common multiple can be extended easily to any number of inputs, e.g., lcm(4, 5, 8, 7) = 280 since 280 is the smallest multiple of 4, 5, 8, and 7. If a permutation consists of a 4-cycle, a 5-cycle, an 8-cycle, and a 7-cycle, then the order of that permutation would be 280. Exercise 11. Find the orders of (398)(14562)(7) and (36748125).

6. Applications of Cycle Structure to the Cube 6.1. The macro (FFRR)3 . Consider the permutation P = (1 2)(3 4 5 6 7), whose order is lcm(2, 5) = 10. What happens if we repeat P five times: what does P 5 look like? The 5-cycle will disappear, since after 5 applications every element in it has cycled back to its starting point. The 2-cycle will have been applied an odd number of times, so it will remain a 2-cycle. Thus P 5 = (1 2). Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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3. RUBIK’S CUBE. PART I

Therefore, although the permutation P by itself moves 7 objects, the permutation P 5 moves only two objects. If the objects 1, 2, . . . , 7 in this example were really cubies in Rubik’s cube, then the net result of 5 repetitions of the operation P would be an operation that moved exactly 2 cubies and left all the others where they were! Hey, wasn’t this called a macro, useful in solving the cube? In Exercises 6-7 we saw an example of this: the operation FFRR moves 13 cubies, but (FFRR)3 moves only 4 – it exchanges two pairs of edge cubies. Let’s see why this is so using our cycle notation. Recall the names for the individual cubies described at the end of Section 1. For example, “DF” indicates the “down-front” edge cubie, “DBR” represents the “down-backright” cubie, etc. Problem 3. Using the individual cubies’ names, find the cycle structure of the move FFRR. Explain why (FFRR)3 is a product of two disjoint 2-cycles. Solution: The cycle decomposition is FFRR = (DF UF)(DR UR)(BR FR FL)(DBR UFR DFL)(ULF URB DRF), i.e., FFRR consists of two 2-cycles and three 3-cycles. Obviously, since there are 13 cubies in the permutation cycle listing, 13 of the cubies are moved by FFRR and the remaining 13 cubies are fixed. But nine of those 13 appear in 3-cycles in FFRR, so the permutation (FFRR)3 leaves those nine cubies fixed, moving only the two pairs of edge cubies that we noticed earlier. Thus,
(FFRR)3 = (DF UF)(DR UR). 6.2. Refining cycle notation in the Rubik’s cube. The Rubik program has a command “Display Permutation” (in the “File” pull-down menu) that will display the permutation required to get to the current cube coloring from the solved cube. Although the notation above appears to describe the permutation, there are a couple of problems with it: 1. If a cubie is left in its position but is rotated (a corner cubie) or flipped (an edge cubie), then there is no way to indicate this. Figure 7a shows the result of the macro LdlfdFUfDFLDlu. (We will learn later where this comes from.) All the cubies are left in their original locations, and all in their original orientations except that the ULF and URF cubies are rotated 1/3 of a turn in opposite directions. 2. Even if the cubies move in a cycle to different positions on the cube, there is again no way to indicate how they are flipped or rotated in their new positions relative to how they were before. The easiest way to indicate the details of a permutation exactly is to list where every facelet of every cubie moves. Assuming that the 6 center cubies stay in place, the remaining 48 facelets can move; hence such a complete description is a lot longer and doesn’t make it quite so obvious which cubies move to which locations.

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6. APPLICATIONS OF CYCLE STRUCTURE TO THE CUBE

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So in spite of its drawbacks, the first form of the notation (as used in Problem 3) is usually more useful. It can be improved slightly to indicate cubies that flip or rotate in place as follows: • (UF) means that the UF edge cubie stays in place, but is flipped. • (URF) means that the URF corner cubie is twisted in place where the up facelet moves to the right facelet, the right to the front, and the front back to the up facelet.

Figure 7. Rotated cubies and Generating a useful macro When you issue the “Display Permutation” command, you will be presented with both the most useful and most accurate descriptions of the permutations. The notation for indicating the movement of cubie facelets requires that each corner cube be assigned the names for its three facelets and each edge cubie needs two. The URF corner cubie has the following three facelets: [Urf], [Ruf], and [Fur]. The three letters indicate which corner it is, and the letter in upper case is the particular cubie facelet. Similarly, the two facelets of the cubie [UF] are [Uf] and [Fu]. The description of the movement that flips the UF and UL cubies in place is (Lu Ul)(Fu Uf). To save you the trouble of counting the number of terms in each cycle, the standard cycle notation is listed below for each permutation, e.g., 4(3) 2(5) 1(6) 1(8) 1(12) means that the particular permutation consists of four 3-cycles, two 5-cycles, and one each of a 6-cycle, an 8-cycle, and a 12-cycle. In the first listing Rubik may also display a certain number of 1-cycles, but these simply represent cubies that stay in place and are either flipped or rotated. Look at the detailed permutation description to see what they are. Cubies that stay in place without being flipped or rotated are not listed as 1-cycles, and neither are the 1-cycles in the face-permutation listing. 6.3. Creating useful macros. Here follows an example where the permutation cycle form can be used to find a macro that would be truly useful for solving the cube, although it contains far too many steps. On the other hand, if you didn’t know any better techniques, this one would work. The example also illustrates one of the shortcomings of the cubie-based cycle notation. Although you apply a 9-cycle nine times, it does not return completely to solved, since those movements have a net effect of flipping some edge cubies. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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3. RUBIK’S CUBE. PART I

If you look at the cubie-facelet permutation, you will see that one of the cycles, in fact, has length 18. Imagine that you’ve experimented with a number of short move sequences and you stumble across this one: FUllR (recall that l = L−1 ). Problem 4. Find the order and the standard cycle structures of FUllR. Which power of FUllR might be a useful macro? Solution: The order of FUllR is 36, but take a look at the cycle notation: (UR UF)(UL UB BR DR FR DF BL FL DL)(RFU)(BRU) (DRF UBL DBR)(DLB ULF)(DFL). We see here cycles of lengths 9, 3, and 2. At first it looks like applying FUllR 9 times might be useful since that would only leave a pair of 2-cycles. But when you try this, you obtain (FUllR)9 = (UR UF)(UB)(UL)(DF)(DR)(DL)(FR)(FL)(BR)(BL)(DLB ULF). In fact, the long cycles also flip cubies when they operate . . . far too much is done by this operation! However, since the order of the macro was 36, not 18, an additional 9 applications will undo the flips and must leave something changed afterwards. As a result, the cycle structure is simply (FUllR)18 = (UF)(UR), which flips two cubies in place. The unfortunate thing is that 18 applications of a 5-step macro, or 90 total twists, are required to do this; but if you have no better options, at least it will do the job. If not fast in practice, this macro is excellent for theoretical purposes of solving the Rubik’s cube.
Figures 7b–c illustrate the results of applying separately FUllR9 and FUllR18 to a solved cube. Notice that the double application flips certain edge cubies twice, leaving them as they were originally.




Problem 5. Using Rubik, investigate the properties of the move FRfr, the order and cycle structure of that permutation. Apply multiples of FRfr that are lengths of the cycles and see if you can generate useful macros from it. Repeat the experiment with other moves that you invent, different from FRfr.

7. Conclusions If we know the cycle structure of a permutation of the facelets of Rubik’s cube, we can repeat the permutation certain numbers of times that are multiples of the individual cycle lengths and use these repeated moves to generate macros that are quite powerful, although sometimes quite long. In a later volume of the book we will learn much more powerful techniques that produce shorter macros, using what we have already learned about permutations in combination with more sophisticated mathematical tools.

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Session 4 Number Theory. Part I Remainders, Divisibility, Congruences and More Zvezdelina Stankova Sneak Preview. Delivered originally in 1998 at the former Oakland Math Circle, this session answered a common demand: elementary concepts associated with numbers, such as remainders and divisibility, pop up everywhere in problem solving and even in research mathematics; yet, our standard knowledge suitable for “grocery store” calculations is vastly insufficient in exploring intriguing problems with numbers. So we shall start here from the very beginning, learn about basic properties of numbers, build up the framework of congruences, and move on to various applications in number theory, ranging from easy exercises to some of the toughest IMO challenges, to be continued in Part II.

1. Wearing the Crown of Mathematics “However impenetrable it seems, if you don’t try it, then you can never do it.”

Sir Andrew John Wiles

1.1. The power of numbers. We can simply, but reasonably correctly, claim that there are three fundamental building blocks of mathematics: numbers, geometric figures and logical reasoning. Just as geometric objects and mathematical statements can be thought of as systematic collections of points and words, so can integers be broken into their indivisible components known as prime numbers. As we have already paid proper respect to geometry and logic, it is high time to turn to the realm of number theory, considered by many as the “queen of all mathematical sciences”. Many a time, upon being “revealed” to be a mathematician at a party, I have been asked the inevitable question: “Wow! What are the largest numbers you have ever multiplied?” By now the reader must be aware that mathematicians prefer, by and large, to be challenged with hard and deep problems rather than to show off extraordinary computational abilities. So, what do mathematicians really do with numbers? Try the following. 63

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Problem 1. Prove that 1n + 2n + 3n + · · · + (n − 1)n is divisible by n for any odd n > 1. Checking small cases is easy enough with a calculator at hand: for n = 3, 5, and 7 the sums 9, 1300, and 376761 are divisible by 3, 5, and 7, respectively, but any further brute force calculation becomes exceedingly cumbersome and less illuminating. A method bypassing an actual computation of the total sum, yet showing that this sum is divisible by n, is what we will be after in this session. The more advanced reader may question the choice of the above example as a representative of the full power of number theory. True. So see how far you can get with Problem 2 (IMO, Australia ’88). If for positive integers a and b the a2 + b2 is an integer, then show that it is a perfect square.1 ratio 1 + ab To better understand the problem statement, plug in a couple of possibilities for a and b. If a = 2 and b = 3, the fraction 13/7 is not an integer, rendering the example irrelevant. But if a = 2 and b = 8, the fraction is 68/17, which is not “just” the integer 4: it is 22 , the square of an integer. Even though Problem 2 looks overall like some sort of a “simple division exercise,” we will see that it incorporates a bundle of mathematical ideas ranging from algebra to number theory, utilizing inequalities and reasoning by contradiction, and very unexpectedly relying on the method of induction. The problem was solved perfectly by only 12 students at the IMO, and very few partial scores were given: it was “all or nothing.” As if to compensate for this level of difficulty, one of the correct solutions written by the Bulgarian Emanuil Atanassov was so elegant and beautiful that he received the IMO brilliancy award for it. In Part II the patient reader will be rewarded with a thorough discussion of Emanuil’s solution by his IMO ’88 teammate, the author of the present session. Still, the well-read reader will argue that much harder problems should “carry the flag” of number theory. No objections. For three and a half centuries, the following question stumped and exhilarated humanity: Theorem 1 (Fermat-Wiles). If n is an integer greater than 2, the equation xn + y n = z n has no positive integer solutions (x, y, z). One can certainly try cases for small values of n, e.g., n = 3, 4, etc., and may even succeed in proving some of them. But, as Andrew Wiles, the British mathematician who finally defeated Fermat’s challenge in 1995, tells us: “Mathematicians aren’t satisfied because they know there are no solutions up to four million or four billion, they really want to know that there are no solutions up to infinity.” 1 A perfect square is the square of an integer, e.g., 49 = 72 .

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1. WEARING THE CROWN OF MATHEMATICS

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The quote refers to numerous cases for the power n which were indeed resolved by hand or by computers; yet no matter how many cases were done, there still remained infinitely more to pursue. Anyone who even remotely knows about Wiles’s actual proof will tell you that it is impossible to describe it in a few easy-to-understand sentences. In fact, it is well beyond the scope of this book or of any undergraduate textbook and becomes accessible only in an advanced graduate course on modular forms.2 Wiles’s proof is truly a product of the 20th century, for it is based on works of a number of modern mathematicians.3.

Figure 1. Andrew Wiles and Pierre de Fermat But Pierre de Fermat (1601–1665) wrote in 1637 that he had “a truly marvelous proof of this proposition which this margin is too narrow to contain.” Fermat’s proof was never found, nor rediscovered subsequently. “The tiny possibility that there does exist an elegant 17th -century proof,” suggests Wiles, is “what has made this problem special for amateurs . . . .” Could anyone, a high school student perhaps, recreate the elusive elementary proof? In this session, we will not attempt to find an alternative proof of Fermat’s Last Theorem. We will instead start from ground zero – we shall learn the basics about numbers, which Wiles and Fermat must have certainly mastered before moving onto “bigger and greater things.” 4 2 Two graduate textbooks representative of the subject’s depth and breadth are the

recent Diamond-Shurman [21] and its 30-year old predecessor Edwards [25]. The real diehards who want nothing but the original must try Andrew Wiles [99], followed by a joint article with his former student Richard Taylor [91], which fixed a gap in the original proof. For the “human side” in Wiles’s spectacular achievement, watch NOVA’s documentary The Proof [84], and for a popular treatment, browse through Singh’s bestseller [83]. Special cases of Fermat’s Theorem are tackled with more elementary tools in Ribenboim [75]; while previous attempts at Fermat’s Theorem can be found in Poorten [74]. 3 Shimura, Taniyama, Weil, Hellegouarch, Frey, Mazur, Serre, Ribet, Taylor, . . . . 4 Speaking of which, Figure 1a shows Wiles addressing the final ceremony at IMO ’01 in Washington, D.C., and bestowing the gold medals. That year the USA tied with Russia for the second place among 83 countries. Half of the USA team was comprised of Berkeley and San Jose Math Circlers: Oaz Nir won a silver medal, Tiankai Liu won a gold medal, while Gabriel Carroll “topped the charts”: he achieved a perfect score.

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1.2. On-going number challenges for humanity. We hope that the session so far has shed some light on the incredible range and difficulty of modern number theory problems. However, if the above does not seem challenging enough – these are, after all, solved problems – try to reshape your destiny by conquering another puzzle: Twin Prime Conjectures. There are infinitely many primes p such that (a) (Euclid, 300 B.C.) p + 2 is also prime; (b) (Germain, 1825) 2p + 1 is also prime. The common definition “a prime number p is an integer divisible only by itself and 1,” is not quite correct since a prime has in fact four distinct divisors: ±p and ±1. The first few (positive) primes are 2, 3, 5, 7, 11, 13, 17, 19, 23. Note that we didn’t include 1 among the primes – it is so special that it has its own name: 1 is a unit in Z. Going back to the Twin Prime Conjectures, examples in part (a) are the pairs of primes (p, p + 2) = (3, 5), (5, 7), and (11, 13); and in part (b): (p, 2p + 1) = (3, 7), (5, 11), and (11, 23). The p’s in part (b) are called Sophie Germain primes, e.g., 3, 5, 11, etc. “Incidentally,” among the many scientific achievements of Marie-Sophie Germain (1776–1831), she made partial progress on Fermat’s Last Theorem: she showed that there were no solutions for the prime exponents less than 100. She further proved that if n = p is a Sophie Germain prime and x, y, and z are integers satisfying xp + y p = z p , then at least one of x, y, or z must be divisible by p (cf. [76]).5 The importance of prime numbers themselves should be quite evident by now. Before we move onto “serious business,” some time for fun is due. To the reader who enjoys an occasional non-traditional humorous angle on math concepts, we offer Carroll’s cartoon.

5 As was not uncommon for those times, Marie-Sophie Germain worked under a male pseudonym in order to gain respect for some of her earlier works. Her paper in mathematical physics on acoustics and elasticity won a research contest at the French Academy of Sciences in 1911 and is applied even today in skyscraper constructions. Germain was finally recognized as a scientist and, apart from wives of other members, she became the first woman to attend sessions at the French Academy of Sciences.

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2. REMAINDERS: WHERE IT ALL BEGAN

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2. Remainders: Where It All Began Even though this section may seem “obvious” or “lacking” much mathematical content to some readers, it is fundamental to our future understanding of number theory properties and full of logical tips and basic problem solving ideas. The reader well versed in remainders and in proofs may skip to Section 3 on congruences “at his/her own risk.” 2.1. Deconstructing mathematical statements. As everyone knows, dividing one integer by another will not always result in a whole number, but we can make the “leftover” as small as possible. For example, dividing 13 by 5 results in 2 with 3 left over, or equivalently, 13 = 2 · 5 + 3. The number 13 is called the dividend, 5 is the divisor, 2 is the quotient and the remainder is 3. Don’t sweat if you can’t remember all of these names. The important thing to notice is that 3 < 5, i.e., we can always make the remainder smaller in magnitude than the divisor. Figure 2a, for example, portrays a folk line dance requiring 5 people in each line; because there are 13 people altogether, 3 “outcasts” will not be able to join the dance. Let us say this more formally.

Figure 2. Dancing in quintuples




Problem 3. If a and d are integers and d = 0, we can divide a by d to obtain a quotient q and a remainder r, i.e., a = q · d + r such that 0 ≤ r < |d|. Furthermore, these integers q and r are uniquely determined by a and d. If you think that we will quickly scribble the proof of Problem 3 and move on, you are mistaken. We shall first internalize the statement well by asking the “right” questions. 2.1.1. Do we notice any concepts whose appearance seems striking or “unjustified”? The absolute value |d| jumps out at us: why do we need it? Well, if we divide by a negative d, we still want to arrive at a non-negative remainder r, but the inequalities 0 ≤ r < d < 0 won’t make much sense! This is remedied by the absolute value |d| > 0. For example, dividing 13 by −5 yields 13 = (−2) · (−5) + 3: the remainder 3 satisfies 0 ≤ 3 < |−5|.

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4. NUMBER THEORY. PART I

2.1.2. Have we seen special cases of the problem before, possibly under a different name? The situation with remainder r = 0 should be familiar: Definition 1. If a = qd for some non-zero integers a, d, and q, we say that d and q divide a, or that d and q are divisors of a, and write d | a and q | a. Further, we say that d | 0 since 0 = d · 0. For instance, 5 | 10, but 5
13. Note that the bar in “d | a” is vertical, not slanted; while 5/10 stands for the usual fraction 1/2, the expression “5 divides 10” means something different, so be careful! Also, another trivial observation: if d | a, it is not necessarily true that d ≤ a; indeed, 7 | (−14) but 7 > −14. To avoid annoying minus signs, we conclude in general that d | a implies |d| ≤ |a| (unless a = 0; why?). 2.1.3. What is the problem really asking us to do? Problem 3 claims two things: that the integers q and r exist and that they are unique. “But this is obvious,” most people would say. “We’ve learned to divide with remainders since childhood and it has always worked!” Instead of taking this experience as unshakable reasoning, we prove here Problem 3 from scratch.



2.2. “Cogito ergo sum” – “I think, therefore I am,” said the French mathematician René Descartes (1596–1650) and thus authored one of the most famous phrases about existence in Western philosophy. But really, how do we prove that certain things exist? PST 24. One way to show existence of a mathematical object is to exhibit a specific example of such an object; another way is to describe a construction leading to such an object. Suppose we blank out and forget how to divide: is there still a way to obtain the remainder when 13 is divided by 5? Yes, instead of division, Euclid would have used only subtraction. Starting with 13, he would keep subtracting 5: 13  13 − 5 = 8  13 − 2 · 5 = 3  13 − 3 · 5 = −2, until he arrived at a negative number (−2) and would declare the preceding, non-negative number 3 to be the remainder r. Exercise 1 (Warm-up). Without using division, find q and r when a is divided by b: (a) a = 1998, b = 600; (b) a = −1998, b = 600; (c) a = −1998, b = −600; (d) a = 1998, b = −600. Do you notice connections between answers in different parts? Hint: Obviously, subtraction won’t do us any good when a = −1998 and b = 600. How about addition? Remember that you are looking for a nonnegative remainder r. ♦ We summarize Euclid’s approach formally in:

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2. REMAINDERS: WHERE IT ALL BEGAN

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Proof of Existence: If both a ≥ 0 and d > 0, consider the differences: a  a − d  a − 2d  · · ·  a − md  · · · . Set r to be the smallest non-negative such difference, say, r = a − qd for some q. Since we start with a positive a and decrease it by d each time, we will eventually plunge under 0: the difference right before this plunge is our r. Rewriting, we have a = qd + r with r ≥ 0. Thus, we have constructed our quotient q and remainder r ≥ 0; we only need to verify that r < d. To the contrary, suppose r ≥ d. For instance, imagine that there were 7 “outcasts,” as in Figure 2b (r = 7, d = 5, and 7 > 5). But then 5 of these “outcasts” could form their own dance line, so there are really only 2 outcasts, not 7. In general, we can subtract another d from r and get an even smaller non-negative difference a − qd = r  a − (q + 1)d = r − d ≥ 0, which contradicts our choice for a − qd as the smallest such difference. We conclude r < d. Hence we have constructed the desired quotient q and remainder r.
Exercise 2. Modify the above proof to show the existence of a quotient q and a remainder r when a and/or d is negative.

 

2.3. To be unique is a desire embedded deeply in human nature – to be different from the others, not duplicable, distinguished. Apparently, a number of objects such as remainders and quotients have similar ambitions. PST 25. To show that a mathematical object is unique, assume that there are two such objects, A and A1 , and prove that they are the same: A = A1 . Proof of Uniqueness: We suppose that in addition to one quotient q and remainder r, there is another quotient q1 and remainder r1 : a = qd + r = q1 d + r1 .

(1)

PST 26. When faced with two equal analogous expressions, regroup the similar terms, and make conclusions based on these new groups of terms. In our situation, we pull the quotients to one side and the remainders to the other side, factor and make conclusions about divisibility: (2) (q − q1 )d = r1 − r

Def. 1

⇒

d | (r1 − r) ⇒ |d| ≤ |r1 − r| or r1 − r = 0.

But now recall that by definition divisors are always greater in magnitude than remainders, e.g., |d| > r and |d| > r1 . Hence the warning bell: is it really possible in (2) for d to divide the difference (r1 − r) of two remainders?      0 r1 r

 |d|

Figure 3. Remainders and divisors: |r1 − r| < |d| Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Whether r or r1 is the larger remainder, Figure 3 illustrates that the distance between them is smaller than |d|, i.e., |r1 − r| < |d|. This seems to contradict (2), but in fact, there is one possibility remaining for d to divide (r1 − r): when r1 − r = 0, i.e., r1 = r and hence q1 = q in (1). Now we are truly done: we have shown that dividing a by d results in a unique quotient q and a unique remainder r.
2.4. Did we waste our time going through the proof of a well-known “obvious” fact? Not at all. In addition to teaching us something about existence and uniqueness proof methods, Problem 3 lays the foundation for the mathematical area of abstract algebra, and hence its proof and consequences are discussed in detail in any such standard college textbook (cf. Hungerford [45] and Gallian [29]).

3. Congruences in Z For simplicity, all “numbers” from now on will be integers and we will always divide by a positive number, unless otherwise stated. 3.1. A world made exclusively of remainders. Different numbers have different remainders when divided by d, so we need a special framework to enable us to work only with these remainders and to forget about “the rest of the world.” Does such a framework exist? Exercise 3 (Warm-up). Pick your favorite positive integer d. If a1 and a2 have remainders r1 and r2 when divided by d, is it always true that (a) a1 + a2 has remainder r1 + r2 when divided by d? (b) a1 − a2 has remainder r1 − r2 when divided by d? (c) a1 · a2 has remainder r1 · r2 when divided by d? (d) a1 /a2 has remainder r1 /r2 when divided by d? The reader should be aware of something “fishy” in the above exercise: adding up two remainders does not always yield a remainder, e.g., if d = 7, r1 = 4 and r2 = 5, then r1 + r2 = 9, which certainly does not qualify as a remainder! Ditto for the other proposed operations. In fact, part (d) poses an even stupider question: not only might r1 /r2 fail to be a remainder, but r1 /r2 and a1 /a2 themselves might not be integers! The new mechanism we are looking for ought to allow r1 + r2 , r1 − r2 , and r1 · r2 to still be used in place of remainders. To make this happen and to avoid cumbersome repetitions, we introduce new notation. Definition 2. Let d be a positive integer. Two numbers a and b are called congruent modulo d if they have the same remainder when divided by d, i.e., a = q1 d + r and b = q2 d + r. We denote this by a ≡ b (mod d). Exercise 4. Explain why 5 ≡ 9(mod 4), but 5 ≡ 7(mod 4). Are 2007 and 1707 (Euler’s year of birth) congruent (mod 13); how about mod 300? Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

3. CONGRUENCES IN Z

71

Note that modulo is another word for “when divided by.” As opposed to geometry, where two figures are congruent if all of their features are the “same,” note that two numbers are congruent modulo d if they just share the same remainder (mod d). 3.2. Ecologically friendly congruences. What does the world modulo d really look like? For d = 5, the world consists of 5 trees (cf. Fig. 4). Each tree is named after one of the 5 possible remainders. Tree “3,” for instance, hosts all numbers a with remainder 3, i.e., a ≡ 3 (mod 5). Note that the numbers on different trees do not overlap, and each number resides on exactly one tree. 2006 −9

11

−10 2005 −5

10

0

5

14

6

−4

1

1

−8

2007

12

−3

2

7

0 2

−2

−7

8

−1

−6

13

9

2009

4

2008

3

4 3

Figure 4. The world modulo 5

Exercise 5. Draw the world modulo d for d = 2, 3, 4, and 6. Is there a connection between the world modulo 2 and the world modulo 4, that is, can you use the trees (mod 4) to obtain the trees (mod 2)? How about connections between the trees (mod 3) and the trees (mod 6)? The world modulo 2 should be familiar to everyone: one tree consists of all even numbers, and the other tree of all odd numbers. As you may have seen, the even numbers can be written as 2k, and the odd numbers can be written as 2k + 1, where k is any integer. In fact, all numbers residing in a single tree (mod d) can be encoded in a similar fashion. Exercise 6. For each tree in the world modulo 5, find formulas describing all numbers residing in this tree. Repeat the exercise for a general d. Partial Solution: In the case of d = 5, let us describe all numbers in the tree corresponding to remainder 3, i.e., all a such that a ≡ 3(mod 5). By definition, when we divide a by 5, we must get remainder 3, i.e., a = q · 5 + 3. It is customary to rewrite this formula as a = 5k + 3 where k is any integer. More generally, the tree corresponding to remainder r(mod 5) is described by a = 5k + r for any integer k. ♦ Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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4. NUMBER THEORY. PART I

3.3. Translating from English to English. Observe that the tree corresponding to remainder 0 in Figure 4 is special: it hosts all numbers divisible by 5. More generally, Lemma 1. Let a and b be any numbers. In the world mod d (a) d | a iff a ≡ 0 (mod d); (b) a is congruent to its own remainder (mod d); (c) a ≡ b (mod d) iff d |(a − b). For instance, • 2000 ≡ 0 (mod 5) because 5 | 2000; • 2007 ≡ 2 (mod 5) because the remainder of 2007 (mod 5) is 2; • 2006 ≡ 11 (mod 5) because 5 | (2006 − 11) = 1995. Note that substituting b = 0 or b = r in part (c) yields parts (a) and (b). Proof of Lemma 1(c): Before we tackle the equivalence sign iff, we are allowed to make general assumptions about a and b. Let a = qd + r and b = q1 d + r1 where r and r1 are the corresponding remainders (mod d). Then a − b = d(q − q1 ) + (r − r1 ). By comparing the LHS and RHS, we see that d | (a − b) exactly when d | (r − r1 ) (why?). But the proof of remainder uniqueness (in Problem 3) landed us in an identical situation: d | (r − r1 ) exactly when the two remainders are equal, i.e., r = r1 . By definition of congruence, this means a ≡ b (mod d).
One can use Lemma 1(c) as a translation device between the language of congruences and the language of divisibility. But what is the advantage of congruences over divisibility, if we can freely translate back and forth between the two? The main disadvantage of “d | (a − b)” is that it is a statement: it is not always obvious how to “manipulate algebraically” several such statements, while it is often relatively straightforward to do so with the equations representing congruences.

4. Properties of Congruences




Let us consider the following “innocent looking” Exercise 7. Find the remainder of 21998 (mod 3) and of 31998 (mod 5). Obviously, no one expects us to raise 2 to the power 1998 and then to divide the resulting horrendous number by 3 in order to find its remainder: there must be a shortcut for doing such things. One path to the discovery of such a shortcut will rely on knowing properties of congruences. In this section, we will study the most basic properties of congruences. Some will be fairly transparent, but others will make us work to prove them. Except for the very last property, which combines congruences by different moduli, we will be using the same “modulus d ” throughout: the properties will fail in general if we do not restrict them to a single “modulus d.”

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4. PROPERTIES OF CONGRUENCES

73

4.1. Congruence as an equivalence relation. One can use the sign “≡” for congruent numbers very much like the sign “=” for equal numbers. For instance, recall the following trivialities on equal numbers: a = a for all a; • Reflexivity: if a = b then b = a; • Symmetry: • Transitivity: if a = b and b = c then a = c. The same properties are true for congruent numbers: Lemma 2. If a, b and c are any numbers, then (a) a ≡ a (mod d) for all a; (b) if a ≡ b (mod d) then b ≡ a (mod d); (c) if a ≡ b (mod d) and b ≡ c (mod d) then a ≡ c (mod d). Partial Solution: There is almost nothing to prove here if one correctly uses the definition of congruence. For instance, in part (b), if ra and rb are the remainders (mod d) of a and b, respectively, we reason as follows: def

sym

def

a ≡ b (mod d) ⇒ ra = rb ⇒ rb = ra ⇒ b ≡ a (mod d). In other words, the symmetry of “=” implies the symmetry of “≡.”




The reflexive, symmetric and transitive properties of “=” are so ingrained in our brains since early childhood that one is tempted to assume them for any relation. Big “no-no!” Take for instance the relation of “smaller than,” i.e., “a < b.” It satisfies the transitivity property – if a < b and b < c then a < c – but fails the other two (check it!). Similarly, the relations “a ≤ b” and “a | b” fail one of the three properties (which one?). In fact, relations like “a = b” and “a ≡ b (mod d)” that satisfy reflexivity, symmetry and transitivity are so important that they have a special name: equivalence relations.6 4.2. Algebraic properties of congruence. Given a couple of equalities a = b and c = d, we can do a bunch of operations on them. For instance, we can add, subtract, and multiply them, and, if no zeros are involved, we can even divide them: a + c = b + d, . . . , a/c = b/d. Let’s see which operations are possible on congruences. Lemma 3. If a ≡ b (mod d) and c ≡ e (mod d), then (a) a + c ≡ b + e (mod d); (b) a − c ≡ b − e (mod d); (c) a · c ≡ b · e (mod d). Proof of (a): The key point here is to translate back and forth between congruences and divisibility, using Lemma 1(c). Thus, we are given (3)

d | (a − b) and d | (c − e).

6 For the advanced reader: the fact that congruence is an equivalence relation is what made it possible to represent the world (mod d) as d non-overlapping trees.

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4. NUMBER THEORY. PART I

But then d divides the sum of the two numbers (why?), i.e.,     (4) d | (a − b) + (c − e) = (a + c) − (b + e) . By Lemma 1(c) again, this means that a + c ≡ b + e (mod d).
We leave the subtraction property (b) for the reader to prove in a very similar way, and we move directly to the more difficult part (c).



Proof of (c): Again we are given d | (a − b) and d | (c − e), and by Lemma 1(c) we need to show that d ? | (ac − be). But nothing obvious works: multiplying the given divisibilities yields d | (a−b)(c−e) = (ac+be−ae−bc), which is true but unrelated to the desired expression ac − be. Instead, PST 27. We can force given expressions into the desired expression by adding and subtracting the same quantity. Let’s see how this works in practice. The “given” expressions (about which we have some information) are (a − b) and (c − e); the “desired” expression (about which we want to obtain information) is (ac − be). It takes a moment of thought to see that the “magic” quantity to be subtracted and added to (ac − be) is ae: (5)

ac − be = ac − ae + ae − be = a(c − e) + (a − b)e.

The last is certainly divisible by d (why?); hence the first ac − be is also divisible by d. Thus, d | (ac−be) and by Lemma 1(c), a·c ≡ b·e (mod d).
If the idea of adding and subtracting in (5) seems too tricky, we offer in the Hints section an ugly but more straightforward alternative. A few more algebraic manipulations common for equations are also valid for congruences: adding or subtracting the same number to/from both sides of the equation, multiplying both sides by the same number, or raising both sides to the same natural number power. Corollary 1. If a ≡ b (mod d), then for any integers h and k > 0: (a) a + h ≡ b + h (mod d), a − h ≡ b − h (mod d) and ah ≡ bh (mod d); (b) ak ≡ bk (mod d). Hint: We can view each property as a special case of Lemma 3. For instance, Corollary 1(b) is a repetitive application of Lemma 3(c). Indeed, if you want a6 ≡ b6 (mod d), all you need to do is multiply 6 copies of the same congruence a ≡ b (mod d). ♦ 4.3. Division is bad for congruences for two reasons: as we mentioned earlier, the resulting ratios might not be integral; or escaping this fate, the new congruences can be just plain false. For example, 6 ≡ 4 (mod 2) and 2 ≡ 2 (mod 2). Obviously, it doesn’t make sense to divide the first congruence by 3 (why?), but we can try dividing the first by the second congruence: 6 ? 4 2 ≡ 2

(mod 2) is false since 3 ≡ 2 (mod 2). Hence, as far as congruences are concerned, division at this point is outlawed, so restrain yourselves appropriately in this session. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

5. REMAINDERS LEARN TO RIDE A BIKE





75

4.4. Bridging two different modular worlds. So far each congruence property was restricted to a single modulus d at a time. Our last congruence property cautiously brings together two moduli d1 and d2 . Lemma 4. Let d1 and d2 be relatively prime numbers. If a ≡ b (mod d1 ) and a ≡ b (mod d2 ), then a ≡ b (mod d1 d2 ). To start with, d1 and d2 are called relatively prime if they have no divisors in common other than ±1. For example, 4 and 9 are relatively prime, while 4 and 6 are not (why?). Note that being “relatively prime” has nothing to do with being “prime.” The question is, why should relative primeness of d1 and d2 matter in Lemma 4? Take, for instance, d1 = 4 and d2 = 6. It is true that 5 ≡ 17 (mod 4) and 5 ≡ 17 (mod 6), but 5 ≡ 17 (mod 4·6), showing that Lemma 4 can fail when the moduli are not relatively prime. Setting a − b = c and using Lemma 1(c), Lemma 4 can be rephrased as Lemma 4 . If d1 and d2 are relatively prime, d1 | c and d2 | c, then d1 d2 | c. Lemma 4 may look obvious to most people, and yet its proof is highly non-trivial: it is based on the so-called Euclidean algorithm for finding greatest common divisors, which can be found in any standard abstract algebra textbook (cf. [45, 29]). We shall skip the proof here in favor of finally applying the framework of congruences to problems.

5. Remainders Learn to Ride a Bike We are ready now to start applying congruences. Let’s first tackle Exercise 7. Recall that we were asked to find the remainders of 21998 (mod 3) and 31998 (mod 5). Is there a fast way to find these remainders without having to raise 2 or 3 to the power 1998? 5.1. Negative numbers to the rescue. Finding the suitable congruence property to help us is half the problem. Corollary 1(b) is the obvious “suspect” since it deals directly with powers; so let’s try it. We must start with some congruence 2 ≡ b (mod 3), conclude that 21998 ≡ b1998 (mod 3), and then calculate b1998 (mod 3). But for which b is the calculation of b1998 (mod 3) easier than the original 21998 (mod 3)? On one hand, 2 is congruent (mod 3) to 5, 8, 11, etc., and these numbers evidently can only complicate the power calculation of b1998 . . . . But on the other hand, 2 ≡ −1 (mod 3), which looks perfect! Indeed, since (−1)1998 = 1
and 21998 ≡ (−1)1998 (mod 3), we have 21998 ≡ 1 (mod 3). Isn’t it amazing how a negative number can solve a problem phrased entirely for positive numbers? We should immediately try this approach on 31998 (mod 5). But 3 ≡ −2 (mod 5) does not give us anything helpful since calculating (−2)1998 (mod 5) is not so straightforward! We need a different idea for this power. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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4. NUMBER THEORY. PART I

5.2. Cycling remainders. It seems that the 1998th power of 3 was chosen arbitrarily – why not look for the remainders of 32007 or 3891 instead? PST 28. If certain parts of a problem seem randomly chosen, try generalizing the problem, thereby concentrating on the important features of the situation. Solving the generalized problem may actually be easier than solving the original specific problem. Thus, let’s try to Exercise 8. Find the remainder of 3n (mod 5) for any n ∈ N. Solution: It is intuitive to start from small powers and build up to larger powers, so we construct a list of initial examples: • • • •

31 ≡ 3 (mod 5), ≡ 4 (mod 5), 32 = 31 · 3 ≡ 9 33 = 32 · 3 ≡ 4 · 3 = 12 ≡ 2 (mod 5), 34 = 33 · 3 ≡ 2 · 3 = 6 ≡ 1 (mod 5),

• • • •

35 36 37 38

= 34 · 3 ≡ 1 · 3 ≡ 3 (mod 5), = 35 · 3 ≡ 3 · 3 = 9 ≡ 4 (mod 5), = 36 · 3 ≡ 4 · 3 = 12 ≡ 2 (mod 5), = 37 · 3 ≡ 2 · 3 = 6 ≡ 1 (mod 5).

The reader should have noticed two things. First, we can calculate every remainder by multiplying the previous one by 3: if 3n ≡ k (mod 5), then 3n+1 = 3n · 3 ≡ 3k (mod 5). Second, it seems we are caught up in a pattern of remainders (mod 5): {3, 4, 2, 1, 3, 4, 2, 1, ...}.       Will this pattern persist forever? Yes, on account of the single most important remainder 1: 34 ≡ 1 (mod 5), which causes the sequence of all remainders to repeat itself. Thus, for instance, when the exponents n are divisible by 4, the powers will always have remainder 1: 3n = 34m = (34 )m ≡ 1m = 1 (mod 5). Then any power of the form 34m+1 will be congruent to 3: 3n = 34m+1 = 34m · 31 ≡ 1 · 3 = 3 (mod 5). Similarly, one can show that 34m+2 ≡ 4 (mod 5) and 34m+3 ≡ 2 (mod 5).

312 38 34 3

311

0 3

4

2

1

37 3

9

5

31

35 3

32 36 310

Figure 5. Bicycle of powers 3n (mod 5) Thus, the four non-zero remainders (mod 5) will repeat themselves in a cycle of length 4, as the bicycle in Figure 5 illustrates. The big wheel depicts the Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

5. REMAINDERS LEARN TO RIDE A BIKE

77

repetitious patterns of the powers 3n (mod 5), and the small wheel depicts the auxiliary modulus 4 and the remainders (mod 4) of the corresponding exponents n. Therefore, to find 31998 , we figure out the remainder (mod 4) of its exponent 1998: 1998 = 4 · 499 + 2 ≡ 2 (mod 4), ⇒ 31998 = (34 )499 · 32 ≡ 1499 · 32 ≡ 4

(mod 5).




Note that our new method also applies to our previous problem on 21998 (mod 3). Alternative Solution: The first power of 2 with remainder 1 (mod 3) is 22 : 22 = 4 ≡ 1 (mod 3). Thus, the auxiliary modulus is the exponent 2, so we represent the desired exponent 1998 modulo 2: 1998 = 2·499 ≡ 0 (mod 2), ⇒ 21998 = (22 )499 ≡ 1499 = 1 (mod 3).




Exercise 9. Find the remainders of 52007 (mod 7) and of 171707 (mod 11). 5.3. Further generalizations? Can we use the same method on any power an (mod d)? For instance, Exercise 10. What is the remainder of 426 (mod 14)? Solution: The readers should be able to establish quickly that the powers 4n seem to settle into a pattern (mod 14): {4, 2, 8, 4, 2, 8, . . .}, so our auxiliary       modulus is the length 3 of the pattern. Even though the remainder of 1 is missing from our pattern, it can be shown that the pattern will persist forever. Indeed, suppose that the pattern has repeated itself m times: {4, 2, 8, 4, 2, 8, . . . , 4, 2, 8, ?, ?, ?}            

1 3m−2 4

2

m

43m−1

≡ 4, ≡ 2 and 43m ≡ 8 (mod 14). We are Thus, we know that looking now at the next three powers: 43m+1 = 43m · 4 ≡ 8 · 4 = 32 ≡ 4

(mod 14)

⇒ 43m+2 = 43m+1 · 4 ≡ 4 · 4 ≡ 2 (mod 14) ⇒ 43m+3 = 43m+2 · 4 ≡ 2 · 4 ≡ 8 (mod 14). Thus, the pattern {4, 2, 8} repeats for the (m + 1)th time. By induction (cf. Session 6), we conclude that it will repeat forever. Thus, to finish the problem, we simply find out 26 ≡ 2 (mod 3), so that 426 would land in the second place in the pattern: 426 ≡ 42 ≡ 2

(mod 14).




For further practice, calculate the following: Exercise 11. Find the remainders of 61939 (mod 22) and of 31945 (mod 21). Exercise 12. Calculate the last digit of 21999 , 31999 , 61939 , and 71945 . Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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4. NUMBER THEORY. PART I

What do last digits have to do with remainders? If you don’t suspect the answer already, the following PST will be helpful. PST 29. The last digit of a number n is n’s remainder (mod 10) (why?). Hence reformulate problems about last digits in terms of mod 10. 5.4. A look ahead. Part II of Number Theory and future sessions on abstract algebra will explain why the method of cyclic remainders described above always works (with appropriate modifications). In these sessions we will get acquainted with Fermat’s Little Theorem: ap−1 ≡ 1 (mod p) for any prime p and any a relatively prime to p, and its generalization, Euler’s Theorem: aφ(n) ≡ 1 (mod n) for any relatively prime a and n, where φ(n) counts how many numbers between 1 and n are relatively prime to n (e.g., φ(10) = 4).

6. Twists in the Brute-Force Approach 6.1. Case-chasing remainders. Very often problems ask to establish divisibility by certain numbers, e.g., Exercise 13. Prove that n2 − 1 is always divisible by 3, provided that n itself is an integer not divisible by 3. The “smart” proof would involve noticing that among any 3 consecutive numbers {n−1, n, n+1} exactly one is divisible by 3. Since 3 does not divide n, then 3 must divide exactly one of n − 1 or n + 1. Since we can factor
n2 − 1 = (n − 1)(n + 1), this product is divisible by 3. But what if we didn’t see how the “trick” of factoring would be helpful in this problem, and what if our expression was not factorable? Is there another way to reach our divisibility conclusion? Alternative Proof: The key idea is to use the world (mod 3). Every number n resides in one of the three trees (mod 3): n ≡ 0, n ≡ 1 or n ≡ 2 (mod 3). Since 3
n, the case n ≡ 0 (mod 3) is irrelevant. Let’s consider the next case n ≡ 1 (mod 3). By Corollary 1, we can square both sides of the congruence and subtract 1: n ≡ 1 (mod 3) ⇒ n2 ≡ 12 (mod 3) ⇒ n2 − 1 ≡ 12 − 1 = 0 (mod 3). The last congruence means that 3 | (n2 − 1). We perform analogous calculations when n ≡ 2 (mod 3): n ≡ 2 (mod 3) ⇒ n2 ≡ 22 (mod 3) ⇒ n2 − 1 ≡ 22 − 1 = 3 ≡ 0 (mod 3). We conclude again that 3 | (n2 − 1). Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

6. TWISTS IN THE BRUTE-FORCE APPROACH

n2

79

We have exhausted all possibilities for n and hence shown in general that − 1 is divisible by 3 when n itself is not divisible by 3.


6.2. The table-mod approach. It is important to understand what happened in the above alternative solution, so that we can quickly apply the same method in harder problems. A table of all possible remainders (mod 3) of n and of n2 − 1 is shown below: n (mod 3) 0 1 2



n2 − 1 (mod 3) 02 − 1 ≡ 2 12 − 1 = 0 22 − 1 ≡ 0

3|(n2 − 1)? No Yes Yes

In other words, Corollary 1 allow us to substitute all possible remainders (mod 3) for n into the expression n2 − 1 and, based on the outcome, to conclude whether 3 divides n2 − 1 or not. Obviously, we can apply the same method to much more complex expressions. PST 30. Let E be an algebraic expression involving only addition, subtraction, multiplication and raising to natural powers. If n ≡ m (mod d) and n participates in E but not as an exponent (as in xn ), you can replace n by m without changing the overall remainder of E (mod d). Thus, for instance, if E = 5n3 − 2n and n ≡ 2 (mod 7), then E ≡ 5 · 23 − 2n ≡ 5 − 2n (mod 7). However, PST 30 warns us that E may not be congruent to 5 · 23 − 22 ≡ 1 (mod 7). Indeed, if n = 9, E = 5 · 93 − 29 = 3133 ≡ 4 ≡ 1 (mod 7). 6.3. Why are exponents off-limits? Why can’t we replace xn by xm when n ≡ m (mod d)? For starters, we do not have a lemma implying xn ≡ xm (mod d)! Besides, the previous calculation shows that this is false in general. If you are not convinced yet, here is another counterexample. We know that 2 ≡ 5 (mod 3), but note how replacing the exponent 2 by 5 in n2 − 1 sometimes changes the overall remainder (mod 3): n (mod 3) 0 1 2

n2 − 1 (mod 3) 02 − 1 ≡ 2 12 − 1 = 0 22 − 1 = 0

n5 − 1 (mod 3) 05 − 1 ≡ 2 15 − 1 = 0 25 − 1 ≡ 1

The last line means that n2 − 1 ≡ n5 − 1 (mod 3) when n ≡ 2 (mod 3). Thus, changing exponents is a dangerous thing to do in divisibility problems. To add some more “spice to the dish,” recall that in the method of “cycling remainders” we manipulated exponents by resorting to a new modulus altogether: if the original problem for powers was formulated with mod 5, the exponents required calculations under mod 4. As mentioned earlier, this phenomenon shall be explained in Part II via Fermat’s Little Theorem, and until then we shall stay clear of messing around with exponents. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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4. NUMBER THEORY. PART I

Try the table-mod approach in the following two problems. Exercise 14 (Warm-up). If n is any number, prove that (a) 17n2 + 1 is divisible neither by 4 nor by 5; (b) n(n + 1)(2n + 1) is divisible by 6. Hints: In part (a), obviously we need two separate tables. As for part (b), the reader may have seen a familiar expression for the sum of the squares (to be proved in the Induction session): 12 + 22 + · · · + n2 =

n(n+1)(2n+1) · 6

This gives an alternative roundabout way for (b): since the LHS is an integer, then the RHS is too, i.e., 6 divides n(n + 1)(2n + 1). However, this clever fast-track solution does not excuse anyone from practicing congruences via constructing a table (mod 6) for E = n(n + 1)(2n + 1). ♦ 6.4. Table-mod refinements. Should we blindly apply the table-mod approach from now on? It always works after a finite amount of calculations – just separately plug in each remainder (mod d) – so what’s the big deal? Well, the more often mindless calculations appear in a solution, the less enlightenment we gain from such a solution and the more prone to calculation errors we become. Exercise 15. If n is any number, prove that (a) n5 + 29n is divisible by 30; (b) n2 − 1 is divisible by 24, provided neither 2 nor 3 divides n.

 

The image of a big ugly table for all 30 remainders (mod 30) should make anyone frown! There’s got to be a better reason for 30 to divide n5 + 29n. PST 31. To show that a number d divides an expression E, it may be advantageous to split d as a product of two or more relatively prime factors: d = d1 · d2 · · · dk , and prove separately that each of these factors dj divides E. Lemma 4 would then imply that the product d also divides E. Partial solution to Exercise 14(a): Since 30 = 2 · 3 · 5 and 2, 3, and 5 are relatively prime, we can solve instead 3 simpler problems – one (mod 2), another (mod 3), and a third (mod 5). As an example, let’s show why 5 divides n5 + 29n. Before we brute-force all remainders (mod 5) into n5 + 29n, one further “trick” is due: PST 32. Instead of the remainders 0, 1, 2, 3 and 4 (mod 5), it is often more convenient to use slightly different representatives of the world (mod 5), e.g., {−2, −1, 0, 1, 2} (and analogously, for any (mod d)). These numbers are smaller in size, hence simpler to compute with. Similarly, not only letters like n, but also any numbers like 29 in Exercise 14(a), can be replaced by smaller congruent siblings: 29 can be replaced

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by 4 (mod 5), but even better by −1 (mod 5). For instance, why would we calculate 45 + 29 · 4 (mod 5), when we can do instead (−1)5 + (−1)(−1) = 0? The use of negative numbers should come as no surprise to us now, as we’ve seen them before in calculating 21998 (mod 3). So, here is the situation (mod 5) in Exercise 15(a): n (mod 5) 0 1 −1 2 −2

n5 + 29n ≡ n5 − n (mod 5) 05 − 0 ≡ 0 15 − 1 ≡ 0 (−1)5 − (−1) ≡ 0 25 − 2 ≡ 0 (−2)5 − (−2) ≡ 0

5|(n5 − n)? Yes Yes Yes Yes Yes

These simple calculations imply that 5|(n5 + 29n). The reader can similarly ♦ show that 2 and 3 divide n5 + 29n. By Lemma 4 , 30 | (n5 + 29n). Note that the total number of remainder checks that will be performed in the above solution is 5 + 2 + 3 = 10 (counting all of the moduli 5, 2 and 3). Compared to using the original 30 remainders (some of which are very large), our solution presents a substantial shortcut.

7. Pairs and Divisibility






7.1. Divisibility promise fulfilled. In Section 1, we posed Problem 1. Prove that 1n + 2n + 3n + · · · + (n − 1)n is divisible by n for any odd n > 1. We observed that as n grows, it becomes exceedingly hard to calculate the total sum S = 1n + 2n + 3n + · · · + (n − 1)n and check its divisibility by n. With our newly acquired knowledge in this session, it should only be natural to show instead that S ≡ 0 (mod n). Let’s try this for a small n first. When n = 7, we have S = 17 + 27 + 37 + 47 + 57 + 67 , but we refuse to calculate S or any individual summand k 7 . Let’s look around at the various summands: are they somehow related? For instance, what is the remainder of 67 (mod 7)? The number 6 is relatively large to raise to power 7, but we can replace it with the smaller −1: 6 ≡ −1(mod 7) ⇒ 67 ≡ (−1)7 ≡ −1 (mod 7). Then the first and the last summands will kill each other’s remainders in S: 17 + 67 ≡ 1 + (−1) = 0 (mod 7). This gives us an idea: PST 33. In divisibility problems involving sums, try to pair up as many summands as possible so that the remainders in each pair add up to 0 (mod d). 37

Thus, 27 + 57 ≡ 27 + (−2)7 = 27 − 27 = 0 (mod 7) and 37 + 47 ≡ + (−3)7 = 37 − 37 = 0 (mod 7). Overall, S = (17 + 67 ) + (27 + 57 ) + (37 + 47 ) ≡ 0 + 0 + 0 ≡ 0 (mod 7) ⇒ 7|S.

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Now let’s write the formal solution to the whole problem. Note how this will require more notation to encode our idea above. Proof of Problem 1: Group the (n − 1) summands in pairs taken symmetrically from the beginning and the end of the sum: {k n , (n − k)n }. Note that the remainders (mod n) in each pair add up to 0: n odd

n − k ≡ −k (mod n) ⇒ (n − k)n ≡ (−k)n ≡ −k n (mod n) ⇒ k n + (n − k)n ≡ k n − k n = 0 (mod n). Thus, the remainder of the total sum can be calculated by regrouping and adding a bunch of 0’s:   S = (1n + (n − 1)n ) + (2n + (n − 2)n ) + · · · + ( n2 − 1)n + ( n2 + 1)n ≡ 0 + 0 + · · · + 0 = 0 (mod n). We conclude that n divides S.




This proof is quite succinct: if read without the preceding discussion, the idea of grouping summands will seem to come from out of the blue. Thus, whenever possible, solving simple cases at first (as we did for n = 7) is a good starting point to provide a feeling for the situation. 7.2. The difficulty bar is raised in problems which don’t easily allow us to check simpler cases, as such “simpler cases” may not exist or may be hard to identify. Take, for instance,




Problem 4 (USAMO ’98). Suppose that the set {1, 2, . . . , 1998} has been partitioned into disjoint pairs {ai , bi } (1 ≤ i ≤ 999) so that for all i, |ai − bi | equals 1 or 6. Prove that the sum |a1 − b1 | + |a2 − b2 | + · · · + |a999 − b999 | ends in the digit 9. At first glance, it is unclear if and how much the problem depends on the number 1998. For instance, we could properly pair up the numbers {1, 2, 3, 4, 5, 6, 7, 8} as {1, 7}, {2, 8}, {3, 4}, {5, 6}, but the sum |1 − 7| + |2 − 8| + |3 − 4| + |5 − 6| = 6 + 6 + 1 + 1 = 14



ends in the “undesirable” digit 4. Thus, we can’t easily generalize from smaller cases: something about 1998 must be essential in our problem! But how can we approach the original situation without knowing how many pairs with |ai − bi | = 6 and how many pairs with |ai − bi | = 1 there are? PST 34. As soon as you label mathematical concepts with letters (instead of using specific numbers only), many situations are clarified and seemingly impossible problems become approachable. Thus, let m be the number of pairs with |ai − bi | = 6. There are 999 − m of the remaining pairs (those with |ai − bi | = 1), and the total sum is |a1 − b1 | + · · · + |a999 − b999 | = 6m + (999 − m) = 5m + 999.

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83

On the surface, it seems impossible to find the last digit of 5m + 999 without knowing m, but PST 29 now kicks in to help us reformulate the problem in terms of mod 10: ?

?

?

5m + 999 ≡ 9 (mod 10) ⇔ 5m ≡ −990 (mod 10) ⇔ 5m ≡ 0 (mod 10).



But that’s the same as asking why 10 divides 5m or why 2 | m. In other words, we don’t need to know the exact value of m: we just need to find out why m must always be even. For a beginner, the answer to this question may be very hard to obtain. However, a seasoned problem solver would immediately realize that PST 35. If the parity of one number is questioned (is m even?), then the parities of all numbers (in our case {1, 2, . . . , 1998}) might be involved in the solution. Therefore, counting how many odd and how many even numbers appear in total may be useful. We know that there are 999 odd and 999 even numbers in {1, 2, . . . , 1998}. The idea is to count these in a different way and to compare the results. • In the pairs (ai , bi ) with |ai − bi | = 1, exactly one number is even and the other is odd (why?). Hence these pairs contribute a total of 999 − m odds and 999 − m evens. • In the pairs (ai , bi ) with |ai − bi | = 6, either both numbers are even or both are odd (why?). Say, k pairs have odd ai and bi , and the remaining m − k pairs have even ai and bi . Hence, these pairs contribute overall 2k odds and 2(m − k) evens. Summing up, we have 999−m+2k odds and 999−m+2(m−k) = 999+m−2k evens. Since there are exactly 999 odds, 999 − m + 2k = 999 so that m = 2k. Wrapping up the whole solution, we have established that there are an even number of pairs with |ai − bi | = 6, namely m = 2k, and hence the given sum ends in the digit 9.
Let’s go back to the beginning of our discussion: what properties of 1998 made the problem true? The reader should try to redo the proof with an arbitrary number 2l instead of 1998 and show that




Problem 5 (General USAMO ’98). For a fixed l, no matter how the set {1, 2, . . . , 2l} is partitioned into pairs {ai , bi } with |ai − bi | = 1 or 6, the sum |a1 − b1 | + |a2 − b2 | + · · · + |al − bl | always ends in the same digit. For instance, in our earlier example with l = 4, we partitioned the set {1, 2, . . . , 8} so that the resulting sum was 14. If we partition the set into adjacent pairs {1, 2}, {3, 4}, {5, 6} and {7, 8}, the sum 1 + 1 + 1 + 1 = 4 will have the same last digit as 14. This is not coincidence, as Problem 5 claims.

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4. NUMBER THEORY. PART I

8. Hints and Solutions to Selected Problems Exercise 1. (a) q = 3, r = 198; (d) q = −3, r = 198.

(b) q = −4, r = 402; (c) q = 4, r = 402; ♦

Exercise 2. Just change decrease to increase and subtract to add in the given proof. ♦ Exercise 4. 5 = 4 × 1 + 1 and 9 = 4 × 2 + 1, but 7 = 4 × 1 + 3. 2007 and 1707 have different remainders when divided by 13, so 2007 and 1707 are not congruent (mod 13). However, they both have remainder 207 when divided by 300, so 2007 ≡ 1707 (mod 300). ♦ Exercise 5. The world modulo 4 consists of trees {. . . , −8, −4, 0, 4, 8, . . .}, {. . . , −3, 1, 5, . . .}, {. . . , −2, 2, 6, . . .}, and {. . . , −1, 3, 7, . . .}. The union of the 1st and 3rd trees and the union of the 2nd and 4th trees are the “even” and “odd” trees in the world modulo 2, respectively. Similarly, the 6 trees that comprise the world modulo 6 can be paired into the 3 trees of the world modulo 3. ♦ Exercise 6. In the world modulo d the general formulas describing the residents of each tree are: a = dk + r where r runs over all remainders (mod d), i.e., r = 0, 1, 2, . . . , d − 1, and k is any integer. ♦ Proof of Lemma 3(b): Instead of addition, use subtraction to obtain a statement similar to (4). ♦ Second Proof of Lemma 3(c): Let’s divide all a, b, c, and e by d and denote their respective remainders by r1 , r1 , r2 , and r2 , i.e., a = q1 d + r1 , b = q2 d + r1 , c = q3 d + r2 , and e = q4 d + r2 . Then we can multiply through and factor out d: ac = (q1 d + r1 )(q3 d + r2 ) = d(q1 q3 d + q1 r2 + r1 q3 ) + r1 r2 = dQ + r1 r2 , be = (q2 d + r1 )(q4 d + r2 ) = d(q2 q4 d + q2 r2 + r1 q4 ) + r1 r2 = dP + r1 r2 . Subtracting, we obtain ac−be = d(Q−P ), and thus d | (ac−be). Translating into congruences, ac ≡ be (mod d).
Exercise 9. 51 ≡ 5 (mod 7), 52 ≡ 4 (mod 7), 53 ≡ 6 ≡ −1 (mod 7). Note that −1 is almost as nice as 1 when working with powers in congruences, so 52007 = (53 )669 ≡ (−1)669 ≡ −1 ≡ 6 (mod 7). Using the same idea in the next problem, we find 65 ≡ −1 (mod 11). Then 61707 = (65 )341 · 62 ≡ −1 · 3 ≡ 8 (mod 11).
Exercise 11. The powers of 6 have remainders (mod 22) that cycle through the ten numbers in the pattern {6, 14, 18, 20, 10, 16, 8, 4, 2, 12}. Thus, our auxiliary modulus is 10, so 1939 = 193 × 10 + 9, and the remainder of 61939 will land on the 9th number in our pattern, namely, 61939 ≡ 2 (mod 22).
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8. HINTS AND SOLUTIONS TO SELECTED PROBLEMS

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For the other part, check that the remainders of 3n (mod 21) form a cycle of length 6, namely, {3, 9, 6, 18, 12, 15}, so that our auxiliary modulus
is 6. Thus, 1945 = 6 × 324 + 1 and 31945 ≡ 31 ≡ 3 (mod 21). Exercise 12. By PST 29, the problem is equivalent to finding the remainders (mod 10) of the given numbers. For example, the powers of 2 divided by 10 have remainders that form a cycle {2, 4, 8, 6}. Since 1999 = 4 · 500 − 1, use the next to last element in the cycle, 8, to find 21999 ≡ 8 (mod 10). Similarly, we find the last digits of 31999 , 61939 , and 71945 are 7, 6, and 7, respectively. ♦ Exercise 14(b). Here is a mod-table to show that 6 | n(n + 1)(2n + 1): n (mod 6) 0 1 2 3 4 5

n(n + 1)(2n + 1) (mod 6) (0) (1) (1) = 0 (1) (2) (3) = 6 (2) (3) (5) = 6 · 5 (3) (4) (7) = 6 · 14 (4) (5) (9) = 6 · 30 (5) (6) (11) = 6 · 55

≡ (mod 6) 0 0 0 0 0 0

Exercise 15(b). Since n is not divisible by 2 or 3, the only numbers to be tested are {1, 5, 7, 11, 13, 17, 19, 23} ≡ {1, 5, 7, 11, −11, −7, −5, −1} (mod 24). Since n2 − 1 = (−n)2 − 1, the only numbers that need to be checked are 1, 5, 7, and 11. But the values of n2 − 1 for these four numbers are 0, 24, 48, and 120, which are all multiples of 24.
2 Alternatively, 24 = 3 · 8, so it is enough to show that n − 1 is divisible by 3 and by 8. Since 3
n, the only numbers to check (mod 3) are {1, 2} ≡ {1, −1}, while (mod 8) requires only {1, 3, 5, 7} ≡ {1, 3, −3, −1} (as n is odd). By the symmetry argument above, it suffices to check only 1 for (mod 3) and only 1 and 3 for (mod 8). These remainders are trivial to plug
into n2 − 1, as they require just 1-digit calculations. Problem 5. Replacing 999 by l in the solution of Problem 4, the total sum is now S = 5m + l and there are exactly l − m + 2k odds among the given 2l numbers. Thus, l − m + 2k = l, yielding again m = 2k. Consequently, S = 10k + l ≡ l (mod 10), i.e., the last digit of S coincides with the last digit of l: this is independent of how the 2l numbers are paired up.


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Session 5 A Few Words About Proofs. Part I based on

Mira Bernstein’s sessions

Sneak Preview. A proof is nothing more than a water-tight logical argument. There is no special way a proof has to look on a page. (Don’t worry about the “two column” format they teach in high-school geometry.) As long as you state precisely what you’re trying to show and then make your case logically and coherently, it counts as a proof. In this session, we’ll explore some of the basic and most common types of proofs. We’ll show that Russian samovars invariably produce √ the best proportions of milk and tea (half kidding!), contradict the claim that 2 is rational and thereby enter the exclusive Pythagorean society, ponder if it is possible to prove that something is impossible, coordinate handshakes at parties and deny entrance to witches into the BMC, play “hide-and-seek” with invariants, cook up counterexamples when they suit us, and chart escape routes for n-dragons on infinite chessboards . . . . And all the time, we’ll watch like hawks for common pitfalls and non-proofs.

1. Why Prove Things? Often, what mathematicians consider a proof is quite different from what the general public will be satisfied with. How can you tell if an argument is good enough to be a proof ? Imagine that your annoying (but clever) little sister is looking over your shoulder, always nagging at you: “And why is this? And how do you know that?” Whenever you say that something is obvious, she’ll ask, “How come?” Whenever you are tempted to be a little vague, she’ll say, “I don’t get it – what do you mean by that?” If your argument passes the annoying little sister test, then it’s a proof! But why prove things? Before going any further, try your hand at




Problem 1. Suppose you have a white box with 60 white balls and a black box with 60 black balls (cf. Fig. 1). You take 20 balls from the white box, put them into the black box, and mix everything up thoroughly. Now you take 20 balls (most likely some white, some black) from the black box and put them into the white box. In the end, which is larger: the number of black balls in the white box or the number of white balls in the black box? 87

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88

5. A FEW WORDS ABOUT PROOFS. PART I

20

20

?

?

Figure 1. White and black balls If you play around with this problem and try a few cases, it shouldn’t take you long to guess the answer. (If you haven’t done it yet, do it now!) But proving that your answer is correct requires that you look at the problem in a whole new light. It’s no longer enough to say, “The numbers always seem to work out this way, so it must be true.” You are forced to ask the question: “What is really going on here?” A proof is an answer to this question. Sure, it enables you to convince other people that your solution is correct, for instance, your annoying little sister . . . . But even more importantly, it gives you a deeper understanding of the problem, which may help you solve other, harder problems in the future. In the words of the Russian mathematician Yuri Manin, “A good proof is one that makes us wiser.” Try it now with Problem 1. See if you can explain why the answer must always come out the way it does. Whatever argument you come up with (even if you’re not very happy with it), write it down! Really: take out paper and pencil, and do it. You may find that it’s quite hard to pin down your thoughts with words. Often an argument seems much more convincing in your head than it does on paper. One of the challenges in writing proofs is to take your vague hunches and intuitions and to make them precise. This way, not only can you communicate your thoughts to someone else, but you actually end up understanding them better yourself. Don’t read any further until you’ve written down some attempt at a proof!

2. Proofs versus Non-proofs You have probably figured out by now the Answer to Problem 1: “The number of white balls in the black box is always the same as the number of black balls in the white box.” 2.1. Non-proofs pretending to be proofs. Here are a few invalid proofs of this result. Compare them to your own proof and see if the comments help you improve your argument. “Proof” 1: The two numbers are the same, because however many black balls we move to the white box, that’s how many white balls have to stay in ? the black box. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

2. PROOFS VERSUS NON-PROOFS

89

The sentence above might feel like an explanation, but if you read it carefully, you’ll see that it’s simply restating the claim that we’re trying to prove! Beware of “proofs” that rephrase the conclusion in words that sound more convincing, but that offer no actual argument. “Proof” 2: When we first move the 20 white balls to the black box, the fraction of white balls in the black box is 1/4 (= 20 : 80). So of the 20 balls that we move from the black box to the white box, one quarter (i.e., 5 balls) will be white and three quarters (i.e., 15 balls) will be black. Thus, in the end, the white box will have 15 black balls and the black box will have 15 ? white balls (since we originally put in 20 and then moved 5 back). This “proof” introduces an irrelevant consideration: probability. Just because a quarter of the balls in the black box are white, we can’t count on getting exactly 5 white balls in our batch of 20. It’s true that 5 is the most likely number of white balls, but anything from 0 to 20 is theoretically possible. Unless otherwise indicated, a problem always wants you to consider not only the most probable case but all possible cases, however unlikely. “Proof” 3: If the white box ends up with fewer than 20 black balls, then some of the balls that we moved to it must have been white. So the black box will have fewer than 20 white balls left, since some of them got returned to the white box. The numbers must match up, since we moved 20 balls each time, so the final number of wrong-colored balls in each box is the same. ? Such an argument may be on the right track, but it’s too vague for a proof. Just because both boxes will have fewer than 20 wrong-colored balls in them, why must they have the same number of wrong-colored balls? There’s no explanation given for why “the numbers have to match up”; it’s not even clear what that means. “Proof” 4: The two numbers are the same because, however many black balls we move to the white box, that’s how many white balls have to stay in the black box. For example, say the batch of balls that we moved to the white box had 13 black balls and 7 white ones. Then the black box will have 13 white balls left (since we originally put in 20 and then moved 7 back). In ? this way, the two numbers always work out to be the same. You might feel that this explanation conveys a pretty good sense of what is going on in the problem. The trouble is that it relies on one specific example. Although an example can be great for conveying intuition and suggesting ideas, it won’t satisfy our annoying little sister. “I can see that the numbers come out to be the same in this one case,” she’ll say. “But how do you know they’ll always be the same? You assert this without any evidence.” It may seem like nitpicking, but the little sister has a point: the example feels representative; yet the fact that we have to rely on this feeling shows that we still haven’t gotten to the bottom of things. Still, it feels like we’re very close, and we are. Before you go on, see if you can modify “Proof” 4 to make it valid! Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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5. A FEW WORDS ABOUT PROOFS. PART I

2.2. The real thing. We now present two valid proofs for Problem 1. Proof 1: Suppose the batch of balls that we moved to the white box had k black balls and 20−k white ones. Then the white box will have k black balls, while the black box will have 20 − (20 − k) white balls: we originally put in 20 white balls and then moved 20−k white balls back. But 20−(20−k) = k, so the number of “wrong-colored” balls in the two boxes is the same.




Notice how short the proof is! This often happens: once you grasp the essence of the problem, long explanations become unnecessary. Further, this proof is very similar to the example in “Proof” 4. The only difference is that instead of picking specific numbers (13 black balls and 7 white ones), we used a variable to represent the general case (k black balls and 20 − k white ones). An example is not a proof, but sometimes it’s very easy to turn it into one. Our proof illustrates a very useful mathematical technique: PST 36. When faced with an unknown quantity (such as the number of black balls in the white box), give it a name! A name is a handle that you can grab and use to your advantage. Once you’ve named one of your unknown quantities (say, k black balls), you may be able to find out the other unknown quantity (i.e., 20 − (20 − k) = k white balls). You could probably write down the same proof without giving the unknown quantity a name, but it would be harder to make the idea precise. Proof 2: There are a total of 120 balls: 60 white and 60 black. In the end, 60 of these balls are in the white box, and 60 in the black box. If the white box contains k black and 60 − k white balls, then the remaining 60 − k black and k white balls must end up in the black box. Thus the number of “wrong-colored” balls in both boxes is k.




This proof is even better (if only slightly shorter) – it definitely “makes us wiser”. It tells us that PST 37. Occasionally it is advantageous to forget the particular way a process is developing (e.g., how and which balls got moved around above) and just to focus on the end result. Thus, another way to think of what’s happening is to “pour” all white and black balls into a Russian samovar and then let all balls “dribble” in equal amounts from the samovar back into the two boxes. So, we not only don’t have to keep track of the particular ball movement, but we now understand that moving any number of balls any number of times back and forth between the two boxes will always result in the same number of “wrong-colored” balls, as long as we insist at the end on an equal total number of balls in the two boxes.

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3. PROOF BY CONTRADICTION

91

The new approach of PST 37 turns out to be extremely powerful in many problems and in many branches of mathematics. We’ll see more examples of it later on. But for now, let’s finish this section with another popular version of our problem. Since no one really uses a samovar for balls, let’s talk about milk and tea. ?

?

Figure 2. Milk and tea Exercise 1. Suppose that a white cup contains some milk and a black cup contains the same amount of tea. We take 3 spoons of milk from the white cup and pour in into the black cup (cf. Fig. 2), stir the black cup thoroughly and then pour 3 spoons of its mixture back into the white cup. Which is the higher ratio: tea to milk in white cup or milk to tea in black cup? How does the answer change if we repeat the above process once again?

3. Proof by Contradiction OK, so a proof is just a logical argument. But how do you come up with such an argument when you are faced with a problem and yet have no idea where to start? There are several tricks that mathematicians use. One of the oldest, and still the most common, is called proof by contradiction. It goes like this: • Suppose that the claim you want to prove is false. • By a chain of logical arguments, show that this supposition leads to an impossible result – a contradiction! • Conclude that your supposition must have been wrong, so the claim that you want to prove must be true.




Let’s illustrate this technique with two of the most famous theorems and their proofs from ancient Greece. Theorem 1. There are infinitely many prime numbers. Proof (Euclid): Suppose the claim is false and there are in fact only finitely many primes. Then we can write them in a list: p1 , p2 , . . . , pk . Form a new number N by multiplying all the primes on the list and adding 1: N = p1 p2 . . . pk + 1. If we divide N by any of the numbers on our list, the remainder will be 1. Thus N is not divisible by any prime on our list. But N > 1 so it must be divisible by some prime number, as every integer has a prime factorization1. 1 The Prime Factorization Theorem will appear prominently in the Induction session.

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5. A FEW WORDS ABOUT PROOFS. PART I

Our list was supposed to contain all the primes, yet there is a prime number which is not on it, namely, this prime divisor of N , possibly N itself. We have thus reached a contradiction. Therefore our starting premise – that there are finitely many primes – must have been wrong: there must be infinitely many primes.





The specific idea to multiply “all” primes and add 1 is the brilliant piece which makes Euclid’s proof immortal. His argument also presents us with a more “mundane” but frequently used idea: replacing infinitely many by finitely many objects. In search of a contradiction, the solution below also avoids exploring infinitely many cases by concentrating on a single fraction. √ Theorem 2. 2 is irrational: it cannot be written as a ratio of two integers. √ Proof (Aristotle): Suppose the claim is false and 2 is in fact a rational number. Write this fraction in lowest terms: √ a 2= , b where a √ and b are integers with no common factors. To get rid of the incon, raise both sides to the second power. Then 2 = a2 /b2 and venient a2 = 2b2 . The RHS is even, so a2 must be even too. But then a itself must be even (for odd numbers have odd squares), i.e., a = 2k for some integer k. Rewriting the equation in terms of k, we obtain (2k)2 = 4k 2 = 2b2 , so that 2k 2 = b2 . Now, by the same argument, b2 is even, so b is even. But if both a and b are even, this contradicts our assumption that the fraction ab was in lowest terms, √ as we can cancel a 2 from top and bottom! Thus our starting premise – that 2 can be written√as a fraction – must√ have been wrong. Therefore,
it is impossible to write 2 as a fraction, i.e., 2 is irrational.2 The great thing about a proof by contradiction is that it gives you a foundation to build on. Imagine trying to devise a direct proof that there are infinitely many primes: where could you possibly start? Euclid’s proof by contradiction starts from a very concrete premise: “There are finitely many primes”. Now he can list them, name them, compute with them, √ and get somewhere! Similarly, in Aristotle’s proof of the irrationality of 2, he 2 Although the proof above was hinted at by Aristotle, some (possibly other) proof seems to have been found earlier by Hippasus, a member of the Pythagorean society. The discovery was devastating to Pythagoras (569–500 B.C.) and his followers, as they had hitherto based their mathematical, philosophical, and religious beliefs on the premise that all numbers are ratios of the whole numbers. The members of the society were sworn to secrecy, but apparently Hippasus leaked the existence of irrationals to the wider community. The punishment was swift: as the legend goes, he was √ strangled or drowned by other members. Who could have thought that the irrationality of 2 alone could inspire such mysticism and brutality?! Isn’t it a relief that we can talk freely about irrational (let alone transcendental and complex) numbers nowadays?

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√ started with the premise “ 2 = ab ”, which gave him specific integers a and b to work with. In general, PST 38. To see if a proof by contradiction might work for a particular problem, ask yourself: “If I assume the claim is false, will this give me something concrete to build on, a good place to start my argument? ” In Problem 1, for instance, the answer would clearly have been No – so proof by contradiction would not be a useful technique there. In contrast, as we saw above, replacing infinitely by finitely many objects, or even by just a single object, gave us definite advantages. We will see more examples of proofs by contradiction when we talk about impossibility proofs. For now, try the following three problems. √ √ Exercise 2. Prove that 3 is irrational. How about p for a prime p? Exercise 3. Prove that there is no smallest positive rational number. Exercise 4. Prove that the sum of a rational and an irrational numbers is irrational. How about the sum of two irrational numbers?

4. Proofs of Possibility and Impossibility 4.1. Lead by example? Earlier, we said that an example does not constitute a proof. That’s actually not quite true: it depends on what you are proving. If you are trying to show that something is always true, then you must provide a general argument. For instance, in Problem 1 we needed to show that the number of wrong-colored balls in the two boxes is always the same. No single example can demonstrate this; we had to prove it for every possible case, using the variable k. On the other hand, to prove that something is possible, all it takes is one example demonstrating that it can be done. For instance, if you want to win a game for sure, you only have to win it in one way, i.e., construct for yourself one example of a winning strategy. (There may be other valid strategies, but no one cares about them after you win.) Such a proof is called a constructive proof. Here is another situation asking for this kind of proof: Exercise 5. There are 8 people at a party. Prove that it is possible for each of them to shake hands with exactly 3 other people. Proof: Arrange the 8 people in a circle and let each person shake hands with her neighbors and the person directly across from her (cf. Fig. 3a). In this way, each person shakes hands with exactly 3 people, as required.
Note that to prove the statement, you just need to exhibit one specific case in which each person shakes 3 hands. Our proof does this, so it is a valid constructive proof. There are other ways to arrange for everyone to shake hands with 3 people, but there is no need to look for them. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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4.2. Counterexamples are not necessarily “bad”. There is one other important context in which a single example constitutes a proof. Suppose you want to show that a certain statement is not always true. This is the same as showing that it’s possible for the statement not to be true, so all you need is a single example where the statement is false. Such an example is called a counterexample. Exercise 6. Prove or disprove: every system of n equations in n variables has a unique solution. Proof: The claim is false. A counterexample is the system of two equations x − y = x and x + y = x, which has two variables and infinitely many solutions: all pairs (x, 0) for any x.
There are, of course, many other instances where the claim fails, but again, there is no need to look for them. A single system of equations for just one value of n serves as a counterexample to the whole claim. When asked to disprove a claim, people often come up with elaborate explanations of why the claim does not have to hold. Usually these explanations don’t actually prove anything: they only show that the claim might be false, not that it actually is false. The only way to be sure is to find a counterexample – and this makes all further explanations unnecessary.




4.3. Is it possible to prove that something is impossible? To prove that something is possible, you just show how to do it. But how do you prove that something is impossible? Before going on, try it on your own: Problem 2. There are 9 people at a party. Prove that it is impossible for each of them to shake hands with exactly 3 other people. How can we do this, short of trying all possible arrangements of handshakes for 9 people? First, here is one way not to do it: “Proof”: Arrange the 9 people in a circle and have each person shake hands with the two people next to him. Now each person needs one more handshake (cf. Fig. 3b.) However, since every handshake involves two people ? and there are an odd number of people in total, this is impossible. 5 6 7

7 3

8

2 1

5

6

4

4

?

8

3 9

1

2

Figure 3. 8 or 9 people searching for 3 handshakes This “proof” shows only that the approach which succeeded in Exercise 5 for 8 people fails for 9 people. However, proving that a particular approach does not work is not enough. The fact that there is no solution which begins Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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with “everyone standing in a circle” doesn’t mean that there is no solution of some other sort. Beware of impossibility “proofs” that make assumptions about what a solution must look like! Although it is incorrect, this “proof” is onto something: the fact that 9 (unlike 8) is odd is likely to be relevant, as is the fact that every handshake involves two people. Both of these observations are true regardless of the particular arrangement of handshakes. PST 39. Instead of going through lots of separate cases and showing that none of them work, the key to most impossibility proofs is to step back and look at the big picture. Find a fundamental aspect of the problem – a pattern, a number, a parity – that cannot change no matter how the details work out. This is called an invariant.3 Then prove that no solutions exist, by showing that any solution would require the invariant to change. Note that this is a special case of a proof by contradiction. Here’s what PST 39 looks like for our 9 people searching for 3 handshakes: Proof of Problem 2: Imagine that each person puts a check mark in her notebook every time she shakes hands with someone. At the end, what is the total number of marks in all the notebooks? If each of the 9 people shook hands with three others, there would have to be 27 marks. But each handshake produces two marks (one for each participant), so the total number of marks must be even. Thus it is impossible for each person to shake hands with three others.





The invariant here is the parity of the number of check marks (the latter can be thought of as “half-handshakes”). Regardless of the particular arrangement of handshakes, the parity must always be even, yet an arrangement when each of the 9 people shakes 3 hands would force it to be odd. Problem 3 below continues the theme of parity, but it also teaches us something else. Although we introduced invariants as a technique for proving that something is impossible, they can also be used in other situations. Problem 3. If every room in a house has an even number of doors, show that the number of outside doors must be even as well. Hint: Every door has . . . two sides!

♦

4.4. Hide-and-seek with invariants. Proofs involving invariants often require a lot of creativity, because the right invariant can be tricky to spot. A good place to start is on a chessboard. It is easy to see how one would cover a regular 8 × 8 board with 32 dominoes, with each domino covering two adjacent squares. But what if we used trominoes (tiles made of three squares in an L-shape) or 4 × 1 tetrominoes (tiles made of four squares in a line)? When we place them on the board, we can rotate or flip our shapes however we want, but they must always cover 2, 3, or 4 squares, respectively. 3 We shall revisit invariants in depth in the Stomp session.

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Problem 4. Can you cover by non-overlapping




(a) trominoes a 6 × 6 chessboard with one square removed? (b) dominoes an 8 × 8 chessboard with two diagonally opposite corner squares removed? (c) 4 × 1 tetrominoes a 10 × 10 chessboard? Hints: Part (a) is really easy. Each domino in part (b) covers two squares: what are their colors? The usual chessboard coloring doesn’t help in part (c); think of a new coloring (cf. Problem 8 for a more general version). ♦




Problem 5. A class of 25 students has their desks arranged in a 5×5 square. No more than one student can fit at a single desk. The teacher asks every student to move to an adjacent desk (directly in front, behind, to the left, or to the right of her current desk). Show that this is impossible! Hint: Imagine that the desks are arranged on a chessboard with the standard coloring (cf. Problem 9 for a generalization). ♦




Problem 6. Three witches are hovering over Berkeley, always keeping at the same height. At every instant, only one of them can move; she can go as far as she wants, but only in a direction parallel to the line connecting her two sisters. If the first witch starts out directly over Evans Hall (where BMC meets), the second 2 miles north, and the third 4 miles east, is it possible that after some time they will end up with the first witch again over Evans, the second one 3 miles northeast, and the third 3 miles southeast?

2 4

? 3

3

Hint: As we can see, not all invariant problems involve chessboards or parity. Think of a quantity that doesn’t change when the witches move. ♦ 4.5. Summary of PST’s. In this section we learned that to prove something is true, possible, not true, or impossible, you have to employ different PST’s. A summary of these PST’s appears in the table below.



A claim is

Meaning

To prove, provide

1. true

always true, never false

2. possible

true in at least one instance, not always false false in at least one instance, not always true always false, never true

general argument (e.g., the black and white balls proof) example (constructive proof) or existential proof (e.g., remainders) counterexample

3. not true 4. impossible

general argument (e.g., proof by contradiction via invariant)

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5. SOME PROBLEMS NEED TWO PROOFS!

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5. Some Problems Need Two Proofs!




5.1. Isn’t one proof plenty? Suppose a problem asks you to determine which numbers n satisfy a certain condition. For example: Problem 7. At a party attended by n people, each person shakes hands with exactly 3 others. For what values of n is this possible? (Sound familiar?) PST 40. A complete solution to such a problem consists of an answer (a set of numbers) and two proofs: • a proof that all the numbers in your set do satisfy the conditions of the problem; • a proof that other numbers don’t satisfy the conditions. Even if one of the two proofs seems easy and obvious, you must still include it for the sake of logical completeness. Often, however, both proofs are interesting as they generally require completely different techniques. For instance, in Problem 7, one is a proof of possibility and the other is a proof of impossibility. As we saw in the previous section, these two kinds of proofs use very different methods. A complete solution to Problem 7 therefore requires a combination of the proofs from Exercise 5 and Problem 2. Neither half can be omitted! Answer to Problem 7: Each person can shake hands with exactly 3 others if and only if n is even. Proof: If n is even, we can arrange all the people in a circle and let each person shake hands with her neighbors and with the person directly across from her ( n2 people away). In this way, each person shakes hands with exactly 3 people, as required. On the other hand, if n is odd, imagine that each person puts a checkmark in her notebook every time she shakes hands with someone. Because each handshake produces two marks (one for each participant), the total number of marks must be even. But if each of the n people were to shake hands with three others, there would be a total of 3n marks, which is an odd number, a contradiction. Thus it is impossible for each person to shake hands with 3 others when n is odd.
5.2. A prime example of “math shorthand”. The phrase “if and only if ”, which we used in the answer to Problem 7, is a very common one in mathematics – so common that it has its own abbreviation: iff. When we say “A if and only if B ”, we mean “A is true whenever B is true, and A is false whenever B is false”. An “iff ” statement is considered the gold standard in mathematics: it offers a complete solution to a problem, since it tells you exactly what works and what doesn’t. Very often, “iff ” is secretly present, and your role is to draw it out explicitly.

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5. A FEW WORDS ABOUT PROOFS. PART I

Here are three more problems whose answers should contain an “iff ” and whose proofs should be in two parts. Problem 8. For what values of m and n can an m × n rectangle be covered by non-overlapping 4 × 1 tetrominoes (cf. Problem 4(c))? Solution: Obviously, the number of squares in our m × n board must be divisible by 4 in order to tile it by 4 × 1 tetrominoes. Thus, either 4 divides m or n, or both n and m are even. But in Problem 4(c) we saw that it is impossible to cover a 10 × 10 board; there 4 did not divide m = n = 10. So, let’s see what goes wrong in general when both n = 2k and m = 2l are even, yet neither is divisible by 4. Divide your 2k × 2l board into kl 2 × 2 blocks and color each block black or white, in a checkerboard pattern (cf. Fig. 4a). If k and l are both odd, then there will be an unequal number of black and white squares. On the other hand, each tetromino must cover exactly 2 black and 2 white squares (cf. the two grey tetrominoes in Fig. 4a), so our coloring must be color-balanced, a contradiction.

|

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Figure 4. Impossible and possible tilings by 4 × 1 tetrominoes The question of coloring a board when 4 divides m or n is rather trivial. For example, if 4 | n, you can partition the board into batches of 4 columns, and in each such batch arrange your 4 × 1 tiles horizontally (cf. Fig. 4b). Thus an m × n board is tileable by tetrominoes iff 4 divides m or n.
Note that the above solution used a couple of our proof methods: • to show that certain boards were not tileable, we used an invariant (born of our balanced 2-coloring) and proof by contradiction; • to show that certain boards were tileable, we just found a specific tiling example. See the Hints section for an alternative solution which is generalizable to other n × 1 polyominoes.




Problem 9 (BAMO ’06). Let n2 students be seated at desks arranged in an n × n square. For what values of n can the students fulfill a teacher’s request: each student should move to an adjacent desk (directly in front, behind, to the left, or to the right of her current desk) so that no more than one student sits at a single desk (cf. Problem 5)?

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5. SOME PROBLEMS NEED TWO PROOFS!






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Problem 10. As you probably know, a knight in chess moves in an “L” shape: 1 square in one direction and 2 squares in a perpendicular direction. What you probably don’t know is that this knight is chasing an n-dragon: a creature that moves 1 square in one direction and n squares in a perpendicular direction. On an infinite chessboard, which n-dragons can go from any square to any other square? Partial Solution: A 1-dragon will move simply by one square in each of the four diagonal directions (cf. Fig. 5a). Using the usual chessboard coloring, it is evident that this dragon is limited to visiting only part of the board (why?). But wouldn’t the same limitations apply to any odd-dragon? Even-dragons are more capable. For example, a 2-dragon moves exactly like an ordinary knight (cf. Fig. 5b). If you have played chess, at some point you may have tried and succeeded in covering the whole chess board by the moves of a single knight. However, it is not necessary to prove separately that each and every square can be reached by your knight. PST 41. To show that a figure can travel from any square to any other square on an infinite chessboard, (a) it is necessary and sufficient to show that it can move one square to the right, left, up, and down; (b) if the set of original moves is symmetric with respect to the four directions (right, left, up, and down), it is necessary and sufficient to show that the figure can move one square to the right.

Figure 5. Moves of 1-Dragon, Knight, and 6-Dragon For instance, if you take any of the original knight’s moves and reflect it across a vertical or a horizontal axis, you will obtain another of his original moves (cf. Fig. 5b); thus PST 41(b) kicks in and reduces your problem to showing that a knight can move one square to the right. Ditto for evendragons. Figure 5c demonstrates one way for a 6-dragon to move one square to the right. Can you generalize this construction to any even-dragon? ♦ Exercise 7. Prove that an odd-dragon standing on a black square can reach any other black square on an infinite chessboard. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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5. A FEW WORDS ABOUT PROOFS. PART I

6. Hints and Solutions to Selected Problems Exercise 1. The answer, of course, does not depend on how many times and how many spoons we move back and forth between the cups, as long as the two cups contain the same total amount of liquid at the end. The two desired ratios will be equal and the explanation goes along the lines of Proof 2 of Problem 1: just replace 60 and k balls by n and k ounces. ♦ √ as a fracExercise 2. Suppose 3 is in fact a rational number. Write it √ √ tion 3 = ab in lowest terms, and proceed the same way as for 2, using divisibility by 3 instead. ♦ Exercise 3. To the contrary, suppose a is the smallest positive rational number. But then isn’t the rational a/2 even smaller? ♦ shows that the Exercise 4. The usual formula for fractions ab ± dc = ad±bc bd sum and the difference of two rational numbers are also rational. Now, let x be rational and y be irrational; if the sum x+y = z is rational, then y = z −x is also rational as a difference of two fractions, a contradiction. Finally, √ √ the 2+ 2 = sum of two irrational numbers may either be rational or irrational: √ √ √ ♦ 2 2 is irrational and (2 + 2) + (2 − 2) = 4 is rational. Problem 3. Add another “room” to the house, namely the outside. Now count the doors in all rooms: this number will be even since each door has been counted twice. But the contribution from “genuine” rooms inside the house is even since each such room is given to have an even number of doors. Therefore the contribution of the “outside” room must also be even, i.e., there are an even number of doors to the outside.
Problem 4. (a) 6 × 6 − 1 = 35 is not divisible by 3, but we want it to be! ♦ (b) In the classic chessboard coloring the two diagonal squares are the same color, say, black, so their removal yields two more white than black squares. But each domino covers one square of each color! ♦ (c) Color each row with “colors” 1,2,3,4,1,2,3,4,1,2, cycling the colors so that the second row is colored 2,3,4,1,2,3,4,1,2,3, etc. A count of the squares with each color reveals that 26 squares get one color, but only 24 get another color! Yet the 4 × 1 tetromino covers one square of each color! ♦ Problem 5. Using the standard chessboard coloring, there will be 13 squares of one color and 12 squares of the other color. Since a move consists of changing from one color to the other, the student exchange is impossible without two students ending up at the same desk. ♦ Problem 6. The area of the triangle formed by the witches does not change: it is an invariant. Indeed, given ABC and a line l through C parallel to AB, then every ABD with D on l has the same area. (The line l charts the course of the witch “C ” as described in the problem.) Since the original triangle has area 4 and the final triangle has area 4.5, the final witch-configuration is not achievable. ♦ Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

6. HINTS AND SOLUTIONS TO SELECTED PROBLEMS

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Problem 8. This solution differs from the one in the text only in the way it shows that an m × n board is not tileable when both m and n are even, yet neither is divisible by 4. Place “colors” {1, 2, 3, 4, 1, 2, 3, 4, . . .} in row 1, “colors” {2, 3, 4, 1, 2, 3, 4, 1, . . .} in row 2, “colors” {3, 4, 1, 2, 3, 4, 1, 2, . . .} in row 3, “colors” {4, 1, 2, 3, 4, 1, 2, 3, . . .} in row 4 (cf. Fig. 6), and then repeat the row 1 pattern in row 5, the row 2 pattern in row 6, etc. Now, partition the board into batches of 4 rows, excluding the last two rows which won’t fit in a batch. Each batch of 4 rows will have an equal amount of all four “colors”, so no problem here. As for the last two rows, let’s partition them into batches of 4 columns: each of the newly formed 4 × 2 sub-boards has an equal amount of the four “colors”. However, the remaining ungrouped 2 × 2 board causes trouble: its “colors” are {1, 2, 2, 3}, yielding one extra square in “color” 2 (and no square in “color” 4). Thus our overall coloring is unbalanced. But regardless of how each 4 × 1 tetromino is placed on the board, it covers exactly one square of each of the four “colors”, so our board must be color-balanced in order to be tiled, a contradiction. Hence, it is impossible to tile a board m × n if neither m nor n is divisible by 4.
1 2 3 4

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3 4 1 2

4 1 2 3

1 2 3 4

2 3 4 1

3 4 1 2

4 1 2 3

... ... ... ...

{z

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2 3 4 1

} 1 2 3 4 1 2 3 4 ... 2 3 4 1 2 3 4 1 ... | {z } | {z } | {z 4 4 4

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{z 4

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1 2 2 3

Figure 6. Impossible tiling by 4 × 1 tetrominoes The two proofs of non-tileability demonstrated that there could be more than one invariant providing the necessary contradiction. The more you experiment with invariants, the more readily you will be able to find them. Problem 9. If n is even, then exchange adjacent students in each row: the student in an odd-numbered desk k exchanges places with the student in desk k +1. If n is odd, imagine that the desks are colored as on a chessboard. There will be one more desk of one color than the other. Since a valid move requires a student to go to a desk with a different color, one of the students will have nowhere to go and cannot follow the directions.
Exercise 7. By a version of PST 41, it is enough to show that an n = (2k + 1)-dragon can move one square northeast. Let R, U, D, and L stand for right, up, down, and left, respectively; adding a superscript indicates how many squares in the corresponding direction the dragon moves, e.g., U n means “n squares up”. Then (R1 U n ) followed by (D1 Rn )(D1 Ln ) for k times ♦ will land the n-dragon in the R1 U 1 square.

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Session 6 Mathematical Induction based on

Quan Lam’s sessions

Sneak Preview. One of the first concerns that had to be addressed at BMC was the poor training most American students get in writing proofs. The present article has combined several sessions under the topic of mathematical induction (MI ). The latter is a particular method of proof in whose validity it is easy to believe and which becomes, with a little practice, a handy tool for the problem solver. That is not to say that all proofs by MI are easy to create, but the structure of MI is so simple that one can concentrate on the mathematical ideas without concern for the logic behind the proof: the logic will be validated by the method itself. Not all problems succumb to MI ; yet, when they do, the results are very satisfying: not only does our problem hold for the first ten or the first ten billion cases; it holds for all of the infinitely many possibilities. So, dive headlong into an “inductive” pool of sums, inequalities, and partitions, symbolic algebra and the ubiquitous Fibonacci numbers; design with inspiration tilings and attack without fear the Towers of Hanoi . . . . Just make sure you don’t get carried away and prove that all dogs are the same color!

1. Examples and Conjectures 1.1. Conjecturing is definitely fun! Let us explore several introductory examples to sharpen your sense for patterns. Each example starts with the first few cases of a mathematical “event”. Your task is to find some common feature and formulate a conjecture which you believe holds true in all cases. Exercise 1 (Sums). Observe the sums of consecutive odd numbers, starting always with 1. Can you guess a formula for this sum which will work in all cases?

1 1+3 4 1+3+5 9 1 + 3 + 5 + 7 16 5 1+3+5+7+9 2

After a brief observation, your response will most likely be Conjecture 1 (Sums). The sum of the first n odd natural numbers is n2 . 103

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6. MATHEMATICAL INDUCTION

Now try the following exercises on your own. Possible conjectures are given afterwards, but you should answer the questions without peeking. Exercise 2 (Inequalities). Which is larger: (a) the natural numbers n or the powers of 2, 2n ? (b) the cubes of natural numbers, n3 , or the powers of 2, 2n ? Exercise 3 (Divisors). If n is any natural number, find a common divisor (a) for all differences n3 − n; (b) for all sums 2n+2 + 7n . Exercise 4 (Partitions1). Which natural numbers n can be written as (a) the sum of 4’s and 5’s only? (b) the sum of two (positive) primes? Exercise 5 (Prime Factorization). Which natural numbers can be completely split into a product of primes? 1.2. How does one come up with conjectures? For starters, list the data from the first several possible cases and look for common features, repetitions, patterns, any relations, etc. In Exercises 2–3, for example, we construct the following tables for n = 1, 2, . . . , 6: n 2n 1 2 2 4 3 8 4 16 5 32 6 64

n3 1 8 27 64 125 216

n n3 − n 2n+2 + 7n 1 0 15 2 6 65 3 24 375 4 60 2465 5 120 16935 6 210 117905

Figure 1. Inequalities and Divisors Comparing the columns in Figure 1a, we see that 2n is overwhelmingly larger than n; and except for a “slow start” at n = 1, the cube n3 seems to win over 2n with a larger and larger margin. So, we can form Conjecture 2 (Inequalities). Let n be any natural number. Then (a) n < 2n for all n ≥ 1; (b) 2n < n3 for all n ≥ 2. Similarly, the columns in Figure 1b are very suggestive: the numbers in the second column are obviously all divisible by 6, while the numbers in the third column – by 5. Without delay, we formulate Conjecture 3 (Divisors). Let n be any natural number. Then (a) n3 − n is divisible by 6; (b) 2n+2 + 7n is divisible by 5. 1A partition of a number n refers to expressing n as a sum of positive integers.

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Let’s check what’s going on in Exercise 4(a), where we are asked to write natural numbers as sums of 4’s and 5’s. While 4, 5, 8, 9, and 10 can be easily written as sums of 4’s and 5’s, it is obvious that 1, 2, 3, 6, and 7 cannot. With a little experimenting, you can convince yourselves that 11 joins the last bunch of “impossible” numbers. So, we get the impression that about half of the numbers can – and about half cannot – be written in the desired form. But look what happens for larger numbers in Figure 2: everything between 12 and 18 is listed affirmatively! n 12 13 14 15 16 17 18  4 s+5 s 4+4+4 4+4+5 4+5+5 5+5+5 4+4+4+4 4+4+4+5 4+4+5+5 

Figure 2. Partitions into 4’s and 5’s The evidence is so strong that we are practically forced to declare Conjecture 4(a) (Partitions into 4’s and 5’s). Every natural number greater than 11 can be written as a sum of 4’s and 5’s. Now we move on to writing n as a sum of two (positive) primes p and q, as requested by Exercise 4(b) (cf. Fig. 3). You can easily see that among the first 16 natural numbers only 1, 2, 3, and 11 cannot be written in such a way; let’s call such numbers “bad”. Continuing, you should verify that the next “bad” numbers are 17, 23, 27, 29, etc. Overall, it looks like all even numbers except 2 are “good”, while some odd numbers are “good” and some are “bad”. n 4 5 6 7 8 9 10 12 13 14 15 16 p+q 2+2 2+3 3+3 2+5 3+5 2+7 5+5 5+7 2+11 7+7 2+13 3+13

Figure 3. Partitions into two primes Naturally, a problem solver would ask: when can an odd n be written as a sum p + q? First of all, there is only one even (positive) prime, namely 2, and the rest of the primes are odd. Well, if the sum n = p + q is odd, one of p or q must be even and the other must be odd, say, p = 2 and q is odd. Summarizing, an odd n is “good” exactly when it is 2 more than an (odd) prime. That’s why, for instance, n = 5, 7, 9, 13, 15, 19, 21, and 25 are the first “good” odd numbers: they are 2 more than the corresponding primes 3, 5, 7, 11, 13, 17, 19, and 23. However, any guesses about the universal “goodness” of even numbers are still just conjectures: Conjecture 4(b) (Partitions into Primes). Every even natural number except 2 can be expressed as the sum of two (positive) primes. Finally, let’s turn our attention to splitting a natural number n into a product of primes. Contrary to our earlier advice to start with small cases, Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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let’s pick a random large number, e.g., n = 1092, and successively split it into smaller and smaller pieces until we cannot split anymore: 1092 = 2 × 546 = 22 × 273 = 22 × 3 × 91 = 22 × 3 × 7 × 13. In fact, our intuition suggests that if we pick any n and start splitting it into smaller and smaller factors, the process must stop at some point and yield a product of “indecomposable pieces”, which we call primes. Furthermore, if we agree to use only positive factors (so 1092 = 22 × (−3) × (−7) × 13 is not allowed), it seems that no matter how we do the splitting of n, we will end up with exactly the same prime factorization. For instance, let’s try to factor 1092 in a different way: 1092 = 14×78 = (2×7)×(6×13) = (2×7)×(2×3×13) = 22 ×3×7×13. As expected, this calculation gave us nothing new! We conclude Conjecture 5 (Prime Factorization). Every natural number n greater than 1 can be written as the product of primes. If only positive primes are allowed as factors, then this factorization is unique (up to reordering).

2. Mathematical Induction and Proof 2.1. Sequence of problems in disguise. A good sign of creativity in a mathematician is that he or she can “guess” answers from a few initial cases, just as in the eight examples above. Of course, the “guess” may be correct sometimes and incorrect at other times. Hence, the ability to prove or disprove these “guesses” is an essential part of being a good mathematician. Notice that in all of the above examples we are trying to draw a conclusion about a sequence of cases for all odd, for all even, or for all natural numbers. In other words, our problem is not just “one” statement; but, rather, it is comprised of infinitely many statements grouped in a single formulation. For instance, Conjecture 1 can be “broken” up into the following sequence of individual problems Pn for n = 1, 2, 3, . . .: P1 : 1 = 12 , P2 : 1 + 3 = 22 , P3 : 1 + 3 + 5 = 32 , .. .

Pn : 1 + 3 + 5 + 7 + · · · + (2n − 1) = n2 .    n

Let’s not get carried away: not every problem is set to be broken up into such a sequence of “smaller” problems, e.g., “Harry Potter defeated the Onewho-cannot-be-named” or “A rectangle has diagonals of equal length” better stay as single statements. However, once you do recognize a sequence of mathematical events {Pn }, you are on the right track: you would naturally look for some kind of sequential proof to verify your conjecture. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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2.2. Will it rain forever? Imagine the following unbelievable situation. Upon being defeated and forced to abandon his castle forever, an evil magician casts a spell: if it rains one day over the castle, then it will rain the next day too. It happens to rain there today. Will it rain forever over the castle? The answer, of course, is Yes: if it rains today, then the spell will make it rain tomorrow, and again by the spell it will rain the day after tomorrow, and so on and so forth without stopping. To formalize mathematically, we break the desired statement “It will rain forever over the castle” into a sequence of statements Pn for n = 0, 1, 2, . . .: P0 : “It rains today over the castle.” P1 : “It rains tomorrow over the castle.” P2 : “It rains the day after tomorrow over the castle.” .. . Pn : “It rains on the nth day into the future over the castle.” The magic spell can be encoded by the implication Pn ⇒ Pn+1 for all n ≥ 0; in words: “if it rains on the nth day, then it rains on the (n + 1)st day.” Now, even with the magic spell, until it happens to rain on some particular day, we won’t be able to start off our argument. What triggers the infinite sequence of events {Pn } is the extra information that “It rains today”, i.e., that P0 is true. Mathematicians refer to P0 as the basis case; the magic spell “Pn ⇒ Pn+1 for all n ≥ 0 ” is the inductive rule; and the method justifying that all Pn ’s are true is called 2.3. The Principle of Mathematical Induction2 (MI). Suppose we have managed to break up our problem into a sequence of propositions3, and let Pn represent the nth such proposition. A proof by MI shows that Pn is true for every claimed value of n (e.g., every natural n, every n ≥ 3, etc.). The proof itself consists of two steps: Basis Step: Show that Pn is true in the first instance of the sequence of events. For example, P0 is true in our “rainy” problem or P1 is true in Conjecture 1 (Sums) or P12 is true in Conjecture 4(a) (Sums of 4’s and 5’s). 2 For an introduction and motivation to the method of induction, see Sominsky [87]. 3 As you may have noticed, the words “proposition”, “statement”, “claim”, and “sen-

tence” are often used interchangeably to denote a mathematical event.

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Inductive Step: Assuming that Pn is true for some n, prove that the next statement Pn+1 is also true. The value of n here is fixed but unspecified, i.e., you can’t just assume, say, P5 is true and show P6 is true! In the setting of the “rainy” problem, this means that the “magic spell” must work on all days, not just on your favorite day. Some terminology: “Pn is true for some n” is the inductive hypothesis (IH ), and “Pn+1 is true” is the inductive conclusion. If you complete successfully both basis and inductive steps, then you have proved that all Pn ’s are true by the Principle of Mathematical Induction.

3. Mathematical Induction in Action Let us try to put the idea of MI to use on our initial conjectures from Section 1. Since we will be dealing constantly with the LHS and the RHS of the statements Pn , we will denote them for short by Ln and Rn . 3.1. Sums via MI. As we saw, Conjecture 1 breaks up into the statements ?

Pn : 1 + 3 + · · · + (2n − 1) = n2 for all n = 1, 2, 3, . . . . Basis Step. P1 : 1 = 12 is obviously true (n = 1).



Inductive Step. Assume that Pn is true for some n ≥ 1, i.e., 1 + 3 + · · · + (2n − 1) = n2 . We need to prove that Pn+1 is also true, i.e., ? 1 + 3 + · · · + (2n + 1) = (n + 1)2 . How do we do that? Obviously, we need to somehow use Pn . PST 42. Finding a relationship between Pn+1 and Pn is the most important part of the inductive step. If sums are involved, usually Pn+1 has one or more terms than Pn , and we can write Ln+1 = Ln + these additional terms. In the context of Conjecture 1, Ln+1 adds the next odd number (2n + 1) to Ln , and we have assumed (by IH ) that Ln = n2 . We calculate   Ln+1 = 1 + 3 + · · · + (2n − 1) + (2n + 1) = Ln + (2n + 1) = n2 + (2n + 1) = (n + 1)2 = Rn+1 .

IH

We just established Ln+1 = Rn+1 , which means that Pn+1 is true and completes the inductive step. The method of MI now allows us to conclude
automatically that all statements Pn are true for all n = 1, 2, 3, . . . . Let’s recapitulate how MI worked above: we proved that if statement Pn is true for some n, then Pn+1 is also true. Since P1 is true, this proves that P2 is true. Since P2 is true, this proves that P3 is true . . . and so on forever. The reader should try to prove by MI in a similar fashion the following formulas for summing up natural numbers, their squares or cubes. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Exercise 6. Let n ∈ N. Prove that (a) 1 + 2 + 3 + · · · + n = n(n + 1)/2; (b) 12 + 22 + 32 + · · · + n2 = n(n + 1)(2n + 1)/6; (c) 12 + 32 + 52 + · · · + (2n − 1)2 = (4n3 − n)/3; (d) (1 + 2 + 3 + · · · + n)2 = 13 + 23 + 33 + · · · + n3 . 3.2. Anti-induction detour. Exercise 6(a)–(c) should have raised some eyebrows: how did the elegant RHS of these formulas come about in the first place? “Guessing” in such problems is an increasingly tricky affair. For instance, try to figure out the RHS for part (b) by looking at the first 10 cases: the answer won’t be obvious at all! Or, try to create a formula for the sum of the 5th powers of the first n natural numbers! Indeed, one big disadvantage of mathematical induction is the need to come up with an explicit conclusion (e.g., a formula) that can be stated precisely. Another disadvantage is that MI doesn’t really explain why a certain statement is true: MI supplies the logical proof, yes, but it may still leave a gap in our understanding of the underlying reasons. Therefore, in many situations it is preferable to discover a direct non-MI proof. Even though this session emphasizes proofs by MI, we encourage you also to try to solve as many problems as possible without MI.



Let us demonstrate such an MI -free solution. Alternative Proof of Conjecture 1: An old idea4 is to group the terms so that each group has the same sum. To this end, we pair up the first and the last terms, the second and the second to last terms, etc., e.g., 1+3+5+7+9+11 = (1+11)+(3+9)+(5+7) = 12+12+12 = 3·12 = 36 = 62 . Thus, if we start with the first 6 odd numbers, we can group them in 3 pairs, each with a sum of 12. We need to modify this idea when we add up, say, the first 5 odd numbers: 1 + 3 + 5 + 7 + 9 = (1 + 9) + (3 + 7) + 5 = 10 + 10 + 10/2 = 10 · 5/2 = 52 . To avoid having two separate cases, we add up the desired sum S twice and pair up everything into exactly n pairs, each with sum 2n: S= 1 + 3 + · · · + (2n−2) + (2n−1) + S = (2n−1) + (2n−2) + · · · + 3 + 1 2S =

2n

+

2n

+ ··· +

2n

+

2n.

The last row contains exactly n terms 2n, one for each sum k + (2n − k).
Therefore, 2S = 2n · n, from which S = n2 . The intermediate-to-advanced reader is encouraged to do Exercise 6 in two ways: by MI and by non-MI methods. Part (a) will succumb to both 4 Attributed to Gauss as a child and already used in the Number Theory session.

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methods easily, but parts (b)–(d) will require quite some ingenuity in the search of non-MI proofs (cf. Hints section). 3.3. MI vs. telescopes. You may have had a hard time coming up with a formula for the sum of the consecutive cubes in Exercise 6(d), but let’s see if you are “luckier” with the following: Exercise 7. Find the sum

1 1·2

+

1 2·3

1 3·4

+

+ ···+

1 n·(n+1)

for n ≥ 1.

Hint: Since an explicit formula for the sum is not given, you have to “guess” it by searching for a pattern. n 1 2 3 4

n

sum

result

1 1·2 1 1·2 1 1·2 1 1·2 1 1·2

1 2

+ + + +

1 2·3 1 2·3 1 2·3 1 2·3

+ + +

1 3·4 1 3·4 1 3·4

+

1 4·5

+ ···+

2 3

3 4

4 5

1 n·(n+1)

n n+1 ?

♦

It remains to use induction to prove the conjecture above.

Was it coincidental that we came up so easily with such a simple formula in Exercise 7? Or is there another, non-MI explanation for this phenomenon? Check the identity of fractions for any k ≥ 1: 1 k·(k+1)

=

1 k

−

1 k+1 ·

Thus, every term in our sum splits as a difference of two fractions:



      1 1 − n1 + n1 − n+1 1 − 12 + 12 − 13 + 13 − 14 + · · · + n−1         1 = 1 + − 12 + 12 + − 13 + 13 + − 14 + 14 + · · · + − n1 + n1 − n+1 = 1−

1 n+1

=

n n+1 ·

1− 1 2

Above we used the so-called telescoping method for adding up series. The end of each bracketed expression cancels with the beginning of the next, reminding us of adjacent sections of a telescope, sliding towards and “canceling” parts of each other. Try to apply now both MI and the telescoping methods to

1 3−1 4

1 1 2 −3

1 n−1 − 1 n

1− 1 n

Exercise 8. Find the sums for all n ≥ 1: (a) 4·122 −1 + 4·222 −1 + 4·322 −1 + · · · + 4n22−1 ; 1 1 1 1 + 2·5 + 3·6 + · · · + n·(n+3) · (b) 1·4




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3.4. Inequalities via MI. With our newly-gained experience, we recognize that Conjecture 2(a) is conveniently broken up into a sequence of statements: n < 2n for all n ≥ 1. We immediately suspect that an MI -proof is waiting to be discovered. Basis Step. 1 < 21 is obvious. Inductive Step. Assume that n < 2n for some n ≥ 1. We need to prove the next proposition: why is (n + 1) < 2(n+1) ? Indeed, n + 1 < 2n + 1 < 2n + 2n = 2(2n ) = 2n+1 . Along the way, we used that n < 2n (by IH ) and 1 < 2n (for n ≥ 1). Thus, Pn+1 : n + 1 < 2n+1 is proven, and the inductive step is completed. By MI,
we conclude that n < 2n for all natural numbers n. Not so fast! How did we know to use the extra inequality above? PST 43. To prove A < B, it is often easier to inB sert some C in between and prove instead two other inequalities: A < C and C < B. Then the chain of C inequalities A < C < B implies the desired A < B. This technique is called a sandwich: the bottom and top “bread slices” are A and B, and the “lettuce” is C. A To prove Conjecture 2(a) above, we sandwiched C = 2n + 1 between A = n+1 and B = 2n+1 . Obviously, if more intermediate inequalities are needed, e.g., A < C < D < E < B, you add “eggs”, “cheese”, and other favorites. Equipped with our new Sandwich PST, we are now eager to attack another inequality, Conjecture 2(b): 2n < n3 for n ≥ 2. Basis Step. 22 < 23 certainly doesn’t cause trouble. Inductive Step. Assume that 2n < n3 for some n ≥ 2. We need to prove the next proposition: why is 2n+1 < (n+1)3 ? Note the relationship Ln+1 = 2Ln : IH

?

2n+1 = 2 · 2n < 2 · n3 < (n + 1)3 . So the bottom and top slices are A = 2n+1 and B = (n + 1)3 , the lettuce is C = 2n3 , and we hope to show that C < B. Let’s try: ?

?

?

2n3 < (n + 1)3 ⇔ 2n3 < n3 + 3n2 + 3n + 1 ⇔ n3 − 3n2 − 3n < 1. But the last inequality is false for n ≥ 4, as you can convince yourselves in a number of ways. For instance, you can factor the LHS and substitute n = 4:     n3 − 3n2 − 3n = n n(n − 3) − 3 ≥ 4 4(4 − 3) − 3 = 4 > 1. It seems that our “lettuce” C = 2n3 is above the top slice B = (n + 1)3 and should more appropriately be called “the olive” on top of the sandwich! In the aftermath of our attempt, we realize that we have shown5 A < C > B . . . . The sandwich technique didn’t work! Now what? 5 Let’s use the very non-standard notation “A < C > B ” to mean “A < C and C > B.”

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The fact that we failed to produce a proof with a certain method doesn’t necessarily mean that Conjecture 2(b) is false. But our unsuccessful attempt should at least make us suspicious about the truthfulness of the conjecture. So, we go back and calculate a few more cases for comparison (cf. Fig. 4a): n 2n 7 128 8 256 9 512 10 1024 11 2048 12 4096

n3 343 512 729 1000 1331 1728

y

2x

x3

x3 2x a

b x

Figure 4. More on Conjecture 2(b) Aha! At n = 10, the conjecture falls apart: 2n overpowers n3 and seems to break even further away afterwards! No wonder we failed to complete the inductive step earlier: we were trying to prove a faulty conjecture! Note that a single counterexample (e.g., 210 > 103 ) is all we need to pinpoint an incorrect conjecture. Should we stop here? To the contrary, the new evidence gives us a new idea: couldn’t the opposite inequality be true for large enough n’s? We leave the reader to rearrange the previous arguments to settle just that. Exercise 9. Prove that 2n > n3 for n ≥ 10. If you are familiar with graphing, Figure 4b visually explains what is going on. For x > 0, the graphs of the functions 2x and x3 intersect in two places: a ≈ 1.38 and b ≈ 9.94. Originally, we were fooled by the integer values n = 2, 3, 4, 5, 6 since they all lie inside the interval (a, b), where x3 is larger than 2x . However, 2x eventually catches up with x3 and stays ahead for all x > b. Some readers may have heard that “exponential functions increase faster than polynomials”; those skilled in calculus can use derivatives to show that 2x > x3 for x ≥ 10, thereby discovering a non-MI proof of Exercise 9. Another example which led us to believe in a very convincing but nevertheless faulty conjecture is Problem 1 from the Combinatorics session: its sequence of number of regions started with 1, 2, 4, 8, 16. Is the next term 32? To find and prove the correct conjecture in this situation requires, along with MI, more advanced techniques and problem solving ideas. 3.5. Divisibility via MI. Let’s turn now to Conjecture 3(a): “n3 − n is divisible by 6 for all n ≥ 1.” Basis Step. 13 − 1 = 0 is obviously divisible by 6. Inductive Step. Assume n3 − n is divisible by 6 for some n ≥ 1. We want to Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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prove that (n + 1)3 − (n + 1) is also divisible by 6. Indeed, (n + 1)3 − (n + 1) = (n3 + 3n2 + 3n + 1) − (n + 1) = (n3 − n) + (3n2 + 3n) = (n3 − n) + 3n(n + 1). In the last sum observe that, by IH, the term n3 − n is divisible by 6. For the second term 3n(n + 1), note that one of n and n + 1 is bound to be even (why?). Thus, n(n + 1) is always divisible by 2, and hence 3n(n + 1) is divisible by 6. Adding two numbers divisible by 6 yields a sum also divisible by 6. We conclude that (n + 1)3 − (n + 1) is divisible by 6, and the method of MI automatically completes the solution for us.
As we discussed earlier, a good problem solver shouldn’t be satisfied with just finding an MI proof. For instance, one can use the table-mod approach from the Number Theory session to show that n3 − n ≡ 0 (mod 6). A third direct approach is also possible due to the convenient factorization n3 − n = n(n2 − 1) = (n − 1)n(n + 1). Of three consecutive numbers, exactly one is divisible by 3 and at least one is even. Thus, (n − 1)n(n + 1) is divisible by 3 and by 2, hence by 6.
Let’s practice further with MI on Conjecture 3(b): “2n+2 + 7n is divisible by 5 for all n ≥ 1.” Basis Step. 23 + 71 = 15 is indeed divisible by 5. Inductive Step. Assume 5 divides Ln = 2n+2 + 7n for some n ≥ 1. To show that 5 also divides Ln+1 = 2n+3 + 7n+1 , PST 42 suggests finding a relationship between Ln+1 and Ln . This requires some algebraic manipulations: Ln+1 = 2n+3 + 7n+1 = 2n+2 · 2 + 7n · 7 = 2n+2 · 2 + (Ln − 2n+2 ) · 7 = 2n+2 (2 − 7) + 7Ln = 7Ln − 5 · 2n+2 . Since 5 | Ln (by IH ), the last sum is obviously divisible by 5.




Now, we won’t sit on our accomplishment, but we’ll find a non-MI proof too. Using 7 ≡ 2 (mod 5), we derive the congruences 2n+2 + 7n ≡ 2n · 22 + 2n = 2n (22 + 1) = 5 · 2n ≡ 0 (mod 5).
Here is an exercise for you to try using both MI and congruences: Exercise 10. Prove that for all n ≥ 1, 6n+2 + 72n+1 has 43 as a factor. 3.6. The burden of proof. The difference between our two partition Conjectures 4(a) and 4(b) lies in the relative difficulty in proving them. As Figure 2 clearly demonstrates, there is a simple pattern in writing n ≥ 12 as a sum of 4’s and 5’s: we shall investigate it in the next section on the strong form of MI. The situation with Conjecture 4(b) is entirely different. Figure 3 unequivocally suggests that all even numbers except 2 should be expressible as Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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a sum of two primes, but we know better than to take this on faith. The conjecture was made in 1742 in a letter from Goldbach to Euler: it remains a conjecture to this day. No one has been able to prove it by either MI or any other mathematical method, even though the evidence is overwhelming: the statement has now been confirmed for all n < 1018 . Such a phenomenon of a long string of verifications which, nevertheless, is unacceptable as a justification is a characteristic of mathematics and is not often found in other disciplines. Outside of mathematics, the fact that the first quintillion cases are all true is likely to be considered sufficient as a “proof.” However, mathematicians always insist on finding an indisputable logical proof and do not rely on empirical evidence. For example, the most celebrated of mathematical conjectures, the Riemann Hypothesis,6 was verified in 2004 for the first 1013 zeros. Notwithstanding, experts in the field are still divided on whether they believe it or not. And the story doesn’t end here. One way to prove the Riemann Hypothesis would have been to prove the so-called Mertens Conjecture, which was known to be true for the first 10 billion integers. Yet, in 1985, a century after the conjecture was posed, it was shown to be false, amazingly, without providing an explicit counterexample (cf. Odlyzko and Riele [70]). The moral of the story is clear: even though conjecturing is fun, it carries a lot of responsibility as it places the burden of proof on humanity. But let’s slow down a bit and consider a couple of “more approachable” problems. Exercise 11 (Euler, 1772). Prove or disprove that n2 + n + 41 is a prime number for n ≥ 0. To complete the next exercise, one would normally require a computer program that does symbolic mathematics (e.g., Mathematica, Maple, etc.). Curiously enough, the person who first solved the problem did so before the advent of computers.




Exercise 12 (Ivanov, 1941). Prove or disprove that, when xn − 1 is completely expanded into irreducible factors with integral coefficients, all of the coefficients are 1 or −1. Let’s experiment a bit with the first cases to make sure that we understand the question. Everything is clear with x−1 and x2 −1 = (x−1)(x+1), but x3 −1 requires some thought. If we factor it as x3 −1 = (x−1)(x2 +x+1), we observe that all coefficients are ±1. One can show that any further factorization would involve the roots of x2 + x + 1, which are not real numbers, much less integral, so such factorizations are not allowed. Similarly, the only integral complete factorization of x4 − 1 is (x − 1)(x + 1)(x2 + 1), all of whose coefficients are exclusively ±1. Does this pattern continue to hold forever? Good luck in making up your mind and proving your conjecture. 6 The Riemann Hypothesis states that all of the non-trivial zeros of a particular funcP 1 tion, ζ(z) = ∞ n=1 nz , lie on the line z = 1/2 in the complex plane.

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4. Strong Induction 4.1. The Second Principle of Mathematical Induction. Conjecture 4(a) and Conjecture 5 pose a different type of difficulty. It turns out that our original version of proof by MI is not “strong” enough for many math problems. Thus, mathematicians have come up with a “strong” induction, or the second principle of mathematical induction (M I s ). The method of MI we described earlier is referred to as the first principle. These two forms of mathematical induction can be proved to be equivalent. The second principle uses slightly different, stronger, basis and inductive steps. Strong Basis Step: Show that Pn is true in the first one or more instances of the desired event; exactly how many initial cases you need to show will depend on your argument in the strong inductive step. Strong Inductive Step: Assuming that all of P1 , P2 , P3 , . . . , Pn are true for some n, prove that the next statement Pn+1 is also true. If P0 , P12 or some other Pk is the first case, start from it and assume everything up to Pn , inclusive. We denote the strong induction hypothesis by IH s . The IH s may seem like a very strong assumption, but it isn’t hard to see why it is reasonable. If, for instance, we know that P1 is true and we prove that P2 is also true, then we will have that {P1 , P2 } are true. If we can now show that P3 is true, we will have {P1 , P2 , P3 } are true. If P4 can now be proven to be true, it joins this sequence too, and so on. This way we can enlarge the sequence {P1 , P2 , . . . , Pn } of truthful statements to an arbitrary n and thus cover any individual Pn . 4.2. The power of M I s . As promised, we attack Conjecture 4(a) with M I s . Figure 2 displays a certain similarity in the partitions into 4’s and 5’s of 12 and 16, of 13 and 17, and of 14 and 18. Can you see it? This “cycle” of length 4 prompts us to prove 4 initial cases in our strong basis step. Strong Basis Step. We know that 12 = 4+4+4, 13 = 4+4+5, 14 = 4+5+5, and 15 = 5 + 5 + 5, so {P12 , P13 , P14 , P15 } are all true. Strong Inductive Step. Assume that P12 , P13 , P14 , P15 , . . . , Pn are all true for some n ≥ 15, i.e., all numbers from 12 through n can be written as sums of 4’s and 5’s. To get hold of n + 1, we notice that n + 1 ≥ 16, so that (n + 1) − 4 = n − 3 ≥ 12. By IH s , n − 3 is a sum of 4’s and 5’s. Adding an
extra 4 to this sum yields n + 1 and shows that Pn+1 is also true. To summarize, to get Pn+1 we add a 4 to the sum in Pn−3 , e.g., 15 = 5 + 5 + 5  19 = (5 + 5 + 5) + 4, etc. One can actually get by without M I s , but the proof might be messier than the above solution, as it requires case-chasing. See if you can find such a non-MI proof for Conjecture 4(a). Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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4.3. The Fundamental Theorem of Arithmetic is our Conjecture 5: “Every integer n > 1 can be written uniquely as a product of primes.” A proof by the regular MI is inconceivable here because the factorizations of two consecutive numbers n and n + 1 have nothing to do with each other! For instance, we found earlier that 1092 = 22 · 3 · 7 · 13; at the same time, the next number 1093 is prime, and hence it shares no common divisor with 1092 (other than 1). Thus, it would be ludicrous to expect to prove Pn+1 from Pn alone; in fact, we will need the help of all P2 , P3 , . . . , Pn to make progress on Pn+1 . Hence, M I s . Basis Step. 2 = 2 is a product of one prime, itself. Strong Inductive Step. For some n > 1, we assume that each of 2, 3, . . . , n is a product of primes. We want to prove that n + 1 is also a product of primes. To that end, we divide the proof into two cases. Case 1: n + 1 is prime itself, and we are done. Case 2: n + 1 is composite, i.e., n + 1 = ab for some natural numbers a and b, each between 2 and n. By IH s , both a and b are products of primes: a = p1 p2 · · · pk and b = q1 q2 · · · qm . Multiplying, we obtain that n + 1 = ab = (p1 p2 · · · pk )(q1 q2 · · · qm ), which is also a product of primes. This completes the inductive step and the overall proof by M I s .






As we witnessed, Case 2 used both Pa and Pb to prove Pn+1 . This proof wouldn’t have been possible with the regular MI. In addition to M I s , another important idea underpinned the whole solution. PST 44. In divisibility and other problems with integers, it is often useful to split all numbers into prime and composite and investigate each case separately. These two cases may require different reasoning, one usually much more straightforward than the other. As for uniqueness of the prime factorization, a different method and more subtle argumentation are needed. A complete proof can be found in standard abstract algebra textbooks (cf. [45, 29]). 4.4. Recursive sequences are born for strong induction. Take a look at the so-called recursive sequence in Exercise 13 below: each term an+1 is determined by the preceding two terms an and an−1 . If we know that, say, both an and an−1 are divisible by 5, we would be able to use the given recursive formula to conclude that an+1 is also divisible by 5. In general, every term in a recursive sequence can be calculated by using one or more of the previous terms, and thus we can relate known properties of previous terms to the next terms. This should remind the reader of the strong inductive hypothesis of M I s .

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Furthermore, to start off calculating the terms of a recursive sequence, an appropriate number of initial terms must be given. Again, in Exercise 13 this would mean knowing the first two terms, a0 and a1 . Without doubt this situation is related to the strong basis step of M I s . In conclusion, it should be no surprise that the method of M I s persistently appears in problems with recursive sequences. Exercise 13. Let a0 = 8, a1 = 13, and an+1 = 7an − 6an−1 for n ≥ 2. Prove that an = 6n + 7 for n ≥ 0. Solution: The statement is clearly true for n = 0 and n = 1. Assuming that an = 6n + 7 and an−1 = 6n−1 + 7 for some n ≥ 1, we will show that an+1 = 6n+1 + 7. By definition we have IH s

an+1 = 7an − 6an−1 = 7(6n + 7) − 6(6n−1 + 7) = 7 · 6n + 49 − 6n − 42 = 6 · 6n + 7 = 6n+1 + 7. The inductive step is completed, and the statement is proven by M I s .




In the above inductive step, it was enough to assume that only Pn and Pn−1 were true, which enabled us to derive Pn+1 . If we wanted to follow IH s to the letter, we should have assumed that all P0 , P1 , P2 , . . . , Pn were true, but that would have been overkill without giving us any particular advantage. Now that we are done with our preliminary example, we can get down to business. No treatment of recursive sequences would be complete without mention of the most famous sequence of all, the Fibonacci sequence {Fn }. Every term Fn is defined as the sum of the previous two terms: Fn = Fn−1 + Fn−2 for n ≥ 2, F0 = 0, F1 = 1.




The Fibonacci sequence looks so simple; yet, it has been the focus of attention and study for many centuries. Its combinatorial properties are numerous, and the sequence itself is ubiquitous throughout mathematics. As is true for any recursive sequence, {Fn } is inextricably intertwined with M I s . Below you will derive two of its properties: the first is yet another recursive relation, which will require a twist in the M I s method, while the second is the best we can hope for in a recursive sequence – a direct formula for its terms. Exercise 14. Prove that Fn+m+1 = Fn Fm + Fn+1 Fm+1 for all n, m ≥ 0. This is our first problem with two variables – the indices m and n. But M I s is phrased with one variable only! So we need to reduce this new situation to one where we can apply M I s . One way is to consider a lattice whose points are all pairs of nonnegative integers (n, m). To visualize better, imagine an infinite garden planted with lettuce as in Figure 5a, where each lettuce represents a lattice point and hence corresponds to one case of the proposed relation in Exercise 14. The

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“lettuce” (4, 2), for example, corresponds to the identity F7 = F4 F2 + F5 F3 , which is certainly true: 13 = 3·1 + 5·2. m P4

P4 m P3

P3

P2

P2

(4, 2)

P1

P1 P0

P0 n

0

0

n

Figure 5. Induction by rows and by diagonals Now, to prove the desired identities separately for each (n, m) will be too much and too chaotic: we need some reorganization to streamline this process. As every good gardener will group the lettuce in row-beds, so we will group our identities too. Our propositions P0 , P1 , P2 , . . . , Pm , . . . will correspond to the row-beds. For instance, P2 will combine all desired identities represented by “lettuce” in the 2nd row-bed, i.e., for which m = 2: ?

P2 : Fn+3 = Fn F2 + Fn+1 F3 for all n ≥ 0. To demystify the “row-bed” P2 , let’s verify it directly: Fn F2 + Fn+1 F3

=

Fn · 1+Fn+1 · 2 = (Fn +Fn+1 )+Fn+1

def

Fn+2 +Fn+1 = Fn+3 .

=

def

Thus, our (strong) induction will move from row to row: we say that we will induct on m. Within each statement Pm , we fix m and vary n so as to encompass all the “lettuce” in the mth row: (0, m), (1, m), (2, m), etc. Since the Fibonacci sequence requires two initial values for F0 and F1 , it shouldn’t be surprising that we will need two basis cases: P0 and P1 . Strong Basis Step. P0 is a rather trivial statement: ?

Fn+1 = Fn F0 + Fn+1 F1 = Fn · 0 + Fn+1 · 1 = Fn+1 . P1 , on the other hand, is the defining relationship for the Fibonacci sequence: ?

Fn+2 = Fn F1 + Fn+1 F2 = Fn · 1 + Fn+1 · 1 = Fn + Fn+1 . Strong Inductive Step. Assume that Pm and Pm−1 are true for some m ≥ 1, i.e., all “lettuce” in rows m and m − 1 corresponds to true identities: (1) (2)

Pm : Fn+m+1 = Fn Fm + Fn+1 Fm+1 for all n ≥ 0, Pm−1 : Fn+m = Fn Fm−1 + Fn+1 Fm for all n ≥ 0.

We need to show that the (m + 1)st row is planted with “truthful lettuce”: ?

Pm+1 : Fn+m+2 = Fn Fm+1 + Fn+1 Fm+2 for all n ≥ 0. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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The defining relation Fn+m+2 = Fn+m+1 + Fn+m suggests adding (1) and (2) and simplifying. We leave the algebraic “joy” of such manipulations to the reader. Consult the Hints section if you run into trouble. ♦



The above row-induction technique deserves its own PST, as it is an important and useful way to apply M I s . PST 45. To prove statements {Pn,m } for all n, m ≥ 0, you can fix one of the variables, m, and vary the other variable, n. This way you will construct a new sequence of propositions, {P0 , P1 , P2 , . . . , Pm , . . .}, where each Pm combines a whole “row” of statements: Pm = {P0,m , P1,m , P2,m , . . . , Pn,m , . . .}. Applying M I to this new sequence {Pm } is referred to as inducting on m. s

Evidently, one can induct on n instead: this would correspond to using column-beds. In fact, nothing can stop us from grouping the lettuce in any way we like, as long as each lettuce appears in some group, and the groups themselves form a sequence to which induction can be applied. For example, we can form a sequence {Pm } out of the diagonals as in Figure 5b.




We shall stop here and refer the reader for further applications of these ideas and an extensive study of Fibonacci identities to Proofs that Really Count, a uniquely written book about combinatorial proof by Arthur Benjamin and Jennifer Quinn [6]. Meanwhile, we’ll end this subsection with a discussion of the promised direct formula for {Fn }. √ n √ n

1+ 5 − 1−2 5 for any n ≥ 0. Problem 1. Prove that Fn = √15 2 This formula is known as Binet’s formula (1843), but it was actually discovered by Euler a century earlier. To the novice, it looks √ rather forbidding, √ maybe even mystical, as it contains the irrational number 5: how could 5 have anything to do with the integral Fibonacci numbers? It is unreasonable to expect anyone to “guess” the formula with his bare hands – there must be some reasonable explanation behind this lofty expression! Indeed, in a good undergraduate course in linear algebra, you will become acquainted with the theory of matrix diagonalization and Jordan form; in particular, it will become clear to you how to translate the Fibonacci recursive relation into the simple equation x2 = x + 1 and why the two roots of this equation should appear prominently √in the desired formula. √ These roots are the golden ratio φ = 1+2 5 and its “evil twin” φ = 1−2 5 . With their help, Exercise 1 can be rewritten as ? 1 (3) Fn = √ (φn − φ n ) for all n ≥ 0. 5 s The beauty of M I is that, once you are given a formula as a goal, you don’t need to know how or where this formula came from – everyone has a fair ♦ chance of proving it by using M I s . So, give (3) your best shot.

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5. Mathematical Induction in Other Areas




5.1. MI as a tiling specialist. Mathematical induction is not simply used with numbers and algebraic formulas; it can also be used in combinatorial geometry, as the following problem demonstrates. Problem 2. Let n be a natural number. Show that any 2n × 2n chessboard with one square removed can be tiled using L-shaped pieces (as in Figure 6a).  Figure 6. Tiling of a 2 × 2 board Basis Step. When n = 1, we can remove a square from a 2 × 2 chessboard in one of four ways (cf. Fig 6b–e). Obviously, each configuration can be covered by a single, suitably rotated L-shape. Inductive Step. Assume each 2n × 2n board with one square removed can be tiled by L-shapes. By dividing in half horizontally and vertically, we can partition a 2n+1 × 2n+1 board into four smaller 2n × 2n sub-boards. By symmetry, we can assume that our removed square is in the lower right such sub-board (cf. the dotted square in Figure 7a).

→



Figure 7. 2n+1 × 2n+1 board Now place a single L-shape in the center of the big board so that it covers one (corner) square from each of the remaining 3 small sub-boards (cf. Fig. 7b). If we temporarily remove the 3 squares covered by this Lshape, each of our four sub-boards will be missing one square. By IH, these 2n × 2n sub-boards can each be tiled by L-shapes. We can finally return the temporarily removed L-shape and cover the 3 “missing” central squares. Thus, we have tiled with L-shapes our original
2n+1 × 2n+1 board (with one square removed). Figure 7c demonstrates the tiling of a 16 × 16 board induced by our construction. The shading of the L-shapes changes: the darker shapes are placed earlier, and the lighter shapes are placed later in the inductive process. Thus, the original place L-shape is black, followed by 4 dark grey L-shapes, then by 16 grey L-shapes, and finally by 64 light grey L-shapes. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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5.2. MI climbs the Towers of Hanoi. As we couldn’t forgo the Fibonacci numbers when talking about recursive sequences, so it is inconceivable to discuss mathematical induction without including the following classic:




Problem 3. A toy puzzle called the Towers of Hanoi consists of three pegs and a set of graduated disks (cf. Fig. 8a). The task is to move the pile of disks from one peg to another while obeying the rules: 1. Only one disk at a time may be moved from one peg to another. 2. No disk may ever be placed over a smaller disk. Prove that this is always possible, regardless of how many disks there are. Basis Step. As it is, there is no problem in transferring 1 disk; P1 is verified. If we have 2 disks, we move the top one to an empty peg, move the second disk to the third free peg and cover it with the smaller first disk; P2 is verified. For 3 disks, we move the top two disks to another peg (by P2 ); then move the third disk to the remaining free peg, and finally (again by P2 ), move the first two (smaller) disks to cover the third disk; P3 is verified.

Figure 8. Towers of Hanoi with 5 disks Inductive Step. From the basis step above, the pattern has now emerged. If we know how to move n disks (assume Pn ), we can transfer n + 1 in the following way (we’ll prove Pn+1 ). First move the n top disks to another peg by Pn (cf. Fig. 8b); then move the (n + 1)st disk to the third peg (cf. Fig. 8c), and finally, again by Pn , move the n smaller disks to cover the largest one on the third peg (cf. Fig. 8d). We have verified Pn+1 . So it is possible to move any number of graduated disks from one peg to another without violating the rules.





Strictly speaking, we didn’t need to prove P2 and P3 as basis steps; however, working them out prompted the general inductive construction Pn ⇒ Pn+1 . As with any construction, we can ask ourselves if it is the most efficient one for moving all these n disks. Problem 4. For any n ≥ 1, prove that 2n − 1 is the smallest number of moves required to move a stack of n disks from one peg to another. Hint: You need to prove two things: that you can move n disks in 2n − 1 moves and that you cannot do it in fewer moves. MI is helpful in both. ♦

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6. A Word of Caution We must be careful when applying MI to be sure that both steps are completely proved. The following examples show what absurd claims can be “proven” if we slack off and relax either one of these two steps. Claim 1. All natural numbers are even. Inductive Step. Assume that any natural number up to n (inclusive) is even. We want to prove that n + 1 is also even. Consider n − 1. If n − 1 ≥ 1, then it is even by IH ; if n − 1 = 0, then again it is an even number; in any case, n − 1 = 2m for some integer m ≥ 0. Thus, n + 1 = (n − 1) + 2 = 2m + 2 = ? 2(m + 1), so n + 1 is also even. The “proof” is of course faulty as the basis step is not true: 1 is odd. Claim 2. All dogs are the same color. To simplify notation, if two dogs are named a and b, we shall mean by ‘a = b’ that they are the same color. Further, we need a sequence of statements to which MI will be applied. For any n ≥ 1, set Pn : “In any group of n dogs, these dogs are the same color.” Basis Step. If we have 1 dog, then obviously it is the same color as itself.

Figure 9. P2 ⇒ P3 : Any 3 dogs are the same color Inductive Step. Suppose Pn is true: any n dogs are the same color. Consider any group of n + 1 dogs: {a1 , a2 , a3 , . . . , an , an+1 } (cf. Fig. 9). If we take an+1 out of this group, then {a1 , a2 , a3 , . . . , an } are the same color by IH. And if we take a1 out of the original group, then the remaining n dogs {a2 , a3 , . . . , an , an+1 } also are the same color by IH. Since a1 = a2 and a2 = ? a3 = · · · = an+1 , then a1 = a2 = · · · = an = an+1 and Pn+1 is proved. All right, there is a problem in the above “proof”, but where? This time we did show that P1 is true. If you go carefully through the inductive step again, you’ll see that the argument there needs n ≥ 2 so that the two subgroups of n dogs, {a1 , a2 , . . . , an } and {a2 , a3 , . . . , an+1 }, have an overlap, namely a2 . Without such an overlap, we cannot conclude that the two groups are the same color. Thus, even though the inductive step correctly proves the implications P2 ⇒ P3 ⇒ P4 ⇒ · · · ⇒ Pn ⇒ Pn+1 , Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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it doesn’t establish P1 ⇒ P2 . The remedy is to prove P2 independently as a basis case, but unfortunately P2 is false: simply take one white and one black dog and you’ll have your counterexample. Claim 3. Prove that an = 1 for all integers n ≥ 0 and all real a = 0. Basis Step. For n = 0 we have a0 = 1. Inductive Step. Assume ak = 1 for all non-negative integers k ≤ n. Then an+1 = an · a1 = an ·

an 1 = 1 · = 1. n−1 a 1

?

What is wrong with this reasoning?

7. Hints and Solutions to Selected Problems Exercise 6. Use algebra to work out the inductive steps in parts (b)–(c): n(n+1)(2n+1) + (n 6 n(2n+1)(2n−1) + (2n 3

?

+ 1)2 = ?

+ 1)2 =

(n+1)(n+2)(2n+3) , 6 (n+1)(2n+1)(2n+3) · 3

The result of part (a) transforms the LHS of (d) into tion step in (d) only requires that you show n2 (n+1)2 4

? (n+1)2 (n+2)2 4

+ (n + 1)3 =

n2 (n+1)2 , 4

so the induc-

·

♦

For a non-MI method that finds the sum of consecutive squares in (b), we consider the sum of consecutive . . . cubes! You may think that this can only complicate things, but observe the essentially telescoping phenomenon: (1 + 1)3 (2 + 1)3 (3 + 1)3 + .. . (n + 1)3

= 13 + 3 · 12 + 3 · 1 + 1 = 23 + 3 · 22 + 3 · 2 + 1 = 33 + 3 · 32 + 3 · 3 + 1

= n3 + 3 · n2 + 3 · n + 1   (n + 1)3 = 13 + 3 k2 + 3 k + n = 1 + 3S + 3 n(n+1) + n. 2

 After adding, we cancelled almost all cubes, used the for k from  formula (a), and denoted by S our desired sum of squares k 2 . Solving now for S easily yields the formula in (b). This is very similar to the way Archimedes found the sum 2000 years ago (cf. Stein [90]). Analogously, you can use fourth powers k 4 to extend this approach and find the sum of cubes in (d): 13 + 23 + 33 + · · · + n3 =

n2 (n+1)2 · 4

For a non-MI proof of (c), let On , En , and An denote, respectively, the sums of the squares of the first n odd, n even, and n natural numbers. Thus, Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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the LHS of (b) is An , the LHS of (c) is On , and On + En = 12 + 22 + · · · + (2n)2 = A2n . So, En = 22 + 42 + · · · + (2n)2 = 4(12 + 22 + · · · + n2 ) = 4An (b) 2n(2n+1)(4n+1) 6

⇒ On = A2n − 4An =

− 4 n(n+1)(2n+1) = ··· = 6

4n3 −n 3 ·

For some geometric insight into why the formulas are true, there is a very good source – The Book of Numbers by Conway and Guy [17]. A column on “Look-See” diagrams by Martin Gardner in Scientific American (cf. [30]) shows a way to visualize three times the sum of the squares as  a rectangle: 3 k 2 = (2n + 1) × n(n+1) . The crucial ingredient here is 2 Exercise 1: each square k 2 is the sum of the first k odd numbers (cf. the white dots in Figure 10). In Chinese texts, formula (b) first appeared in the 13th century, but was not proved until the 17th century using a threedimensional geometric interpretation. For many more proofs of this fact and a variety of other similar formulas, check out Proofs Without Words by Nelsen [66]. ♦ 

12 22



32 2

4

n(n+1) 2



 2n+1



52 

Figure 10. Old Chinese proof of 12 + 22 + · · · + n2 = 2 = 1 − 1 and 4n2 −1 2n−1 2n+1 1 1 1 1 and six n(n+3) = 3 n − n+3

Exercise 8. In (a), 1 3



In (b), 1+

1 2

+

1 3

−

1 n+1

−

1 n+2

−

1 n+3



n(n+1)(2n+1) 6

the sum telescopes to

2n 2n+1 ·

terms survive the telescoping:

, to be gathered into one fraction.

♦

Exercise 9. By taking cube roots on both sides of 2k 3 > (k + 1)3 , we see 1 > 3. So the “sandwich” in the that the inequality is true for k > √ 3 2−1

inductive step works:

2n+1

=2·

2n

IH

> 2 · n3 > (n + 1)3 for n ≥ 10 > 3.

♦

Exercise 10. For n = 1 in the basis step, we have 63 + 73 = 216 + 343 = 559 = 43 · 13. Now assume that the statement is true for some natural number k, i.e., 6k+2 + 72k+1 is a multiple of 43. Then 6k+3 + 72k+3 = 6 · 6k+2 + 72 · 72k+1 = 6(6k+2 + 72k+1 ) + 43 · 72k+1 . Both terms are divisible by 43, and so is their sum.
For a non-MI proof, notice that 72 ≡ 6 (mod 43). Then 6n+2 + 72n+1 ≡
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Exercise 11. For n ∈ {0, 1, . . . , 39} the value of n2 + n + 41 is a prime number. However, for n = 40 the value is 412 , which is not a prime, so the statement is false. Perhaps a more obvious counterexample is obtained when
n = 41: 412 + 41 + 41 = 41(41 + 2) is a multiple of 41. Exercise 12. By checking the coefficients of the factors (e.g., via Mathematica), one finds that the statement is true for n < 105. However one of the factors of x105 − 1 is 1+x+x2 −x5 −x6 −2x7 −x8 −x9 +x12 +x13 +x14 + x15 +x16 +x17 −x20 −x22 −x24 −x26 −x28 +x31 +x32 +x33 +x34 +x35 +x36 −x39 − x40 −2x41 −x42 −x43 +x46 +x47 +x48 . The coefficients of x7 and x41 are −2, so the statement is shown to be false.
Exercise 14. Adding equations (1) and (2) yields Fn+m + Fn+m+1 = Fn (Fm−1 + Fm ) + Fn+1 (Fm + Fm+1 ) ⇒ Fn+m+2 = Fn Fm+1 + Fn+1 Fm+2 .




Conjecture 4(a). For a non-MI proof, use remainders (mod 4) to express n as 4k, 4k + 1, 4k + 2, or 4k + 3. If n ≥ 12, then k ≥ 3 and these expressions can be written as 4k, 4(k − 1) + 5, 4(k − 2) + 5 · 2, or 4(k − 3) + 5 · 3; in other words, n consists of k 4’s, (k − 1) 4’s and one 5, (k − 2) 4’s and two 5’s, or (k − 3) 4’s and three 5’s, respectively.
Problem 1. The statement is easy to verify for n = 0 and n = 1. The two numbers, φ and φ, are solutions of the quadratic equation x2 = x + 1, so φ 2 = φ + 1 and φ 2 = φ + 1. Use this fact, strong induction, and the ♦ definition Fn = Fn−1 + Fn−2 to prove the inductive step. Problem 4. To move n disks to another peg, we used exactly 2n − 1 moves in our solution. Indeed, for 1 disk, obviously 21 −1 = 1 moves is all it took to move it to another peg. By induction, if we use 2n−1 −1 moves for n−1 disks, then for n disks our solutions will use exactly (2n−1 − 1) + 1 + (2n−1 − 1) = 2n − 1 moves. To show that we can’t do better than 2n − 1 moves for n disks, not surprisingly, we again employ induction. For n = 1 there is nothing to talk about. Assume that the minimal number of moves required for n − 1 disks is 2n−1 − 1. Now consider the puzzle with n disks. At some point the top n − 1 disks must be stacked on a separate peg so that the bottom disk can be moved to the remaining third peg; by our IH, this takes a minimum of 2n−1 − 1 moves, and the bottom disk adds at least 1 more move. When the bottom disk settles on a peg and won’t move anymore, we need to bring the smaller n − 1 disks on top of it. Notice that, while the bottom disk makes this final move, say, from peg A to peg C, all other n − 1 disks must be stacked on peg B. Again by IH, in order to bring all n − 1 smaller disks from peg B to peg C, we need at least 2n−1 − 1 moves. Adding up, in essence we have the same calculation as above: the minimal number of moves required for n disks is (2n−1 − 1) + 1 + (2n−1 − 1) = 2n − 1. By MI, we are done.


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Session 7 Mass Point Geometry Tom Rike Sneak Preview. This chapter will introduce a technique that simplifies calculations of ratios in geometric figures in an intuitive way by merging algebra, geometry, and basic physics, and that is often far faster than the standard techniques of vectors and area addition. The method is as simple as balancing a seesaw, so it can be understood by young students, but it will also present a new tool to older students since this subject will not be found in standard school curricula. Apart from its powerful applicability to olympiad problem solving, the original approach to the theory of mass points (by Hausner) also provides an introduction to the axiomatic method – the foundation of Euclidean geometry and modern mathematics. Natural extensions from this topic are vector spaces, the more general barycentric coordinates, and projective geometry.

1. Introduction 1.1. The power of the mass point technique. Let us begin with an old geometry concept. Given a triangle, a cevian is a line segment from a vertex to an interior point of the opposite side. (The ‘c’ is pronounced as ‘ch’.) Figure 1a illustrates two cevians AD and CE in ABC. Cevians are named in honor of the Italian mathematician Giovanni Ceva, who used them to prove his famous theorem in 1678 (cf. Theorem 6). Problem 1 below is not that famous, but it certainly presents a situation that you may stumble upon in everyday problem solving. D B E 4 A

3

5 F

F G

D 2 C

A

E J

C

H B

Figure 1. Two cevians and Centroid of a tetrahedron 127

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Problem 1. In ABC, as shown in Figure 1a, side BC is divided by D in a ratio of 5 : 2, side BA is divided by E in a ratio of 3 : 4, and BC intersects BA in F . Find the ratios in which F divides the cevians AD and CE, i.e., find AF : F D and CF : F E. PST 46. A very powerful problem solving tool is given by area addition: when a planar region is broken into non-overlapping parts, you can find the total area by adding the areas of these parts. Further, when a figure has segments with given lengths or ratios, applying areas or another powerful tool – vectors – is often the key to an easy solution.1 It is true that Problem 1 can be successfully attacked by both areas and vectors, thereby reducing it to elementary algebra. Yet, we shall see in this session that a much easier and more intuitive solution will result from assigning masses to the vertices of ABC in such a way that F becomes the balancing point, and the problem is reduced to elementary arithmetic. Examples illustrating the power of our mass point technique do not need to be constrained to the plane. The next problem, for instance, is set in 3-dimensional space. Recall that a tetrahedron is the mathematical term for what we know as a pyramid: a polyhedron with four triangular faces. Review also the definition of a centroid of a triangle as the intersection of the three medians in the triangle (cf. the Inversion session).




Problem 2. Consider tetrahedron ABCD and mark the centroids of its four faces by E, F , G, and H (cf. Fig. 1b). Prove that the four segments connecting vertices to the centroids of the opposite faces are concurrent, i.e., that AE, BF , CG, and DH intersect at a point J, called the centroid of the tetrahedron. In what ratio does this centroid J divide the four segments, i.e., what is DJ : JH? Undoubtedly, this problem reminds the reader of the Centroid Theorem, cited in the Inversion session. Yet, how do we solve Problem 2 in space? Vectors are often an option for those who are skilled with them. But mass points, when applied appropriately, will empower us to complete the problem almost in our heads! Thus, if you are intrigued by our new technique and would like to master it, then continue reading this session. 1.2. Archimedes’ lever. The underlying idea of the mass point technique is the principle of the lever, which Archimedes used to discover many of his results. You may have heard the boast of Archimedes upon discovering the lever, “Give me a place to stand on, and I will move the earth.” This is beautifully portrayed by an engraving in Mechanics Magazine [62], London 1824 (cf. Fig. 2). Although Archimedes knew his results were correct based on reasoning with the lever, such justification was unacceptable as proof in Greek mathematics; so he was forced to think of very clever proofs using 1 You will see the method of vectors in action in the next volumes.

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1. INTRODUCTION

129

Euclidean geometry to convince the mathematical community at the time that his results were correct. These proofs are masterpieces of reasoning, and it is recommended that the reader study some of them to appreciate the elegance of the mathematics (cf. [43] and the History and Sources section).

Figure 2. Archimedes moves the Earth using a lever However, in this session we are going to use Archimedes’ lever. The basic idea is that of a seesaw with masses at each end. The seesaw will balance if the product of the mass and its distance to the fulcrum is the same for each mass. For example, if a baby elephant of mass 100 kg is 0.5 m from the fulcrum, then an ant of mass 1 g must be located 50 km on the other side of the fulcrum for the seesaw to balance (cf. Fig. 3, not drawn to scale): distance × mass = 100 kg × 0.5 m = 100, 000 g × 0.0005 km = 1 g × 50 km.

E

   F 0.5 m

 50 km

A

Figure 3. Elephant and ant balance on a seesaw As we shall see, the uses of the lever are more far-reaching than one might imagine. For instance, here is another problem for you to consider. It extends our mass point technique to transversals: a transversal of lines l and m is a line that joins a point on l and a point on m. Note that a transversal of two sides in a triangle is a generalization of a cevian. Thus, try Problem 3 first on your own; then learn the definitions and properties of the mass point technique in Section 2; and finally compare your solution with our elegant, organized attack on it in Section 3. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Problem 3. In Figure 4, transversal ED joins points E and D on the sides of ABC so that AE : EB = 4 : 3 and CD : DB = 2 : 5. Cevian BG divides AC in a ratio of 3 : 7 counted from vertex A and intersects the transversal ED at point F . Find the ratios EF : F D and BF : F G. B 3 E

5 F

D

4 A

3

G

7

2 C

Figure 4. Using mass points with a transversal

2. Definition and Properties of Mass Points Many students are confused by mathematics until they realize that the definitions, postulates, and theorems are the key to everything; at that point they start making real progress. Ultimately, success still depends on how clever one is in using them to arrive at conjectures and prove more theorems. But if one doesn’t understand these fundamentals completely, he/she will not go very far in mathematics. With that in mind, this session is designed after Hausner’s axiomatic approach to mass points in a 1962 paper [39]. 2.1. Objects of mass point geometry. When a new theory is being developed, the objects in it must be clearly defined, so there are no ambiguities later on. For example, in ordinary high school geometry, you defined what triangles are and explained what it means for two of them to be congruent. In the slightly more advanced setting of coordinate geometry, it becomes necessary to define even more basic objects such as “point” – a pair of numbers (x, y) called the coordinates of the point – and “line” – the set of points which are solutions to a linear equation Ax + By = C; one would also define what it means for two points P and Q to be the “same”: P = Q if and only if their corresponding coordinates are equal. In this vein, we define below the main objects of our new mass point theory. Definition 1. A mass point is a pair (n, P ), also written as nP , consisting of a positive real number n, the mass, and a point P in the plane or in space. The reader can think of a mass point (n, P ) as the “fat point” P of mass n. We will also use the convention from algebra that 1P = P , so that (1, P ) can be written as P when it is clear that a mass point is being used. Definition 2. We say that two mass points coincide, nP = mQ, if and only if n = m and P = Q, i.e., they correspond to the same ordinary point with the same assigned mass. Note that this is just the usual definition of equality for ordered pairs, (n, P ) = (m, Q), as in the setting of coordinate geometry mentioned above. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

2. DEFINITIONS AND PROPERTIES

131

2.2. Operations in mass point geometry. Our theory is interesting and powerful because it combines objects and ideas from both geometry and algebra. However, this makes it necessary to define operations on mass points from scratch. Just as you defined what it means for two numbers to be added in an algebra class, we would want to know what it means for two mass points to be added. Definition 3 (Addition). Given two mass points nE and mA with E = A, we set their sum to be nE + mA = (n + m)F where F lies on segment EA and EF : F A = m : n. If E = A, we set nE + mE = (n + m)E. In either case, the sum is called the center of mass of the two mass points nE and mA. This is the crucial idea: adding two mass points nE and mA results in a mass point (n + m)F so that (a) F is located at the balancing point of the masses on segment EA (which is why we use the phrase “center of mass”) and (b) the mass at this location F is the sum n + m of the two original masses. The physical model of the situation gives motivation for this definition. For part (b), think of a stick EA with weights at both ends. If you hold the stick at some point F to make it horizontal, the sum total of the two weights will push down at F ; not surprisingly, F is called the balancing point of the two weights. Part (a) requires more reflection to be justified. By the principle of the lever, n |EF | = m |F A|; hence, for example, an elephant E with a mass of n = 100 kg and an ant A with a mass of m = 1 g will balance if the fulcrum is placed at point F as in Figure 3. In other words, adding two mass points nE and mA corresponds to finding the point F on segment EA that divides it in a ratio of m : n. A classic example of mass point addition is obtained when both masses are 1: the physical center of mass is then located at the midpoint of EA with a mass of 2. The special definition of adding two mass points when they reside in the same location will allow us later to split a mass point into two or more parts; and, conversely, to add the same mass point nP several times to itself. Indeed, for the latter, check that kP + (mP + nP ) = (k + m + n)P = (kP + mP ) + nP , so that when k = m = n: nP + nP + nP = (n + n + n)P = (3n)P = (3n, P ), which can be written for short as 3nP . But now comes the inevitable question: √ how about if we want to add (n, P ) to itself “2/3 times,” or even worse: “ 2 times”? The obvious way to settle this is to introduce Definition 4 (Scalar Multiplication). Given a mass point (n, P ) and a real number m > 0, called a scalar, we define m(n, P ) = (mn, P ). Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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2.3. Basic properties in mass point geometry. In elementary school you probably discussed, at some point, properties of the operations on numbers, e.g., that numbers commute2 under addition: 7 + 4 = 4 + 7, and that it does not matter where one puts parentheses when adding three numbers, as the result will be the same: (7 + 4) + 9 = 7 + (4 + 9) = 20. The latter property is called, as many will remember, associativity of addition. In our setting, can we find similar properties for addition and scalar multiplication of mass points? Property 1 (Closure). Addition produces a unique sum. Property 1 says that when two mass points are added together, the sum is another unique mass point. The elementary geometric construction of a point on a segment that divides it in a given ratio provides both the existence and uniqueness of the mass point sum. See Stein’s book on Archimedes [90] for a “proof” of the physical property that the center of mass is unique. Property 2 (Commutativity). nP + mQ = mQ + nP . To justify Property 2 physically, just step through the plane and look at the seesaw from the “other side.” Algebraically, the justification comes from the commutativity of addition of real numbers. Property 3 (Associativity). nP + (mQ + kR) = (nP + mQ) + kR. Thus, we can drop the parentheses and denote the sum simply as nP + mQ + kR. The property of associativity roughly says that a system of masses n, m, and k placed at points P , Q, and R, respectively, is equivalent to placing a mass of n+m+k at M : the sum nP +mQ+kR = (n+m+k)M locates the center of mass of the system of mass points nP , mQ, and kR at M . Again, a classic example of this is obtained when the masses are all 1; as we will see later in Exercise 5, the center of mass in this situation is located at the usual centroid of P QR. Property 3 turns out to be equivalent to a famous theorem of Menelaus (cf. Theorem 7 and Problem 6) and is not trivial to prove from scratch (cf. Hausner [39] or [40]). Property 4 (Distributivity). k(nP + mQ) = knP + kmQ. Property 4, which is the distributive property of scalar multiplication over mass point addition, just means that multiplying all of the masses in a system by a number will multiply the total mass in the system by that number but will not change the center of mass. The truth of this property relies on the basic arithmetic fact that multiplying the numerator and denominator of a fraction by any positive number results in an equivalent fraction. The case where P = Q must be considered separately and depends on the distributive property of real numbers. Property 4 will allow for simplifying fractions in mass point computations, among other things. 2 For comparison, see the Rubik’s session on commutativity of some cube operations.

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2. DEFINITIONS AND PROPERTIES

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Finally, note that with very little extra effort, we can formally define mass points with mass m = 0 and then easily recover Definition 4, k(nP ) = knP , from Property 4. 2.4. More operations on mass points? After you mastered addition of numbers, your teacher probably introduced subtraction of numbers: a − b is the number c such that c + b = a. In other words, subtraction is the inverse operation of addition on numbers. In any mathematical system with a given operation, it is natural to ask if there is an inverse operation that makes sense in the system. Obviously, the operation of subtraction is not welldefined within the positive numbers (what is 7 − 9?), but it is well-defined in the set of all (real) numbers (7 − 9 = −2). Thus, the question to answer in the set-up of our mass point theory is the following: when does subtraction make sense, i.e., when is nP − mQ a legitimate mass point xX (with x > 0)? Property 5 (Subtraction). If n > m then nP = mQ + xX may be solved for the unique unknown mass point xX. Namely, xX = (n − m)R and either P = Q = R = X or P is on segment RQ so that RP : P Q = m : (n − m). The reader should check that the mass point (n − m)R can be substituted for xX to make the equation nP = mQ + xX work. We conclude that nP − mQ = (n − m)R. 3Q

5P

xR

Figure 5. Subtraction property and principle of the lever Exercise 1. Given mass points 3Q and 5P , find the location and mass of their difference 5P − 3Q (cf. Fig. 5). Solution: By definition of subtraction, 5P − 3Q = (5 − 3)R = 2R, where P is the balancing point of the mass points 3Q and 2R. This means 3|QP | = 2|P R|, so that R will be on the other side of P at a distance of 32 |QP |.
2.5. What was proved and what was “swept under the rug”? We defined the mass point (n, P ) using our old familiar point P in Euclidean geometry (i.e., in the usual plane or in space). We also explained how we can add or subtract such mass points and multiply them by a scalar. We further listed five properties, most of which can be justified with a little thought. Property 3 still remains “at large” as far as its proof is concerned. The beginner problem solver is advised on a first reading to accept these properties and skip to their applications in the next sections. The more advanced reader could attempt to prove them or follow up on the suggested references. In either case, Property 3 shall be rigorously proved at the end of Section 6. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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3. Fundamental Examples As promised, we now attack Problem 1 of this session. Solution to Problem 1: In order to have D as the balancing point of BC, we assign a mass of 2 to B and a mass of 5 to C. Now, to have E as the balancing point on side BA, we assign 2 · 3/4 = 3/2 to A. Then at the balancing points on the sides of the triangle, we have 2B + 5C = 7D and 2B + 32 A = 72 E (cf. Fig. 6a). 2B 7 2E

4 3 2A

3

1B 5 F

y

2N

7D

x 3G

y

2 5C

1A

z

2L x z

2M

1C

Figure 6. Two cevians problem solved and Medians proved concurrent The center of mass 8.5X of the system { 32 A, 2B, 5C} is located at the sum 32 A + 2B + 5C. The latter can be calculated in two ways according to our associativity property: 7 2E

+ 5C = ( 32 A + 2B) + 5C = 8.5X = 32 A + (2B + 5C) = 32 A + 7D.

Thus, by definition of addition, X is located, on the one hand, on segment EC, and, on the other hand, on segment AD, i.e., at their intersection point F . Hence F is the fulcrum of the seesaw balancing 32 A and 7D and of the seesaw balancing 5C and 72 E. This means that DF : F A = 3/2 : 7 = 3 : 14 and EF : F C = 5 : 7/2 = 10 : 7.


 

Note that all of the above calculations can be written down on the figure in a matter of seconds. An important part of this technique is to PST 47. Have a figure at hand. As in inversion, an accurate drawing may not be necessary: only the relative locations of points and lines are needed. It is true, however, that in other situations an accurate drawing may provide the clues for solving the problem. Another mass point “trick” worth remembering is to PST 48. Assign masses at the vertices of ABC in such a way that the intersection point F of the cevians becomes the center of mass of the resulting mass point system. This allows for calculations based on the seesaw principle and our five listed properties of mass points. Many of the following exercises in this section are based on those in an article by Sitomer and Conrad [85], no longer in print and not found in most libraries. This will make available to the reader some of the examples from their presentation which expanded my understanding of the above technique.

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3. FUNDAMENTAL EXAMPLES

135

Exercise 2 (Warm-up). If G is on BY , find x and BG : GY provided (a) 3B + 4Y = xG; (b) 7B + xY = 9G. Exercise 3. In ABC, D is the midpoint of BC and E is the trisection point of AC nearer to A (i.e., AE : EC = 1 : 2). Let G = BE ∩ AD. Find AG : GD and BG : GE. Solution: Draw the figure! Assign mass 2 to A and mass 1 to each of B and C. Then 2A + 1C = 3E and 1B + 1C = 2D. Similarly as before, the center of mass of the system is 2A + 1C + 1B = 3E + 1B = 2A + 2D = 4G, so that AG : GD = 2 : 2 = 1 : 1 and BG : GE = 3 : 1.
Exercise 4 (East Bay Mathletes ’99). In ABC, D is on AB and E is on BC. Let F = AE ∩ CD, AD = 3, DB = 2, BE = 3, and EC = 4. Find EF : F A in lowest terms. Exercise 5. Show that the medians of a triangle are concurrent in a point which divides each median in a ratio of 2 : 1 counted from the vertices. Hint: Assign a mass of 1 to each vertex (cf. Fig. 6b).



♦

We discuss below a follow-up to Exercise 5. The solution uses the area addition PST 46 along with another area PST that shows up whenever cevians are present. PST 49. If two triangles have the same altitudes but possibly different bases, it is useful to form the ratio of their areas – it equals the ratio of the bases. Of course, if the triangles also have equal bases, then their areas are equal. Exercise 6. Show that all six regions obtained by slicing a triangle via its three medians have the same area. Solution: In ABC, let AL, BM , and CN be the medians intersecting at G (cf. Fig. 6b). The notation [XY Z] will represent the area of XY Z. By PST 49, we have x = [CGL] = [BGL], y = [BGN ] = [AGN ], and z = [AGM ] = [CGM ], since each pair of triangles has equal bases and equal altitudes. By Exercise 5, the ratio AG : GL = 2 : 1, which implies (by PST 49) that [AGB] = 2[BGL] = 2[CGL] = [AGC], or 2y = 2x = 2z. Thus, y = x = z and all six areas are equal.
Exercise 7 (Varignon’s Theorem). Show that the four midpoints of the sides of any quadrilateral are the vertices of a parallelogram. Hint: Assign mass 1 to each vertex of the original quadrilateral and find the center of mass in two ways. Show that this center bisects the line segments joining midpoints of opposite sides. ♦ Exercise 8. In quadrilateral ABCD, let E, F , G, and H be the trisection points of AB, BC, CD, and DA nearer to A, C, C, and A, respectively. Show that EF GH is a parallelogram. Repeat the exercise when the points divide the sides in corresponding ratios of m : n. Hint: Use point K = EG ∩ F H, as in Exercise 7.

♦

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4. Angle Bisectors and Altitudes For this section, you should review the concepts of angle bisector, incenter, circumradius, circumcenter, altitudes, perpendicular bisectors and a bit of trigonometry, although the text will provide some information about them. 4.1. Using angle bisectors. To start with, you will need the following famous theorem which you may have seen in a high school geometry class:




Theorem 1 (Angle Bisector). An angle bisector in a triangle divides the opposite side in a ratio equal to the ratio of the other two sides. More −−→ precisely, in ABC, if ray BD bisects ∠ABC then CD : DA = CB : BA. B

E c

B

a O

A A

D

C

C

b D

Figure 7. Angle Bisector Theorem and Law of Sines Proof: Construct a line through point A, parallel to BD, and intersecting the extension of CB at a point E (cf. Fig. 7a). Notice that ABE is isosceles with BE = BA. (Why? Chase some equal angles.) Segment BD, which is parallel to AE, divides the sides of CAE proportionally ⇒ CD : DA = CB : BE = CB : BA.




Use the Angle Bisector Theorem to do the following exercise. Exercise 9 (Incenter). Let AB = c, BC = a, and CA = b in ABC. Assign a mass to each vertex equal to the length of the opposite side, resulting in mass points aA, bB, and cC. Show that the center of mass (a + b + c)I is located on each angle bisector, and conclude that the three angle bisectors are concurrent at this point I (called the incenter of ABC). Those who know the definition of sin θ may recall the well-known




Theorem 2 (Law of Sines). Let AB = c, BC = a, and CA = b in ABC, and let R be the circumradius of the triangle. Then b c a = = = 2R. sin A sin B sin C The suggested sketch of the proof does not involve mass points, but the Law of Sines is such a useful statement that it is worth trying to prove it here and now.

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4. ANGLE BISECTORS AND ALTITUDES



137

Sketch of Proof: Let O be the circumcenter of ABC (cf. Fig. 7b). Draw the circumcircle and its diameter BD. Notice that ∠BDA = ∠BCA (they intercept the same arc) and ∠BAD is a right angle (why?). Finally, use the definition of sin ∠BDA = BA/BD. If ∠BCA is obtuse, then it is supplementary to ∠BDA instead of equal to it, but still has the same sine. ♦ PST 50. When angles rather than sides are known, use the Law of Sines to assign masses to the vertices. Exercise 10. In ABC, the bisector of ∠B intersects AC at D. (a) Show that AD : DC = sin C : sin A, i.e., AD sin A = DC sin C. (b) Let sin A = 4/5 and sin C = 24/25. The bisector BD intersects median AM at point E. Find AE : EM and BE : ED. 4.2. Using altitudes. Let BD be an altitude of acute ABC. Note the obvious equality AD · (DC/BD) = DC · (AD/BD). So the appropriate masses to assign to A and C are DC/BD and AD/BD, respectively, in order to have D as the balancing point of AC. Those of you who know some trigonometry will recognize DC/BD = cot C and AD/BD = cot A. Therefore, assigning masses proportional to cot C and cot A to the points A and C, respectively, will balance the side AC at the foot D of the altitude. Exercise 11. Let ABC be a right triangle with AB = 17, BC = 15, and CA = 8. Let CD be the altitude to the hypotenuse, and let the angle bisector at B intersect AC at F and CD at E. Show that BE : EF = 15 : 2 and CE : ED = 17 : 15. Recalculate with AB = c, BC = a, and CA = b. Answer: In the general case, BE : EF = a : (c − a) and CE : ED = c : a. ♦ As shown in Exercise 9, the three angle bisectors meet in the incenter of a triangle. The lines containing the three altitudes also meet in a point, called the orthocenter of the triangle. In the following exercise and theorem you will prove this statement in two stages: first for acute triangles, where the mass point technique is quite suitable; and then in general for all triangles, where a different method is called for. Exercise 12. Prove that the altitudes of an acute triangle are concurrent.




Now try to prove the more general statement that Theorem 3 (Orthocenter). In any triangle, the lines containing the altitudes are concurrent. The following is a very clever proof, which uses an extra geometric construction and depends on the fact that the perpendicular bisectors of the three sides in a triangle are concurrent: they intersect in the circumcenter of the triangle (cf. Fig. 11c on p. 12 of the Inversion session).

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Sketch of Proof: Form a larger triangle by constructing a line through each vertex of the given triangle parallel to the opposite side (cf. Fig. 8a). Notice that the vertices of the original triangle form a parallelogram with each of the vertices of the larger triangle. Using the fact that opposite sides of a parallelogram are equal, argue that the vertices of the original triangle are the midpoints of the sides of the larger triangle. Now notice that the perpendicular bisectors of the large triangle, which are known to be concurrent, contain the altitudes of the original triangle. Thus, the circumcenter of one triangle is the orthocenter of the other. ♦ B F E A

D

C

Figure 8. Orthocenter Theorem and Problem 4



This solution does not use mass point geometry, but because the method is so elegant, I never waste an opportunity to share it. Its main idea is to recognize that the altitudes in one triangle are the perpendicular bisectors in another triangle. Thus, the proof of Theorem 3 illustrates an essential ingredient of mathematical thinking: PST 51. Try to look at the information in a problem from different viewpoints. Are the objects in the problem also something else for which you know certain facts? Whatever the conclusions are for this other something, they must also be true for the original objects, and this could completely solve your initial problem. 4.3. Angle bisectors and altitudes meet again. Let us combine now the knowledge we have acquired about angle bisectors, altitudes, and mass points, and even add a little extra on top of it.




Problem 4. The sides of ABC are AB = 13, BC = 15, and AC = 14. Let BD be an altitude of the triangle. The angle bisector of ∠C intersects side AB at F and altitude BD at E. Find CE : EF and BE : ED. Naturally, you will attempt to assign weights to vertices A, B, and C in order to have point E as the centroid of the resulting mass point system. As you use the properties of the angle bisector CF and the altitude BD to balance, respectively, side AB at F and side AC at D, you will soon discover that the solution rests on calculating a single ratio, AD : CD (cf. Fig. 8b). One way to proceed is to find the length of BD and then to apply the Pythagorean Theorem to BDC and to BDA to calculate AD and CD.

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5. AREAS, SPACE, AND SPLITTING MASSES

139

The altitude BD itself can be handled via the classic area formula: [ABC] = 1 2 AC · BD = 7BD. But how about [ABC]? The only given information about ABC is its sides. Is there a formula to patch together these sides into a nice, concise expression for [ABC]? Theorem 4 (Heron’s formula). In ABC with sides a, b, c and semiperimeter s = (a + b + c)/2, the area of ABC is given by [ABC] = s(s − a)(s − b)(s − c). The reader is welcome to try to prove Heron’s formula; however, the methods that will succeed are likely to be beyond our mass point treatment.

5. Areas, Space, and Splitting Masses This section will extend the fundamental idea of mass points in several new directions. First, mass points will join forces with areas; then they will fly into space; and finally they will split into pieces providing the key to solutions of even more challenging problems. 5.1. Combining mass points and areas. As you attempt to solve the following problem, keep in mind that it requires a combination of mass points and area addition. Have fun! Problem 5. In ABC, D, E, and F are the trisection points of AB, BC, and CA nearer to A, B, and C, respectively (cf. Fig. 9a).




(a) If BF ∩ AE = J, show that BJ : JF = 3 : 4 and AJ : JE = 6 : 1. (b) Let CD ∩ AE = K and CD ∩ BF = L. Extend part (a) to show that DK : KL : LC = EJ : JK : KA = F L : LJ : JB = 1 : 3 : 3. (c) Use parts (a) and (b) to show that the area of JKL is 17 the area of ABC. (d) Generalize this problem for points which divide the sides in a ratio of 1 : n (in place of 1 : 2). In particular, show that the ratio of the area of JKL to the area of ABC is (n − 1)3 : (n3 − 1). B

B 1

2 D 1 A

E

J K 2

1

l D

2 L

J

m

K

1 F

1

C

A

E L

n

F

1

C

Figure 9. Area addition and Routh’s Theorem My interest in part (c) started with a problem in Jacobs’ Geometry (cf. [47], p. 534), in which the areas of the three shaded quadrilaterals in Figure 9a were 5 acres and the areas of the three triangles in the corners Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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7. MASS POINT GEOMETRY

were 1 acre. When I found out that it was not necessary to be given any of these areas to find the ratio of the area of the central JKL to that of the whole triangle, my interest in the problem grew. See Mathematics as Problem Solving by Soifer [86] for solutions to parts (c) and (d) using area addition and PST 49. Soifer used the result of part (c) in 1971 to create the following problem: Partition an arbitrary triangle by six straight cuts into parts from which it is possible to put together seven congruent triangles. His solution is a beautiful figure with only one word: “See!” Part (d) can be generalized even further using different ratios on each side. This generalization is known as Routh’s Theorem (cf. Fig. 9b). Theorem 5 (Routh). If the sides AB, BC, and CA of ABC are divided at D, E, and F in the respective ratios of 1 : l, 1 : m, and 1 : n, then the cevians CD, AE, and BF form a triangle whose area is (lmn − 1)2 (lm + l + 1)(mn + m + 1)(nl + n + 1) times the area of ABC. For example, the reader should check that when the ratios are all equal, l = m = n, Routh’s formula yields the answer in Problem 5(d). The proof of Theorem 5 is beyond the scope of the present session, but the reader is certainly encouraged to crack it with other methods. See Coxeter [18], Niven [67], and Klamkin-Liu [52] for various approaches. The proof by Coxeter uses a generalization of mass points called areal coordinates, or normalized barycentric coordinates. It is only four lines long. 5.2. Mass points in space. Going in another direction, an extension of the mass point technique can be used to solve problems in space, as illustrated by Problem 2 from the Introduction of this session and Exercise 13 below. To this end, we shall assume the same definitions and properties of addition, subtraction, and scalar multiplication of mass points in space as those in the plane. Solution to Problem 2: At this late point in the session, the reader should not be surprised that we assign a mass of 1 to each vertex of the tetrahedron ABCD (cf. Fig. 1b) and denote by J the center of mass of the resulting system, i.e., A + B + C + D = 4J. Since the centroid H of face ABC satisfies A+B +C = 3H (why?), then 3H +D = 4J. Thus, J lies on segment DH and divides it in a ratio of 3 : 1, and similarly for the other faces of ABCD. We conclude that AE, BF , CG, and DH are concurrent at J and divided in a ratio of 3 : 1 (counted from the vertices of the tetrahedron).
As the Centroid Theorem generalizes to a spatial version in the tetrahedron Problem 2, Varignon’s Theorem (Exercise 7) also has a 3-dimensional variation. Note that in a tetrahedron, opposite edges are pairs of edges with no vertex in common. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

5. AREAS, SPACE, AND SPLITTING MASSES

141

Exercise 13. Show that the three segments joining the midpoints of opposite edges of a tetrahedron bisect each other (cf. Fig. 10a). 4B +

18 35 B

3 7E

F

5 9 5D

4 3A

2

3

30 7 G

7

9 7C

Figure 10. Exercise 13 and Problem 3 5.3. Splitting masses. Now let’s take a look at Problem 3. This situation is not as intuitive as the two-cevian Problem 1. But once we have mastered the solution, we can then apply the mass-splitting technique used to answer a whole new class of questions. So let’s do it! Solution to Problem 3: Splitting the mass points as in (m + n)P = mP + nP is the technique to use when dealing with transversals. The actual assignment of masses works as follows. As a first approximation, assign 4 to B and 3 to A to balance AB at E. Then to balance AC at G, assign 97 to 18 C. To balance BC at point D, 18 35 B is needed. So we now have (4 + 35 )B. 44 This gives 5 F as the center of mass for the masses at A, B, and C. Indeed, using associativity and commutativity of addition, 30 7 G

+ (4 +

18 35 )B

= (3A + 97 C) + (4B +

= (3A + 4B) + ( 18 35 B

18 35 B) + 97 C)

= 7E + 95 D

shows that the center of mass lies on both ED and BG, i.e., it is located at point F . The sought-after ratios can now be read directly from the diagram: EF : F D = 9/5 : 7 = 9 : 35 and BF : F G = 30/7 : 158/35 = 75 : 79.
Here is another example. Try solving it by yourself. Use your paper to cover the solution. If you get stuck, lower your paper one line to reveal the first line of the solution. See if you can find the solution now. After spending five minutes, if you cannot proceed, then reveal one more line of the solution and try once more to complete the solution. Continue this way until you have a solution. If you arrive at a solution before revealing all of the lines, be sure to compare it with the one given here. Exercise 14. In ABC, let E be on AB such that AE : EB = 1 : 3, D on BC such that BD : DC = 2 : 5, and F on ED such that EF : F D = 3 : 4. −−→ Finally, let BF intersect AC at G. Find AG : GC and BF : F G. Solution: Draw the figure! Assign mass 2 to C and mass 5 to B so that 2C + 5B = 7D. It is now necessary to have mass 28/3 at E to balance ED at F . To balance AB at E, we calculate (3t)A + (1t)B = 28E/3, i.e., Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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7. MASS POINT GEOMETRY

t = 7/3 and 21A/3 + 7B/3 = 28E/3. So assign another mass of 7/3 to B for a total mass of 22/3 at B, and assign a mass of 21/3 = 7 to A. Then on side AC we have 7A + 2B = 9G. As in Problem 3, the center of mass of ABC is now F , and the desired ratios can be read directly from the figure: AG : GC = 2 : 7 and BF : F G = 9 : 22/3 = 27 : 22.
Practice the mass-splitting technique in a similar exercise. Exercise 15. With the same configuration as in Exercise 14, AE : EB = 3 : 1, BD : DC = 4 : 1, and EF : F D = 5 : 1. Show that AG : GC = 4 : 1 and BF : F G = 17 : 7.

6. Ceva, Menelaus, and Associativity of Addition In this section we discuss two famous theorems of Ceva and Menelaus. In particular, two completely different and independent proofs of Menelaus’s theorem will lead the way to finally verifying associativity Property 3, whose justification has remained “at large” until now. 6.1. Associativity implies Ceva’s and Menelaus’s Theorems.




Theorem 6 (Ceva). Three cevians AD, BE, and CF in ABC are concurrent if and only if the product of the three ratios (going around ABC in one direction) into which the cevians divide the opposite sides is 1, i.e., AF BD CE F B · DC · EA = 1, or equivalently as seen in Figure 11a, abc = xyz. B x a A

x

b

F

G z

E

D c

B b

F

D y

a

y C

A

z−c

C

c

E

Figure 11. Ceva’s and Menelaus’s Theorems Sketch of Proof: Set the lengths of the six segments cut out by the cevians along the sides of ABC as shown in Figure 11a; for example, AF = a, etc. Let two of the cevians BE and CF meet in G. Assigning masses x to A, a to B, and xz/c to C balances sides AB and AC at F and E, respectively, thus making G the center of mass. By associativity Property 3, G = xA + (aB + xz c C); therefore, the third cevian AD will also pass through G iff BC is balanced at D, i.e., iff ab = xz c y, or abc = xyz, which can be easily rewritten as the desired product of three ratios. ♦ The situation when three lines (cevians) intersect in a point is in some sense dual to the situation when three points lie on a line. Thus, replacing the concurrency condition in Ceva’s Theorem by a collinearity condition leads us to the next challenge. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

6. CEVA, MENELAUS, AND ASSOCIATIVITY




143

Theorem 7 (Menelaus). Let points F , D, and E lie, respectively, on sides AB and BC and the extension of side CA of ABC. Then the points F , D, and E are collinear if and only if the product of the three ratios (going around ABC in one direction) into which the three points divide the sides CE is 1, i.e., FAFB · BD DC · EA = 1, or equivalently, as seen in Figure 11b, abc = xyz. It may seem that the products of ratios for both Ceva’s and Menelaus’s Theorems are identical! Of course, this is not so: while the ratio CE EA in Ceva’s Theorem refers to point E dividing side AC internally, Menelaus’s Theorem requires that E be external to side AC. For the following proofs, set the lengths of the six segments determined by F , D, and E along the sides of ABC as shown in Figure 11b, e.g., AF = a, . . . , EA = z. Proof of Menelaus’s Theorem (⇐): Assume first that abc = xyz. Assigning masses x to A, a to B, and ay/b to C balances AB at F and BC at D. Thus, in order to balance AE at C, we have no choice but to assign mass (ab/y − x) to E ; indeed, the desired balancing is equivalent to ?

( ab y − x)c = (z − c)x ⇔



abc y

?

?

− xc = zx − cx ⇔ abc = xyz,

which has been assumed. Therefore, ABE is balanced at the intersection D of segments EF and BC, which divides BC in a ratio of ab y : a = b : y.  But point D divides BC in that same ratio, so D = D and D lies on EF .
To prove the forward direction, we will utilize a slick, new PST 52. Suppose you have already shown S1 ⇐ S2 . To prove the opposite direction S1 ⇒ S2 , you could introduce new objects to force the situation of statement S2 , conclude then S1 , and finally observe that the “new” and “old” objects in the problem coincide. We already saw traces of this PST in the above proof, where D and D  coincided; however, now we will apply PST 52 in its full force.




Proof of Menelaus’s Theorem (⇒): Assume that D, E, and F
CE  are collinear. Pick a point D  on side BC so that FAFB · BD D
C · EA = 1, i.e., D
F B EA xz divides BC in the ratio BD D
C = AF · CE = ac . By the backwards direction of Menelaus’s Theorem (just proved above), we conclude that F , D  , and E are collinear. This means that D  lies on the intersection of EF and BC, and as such, D  coincides with D. Therefore, substituting D for D  above CE
implies the desired ratio FAFB · BD DC · EA = 1. Another version of Menelaus’s Theorem allows the transversal EDF to intersect all three sides of ABC externally: prove this via mass points too. 6.2. Turning the tables around. Even though associativity Property 3 for mass points may seem very natural and, therefore, “automatic,” in reality this is not so. We shall see below how the forward direction of Menelaus’s Theorem is used to prove this property.

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7. MASS POINT GEOMETRY

Theorem 8 (Associativity Property). If aA, bB, and c C are mass points, then aA + (bB + c C) = (aA + bB) + c C. D

bB (a + b)D

aA

F

F

(b + c)E

cC

G A

F

B G

E E

C

Figure 12. Menelaus’s Theorem implies Associativity Property Proof: Let (a + b)D balance AB and (b + c)E balance BC (cf. Fig. 12a). CF DA Menelaus’s Theorem for BCD with transversal EF A gives 1 = BE EC · F D · AB c CF b CF a+b = b · F D · a+b ⇒ F D = c . Thus, CD is balanced at F (why?), and (aA + bB) + c C = (a + b)D + c C = (a + b + c)F. Analogously, Menelaus’s Theorem for BAE with transversal DF C gives aA+(bB +c C) = aA+(b+c)E = (a+b+c)F . Putting everything together, ? we conclude that aA + (bB + c C) = (aA + bB) + c C. 6.3. Are we there yet? “The beginning student often scoffs at degenerate cases in geometry,” writes Hausner in [39]. Yet, the last detail not covered by Menelaus’s Theorem is precisely the degenerate case where the three points are collinear (instead of forming a triangle). In terms of mass points: Lemma 1. If aA, bB, and c C are mass points with A, B, and C collinear, then (aA + bB) + c C = aA + (bB + c C). Proof: We will project the non-degenerate case in Theorem 8 onto a line to demonstrate the degenerate case of Lemma 1. First select a point D not on line ABC (cf. Fig. 12b) and assign mass b to it. Now ADC is nondegenerate; so (aA + bD) + c C = aA + (bD + c C). Let (a + b)F and (c + b)E balance, respectively, sides AD and DC, and let CF and AE intersect in G. Consider the parallel projection of points onto line ABC that takes D to B: points F , G, and E are projected to F  , G , and E  . Since parallel lines cut proportional segments on any transversal, the projection of (a + b)F = aA + bD is (a + b)F  = aA + bB; in other words, F  balances AB. Similarly, CF  , AE  , and BC are balanced at G , G , and E  , respectively. Therefore, (aA + bB) + c C = (a + b)F  + c C = (a + b + c)G = aA + (b + c)E  = aA + (bB + c C).




6.4. Independence is always desirable. After solving a problem one way, how many times have you searched for another way? But why would anyone need two different solutions to the same problem? Once a problem is solved, well, it is solved forever! Yet, finding two or more sufficiently different solutions to a problem may even serendipitously prove new statements! Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

7. EXAMPLES OF CONTEST PROBLEMS




145

Indeed, combining everything in this section so far, we can say that the associativity property of mass points and Menelaus’s Theorem imply one another; in other words, they are equivalent. But this doesn’t mean we have shown either of them! An independent proof of one statement is necessary to establish the other statement. Problem 6. Prove the forward direction of Menelaus’s Theorem without mass points – use classic geometry methods, e.g., similar triangles.3 Sketch of Proof: Let ABC and transversal F DE be as in Figure 11b. Construct a line through C parallel to AB intersecting DE at G. Let CG = w. Since CGD ∼ BF D and CGE ∼ AF E, deduce that w/x = y/b and w/a = c/z. Thus, xy/b = w = ac/z and abc = xyz. ♦ Since associativity Property 3 was proved using part of Menelaus’s Theorem, it was necessary to prove that part of Menelaus’s Theorem with methods other than mass points. With the completion of Problem 6, we can claim to have actually proved Property 3 without getting trapped in circular reasoning. We are finally there!

7. Examples of Contest Problems 7.1. Math contests versus research mathematics. You may still be thinking that the type of problem yielding to a mass point solution is rare, and that it is more like a parlor trick than an important mathematical technique. In the collection of problems below you will encounter problems that can often be solved by mass point geometry more readily than with the official contest solution. They come from a wide variety of contests and were often found difficult by the contestants. Part of the fun of such contests is knowing that a solution that could take you between 5 and 15 minutes exists and trying to find it. Most real mathematics, on the other hand, cannot be done racing the clock. Mathematicians take days, months, and sometimes years to solve their problems, which require long periods of profound, concentrated thought. Time must also be spent relaxing and giving your mind an opportunity to work unconsciously on the problem. It is surprising how many times the solution to a problem will ‘pop’ into one’s head from nowhere upon awakening in the morning or when doing something completely unrelated. 7.2. The contests surveyed for this collection are city-wide, regional, and national: New York City Mathematics League (NYCML), American Regional Mathematics League (ARML), American High School Mathematics Examination (AHSME, now the American Mathematics Competition 12 3 For a more complete discussion of Ceva’s and Menelaus’s Theorems see the upcoming session Classical Theorems in Geometry in Volume II.

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7. MASS POINT GEOMETRY

(AMC 12)), and American Invitational Mathematics Examination (AIME). For collections of problems from these contests see [5, 10, 78, 81, 102]. If you time your solutions, then use 5–10 minutes per problem. The selected problems below are grouped in subsections according to the suggested techniques, which will help you assign appropriate masses to force the center of mass at a convenient location. As before, try them first without looking at the comments, hints, or solutions. (And even if you do peek, you are expected to justify any statement made there!) An important part of the solving process is drawing the figure from the given data, as this helps you internalize the problem in a way that just reading the words does not accomplish. American geometry (and calculus) books fail miserably on this account by providing far too many figures. 7.3. Using the fundamental mass point technique. Exercise 16 (AHSME ’65). Points E and D are selected on sides AB and BC of ABC in such a way that AE : EB = 1 : 3 and CD : DB = 1 : 2. AF The point of intersection of AD and CE is F . Find EF FC + FD. Exercise 17 (NYCML ’75). In ABC, C  is on side AB such that AC  : C  B = 1 : 2, and B  is on side AC such that AB  : B  C = 3 : 4. If BB  and CC  intersect at P and ray AP and BC intersect at A , find AP : P A . Exercise 18 (NYCML ’75). In ABC, D is on AB and E is on BC. Let CD and AE intersect at K, and let ray BK and AC intersect at F . If AK : KE = 3 : 2 and BK : KF = 4 : 1, then find CK : KD. 7.4. Using the angle bisector theorem and the transversal method. Exercise 19 (ARML ’89). In ABC, angle bisectors AD and BE intersect at P . If the sides of the triangle are a = 3, b = 5, and c = 7, with BP = x and P E = y, compute the ratio x : y.





Problem 7 (AHSME ’75). In ABC, M is the midpoint of side BC, AB = 12 and AC = 16. Points E and F are taken on AC and AB, respectively, and lines EF and AM intersect at G. If AE = 2AF , find EG : GF . Problem 8 (ARML ’92). In ABC, points D and E are on AB and AC, respectively. The angle bisector of ∠A intersects DE at F and BC at T . If AD = 1, DB = 3, AE = 2, and EC = 4, compute the ratio AF : AT . 7.5. Using ratios of areas via PST 49.




Problem 9 (AHSME ’80). In ABC, ∠CBA = 72◦ , E is the midpoint of side AC, D is a point on side BC such that 2BD = DC, and AD and BE intersect at F . Find the ratio of the area of BDF to the area of quadrilateral F DCE.

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7. EXAMPLES OF CONTEST PROBLEMS 1A

F

1

1 3D

A

2E F

1 2B

147

1 2

1C

B

1

x

1

E 3x

2x D

2

C

Figure 13. Solving Problem 9 using area addition Solution: The angle measure is a red herring and is not needed to solve the problem. Assigning appropriate masses to the vertices gives 1A + 1C = 2E and 2B + 1C = 3D so that BF = F E (why?; cf. Fig. 13a). Since there is no general formula for the area of a quadrilateral in terms of its sides, it is often useful to draw one of its diagonals, thereby dividing it into two triangles; the diagonal CF of F DCE will do the job (cf. Fig. 13b). Let [BDF ] = x. Then by PST 49, [CDF ] = 2x and [CEF ] = [CF B] = [BDF ]+ [CDF ] = 3x. Therefore, [BDF ] : [F DCE] = x : (2x + 3x) = 1 : 5.







Problem 10 (AIME ’85). In ABC, cevians AD, BE, and CF intersect at point P . The areas of P AF, P F B, P BD, and P CE are 40, 30, 35, and 84, respectively. Find the area of ABC. Solution: By PST 49, AF : F B = 4 : 3 and AP : P D = 2 : 1. Using mass points 3A, 4B, and 6D, we balance AB at 7F and AD at 9P . To make P the center of mass of ABC, we assign some weight x to C so that x : 7 = P F : P C and (3A + 4B) + xC = 7F + xC = (7 + x)P . By the associativity property, 3A + (4B + xC) = (7 + x)P . Since P lies on cevian AD, we must have 4B + xC = yD for some mass y = 4 + x, i.e., (7 + x)P = 3A + (4 + x)D. But we already know that AP : P D = 2 : 1, so that 3 · 2 = (4 + x) · 1, from which x = 2. This is no surprise, as 2C + 4B = 6D, so that BD : DC = 1 : 2. Again by PST 49, we have [ABC] = 3[BAD] = 3(40 + 30 + 35) = 315.
Note that the information [P EC] = 84 was superfluous! We did not use it, although we can certainly derive it from our calculations: as 3A+2C = 5E and 5E + 4B = 9P , we have EP : P B = 4 : 5; thus, [P CE] = 45 [P BC] = 4 12 5 · 3[P BD] = 5 · 35 = 84. Another, even more important, observation is that a problem-solving principle was demonstrated by the above solution: PST 53. Once three masses are correctly assigned to some three non-collinear points among the vertices of a triangle and the feet of the cevians, then a unique choice of masses for the remaining three points forces the intersection point of the cevians to be the center of mass of the resulting system. In Problem 10, we correctly assigned masses to vertices A and B and to cevian foot D, implying a unique choice of masses for vertex C and cevian feet F and E that forces P to be the center of the resulting system.

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7.6. Changing your viewpoint.




Problem 11 (NYCML ’76). In ABC, D is on AB such that AD : DB = 3 : 2, and E is on BC such that BE : EC = 3 : 2. If ray DE and ray AC intersect at F , then find DE : EF .




Problem 12 (NYCML ’77). In a triangle, segments are drawn from one vertex to the two trisection points of the opposite side. A median drawn from a second vertex is divided by these segments in the continued ratio x : y : z, counted from the vertex. If x ≥ y ≥ z then find x : y : z.

Hint: Do not focus on ABC. Draw BF to form ABF with cevians BC and F D. Assign a mass of 6 to B and deduce that DE : EF = 1 : 2. Alternatively, two applications of Menelaus’s Theorem will do the job. ♦

7.7. Using special triangles and topics from geometry. Two special right triangles are naturally studied in a first course in geometry by dividing in half the two most basic symmetric shapes: the square and the equilateral triangle. These two triangles are the isosceles right triangle and the 30◦ – 60◦ –90◦ triangle. Using the Pythagorean Theorem, the reader can derive the relative lengths of the sides of each triangle, as shown in Figure 14. 1 45◦ 45◦

√ 1

30◦ 30◦

2

45◦ 45◦




1 60◦

1




1

√ 3 2

1 2

1

1 2

60◦

Figure 14. Isosceles right triangles and 30◦ –60◦ –90◦ triangles Problem 13 (NYCML ’78). In ABC, ∠A = 45◦ and ∠C = 30◦ . If altitude BH intersects median AM at P , then AP : P M = 1 : k. Find k. Hint: First draw the figure! Use the ratios of sides of 30◦–60◦–90◦ triangles √ 3. A and isosceles right triangles (cf. Fig. 14) to find AH : HC = 1 : √ 3A, 1C, 1B, and 2M , which leads direct application of mass points uses √ √ ♦ to AP : P M = 2 : 3 = 1 : k and k = 3/2. What is BP : P H? In addition to Heron’s formula (cf. Theorem 4), the following formula for the area of a triangle is often missing from first geometry courses: Theorem 9. In ABC with sides a, b, and c, inradius r, and semiperimeter s = (a + b + c)/2, the area of ABC is given by rs (cf. Fig. 15a). Proof: Recall that the incenter I is the point where the angle bisectors meet (cf. Exercise 9), so that I is equidistant from the sides of the triangle. By area addition, [ABC] = [BIC] + [CIB] + [AIB] = ra/2 + rb/2 + rc/2 = r(a + b + c)/2 = rs (cf. Fig. 15a).


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7. EXAMPLES OF CONTEST PROBLEMS 3B B

2

c a

r I r

9D q F

r C

149

p

1 A

b

6C 1 E

3

2A

Figure 15. [ABC] = rs and Problem 14




Problem 14 (NYCML ’77). A circle is inscribed in a 3–4–5 triangle. A segment is drawn from the smaller acute angle to the point of tangency on the opposite side. This segment is divided in the ratio p : q by the segment drawn from the larger acute angle to the point of tangency on its opposite side. If p > q then find p : q. Hint: In this problem Theorem 9 becomes 6r = 6 ⇒ r = 1. Therefore the legs are divided in ratios of 1 : 3 and 1 : 2 by the points of tangency. Using a mass of 6 at the right angle leads directly to p : q = 9 : 2 (cf. Fig. 15b). ♦




Problem 15 (AHSME ’64). The sides of a triangle are of lengths 13, 14, and 15. The altitudes of the triangle meet at point H. If AD is the altitude to the side of length 14, what is the ratio HD : HA? Hint: This triangle is a favorite for contest problems since it has some nice “integral” properties: its area is 84, while the altitude to the side of length 14 is 12 and can readily be seen to divide the original triangle into two triangles with integral sides. ♦ 7.8. Some especially challenging problems. Although the next problem is one of the more difficult problems on the 1992 AIME, it is quite easy to solve using mass point geometry.




Problem 16 (AIME ’92). In ABC, A , B  , and C  are on sides BC, AC, and AB, respectively. If AA , BB  , and CC  are concurrent at the point AO BO CO AO BO CO O and if OA
+ OB
+ OC
= 92, find the value of OA
· OB
· OC
.




Theorem 10. In ABC, if cevians AD, BE, and CF meet at P , then PF PD PE + + = 1. AD BE CF Proof: Draw the figure. Let BD : DA = p : q and BD : DC = r : s. Form mass points psA, qsB, and qrC. Then balancing along the sides at the feet of the cevians gives mass points (qs + qr)D, (ps + qr)E, and (ps + qs)F . Reading the ratios from the triangle nails down the problem:

The next problem is based on an interesting property about concurrent cevians in a triangle, which can also be readily proved using mass points.

PD AD

+

PE BE

+

PF CF

=

ps ps+qs+qr

+

qs ps+qs+qr

+

qr ps+qs+qr

= 1.




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150

7. MASS POINT GEOMETRY




Problem 17 (AIME ’88). Let P be an interior point of ABC, and extend lines from the vertices through P to the opposite sides. Let AP = a, BP = b, CP = c, and let the extensions from P to the opposite sides all have length d. If a + b + c = 43 and d = 3, then find abc.




Problem 18 (AIME ’89). Point P is inside ABC. Line segments AP D, BP E, and CP F are drawn with D on BC, E on CA, and F on AB. If AP = 6, BP = 9, P D = 6, P E = 3, and CF = 20, find the area of ABC. Sketch of Solution: Start with 6A and 3B, yielding 6D, 9F , and 9E. Subtraction gives 3C, which implies that CP : P F = 3 : 1 with CP = 15 and that D is the midpoint of BC. However, to solve the problem, a stroke of genius is still necessary. Draw a segment from D to the midpoint, M , of BP . This is a midsegment of BCP and is therefore parallel to CP and half as long. Now notice that the sides of DP M are 9/2, 6, and 15/2, which are each half of 9, 12, and 15. Hence, BCP is a right triangle with area 27/2 ! By repeated application of PST 49, we conclude that [BP D] = [BP A] = [AP C] = [CP D] = 27; so we find that [ABC] = 4[BP D] = 108. ♦




Problem 19 (Larson). In ABC, let D and E be the trisection points of BC with D between B and E. Let F be the midpoint of AC, and let G be the midpoint of AB. Let H be the intersection of EG and DF . Find the ratio EH : HG. While comparing the method of applying vectors to the mass point geometry method, I came across the above somewhat difficult and quite interesting problem in Problem Solving Through Problems by Larson [56]. The solution in the book contains a full page of vector equations, which finally simplify to an elementary system of equations. 2B D 4G K 2A

3D H

4G

F

K 3C

6E 3C

Figure 16. Problem 19 from Larson, solved with mass points The mass point geometry solution is straightforward. Sketch of Solution: Draw the figure without EG, and draw GC intersecting DF at K (cf. Fig. 16a). Find CK : KG = 4 : 3 using the transversal method by assigning a mass of 2 to A and a mass of 2 to B with a split mass of (2 + 1) to C. Now use the two-cevian method on CDG (cf. Fig. 16b). Assign a mass of 3 to C, a mass of 4 to G, and a mass of 3 to D. Then 3D + 3C = 6E and EH : HG = 4 : 6 = 2 : 3. ♦ Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

9. HINTS AND SOLUTIONS TO SELECTED PROBLEMS

151

8. History and Sources Mass points were first introduced by Augustus Ferdinand Möbius in 1827. They didn’t catch on right away. Cauchy was quite critical of his methods, and even Gauss in 1843 confessed that he found the new ideas of Möbius difficult (cf. a note by Dan Pedoe in Mathematics Magazine [72]). I first encountered the idea about 30 years ago in a math workshop “Teeter-totter Geometry” by Brother Raphael from Saint Mary’s College of California in Moraga. He taught one of his courses each year using only original sources, and that year he was reading Archimedes with his students. It was Archimedes’ “principle of the lever” that he used in the workshop to show how mass points could be used to make deductions about triangles. For a very readable account of the assumptions Archimedes makes about balancing masses and locating the center of gravity, I recommend the new book Archimedes: What Did He Do Besides Cry Eureka? [90] by Sherman Stein of U.C. Davis. About 25 years ago Bill Medigovich, who was then teaching at Redwood High School in Marin County, California, sent me a 30-sheet packet [63] that he used for a presentation to high school students. I also found the topic discussed in the appendix of The New York City Contest Problem Book 1975-1984 [81], with a further reference to an article The Center of Mass and Affine Geometry [39] by Melvin Hausner in 1962. Recently, Dover Publications reissued a book by Hausner [40] published in 1965, which extends mass point geometry to vector spaces as a vehicle to study geometry. As I was preparing for the BMC session in 2002, I found a key paper by Harry Sitomer and Steven R. Conrad on this subject: Mass Points [85], originally written for the NYC Senior ‘A’ Mathletes and providing what I considered to be the most attractive way to present these ideas. I discovered the article in an old, no longer available issue of Eureka. The latter is a wonderful journal for problem solvers from the Canadian Mathematical Society, presently published as Crux Mathematicorum with Mathematical Mayhem. A few other articles used in preparing for the BMC talk are BoydRaychowdhury [12], Honsberger [44], and Pedoe [71].

9. Hints and Solutions to Selected Problems Exercise 2. Answers: (a) x = 7 and BG : GY = 4 : 3; (b) x = 2 and BG : GY = 2 : 7. ♦ Exercise 4. Assign 2 to A, 3 to B, and 9/4 to C. Then 3B + 94 C = Reading from the figure, EF : F A = 8 : 21.

21 4 E.

♦

Exercise 8. Assign m to a pair of opposite vertices and n to the other pair so that each side balances at its division point with the same mass of m + n. Then EF GH is also a parallelogram since its diagonals bisect each other. ♦ Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

152

7. MASS POINT GEOMETRY

Exercise 9. The unique sum of the mass points is aA+bB+cC = (a+b+c)I. But I is on each of the cevians (angle bisectors). ♦ Exercise 10. (a) By the Law of Sines and the Angle Bisector Theorem sin C AB AD ♦ sin A = BC = DC .

24 (b) Assign 45 to A, 24 25 to C, and 25 to B. With 12 : 5 and with 44 25 at D, BE : ED = 11 : 6.

48 25

at M , AE : EM = ♦

Exercise 12. Assign to A, B, and C masses of cot B cot C, cot A cot C, and cot A cot B, respectively. Then each altitude is seen to balance at the same mass point, (cot B cot C + cot A cot C + cot A cot B)H. ♦ Problem 4. Assign 15 to A and 14 to B by the Angle Bisector Theorem. Using the hint in the text gives BD = 12 and AD : DC = 5 : 9. Now assign 15 · 5/9 = 25 3 to C. A combined mass of 29 at F shows CE : EF = 87 : 25. ♦ A combined mass of 70 3 at D results in BE : ED = 5 : 3. Problem 5. (a) Assign 1, 4, and 2 to A, B, and C, respectively, so that 4B + 2C = 6E and 1A + 2C = 3F . ♦ (b) By symmetry AK : KE = BJ : JF = 3 : 4, and from part (a), AJ : JE = 6 : 1; so K must be the midpoint of AJ and AK : KJ : JE = 3 : 3 : 1. ♦ (c) Draw AL, BK, and CJ. Using the ratios from parts (a) and (b) and PST 49, show that ABC is composed of 3 small ’s in each corner with area x, 3 ’s with area 2x, and 4 ’s with area 3x, for a total area of 21x. Then 3x/21x = 1/7. ♦ Exercise 13. Assign mass 1 to each vertex. The sum of the four masses can be calculated by balancing any of the segments joining the midpoints of opposite edges with a mass of 2 at each endpoint. Thus the unique sum of the four vertex masses is the mass point 4G located at the midpoint of each segment joining opposite midpoints. ♦ Exercise 15. Assigning masses so that 4C + 1B = 5D forces E to have mass 1. The latter must be split in a ratio of 3 : 1; so assign 14 to A and ♦ another 34 to B. Exercise 16. Draw the figure, assign the masses, and read the answer from ♦ the figure: 12 + 11 = 32 . Exercise 17. Assign a mass of 4 to A, and use the given ratios to find the remaining masses that make P the balancing point. The answer is 5 : 4. ♦ Exercise 18. Use the ratio on the cevians to assign masses to balance at K. This will yield mass points 3D and 2C and will eventually result in CK : KD = 3 : 2. ♦ Exercise 19. This is a standard angle bisector problem. Assign masses to each vertex equal to the lengths of the opposite sides. Then x : y = 2 : 1. ♦ Problem 7. While the official solutions include several auxiliary lines and two pages of computations, the reader is advised to use the transversal Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

9. HINTS AND SOLUTIONS TO SELECTED PROBLEMS

153

method from Problem 3. Let AF = x and AE = 2x; then F E = 12 − x and EC = 16 − 2x. Now assign x to B and C. The split mass at A will be (12 − x) + (8 − x), with mass points of 8E and 12F along the sides, giving 3 ♦ 2 for the desired ratio. Problem 8. This is an angle-bisector problem in combination with a transversal problem. Assign split masses of 9 and 4 at vertex A. This gives 3B + 2C = 5T ; so AF : F T = 5 : 13 and AF : AT = 5 : 18. ♦ Problem 12. Let ABC have trisectors AS and AT and median BM . Use one trisector at a time, ignoring the other. First show x + y : z = 4 in ABC with cevian AT where AT : T B = 2 : 1. Also deduce that the median divides AT in a ratio of 3 : 2. Now use ABT with median AS to show that x : y = 5 : 3. Combine the findings as x : y : z = 5 : 3 : 2. ♦ Problem 15. The altitude divides the side of length 14 into two parts of length 5 and 9. So place masses of 9 and 5 at the endpoints with a sum of 14 at the point D. Now let the altitude to the side of length 15 divide it into a ratio of a : (15 − a). The Pythagorean Theorem gives 132 − (15 − a)2 = 142 −a2 ⇒ a = 42/5 and 15−a = 33/5. In order to balance the side of length 15, the mass assigned to the third vertex must be 5(42/5)(5/33) = 14(5)/11. Now the required ratio can be read from the triangle as 5 : 11. ♦ Problem 16. Draw the figure. Begin by assigning arbitrary masses at the vertices to form the mass points xA, yB, and zC. Using the fundamental technique of adding the mass points at the vertices, show that the sides balance at the feet of the cevians with mass points (x + y)B  , (y + z)A , and y+z x+y x+z AO BO CO (x + z)C  . So OA
· OB
· OC
= x · z · y . Expand the numerator and divide each expression by xyz to give y+z x+y x+z + + + 2 = 92 + 2 = 94. ♦ x z y Problem 17. Use Theorem 10 and substitute d = 3 into the equation d d d 3 3 3 + b+d + c+d = a+3 + b+3 + c+3 = 1. Multiply by the to arrive at a+d common denominator and do some algebraic manipulation, isolating abc, to get abc = 9(a + b + c) + 54 = 9(43) + 54 = 441. ♦

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Session 8 More on Proofs. Part II Mira Bernstein and Zvezdelina Stankova Sneak Preview. The reader must have realized by now that proofs are the backbone of serious mathematics. In Part I we learned about proofs by contradiction, proofs of possibility and impossibility, including constructive proofs, devising counterexamples, and seeking out invariants. We also examined situations with iff, which require two types of proofs. In this session we add a few more sophisticated methods of proof to our tool kit and deepen our understanding of previously learned methods. Along the way, we inductively divide a treasure among pirates while helping one toddler against another in a fierce game of raisins, play a game of solitaire until we see how to win it or when to not even attempt it, delve into the geometry of polygons with large numbers of right angles, and stuff just about anything in sight into pigeonholes until we succeed in solving our problems.

1. Proof by Induction Again Yes, you have seen this method before: a whole session in this book is devoted to induction (MI ). However, seeing it among other methods of proof will give you more perspective and deepen your understanding of MI. 1.1. Pirates and loot: what else did you expect? Here is an old-time favorite problem. Try solving it yourself before you read on!




Problem 1 (BAMO ’03). Six mathematically talented pirates, Awful, Bad, Crummy, Dreadful, Evil, and Foul, must divide a loot of 100 gold doubloons. According to pirate law, Foul, the leader, can distribute the coins as he sees fit. But when he is done, all six pirates (including Foul himself) take a vote: if more than half of them are unhappy with the distribution, Foul has to walk the plank (leaving his share of the gold behind). The remaining five pirates, with Evil as their new leader, then repeat the process: Evil divides the loot, they all vote, and if more than half are unhappy, then Evil is done away with and Dreadful takes charge of the gang. This process continues until the treasure is successfully distributed. How should Foul divide the money to ensure himself as large a share of the loot as possible? 155

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156

8. MORE ON PROOFS. PART II

A few reasonable assumptions are in order here. • Every pirate completely understands the strategy behind the process and will always vote in such a way as to maximize his share. • Furthermore, each pirate is bloodthirsty and will vote to kill the current leader if this won’t decrease his ultimate share of the loot. Hint: When faced with a problem like this, it is always good to try small cases first. What if there are only two pirates? Three? Four? Can you see a pattern, formulate a general strategy, and prove that it works? ♦ Our intuition says that the captain has a great chance of losing his life if he doesn’t give a considerable share of the loot to half of his crew. Let’s see if this is indeed what happens for a small number of pirates. Solution: Rank the pirates from 1 to n, with Pirate n as the captain. 2 pirates: #2 keeps all the money; it takes a majority to kill, so #1 can’t kill him. 3 pirates: #3 needs one vote besides his own to survive. He needs to bribe someone to vote for him, using the smallest possible amount of money. #2 cannot be bribed: if #3 dies, he’ll get everything. But #1 can be bribed with just 1 coin because if #3 dies, he’ll get nothing. Thus #3 should give 1 coin to #1 and keep the rest.

4 pirates: #4 needs one vote besides his own to survive. The easiest person to bribe is the person who gets the least if #4 dies. Checking above, we see that it’s #2, who gets nothing when #3 is captain. Thus #4 should give 1 coin to #2 and keep the rest. 5 pirates: #5 needs two votes besides his own to survive. He should bribe #1 and #3, as they will get nothing if #4 is captain. Thus #5 should give 1 coin to #1, 1 coin to #3, and keep the rest.

We summarize our findings so far in the following table; every column lists the number of coins the corresponding pirate will end up with ($ = 100). ↓ Captain ↓ #6 #5 #2 #3 #4 #5 $−2 #6 $−2 0

#4

#3

$−1 0 1

$−1 0 1 0

#2 $ 0 1 0 1

#1 0 1 0 1 0

By the same logic, for 6 pirates as in the original problem, #6 (Foul) should give 1 coin each to #2 (Bad) and #4 (Dreadful) and keep the rest.
Wow! This is totally contrary to our intuition! It seems that the captain can take almost the whole treasure for himself and still walk away alive . . . . 1.2. A recap of MI: robots to the rescue! What if you start with more than six pirates, say, n pirates, with Pirate n as the captain? Can you formulate a general answer based on the pattern above? Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

1. PROOF BY INDUCTION AGAIN

157

Once we have the answer (or think we do), how do we go about proving it? Let’s take our cue from the pirates’ own logic. To decide how to vote when there are n pirates, each pirate first has to consider what would happen if there were n − 1 pirates (i.e., how much money he can expect to get if he votes to kill Pirate n). Thus, to analyze what happens with n pirates, we must already know what happens with n − 1 pirates. For that, in turn, we must know what happens with n − 2 pirates. And for that, we must know what happens with n − 3 pirates. And for that . . . . This might sound like a nightmare, but, fortunately, there is a form of mathematical proof that is perfectly suited to this kind of problem: MI. Recall that the basic form of induction is designed to establish that some statement is true for all natural numbers n. (In our case, n is the number of pirates.) Instead of proving the statement for all n at once, we do it in two stages: the basis and the inductive steps. One way to see why MI works is to imagine our proof as an infinite ladder. To get to the k th step, we have to show that our claim is true for the number k. We might start by trying to climb the first few steps of the ladder to figure out how it works. But after a while we decide that we have better things to do than climbing an infinite ladder, so we delegate the task to a robot. We give the robot instructions for getting onto the first step of the ladder (the base case) and for going from one step to the next (the inductive step). Then we leave him to his (infinite) task while we go home and eat dinner. As long as our instructions were valid, we can be absolutely sure that eventually the robot will reach the 7th step, and the 77th , and the 777th . Since every step will eventually be reached and proved, the statement must be true for all n. In practice, the base case is usually very easy (often ridiculously easy). All the work goes into the inductive step. Here, we assume that our statement has already been proved for some n and use this assumption to prove that the statement is also true for n + 1. This might seem like a circular argument: aren’t we assuming what we are trying to prove? But, remember, in our instructions to the robot we just need to tell him how to go from step n to step n + 1. We can therefore imagine (assume) that he is already at step n – we don’t need to worry about how he got there.




1.3. Pirates use inductive reasoning. Let’s see how this all plays out in the case of our democratic pirates. Claim: If there are n pirates, then the captain (Pirate n) will give one coin each to Pirates n − 2, n − 4, . . . and keep the rest of the money himself. Proof: Basis Step. A lone pirate won’t give anyone any money, obviously! Inductive Step. Assume the claim holds for n pirates, and consider what happens with n + 1 pirates. The captain, Pirate n + 1, needs at least half

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158

8. MORE ON PROOFS. PART II

the votes, including his own, to survive. So he needs to bribe n/2 other pirates to vote for him.1 If he wants to give away as little money as possible, it makes sense for him to bribe the n/2 pirates who would get the least money if he is killed: he has to give them one more coin than they would otherwise get, so the worse off they expect to be, the less money he needs to give them to secure their votes. Because we assumed that the claim is true for n (the inductive hypothesis, IH ), we know exactly whom Pirate n + 1 should bribe. If Pirate n becomes captain, we know that Pirates n − 1, n − 3, . . . will get nothing at all. There are precisely n/2 of these anxious pirates. Pirate n + 1 can bribe them with one coin each, and he’s done! Thus, when there are n + 1 pirates, the captain will give Pirates (n + 1) − 2, (n + 1) − 4, . . . one coin each and keep the rest. This is exactly what our claim says for n + 1 pirates. We have thus shown that if the claim is true for n, it will be true for n + 1 as well.


1



5

3

7 9 11

$

2 10 6 8 4 12

We can think of the answer to the pirate problem in terms of two “clans” within the crew: one clan consists of the odd-ranked pirates, and the other – of the even-ranked pirates. Thus, the captain gives bribes only to his clan, who will always vote for him, while the other clan remains doubloonless.

MI is an awesome tool. It goes hand-in-hand with the equally powerful: PST 54. Always start by playing with small cases and work your way up. As you examine larger and larger cases of your problem, you will often notice how the solution for size n helps you find the solution for size n + 1. That’s your cue to use induction in your proof! This is how we originally discovered the answer to the pirate problem, and the same approach works for many other problems as well. You don’t have to be able to construct – or prove – the general solution for all n from scratch. As long as you can show that it can be built up step by step, you’re all set.




We finish the pirates discussion by finessing our previous solution in extreme cases. You might have noticed a flaw in our analysis of the pirate problem: we didn’t consider the possibility that the captain runs out of coins with which to bribe his crew! This first happens with 203 pirates: the captain needs 101 votes (besides his own); but with 100 coins, he can only afford 100 bribes. So with 203 pirates, the captain dies! Problem 2. What happens when there are more than 203 pirates? Hint: It’s not as simple as you might think! Find the pattern and prove it by induction. ♦ 1x is the greatest integer less than or equal to x: 4.7 = 4, −4.7 = −5, 2 = 2.

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1.4. Games and strong induction. Most of the time, the basic form of induction is sufficient, but as we saw in the Induction session, occasionally, you will need the more powerful variant: strong induction, M I s .




Problem 3. Alice and Zoe are playing a game with two piles of raisins. On her turn, a player eats one of the piles and splits the other pile into two smaller piles. Whoever can’t move loses. Alice goes first. Show that if at least one of the piles is even, she can always win. Proof: Let one of the piles be even, with 2n raisins. We use M I s on n. Basis Step. One of the piles has 2 raisins (n = 1). Alice eats the other pile and splits the pile with 2 raisins into two piles of 1 raisin apiece. Zoe can’t split either of these piles, so she can’t move. Alice wins! Inductive Step. Assume that Alice can win when one of the piles is even with fewer than 2n raisins. We must use this to show that she can win when one of the piles has 2n raisins. Here is one possible winning strategy for Alice: split the pile of 2n raisins into piles of 1 and 2n − 1, and eat the other pile. Now it’s Zoe’s turn. She can’t split the pile of 1, so she must eat it and split the pile of 2n − 1 (unless 2n − 1 = 1, in which case she has already lost). Since 2n − 1 is odd, no matter how Zoe splits it, one of the new piles will be even with fewer than 2n raisins. By the inductive hypothesis, Alice can now win!
Z A

A

A

Z

Figure 1. Alice wins a game of (5, 8) raisins Figure 1 demonstrates one possible game with initial piles of 5 and 8 raisins (if a pile is crossed out, it means it is eaten at the next move); you can check that Alice follows our inductive strategy and wins this game. By the way, why doesn’t regular induction (MI ) work for our general proof? To further understand the solution, try solving the opposite situation: Exercise 1. If initially both piles have an odd number of raisins, is there a strategy for one of the girls to win the game? Prove your claim. 1.5. Choosing appropriate induction. At times, inducting on the most obvious quantity is not the best choice, as we will see in the following solitaire game. Think about how, when, and why you can win the game. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Problem 4. A game of solitaire starts by laying out n cards in a row, some face up and some face down. A turn consists of removing a face-up card and turning over its immediate neighbors, if any. (Two cards are not considered immediate neighbors if there used to be a card between them that got removed on an earlier turn.) You win if you can remove all the cards. Show that you can win if you start with an odd number of face-up cards, but not if you start with an even number (cf. Fig. 2).

···

Figure 2. Lose or win in Problem 4



The problem asks for two separate proofs; while both involve strong induction, we shall see that they are otherwise quite different. For short, we shall call a card layout even or odd depending on the parity of the face-up cards in it. Now, the removal of a face-up card F splits the original layout into two shorter layouts: LF to the left and RF to the right of F . We denote by lF and rF the left and right neighbors of F (before F was removed). PST 55. Since you want to win for any odd layout and show that no matter how you play you will lose for any even layout, (a) you must find a strategy which splits any odd layout into two shorter odd layouts LF and RF and then apply (by induction) your winning strategy to those shorter layouts; and (b) you must show that, no matter which face-up card F you remove from an even layout, one of the two resulting shorter layouts LF or RF is even, and hence you will lose on it (by induction). Even though this PST is phrased in the specifics of Problem 4, it can be easily modified to reflect a large type of games where one seeks a winning strategy under certain initial conditions and a proof that victory is impossible under other conditions. For instance, try to restate PST 55 in the set-up of the Alice-Zoe game of raisins (Problem 3 and Exercise 1). Back to our solitaire. It is tempting to use induction on the number of face-up cards, but this does not work out too nicely, since the number of face-up cards does not always decrease after one turn. While this can be taken care of with a double induction, it is better to induct on a quantity that does always decrease: the total number of cards in the layout. Proof (odd layouts): An odd layout with one card is trivially winnable. Using strong induction, assume that we can win any odd layout with a total of n or fewer cards. Consider an odd layout with a total of n + 1 cards of which m are faceup. Let F be the rightmost face-up card in this layout. As we observed before, removing F leads to two shorter layouts LF and RF . Since card rF

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161

is face-down or does not exist, RF either has one face-up card or is empty. Depending on whether card lF is face-down, face-up, or does not exist, LF has m or m − 2 face-up cards or is empty. (Figure 3 illustrates the situations when F is not at an end.) In all cases, we obtain one or two shorter odd layouts, each of which is winnable by the induction hypothesis.
··· ··· l F F rF

···

···

···

···

···    LF

···    RF

Figure 3. Breakdown of an odd layout when F is not at an end Proof (even layouts): If there are no face-up cards, no moves can be made, so we lose. Using strong induction, assume that no even layout with n or fewer cards is winnable. Consider an even layout with a total of n + 1 cards of which m are faceup. Now, we cannot remove a specifically chosen face-up card (as we did for odd layouts) since we must show that, no matter what we do, we will lose. So, we remove an arbitrary face-up card F ; this alone decreases m by 1. The flip of card lF (if it exists) either increases or decreases m by 1, and ditto for card rF . Thus, (a) If F is not at an end (i.e., both lF and rF exist), then overall m increases by 1, or decreases by 1 or 3. Therefore, the number of face-up cards changes its parity from even to odd. But then an odd number of faceup cards are split among the two resulting shorter layouts LF and RF , i.e., one layout is odd and the other is even. (b) If F is at an end, say, at the left end (i.e., lF does not exist), then overall m stays the same or decreases by 2, i.e., keeps its even parity. This means that the remaining shorter layout RF is even. In all cases, we obtain a shorter even layout, which is not winnable by IH.


 

For both odd and even layouts, our analysis depended on the particular positioning (face-up, face-down, or does not exist) of the cards lF and rF . PST 56. If the conclusion you attempt to draw depends on variations in the given situation, then study separately each of these variations. In other words, apply case-by-case analysis, aka “case-chasing”. It is instructive to try the same proof using induction on the number m of face-up cards. If you do this, you will see that the argument becomes much more complicated! As we saw, m may decrease, but it may stay the same or even go up. Thus, PST 57. In game analysis, it is almost always preferrable to induct on a quantity which decreases at each step (or increases at each step). If you cannot find such a quantity, you may have to modify the method of induction to apply it to two or more quantities in the same argument.

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2. Extremes Are Naturally Demanding




2.1. “Two-headed” proofs. In Part I of Proofs we encountered problems with iff, forcing us to come up with two different types of arguments for a complete solution. Another category of problems requiring two proofs asks us to find the smallest (or largest) number with some property. For instance, in the Towers of Hanoi problem from the Induction session, you were asked to prove that 2n − 1 was the smallest number of moves necessary to solve the puzzle with n disks. Thus, you had to show that it is possible to do it with 2n − 1 moves, but that fewer moves wouldn’t have worked. PST 58. When you answer that the smallest number satisfying a certain condition is n, you are really making two claims: • n satisfies the condition and • no number smaller than n satisfies the condition. As before, the proofs of these two assertions will often entail completely different approaches. Even if one of the two claims seems obvious, you should always include both proofs, for logical completeness. 2.2. Extremes in geometry. For the rest of this section, we shall try to conquer a single question: Problem 5. Let k be an integer, k ≥ 5. What is the smallest integer n for which there is an n-sided polygon with k interior 90◦ angles? C D A

F

B E

Figure 4. 12-gon, Self-intersecting polygon, and Quadrilateral




Note that the polygon might not be convex, so we need to distinguish between 90◦ and 270◦ angles. For instance, the polygon in Figure 4a has eight 90◦ and four 270◦ interior angles. Further, for “interior” angles to make sense, our polygons cannot be self-intersecting: as Figure 4b warns us, it is not clear which angles of ABCDEF are “interior” and which are “exterior”. With this said, to solve Problem 5 it is imperative to find some relationship between the two variables involved, k and n. It is well known that the sum of the (interior) angles in a triangle is 180◦ , and in a quadrilateral this sum is 360◦ . (Why? Figure 4c is very suggestive!) Can you guess this sum in a pentagon? A hexagon? Is induction lurking behind this situation too? Problem 6. Show that the sum of the interior angles in an n-gon (which does not intersect itself) is 180◦ (n − 2).

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2.3. Inequalities in search of extremes. The formula in Problem 6 still doesn’t have anything to do with the variable k, so it is high time to bring k into the picture. You know that there are at least k interior 90◦ angles, but you don’t know the remaining angles, so name them αk+1 , αk+2 , . . . , αn . Putting everything together, (1)



180◦ (n − 2) = 90◦ k + αk+1 + αk+2 + · · · + αn . Here comes the key observation:

PST 59. When looking for the maximal or minimal value of a variable n, it is advantageous to turn equalities into inequalities, solve the latter for n, and thereby come up with possible extreme values of n. OK, but how do we turn (1) into an inequality? All we know is that each angle αi < 360◦ . . . . Replacing all αi ’s by 360◦ may seem like a crazy thing to do, but in reality it produces a good upper bound for the LHS: (2)

180◦ (n − 2) < 90◦ k + 360◦ (n − k).

One catch: if n = k, i.e., if all the angles in our polygon are right angles, then we don’t have any extra angles αi , so (2) should really be an equality! Does this ever happen? Exercise 2. Show that all polygons satisfying the conditions of Problem 5 have at least one non-right angle, i.e., n > k. 90◦ (3)

Exercise 2 implies that (2) is a strict inequality. We divide both sides by and finally solve for n: 2(n − 2) < k + 4(n − k) ⇒ 3k − 4 < 2n ⇒ n>

3k 2

− 2.

For instance, when k = 6 then n > 7, i.e., n ≥ 8; and when k = 5 then n > 5 12 , i.e., n ≥ 6. In principle, inequality (3) is good enough to conjecture a minimal value of n for any k, but it involves calculations with fractions and rounding up for odd k. It is customary to come up with a unified formula for all cases, regardless of whether k is even or odd. This can be achieved via the ceiling function x. 2 Exercise 3. Using the fact that n is an integer, show that inequality (3) can be rewritten equivalently as

 3(k−1)  3 (4) n ≥ 3k . 2 − 2 = 2



As  an example, if k = 7 then n ≥ 9 = 9, while if k = 8 then n ≥ 10.5 = 11. We conclude that the minimal possible value of n for a non  . self-intersecting n-gon with k right angles is nk = 3(k−1) 2 2x is the smallest integer greater than or equal to x: 4.7 = 5, −4.7 = −4, 2 = 2.

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2.4. Art and induction. At this point, many beginners may think we are done – but according to PST 58, we have only done half the work! We have shown that n can’t be any smaller than a certain value, but not yet that this value can actually be achieved. A creative part is still to follow: for each k ≥ 5, it “only” remains to construct an nk -gon with k (interior) right angles. Since there are infinitely many values of k ≥ 5 and correspondingly infinitely many cases to draw pictures for, we must be clever in doing this. Doesn’t induction describe infinitely many steps in finitely many words? Try the first two cases. The top row of Figure 5 exhibits a hexagon with 5 right angles (n5 = 6) and an octagon with 6 right angles (n6 = 8). A B

C

n5 = 6

n6 = 8 A F

D E G

n7 = 9

n9 = 12

C

n8 = 11

n10 = 14

Figure 5. Adding two more sides and one more right angle We need an inductive construction to turn, say, the hexagon into a 9-gon with 7 right angles (n9 = 7) and the octagon into an 11-gon with 8 right angles (n8 = 11). Note that in both cases we will be adding three more sides and two more right angles. One possible general algorithm of making this inductive step is represented by the middle “diagonal” in Figure 5: 1. Pick a “big” (interior) ∠ABC in the n-gon, i.e., ∠ABC > 180◦ . 2. Select two points D and E on segments AB and CB, respectively. 3. Draw perpendiculars DF and EG to AD and CE, respectively, both going towards the inside of the n-gon. 4. Replace vertex B with new vertices D, E, F , and G, and replace segments DB and EB with new sides DF , F G, and GE. Check that this construction does indeed add three more sides and two more right angles; plus, we will never run out of the necessary “big” angles because of the new (interior) ∠DF G > 180◦ and ∠EGF > 180◦ , i.e., the construction will add two more “big” angles. If you choose the points D and E close enough to B and make the perpendiculars DF and EG short enough, you will keep the new (n+3)-gon from becoming self-intersecting. With this said, we leave the formal proof by induction to the reader. ♦ Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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2.5. After such a multi-step solution, a good problem solver should wonder about (at least) three things. Question 1. Why did we need two base cases, for k = 5 and k = 6? To have one base case requires an algorithm that adds precisely 1 new right angle and (nk+1 −nk ) new sides, i.e., either 1 side (for k even) or 2 sides (for k odd). Instead of finding two different algorithms to address each case, we opted for one algorithm, which we applied separately to two sequences of polygons: those with an even and those with an odd number of right angles. Question 2. Why was the restriction k ≥ 5 imposed in the first place? Nothing prevents us from trying to solve the same problem for k ≤ 4. If k = 4, it is obvious that a rectangle provides the required minimal example, i.e., n4 = 4. Check that n3 = 4 = n2 and n1 = 3. None of these cases fits our

 , so maybe a desire to keep the final answer uniform formula nk = 3(k−1) 2 and elegant was the reason for “kicking out” the initial values of k from the original statement. Question 3. How does one come up with inequality (3), leading to the actual minimal possible value of n, and with the inductive algorithm in Figure 5 for constructing extreme nk -gons?




These are good questions, without equally good answers. Previous experience with similar problems, familiarity with methods of proof involved, and the determination to go through several initial cases until a pattern develops are probably the most helpful factors. Specifically, to find the minimal values nk requires establishing a relationship between the two variables involved, n (sides) and k (right angles), and being bold enough to turn the resulting equality into an inequality. The ceiling function is only a fancy gadget to make our final answer look sleek, but we could have done just as well without it. Note that we have given only one of many possible algorithms for constructing an optimal nk -gon. The value nk is unique, but there are many different nk -gons that work and many ways to construct them (some without using induction). The reader is welcome to apply his/her own mathematical creativity in designing other algorithms for constructing optimal nk -gons.

3. The Pigeonhole Principle 3.1. The need for a new method. We saw earlier that, to prove something is possible, you just have to give one example of how to do it. But what about a problem like this:




Problem 7. 15 chairs are placed around a circular table. On the table are name cards for 15 guests. After the guests sit down, it turns out that none of them is sitting in front of his own card. Prove that the table can be rotated so that at least 2 guests are simultaneously correctly seated.

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The problem asks for a possibility proof, but we can’t just construct a single example: we have to show that a solution exists no matter how the guests are seated. We need some way of getting past the details of the initial seating arrangement – something like an invariant, but not quite. The trick for solving this kind of problem is: Theorem 1 (Pigeonhole Principle, PHP). (a) Basic version: If you put n + 1 or more pigeons into n holes, then at least one hole will contain more than one pigeon. (b) Generalized version: If you put kn + 1 or more pigeons into n holes, then at least one hole will contain more than k pigeons. 3.2. Is it OK to question PHP? If you have never encountered PHP before, you are probably asking yourself at least three questions right now: Question 4. Why bother stating this as a principle – isn’t it obvious? Question 5. What does this have to do with Problem 7? Question 6. Why pigeons? To answer Question 4: yes, PHP is obvious, but we are about to see that it is the basis for a powerful problem solving technique. Formulating it as an explicit principle reminds us that we have this technique in our tool kit. To answer Question 5: the challenge in using PHP is figuring out what the pigeons are, what the holes are, and what it means to put a pigeon in a hole. Since Problem 7 is not exactly a warm-up exercise, beginners should try the pigeon/hole identification first in: Problem 8. An ice cream shop serves 4 flavors of ice cream. 7 friends show up, and each of them orders a cone with 2 different flavors. Prove that there must be at least 2 people who ordered the same combination of flavors. Hint: How many possible combinations of flavors are there in total?

♦

Proof: According to the Combinatorics session, there are 6 possible com binations of 2 flavors: recall that 42 = 6 (cf. Fig 6). If each of 7 people get one of 6 combinations, then by PHP at least 2 people must have the same combination: the people are the 7 “pigeons” and the combinations are the 6 “holes”.


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167

Note that the “pigeons” do not necessarily spread out in the holes as evenly as possible. In Figure 6, for example, the third flavor combination got its 2 “pigeons”, the last combination is the most popular with 3 “pigeons”, while no one liked the second or fourth combination. PST 60. You should always start a pigeonhole problem by asking yourself, “What are the pigeons, what are the holes, and how do we assign pigeons to holes?” However, the final proof should not be formulated in terms of pigeons and holes, but in terms of the actual objects of the problem (in the case of Problem 8, people and flavor combinations). As to Question 6 (why pigeons?): the name “Pigeonhole Principle” originated in Britain, where “pigeonhole” is commonly used to mean a small compartment or a cubby. Thus, the name was supposed to make you think of distributing objects into slots. But in America, where the word “pigeonhole” is rarely used, it tends to make us think of pigeons! This image has been passed down from generation to generation of American mathematics students, and it seems to have stuck. 3.3. More sophisticated pigeons and holes. Identifying the pigeons and the holes is not always obvious; but, when successfully done, it accounts for more than half of the solution. So, let’s go back to the original Problem 7 and solve it in stages. Hint for Problem 7: The wording “at least two guests” in the problem is suggestive of “at least two pigeons” in PHP, and hence the “pigeons” should probably be the guests. The choice for “holes” requires some thinking. The problem asks you to consider all possible ways of rotating the table – so maybe we can let the “holes” be all the possible table positions? How would we then decide which “pigeons” go in which “holes”? What would it mean for a “hole” to hold two “pigeons”? There is just one problem with this approach: the way we have set it up, there are 15 holes for 15 pigeons, so PHP does not seem to apply at all! We can fix this by taking the “hole” corresponding to the initial position of the table out of consideration: we already know it will contain no pigeons (why?). Thus, the 15 pigeons must be distributed among the remaining 14 holes, and PHP applies. ♦

 

Before reading further, try to think of at least two new PST’s which were employed in the above hint. PST 61. Often, you can figure out what the pigeons and holes should be by working backwards: match the data in the problem with the PHP setting. PST 62. If it seems that you have too many holes in a pigeonhole problem, see if any of them can be excluded to bring the number of holes down so that PHP applies. Let us analyze the hint via a specific example. For simplicity, consider a situation with only 5 guests, A, B, C, D, and E, corresponding to name cards

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8. MORE ON PROOFS. PART II D b aE A c d eB C

D a e E A b c d B C C E  72◦

D E e d A a b cB C A  144◦

D d c E A e a b B C

D E c b A d e a B C

D

B  216◦

 288◦

Figure 7. Table positions for 5 “pigeons” a, b, c, d, and e. Figure 7 displays the 5 (rotated) table positions, starting with the initial position on the left. In each position, the shaded sectors of the table indicate a match between a guest and his name card; for instance, in the 3rd position only guest A is facing his own name card. Since we know that no one is sitting in his proper place in the initial position of the table, we do not count this position among the “holes”. Underneath each of the remaining table positions, we record the “pigeons” who get assigned to this “hole” (the guests who are facing their own cards in this position). Thus, the 4th table position (obtained by a 216◦ -rotation from the original) contains two, incidentally kissing, pigeons D and B. In summary, there are 5 “pigeons”: A, B, C, D, and E, and 4 “holes”: the nd 2 , 3rd , 4th , and 5th table positions; two of these “holes” contain two pigeons each, namely, the 2nd and 4th table positions each satisfy the requirement of the problem. After you have thoroughly considered the above hint and example, here is one way to write the formal proof. Proof of Problem 7: Let us call a guest “satisfied” if he is sitting in front of his own place card. For each guest, there is exactly one position of the table that will satisfy him. We need to prove that there is a position which simultaneously satisfies at least two of the guests. There are 15 possible positions. However, we are told that the initial position satisfies no one, so the remaining 14 positions must satisfy all 15 guests. Thus, by PHP, there must be a position that satisfies at least two guests at once.





Here is another example of this kind of reasoning: Problem 9. If we choose 6 integers between 1 and 10, prove that at least two of them must be consecutive. PHP says: “at least two pigeons must be in the same hole”. The problem says: “at least two of the numbers must be consecutive”. This suggests that the numbers should be the “pigeons” and that being consecutive should imply being in the same “hole”. Since we have 6 “pigeons”, we need to have 5 (or fewer) “holes” for PHP to apply. Before you read the proof: can you think of what the “holes” should be?

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169

Proof: Consider the pairs of consecutive numbers {1, 2}, {3, 4}, {5, 6}, {7, 8}, and {9, 10}. Each of our 6 numbers belongs to exactly one of these 5 pairs. By PHP, two of the 6 numbers must belong to the same pair, so they are consecutive.
PST 63. Look for the very common situation: the “pigeons” are the objects of the problem and the “holes” are sets of possible objects so that any two objects in the same set satisfy the required condition. In Problem 9, for example, the objects were the numbers, the sets were the pairs of numbers, and the required condition was to be consecutive, i.e., to belong to the same set. The sets themselves were the same size (pairs), but in general they don’t have to be.



3.4. PHP is everywhere! An important part of learning to use PHP is becoming skillful in recognizing the kinds of problems to which PHP can be applied. Below are several examples that an experienced problem solver would immediately label as “pigeonhole problems”. Read them through once to get a sense for the kind of problems we’re talking about. Then try to solve them; be sure to think in terms of pigeons and holes! Problem 10. Let S be any set of 20 distinct integers chosen from the arithmetic progression3 1, 4, 7, . . . , 100.




(a) Show that there are two numbers in S that sum to 101. (b) Must there be two numbers in S that sum to 107? 110? 116? Give a proof or a counterexample. (c) For which N is it guaranteed that there will be two numbers in S that sum to N ? Hint: For a desired sum N , there could be some “bad” numbers in S: those that do not pair up with anything else to give the sum N . Find these “bad” numbers, as they won’t qualify to be “pigeons”, and count how many “good” numbers are left. ♦ Problem 11. Consider any five points in the interior √ of a square of side 1. Show that two of these points are no more than 1/ 2 apart. Hint: Obviously, the pigeons are the 5 points, but what are the holes? We need at most 4 “holes” such that √ if two points land in the same “hole”, then they must be no more than 1/ 2 apart. Can you partition the square into 4 such regions? ♦




Problem 12. Show that in a class of n students, there are at least two students with the same number of friends in the class. (Assume friendships are always mutual.) 3Every member of an arithmetic progression (or series) is obtained from the previous member by adding the same number d. In our case, d = 3.

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Proof: At first glance PHP doesn’t seem to work since there are n possibilities for the number of friends a person can have: {0, 1, 2, . . . , n − 1}. The insight is that, if someone is friends with everybody, then everyone has at least 1 friend. In other words, you can’t simultaneously have a person with no friends and a person with (n − 1) friends. (Note that we just applied PST 62 and reduced the number of “holes”.) Thus, there are only (n − 1) possibilities for the number of friends a student in this class can have: either {0, 1, . . . , n − 2} or {1, 2, . . . , n − 1}. By PHP, at least two of the n students must have the same number of friends.





3.5. Pigeons like to paint. PHP often comes to the rescue in coloring problems. Imagine a painter randomly splashing several colors on his white canvas. What kind of unicolor figures4 are we always guaranteed to find? Problem 13. Let every point of the plane be painted white or red. Show that there is a rectangle whose vertices are all the same color. Hint: This problem is trickier than the others, requiring PHP twice.

♦

Let’s see how we can find a unicolor rectangle. The vertices of any rectangle lie on two sets of parallel lines, so for starters, let’s draw two horizontal lines h1 and h2 . It would be nice if we could find the four vertices of our unicolor rectangle along h1 and h2 , but this may not be possible; for all we know, h1 could be all white and h2 could be all red . . . . So, we need at least one other horizontal line, h3 . Now we draw in several vertical lines v1 , v2 , . . . , vn , each of which produces 3 intersection points with h1 , h2 , and h3 . The question is, how many vertical lines do we have to draw to ensure that among the resulting intersection points four will form a unicolor rectangle? PHP will answer this question. h3 h2

Ci Bi

h1

Ai v1 v2 

v i

v2 v6 v 6

v9

Figure 8. 9 vertical lines and PHP Proof: Let v1 , v2 , . . . , vn be vertical lines, and let line vi intersect the horizontal lines h1 , h2 , and h3 in points Ai , Bi , and Ci , respectively (cf. Fig. 8a). 4In this subsection, unicolor means “having vertices of the same color”.

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The triplet of points (Ai , Bi , Ci ) on vi can be colored in 8 different ways, as there are two color choices for each point. So, if we draw 9 vertical lines v1 , v2 , . . . , v9 , by PHP, the triplets of points (Ai , Bi , Ci ) for at least two of these lines will be identically colored (cf. Fig. 8b). Without loss of generality (WLOG), assume that v2 and v6 are these two lines, so that A2 ∼ A6 , B2 ∼ B6 , and C2 ∼ C6 , where ∼ means “same color”. (In Figure 8, black, white and grey dots stand for unknown, white, and red colors, respectively.) Finally, concentrate on one of the two identical triplets, say, (A2 , B2 , C2 ) along v2 . We have 2 colors (the “holes”) and 3 points (the “pigeons”). Again by PHP, two of these points are colored the same way. WLOG, let B2 and C2 both be red. We conclude then that the rectangle formed by intersecting v2 and v6 with h2 and h3 has four red vertices, namely, C2 ∼ B2 ∼ B6 ∼ C6 .





Here is another popular coloring problem, where, instead of a unicolor rectangle, we are looking for a unicolor segment of a certain length. Problem 14. Let every point of the plane be painted white or red. Show that there are two points colored the same way and exactly 1 inch apart. We first discuss a solution that is neither the shortest nor the most clever. Its strong point is that it follows a natural path of reasoning without involving any ideas out of the blue. Proof 1: By contradiction, suppose that there are no two unicolor points exactly 1 inch apart. In other words, for any two points 1 inch apart, one is white, and the other is red. Pick two such points A and B (AB = 1 inch) and WLOG assume that A is white and B red. But then, by our supposition, all points exactly 1 inch apart from A must be red, i.e., the circle k1 with center A and radius 1 inch is all red. Similarly, circle k2 with center B and radius 1 inch is all white. Where is the contradiction? C k1

k2 A

1

B

D

Figure 9. Unicolor segment of unit length Look at Figure 9. Are there any new special points that require our attention? Well, the circles k1 and k2 intersect in two points C and D, each of which is attempting to be both white and red at the same time: this is against the rules and hence provides our contradiction.
A key moment in Proof 1 was to find out exactly what happens to all points 1 inch away from A and from B. For this, it was necessary to Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.



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PST 64. Paraphrase the negative supposition “There are no two such-andsuch points . . . ” into a positive statement “For any two points so-and-so happens . . . .” This little trick is especially useful in proofs by contradiction because the positive reformulation usually tells us what is really happening and gives us a base to step on in our constructions and analysis. Now let’s try to condense our solution by extracting only the relevant information from it. What caused the contradiction? Obviously, we didn’t need all the points on circles k1 or k2 : the contradiction would still occur if we considered only the four points A, B, C, and D, which form a rhombus of side length 1 inch. Now that we think about it, we didn’t really need point D either, as one contradiction provided by point C would have sufficed. So, what is left from our original solution? Proof 2: Let ABC be equilateral with side length 1 inch. We have two colors (the “holes”) and three vertices (the “pigeons”) to color. By PHP, two of the vertices will be colored the same way.
Wow, that was short! But remember that coming up with such an elegant “simple” solution may have to be preceded by a longer path of initial investigation and possibly other longer solutions. While we are on the topic, let’s answer a related question. We assumed that the whole plane was colored in white and red, but evidently we didn’t need the whole plane to carry out the solutions above. So, we can restrict ourselves to a square canvas. But too small a canvas wouldn’t work, e.g., a tiny 1/2 × 1/2 in2 canvas wouldn’t even have two points 1 inch apart! Problem 15 (Intermediate). Let each point on a square canvas be colored white or red. What is the smallest size for such a canvas to guarantee that there will still be two unicolor points exactly 1 inch apart? Does this type of question ring a bell? It asks for an extremal value – the smallest size of a canvas with a certain property. The experienced reader should know by now that the solution must consist of two parts: (a) Conjecture the minimal size of the canvas, and show that any canvas of smaller size is no “good”, i.e., exhibit an example of a white-red coloring of the smaller canvas in which there are no two unicolor points exactly 1 inch apart. (b) Prove that any canvas of size equal or larger than your conjectured minimal size will satisfy the conditions of the problem, i.e., regardless of the white-red coloring, there will be two unicolor points on the canvas exactly 1 inch apart. Now let’s raise the stakes and ask if Problem 14 is still true for 3 colors:




Problem 16 (Intermediate). Let every point of the plane be painted white, red, or blue. Show that there are two points colored the same way and exactly 1 inch apart.

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Hint: Start first with a proof by contradiction. You will arrive again at the rhombus ABCD from our first solution above, with the difference that, say, A would be white, B red, while C and D blue . . . no contradiction yet. √ However, CD = 3 inches, and C and√D are unicolor! Could this be used somehow? If you can find two points 3 inches apart which are differently colored, you can use them in the roles of C and D, reconstruct the rhombus ABCD, and achieve your contradiction. Thus, you enter stage√two of the solution. Again for contradiction, suppose that any two points 3 inches apart are unicolor. Is this really possible? What can you say about an isosceles XY Z with base XY = 1 inch and √ sides XZ = Y Z = 3? Wouldn’t you get that X and Y are unicolor and 1 inch apart? What exactly does this contradict? ♦




It will take some sophistication and skills in logical argumentation to really understand the above “contradiction-within-contradiction” hint and to turn it into a formal solution. Your next task is to construct a short elegant second solution: extract the minimal configuration of points from the above solution to imply the existence of two unicolor points 1 inch apart. The die-hard problem solver will not stop here. Problem 17 (Advanced). What is the minimal number of colors needed to color the plane so that no two points 1 inch apart are unicolor?







This minimal number of colors is denoted by χ(E2 ) and referred to as the chromatic number of the plane.5 If you think about it, Problem 16 showed that 3 colors do not suffice, i.e., χ(E2 ) ≥ 4. On the other hand, Problem 18 (Intermediate). Find a coloring of the plane with 7 colors such that any two points 1 inch apart are colored differently. Thus, as far as the world knows, 4 ≤ χ(E2 ) ≤ 7. If you would like to make history, try to resolve the grey area of 4, 5, and 6-colorings to find the exact chromatic number of the plane. 3.6. Generalized PHP. Finally, here is a classic problem which requires the generalized pigeonhole principle (GPHP). As in previous solutions, GPHP provides only one step of the proof; another step may demand a different type of reasoning. In this subsection, a “unicolor” figure will have unicolor sides. Problem 19. Color the diagonals and sides in a (convex) hexagon red or blue. Show that the diagram contains an all-red or an all-blue triangle. Hint: Consider the 5 edges coming out of your favorite vertex. What does GPHP say about their colors? Now think about that tricky second step. ♦ 5 The Greek letter χ is called chi (pronounced like “kite” without the final “t”), and E2 refers to the (Euclidean) plane.

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Proof: By GPHP, among the 5 = 2 · 2 + 1 segments coming out of a vertex A, at least 2 + 1 = 3 must have the same color. WLOG, let these segments be AC, AD, and AE, all colored red. What can we say now about CDE? If its sides are all the same color, we are done (cf. Fig 10a). If not, then at least one of them is red, say, CE, so that ACE is all red (cf. Fig 10b).
B

B

C D

A

G A BC

C D

A E F

E

D E F

F

Figure 10. Unicolor triangle in Problems 19–20 Do we really need a hexagon to guarantee a unicolor triangle? Exercise 4. Color the diagonals and sides of a (convex) pentagon in red and blue in such a way that no unicolor triangle is formed.




Together, Problem 19 and Exercise 4 show that 6 is the minimal value of n for which an n-gon colored with 2 colors must have a unicolor triangle. What about 3 colors? How many vertices do we need in our polygon to guarantee a unicolor triangle? Problem 20. Color the diagonals and sides in a convex 17-gon red, blue or green. Show that the diagram must contain a unicolor triangle. Partial Solution: By GPHP, 6 of edges emanating from a vertex G must be the same color, say, green (cf. Fig. 10c). The other 6 vertices of these green edges form a convex hexagon H. If any of the diagonals or sides of H is green, it will complete an all-green triangle formed by the corresponding green segments leading to G. But if none of H’s diagonals or sides is green, then they are all red or blue, and we are back in the set-up of Problem 19. ♦ Thus, any 17-gon colored with 3 colors must have a unicolor triangle. But is 17 the minimim? The real challenge is to color a (convex) 16-gon in 3 colors without forming a unicolor triangle. This can be done, but the picture is very complicated! If you are interested, you can look it up online in Wikipedia under “Ramsey’s Theorem.” For the readers familiar with recursive sequences, we pose Problem 21 (Advanced). Let each diagonal and side of a convex n-gon be colored in one of k colors. What is the minimal number of vertices n to guarantee that the resulting diagram contains a unicolor triangle? 3.7. Should PHP be proven or taken for granted? A fitting end to our discussion would be to put PHP on a solid foundation by . . . actually proving it! Note that the basic version of PHP is a special case of the generalized PHP (set k = 1). The proof of GPHP itself is so elementary

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4. HINTS AND SOLUTIONS TO SELECTED PROBLEMS

175

that it seems like an “anticlimax” after the complex problems we’ve just been through. Yet, something can still be learned from it: the powerful and ever-present PHP is nothing but a simple statement proven by contradiction.

Figure 11. GPHP for 16 pigeons and 5 holes Proof of Theorem 1(b). Suppose that the generalized PHP is false, i.e., for some n and k ≥ 1 we can put kn + 1 pigeons into n holes without crowding any hole with more than k pigeons. But how many pigeons can be accommodated this way? At most kn since we have n holes each containing at most k pigeons. Thus, the remaining (kn + 1)st pigeon is not really put into a hole, a contradiction. Therefore, the generalized PHP is valid.
Figure 11 illustrates GPHP for 16 = 5 · 3 + 1 pigeons and 5 holes. By putting at most 3 pigeons into each hole, we can accommodate at most 15 pigeons, leaving the last, 16th , pigeon to hover outside. PHP is so ubiquitous throughout mathematics, molding solution after solution into its elegant frame, that even though it is “just” a theorem, PHP is often perceived as a problem solving technique or a method of proof.

4. Hints and Solutions to Selected Problems Problem 2. The table below summarizes the cases when Pirates #201 through #208 are captains, with the following notation: • E and O are, respectively, all even and all odd pirates among the first 200 pirates; $ and 0 mean that every pirate in the corresponding group gets 1 coin or no coins; • Y and N stand for votes “Yes” or “No”; • indicates that the captain dies due to lack of votes; • “–” shows that no matter how much money these pirates are offered, they won’t get it since the current captain will die and the situation will move to the line above. ↓ Captain ↓ 200 201 202 203 204 205 206 207 208

208

Y

207

Y

206

Y Y

205

Y N Y

204

Y N N N N

203

202

201

Y N N N N

Y N N N N N N

Y N N N N N N N

E $ 0 $ – 0 – – – $

O 0 $ 0 – $ – – – 0

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The table suggests that in all cases of captains #n > 200 • no pirate of ranking > 200 will get any money at all: the treasure will be split evenly among the pirates in group E or in group O; • a pirate of ranking > 200 will vote “Yes” only to avoid dying in a situation described on some previous line; • captain #n will survive only if half of the pirates of ranking > 200 are in danger of dying should #n die. For example, as we discussed in the text, #203 will die as a captain because #202 is assured to live as a captain and #201 will gain no money in either case so he has no incentive to vote for #203. However, #204 will live with the vote of #203 (who is trying to avoid becoming a captain and dying for sure) and the votes of O = {#1, #3, . . . , #199}, among whom the treasure will be equally distributed. Next #205, #206, and #207 will die for lack of votes from the higher ranking pirates. But #208 will live due to exactly the same reason: #205, #206, and #207 will vote for him, knowing that if #208 dies, they’ll die too; note that the treasure will have to be distributed among E here (why?). Check that the next captain to live is #216. A pattern develops: the only captains surviving the ordeal are ranked in “powers of 2”, i.e., #201, #202, #204, #208, #216, . . . , #(200+2n ), . . . . This calls for a proof by induction. For the inductive step assume that the pattern and all of our conclusions above hold up to the surviving captain #(200+2n ), inclusively. So captain #(200 + 2n+1 ) will get 100 votes from among the first 200 pirates (which ones?) and 2n votes among the last 2n+1 pirates (including himself), giving him half the votes and ensuring his survival. ♦ Exercise 1. The second girl, Zoe, can always win. The first girl, Alice, must eat one of the piles and break the other odd pile into two piles, one of which is even. Zoe now just uses the previous strategy developed for Alice in Problem 3: Zoe splits the even pile of 2n raisins into piles of 1 and 2n − 1 and eats the other pile.
Problem 6. As the text suggests, let’s use strong induction on n for our ngons (which will never be self-intersecting). For n = 3, the proposed sum of angles in a triangle is 180◦ , which we know to be true. For n = 4, Figure 4c suggests to split a quadrilateral into two triangles, so the total sum of its angles is 180◦ + 180◦ = 360◦ ; this equals the proposed sum of 180◦ (4 − 2). Note that a splitting into two triangles is possible even if the quadrilateral is not convex (cf. Fig. 12a). G

G G 

H

H H

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For the inductive step, we assume that for k = 3, 4, . . . , n the sum of the (interior) angles for any k-gon is 180◦ (k − 2). Now, if an (n + 1)-gon G is convex, any diagonal obviously splits G into two smaller polygons, a k1 -gon G  and a k2 -gon G  (cf. Fig. 12b). But one has to be careful about which diagonal to choose in a non-convex (n + 1)-gon H; in Figure 12c, for instance, only 3 of the 9 “diagonals” would split H into two smaller polygons, a k1 -gon H and a k2 -gon H (cf. [51], §82). In any case, k1 + k2 = n + 3 with k1 , k2 ≤ n (why?), so that induction applies to the two smaller polygons G  and G  , or H and H . Thus, the total angle sums in G and in H equal ♦ 180◦ (k1 − 2) + 180◦ (k2 − 2) = 180◦ (k1 + k2 − 4) = 180◦ (n − 1). Exercise 2. Note that n ≥ k, as the total number of angles is n while the number of right angles is k. If n = k, there are no αi ’s in (1), and 180◦ (n − 2) = 90◦ k, i.e., 2n − 4 = k ≤ n ⇒ k ≤ n ≤ 4. But k ≥ 5 in Problem 5, so the case n = k is not under consideration. Thus, n > k.
Exercise 3. If k is even, then inequality (3) implies n ≥ 1 3k 3 while if k is odd, then n ≥ 3k 2 − 1 2 =  2 − 2 .

3k 2

1 −1 =  3k 2 − 1 2 , ♦

Problem 10. (a) Write down the 17 pairs that sum to 101: {1, 100}, {4, 97}, {7, 94} . . . , {46, 55}, {49, 52}. Each of our 20 numbers belongs to one of these 17 sets. By PHP, two of the 20 numbers must belong to the same set, so their sum must be 101.
(b) For a sum of 107, check that only 1 and 4 do not pair so that there are 16 pairs altogether. This gives 16 pigeonholes for the 18 (= 20 − 2) remaining numbers. Similarly, for a sum of 110, only 1, 4, 7, and 55 do not pair, yielding 15 pigeonholes for the remaining 16 numbers, just enough. ♦ However, the sum of 116 is troublesome: 1, 4, 7, 10, 13, and 58 do not pair, leaving only 14 pairs for the remaining 14 numbers: PHP is not applicable. In fact, here is a counterexample: select 1 number from each of the 14 pairs plus the “lonely” six numbers above, e.g., {1, 4, 7, 10, 13, 16, . . . , 55, 58} are 20 numbers from S no two of which add up to 116.
(c) Answer: N = 89, 92, 95, 98, 101, 104, 107, 110, 113. ♦ Problem 11. Divide the square into four smaller squares by its horizontal and vertical middle lines. By PHP at least 2 of the 5 points must have landed in one of the small squares (or on its boundary). Since each of the small squares has a side length 1/2, √ the greatest distance between 2 points
in it is the length of the diagonal, 1/ 2. √ Problem 16. Suppose that any two points 3 inches apart are unicolor. Draw an isosceles ECD with base ED = 1 inch and sides EC = DC = √ 3 inches (cf. Fig 13a). By assumption, E and C are unicolor, as well as C and D. So E and D are unicolor and exactly 1 inch apart. We √ are done. Now suppose the opposite: some two points C and D are 3 inches apart and colored differently, say, C is red and D green. Construct a rhombus ADBC with sides 1 inch whose long diagonal is CD (cf. Fig 13b). Then its Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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8. MORE ON PROOFS. PART II C

C

A

C

B

G

B A F

E

D

D

E

D

√ Figure 13. Dashed segments  3 in., solid segments  1 in. short diagonal AB is also 1 inch long (why?). If one of A or B is colored red or green, then that point, along with C or D, will form the desired unicolor segment of length 1 inch If this isn’t so, then both A and B must be colored in the third color, blue, i.e., AB would be the desired unicolor segment of length 1 inch This exhausts all possibilities and completes the proof.
Note that we have managed to avoid using contradiction at all. The concise second solution extracts from the above argument a 7-point configuration, as in Figure 13c, consisting of √ • EDC with base ED = 1 in. and sides EC = DC = 3 in., and • two rhombuses ADBC and EF CG with sides and small diagonals 1 in. whose long diagonals are CD and CE, respectively. If any two points 1 inch apart are colored differently, then each of the 4 equilateral triangles (making up the rhombuses) have vertices colored in the 3 given colors. Starting with point D, say, colored in green, you can easily move through these 4 triangles and conclude that C and E are also green. But then segment ED is unicolor and of length 1 inch, a contradiction.
Problem 18. Tessellate the plane with regular hexagons of diameter < 1. Color a hexagon and the interiors of the 6 surrounding hexagons with the 7 colors. Use this coloring scheme throughout the plane. There are several ways to choose the colors for the edges and vertices. A complete discussion of the problem can be found in Halmos [38] and Klee and Wagon [53]. ♦ Exercise 4. A simple and symmetric way to do this is to color all sides of the pentagon in one color and all diagonals in the other. Note that of the 4 segments emanating from each vertex, 2 are red and 2 are blue: we cannot afford to color 3 of those segments in the same color, or else we will land in the middle of Problem 19’s solution.
Problem 21. Using induction on GPHP solutions of Problems 19–20, one can come up with the recursive sequence: bk = 1 + kbk−1 , b1 = 2, where n = bk + 1 vertices will guarantee that an n-gon colored in k colors will have a unicolor triangle. For instance, b2 + 1 = 6 and b3 + 1 = 17 correspond to the 6-gon and 17-gon in Problems 19–20. It is a different matter altogether to decide if bn + 1 is the desired minimal number of vertices. For n > 3, this is an open problem! The branch of mathematics that deals with problems like this is called Ramsey theory. Look it up online! ♦

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Session 9 Complex Numbers. Part I based on Tatiana Shubin’s session Sneak Preview. What are complex numbers? Are they numbers? Do they really exist? Where do they come from and why do we need them? In this session you will find answers to these questions. We will introduce and motivate most of the basic operations on complex numbers: addition, subtraction, multiplication, conjugation, and taking the modulus; present side-by-side three forms of complex numbers: Cartesian, polar, and Euler ; and among other things, interpret the standard transformations in the plane via the interplay between geometry and algebra. While trigonometry is not an absolute requirement in this session, a small amount of it will pop up in the last section on polar form, to urge us to learn and understand it more deeply. For instance, we will observe how complex numbers trivialize the proofs of the famous formulas for cos(α+β) and sin(α+β). For more advanced theory and applications, see Parts II–IV in future volumes.

1. A Problem from Geometry An intriguing aspect of mathematics in general is how a problem in one field turns out to have a solution using concepts in a seemingly unrelated field. For example, consider the following amazing plane geometry problem. A0

A1

A8

A2

A7 A6

A3 A4

A5

Figure 1. Nonagon problem via complex numbers Problem 1. Let A0 A1 A2 . . . A8 be a regular nonagon (9-sided polygon) inscribed in a unit circle (cf. Fig. 1). Prove that the product of the distances from one vertex to each of the other eight vertices is 9: |A0 A1 | · |A0 A2 | · |A0 A3 | · |A0 A4 | · |A0 A5 | · |A0 A6 | · |A0 A7 | · |A0 A8 | = 9. 179

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How can this problem have anything to do with complex numbers? The more advanced readers will “whip out” their analytic geometry skills; set up a coordinate system; use the Pythagorean Theorem or Law of Sines to calculate the lengths of the 8 segments above; and if successful up to this point, will multiply the results; and then get stuck in a horrendous mess of calculations, which will be almost impossible to simplify to the actual answer, 9. The expected failure of this attempt stems from using the “wrong” (Cartesian) coordinates: instead, the polar coordinates of complex numbers and roots of unity will solve this problem efficiently when we revisit it in a future Part II. Thus, as you work your way through the current session, keep this problem in mind: we are setting up the necessary background for its slick solution, along with numerous applications of complex numbers to many other areas of mathematics. By the way, when we eventually do solve Problem 1 via complex numbers, the method will be so powerful and illuminating that the generalization to any n-gon will become a routine exercise.

2. Some History 2.1. Impossible quadratics. For quite some time in the past, it had been well known that some equations do not have solutions. Take, for example, the equation x2 + 1 = 0. Finding its solutions is equivalent to finding the x-intercepts of the graph of the function f (x) = x2 + 1. But the graph is completely above the x-axis (cf. Fig. 2a), and hence there are no solutions. End of story! 2.2. Friendly cubics. But what if we are trying to solve a cubic equation? After some thought, the diagram in Figure 2b should convince you that any equation of the form x3 = bx + c must have at least one solution: this is because any line y = bx + c clearly intersects the graph of g(x) = x3 in at least one point. For example, Figure 2b indicates that each member of the family of parallel lines y = 23 x + c (where c is any real number) intersects the graph of x3 in 1, 2, or 3 points. y

y

y = x3

y = x2 + 1 x y = 23 x − 1 x

Figure 2. f (x) = x2 + 1

g(x) = x3 and lines y = 23 x + c

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Exercise 1 (Warm-up). Draw another family of parallel lines each of which intersects the graph of x3 in exactly 1 point. Can you draw a family of parallel lines some of which intersect the graph in 1 point and some in 3 points, but none in 2 points? Convince yourself that no matter which line you draw, it will intersect the graph of y = x3 in at least 1 point, and hence x3 = bx + c always has at least 1 solution. However, the cubic equation x3 = bx + c is of special form: it lacks the quadratic term x2 . Yet, it is not hard to see that every cubic equation will also have a real solution. Can you see why? To the reader skilled in algebraic manipulations, we offer


 

Problem 2 (Intermediate). Starting with a cubic equation x3 = ax2 + bx + c, “kill” the quadratic term. Specifically, find a substitution x → x + k such that the new equation is still cubic but has no quadratic term. Explain why the new and the old cubic equations have the same number of solutions. Apply this method to x3 = −6x2 −12x−6 and x3 = 3x2 +6x−18 to get rid of their x2 -terms. Can you guess some “obvious” roots of these equations? Here we used two very old “tricks”: PST 65. In order to eliminate an undesirable power in a polynomial, make a suitable linear change of the variable: x → dx + k. PST 66. In an equation, linear changes of the variable only shift the solutions but do not change the number of these solutions. In particular, a linear change of the form x → x + k shifts the involved graphs to the left or right and hence preserves the number of x-intercepts. To summarize, starting with a general cubic equation x3 = a1 x2 +b1 x+c1 , we can reduce it via Exercise 2 to a cubic equation of the form x3 = bx + c with the same number of solutions. Exercise 1 then tells us that any cubic equation will have at least one real solution. Always! 2.3. The famous cubic formula. Let us take a look at a particular equation of our special type: x3 = 15x + 4. An “obvious” solution is x = 4, The welli.e., we can factor out (x − 4): x3 − 15x − 4 = (x − 4)(x2 + 4x + 1). √ known√quadratic formula then yields two more solutions: x = −2± 4 − 1 = −2 ± 3. Yet the Cardano-Tartaglia formula, discovered independently by Scipione del Ferro and Niccolo Tartaglia and published by Girolamo Cardano in 1545 in his great book Ars Magna, tells us that   3 3 2 3 (1) x = q + q − p + q − q 2 − p3 is a solution of the equation x3 = 3px + 2q. Here, the relabeling of the coefficients b = 3p and c = 2q is done so that (1) is free of unsightly denominators.

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Applying this formula to our equation x3 = 15x + 4, with p = 5 and q = 2, we obtain a very strange expression:   √ √ 3 3 (2) x = 2 + −121 + 2 − −121. Is this yet another solution? But how can this be possible – we all know that a cubic equation cannot have more than 3 roots! What’s going on? 2.4. The debut of complex numbers. In 1569, Rafael Bombelli intro√ duced an “imaginary number,” −1, that should obey all of the familiar algebraic rules. He guessed that   √ √ √ √ 3 3 2 + −121 = a + b −1, while 2 − −121 = a − b −1. He then found that a = 2 and b = 1 do indeed satisfy these equations. Exercise 2 (Leap of faith). Assuming the usual algebraic rules, prove that √ √ √ √ (2 + −1)3 = 2 + −121 and (2 − −1)3 = 2 − −121, and conclude that it is reasonable to write   √ √ √ √ 3 3 2 + −121 = 2 + −1 and 2 − −121 = 2 − −1. √ √ Therefore, Bombelli concluded x = (2 + −1) + (2 − −1) = 4 in (2), which is not new to us: it is our old solution 4 of the equation x3 = 15x + 4. Yet, along the way, something new had been born in the 16th century: the complex numbers appeared on the scene of mathematics. However they remained somewhat “unreal” until Gauss (almost two and a half centuries later) introduced the idea of treating them as points in the plane. Meanwhile, Caspar Wessel and Jean-Robert Argand had published papers relating their independent discovery of the geometric representation of complex numbers slightly earlier than Gauss, but these papers passed unnoticed by the mathematical community. It took the authority of the “Prince of Mathematicians,” Carl Friedrich Gauss (1777–1855), to finally make these numbers completely acceptable and widely useful. So useful, in fact, that they led Riemann in 1859 to formulate his famous conjecture, the Riemann Hypothesis, which might be the most famous unsolved problem in mathematics today. In turn, the Riemann Hypothesis guided Hadamard and Vallée-Poussin to complete the proof of the Prime Number Theorem at the end of the 19th century. The reader who has successfully completed Problem 2 and who wants to independently try out the ideas presented here is further challenged with




Problem 3 (Advanced). Apply the Cardano-Tartaglia formula to the special cubic equations x3 = 2 and x3 = 9x − 10, and simplify the answers as much as possible. Using your newly found roots, factor these equations and calculate the remaining two roots via the quadratic formula. Translate all this to find all roots of the original equations x3 = −6x2 − 12x − 6 and x3 = 3x2 + 6x − 18 from Problem 2.

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183

(1.2, 2.3) = 1.2 + 2.3 i

2i i −2 −1 0 −i y −2i

x 1

2

3

z = (x, y) = x + yi

−3i

Figure 3. Bombelli’s L’Algebra and Definition of complex numbers

3. Complex Numbers via Geometry It is time to explain clearly what we mean by complex numbers. The following definition may be somewhat different from the one you may have encountered in school. Definition 1 (Complex Numbers). A complex number is a point z = (x, y) in the Cartesian plane where the x-axis is measured in ordinary, real units 1 and the y-axis is measured in different, imaginary units i (cf. Fig. 3b). The real number x is called the real part of z, and the real number y is called the imaginary part of z. We write, respectively, x = Re(z) and y = Im(z). For example, Figure 3b depicts the complex number z = (1.2, 2.3) with Re(z) = 1.2 and Im(z) = 2.3. Notice that every point on the x-axis is of the form (x, 0) and we can identify it with a real number x. For example, (1, 0) = 1. Likewise, (0, 1) = i, as i is the unit measure on the y-axis.

4. Basic Operations on Complex Numbers What can we do with complex numbers? Let’s learn first how to operate on them. The operations on complex numbers can be defined from algebraic or geometric viewpoints. The operations of addition and scalar multiplication, of taking the modulus and conjugation are presented from both viewpoints in tandem. In any given situation, one view might be more revealing than the other; so facility with both is a real asset. 4.1. Addition and scalar multiplication. Let z and w be two complex numbers. To define their sum z+w, construct a segment zv which is a parallel translate of the segment Ow (cf. the two thick segments in Figure 4a), i.e., shift Ow in a direction parallel to Oz until its beginning O coincides with z. Definition 2 (Addition). The sum z + w of two complex numbers z and w is defined to be the complex number v to which w has been shifted, as described above. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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−→ −→ When rays Oz and Ow are not collinear, the sum v = z + w can be thought of as the fourth vertex of the parallelogram Ozvw, as in Figure 4a. Note that v is located in the interior of the angle formed by the two rays. −→ −→ When the rays Oz and Ow happen to lie on the same line l (through O), the parallelogram Ozvw is degenerate – all collapsed onto l – but the sum z + w is still the complex number v on l, as defined earlier. (Draw the picture.) If z is a complex number and k is a real number, the product kz is a point on −→ the ray Oz and k times as far from the origin as z (cf. Fig. 4b). If k = 0, the product is the point at the origin; and if k < 0, the product is (−k) times as −→ far from the origin, and it is located on the ray opposite Oz. v =z+w t y O

←− Geometric −→ viewpoint

w z s

x

z = (x, y), w = (s, t) ⇒ z + w = (x + s, y + t)

←− Algebraic −→ viewpoint

kz

ky y

z

O

x

kx

z = (x, y), k real ⇒ kz = (kx, ky)

Figure 4. Addition and Scalar multiplication on complex numbers In particular, for real numbers x and y, we can use the above geometric definitions to verify readily the scalar multiplications x(1, 0) = (x, 0) = x and y(0, 1) = (0, y) = yi and the addition z = (x, y) = x(1, 0) + y(0, 1) = x + yi. We can also see that the two operations can be defined coordinatewise: if z = (x, y) and w = (s, t), then z + w = (x + s, y + t) and kz = (kx, ky). (Can you see why? Use Figure 4 and congruent/similar triangles to prove it.) Thus, for instance, (2, 3) + (−1, 4) = (1, 7) and 5(2, 3) = (10, 15). The set of complex numbers is universally denoted by C: ⎧ ⎫ ⎨ (x, y) : x, y ∈ R, ⎬ ordered pairs ⎩ ⎭ of real numbers

= C

⎧ ⎨

⎫ x + yi : x, y ∈ R, ⎬ sums of a real number = · ⎩ ⎭ and a real number times i

The second equality follows from our earlier observation that 1 = (1, 0), i = (0, 1) and (x, y) = x + yi. Corollary 1 (Cartesian form). For any real x and y, we can write z = (x, y) = x(1, 0) + y(0, 1) = x + yi. The notation “x + yi” is a source of confusion to many beginners. Recall that the real part x and the imaginary part y are both real numbers. Thus, Re(x + yi) = x, but Im(x + yi) = y, not yi. So, contrary to our intuition, the imaginary part y is not an imaginary number, it is a real number. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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185

If you have been paying attention in a number of previous sessions, it will come as no surprise that we would like to define the inverse operation of addition on complex numbers. Definition 3 (Subtraction). The difference z −w of two complex numbers z and w is defined in terms of addition and scalar multiplication by z − w = z + (−1)w. Exercise 3. If z and w are complex numbers, then what is z −w? Give both a geometric and an algebraic interpretation for z − w, just as in Figure 4a. The reader familiar with vector spaces 1 can readily verify that the operations of addition and scalar multiplication defined above satisfy all laws governing vector spaces over R; and hence we can think of complex numbers as vectors. More precisely, the set of complex numbers C can be thought of as equivalent to the real vector space R2 . 4.2. Modulus and conjugate. You may recall the absolute value of a real number, |x|, which measures the distance from x to 0, e.g., | − 6| = |6| = 6. Can this notion be extended in a reasonable way to complex numbers too? Definition 4 (Modulus). The modulus of a complex number z is the distance from the point z to the origin, and it is denoted by |z|. z y

|z| =  x

x2 + y 2

O

z ←− Geometric −→ viewpoint

O z

z = x + yi ⇒ |z| = x2 + y 2

←− Algebraic −→ viewpoint

z = x + yi ⇒ z = x − yi

Figure 5. Modulus and Complex conjugate It is well known that the equation |x| = 1 has only two real solutions: x = 1 and x = −1. What happens in the complex situation? Exercise 4. Find at least 4 complex solutions to |z| = 1. Now draw a graph with all solutions in C. How many solutions did you find? Following the definition of modulus, the reader will discover that the solutions are all the points on the circle of radius 1 centered at the origin: this means there are infinitely many solutions! If we want to be more algebraic, we need to create a formula for the modulus, |z|. By the Pythagorean Theorem, 1 See a session on vectors in a later volume.

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9. COMPLEX NUMBERS. PART I

for any particular point z = (x, y), we compute |z|2 = x2 + y 2 (cf. Fig. 5a) and arrive at the standard distance formula (3) |z| = x2 + y 2 . The question below will check your understanding of the new concepts and develop some basic formulas using moduli (the plural of modulus). Exercise 5. Let z, w ∈ C. (a) Calculate |3 − 4i|. Find at least 4 complex solutions to the equation |z| = 5. Then, write a general formula describing all complex solutions to this equation. (b) In Exercise 3, you interpreted z − w. Now, knowing the modulus, describe |z − w| both in algebraic and in geometric terms. (c) If k ∈ R, then prove |kz| = |k||z|. It is time to draw the reader’s attention to a simple but important fact: real numbers are also complex, e.g., 8 = (8, 0) = 8 + 0 i. Thus, R ⊂ C. Geometrically, the set of real numbers R can be thought of as the horizontal x-axis inside C. While this may help you solve parts of Exercise 6 faster, it will, more importantly, deepen your understanding of the next operation. Definition 5 (Conjugation). The conjugate of a complex number z = (x, y) is the point (x, −y), denoted by z. In other words, x + yi = x − yi: to form the conjugate of a complex number, replace the imaginary part y with its opposite −y. Geometrically, the conjugate of a complex number z is the reflection of the point z through the x-axis (cf. Fig. 5b). Exercise 6. Consider the operation of conjugation. (a) Find at least 4 complex numbers which remain fixed under conjugation. Describe now all z ∈ C such that z = z. (b) Is it true that |z| = |z| for all z ∈ C? Explain. (c) Find at least 4 complex numbers which go to their negatives under conjugation. Describe now all z ∈ C such that z = −z. (d) Prove that z + w = z + w and z − w = z − w for any z, w ∈ C. (e) If k is a real number, prove that kz = kz. It is easy to see that conjugation preserves the modulus (part (a)), fixes the real numbers (part (b)), and negates the numbers of the form (0, y) = yi (part (c)). The numbers yi are the points on the vertical y-axis and are called purely imaginary. While addition, scalar multiplication, and even modulus are natural operations to define on C – for one, we’ve seen them before in R – it is not immediately obvious why anyone would need conjugation. It seems out of place. Indeed, as we saw in Exercise 6(a), conjugation acts trivially on real Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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187

numbers: r = r for any r ∈ R, hence we don’t “see” conjugation on R. However, conjugation acts non-trivially on the complex numbers and it is widely applicable both in theory and in problem solving. For example, Definition 6 (Respect). Consider a set S with a binary operation ∗; i.e., to every pair (s1 , s2 ) of elements of S, we can associate an element s = s1 ∗s2 in S. Suppose f : S → S is a function such that f (s1 ∗ s2 ) = f (s1 ) ∗ f (s2 ) for every s1 , s2 ∈ S. Then f is said to respect 2 the operation ∗ on S. Thus, if f (z) = z is conjugation, Exercise 6(d) says that conjugation respects addition and subtraction on C: for all z, w ∈ C, f (z + w) = z + w = z + w = f (z) + f (w) and f (z − w) = z − w = z − w = f (z) − f (w). Not surprisingly, this can be shown in two ways: using the algebraic definition of conjugation and using the geometric interpretation of conjugation. We shall see in this session and later ones that conjugation also respects multiplication and division of complex numbers.

5. Complex Multiplication 5.1. Why talk about this operation? Didn’t we define it earlier already? Let us be careful: what we defined before was scalar multiplication, which involved a real and a complex number, i.e., kz for k ∈ R and z ∈ C. Now we want to be able to multiply two complex numbers z and w, while obeying the well-known algebra operations on reals, so that C would be as similar as possible to R. 5.2. Is componentwise multiplication “wise”? For example, how should we multiply z = (2, 3) = 2 + 3i and w = (4, 5) = 4 − 5i? Following suit from addition, subtraction, and scalar multiplication, we could try defining the product componentwise: (4)

?

(2, 3) ∗ (4, 5) = (2 · 4, 3 · 5) = (8, 15) = 8 + 15i.

But does this give us what we really want? For example, under this rule, we would get 2 ∗ (3i) = (2, 0) ∗ (0, 3) = (0, 0) = 0: oops, the nonzero complex numbers 2 and 3i multiply to 0? Such a thing never happens in R – can we avoid it in C? Moreover, if we obey in C the usual distributivity, associativity, and commutativity laws of R, we would get two different answers for the proposed product (4): 8+15i = (2, 3)∗(4, 5) = (2+3i)∗(4−5i) = 8−10i+12i−15i2 = 8+2i−15i2 . 2 Despite the seeming irrelevance of Definition 6, it leads to the key concept of ho-

momorphism in just about all areas of mathematics, e.g., in abstract algebra, analysis, topology, differential geometry, algebraic geometry, etc.

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9. COMPLEX NUMBERS. PART I

Canceling stuff from the beginning and the end above results in 13i = −15i2 , i.e., i = −13/15. But then all complex numbers would be of the form z = x + yi = x − 13 15 y. . . . They would be real! This really doesn’t make any sense: if C = R, why did we bother with this session at all? 5.3. Where does i fit in the picture? OK, we are convinced by now that the componentwise product doesn’t do us any good in C. We need a better definition of complex multiplication. So, let’s start all over again with the root of the problem. We should be able to multiply i and i, but there is nothing in R to tell us what i2 should be! Returning to the very first question in this session, solve x2 + 1 = 0, we get the idea: we want i2 = −1 so that i would be one solution to this “impossible” quadratic. As mentioned above, we also want the usual algebraic rules to still hold: associative and commutative laws for both addition and multiplication, and also, the distributive law. Back to our example, multiplying out yields: (2 + 3i)(4 − 5i) = 8 + 12i − 10i − 15i2 = 8 + 2i − 15(−1) = 23 + 2i. Now the algebraic definition of complex multiplication can be easily derived. Definition 7 (Multiplication). The product of two complex numbers z = x + yi and w = s + ti is given by (5)

zw = (x + yi)(s + ti) = (xs − yt) + (xt + ys)i.

Verify that (5) is indeed what you will get if you multiply (x + yi) and (s + ti). The true challenge is to find a geometric interpretation of this definition, just as we did for the previous five operations. 5.4. Visualizing multiplication by i . Let us consider first the special case when one of our numbers is just i, and try to visualize the product of z = x + yi and w = i. Exercise 7. Calculate iz for z = 2 + 3i and plot both numbers z and iz on the same graph. Can you visualize a geometric transformation that can be performed on point z in order to get the point iz? Describe this transformation in words. In the general case, we calculate iz = i(x + yi) = −y + xi and plot everything as in Figure 6. Notice that the two dashed rectangles are congruent. Indeed, under a 90◦ rotation in the counterclockwise direction about the origin, the rectangle in the first quadrant is rotated into the rectangle in the second quadrant (while the origin remains fixed), and x is carried to xi as corresponding vertices (as well as yi → −y and z → iz). We see that Corollary 2. Multiplying a complex number z by i rotates z counterclockwise by 90◦ about the origin. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

5. COMPLEX MULTIPLICATION

iz

189

xi z

yi

−y

O

x

Figure 6. Geometric view of complex multiplication: iz The reader should check that Corollary 2 holds true even if z is in the second, third, or fourth quadrant. We can now extend Exercise 5(c) to include multiplication by i:



Exercise 8. Let z ∈ C and t ∈ R. Prove that |iz| = |z| and, more generally, |tiz| = |t||z|. If you have trouble with this or any other exercise in this session, PST 67. Plug in several specific numbers and verify that the exercise is true for your numerical examples. Then switch from numbers to letters and try to repeat the same verification in the general case, thereby solving the problem. For instance, in Exercise 8 we could plug in z = 3 + 4i and t = −7, and verify algebraically: |iz| = |i(3 + 4i)| = |3i − 4| = (−4)2 + 32 = 5 = 32 + 42 = |z|, and |tiz| = | − 7i(3 + 4i)| = | − 7 · 3i + 7 · 4| = (−7 · 3)2 + (7 · 4)2 = 72 (32 + 42 ) = 7 32 + 42 = | − 7| · 5 = |t||z|. You shouldn’t be deceived that this is the end: now replace 3, 4, and −7 by a, b, and t, and verify that the calculations go through without any trouble. However, the more insightful approach to solving Exercise 8 is the geometric interpretation of multiplication by i: when rotating about the origin (by whatever angle), the distance to the origin is preserved, and hence the modulus does not change, i.e., |iz| = |z|. Similarly, rescaling by a real number t and rotating about the origin yields the same modulus as rescaling by |t| does, i.e., |tiz| = |t||z|.
5.5. The challenge of multiplying any two complex numbers. Here are two “arbitrary” complex numbers: z = 3 + 2i and w = 4 + 5i. Let’s multiply them as an example. Draw everything as we go along. We can break down the total product as wz = (4 + 5i)z = 4z + 5iz. We know how to draw both 4z and 5iz: rescale z by 4, or rescale z by 5 and rotate 5z by 90◦ counterclockwise about the origin. We also know how to add

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9. COMPLEX NUMBERS. PART I

4z + 5iz: the parallelogram with vertices O, 4z, 4z + 5iz, and 5iz is actually a rectangle (why?) one of whose diagonals passes through O and 4z + 5iz. The question is: how do we describe wz = 4z +5iz in simple geometric terms that involve symmetrically z and w? The calculation below finds such an interpretation. Its argument constitutes the heart of this section, and it may be hard to follow by the beginner without initial guidance. Take two truly arbitrary complex numbers, z = x + yi and w = s + ti, and consider the product wz. Break it apart just as in the example above: wz = (s + ti)z = sz + t(iz) (with s, t ∈ R). Label the involved points by z = P , w = Q, sz = M , tz = N , tiz = R, and wz = T (cf. Fig. 7). Since ∠RON is right (by Corollary 2) and ROM T is a parallelogram (as T = R + M ), conclude that ROM T is a rectangle. T = wz R = tiz

O

tz Q = w sz N z M P X

Figure 7. Geometric view of complex multiplication: wz Exercise 9. Let X be the foot of the perpendicular from Q to the x-axis, i.e., X corresponds to s = s + 0i ∈ R. Prove that T M O ∼ QXO. ∼ ∠QXO since they are both right angles. Next, Proof: Note that ∠T M O = look for equal ratios of adjacent sides. Taking into account that M T = OR, OR |tiz| Ex. 8 |t||z| |t| QX MT = = = = · = OM OM |sz| |s||z| |s| OX The last two fractions measure the slope of the line OQ and take into account that w = s + ti. By SAS, T M O ∼ QXO.
Reaping the effects of this similarity, we obtain that ∠M OT ∼ = ∠XOQ. −→ The latter is the angle which Ow makes with the x-axis. Since OM = s∈R |sz| = |s||z| = OX|z|, the scaling factor between the two triangles is |z|. Therefore (6)

|wz| = OT = OQ|z| = |w||z|.

Equation (6) means that to get from the modulus of z to the modulus of wz, one needs to rescale by the modulus of w. Coincidentally, this extends Exercise 8 to any two complex numbers. We summarize these findings as follows. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

5. COMPLEX MULTIPLICATION

191

Corollary 3. Let z, w ∈ C, and let w make angle µ with the x-axis. Then (a) |wz| = |w||z|, i.e., modulus respects complex multiplication; (b) the product wz can be obtained geometrically by rotating z about the origin counterclockwise through µ and rescaling by a factor of |w|. Exercise 10. Let z = 2 + 3i and w = 4 + 5i. Verify part (a) of Corollary 3 algebraically and part (b) geometrically. Exercise 11. Let w = (1 + i)(1 + 2i)(1 + 5i). (a) Write w in the Cartesian form w = x + yi for some x, y ∈ R. (b) Calculate the modulus |w| and the conjugate w in two ways: using the original expression for w and then (a). Compare your answers.




Problem 4. We say that segment AB is an SOS hypotenuse 3 if it is the hypotenuse √ of a right triangle whose legs have integer lengths; equivalently, |AB| = n2 + m2 for some n, m ∈ N. Show that multiplying two SOS hypotenuses yields another SOS hypotenuse. √ √ Sketch of Proof: Let |AB| = a2 + b2 and |CD| = c2 + d2 for some a, b, c, d ∈ N. Consider the complex numbers z = a + bi and w = c + di: |AB| · |CD| = |z||w| = |zw| = |(a + bi)(c + di)| = |(ac − bd) + i(ad + bc)| (ac − bd)2 + (ad + bc)2 = |OE| = where E = zw = (ac − bd) + i(ad + bc). Thus, |OE| is an SOS hypotenuse.

♦

We can get extra mileage from the above solution. Taking into account that |zw| = |zw| is also true (why?), derive the formulas: Exercise 12. For any real a, b, c, d: (a2 + b2 )(c2 + d2 ) = (ac − bd)2 + (ad + bc)2 = (ac + bd)2 + (ad − bc)2 . We could use brute force to do the above proofs by ordinary multiplication, but the meaning of these formulas is really best encoded by the fact that conjugation preserves modulus, and modulus respects complex multiplication.




Problem 5 (For the die-hards). Find the modulus of 2007 
1 + k + k 2 + i 
1 + i 
3 + i 
7 + i 
4030057 + i  ··· . = w= 1 + k2 1 2 5 4028050 k=0

Hint: Assuming that 2007 has no special meaning in this problem, replace it by n, find a pattern for the product w as n = 0, 1, 2, 3, etc., form a conjecture, and prove it by induction. Then you can find the modulus |w| in two different ways: using your simplified formula for w or using the original formula for w and |z0 z1 · · · zn | = |z0 ||z1 | · · · |zn |. For an extra challenge, do it both ways and make sure that your answers match. ♦ 3 SOS is short for “sum of squares,” not a distress call for help.

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9. COMPLEX NUMBERS. PART I

6. Another Form of Complex Numbers Corollary 3(b) described the product wz geometrically, but not quite symmetrically with respect to w and z. Another form of the complex numbers will complete this task beautifully. 6.1. Polar coordinates. Complex numbers are points in the plane, so we need two coordinates to locate them. So far we have used the usual Cartesian coordinates x and y. However, there are different ways to choose such coordinates. For instance, a point P = z can be located alternatively −−→ by its distance r = |z| = OP from the origin and the angle θ which OP makes with the positive real axis (cf. Fig. 8a). Exercise 13. Locate the complex number z1 if r = 5 and θ = 120◦ . Conversely, find r and θ if z2 = 3 + 3i. Draw pictures in both cases. Definition 8 (Polar form). The pair (r, θ) is called the polar coordinates of the complex number z.

P = (r, θ) µ

r = |z| 

w = (1, β) z = (1, α)

wz = (1, α + β)

y = r sin θ

θ X ↑ O x = r cos θ

O

Q = (r1 , µ)

Figure 8. Polar form of complex numbers Figure 8a depicts two points P (r, θ) and Q(r1 , µ) in the first and third quadrants, respectively. While the distance r is measured in a unique way, the angle θ can be measured in many different ways, as Figure 8a shows for point P : θ or θ + 2π. Similarly, in Exercise 13 for z2 = 3 + 3i, θ could be any of the angles 45◦ + 360◦ k = π4 + 2πk (for any integer k), since any of −−→ these angles will determine the same ray OP . Let arg(z) signify any of these angles, measured in radians 4. For example, arg(3 + 3i) = π4 + 2kπ for any k ∈ Z. Formally, we write θ ≡ arg(z) (mod 2π) to mean that θ and arg(z) are defined and equal “up to an integral multiple of 2π radians.” Thus, z in polar coordinates is written as z = (|z|, arg(z)). 4 2π radians correspond to 360◦ degrees; ‘arg’ is short for argument.

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6. ANOTHER FORM OF COMPLEX NUMBERS

193

8π 4π Exercise 14. Explain why (5, 2π 3 ), (5, 3 ), and (5, − 3 ) are equivalent polar representations of the same complex number z1 from Exercise 13.

Using the polar form, we can now rewrite the geometric interpretation in Corollary 3(b) in a symmetric form: Corollary 4. The product of z = (|z|, θ) and w = (|w|, µ) equals zw = (|z||w|, θ + µ). In other words, in polar coordinates complex multiplication multiplies the moduli: |zw| = |z||w|, and adds the arguments: arg(zw) = arg(z) + arg(w) (mod 2π). Exercise 15. Write z1 and z2 from Exercise 13 in polar form, calculate z1 z2 in polar form, and verify that Corollary 4 describes this product z1 z2 correctly. Draw a picture as usual. 6.2. Trigonometry detour. Having two coordinate representations of a complex number z, the Cartesian and the polar, we would like to move freely from one to the other. This can be done with a little trigonometry. Let (r, θ) be the polar coordinates for P = z = x + yi. As before, let X be the foot of the perpendicular from P to the x-axis (cf. Fig. 8a). Then r = |z| = x2 + y 2 is the hypotenuse of OP X, while the legs OX = x and XP = y enter in the well-known ratios: y x (7) ⇒ x = r cos θ and y = r sin θ cos θ = and sin θ = r r (8) ⇒ z = x + yi = r cos θ + ir sin θ = r(cos θ + i sin θ). Thus, the modulus formula r = x2 + y 2 and the first half of (7) transform Cartesian to polar coordinates, while the second parts of (7) and (8) do just the opposite. Note that multiplying any complex number z by 1(cos θ + i sin θ) results in rotating z through an angle θ about the origin. √

Exercise 16. Consider the pair ( 23π , π2 ). If this pair represents (a) the polar coordinates of z1 ∈ C, find the Cartesian coordinates of z1 ; (b) the Cartesian coordinates of z2 ∈ C, find the polar coordinates of z2 . Compare the numbers z1 and z2 . Are they equal? Explain and draw pictures. Conjugation and polar form mesh together well. For any complex number z = r(cos θ + i sin θ), we have z = r(cos(−θ) + i sin(−θ)) = r(cos θ − i sin θ). Therefore, conjugation in polar form can be written as (r, θ) = (r, −θ), which coincidentally looks identical to the Cartesian form of conjugation, (x, y) = (x, −y). The polar form of conjugation turns out to be useful in problem solving. Exercise 17. Prove that conjugation respects complex multiplication: zw = z w for any w, z ∈ C. Use first Cartesian and then polar coordinates. Is there a fast geometric proof? Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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9. COMPLEX NUMBERS. PART I

6.3. Applications of complex numbers to trigonometry. If you have ever wondered how to prove the fundamental formulas for sine and cosine of a sum of two angles, you can now derive these formulas as direct consequences of our geometric interpretation of complex multiplication. Problem 6. For any α, β ∈ R, show that cos(α+β) = cos α cos β−sin α sin β and sin(α + β) = sin α cos β + cos α sin β. Proof: Take points z and w on the unit circle such that z = (1, α) and w = (1, β) in polar coordinates (cf. Fig. 8b). Multiplying them yields point zw = (1, α + β) on the unit circle. Alternatively, in Cartesian coordinates: z = cos α + i sin α, w = cos β + i sin β, and zw = cos (α + β) + i sin (α + β).

(9)



But the algebraic definition of complex multiplication in (5) computes this product differently: zw = (cos α + i sin α)(cos β + i sin β), and hence (10)

zw = (cos α cos β − sin α sin β) + i(sin α cos β + cos α sin β).

PST 68. Two complex numbers are equal if and only if they have equal real parts and equal imaginary parts. Thus, equating the real and imaginary parts for zw in the two expressions (9)–(10) yields the desired trigonometric formulas.
6.4. Euler form. There is yet a third “coordinate” form of complex numbers. In the advanced sessions, we will show that Theorem 1 (Euler’s formula). For any real number r and any angle θ, r(cos θ + i sin θ) = reiθ . If you haven’t met e before, for now it suffices to say that e is an irrational number like π and just as useful and famous as π. The above formula requires, among other things, an explanation of what it means to raise the real number e to a complex power iθ. If you are uncomfortable with the Euler form, you can successfully skip it until later volumes. Euler’s formula proves in yet another way the geometric interpretation of complex multiplication. If z = |z|eiθ and w = |w|eiµ , then zw = |z||w|eiθ eiµ = |z||w|ei(θ+µ), i.e., zw = (|z||w|, θ + µ) in polar coordinates.


7. Summary: What Have We Learned? Let’s compare complex multiplication in the three coordinate forms: Variable z w zw

Cartesian form x + yi s + ti

↔ ↔

(xs − yt) + (ys + xt)i ↔

Polar form (|z|, θ) (|w|, µ)

↔ ↔

Euler form |z|eiθ |z|eiµ

(|z||w|, θ + µ) ↔ |z||w|ei(θ+µ)

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7. SUMMARY: WHAT HAVE WE LEARNED?

195

Here is a question to test your understanding of the operations on C.




Problem 7. Interpret the four well-known transformations in the plane – translation, rotation, dilation, and reflection – in two ways: (a) geometrically, as a composition of basic operations on C; (b) algebraically, as a formula involving complex numbers. Partial Solution: By definition, a complex number z = x + yi corresponds to a point Z = (x, y) in the plane, so we can write Z = z. To get you started on the right foot, let’s demonstrate how to do this problem for a rotation ρ about some point P = p through angle α. We denote by ρ(z) the image of a point Z = z (cf. curvy dashed arrow in Figure 9). Applying ρ directly does not correspond to a basic operation on C because the center of rotation P may not be the origin. To correct this, we proceed in 3 steps. −−→ Step 1. Translate everything in the direction of P O so that P goes to O. Denote this translation by t (cf. straight thick arrows in Figure 9). Algebraically, t(z) = z − p (why?) and t(p) = p − p = O. Step 2. Rotate about the origin O through α. Denote this rotation by ρα (cf. curvy solid arrow). We have seen that ρα is simply multiplication by u = cos α + i sin α, so ρα (t(z)) = u(z − p) = v as in Figure 9. −−→ Step 3. Translate back along OP to send O to P . Since this translation is the inverse of t, we denote it5 by t−1 . Thus, t−1 (w) = w+p for every point w. The straight dashed arrow corresponds to t−1 (v) = v + p. ρ(z) p

ρ t−1

t

z

v t(p) = O

t

ρα t(z)

Figure 9. Rotation using complex multiplication Putting everything together, our original rotation ρ is the composition of t, ρα and t−1 : t

ρα

t−1

z −→ z − p −→ u(z − p) −→ u(z − p) + p = ρ(z). Formally, ρ = t−1 ◦ ρα ◦ t, and the desired formula is ρ(z) = u(z − p) + p. ♦ To describe this type of algebraic formula more clearly, we rewrite it as ρ(z) = uz + p(1 − u) = uz + c (with c = p(1 − u)), and summarize: 5 This is in line with our notation from the session Rubik’s Cube, Part I.

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196

9. COMPLEX NUMBERS. PART I

Corollary 5. Rotations in the plane are given by linear polynomials f (z) = uz + c with complex coefficients u and c, where |u| = 1. If u = 1, then the center of rotation is p = c/(1 − u) and the angle is arg(u). If u = 1, then c = 0 and the rotation is the identity f (z) = z. The above solution already contains the answer for translations in Problem 7 (what is it?). The reader should continue in the same manner and interpret dilations (with arbitrary center P and ratio k ∈ R) and reflections (across an arbitrary line l). Drawing pictures along the way is essential. As we move through later volumes of the book, the numerous applications of complex numbers will start popping up more often and quite unexpectedly in many areas of mathematics. Parts II–IV, for instance, will demonstrate how hard problems in geometry and algebra easily “fall under the spell” of C.

8. Hints and Solutions to Selected Problems Exercise 1. The answer to the given question is No. Geometrically: within a family of parallel lines, the transition from the lines intersecting x3 in three points to the lines intersecting x3 in a unique point always goes through a line that has 2 common points with x3 (just as Figure 2 illustrates). The formal proof of this fact requires more advanced concepts of calculus, but your geometric intuition should be helpful in visualizing the situation. ♦ Problem 2. We are looking for the most convenient real number k to do the required job. So we make the change x → x + k in x3 = ax2 + bx + c, expand, and collect like terms: (x + k)3 = a(x + k)2 + b(x + k) + c ⇒ x3 + (3k − a)x2 + x +  = 0, where ’s are some constants, which can be calculated but are of no importance to us at the moment. To “kill” the x2 -term, we must have its coefficient 3k − a = 0, i.e., k = a3 . Thus, the substitution x → x + a3 converts the given general cubic to the special form x3 = b1 x + c1 . Each solution r of this simplified equation gives rise to a solution x = r + a3 of the original equation. Geometrically, the replacement of x → x + a3 in the polynomial f (x) = x3 − ax2 − bx − c translates the graph of f to the left by a3 . As for the equation x3 = −6x2 − 12x − 6, the change x → x − 2 simplifies it to x3 = 2; similarly, x → x+1 simplifies x3√= 3x2 +6x−18 to x3 = 9x−10 (verify!). One obvious real root of x3 = 2 is 3 2, which translates√to the root √ 3 2 − 2 of the original equation; if you don’t believe it, plug 3 2 − 2 into x3 = −6x2 − 12x − 6 and check that it is indeed a root of it. Guessing a root of x3 = 9x − 10 is not as obvious, but checking small integer possibilities for x will quickly show that 2 is a root; translating back to the original equation, 2 + 1 = 3 is a root of x3 = 3x2 + 6x − 18. (Check that it is a solution!)
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8. HINTS AND SOLUTIONS TO SELECTED PROBLEMS

197

Exercise 2. The formula (a + b)3 = a3 + 3a2 b + 3ab2 + b3 is useful to prove and remember. It holds for √ any numbers a and b, real or complex. As a 2 2 “leap of faith,” assume that ( √ √−1) = −1 (we write this as i = −1). Hence √ √ √ 3 2 ( −1) = ( −1) −1 = −1 −1. Now, substituting a = 2 and b = −1 into the formula for (a + b)3 gives: √ √ √ √ (2 + −1)3 = 23 + 3 · 22 −1 + 3 · 21 ( −1)2 + ( −1)3 √ √ = 8 + 12 −1 + 6(−1) + (−1) −1 √ √ = 2 + 11 −1 = 2 + −121. √ √ After taking cube roots, we conclude that 2 + −1 = 3 2 + −121. √ √ To show the second statement, 2 − −1 = 3 2 − −121, proceed in a similar fashion by proving and using a corresponding formula for (a − b)3 . ♦ Problem 3. For √ x3 = 2, the Cardano-Tartaglia formula (with p = 0 and 3 q = 1) yields x = 2, which was our obvious guess in Problem 2. To factor √ 3 (x − 2) from x3 − 2√= 0 requires the formula a3 − b3 = (a − b)(a2 + ab + b2 ) with a = x and b = 3 2: √ √ √ √ 3 3 3 3 x3 − 2 = x3 − ( 2)3 = (x − 2)(x2 + 2x + 4). √

√ The quadratic formula6 then yields two more roots: x = 3 2 − 12 ± 23 i .

by subThe roots of the original equation x3 = −6x2 −12x−6 are obtained √

√ √ 3 3 1 tracting 2 from the newly-found roots, giving 2−2 and 2 − 2 ± 23 i −2. The situation with x3 = 9x − 10 is a bit more interesting: the CardanoTartaglia formula (with p = 3 and q = −5) yields   √ √ 3 3 x = −5 + −2 + −5 − −2.

Is this different from the guessed root x = 2 in Problem 2? It turns out to be the same. Indeed, just as in Exercise 2, verify the simplifications √ √ √ √ (1 + 2i)3 = −5 + 2i and (1 − 2i)3 = −5 − 2i, √ √ √ √ 3 3 −5 − 2i = 1 − 2i, and derive write −5 + 2i = 1 + 2i and   √ √ √ √ 3 3 x = −5 + −2 + −5 − −2 = (1 + 2i) + (1 − 2i) = 2. Now we need to factor (x − 2) from x3 − 9x + 10. Those who remember how to do long division with polynomials will be able to write x3 − 9x + 10 = out two more (x − 2)(x2 + 2x − √ 5). The quadratic formula then will spit 3 2 roots: x = −1 ± 6. The roots of the original √ equation x = 3x + 6x − 18
are obtained by adding 1, i.e., they are 3, ± 6. Exercise 3. To find z − w geometrically, first locate −w (the opposite of w) and then z − w = z + (−w) (cf. Fig. 10a). By the definition of addition, this is the fourth vertex of the parallelogram with vertices at z, 0, and −w, 6 In Part II, de Moivre’s formula will quickly and painlessly yield the roots of x3 = 2.

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9. COMPLEX NUMBERS. PART I

−−−−→ −→ located in the interior of the angle formed by Oz and O(−w) (cf. the dotted segment in Figure 10a). To view the difference z−w algebraically, let z = (x1 , y1 ) and w = (x2 , y2 ). Then z −w = (x1 −x2 , y1 −y2 ): the difference is taken componentwise.
z z−w O −w

i

w −1

1 O −i

Figure 10. Subtraction z − w and Solutions to |z| = 1 Exercise 4. Four fairly obvious solutions are 1 = (1, 0), i = (0, 1), −1 = (−1, 0), and −i = (0, −1) (cf. Fig. 10b). Let z = x + iy. Then |z|2 = 1 = x2 + y 2 , which is the set of all the points on the unit circle centered at the origin.
Exercise 5. (a) By symmetry, in addition to 5, −5, 5i, and −5i, there are four more obvious solutions: (3, 4), (3, −4), (−3, 4), and (−3, −4). Also, interchanging the x- and y-coordinates does not change the modulus, i.e., (4, 3), (−4, 3), etc., are also solutions. In general, the circle x2 + y 2 = 25 describes all solutions.
(b) Geometrically, |z − w| is the length of the segment joining z and w (cf. Fig. 10a). This segment is parallel and equal in length to the segment joining O and w − z. Algebraically, if z = (x1 , y1 ) and w = (x2 , y2 ), then |z − w| = (x1 − x2 )2 + (y1 − y2 )2 . We recognize this as the distance formula for two points in the plane.




(c) If k ∈ R then √the modulus |k| is the same as the absolute value of k, and, as such, |k| = k 2 . For z = (x1 , y1 ), we have |kz| = |(kx1 , ky1 )| = (kx1 )2 + (ky1 )2  √  = k 2 (x21 + y12 ) = k 2 x21 + y12 = |k||z|.
Exercise 6. Reasoning geometrically is instructive here. (a) The conjugate of z is the reflection of z through the real axis. Thus, every real number is its own reflection, i.e., z = z for all z ∈ R. In particular, the origin O is fixed.
(b) Since reflections preserve distances between points, for any z ∈ C the
distance from O to z equals the distance from O to z, i.e., |z| = |z|. (c) If z is purely imaginary, it lies on the y-axis, and hence conjugation
will carry it to its opposite −z, i.e., z = −z. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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199

(d) We reflect z, w, and z + w through the real axis, to land on z, w, and z + w, respectively. Figure 11a depicts the two resulting parallelograms, which are reflections of each other through the real axis. The diagonal of the lower parallelogram can be obtained in two ways: either reflect the diagonal of the upper parallelogram (z + w → z + w), or equivalently, add the sides of the lower parallelogram (z + w). We conclude that z + w = z + w and conjugation respects addition. You can prove part (e) in a similar way.
The reader is encouraged to verify all five of the given statements algebraically, using the definition of conjugation: x + yi = x − yi. (4 + 5i)z = wz z+w w z O

i5z

z w

4z

5z

θ

w z+w =z+w

µz

O

Figure 11. Conjugating a sum and Multiplying complex numbers Exercise 7. i(2 + 3i) = 2i + 3i2 = −3 + 2i. Notice that (2, 3) and (−3, 2) have the same modulus and that (−3, 2) is a 90◦ -rotation of (2, 3). Consult Figure 6 for the general case.
√ √ = 533. Exercise 10. (a) |wz| = |(4 + 5i)(3 + 2i)| = |2 + 23i|√= 4 + 529 √ Take √ the modulus √ of each factor√separately: √ √ |w| = 16 + 25 = 41 and
|z| = 9 + 4 = 13. So |w||z| = 41 13 = 533 = |wz|. (b) Graph z = (3, 2) and w = (4, 5). Now graph 4z = (12, 8), 5z = (15, 10), and 5zi = (−10, 15). Construct 4z + 5zi by completing the parallelogram (a rectangle, since 5zi is a 90◦ -rotation of 5z). The ratio of its sides is |4z| : |i5z| = 4 : 5, which is the same as the ratio of the sides formed by dropping perpendiculars from w = (4, 5) to the axes. This means that these two rectangles are similar with a scaling factor |z| (why?). So the angles formed by a diagonal and a corresponding side in each rectangle are equal: θ = µ (as depicted in Figure 11b) and |wz| = |z||w|.
Exercise 11. (a) w = (1 + i)(1 + 2i)(1 + 5i) = −16 − 2i.
√ √ √ √ 2 2 (b) |w| = (−16) +(−2) = 260 = 2 5 26 = |1+i||1+2i||1+5i|.
Exercise 12. As in Problem 4, |z||w| = |zw| = (ac − bd)2 + (ad + bc)2 square ⇐⇒ (a2 + b2 )(c2 + d2 ) = (ac − bd)2 + (ad + bc)2 . Replace |zw| with |zw|: |(a − bi)(c + di)| = |(ac + bd) + (ad − bc)i| = (ac + bd)2 + (ad − bc)2 , Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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9. COMPLEX NUMBERS. PART I

which leads to (a2 + b2 )(c2 + d2 ) = (ac + bd)2 + (ad − bc)2 , the other formula. To paraphrase, if each of two numbers is a sum of two squares, then the product of these two numbers can also be written as a sum of two squares (possibly in more than one way).
Problem 5. Following the hint, let wn correspond to the product with 2007 replaced by n. Then brute-force calculation yields w0 = 1 + i, w1 = 1 + 2i, w2 = 1 + 3i, w3 = 1 + 4i. Evidently, we should conjecture that in general the product is wn = 1 + (n + 1)i. The method of induction kicks in here: we assume that wn−1 = 1 + ni for some n ≥ 1. We need to show that wn =1 + (n + 1)i. Indeed, 



 3+i 7+i 1+i 1 + n + n2 + i ··· wn = 1 2 5 1 + n2   


wn−1


 1 + n + n2 + i n+i = (1 + ni) 1 + = wn−1 1 + n2 1 + n2 i(1 + n2 ) n + i + n2 i − n = 1 + ni + = 1 + (n + 1)i. = 1 + ni + 1 + n2 1 + n2 In our problem, n = 2007 so the product is w2007 = 1+2008i. From here, the reader should be able to compute |w| and complete the problem. ♦ Exercise 13. Let z1 = (x, y). If √r = 5 and θ = 120◦ , then x = 5 cos 120◦ and y = 5 sin 120◦ . So z1 = (− 52 , 5 2 3 ). Converting z2 = 3 + 3i to polar form, √ √ we get r = x2 + y 2 = 32 + 32 = 3 2 and θ = 45◦ , since an isosceles right triangle is formed (draw the picture!).
Exercise 14.

2π 3

+ 2π =

8π 3

and

2π 3

− 2π = − 4π 3 .




√ √ Exercise 15. z1 = (5, 120◦ ) and z2 = (3 2, 45◦ ), so z1 z2 = (15 2, 165◦ ).
Exercise 16. Converting z1 = ( √ 3π 2 )

√ 3π π 2 , 2)

from polar to Cartesian coordinates √

(why so easily?). Converting z2 = ( 23π , π2 ) from Cartesian to gives (0, polar coordinates takes a bit more effort:
 √   3π 2  π 2 + 2 , arctan √13 = (π, π6 ). 2 If you are unfamiliar with the function arctan x, draw a picture of z2 in Cartesian coordinates and verify that you get a 30◦- 60◦- 90◦ triangle, which leads to arg(z2 ) = π/6. Since each point in the plane has unique Cartesian coordinates, it is clear that z1 = z2 . Moreover, z1 is purely imaginary, while ♦ z2 is not. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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201

Exercise 17. Let z = (a, b) and w = (c, d) in Cartesian coordinates. Then zw = (ac − bd, ad + cb) = (ac − bd, −(ad + cb)) = (ac − bd, −ad − cb) ⇒ z w = (a, −b)(c, −d) = (ac − bd, −ad − cb) = zw. For a more inspiring solution, let z = (|z|, θ) and w = (|w|, µ) in polar coordinates. Then zw = (|z||w|, θ + µ) = (|z||w|, −(θ + µ)) ⇒ z w = (|z|, −θ)(|w|, −µ) = (|z||w|, −θ−µ) = zw. This leads to the geometric observation that zw is on the (θ + µ)-ray and z w is on the (− θ − µ)-ray. Since these two rays are reflections of each other through the real axis and since |z||w| = |z||w|, then zw = z w: a much more satisfying explanation.
Problem 7. The answer for translations was already given, more or less, in the solution on rotations. Suppose we want to translate the plane along the −−→ direction P Q; in other words, we want point P = p to move to Q = q. If w = q − p, then every point z will be translated to t(z) = z + w.
As for dilations, suppose we want to dilate with center P = p and ratio k ∈ R, i.e., we want to send point Z = z to d(Z). Translate everything along −−→ P O so that P lands on O, then dilate with center O and ratio k, and finally −−→ translate back along OP . Algebraically, t

∗k

t−1

z −→ z − p −→ k(z − p) −→ k(z − p) + p = kz + p(1 − k).




Corollary 6. Translations in the plane are given by the linear functions t(z) = z + w for any fixed w ∈ C. Dilations in the plane are given by the linear functions d(z) = kz + p(1 − k) for any fixed p ∈ C (the center ) and for k ∈ R (the ratio). Reflections across lines are a bit more intricate, and we leave them for the reader to wrestle with. We shall revisit reflections in a future Part IV. ♦

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Session 10 Stomp. Games with Invariants based on

Sam Vandervelde’s session

Sneak Preview. The reader will encounter games and puzzles with exotic names like Gopher Gun, Escape of the Clones, and Pointless Machine, which are connected by the fundamental problem solving technique of invariants. When organized and led suitably, the material here can provide a highly interactive environment where group work, individual investigation, and class discussions flow seamlessly from one to the other; and indeed the live BMC Stomp session involved with equal enthusiasm everyone from the youngest middle schoolers to the high school seniors. Although the concept is not new, the game of Stomp was created specifically for the BMC session, along with all the related problems. Stomp has appeared on a few other occasions – most notably at the EPGY Summer Programs ’05–06 [27] and the Julia Robinson Math Festival ’07 [77], while Gopher Gun originated at Mathcamp ’05 [14].

1. Warm-up Classics This introductory section1 departs from the style of other sessions: it contains seemingly random yet carefully selected number theory and geometry questions, intended to accommodate the late-comers, warm up the audience, and prepare the math arena for the upcoming treatment of invariants. The more advanced reader can skip this section without harm and proceed to the following harder problems in Section 2. 1.1. Numerical warm-ups. Calculating with numbers may be easy, but how about calculating with “letters” when we don’t know the numbers? Exercise 1 (Warm-up). Can you think of two numbers that have the same sum as their product? Yes, 2 + 2 = 2 · 2. No, the two numbers don’t have to be different. Can you think of another pair? Can you find two distinct numbers with the same sum and product? 1 Editors’ comment: It is the author’s signature-style, used by him to start every circle. 203

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Solution: Well, there are plenty of numbers with the above property, e.g., 0 + 0 = 0 · 0 certainly works. To find more, we make a mathematical model of the situation, i.e., we write an equation a · (1) a + b = ab ⇒ a = (a − 1)b ⇒ b = a−1 Indeed, there are plenty of answers as long a = 1, e.g., a = 3 and b = 1.5, √ as √ √ or something uglier like a = 3 and b = 3/( 3 − 1). We may be very slick and rewrite the solutions in the form a = 1 + t, b = 1 + 1/t. (Check that these work for t = 0.)
But that’s not really the point of the original question, is it? By now an experienced problem solver should suspect that we want to find only natural numbers a and b with equal sum and product. Exercise 2. Prove that the only two natural numbers with the same sum and product are 2 and 2.

 

Now, that’s our type of a non-trivial problem! Let’s apply a well-known PST 69. Check small cases first and make your conclusions. Checking (1) for a = 1, 2, 3, 4, we quickly see that only a = 2 works, yielding b = 2. Hm, should we continue checking, or is it time for a general argument to convince anyone on Earth that it is pointless to plug in larger and larger values for a? Let’s do the latter. PST 70. In order to eliminate certain solutions of equations, form related inequalities. Solution to Exercise 2: Suppose that our equation (1) worked for two “big” numbers a and b, i.e., a ≥ 2 and b ≥ 2. Then inequalities kick in: a + b = ab ≥ 2b ⇒ a ≥ b, and a + b = ab ≥ 2a ⇒ b ≥ a. Hence a ≥ b ≥ a, and, after all, a must be equal to b ! But then our equation reads 2a = a2 , i.e., a = 2 (or a = 0, which is illegal!) so that a = 2 = b is the only solution in natural numbers.
Check the Hints section for a fast solution involving relatively prime numbers. In any case, should we stop here? Are mathematicians satisfied completely by solving a problem, or do they look further for possible generalizations? So, we become more ambitious: Problem 1. What about three numbers? Are there three or more natural numbers whose sum and product are the same? How about if we fix our favorite sum and product and ask the same question: find all natural numbers a, b, and c whose sum and product are each equal to, say, 5? to 2005? to n?

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The reader may find that 1+2+3 = 1·2·3, 1+1+2+4 = 1·1·2·4, and that (1, 1, 1, 2, 5), (1, 1, 1, 3, 3), and (1, 1, 2, 2, 2) are solutions for five numbers. Hm, suspiciously many 1’s are popping up . . . . Is there any pattern to these solutions? Can we find all solutions? (Cf. Moscow Puzzles [55] for this and other similar puzzles.) Solving these questions will take us far beyond the scope and topic of this session: we are supposed to be playing games here, yes? Yet a truly inquisitive mind views every solution only as a step to harder and more interesting problems. More often than not, every solved problem sprouts multiple new questions and that’s how the ocean of mathematics expands. Thus, we leave the above questions open to the reader to think about, and we move to other challenges. 1.2. Warm-ups with visual puzzles. Symmetry and congruent figures will play an important role later in the session. Here are a couple of classical puzzles to remind you what is meant by congruent. Exercise 3 (Warm-up). Can you take away three matches from Figure 1a and leave four congruent squares? Can you take away three matches and leave three congruent squares?

Figure 1. Remove 3 matches, and Divide into congruent parts Exercise 4 (Warm-up). Can you divide the region in Figure 1b into two congruent figures? Can you do it without using a mirror symmetry? These questions are such old entertaining classics that we shall not spoil the reader’s fun by giving away any hints at this point. However, a general comment is due. What feature will distinguish your solutions when the answer is Yes from those when the answer is No? If you believe that something is possible, then you can go ahead and show us how to do it. But if you believe that something is impossible, how do you convince us that it can’t be done?2 For example, in Exercise 2, it wouldn’t do to keep plugging larger and larger values for a and b; we created a general argument showing that larger numbers don’t work. Similarly, if you believe that the only way to divide Figure 1b into two congruent parts would be via the vertical line, you must come up with a general way to explain why no other division would accomplish the task. 2 Recall the analogous discussion from Session 5 on Proofs.

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2. Invariants with Numbers Now we get to the heart of the session! No more skipping – you should try to follow everything in the given order. The exercises in this section are “dressed-up” versions of problems from Fomin, et. al. [28]. They show how collections of problems can be personalized and presented in “interactive mode” to excite the audience. The basic concept we wish to examine is the technique of finding an invariant. In many situations, we are interested in the possible outcomes of some process. An invariant is a quantity that does not change no matter how the process plays out. In keeping with this idea, there is a little platitude that you might be familiar with: “The more things change, the more they stay the same.” 2.1. Erasing and parity. Here is a little activity for you to engage in with your neighbor or friend. Everyone should pair off and spend a few minutes discussing the results obtained. If you think you understand what is happening, change the initial number of 1’s and 0’s and see if you can predict the outcome. Exercise 5. Write down six 0’s and five 1’s on a piece of paper. Then begin crossing out pairs of digits: either two 1’s, two 0’s, or a 1 and a 0. If the digits crossed out are the same, then write a new 0. If the digits crossed out are different, then write a new 1. Continue in this way until there are no more digits to cross out, i.e., only one digit remains. What do you see?



What should we pay attention to while playing this game? Students suggested during the live BMC session • the sum of all digits; • the parity of the number of 0’s; • the parity of the number of 1’s; • the number of choices at each stage; • the average of the digits at any time. PST 71. Look for an invariant: it is often the secret to understanding what is really happening. The above ideas work on various problems, and there is no way of telling which would work on the current problem unless you try all of them out. As it turns out, in Exercise 5 you should consider the parity of the sum, or equivalently, the parity of the number of 1’s, and argue that it is always odd. Solution to Exercise 5: Every operation either preserves the number of 1’s or reduces it by two. Hence, if you start with an odd number of 1’s, you will remain with an odd number of 1’s throughout the game, and the final number will be a 1.


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2. INVARIANTS WITH NUMBERS



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The invariant we found and used allowed us to actually solve a problem more general than the original: we came up with the final answer for any initial combination of 0’s and 1’s. PST 72. A simple but powerful technique to analyze a problem with integers is to consider their parity: the fundamental property of being even or odd. 2.2. Chameleons and invariants with remainders. Here is another similar activity, which is, however, a little more difficult to analyze.




Problem 2. On a planet far away, the only inhabitants are chameleons. They come in three colors: green, yellow, and red. Further, on this planet there is a law which governs how chameleons can change their color. Whenever two chameleons of different colors meet, they change to the third color. Given the initial numbers of chameleons of each color, is it possible to have all chameleons change to the same color? When we attempted to solve this problem at the BMC, there were initially 14 participants, and they divided into 3 groups: 4 green, 5 yellow, and 5 red. All “chameleons” went to the front and formed their groups. If one “chameleon” shook hands with a member from another group, then the two left their groups and joined the third group. In nature, chameleons cannot talk or communicate in signs, so students tried to fast-forward the task of changing the entire class to one color by observing and modeling a specific procedure. After a few false starts, they were able to accomplish the task in quick order. Exercise 6. Can you think of a strategy to turn the (4, 5, 5)-configuration into one with chameleons of only one color? (Configurations with only one color will be called unicolor.) Can you come up with other “less obvious” triples (g, y, r) which can also be turned into unicolor triples?

Figure 2. Recoloring chameleons from initial (4, 5, 6)-configuration Now a late arrival entered the room. Since the students did not want to leave this person out, they formed 3 new groups: 4 green, 5 yellow and 6 red, and again attempted to have all the “chameleons” change into the same color. After trying for awhile to accomplish this, some students became discouraged and voiced the opinion that it was “impossible.” Everyone sat Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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down and tried to figure out what made the (g, y, r) = (4, 5, 6) situation different from the (g, y, r) = (4, 5, 5) situation. Can invariants help prove that the (4, 5, 6)-configuration is impossible to solve? After some period of time, as the first hour was drawing to a close, one student noticed that in a chart of possible positions arrived at from the (4, 5, 6)-configuration, the difference r − y invariably fell into the set {−14, −11, −8, −5, −2, 1, 4, 7, 10, 13}. What do these numbers have in common? They are all 1 more than a multiple of 3. Check that this is indeed so in Figure 2, and do some examples on your own. Exercise 7. Prove that after each recoloring, the difference r − y either remains unchanged or increases/decreases by 3. Conclude that the remainder, when r − y is divided by 3, remains the same throughout. Note that just compiling a list of values for r − y which are all 1 more than a multiple of 3 does not prove that other numbers are impossible. The proof must consider a general case and show that this will always be so, as in Exercise 7. The reader may have realized by now that Exercise 7 extends the parity technique of Exercise 5, since it divides the integers into three classes according to their remainder when divided by 3. While this technique is the main subject of an entire session on number theory in this book (cf. Session 4), in our present case it provides a method for proving that the (4, 5, 6)-configuration is impossible to solve. Do the following exercise and study the given solution afterwards. Exercise 8. If the (4, 5, 6)-configuration could be turned into one color, explain why r − y = −15, 0, or 15 in the end. How does this contradict Exercise 7? Partial Solution: Since the initial r − y = 6 − 5 = 1, Exercise 7 shows that the difference r−y would never be divisible by 3, while Exercise 8 claims that r − y must be divisible by 3. This contradiction proves that unicolor configurations will never be obtained from the initial (4, 5, 6) situation! ♦ Let us now generalize this discussion to any initial (g, y, r)-configuration and discover which configurations are solvable and which are not. Exercise 7 remains valid; it can even be extended to say that all three differences g − r, r − y, and y − g retain their respective remainders when divided by 3. Exercise 8 should be modified to reflect the new total number of chameleons n = r + g + y: in case of a unicolor configuration, each final difference g−r, r−y, and y−g would have to equal 0, n, or −n (why?). We summarize these observations in the next problem, with which we challenge the readers familiar with arithmetic of remainders:




Problem 3 (Intermediate). Prove that a (g, y, r)-configuration is solvable (i.e., can be turned into a unicolor configuration) if and only if at least two of the given numbers have the same remainder when divided by 3.

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In other words, the unsolvable (g, y, r)-configurations are precisely those in which g, y, and r have different remainders when divided by 3. It should come now as no surprise that our previous (4, 5, 6)-configuration proved impossible to solve: the corresponding remainders are 1, 2, and 0, which are all different! 2.3. The Pointless Machine and invariants with divisors. Another puzzle arises from a very elementary machine, the Pointless Machine, that operates in three modes on ordered pairs of positive integers (x, y) (cf. Fig. 3). (3, 17)

(4, 18)

(3, 17) (17, 25)

MODE

MODE

1

1

2 Adv.

(4, 18)

2 Adv.

MODE (2, 9)

1 2

(3, 25)

Adv.

Figure 3. Pointless Machine in modes 1, 2, and Advanced • In Mode 1 the machine adds 1 to both coordinates, e.g., (3, 17) produces (4, 18). • If both coordinates are even, then Mode 2 is also available: the machine halves each coordinate, e.g., (4, 18) produces (2, 9). • If for two pairs the second coordinate of one equals the first coordinate of the other, then the machine has an Advanced Mode, which combines the remaining two coordinates, e.g., (3, 17) and (17, 25) produce (3, 25). The machine remembers and can use any of the pairs that have been previously produced or input into the machine. Exercise 9. If initially only (3, 17) is available to the Pointless Machine, prove or disprove that the following pairs can be produced: (a) (87, 101); (b) (20, 27); (c) (7, 13). Remember that if your answer is affirmative, you should show a specific procedure for arriving at the desired pair. However, if your answer is negative, showing that several specific procedures fail to produce the pair is not sufficient; a general argument is necessary to explain what goes wrong while attempting to reach such an impossible pair (cf. Session 5 on Proofs).




Problem 4. Describe all pairs the Pointless Machine can produce if the starting pair is (a) (3, 17); (b) (x, y).

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Hint: Obviously, we are supposed to find some invariant, but what can it be? Consider each mode separately: • In Mode 1, the move (x, y) → (x + 1, y + 1) evidently preserves the difference y − x. • By sending (x, y) → (x/2, y/2), Mode 2 halves the difference y − x (which is even here) and thus preserves any odd divisor of y − x. • It is much less obvious what the Advanced Mode preserves, but given the above, it is not surprising that it preserves any divisor of y − x. Indeed, consider (a, b) and (b, c) whose differences are each divisible by some d, i.e., b − a = kd and c − b = md for some integers k and m. Then the Advanced Mode yields (a, c) with difference c − a = kd + md = (k + m)d, which is divisible by d. What is common for all three modes? They all preserve the odd divisors of the original difference y − x. For example, in the sequence of moves M1

(2)

M2

M1

M1

M1

(3, 17) −→ (4, 18) −→ (2, 9) −→ (3, 10) −→ (4, 11) −→ M1

Adv. M

M2

· · · −→ (9, 16) −→ (2, 16) −→ (1, 8),

the initial difference is y − x = 14, so its (largest) odd divisor is 7. Correspondingly, check that at any step the difference of coordinates is divisible by 7. If you are wondering what happened towards the end of (2): we applied the Advanced Mode to (2, 9) and (9, 16) to produce (2, 16), whose difference is, not surprisingly, again divisible by 7. You may think we have already found our useful invariant, but we are not done yet. There is one more important property shared by all three modes. Exercise 10. If the starting pair (x, y) has x < y, then any pair produced will also have a smaller first coordinate: the Pointless Machine does not change which coordinate is smaller. At this point, the reader can put together the answer to Problem 4. Let d be the largest odd divisor of y − x > 0; formally, y − x = 2r d for some odd d and some integer r ≥ 0. Then the machine can produce only   ordered pairs (a, b) such that d divides b − a and a < b . For instance, any pair reachable from (3, 17) will be of the form (a, a + 7k) for some k > 0. Yet, this is only half of the problem! We must also give a specific algorithm which starts with (x, y) and ends with any of the desirable pairs. As we saw above, even (1, 8) is challenging to obtain from (3, 17): it took us 11 moves to accomplish this task! How can one, for example, obtain (7, 37) from (2005, 2020)? We leave the reader to play with the Pointless Machine and design such specific procedures. ♦ Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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3. Stomp 3.1. The rules of Stomp. This game originated from walking with my son to school in the morning.3 On the way, we would pass an open lot which was inhabited with gophers. If a gopher had its head sticking up, peering out of its hole, my son would stop and stare, refusing to move until the gopher went back down the hole. So that we would not be delayed, I would stomp on the ground to get the gopher to go back down the hole. But gophers are very curious creatures, and gophers that had been underground and close enough to where I stomped the ground would stick their heads up to see what all the commotion was about. This gave rise to the following game, which I have named Stomp. Think of your shoe as the Stomp-piece and the gopher holes as the squares. The squares with dots have gophers looking out of their holes initially (cf. Fig. 4a). You are allowed to place the Stomp-piece anywhere within the grid on each move, as long as it lines up with the grid and stays within the boundary. All the squares covered by the Stomp-piece then change state, so that dots that are covered disappear, while empty squares that are covered sprout new dots, as the gophers stick their heads up. The goal is to obliterate all the dots in as few moves as possible or to prove that it is impossible to do so.

(a)

(b)

Figure 4. Obliterate the dots with a domino, tetromino, or tromino 3.2. Classic Stomp problems often require invariants with parity and coloring. Try out your first Stomp challenges below.




Exercise 11 (Warm-up). Try to solve Figure 4a with the gophers in the four corners using a 2×1 domino stomp piece; then try using a 2×2 tetromino Stomp-piece; and finally try using a 3 × 1 tromino. Problem 5. Repeat Exercise 11 with Figure 4b. If you think any of the six problems is impossible, can you find suitable invariants to prove it? 3 Editors’ comment: This is also a common game which is marketed, for example, as “Lights Out” by Tiger Electronics.

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It is really fun to investigate and play around with these problems. You should try each problem for a reasonable amount of time, and only if you get stuck, read below or go to the hints at the end. In the meantime, we shall discuss two important techniques that pop up in the solutions. For starters, if you try hard enough with the 2×2 tetromino on Figure 4b, you will quickly discover that the solution requires the creation of a special invariant. PST 73. To simplify a Stomp-like problem, try to find some sub-board on which the Stomp-piece acts in an easily describable way, producing an invariant on this sub-board. Hint: For instance, how many dots will there be in the top row of the Stomp-board in Figure 4b when a 2 × 2 piece acts on it? Show that the parity of the dots in the top row remains constant throughout and draw your conclusion. ♦



Now, can we use the same argument for the 3 × 1 tromino on Figure 4b? Nope – the parity of the dots in any row or column changes with the application of the 3 × 1 tromino. (Why? Check this!) This problem is tough: its invariant can be found only after a non-trivial coloring of the board. PST 74. Partition the Stomp-board into several non-overlapping sub-boards such that, no matter how you place the Stomp-piece on the board, it will cover exactly one square from each sub-board. Conclude that in each subboard the parity of the dots will reverse at each step. For a visual effect, color each sub-board in a different color. But how do we find the sub-boards that will work so perfectly? Solution to Problem 5: One idea is to look at the number of squares in the Stomp-piece, e.g., a 3 × 1 tromino will require at least 3 sub-boards. Any placement of the 3 × 1 tromino will tell you that any three adjacent squares on the Stomp-board need to be “colored” differently. With a little experimentation of shifting the Stomp-piece and coloring various squares, you can come up with the following solution. If we color the southwest-northeast diagonals in alternating red, white, and blue stripes (cf. Fig. 6a), then every move of the 3 × 1 tromino will change the number of red (or white, or blue) dots by exactly one, hence reversing its parity. Since there are originally 2 red dots but only 1 blue dot, the number of red and blue dots will always have opposite parity. Yet, a solved board will have 0 red and 0 blue dots: same parity! This contradiction shows that this Stomp-piece fails to solve Figure 4b.





Problem 6. For each of the Stomp-boards in Figure 5, find a way to clear the board in as few moves as possible or else prove that it cannot be done. The little polyomino next to the board represents the “footprint.”

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3. STOMP

213 (a)

(b)

(d)

(e)

(c)

Figure 5. Obliterate the dots with various polyominoes Hint: While Figure 5d will require a parity argument (look at the total number of dots) and Figure 5e can be solved using the diagonal-coloring technique as in Figure 6a, Figure 5a requires yet another fundamental idea which is ubiquitous in just about all areas of mathematics.



Figure 6. Invariants in Problems 5 and Figure 5e PST 75. The ideal situation in a Stomp-like game is to be able to change the state of a single square without affecting the rest of the board. Can we do this in the setting of the four gophers in Figure 4a? The L-tromino provides the means. In any 2 × 2 sub-board, we can place the L-tromino in 4 different ways. Placing it in exactly 3 of these 4 ways will accomplish our task: it will change the state of exactly one of the 4 squares, leaving the rest unchanged (verify this!). Hence one needs only to obliterate the dots one at a time, using 3 moves per dot. ♦ 3.3. Stomp on large boards naturally invokes previously solved problems.




Problem 7. Suppose that you are given a rectangular board whose length and width are each at least two, with any initial arrangement of dots. Explain why it is always possible to clear the board using an L-tromino. If we had not solved Figure 4a with the L-tromino earlier, this problem would have proved hard to crack. PST 76. Reducing to a previously solved problem is a favorite technique of mathematicians. More generally, making connections between statements (problems, theorems, definitions, conjectures, ideas) is arguably the most valuable math skill. Hint: As in the solution to Figure 4a, it is always possible to change the state of a single square without affecting the rest of the Stomp-board. ♦

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Problem 8. Now imagine that you are playing Stomp on an infinitely large board with no boundaries. If dots are initially placed in any two squares, prove that it is possible to clear the board using the T -tetromino (cf. Fig. 5b). This is also a toughie! We need an intermediate result. PST 77. In a game, look for special sequences of moves whose net effect is some simple and desirable move.4 With this in mind, show that Exercise 12. Using the T -tetromino on a 3 × 3 board with a single dot in the top middle square, show how it is possible to end up with a single dot in (a) the central square in five moves; (b) the bottom middle square in only two moves. The net result is that the dot can migrate by one square or by two squares in any direction (down, up, left or right) on the infinite board. Sketch of Solution to Problem 8: It is always possible to force one dot to migrate on top of the other using a combination of the above two moves (how?), thereby making the two dots cancel out. ♦ 3.4. The Gopher Gun. Imagine now that we have a gun which can shoot along a whole row, column, or diagonal of the board, and each gopher in that row, column, or diagonal goes back down or comes up after the gun is shot. This is a cute interpretation of an original problem by Adam Hesterberg in my problem writing class at Mathcamp ’05. It goes like this.

Figure 7. Gopher Gun




Problem 9. The squares of a 6 × 6 chessboard are each empty or contain a single dot (aka gopher). At each step you are allowed to reverse the state of all the squares in a single row, column, or diagonal (including the trivial diagonal consisting of a lone corner square). There are precisely three dots in the initial state of the board, located along the left-hand half of the top row (cf. Fig. 7). Prove that it is impossible to clear the board of dots.5 4 This idea appeared in the construction of useful macros in the Rubik’s Cube session. 5 A 4 × 4 version can be found in Problem Solving Strategies [26].

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Sketch of Proof: The idea is to find a subset S of squares of the board such that any row, any column, and any diagonal intersects S in an even number of squares. One such set would be the eight squares adjacent to the four corners of the board. Thus, if you consider the parity of the number of dots in S throughout the game, you will find out that it is invariant. ♦ Here is a challenging problem for you. Problem 10. In the set-up of Gopher Gun, find (a) all subsets S whose parity of dots is invariant under the operations; (b) all attainable configurations of dots.

4. Tilings and More Invariants No invariant session would be complete without tilings. Recall that to tile a figure with a particular shape means to cover that figure completely with copies of the given shape which do not overlap. For example, it is possible to tile a 4 × 4 board with eight copies of a 1 × 2 domino. In the following, we shall consider two shapes the same (or congruent) if they are obtained from one another by rotating, translating, flipping, or any combination of those. Exercise 13 (Warm-up). Clearly, one can tile a 4 × 4 board with 2 copies of a 2 × 4 rectangle. It turns out that there are 5 other shapes, each built from eight 1 × 1 unit squares, such that 2 copies of each shape can also tile the 4 × 4 board. Find these other 5 shapes and tilings of the 4 × 4 board. Exercise 14. There are 5 distinct shapes which can be constructed using four unit squares on a sheet of graph paper. These are called the tetrominoes. Draw all 5 tetrominoes and decide which ones can be used to tile a 4 × 4 board (one type of tetromino per tiling!).




Both Exercises 13 and 14 call for an intelligent trial and error approach. Obviously, symmetry (across the center of the board, or across a line) will play a part. You should organize your solutions in a systematic way so that they can generalize to larger boards of suitable sizes. Problem 11. Show that it is impossible to cover a 4 × 5 rectangle with a complete set of tetrominoes, i.e., using each tetromino once. Hint: In a chessboard coloring, which tetromino covers an unequal number of black and white squares? Why is the board non-tileable? ♦ Exercise 15 (Warm-up). Draw a rectangle 7 squares wide and 3 squares high, missing the middle square of the top row. Then find a way to tile this figure with one complete set of tetrominoes. Can you find another interesting shape consisting of 20 unit squares which can be built from one complete set of tetrominoes?

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10. STOMP. GAMES WITH INVARIANTS

Problem 12. Consider a 7 × 7 board. Which square can be removed to allow a tiling with 3 × 1 trominoes of the resulting 48-square board? For example, removing a corner square is one answer: find a way to tile the resulting board. What are the remaining answers? Hint: Some other answers are the top middle square and the middle square. By symmetry, you should find all 9 answers, provide a tiling for each, and explain why the removal of no other squares would do. Note that using one of the diagonal 3-coloring schemes will not be sufficient: you will need both of the two possible 3-coloring schemes. ♦ Our last tiling problem is probably the most interesting question here because it leads to some creative coloring schemes.




Problem 13. For each of the 5 tetrominoes, determine whether or not it can be used to tile a 6 × 6 board. In each case, find a way to do the tiling or demonstrate that it cannot be done using an appropriate coloring scheme. Hint: This one is fun, but I’ll let you sort out the details. For example, the L-tetromino is impossible due to a vertical striped 2-coloring. Finding out why the S-tetromino is “obviously” impossible takes a few tries! ♦ If you like these problems, a similar activity on tilings is presented in Ravi Vakil’s book A Mathematical Mosaic [94]; there, for instance, you can find a more interesting 8 × 8 version of Problem 12. We direct the moderately advanced reader to an accessible, beautiful, and deep account of the mathematics of tilings by Federico Ardila and Richard Stanley [4].

5. Escape of the Clones This is a version of a famous puzzle attributed originally to Maksim Kontsevich, which appeared in the Tournament of the Towns and in the Russian journal Kvant in 1981 (cf. [54, 50]). Its solution will require the creation of invariants with infinite series. 5.1. The set-up of the game. Consider the first quadrant in the Cartesian plane divided into unit squares by horizontal and vertical lines at the positive integers. Place 3 dots (clones) in the shape of an L-tromino in the bottom left-most squares, and draw a “barbed wire fence” enclosing the dots and their 3 respective squares: this is the orange fence in Figure 8.

Figure 8. Escape of the Clones

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5. ESCAPE OF THE CLONES

217

5.2. The rules of the game. At each step you can erase a dot and replace it with two copies in adjacent squares, one directly above and the other directly to the right, as long as those squares are currently unoccupied. In other words, when a clone disappears, it sprouts two more clones above and to the right of it. Notice that this is a Stomp-like game • whose board is the infinite first quadrant and • whose “footprint” is an L-tromino allowed to be placed only in the standard orientation of the English letter L, and only when the corner square of L covers a clone while the other two squares of L land on clone-empty spots.






5.3. Freedom for the clones! Problem 14 (Advanced). Prove that it is impossible to free all clones from the prison. Although the problem setting is elementary enough for anyone to play and enjoy the game, the actual solution is hard to come up with and requires knowledge of a useful summation formula. That’s great: while playing games, we will learn some algebra too! 5.4. In search of the invariant. What could be the invariant preventing us from freeing all 3 prisoners? You should try parity, the coloring technique, and any other previous ideas, but soon enough you will realize that the problem is too complicated to succumb to these methods. We need a different, more powerful approach here: PST 78. Assign a suitable number to each square to create an invariant. What could these suitable numbers be? Let us agree to label a square with the coordinates (a, b) of its bottom left corner. Thus, the 3 initially occupied squares are labeled by (0, 0), (1, 0), and (0, 1). If we assign, for instance, the number 1 to square (0, 0), it makes sense to assign 1/2 to each of squares (1, 0) and (0, 1) so that the sum of assigned numbers to occupied squares before a move and after a move remains constant: 1 = 1/2 + 1/2. OK, but then we are more or less forced to assign 1/4 to each of squares (2, 0), (1, 1), and (0, 2) so that this same reasoning works when a clone in (1, 0) or (0, 1) sprouts two more clones: 1/2 = 1/4 + 1/4. Thus, every time we move to the right or up, the assigned numbers get halved! Going around the board in this manner, we soon discover an exact formula for all the desired numbers: 1 · Show that when Exercise 16. To every square (x, y) assign the number 2x+y a move is applied to a clone anywhere, the sums of the numbers assigned to occupied squares before and after the move are equal.

We have found our invariant: the sum of the numbers in occupied squares stays constant throughout the game! How can we capitalize on this invariant? Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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5.5. Arguing by contradiction. Suppose that we can free all 3 clones from the prison. Along the way, they will have sprouted several clones outside the prison. Call the sum of the numbers in occupied squares at the end of the game Sout . Note that our initial invariant sum for the clones in the prison is simply Sin = 1 + 12 + 12 = 2; so we must have Sout = 2. But could this happen? Well, whatever happens, Sout cannot be more than the sum of all numbers assigned to out-of-prison squares. 4 ··· 1 8 1 4 1 2

1

1 4 1 2

··· 1 8 1 4 1 2

··· 1 8 1 4

··· 1 8

1 ···

2

Figure 9. Assignment of numbers and Problem 16




5.6. Algebra detour. For starters, let’s add up all numbers on the infinite board in an orderly fashion. For example, on row 1 we will have 1 1 1 (3) 1 + + + + · · · = 2. 2 4 8 If the reader wonders how we came up so fast with the answer for this sum, note that a much more general formula holds: Problem 15. For any (real) numbers a and r, −1 < r < 1, show that a · a + ar + ar2 + ar3 + · · · + arn + · · · = 1−r The sum is called a geometric series with initial term a and ratio r. We leave the proof of this formula for the Hints section, but we do encourage the reader to try to prove it from scratch: it won’t be a trivial exercise.6 In the case of row 1 (cf. Fig. 9a), the initial term is a = 1, and the ratio is r = 12 . Now, to see what goes on in row 2, note that each number there is half of the number directly below it, i.e., the sum in row 2 will be half of the sum in row 1. Similarly, the sum in row 3 will be half of the sum in row 2, and so on. In general, in the i th row: 
1 1 1 1 1 1 1 1 1 + i + i+1 · · · = i−1 1 + + + + · · · = i−1 · 2 = i−2 · i−1 2 2 2 2 2 4 8 2 2 Adding up the sums in all rows yields the sum of all numbers on the board: 1 1 1 2 + 1 + + + + · · · = 2 + 2 = 4. 2 4 8 All right, we subtract the “in-prison” numbers, 1 + 12 + 12 = 2, to obtain the maximal value that Sout can ever be: Sout ≤ 4 − 2 = 2. 6 This formula appears in just about any course in calculus when discussing sequences and series. We shall dedicate a session to related topics in a future volume.

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6. HINTS AND SOLUTIONS TO SELECTED PROBLEMS

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5.7. The finishing touch: this is a finite game! Recall that, even though we are playing on an infinite board, the prisoners are supposed to be freed in finitely many moves. Thus, at least one outside-of-prison square is clone-free hence the out-of-prison sum is strictly less than 2: Sout < 2. This contradicts the fact that Sout = Sin = 2 and proves that the three clones cannot all be freed from prison.
5.8. Pushing on: generalizations. What is the real story behind this problem? Why did the method of using a summation invariant work so nicely? If we slightly change the initial set-up, would it still be impossible to free the clones?




Problem 16. A single clone is placed in square (0, 0), and the prison encloses (a) the 10 squares (i, j) with i + j ≤ 3 (Kontsevich [54]; cf. Fig. 9b); (b) the 6 squares (i, j) with i + j ≤ 2 (Khodulev [50]; cf. Fig. 9c). Show that there will always be at least one clone in the prison. Hint: Part (a) is like Problem 14, but part (b) requires another insight. ♦ A delightful discussion and generalization of the above problems is presented by Chung, Graham, Morrison, and Odlyzko in a 1995 piece in the Monthly (cf. [15]). Starting with one initial clone in (0, 0), the articles describes all “inescapable prison shapes,” and relates the problem to a new “Ramanujan-like” continued fraction.7

6. Hints and Solutions to Selected Problems Exercise 2. Since b = a/(a − 1) is an integer, (a − 1) must divide a. But (a − 1) and a are relatively prime (as two consecutive numbers); so the only possibility for the denominator (a−1) is to equal 1, i.e., a = 2 and b = 2.
Exercise 3. For the first part remove the top, left, and bottom sides of the bottom leftmost square. For the second part remove two matches in the right upper corner and the middle match from the bottom row.
Exercise 4. Use a quarter-circle curve through the middle to get two congruent axe heads.
Problem 3. Suppose that some (g, y, r)-configuration is solvable. Recall that r − y doesn’t change its remainder (mod 3), and at the end it should equal one of {0, n, −n} depending on whether all chameleons are green, red, 7 The beginning of [15] can be readily followed by anyone. However, as the reader advances into the article, more maturity and experience will be required. For example, familiarity with the notions of recursive sequences, summation techniques, partial derivatives, continued fractions, and asymptotic manipulations will be needed. The advanced reader may consider learning the necessary background for the article as a long-term project.

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10. STOMP. GAMES WITH INVARIANTS

or yellow, respectively. Here n = r + g + y is the total number of chameleons. We model these observations by the congruence relations8 shown below. Note that −2y ≡ y (mod 3) and −g ≡ 2g (mod 3) will simplify calculations: r − y ≡ 0 or r ≡ r + g + y or r ≡ −r − g − y (mod 3) ⇒ r ≡ y (mod 3) or y ≡ g (mod 3) or g ≡ r (mod 3). This proves that in any solvable configurations at least two of the initial groups of chameleons have the same remainder (mod 3). Conversely, suppose two of r, g, and y are congruent (mod 3). WLOG, assume r ≡ y (mod 3) and r ≥ y. By having all y yellow chameleons meet with some y red chameleons, we get rid of all yellow chameleons. We are left with r − y = 3k red chameleons: (r, g, y) → (r − y, g + 2y, y − y) = (3k, g + 2y, 0). If k = 0 or g + 2y = 0, we have an all-green or all-red configuration. If not, we will show how to recolor 3 red to 3 green chameleons without changing anything else. Simply follow the sequence of moves which involves only 3 red and 1 green chameleon: (3, 1, 0) → (2, 0, 2) → (0, 4, 0).

(4)

The net result of (4) is indeed recoloring 3 red to 3 green chameleons. Applying this procedure k times will end in an all-green configuration.
Exercise 9. The first two pairs are possible (construct specific procedures for them), but the third pair (7, 13) is not since its difference 13 − 7 = 6 is not divisible by 7. ♦ Problem 4. We design first some helpful auxiliary Modes as combinations of the given three basic Modes. For example, Mode 1m is an application of Mode 1 m-times: it adds the same positive integer m to both coordinates, i.e., (x, y) → (x + m, y + m). If x < y, we design an Advanced 2 Mode which doubles the difference of coordinates: (x, y)

Mode 1y−x

−→

(x + (y − x), y + (y − x)) = (y, 2y − x)

Adv. Mode

−→

(x, 2y − x).

To triple the difference of coordinates, we create an Advanced 3 Mode: (x, y)

Adv.2 Mode

−→

(x, 2y − x)

Mode 1y−x

−→

(y, 3y − 2x)

Adv. Mode

−→

(x, 3y − 2x).

The reader should be able to create from here a sequence of Advanced m Modes which multiply the difference of coordinates by m: (x, y) e.g., (3, 17)

Adv.5 Mode

−→

Adv.m Mode

−→

(x, my − (m − 1)x),

(3, 5 · 17 − 4 · 3) = (3, 73) multiplies the difference by 5.

8 Review the congruence notation for remainders from the Number Theory session.

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6. HINTS AND SOLUTIONS TO SELECTED PROBLEMS

221

Solution: For the general procedure, start with (x, y), whose difference is y − x = 2r d (where d is odd). By multiple applications of Mode 2 (with occasional Mode 1 when both coordinates are odd), construct a mode which produces a pair with difference d. Thus, assume now y −x = d. To construct the pair (1, 1 + d), observe that if 2k is a power of 2 larger than x, then (x, y) = (x, x + d) k

Adv.2 Mode

−→

Mode 1(2

−→

k)

(2k , 2k + d)

(2k , 2k + 2k d)

Mode 2(2

−→

k)

(1, 1 + d).

Finally, to get to any (a, b) with difference divisible by d and a < b, write b = a + nd and apply (1, 1 + d)

Adv.n Mode

−→

Mode 1a

(1, 1 + nd) −→ (a, a + nd) = (a, b).




Problem 5. Figure 4a can be cleared with all three Stomp-pieces, but only the domino will clear Figure 4b. The square piece, for instance, fails because there will always be an odd number of dots along the top row. ♦ Problem 6. (b) It is possible to solve the puzzle in 5 moves. The moves are hard to describe on paper but not difficult to discover with a little experimentation. ♦ (c) The best possible is 4 moves. Make the first move by placing the tip of the long edge of the L-tetromino in a corner with a dot. The rest of the solution becomes obvious at this point. ♦ (d) The bottom left Stomp puzzle is impossible to complete. We claim that there will always be an odd number of dots on the board. This is the case at the outset, and any move will either obliterate two dots, create two dots, or do one of each. In any case the parity of the total number of dots will remain unchanged, as claimed. ♦ Problem 12. Consider the first diagonal 3-coloring scheme in Figure 10a. Since each placement of a 3 × 1 tromino covers exactly one square of each color, the total numbers of white, red, and blue squares must be equal in a tiling. Since the red squares are 1 more than the blue and the white, only a red square can be removed to equate the colors and allow for a possible tiling. Now the same analysis applies to the other diagonal coloring displayed in Figure 10b.

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10. STOMP. GAMES WITH INVARIANTS

Notice that the squares which are red in both colorings are precisely the 9 squares filled with dots: these are the answers to our problem. It remains to find specific tilings corresponding to the removal of any of these 9 squares. By symmetry, it is sufficient to find such tiling only for the removal of a corner, middle square in an edge, and the central square in the 7 × 7 board. ♦ Problem 13. A tiling with the square-tetromino is obvious, but only the square-tetromino can tile the 6 × 6 board. Proof: For the L- and T -tetrominoes, consider the coloring schemes of Figure 11a-b. Each placement of these tetrominoes covers 3 red and 1 white, or 1 white and 3 red, squares. If a tiling is possible, let x tetrominoes be of the first type, and let the remaining 9−x tetrominoes be of the second type. Then the total count of covered red squares would be odd: 3x + (9 − x) = 9 + 2x, while the total number of red squares is even: 36/2 = 18, a contradiction.

Figure 11. Tiling a 6 × 6 board via tetrominoes A straight tetromino covers exactly one of the 10 red squares in Figure 11c. Yet, that requires too many tetrominoes: we must use exactly 9 of them to cover 36 squares, a contradiction. Finally, the S-tetromino doesn’t even need a coloring. A corner of the board can be covered by it in one of two symmetric ways. Figure 11d displays one of these coverings of the top left corner. But then the S-tetrominoes must be placed in a unique way along the top edge of the board, with the rightmost tetromino sticking out of the board, a contradiction!





The same argument shows that no rectangular board can be tiled by S-tetrominoes, yet a number of rectangular boards can be tiled by the other tetrominoes. Find out which ones! Problem 15. Assume for the moment that the desired sum exists, i.e., (5)

a + ar + ar2 + · · · + arn + · · · = S

for some number S. Multiplying by the common ratio r gives (6)

ar + ar2 + ar3 + · · · + arn+1 + · · · = rS.

Now subtracting the two equations (5) and (6) yields a · a = (1 − r)S ⇒ S = 1−r Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

6. HINTS AND SOLUTIONS TO SELECTED PROBLEMS




223

Thus when a = 1 and r = 1/2, we obtain the special case of the row sum in 1 = 2. the Clones problem: S = 1 + 12 + 14 + 18 + · · · = 1−1/2 To prove that the sum S actually exists is a completely different ballgame, which we leave to the reader to ponder over. You will encounter a complete proof of this in a calculus course, where limits are introduced. ♦ Problem 16. Use the same assignment of 2−i−j to squares (i, j), and compare sums Sin = 1 and Sout , which are supposed to be equal. In part (a), a contradiction is achieved via the inequalities Sout < 4 − 1 − 2 · 12 − 3 · 14 − 4 · 18 = 34 < 1. The same adding technique produces in part (b) Sout < 4 − 1 − 2 · 12 − 3 · 14 = 54 > 1, which does not provide a contradiction. A more refined argument for Sout is needed. Note that in the first row and first column of the board there will always be exactly one clone (why?); hence this row and column will contribute to Sout at most 1/8 each. Add up the rest of the board, and find a smaller upper bound for Sout : ♦ Sout < 4 − (2 − 18 ) − (2 − 18 ) + 1 − 14 = 1.

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Session 11 Favorite Problems at BMC. Part I Circle Geometry based on Ivan Matić’s session Sneak Preview. In addition to an array of advanced mathematical topics in its sessions, the Monthly Contests (MC) and the preparation for the Bay Area Mathematical Olympiad (BAMO) form an integral part of the Berkeley Math Circle. Both exams have 5 essay-proof style problems of varying difficulty. While the 4-hour BAMO is administered annually in schools and at math circles primarily in the greater San Francisco Bay Area, the Monthly Contests are take-home one-month adventures at BMC, encouraging additional research. Ivan Matić, the Monthly Contest Coordinator ’05–’08, a former Olympian on the Serbian IMO team, and a seasoned Olympiad coach at BMC, shares here some of his favorite geometry problems along with his insights into the art of problem solving. In particular, how to locate circles and use their properties in problems will be the main theme throughout this session.

1. The Search for the Missing Circle Many geometry problems have innocent-looking formulations involving only straight lines and angles. Alas, this does not prevent some of them from being so frustrating that any reasonable or intuitive approach seems to lead nowhere. Take, for instance, the following BAMO problem, fresh out of the contest arena: A




Problem 1 (BAMO ’07). In ABC, D and E are two points in the interior of side BC such that BD = CE and ∠BAD = ∠CAE. Prove that ABC is isosceles.

B

D

E

C

Of the 157 attempted solutions, about a third received perfect or near perfect scores. Yet, just about every solution used ideas entrenching its author in unsightly calculations with trigonometry or analytic geometry. 225

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11. FAVORITE PROBLEMS AT BMC. PART I

Very few (if any) students saw glimpses of the truly beautiful and elegant geometry of the problem. What was so hard to see here? PST 79. Difficult geometry problems often require us to notice and prove that several points lie on a circle, i.e., that these points are concyclic. But there is no circle in Problem 1! How can PST 79 be helpful at all in its proof? Hold your horses: as you will see at the end of the session, a circle with four prominent points on it will be constructed, and the solution will then flow unobstructed. For now, let’s discuss a more basic issue. Why should anyone care about circles or concyclic points? Unlike lines, circles are not straight; they are, in fact, “way too round” to partake in some standard simple constructions. For example, one cannot cut them up and reconnect them to form polygons, while this is trivial to do with lines. Even the device for drawing circles is quite advanced and artificial in comparison to the one used for drawing lines. On the other hand, one can achieve a high level of complexity when using circles: Exercise 1. Draw Mickey Mouse’s head with only circles and circular arcs.1 The reader familiar with Eastern European culture may find it amusing to repeat this exercise with Cheburashka (
), a famous Russian character.2 How many circles did you use in your drawings and how many of them have different radii? The motivation for studying circles can be seen only through practice. As we will discover, there are just a small number of properties one can establish for the circle, but they will provide us with powerful mathematical tools to work miracles in problem solving. The advanced geometer can skim through the theory below and move on to subsection 3.2. With the others, we will derive these circle properties from scratch. Meanwhile, if you need more details and further geometric insights, there is no better book to look into than Kiselev’s Geometry [51]. Before we start, let’s make sure we all know the following. Definition 1. Let O be a fixed point in the plane. A circle k with center O is a set of all points in the plane equidistant from O. In other words, if A and B are two arbitrary points on k, then OA = OB. The common length of these segments is called the radius of the circle. 1 Did you know that today’s Disney image of our favorite Mickey Mouse, as round as it is, differs somewhat in appearance and character from the original, more mouse-looking image created in 1928? An easy-to-read analysis of Mickey Mouse’s metamorphosis can be found in Karal Ann Marling’s article [60]. 2 The non-descript, small, cute animal, probably a hybrid between a koala bear and a squirrel and also known as Topple in early English translations, was created in 1965 by the Russian writer Eduard Uspensky [92].

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2. INSCRIBED AND CENTRAL ANGLES

227

2. Inscribed and Central Angles Angles can be “fitted” within a circle in two common ways. Definition 2. Let points A, B, and C lie on circle k with center O. We say that ∠ACB is an inscribed angle in k and that ∠AOB is its corresponding central angle in k. C

C

C

O

O

A

A

O

B A

D

B

B

Figure 1. Inscribed and central angles There is a slight catch here. Points A and B divide k’s circumference into two parts, called arcs. If C lies on one arc, let D be a point lying on the other arc (cf. Fig. 1a). Now, the two arcs ACB and ADB correspond to two different central angles, which are referred to, unfortunately, by the same name, ∠AOB. For example, the thickly dashed ∠AOB in Figure 1a can be identified as the central angle measuring arc ADB , or, equivalently, as the central angle corresponding to inscribed ∠ACB. With this understanding, we state our first property:













Theorem 1 (Inscribed Angle). An inscribed angle is half as large as its intercepted arc and the corresponding central angle. Figure 1a depicts two such examples: inscribed ∠ACB is half of its central ∠AOB, and inscribed ∠ADB is half of its central ∠AOB. Notice that the intercepted arc is the arc of the circle in the interior of the angle. The proof of Theorem 1 rests on this basic fact: Exercise 2 (Exterior Angle Theorem). Given ABC, let point X lie on line AB such that B is between A and X (cf. Fig. 2a). Then (a) ∠XBC = ∠BAC + ∠BCA; (b) ∠XBC is larger than either of ∠BAC and ∠BCA.3 In short, an exterior angle equals the sum of the remote interior angles, and hence is larger than either of them. We leave it to the reader to find/recall the solution to this exercise and return to Theorem 1. 3 This is an obvious consequence of part (a). The advanced geometer might like to

prove (b) without relying on (a), i.e., without using the fact that the angles in a triangle add up to 180◦ . Euclid was able to do this in his Elements, Book I, Proposition 16 [42].

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11. FAVORITE PROBLEMS AT BMC. PART I

Partial proof of Theorem 1: A common pitfall is to consider only one possible scenario for the relative position of the corresponding inscribed and central angles, namely that the center O is in the interior of the inscribed ∠ACB (cf. Fig. 1a). What happens if O is in the exterior or on a ray of ∠ACB, as in Figures 1b–c? PST 80. Pictures are great! Draw ideas from them, but do not depend entirely just on pictures for a complete solution: you may have missed a case, or you may be tackling a non-existent one! C

F

C

k

E O

D

A A



A

B

X

B

E

BC

Figure 2. Exterior angles, Inscribed angles and Equal chords PST 81. List the various possible positions of your geometric objects and consider how each position affects your solution. Resolve first the simplest, the most typical, or the most special situation, whichever you find suitable in the problem. Ideally, all other possibilities will either reduce to this situation or have a similar solution. With this said, Figure 1c attracts our attention, being the least complicated case: the center O lies on the diameter BC. Since ∠AOB is exterior to isosceles ACO, Exercise 2(a) readily implies that ∠AOB = 2∠ACB (how?), so we are done in this case. To reduce Figure 1b to Figure 1c, we extend CO to the diameter CE (cf. Fig. 2b) and realize that inscribed ∠ACE and ∠BCE each fit into the −−→ conditions of Figure 1c. Indeed, O lies on the side CE of both angles ∠ACE and ∠BCE, so the previous discussion implies ∠ACE = 12 ∠AOE and ∠BCE = 12 ∠BOE. Subtracting the two equations yields the desired ∠ACB = 12 ∠AOB. We leave it to the reader to practice on the case in Figure 1a.




As immediate applications of Theorem 1, do the following two exercises using inscribed and central angles and, possibly, congruent isosceles triangles. Note the “if and only if” conditions in both exercises. Exercise 3 (Warm-up). The inscribed ∠BAC in circle k is right if‌f BC is a diameter of k (cf. Fig. 1c). Similarly, ∠BAC is right if‌f OA = OB = OC for some point O on side BC. Exercise 4. Let AB and CD be two chords of the same circle. Then AB = CD if‌f the corresponding inscribed angles are equal (cf. Fig. 2c).

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3. “GOING-IN-CIRCLES”

229

3. “Going-in-Circles” 4 3.1. When are points concyclic? As we discussed in the beginning of the session, plane geometry problems often depend on establishing the fact that several points are concyclic. Below we extend the Cyclicity Theorem 1 from the Inversion session.




Theorem 2 (Cyclicity). Let C and D be on one side of line AB, and let E be on the other side of AB (cf. Fig. 3). Then (a) A, B, C, and D are concyclic iff ∠ACB = ∠ADB; (b) A, B, C, and E are concyclic iff ∠ACB + ∠AEB = 180◦ . D k

C β O α

k

C B

B

A

A E

E E

Figure 3. Conditions for cyclicity Proof of (b): For the forward direction (⇒), we assume that A, E, B, and C lie on some circle k with center O (cf. Fig. 3a). By Theorem 1, ∠ACB = 12 α and ∠AEB = 12 β, as corresponding inscribed and central angles.5 But α + β = 360◦ , so ∠ACB + ∠AEB = 12 (α + β) = 180◦ .



Now, that was relatively painless. Is the converse statement as straightforward? That is, if ∠ACB + ∠AEB = 180◦ , how can we show that A, B, C, and E are concyclic? PST 82. To show that four points are concyclic, draw a circle through three of them and, using whatever techniques are necessary, prove that the fourth point also lies on this circle. For the reverse direction (⇐), PST 82 prompts us to denote by k  the circle through A, B, and C (cf. Fig. 3b). We need to show that E also lies on k  . Suppose not. Then line AE intersects k  in some other point E  . Note that E  and C are on opposite sides of AB (why?). But now we have four concyclic points, A, E  , B, and C, so the freshly-proven forward direction of (b) above applies perfectly here. We conclude that ∠ACB +∠AE  B = 180◦ . Comparing with our initial hypothesis, we deduce that ∠AEB = ∠AE  B = 180◦ − ∠ACB. Can this ever happen? 4 “Going-in-Circles” was a parents group which carpooled students for 2 years to math circles, contests, and events in the San Francisco (SF) Bay Area. 5 As discussed earlier, ∠AOB cannot denote both central angles α and β.

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11. FAVORITE PROBLEMS AT BMC. PART I

Figure 3b depicts one possibility where ∠AE  B is exterior to E  EB, so that the Exterior Angle Theorem implies ∠AE  B > ∠AEB, a contradiction! There is one other possible situation for the relative positions of E and E  : E is inside k  . The reader should draw the picture in this case and similarly derive a contradiction. We conclude that our assumption is false, i.e., E does
lie on k  , and A, B, C, and E are concyclic. For practice, the reader should prove Theorem 2(a) in a similar fashion. 3.2. The search for the missing circle continues. The remainder of this section demonstrates how to apply our cyclicity conditions to several BAMO problems. It is essential that you draw good pictures, understand well what is given and what must be proved, and make connections with previous results. Problem 2 (BAMO ’99). Let k be a circle in the xy-plane with center on the y-axis and passing through points A(0, a) and B(0, b) with 0 < a < b (cf. Fig. 4a). Let P be any other point on the circle, let Q be the intersection of the line through P and A with the x-axis, and let O = (0, 0). Prove that ∠BQP = ∠BOP . Problem 3 (BAMO ’02). In ABC, ∠B is a right angle. Let ACDE be a square drawn exterior to ABC (cf. Fig. 4b). If M is the center of this square, find the measure of ∠M BC. y

C

B

k3

E k A M

F D

P

B

E

k1

P

A

D

k2

A Q

O

x

C

B

Figure 4. Problems 2, 3, and 4 Hints: The solutions to Problems 2 and 3 reintroduce the idea of a missing “phantom” circle (which one?), or equivalently, of showing that four points are concyclic. The latter can be established via right angles, Exercise 3 and Theorem 2. For instance, in Problem 2, ∠QOB = ∠QP B = 90◦ (why?), which forces points Q, O, P , and B to be concyclic, and this in turn implies that the desired angles ∠BQP and ∠BOP are equal (how?). Figures 5a– b are very suggestive, and the reader should refer to them only after solid attempts have been made to solve both problems. ♦ Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

3. “GOING-IN-CIRCLES”

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3.3. Cyclicity implies concurrency? Problem 4 below also reintroduces a concept we’ve seen before. Recall that several figures (e.g., circles or lines) are called concurrent at P if they all meet in a point P .




Problem 4 (BAMO ’00). In ABC, let D be the midpoint of side AB, E the midpoint of side BC, and F the midpoint of side AC (cf. Fig. 4c). Let k1 be the circle passing through A, D, and F ; let k2 be the circle through B, E, and D; and let k3 be the circle through C, F , and E. Prove that k1 , k2 , and k3 intersect in a point. So far we have learned how to establish that points are concyclic. How is this related to Problem 4, where we are already given several circles, and no more circles seem to be needed? PST 83. To show that several figures are concurrent, label by P an intersection point for two of them and prove the others also pass through P .6 y

B

C γ

E

k

k3

A M P

F D k1

B

A Q

O

x

C

P

α A

E k2 β

D

B

Figure 5. Solutions to Problems 2, 3, and 4 Partial solution to Problem 4: For short, denote the angles of ABC by α, β, and γ as in Figure 5c. PST 83 prompts us to label by P the intersection point (other than D) of k1 and k2 and to show that k3 also passes through P , i.e., that C, F , P , and E are concyclic. But since A, D, P , and F are concyclic on k1 and B, D, P , and E are concyclic on k2 , Theorem 2(b) yields ∠DP F = 180◦ − α and ∠DP E = 180◦ − β, which implies ∠DP F + ∠DP E = 360◦ − α − β = 180◦ + γ. The last equation follows from the well-known sum α + β + γ = 180◦ in ABC. On the other hand, the angles about P add up to 360◦ , so ∠EP F = 360◦ − (∠DP F + ∠DP E) = 360◦ − (180◦ + γ) = 180◦ − γ. To summarize, ∠EP F + γ = 180◦ , which again by Theorem 2(b) implies that C, F , P , and E are concyclic. Since C, F , and E already determine the circle k3 , we conclude that k3 also passes through P , and hence k1 , k2 , ? and k3 are concurrent. We are done. 6 Note how PST 83 for concurrent figures resembles PST 82 for concyclic points.

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Really? A solution of the above type would not have received a perfect score at BAMO 2000, because it has landed in the pitfall about which PST 80 warned us – relying solely on pictures for proofs. To complete the solution, we need to investigate the various possible positions of P , as PST 81 advises. Figure 6a depicts one such possibility for P ; can you pinpoint what feature of ABC “spat” P out into the exterior of the triangle? As the reader will discover, the solution in this situation will invoke Theorem 2 three times: part (a) twice and part (b) once. k3 C P F k1

C k2

E A A

D

B

B

Figure 6. Possibilities in Problem 4



The four shaded areas in Figure 6b represent all possible positions for P . Why can’t P fall in the three white sectors emanating from vertices A, B, and C? After answering this question, the reader is encouraged to complete the solution for each of the remaining shaded areas. Yet, even after all this careful case analysis, a question should be burning in the back of our minds: why did the hypothesis of Problem 4 require that the points D, E, and F be midpoints of the corresponding sides of ABC? We never used this fact in the proof! What happened in practice at BAMO 2000 was that many participants tried to use this extra irrelevant information, which gave no particular advantage in finding a correct solution. We draw our conclusions here: PST 84. Watch out for irrelevant information in the hypothesis of a problem. One sure way to determine that certain facts are irrelevant is to come up with a solution which does not depend on them. Trying to use such facts may be misleading, while throwing them out may lead to generalizations with enlightening proofs. Indeed, the statement of Problem 4 can be generalized to




Problem 5. Let D, E, and F be points on lines AB, BC, and CA, different from the vertices A, B, and C of ABC. If k1 is the circle passing through A, D, and F ; k2 the circle through B, E, and D; and k3 the circle through C, F , and E, prove that k1 , k2 , and k3 are concurrent. We finish this section with a challenge for the advanced readers: Question. (Advanced) Is there a deeper reason (possibly in a different area of mathematics), beyond chasing angles in circles, which explains why the three circles in Problem 5 must be concurrent?

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4. GOING “OFF ON A TANGENT”

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4. Going “off on a Tangent” Leads Straight to the Point! 4.1. Basics on tangents. If we consider the lines through a fixed point A on a circle k (cf. Fig. 7a), we will notice that all of these lines intersect k in two points, except for one special line: the tangent t. Definition 3. A line t is tangent to a circle k if t and k have exactly one point in common. B C

k O

C

k

B

k

R E

O

O C

C

t A

D

S

T

D1

A

D

Figure 7. Tangent lines and tlc angles The reader is encouraged to review the following basics on tangents: Exercise 5. Let k be a circle with center O. (a) If A is a point on k, then a line t through A is tangent to k iff t is perpendicular to the radius OA (cf. Fig. 7a). (b) If T is a point outside k, then there are exactly two tangents through T to k (cf. Fig. 7b). The tangent segments have equal lengths, and T O bisects the angle between them, i.e., T S = T R and ∠ST O = ∠RT O. 4.2. Taking inscribed angles to the limit. In addition to inscribed and central angles, tangent lines provide a third way of relating angles to circles. Let us fix two points A and B on k (cf. Fig. 7a). As a third point C on k
moves from B to A along one of the arcs BA, lines AC approach the tangent AD. Hence, the inscribed angles ∠BAC approach the special ∠BAD. Definition 4. Given two points A and B on a circle k and a line AD tangent to k at A, the angle BAD formed by the tangent AD and the chord AB is called for short a tlc 7 angle of k. How do we measure tlc angles? Can we also associate them to arcs on the circle? Since the inscribed ∠BAC (cf. Fig. 7a) measures half of the


corresponding BC and since BC approaches BA (as C approaches A), it is
reasonable to expect that tlc ∠BAD measures half of BA, i.e., ∠BAD = 1 2 ∠BOA. Theorem 3 confirms this intelligent guess. 7 For lack of a standard name, the acronym

tlc was coined by the editors to avoid the cumbersome wording. It reflects the transition discussed above from an inscribed angle to the Tangential Limiting Case and provides a meaningful mnemonic: Tangent Line-Chord. Needless to add, treat these angles with Tender Loving Care. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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11. FAVORITE PROBLEMS AT BMC. PART I

Theorem 3 (TLC Angle). Let ∠BAD be tlc in circle k with B and A on k and AD tangent to k. Let E be another point on k. Then ∠BAD and ∠AEB are either equal or add up to 180◦ : if D and E are on different sides of line AB, then the angles are equal; otherwise they add up to 180◦ . Hint: Figure 7c is almost self-explanatory. It depicts the first situation where E and D are on opposite sides of line AB and draws our attention to two right angles and three equal angles. Follow the diagram to justify that the tlc ∠BAD equals the inscribed ∠AEB. Use this result and Theorem 2 to quickly show the second case with E and D on the same side of line AB. ♦ 4.3. TLC angles make the day! Even though tlc angles may not be as familiar to the general audience as inscribed and central angles, the following two Monthly Contest problems demonstrate that they are just as useful in problem solving.




Problem 6 (MC ’05). In ABC, ∠BAC = 90◦ and AB < AC. If k is the circle with diameter AB, let E be the intersection of k with BC (cf. Fig. 8a). Denote by t the tangent line to k that contains point E, and let D be the intersection of t with AC. Prove that CDE is isosceles. l A D

k

B

E

C D

k1

t

B A

C

k2

Figure 8. Problems 6 and 7 ?

Solution: The picture suggests that DE = DC, or, equivalently, that ? ∠DEC = ∠DCE. How do we get hold of ∠DEC? Note that AE is perpendicular to BC (why? cf. Fig. 9a), and hence AE is the altitude to the hypotenuse of right ABC. As such, AE divides ABC into two triangles EAC and EBA similar to it.8 Since the corresponding angles of similar triangles are equal, we have ∠DCE = ∠BAE. By Theorem 3, ∠ABE = ∠AED. But complements of equal angles are equal, so ∠DEC = ∠BAE. Summarizing, ∠DCE = ∠DEC and CDE is isosceles.
If you would prefer to prove directly that DE = DC, can you find an alternative solution with segments instead of angles (cf. the Hints section)? 8 On a tangential note, this simple construction incidentally proves a generalized ver-

sion of the Pythagorean Theorem: compare with the collage on the cover of Givental’s translation of Kiselev’s Geometry [51] and the discussion on pp. 230–33. See also [37].

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5. WHEN PHANTOM CIRCLES TEAM UP WITH TANGENTS




235

Problem 7 (MC ’05). Circles k1 and k2 intersect at A and B. Line l is a common tangent to these circles, and it touches k1 at C and k2 at D so that B is in the interior of ACD. Prove that ∠CAD + ∠CBD = 180◦ . The desired equation looks so much like the cyclicity condition for points A, B, C, and D from Theorem 2(b), yet we can verify with bare eyes that the four points do not lie on a circle! Theorem 2(b) is not applicable here because its hypothesis is not satisfied: A and B are on the same side of line CD. So, we must come up with a different idea. Hint: The trick is to realize that the common tangent l calls for using two

tlc angles: one with a chord of k1 and another with a chord of k2 . But there are two possibilities for tlc angles at each of C and D, e.g., ∠ACD

and ∠BCD are the two candidates at C . . . . Which of them should we use? l A D

k

C D

k1

t

B

A B

E

C

k2

F

Figure 9. Solutions to Problems 6 and 7 Looking at the problem from the viewpoint of A, how does ∠CAD enter the solution? If we split it in two by AB, we obtain two inscribed angles: ∠BAC in k1 and ∠BAD in k2 . Which of the angles suggested above do these two match? For further hints, look at Figure 9b. ♦

5. When Phantom Circles Team up with Tangents 5.1. Letting yourself free to investigate intelligently. This entire section is devoted to a single problem. The solution is not presented in the typical fashion of “moving forward” from hypothesis to the end-of-proof symbol. Instead, we examine various paths, some of which lead nowhere, revise old techniques, and even “reason backwards”: all in order to develop a sense of survival in tough problem solving situations.




Problem 8 (MC ’05). Given ABC with ∠B = 90◦ , denote by k a circle with center on side BC and tangent to AC. Let AE be the tangent line from A to k different from AC, with E on k. If B  is the midpoint of AC and M is the intersection of BB  and AE, prove that M B = M E.

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11. FAVORITE PROBLEMS AT BMC. PART I

5.2. Learning from our first failed attempts. The given data does not allow for direct comparison of the desired segments M B and M E: how would anyone calculate effectively their lengths (cf. Fig. 10a)? So, we should instead try to show that ∠M EB = ∠M BE. Their supplementary angles are ∠AEB and ∠EBB  , but it’s not evident at all why they should be equal! We could try to link ∠M EB and ∠M BE to angles in k by extending EB until it intersects circle k in some undistinguished point X. While tlc ∠M EB will measure half of unknown EX , exactly what ∠M BE will measure on k is even more obscure. No, this can’t be the “right” way to approach the problem.




A

D D

B

k



C

E M

A B

O

k

C

B E

X



Figure 10. Problem 8 and Revised cyclicity condition PST 85. Do not give up after several unsuccessful attempts because investigating and discarding non-working ideas actually brings you closer to a solution by encouraging you to look for new useful ideas. If everything you try fails, step back and look at the problem as a whole – are you missing something “obvious” ? The experienced reader should ponder over this (and possibly locate a useful phantom circle in Figure 10a), while we reformulate the basic cyclicity condition from Theorem 2(b): Exercise 6 (Warm-up). Let ABCD be a convex 9 quadrilateral, and let E lie on the extension of CB beyond point B (cf. Fig. 10b). Then ABCD is cyclic if‌f ∠ABE = ∠ADC. Armed with the result of Exercise 6, we can now approach our problem. Did you discover the phantom circle in Figure 10a? 5.3. Nailing down the phantom circle cements the foundation. Let O be the center of k. Once you know to look for a phantom circle, with a little experimenting you will likely conjecture that B, O, D, A, and E lie on a circle k  (cf. Fig. 11a). The fact that D lies on k  will turn out to be 9 Convex means that B and D are on opposite sides of line AC, and A and C are

on opposite sides of line BD, just as in Figure 10b. Convexity helps us avoid cases of “funny” quadrilaterals with an interior angle > 180◦ or with intersecting opposite sides (i.e., self-intersecting quadrilaterals). Draw such cases.

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5. WHEN PHANTOM CIRCLES TEAM UP WITH TANGENTS

237

irrelevant to the solution, but it is exciting to witness a situation where five of the given points are concyclic. Prove the latter by noticing that OA will be a diameter of k  : use three right angles. 5.4. Angle chasing in mid-phase of solution. Moving on to Figure 11b to avoid overcrowding, we engage the desired angles. Contrary to our initial impulse to use ∠M EB as a tlc angle of k, it will enter the solution in conjunction with k  instead. Indeed, Exercise 6 concludes that ∠M EB = ∠BOA for the cyclic quadrilateral AEBO. The Exterior Angle Theorem for OCA then implies ∠M EB = ∠BOA = ∠OCA + ∠OAC = ∠BCB  + ∠OAC.

(1) A

k D

A B

E M

F B

O

k D

k

C

B

E M

O

B

k

C

Figure 11. Solution to Problem 8 Now for the least obvious part: what does ∠M BE equal? Unfortunately, we cannot apply Exercise 6 to it since its supplementary ∠EBB  is not easily measurable as inscribed in k  . . . . But ∠EBO is nicely inscribed in k  : is that useful? Yes! Extending CB to point F splits ∠M BE (cf. Fig. 11b) and (2)

∠M BE = ∠M BF + ∠F BE = ∠CBB  + ∠OAE,

where we used vertical angles and Exercise 6 again for cyclic AEBO. To convince ourselves that ∠M EB does equal ∠M BE, it remains to match the angles in the RHS’s of (1) and (2). For starters, ∠OAC = ∠OAE ?

(why? cf. Exercise 5b), which leaves us with ∠BCB  = ∠CBB  . Why should this be true?



5.5. Logical interlude. Moving forward in a proof “as steady as a tank” may not always lead to the victory. . . . PST 86. Reason backwards. To prove a statement P1 , assume it is true and find out another statement P2 that follows from it. Can you now prove P2 independently of P1 ? Does P2 imply P1 ? If you answer Yes to both questions, in effect you will have shown the equivalence P1 ⇔ P2 , proven P2 , and have the right to conclude P1 automatically.

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11. FAVORITE PROBLEMS AT BMC. PART I

Do not be alarmed: you will not be employing circular reasoning here. “Reasoning backwards” is a valid logical technique which helps locate where your partial solution fits in the big picture. 5.6. Well-deserved victory. Let’s utilize PST 86 to prove Statement P1 : {∠BCB  = ∠CBB  }. Assuming that P1 is true, we would conclude BB  = CB  . But B  is the midpoint of AC, so this would mean BB  = B  C = B  A, i.e., Statement P2 : {Point B  is the circumcenter of ABC}. Can we show P2 from scratch? Ah, yes! Proof: Since ∠ABC = 90◦ , the circumcenter of ABC is the midpoint B  of the hypotenuse AC (why?). Establishing P2 ⇒ P1 is now trivial: we have B  C = B  B = B  A and hence ∠BCB  = ∠CBB  . We can finally equate the RHS of (1) and (2): ∠M EB = ∠BCB  + ∠OAC = ∠CBB  + ∠OAE = ∠M BE, and reach the desired conclusion, M E = M B.




5.7. Epilogue. The bad news is that our solution was quite dependent on Figure 11, and as a result we treated only one of several possible cases. What could vary in Figure 11? For example, the placement of points D and B  along AC could be reversed so that D is between B  and C, which in turn can affect the position of M along line BB  : could M land between B and B  ? How does this change our solution? Exercise 7. List all possible relative positions of the points in Problem 8, paying special attention to D, B  , B, and M . Show how to complete the proof in case M is between B and B  . For extra mileage, check if the problem is still true when we allow for k’s center O to wander off along line BC outside segment BC. The good news is that we not only found one phantom circle in Problem 8 (k  going through five points), but another one popped up naturally in the solution (the circumcircle k  of ABC with center B  ; draw it!), and a third one (k) was given in the hypothesis. This makes the problem even more exciting because every time we encounter 3 circles, something special happens: a certain triple of lines10 is always concurrent. For example, Figure 12a illustrates 3 circles intersecting pairwise in 3 chords. These chords, as you will observe, are concurrent! Exercise 8. In the situation of Figure 11, draw the pairwise intersection chords of k, k  , and k  (ED, AB, and some Y Z). Check in your picture that the lines determined by these three chords are concurrent. 10 Called the radical axes, whose concurrence will be proved in Part II of Inversion.

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6. CONSTRUCTING CYCLIC TRAPEZOIDS “OUT OF THIN AIR”

239

?

Figure 12. Comparison of (3,3,7)-configurations We have brought up the phenomenon in Figure 12a because it is a (3, 3, 7)-configuration of 3 concurrent lines, 3 circles in general11 position, and 7 points lying on various intersections of these lines and circles. Didn’t we encounter something similar in Problem 5? We had another (3, 3, 7)-configuration with concurrent circles and lines in general position (cf. Fig. 12b). Question (Advanced). Is there some transformation which takes one of the (3,3,7)-configurations into the other? In other words, can we consider these two configurations in some sense the same (or “duals” of each other)? If the answer is Yes, then the solution to Problem 5 from Figure 12b should carry over to prove the situation in Figure 12a, and vice versa. If No, then we have two essentially different (3, 3, 7)-configurations in the plane. By the way, re-reading the original Problem 8, one wouldn’t expect to deal with statements about concurrent lines in (3, 3, 7)-configurations. We have witnessed how letting ourselves be intellectually free to wander outside the boundary of a problem leads to unexpected ideas, results, and open questions.

6. Constructing Cyclic Trapezoids “out of Thin Air” 6.1. Auxiliary constructions are arguably the hardest and most gratifying part of geometry problem solving. Deep observations, clever reasoning and leaps of intelligent guessing are needed to come up with relevant extra geometric constructions. So far we have drawn extra circles and extended segments until they met other figures. Constructions of extra lines, triangles, rectangles, and especially cyclic trapezoids await us ahead. Just like phantom circles, cyclic trapezoids are initially nowhere to be found in the statements of the problems below: they need to be constructed “out of thin air.” But hold on: to “trap” a trapezoid in a circle is impossible in general! Exercise 9. Any trapezoid inscribed in a circle is isosceles. 11 Pairwise intersecting but not concurrent.

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11. FAVORITE PROBLEMS AT BMC. PART I

With this clarified, the section title is more than suggestive: incorporate cyclic/isosceles trapezoids in your solutions. The overarching PST which we learn in the long run is PST 87. When out of standard ideas, be creative and try to come up with useful extra figures of various types. Be aware that there is no foolproof way of teaching you how to “concoct” these extra shapes, other than making connections with previous problems and applying your imagination. The more experience you have with such problems, the more successful you will be in new geometric situations. 6.2. Isn’t one solution enough? Problem 9 below has a number of different solutions. Two of them which involve auxiliary constructions are presented below. The first one follows the spirit of the earlier text and uses the fine properties of the circle. The second one is based on symmetries specifically related to the problem and introduces new ideas helpful in future situations. In general, two substantially different solutions will usually benefit us more than just one solution.




Problem 9 (MC ’06). Let AB be a diameter of circle k. An arbitrary line l intersects k in P and Q. If A1 and B1 are the feet of the perpendiculars, respectively, from A and B to P Q, prove that A1 P = B1 Q. Before we start, let’s make sure that our pictures represent all cases. Exercise 10. Draw all essentially different 12 configurations for Problem 9. Hint: The relative configuration of the points depends on where line l intersects line AB with respect to points A and B. Be aware that Figures 13a–b depict only two of the several possible configurations. ♦ A1

P Q

A

P B1 B k

l

A1 A

B k

B1 Q l

Figure 13. Two of the cases in Problem 9 Sketch of first solution to Problem 9: Consider the case when segment A1 B1 doesn’t intersect segment AB, as in Figure 13a. In this situation, exactly one of segments AA1 and BB1 intersects k in two points, or 12 Essentially different configurations require separate proofs, as opposed to essentially equivalent ones. The latter are often obtained from one another by some symmetry.

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6. CONSTRUCTING CYCLIC TRAPEZOIDS “OUT OF THIN AIR”

241

both segments are tangent to k (why?). WLOG, assume that segment AA1 intersects k for a second time at A2 (cf. Fig. 14a). Justify the following items: (a) A1 A2 BB1 is a rectangle (∠AA2 B = 90◦ ); (b) A2 BQP is a cyclic trapezoid, hence isosceles; (c) A2 P = BQ and ∠A2 P A1 = ∠BQB1 ; (d) A2 P A1 ∼ = BQB1 . The required statement A1 P = B1 Q now follows immediately from (d). ♦ Exercise 11. Fill in the gap above: what happens when both AA1 and BB1 are tangent to k? Adjust the solution accordingly. A1 A2 A

P Q

B2

P B1 B k

A1

l A

B

O

s B1 A2

k

Q l

Figure 14. First and second solutions to Problem 9 Sketch of second solution to Problem 9: Let A2 and B2 be the second intersections of lines AA1 and BB1 with circle k (cf. Fig. 14b). Let s be the line through the center of k perpendicular to l. Justify the steps: (a) AB2 BA2 is a rectangle (cyclic, AB2 BA2 , ∠AB2 B = 90◦ ); (b) s is an axis of symmetry for rectangle AB2 BA2 ; (c) A1 B1 B2 A is a rectangle, and s is also an axis of symmetry for it; (d) k is also symmetric with respect to s. We conclude that the whole picture is symmetric with respect to s, in par♦ ticular, so are segments A1 P and B1 Q. Thus A1 Q = B1 P . 6.3. Coming full circle. To close this session, we return to the beginning: let us locate the “missing” circle in our initial Problem 1 in order to solve it. We observe that no four of the given points are concyclic, while the circles going through triplets of points (e.g., the circumscribed circle of ABC) do not suggest anything useful (yet!). Thus, not only are we looking for a phantom circle, but some of the points that determine it may not be in the picture yet – they need to be constructed as well! Lacking any other ideas, let us apply PST 85: we step back and look at the problem as a whole. Essentially, the only information we are given is the equality of two angles, ∠BAD = ∠CAE, and of two segments, BD = CE. Can we somehow combine these to land on a circle? The cyclicity condition Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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11. FAVORITE PROBLEMS AT BMC. PART I

in Theorem 2(a) should come to mind: the two equal inscribed angles in Figure 3a (∠C = ∠D) subtend the same chord (AB). Can we force a similar situation in our Problem 1? Yes, if we manage to put the two segments BD and EC on top of each t t other. This can be done via a translation t which sends B → E and D → C. The vertex A is sent by t to a new point A (cf. Fig. 15). In effect, the whole t BDA → ECA so that ∠EA C = ∠BAD = ∠EAC. A

A

t

k

B

D

E

C

Figure 15. Solution to Problem 1 There is only one thing to do now: we conclude that points E, C, A , and A are concyclic (why?). We’ve got our phantom circle k, but where is the cyclic trapezoid ? Recall that in addition to preserving shapes of figures, translations move everything along parallel lines. In our case this means AA BC, so ECA A is indeed a trapezoid inscribed in k and hence isosceles. The sides EA = CA beg to be used, but it is faster to go with the equal t diagonals of ECA A: CA = EA = BA, and hence ABC is isosceles.
6.4. The aftermath.13 One of the editors of this book worked with a group of middle school teachers on Problem 1. At the end, she asked them to summarize the above solution. A teacher eagerly offered: “Beautiful!” And so, we drew a picture of the exotic flower in Figure 16. Its halfshaded leaves are the PST’s we used, and its dark leaves represent the theory involved. The blossom is the main idea of translation, which is undoubtedly the true heart of the solution, its most beautiful and original part. As we mentioned earlier, there are no algorithms for the creation of such extra geometric constructions: only through solving problems can one gain experience and enough confidence to discover such elegant and useful constructions. In retrospect, the solution to Problem 1 is not that hard; in fact, it rests on three elementary facts (the dark leaves). But how one puts together these facts into a full-fledged solution is a much harder process, which requires the half-shaded leaves of the PST’s, as well as the blossom of the main idea of translation. As the saying goes: “Every problem is easy . . . once you have solved it!” 13 Every year, the first BMC session after the BAMO exam is devoted to discussing solutions to the BAMO problems and is appropriately entitled “BAMO Aftermath.”

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7. HINTS AND SOLUTIONS TO SELECTED PROBLEMS

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Main Idea: Translation Theorem: Cyclicity condition PST: Look for cyclic trapezoids

PST: Relate to previous problems PST: Reason backwards

Properties of translations

PST: Look for phantom circles PST: Look at the problem as a whole

Exercise: Cyclic = isosceles trapezoid

PST: Translate back to original problem

Figure 16. Summary of Solution to Problem 1

7. Hints and Solutions to Selected Problems Exercise 2(a): Use the fact that ∠ABC is supplementary to both the external ∠XBC and the sum ∠BAC + ∠ACB. ♦ Theorem 1: In the case O is inside inscribed ∠ACB, draw diameter CD (cf. Fig. 17a). Then by the case in Figure 1b proven earlier, ∠ACD = 1 1 2 ∠AOD and ∠BCD = 2 ∠BOD. Adding the two equations gives ∠ACB = 1 1 1
1 1
2 ∠AOD + 2 ∠BOD = 2 (AD + DB ) = 2 ADB = 2 ∠AOB.










Exercise 3: BC is a diameter of circle k iff the intercepted arc BDC is a semicircle and hence has a measure of 180◦ (cf. Fig. 17b). By Theorem 1, this is equivalent to ∠BAC = 12 · 180◦ = 90◦ . Similarly, ∠BAC is right iff BC is a diameter in the circle k through A, B, and C, which is the same as OB = OA = OC for the center O of k lying on segment BC.
C

B

α β

O 2α

D

k B

O

2β

A

A D

k C

Figure 17. Solutions to Theorem 1 and Exercise 3 Exercise 4: Let AB = CD. If O is the center of k, then ABO ∼ = CDO by SSS–criterion for congruence, so ∠AOB = ∠BOD (cf. Fig. 18a) and Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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11. FAVORITE PROBLEMS AT BMC. PART I








hence CD = AB where the arcs do not contain F and E, respectively. By
the Inscribed Angle Theorem, ∠CF D = 12 CD = 12 AB = ∠AEB. The converse is carried out correspondingly.
C

F E

C k

k D

O

B A

A

E

B

E

Figure 18. Solutions to Exercise 4 and Theorem 2(b) Theorem 2(b)(continued): We suppose E is in the interior of circle k and ∠ACB + ∠AEB = 180◦ (cf. Fig. 18b). Extend AE until it intersects k in E  . Since A, B, C, and E  are concyclic, ∠ACB + ∠AE  B = 180◦ , which combines with our previous equality to yield ∠AEB = ∠AE  B. But ∠AEB is an exterior angle of EBE  , so that ∠AEB > ∠AE  B, a contradiction. This shows that E cannot be inside k. Earlier we showed that E cannot be outside k, which leaves only one possibility: E must be on k, and thus A, B, C, and E are concyclic.
D

D C A

D

F

γ β

D

k

α

k

D

B k

C

B C

A

E

B

A

E

Figure 19. Theorem 2(a)




Theorem 2(a): (⇒) If A, B, C, and D are concyclic then ∠ACB and
∠ADB are equal since each is half the measure of AEB (cf. Fig. 3). (⇐) If ∠ACB = ∠ADB then construct the circumcircle k of ABC. We must now show that D lies on k. Suppose D is in the exterior of k. If both AD and BD are tangent to k, then ∠ACB > ∠ADB (why is α > β > γ in Figure 19a?), a contradiction. So assume that at least one of AD or BD is not tangent to k; for instance, let AD intersect k in point D  (cf. Fig. 19b). By the Inscribed Angle Theorem, ∠AD  B = ∠ACB = ∠ADB. But exterior ∠AD B > ∠ADB in BD D, a contradiction. Hence D is not outside of k. To complete the proof, suppose D is in the interior of k, and extend AD to D on k (cf. Fig. 19c). An almost identical argument leads to a contradiction. Therefore, D must lie on k and A, B, C, and D are concyclic.
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7. HINTS AND SOLUTIONS TO SELECTED PROBLEMS

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Problem 3: It is well known that the diagonals of square AECD are perpendicular, equal, and bisect each other. Thus AM C is an isosceles right triangle with ∠M AC = 45◦ and ∠ABC +∠AM C = 90◦ +90◦ = 180◦ (check this against Fig. 5b). By Theorem 2(b), A, B, C, and M lie on a circle k  . But then ∠M BC and ∠M AC are two inscribed angles in k  intercepting the
same arc, so they are equal: ∠M BC = ∠M AC = 45◦ . Problem 4: If ABC is a right triangle, it is easy to see that P is the midpoint of its hypotenuse; and if the triangle is obtuse, say, ∠A > 90◦ , then P is in the exterior of the triangle in the region bordering side BC (as in Figure 6a). To prove that k3 passes through P , we begin just as in the original case (where ABC was acute) and label by α, β, and γ the angles ∠A, ∠B, and ∠C, respectively. Let ∠F P E = x and ∠DP E = y. By Theorem 2(a), we have β = y, and by Theorem 2(b), we have α = 180◦ − ∠F P D = 180◦ − (x + y). Then 180◦ = α + β + γ = 180◦ − (x + y) + y + γ, from which γ = x and C, F, E, and P are concyclic by Theorem 2(a). We conclude that P is on k3 . The above argument covers analogously all three shaded exterior regions in Figure 6b, and the case in the text covers the interior of ABC. The only remaining regions to consider are the three white regions in Figure 6b. But none of the three circles even goes through these white regions (why?), so all these cases are vacuous and the problem has been solved.
Exercise 5(a): (⇒) If line AD is perpendicular to radius OA, let B be an arbitrary point on line AD. Then OB is the hypotenuse in OAB, so OB > OA (why?). Thus, point B is in the exterior of the circle, and in particular, B does not lie on k. We conclude that the only point on line AD shared with k is A, so by definition, AD is tangent to k.
(⇐) If t is tangent to the circle at A, then every other point B of t is exterior to the circle (why?). Hence OB > OA, so OA is the shortest segment from O to t. But the shortest segment from a point to a line is the perpendicular segment (why?), from which we conclude that OA ⊥ t.
Exercise 5(b): Construct the circle k  centered at the midpoint M of T O with radius M O. Then k  intersects k at two distinct points; call them R and S. Since ∠ORT and ∠OST are inscribed in semicircles in k  , they are right angles, and T R and T S are tangent segments to k. Since OST ∼ = ORT (the same hypotenuse and equal legs OR = OS), then T R = T S and ∠T OR = ∠T OS. Note that if there were a third tangent T Q to k, then ∠OQT = 90◦ would force Q to lie on k  , i.e., k and k  would have 3 distinct points in common – Q, R, and S – impossible (why?). Thus, T R and T S constructed above are the only tangents to k through T .
Theorem 3: Let C be the midpoint of AB (cf. Fig. 7c). When E and D are on opposite sides of AB, ∠BAD and ∠AOC are complements of ∠OAC (why?), so they are equal. In addition, ∠BOC = ∠AOC, so ∠AOC = Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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11. FAVORITE PROBLEMS AT BMC. PART I

1 2 ∠BOA. 1 2 ∠BOA, If D1

Combining, ∠BAD = ∠AOC = 12 ∠BOA. But inscribed ∠BEA = so ∠BEA = ∠BAD, as desired. lies on tangent AD so that D1 and E are on the same side of AB, then ∠BAD1 = 180◦ − ∠BAD = 180◦ − ∠BEA; hence ∠BAD1 + ∠BEA =
180◦ , as desired. Problem 6: Since AB is a diameter, then ∠AEB = 90◦ , and hence ∠AEC = 90◦ (cf. Fig. 9a). Since DA = DE as tangent segments, we find ourselves in a situation similar to Exercise 3: D must be the midpoint of hypotenuse AC of right AEC, and therefore DC = DE.
Problem 7: By Theorem 3, ∠DCB = ∠BAC and ∠CDB = ∠BAD (cf. Fig 9b). In BCD, 180◦ = ∠DCB + ∠CDB + ∠CBD. Substitut
ing gives 180◦ = ∠BAC + ∠BAD + ∠CBD = ∠CAD + ∠CBD. Exercise 6: Since B, E, and C are collinear, ∠ABE + ∠ABC = 180◦ . By Theorem 2(b), A, B, C, and D are concyclic iff ∠ADC + ∠ABC = 180◦ (cf. Fig. 10b), or, equivalently, ∠ABE = ∠ADC.
A

A

k

k

k B



M

Y

D

D

P E C

O

B E

k

k

B Z

Figure 20. Exercise 7 and Exercise 8 Exercise 7: As point O moves from C to B, point M moves from B  along −−→ ray B  B. When O lies on the angle bisector of ∠BAC, M coincides with −−→ B (why?). As O continues toward B, M moves along BB  with B between B  and M . When O coincides with B, lines AE and BB  do not intersect (why?). As O continues past B outside segment BC, M appears again on line BB  , but now B  is between B and M . Show that in all these cases M E = M B still holds. In the case where M is between B and B  (cf. Fig. 20a), we still have three right triangles with hypotenuse AO, which implies A, B, E, O, and D are concyclic. Combining with the fact that AO is still the bisector of ∠DAE and BB  = CB  , we derive a number of angle equalities: ∠CAO = ∠OAE = ∠OBE (1-arc), ∠C = ∠B  BC (2-arcs), and ∠M EB = ∠AOB (3-arcs). Hence ∠M BE is formed by a 1-arc and a 2-arc angle. But so is ∠AOB = ∠OAC + ∠ACO (3-arcs = 1-arc + 2-arcs); thus ∠M EB = ∠AOB = ∠M BE and M EB is isosceles. ♦ Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Exercise 8: For the drawing see Figure 20b, where the circumcircle k  of ABC has been added. As stated in the footnote on p. 238, the concurrency of AB, ED, and Y Z will be proved in Part II of Inversion using the so-called power of a point with respect to a circle. If you can’t wait till Volume II, read the solution to the concurrency problem in Paul Zeitz’ The Art and Craft of Problem Solving [101], pp. 292-93. Embedding the problem in three dimensions leads to a beautiful proof requiring almost no special knowledge: see Alexander Bogomolny’s web page “Cut-the-Knot” [11]. ♦ Exercise 9: Let ABCD be a cyclic trapezoid with AB || DC. Draw chord BD. Note that ∠ABD = ∠BDC, as equal alternate interior angles. By Exercise 4, AD = BC, and hence ABCD is isosceles.
Exercise 10: Besides the two configurations given in Figure 13, there are several degenerate cases, in which some of the given points coincide with each other. For example, if line l passes through B (but not A), then P = A1 and Q = B = B1 ; if l is perpendicular to AB, then A1 = B1 , etc. What happens if l coincides with line AB; if l is tangent to k at A, or at B, or at some other point C? Does this describe all possible degenerate cases? ♦ Exercise 11: When AA1 and BB1 are both tangent to k, then AB is perpendicular to both segments and AA1 B1 B is a rectangle. Now AP QB is an isosceles trapezoid (why?) and the solution proceeds as before. ♦

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Session 12 Monovariants. Part I Mansion Walks and Frog Migrations based on

Gabriel Carroll’s session

Sneak Preview. We learned in Stomp about invariants: features that never change under certain operations. In this session we introduce their close cousins, the monovariants, which can change but only in one direction. We shall (virtually!) break fine china, stuff ourselves with chocolate, walk around mansions, and crowd frogs onto a single lily pad . . . all in search of the elusive monovariants. We shall construct monovariants in numerous examples and focus on one of their major applications: showing that iterative processes end. For tougher problems and more diverse applications, you must wait till future Parts II–III. Here and there, the reader will encounter exercises using basic combinatorial counting techniques. Such exercises are not essential for the monovariant theme and can be omitted on a first reading, but they can serve as good practice of the material introduced in the Combinatorics session.

1. China Shops and Chocolate Bars 1.1. Does breaking fine china lead to any good? In daily life, there are some things that always increase or always decrease. One simple example is your age: as time passes, you can only get older (sadly). For another example, imagine you have a bowl full of peanuts and are gradually eating them. The number of peanuts left in the bowl can only decrease (barring any impolite conduct on your part). Or imagine that you are in a room full of fine china and proceed to drop it on the floor. You can then pick up the shattered pieces and drop them, in turn, on the floor. With this operation, the number of pieces of china can only increase, never decrease. 249

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12. MONOVARIANTS. PART I

These mundane observations can be used to prove certain facts. For example, assuming you can only eat a whole number of peanuts in each mouthful, you must eventually stop eating: otherwise, sooner or later the bowl will be empty, and you can’t very well go into having negative numbers of peanuts. You can’t reassemble a plate from a bunch of shards by repeatedly dropping them on the floor, because the number of pieces can only increase, whereas you need it to decrease. The above discussion naturally gives birth to our new notion of a monovariant: a quantity that either can only increase or can only decrease when some operation is performed repeatedly. Even though our examples were illustrative, they were not very mathematically sophisticated. We need an actual math problem to get a real taste of monovariants. 1.2. Who gets to eat the chocolate? The game of chess is complex and there is no guarantee at the start of a game that a certain player will win. How about a world championship in the following game:




Problem 1. Two players take turns breaking up a 3 × 4 chocolate bar. At each turn, a player picks up a piece and breaks it along a division between its squares. Eventually, the bar will split into 12 squares and the game will end. The player who makes the last break wins and eats the chocolate. Is there a strategy for one of the two players to win?

Figure 1. Chocolate monovariant problem on a 2 × 3 board To fit an example within the page’s confines, Figure 1 demonstrates one possible game on a smaller, 2 × 3 board. The chocolate problem is a classic in the world of problem solving (cf. [28]), so if you know it, skip down to the next section without spoiling the adventure for your friends. If you are new to this type of problems, play out a few games and come to some conclusions. Meanwhile, we proceed in a typical problem solving fashion. 

















 




Figure 2. Staircase of problem solving events

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1.2.1. Ideas. How does one approach such problems? Relate the situation to your experience with the Stomp session and recall a few strategies: (a) First try examples with small numbers. (b) Are there any unchanging features throughout the game? (c) Pay attention to who wins on what size board. 1.2.2. Observations. Very quickly, you will discover that on 2 × 2 and 2 × 3 boards the winner is always the first player, but a 3 × 3 board seems to allow victory only to the second player regardless of how the game is played out! As for invariants (part (b)), you won’t find any, except that the total number of squares is preserved, which isn’t very helpful . . . Or is it? Let’s backtrack and try to think if we have seen something similar before. 1.2.3. The critical connection clicks. Our seemingly useless (besides being violent) example of breaking fine china should spring to mind: the monovariant there was the total number of pieces, and it increased every time we broke another piece. We are in an almost identical situation in the chocolate problem: at every turn, the players create one more chocolate piece! Solution to the chocolate problem: Our hidden monovariant is the total number of pieces at any time. Since we start with 1 piece (the unbroken 3×4 chocolate bar) and end with 12 square pieces, our monovariant starts at 1 and steadily increases by 1, until it reaches 12 and the game ends. Hence the startling conclusion: regardless of clever “strategizing”, the two players will make a total of 11 moves on a 3 × 4 board (why?). Since the first player will make the 1st , 3rd , 5th , . . . , and 11th moves, he/she will win. The game is a fix!
1.2.4. Extensions. Now the reader can (and is encouraged to) answer the chocolate problem for any m×n board, generalize the game to several players or to rectangular “cakes” in three dimensions, venture into four or more dimensions, tweak the rules, and do all sorts of mathematical exploration, which will likely lead to topics in number theory, game theory, linear algebra, combinatorics, and other fields of mathematics. But as far as this session is concerned, we are done with the chocolate problem and ready to dive into understanding monovariants more seriously.

2. Walking around a Mansion




Problem 2 (Mansion). 2000 people reside in the rooms of a 123-room mansion. Each minute, as long as not everyone is in the same room, somebody walks from one room into a different room with at least as many people in it. Prove that eventually all the people will gather in one room. To get a feeling for the problem, you should first play around with examples having a small number of rooms and people and verify that everyone

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12. MONOVARIANTS. PART I

will end up eventually in the same room. If you haven’t done such problems before, you are not expected to see a solution quickly, so read on. If you do see how to do it, read on also. (Otherwise you’ll just stop here, never get through the rest of this session, and miss out on learning how others think about this type of problem – events that should definitely be avoided.)

 

2.1. Simplicity and intuition. There’s a simple way to think about what is happening: each time someone walks into a room with more people, the people become overall more “concentrated” in crowded rooms. We need to pinpoint a concrete way of measuring concentration and then prove that this concentration increases at each step . . . until it can’t increase any more since, supposedly, all people will be in one room at the end. PST 88. Although mathematics requires rigor, don’t be afraid of intuition. You can be guided by vague concepts at first; then look for concrete ways to express them mathematically. The first thing that may come to mind is to add up the number of people in each room . . . . But that just gives us the total number of people, which is always 2000 – an invariant, not a monovariant! And we don’t need an invariant: we expect that all people will end up in the same room, so the fact that the number of people then will still be 2000 is a true but useless observation for now. What we need is to PST 89. Find some measure that gives disproportionately larger weight to rooms with more people. One way to do this is to add up the squares of the number of people in each room. 2.2. Full stop. Where did these squares come from? How could we have thought of them?. . . If you have played a certain card game called “Oh, hell,” you would have appreciated its peculiar scoring. Roughly, if a player takes n tricks in a game, his/her score is recorded as n2 points. So, for example, if one player takes 4 tricks and another takes 5 tricks, their respective scores will be 16 and 25. In2 4 stead of a difference of 1 (= 5 − 4), the second player will have gained the much larger difference in score of 9 (= 25 − 16). The moral: even just a 52 bit of “better playing” leads to a disproportionately better score. While in the card game there is no reason to add up the scores of all players, in our mansion puzzle we should, in order to have some idea of what is happening all over the mansion. Thus, for instance, if there were only 3 rooms with 4, 5, and 1991 people in them, respectively, we would record the sum S = 42 + 52 + 19912 . Obviously, not only does the 1991-people room

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2. WALKING AROUND A MANSION

253

have more people than the other rooms, but its “weight” of 19912 = 3964081 in the overall sum S is way disproportionate to the size of the crowd in it. 2.3. Concentration is a monovariant. We now have very good reasons to define the concentration of the mansion as the sum of the squares of the number of people in each room. That is, if ni is the number of people in room i, then the concentration is n21 + n22 + · · · + n2123 . Exercise 1. Prove that each time someone walks from a room to another room with at least as many people, the concentration increases. To understand what the exercise says, let’s work on our simplified example with (4, 5, 1991)-configuration of people in 3 rooms. If someone moves from the 4-person to the 5person room, the new configuration will be (3, 6, 1991), so we can subtract the old from the new concentration and regroup according to rooms:

(4, 5, 1991) → (3, 6, 1991)

New − Old = (32 + 62 + 19912 ) − (42 + 52 + 19912 ) = (32 − 42 ) + (62 − 52 ) = −7 + 11 = 4 > 0. Aha! The concentration increased! The whole calculation depended actually only on the two rooms where the action occurred: (4, 5) → (3, 6), while the rest of the sum (19912 ) had no effect on the change of concentration. OK, but this doesn’t explain why the concentration increased ! Have we forgotten something? We haven’t really used the fact that people can only move into more crowded rooms. So, it is time to turn numbers into letters and fully surrender to algebra. Solution to Exercise 1: If someone walks from a room with a people to a more crowded room with b people (b ≥ a), the concentrations are • before the move: a2 + b2 + (terms for the other rooms), • after the move: (a − 1)2 + (b + 1)2 + (terms for the other rooms). Since the other rooms are unaffected, the net change in concentration is         (a − 1)2 + (b + 1)2 − a2 + b2 = a2 − 2a + 1 + b2 + 2b + 1 − a2 + b2 = 2(b − a) + 2 ≥ 2 > 0. The condition b ≥ a is indeed necessary for the last inequalities.




What we’ve just shown is that the concentration, defined via squares, always increases (by at least 2) and hence is a monovariant. Now we can use this fact to solve the original mansion problem. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Solution to Problem 2: As we demonstrated, at each step, the concentration increases. However, there are a finite number of people and a finite number of rooms. Therefore, there are a finite number of ways the people can be distributed among the rooms, and so there are a finite number of possible values for the concentration. This means that the concentration cannot go on increasing forever. So eventually the process must stop, which means all the people are in the same room.


3. Finite versus Infinite If you had trouble following the previous paragraph, do not wither: we shall spend this entire section examining it from various angles. Needless to say, the exact number of rooms and people didn’t matter a bit: a related issue will be under close inspection below too. 3.1. Counterexamples sharpen our understanding. Which single word is the most important in the conclusion of the mansion puzzle? “Concentration” and “increases” are excellent candidates. But there is an arguably better one: “finite” appears four times in the solution paragraph above. What is so important about finite? Very simply, the problem would have been false without it. Indeed, let’s think of situations where • the number of rooms or the number of people (or both) is infinite (we are violating one condition of the problem) and • the remaining conditions of the problem are satisfied (e.g., a person moves only to a room at least as crowded as the previous), but • the conclusion of the problem is false: the people keep moving and the process never ends. Such situations are called counterexamples. We will use them not to disprove the original problem (that would be silly!) but to decide if weakening its hypothesis (dropping the “finiteness” condition) leads to a false new problem. Exercise 2. Can you think of counterexamples to Problem 2 with (a) infinitely many people, but finitely many rooms; (b) finitely many people, but infinitely many rooms; (c) infinitely many people and infinitely many rooms? Partial Solution: In (a), “infinitely many people in finitely many rooms” should ring a bell: by the Pigeonhole Principle some room Z has infinitely many people. If all other rooms have finitely many people, eventually all people will end in Z (why? cf. Fig. 3), so this is not a counterexample. Hence, we need at least two infinitely-populated rooms. Verify that two such rooms, Y and Z, will not allow the process to end, regardless of the distribution of people in the remaining rooms (cf. Fig. 4). ♦ Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Z

Figure 3. A single infinitely-populated room Z In part (b), actually the problem is still true! If we ignore the empty rooms, we have finitely many people in finitely many rooms, which is just the original mansion problem. Explain why the process will always end. ♦

Y

Z

Figure 4. Two infinitely-populated rooms For part (c), the counterexample with the two infinitely-populated rooms Y and Z from part (a) works here as well (why?). But it is more satisfying to create a situation where each room has finitely many people, yet the problem is false. Create such an example and argue carefully why the process will never end there. ♦ The above discussion can be summarized in




Problem 3 (Infinite Mansion). The room-migration process described in Problem 2 never ends iff the mansion initially has (a) at least two infinitely-occupied rooms or (b) infinitely many non-empty rooms. Equivalently, the process ends iff at most one room is infinitely-occupied and only finitely many people reside in the remaining rooms. 3.2. Infinity is temporarily outlawed. The reader should be convinced by now that the word “finite” is crucial in our mansion problem, even though it is only implicitly mentioned as “2000 people”.1 In general, observations about “finite versus infinite” are of the essence when using monovariants. PST 90. After locating a monovariant, you must ensure that its increase (or decrease) is bounded by some obstruction, in order to show that the corresponding process will terminate. 1 The fact that there are finitely many (123) rooms also turns out to be irrelevant: recall that there are no counterexamples in Exercise 2(b).

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Let’s apply this PST to our mansion-monovariant, the concentration. The latter was a function f of 123 variables (the rooms) where each variable ni could take only 2001 values – the possible number of people from 0 to 2000 in one room. Namely, ni ∈ {0, 1, 2, . . . , 2000} for all i and f (n1 , n2 , . . . , n123 ) = n21 + n22 + · · · + n2123 . But since there are 2001 possibilities for each of the 123 inputs ni , the function f can obtain at most (1)

123 values. 2001 · 2001  · · · 2001 = 2001  123

Wow, this is one lousy overestimate! But who cares? Sometimes “the lousier the better”, since we can get more quickly and less painfully to our goal. In our case, we deduce that there are only finitely many possibilities for the monovariant f . More precisely, at each step f changes its value (increases by at least 2), so after at most 2001123 steps f will hit its maximum and the process will terminate. This finishes our more detailed explanation of the original mansion solution from page 253.
3.2.1. Combinatorial detour. For the counting technique in (1), you may want to review the “menu process” from the session on Combinatorics. Imagine 3 slots arranged one after the other: ? ? ? . How many numbers can be formed by writing any of the 10 digits 0, 1, . . . , 9 in each of the slots (repetitions allowed)? Obviously, all 3-digit numbers, including those having 0’s in the beginning: these are all integers in [0, 999], i.e., a total of 1000 possibilities. Another fast way to count the same thing is to multiply the number of possibilities in each slot: 10 · 10 · 10 = 103 . Back to the mansion problem: replace 10 by 2001 and 3 by 123 to obtain 2001123 possibilities for f . 3.3. Epilogue. Before we go on to bigger things, let’s make sure we have a really thorough understanding of the mansion problem. Earlier in this section we worked on weakening the hypothesis of the problem. Can we now strengthen the conclusion? Can we refine the result? Exercise 3 (Refined Mansion). Show that the concentration f is always positive and can never be larger than 20002 . Using this fact, prove that all the people will be in the same room after at most 2 million minutes. Proof: To find the largest possible value of f , recall that the monovariant f keeps increasing until all people are in the same room. This simply means that f is maximal in the end of the process: f (n1 , n2 , . . . , n123 ) ≤ 0 + · · · + 0 + 20002 + 0 + · · · + 0 = 20002 .   Since f attains only integer values in the interval 0, 20002 and it increases by at least 2 at each step, after at most 20002 /2 steps f will hit its maximum, ending the process.
The above proof teaches us one more useful idea: Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.



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PST 91. Try inequalities to restrict a monovariant to some interval [A, B]. If the monovariant attains only finitely many values in [A, B], it will reach its minimum (or maximum) in finitely many steps and end the process. Note that, although f becomes 20002 at the end of the process, the value of 0 can never be attained by f (why?). Since the largest possible value of f is attained when everyone is crowded in one room, our intuition says that the smallest possible value of f should be attained when the people are most evenly “spread out” throughout the mansion. Indeed,




Problem 4 (Intermediate). Prove that the smallest possible value for the mansion concentration f is 32544, and that it is attained exactly when 91 rooms have 16 people each and the remaining 32 rooms have 17 people each. The rigorous proof of this fact requires monovariant usage more advanced than the current session intends to introduce, so we defer the solution to the Hints section. But even if you do manage to successfully solve Problem 4, the newly found smallest value for f will not improve substantially the result of Exercise 3: it will take at most (20002 − 32544)/2 = 1,983,728 steps for all people to gather in the same room, which is still on the order of 2 million. To find the best possible bound on the number of steps in the mansion problem is a hard question, which will require even deeper, more advanced investigation (cf. the Appendix at the end of the session). Meanwhile, there is one more question we should address: Are there other ways we could have defined “concentration”? Exercise 4. Instead of defining the concentration as the sum of all the n2i , let us define it as the sum of 1/(ni + 1). Check that the solution still works as before. (Why wouldn’t the sum of all reciprocals 1/ni work?) Can you think of any other definitions we could have used?

4. Monovariant Teamwork A nice property of monovariants is that they don’t have to work alone; they can form tag teams. Once one monovariant stops changing, another one starts changing. Using more than one monovariant in tandem, we can describe what is going on in the mansion problem in a more general way, and perhaps a more intuitive way, because it doesn’t require any sneaky algebraic manipulations. To keep the ensuing discussion concrete and easy to visualize, we shall frequently refer to a simplified example with only 3 rooms and a (3, 4, 5)configuration evolving as follows: (3, 4, 5) → (2, 5, 5) → (2, 6, 4) → (1, 6, 5) → (0, 6, 6) → (0, 5, 7) → (0, 4, 8) → (0, 3, 9) → (0, 2, 10) → (0, 1, 11) → (0, 0, 12). We have underlined the largest entries at each step, allowing for ties. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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4.1. Alternative solution to original mansion Problem 2. Let’s just think of the numbers of people in each room and ignore the individual rooms. Thus we have an unordered collection of 123 numbers (not all different) which evolves according to the familiar rule: some two positive numbers a, b, with a ≤ b, are replaced by a − 1 and b + 1, for as long as two such numbers exist. 4.1.1. Our first tag team monovariant is the largest number L1 in the collection: it corresponds to the number of people in the most crowded room, though this may not be the same room at different points in time. For instance, in the (3, 4, 5)-example, L1 is represented by the underlined entries: 5, 5, 6, 6, 6, 7, 8, 9, 10, 11, 12 (cf. the top curve in Fig. 5b). These numbers refer back-and-forth to people in the 2nd and 3rd rooms. Observe that L1 is non-decreasing here. In general, L1 can only increase (or stay constant) at each step. Indeed, the only way for it to be affected by the operation is if (in the notation above) L1 = b, in which case it is replaced by the even larger value b + 1. Evidently, L1 cannot keep increasing forever (it can’t get above 2000), so eventually it must stop increasing and will stabilize. Figure 5a displays an example of L1 stabilizing at time t1 .

people

people

4.1.2. Our second tag team monovariant. Now look at what happens starting from time t1 , when L1 has stopped increasing. The same operation is performed but only on the remaining 122 numbers. So we can ignore L1 , and by the same reasoning, the largest of these 122 numbers, i.e., the secondlargest number L2 , will also eventually stop increasing at time t2 . Note that the dotted part of L2 in Figure 5a may initially wiggle up and down, but after t1 , the solid part of L2 will be non-decreasing. L1 L2

5 4 3

L3

12

L1 L2 L3

t1

t2

t3 time

10

time

Figure 5. Tag team monovariants: in theory and in reality 4.1.3. The remaining tag team monovariants. Starting from time t2 , when the two largest numbers L1 and L2 have both stabilized, the same argument shows that eventually the third-largest number L3 will stop increasing at time t3 , and so forth. Note that L3 in Figure 5a stabilizes at the same Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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value as L2 : this simply presents a possible tie without contradicting our reasoning. Repeating this argument, we eventually see that after some finite amount of time (t123 ), the entire collection of 123 numbers will stop changing. But this means that the people stop moving. Indeed, there is no way to perform our operation and have the (unordered) collection of entries happen to stay the same: for example, if a and b (with a ≤ b) are replaced by a−1 and b+1, then the number of entries that are > b has strictly increased by 1, which would change the collection instead of keeping it the same. People will only stop moving when they are all in the same room, so we are done.
4.2. Does the tag team technique really work? The solution above might seem a little misleading, because it makes it look as though L1 stabilizes at some time t1 , then the operation continues and L2 stabilizes at some later time t2 , and so forth: this is what happens in theory, as Figure 5a suggests. But in fact, looking back, we can see that L1 won’t stop increasing until everyone is in the same room and the whole process has finished, so all the other numbers Lj have also stabilized at that time. This is what happens in reality in the (3, 4, 5)-example, as illustrated by Figure 5b: at time t = 10, the monovariant L1 stabilizes at value 12, and simultaneously, L2 and L3 stabilize at value 0. However, we don’t know this at the outset. So we do need to look at all the numbers Lj separately and pretend that they might stabilize at different times in order to have a logically complete proof. In this solution, we’ve used not one monovariant but 123 monovariants linked together. This has to be done a little bit carefully. It’s not enough just to show that some monovariant increases at each step. In other words, we can’t do the following: Wrong solution to Problem 2: Each time someone walks from a room with a people to a room with b ≥ a people, the population of the second room increases to b + 1. So the populations of the rooms keep increasing. They cannot increase forever, so eventually the people must stop moving. ? This can only happen when everyone is in the same room.



This solution fails because it leaves open the possibility that when one monovariant is increasing, the others are sneaking back down (cf. the dotted parts of the Lj ’s in Figure 5), which defeats the whole point of monovariants. PST 92. For the tag team technique to work, the monovariants need to be arranged in an order, and we need to know that at each step, some monovariant is increasing and all the earlier ones are staying unchanged. 4.3. Tag teams attack harder problems. The tag team monovariants technique is quite powerful. For example, we can use it to solve a more general version of the mansion problem.

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Problem 5 (Random Mansion). 2000 people are distributed among the 123 rooms of a mansion. Each minute, as long as the people are not all in the same room, the following event happens: one or more people, all of whom were previously in the same room as each other, leave that room. At least one of these people goes to a new room with at least as many people as the original room; the rest may go to any room of the mansion. Prove that eventually all 2000 people will be in the same room. Solution: The alternative solution to Problem 2 goes over almost wordfor-word here. The largest of the 123 rooms’ populations can never decrease, only increase; eventually it must stop changing. After this point, the secondlargest population can never decrease, so it must eventually stop changing too. After the two largest populations have stabilized, the third-largest population can only increase. And so forth. So after some finite number of steps, the whole collection of 123 room sizes has stopped changing. As before, this can only happen if people have stopped moving, which, by assumption, occurs only when everyone is in the same room.
Exercise 5. Look back at the earlier solution to Problem 2. Could the original definition of concentration, or the definition in Exercise 4, or any other you may have come up with, be used to solve Problem 5? That is, is concentration still a monovariant? Hint: Note that any move (a, b) → (a − k, b + k) with a ≤ b involving only 2 rooms can be interpreted as a sequence of k moves in the original mansion problem, so the concentration will increase after such a move (why?). Thus, to find counterexamples, i.e., specific moves which cause the concentration to decrease, you must involve at least 3 rooms in your moves. ♦




Problem 6 (Intermediate). Define concentration in the mansion as the sum of powers 2n1 + 2n2 + · · · + 2n123 . Prove that this concentration is a monovariant in both the original and the random mansion problems. Show that replacing 2 by a larger number will still provide a monovariant (for any number of people), but any base smaller than 2 will not work for large numbers of people.2

5. Women and Men Walking around the Mansion 5.1. Gender balance in the mansion. It’s good to get comprehensive practice by working on lots of similar-looking problems. So, here are a couple variations on the mansion problem. (If you are a class-conscious proletarian and don’t like mansions, you can imagine that it’s a factory instead.) 2 Michael Burks, a 9th grader from the Stanford Math Circle, suggested this creative solution at a math circle demonstration at the MathFest in 2007 in San Jose.

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5. WOMEN AND MEN WALKING AROUND THE MANSION




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Problem 7. 1000 men and 1000 women are distributed among the rooms of a 123-room mansion. They move among the rooms according to the following rules: either • a man moves from a room with more men than women into a room with more women than men (counted before he moves) or • a woman moves from a room with more women than men into a room with more men than women. Prove that eventually the people will stop moving. To solve this problem, following PST 88, first think about what is happening on an intuitive level. The person moving is making both the old and the new room more “gender balanced” than before. Stating this rigorously is pretty straightforward. Define the imbalance of a room with w women and m men as |w − m|. Define the total imbalance of the mansion as the sum of the imbalances of all the rooms: t = |w1 − m1 | + |w2 − m2 | + · · · + |w123 − m123 |. To make sure we are on the same page, verify that in Figure 3 the total imbalance is 3 if we ignore room Z (which is “infinitely imbalanced”). Exercise 6. Check that the total imbalance of the mansion decreases at each step. (In fact, the imbalance of each room affected decreases.) Solution to Problem 7: By Exercise 6, the total imbalance decreases at each step. Being a non-negative integer, the total imbalance must therefore stop decreasing eventually. Hence, the people must stop moving.
Exercise 7. Refine the above result: check that the total imbalance t decreases by exactly 2 at each step. Use this to prove that people will stop moving after at most 1000 steps. Can you improve this bound?



Sketch: Have we seen a similar problem before? Recall the solution to Exercise 3, where we sandwiched the concentration f in an interval. We should be able to do the same for t. The lower bound is easy: t ≥ 0. But how large can the total imbalance t be? Let’s try a heuristic approach. PST 93. Assume your problem is correct, and use it to guess what you are supposed to prove, including numerical data. This is somewhat like walking backwards on the staircase of problem solving in Figure 2. If Exercise 7 is correct, our total imbalance t decreases by 2 at each step and is non-negative. In order for the process to end in at most 1000 steps, it is logical to guess that t must have started at 2000 or below. Proving that t ≤ 2000 is the heart of the problem. Since t is defined via absolute values and since we want an upper bound for t, it is reasonable to use the Triangle Inequality (cf. the Inversion session) for the imbalance in every room: |w − m| = |w + (−m)| ≤ |w| + | − m| = w + m.

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Adding up over all rooms, we obtain t ≤ (w1 + m1 ) + (w2 + m2 ) + · · · + (w123 + m123 ) = 2000.



The reader should now fill in the gaps and conclude that the mansion will be “gender balanced” in at most 1000 steps. ♦ Whether this bound can be improved requires a different type of thinking. PST 94. If you believe that the bound of N steps for a process to terminate can be improved, you could provide a smaller bound K and show that it indeed works. If you believe that N is the best bound, you could provide a specific example of the process terminating in exactly N steps. We will not spoil your fun here. Think for a while about this and make up your mind which path is the correct one in Exercise 7: whether you can find a better bound than 1000 steps or whether you can exhibit a situation in which the mansion becomes “gender balanced” in exactly 1000 steps. In any case, you might be curious to compare this modest 1000-bound with the 67064-step sequence given in the Appendix for the original mansion: having two genders that seek each other is a less “random” and more “directed” process, which apparently makes the population sort itself out much faster and ends the room-migration much sooner. We end this subsection with a “provocative” question. In the genderbalancing mansion, does everyone have to end up in the same room? No? Correct. Describe several essentially different ending positions with total imbalance 0. For those familiar with basic counting techniques, we pose Exercise 8 (Intermediate). If we do not distinguish the individuals beyond their gender, how many ending configurations of total imbalance 0 are there? Consider first the case when all rooms are different (distinguishable), and then the case when all rooms are identical. 5.2. Hybrid Mansion. Here’s a somewhat harder mansion version, which can still be solved using the techniques we’ve seen so far. Problem 8. There are 1000 women and 1000 men in a 123-room mansion, as usual. This time, they walk among the rooms according to the following rules: at each step, either • a man moves from a room with more men than women into a room with more women than men or




• a woman moves from a room with more women than men into another room with at least as many total people. Prove that eventually the people will stop moving.

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Hint: We won’t put the solution right here; you can refer to the Hints section if you need a tip. Just keep in mind that the problem is a hybrid between the original and the gender-balancing mansions, so it will require a clever combination of their solutions. Concentration is increasing, while imbalance is decreasing: can you fuse these two disparate monovariants into a single one? Can tag team monovariants be used instead? ♦

6. Non-numerical Monovariants



All the problems we’ve considered so far were full of numbers that practically cried out to be assembled into monovariants. It would be an injustice to monovariants to leave it at that, because they can be used to solve not only problems explicitly involving numbers but also problems involving “things”. PST 95. When trying to come up with a monovariant, be on the lookout for things that can be counted or things that have an order and so can have numbers assigned to them. 6.1. A classic frog example suddenly brings to mind a previous idea.




Problem 9. Some number of frogs are squatting on a row of 2000 lily pads in a swamp. Each minute, if there are two frogs on the same lily pad and this pad is not at either end of the row, the two frogs may jump to the two adjacent lily pads (in opposite directions). Prove that this process cannot be repeated forever.

k l

m

Figure 6. Jumping frogs in a “monovariant” row of lily pads This resembles the original mansion problem so much that it is our duty to first try the same method. Lilies and frogs obviously correspond to “rooms” 2 where lj is the number of and “people”, so we set L = l12 + l22 + · · · + l2000 th frogs on the j lily and hope that L is a monovariant. Any move incurs a frog-reconfiguration of the form (k, l, m) → (k+1, l−2, m+1), where k, l and m are the numbers of frogs on the involved 3 consecutive lilies (cf. Fig. 6). So, the net change in our sum L will be   (k + 1)2 + (l − 2)2 + (m + 1)2 − (k 2 + l2 + m2 ) = 2(k + m − l + 3). But the relative populations of the 3 lilies will sometimes cause this net change to be positive and sometimes negative (why?), implying that L is not a monovariant! Thus, unfortunately, our old method does not apply here, at least not in a straightforward fashion. A new idea is needed. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Instead of the “number of frogs residing on a given lily”, can we do sort of the opposite: for a given frog, somehow use the lily on which it resides? So, start by numbering the lily pads, 1 through 2000 in order, and consider the location of the frogs: suddenly we have a new collection of numbers ! At each step, two equal numbers l are replaced by l − 1 and l + 1. At this point it’s not hard to find an expression that is a monovariant for the jump operation: that the sum of the squares of the positions will work should be of no surprise to us. Solution: Number the lily pads from 1 to 2000. For each frog, consider the square of the number of the lily pad it sits on. Let S be the sum of all these squares. Each time two frogs jump from some lily pad l to lily pads l − 1 and l + 1, S changes from 2l2 + (other terms) to (l − 1)2 + (l + 1)2 + (other terms). So the net change in S is (l − 1)2 + (l + 1)2 − 2l2 = 2. In particular, S always increases at each step. Now it’s the usual argument: there are a finite number of frogs and a finite number of lily pads, so a finite number of arrangements, and a finite number of possible values3 of S. Hence S cannot keep increasing forever.
It is interesting to note that, on the several occasions when Problem 9 was discussed at Bay Area math circles in 2007, the audience invariably attempted to solve it instead via monovariant tag teams. Here is an idea adapted from a solution of Julian Ziegler Hunts, a 6th grader at the San Jose Math Circle. Exercise 9. For 1 ≤ i ≤ 2000, let ni be the number of frogs on the ith lily pad. Prove that {n1 , n2 , . . . , n2000 } is a tag team of non-decreasing monovariants in the listed order, i.e., n1 is a non-decreasing monovariant; when n1 stops increasing, n2 picks up and becomes a non-decreasing monovariant; when n2 stops increasing, n3 picks up and becomes a non-decreasing monovariant, and so on, until all 2000 numbers ni stabilize and the jumping stops. The reader should be experienced enough by now to ask the inevitable question (to be answered fully only by the combinatorially equipped folks): Exercise 10 (Intermediate). If there are n frogs altogether, describe and count all final frog-configurations in Problem 9. Find an upper bound for the number of steps after which the process must have ended. Can you improve on your bound? 6.2. A circle of frogs. If you think we are done with the frogs, you are very mistaken. One feature of Problem 9 is crucial for its solution, although we didn’t pay any particular attention to it. The lilies are arranged in a row 3 If you are uncomfortable with this “vague” argument on finiteness, you can be equally

sloppy but concrete: if there are n frogs altogether, then n = 12 + · · · + 12 ≤ S ≤ 20002 + · · · + 20002 = 20002 n, which indeed bounds the values of S to finitely many possibilities.

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of finite length, which enables us to identify two special lilies: the end lilies can only receive frogs, without releasing any of their residents! What will happen if all of the lilies are arranged differently, say, in a circle ?




Problem 10. A circular swamp is divided into 2000 sectors. There are 2001 frogs in the swamp. Each minute, some two frogs that are in the same sector jump (in opposite directions) to the two adjacent sectors. Prove that at some point there will be at least 1001 different sectors containing frogs.4 Hint: Obviously, the “linear”-frog and the “circular”-frog problems are closely related, save for the end lilies. Can we induce “end” sectors in Problem 10 and thereby use the result of Problem 9? One way is to suppose that some unpopular sector E is never visited by frogs. Removing E, we can then “unwrap” the remaining 1999 sectors into a linear arrangement whose end sectors are the original neighbors E1 and E2 of E (cf. Fig. 7). Now we can indeed apply Problem 9 and conclude that the process will terminate. At that moment, how many frogs can reside in E1 and in E2 ? How about in each of the other 1997 sectors? Where is the contradiction?

E2 E E1

E2

E1

Figure 7. Circular swamp → “linear” swamp After you have successfully resolved the above situation, you can consider the case when every sector eventually gets visited by some frog (no unpopular sector). Turning your attention now to pairs of adjacent sectors will be helpful, but . . . at this point you are on your own. If necessary, look in the Hints section for more ideas on this tricky problem. ♦ Notice by PHP that there will always be a sector with two frogs, so some jump will always be possible. It is perhaps surprising that a monovariant helps solve this problem, because it’s about a process that never stops. 6.3. A final puzzle for the die-hards. About 1 out of 30 participants solved the 5th and final problem at BAMO ’06. The paucity of any partial scores spoke to the well-known fact: a problem of this type can be a real tough cookie – you’ll either completely solve the problem (i.e., come up with a useful monovariant) or you’ll get nowhere. Successful or not, many students kept discussing this problem long after the contest was over, for it presented them with a real challenge to their creativity. 4 Adapted from a problem on the 1993 Jiangsu Province (China) Math Olympiad.

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Problem 11 (BAMO ’06, Carroll5). We have k switches arranged in a row, and each switch points up, down, left, or right. Whenever three successive switches all point in different directions, all three may be simultaneously turned so as to point in the fourth direction. Prove that this operation cannot be repeated infinitely many times. For the brave reader who would like to take his/her chances with the problem, we offer some help. Hint: Look at the pairs of adjacent switches that are in the same orientation (call them “twin pairs”) and show that their number can never decrease. Once this number has stopped increasing, check that the twin pairs always get “closer together” (cf. Fig. 8). Use an idea similar to the “linear”-frog Problem 9 to make this precise. For example, assign a number to each twin pair to reflect its position in the row, and construct a suitable function to keep track of all these numbers. ♦

   n

   n+3

      n+1 n+2

Figure 8. Pairs of adjacent switches Even with the above hint, the problem is hard, requiring substantial knowledge and practice of the necessary techniques and ideas, along with a stroke of creativity. It’s no wonder it was cracked by so few. Since we will need to refer later to its solution, you can find it below. Solution sketch: Let’s number the switches by 1, 2, 3, . . . , n and label the position of each twin pair (k, k + 1) as k (for its left switch). In this notation, a triplet of adjacent differently oriented switches (k +1, k +2, k +3) creates two new twin pairs in positions k + 1 and k + 2 but may potentially destroy twin pairs in positions k and k + 3 (as in Figure 8). In any case, the total number P of twin pairs is non-decreasing (why?). Well, P can’t increase forever since there are n − 1 adjacent pairs altogether. Once P stabilizes, essentially the only possible moves are depicted in Figure 8: twin pairs in positions k and k + 3 are replaced by twin pairs in positions k + 1 and k + 2 (why?). Now we proceed as usual: set the monovariant S to be the sum of squares of all twin pair positions. (If no such pairs exist, set S = 0.) The net change in S at every step is of the form (k + 1)2 + (k + 2)2 − k 2 − (k + 3)2 = −4, so the sum S strictly decreases. But S is non-negative, so it must stabilize at some point and the process ends. ♦ To recapitulate, we used a tag team of two monovariants: the increasing number P of twin pairs, and after it stabilized, the decreasing sum of squares 5 Editors’ comment: Gabriel Carroll won 3 consecutive grand BAMO prizes himself and has created a number of BAMO problems, including this problem.

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6. NON-NUMERICAL MONOVARIANTS

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S. Why didn’t we use S from the very beginning? Can you think of moves which increase S? But the more burning question is whether we can get away with only one monovariant throughout the whole process. Here is the author’s original BAMO ’06 solution: Exercise 11. In the set-up of Problem 11, consider the “height” h of a configuration to be the product of all twin pair positions (or 1 if no such pairs exist). Show that h increases with every operation and conclude that it eventually causes the process to√terminate. Repeat the exercise for another height h defined as the sum of k for all twin pairs (k − 1, k) (or 0 if no such pairs exist). In fact, the nine contestants who solved the problem used a variety of monovariants. In a future Part II, the so-called convex functions will demystify to some extent the monovariant creation process and show how they help us cook up monovariants such as those above and plenty more.



6.4. When and how to use monovariants. There’s one basic idea behind all the problems we’ve seen, which should be apparent by now: PST 96. To prove that some repeated process must eventually terminate, use a monovariant. It should be emphasized that this PST is not just something to try if it happens to cross your fancy. Almost the only way to show that a repeated process terminates is to use a monovariant. So when you need to prove something of that sort, an alarm should go off telling you to look for a monovariant; it may be very easy to find, or it may require some cleverness. Let’s briefly think about what it takes to apply PST 96. There are, in fact, several tasks involved: you must (1) identify the steps; (2) identify the monovariant; (3) show that the monovariant either always increases or always decreases with each step (see “monotonicity ” in Figure 9); and (4) show that it cannot increase/decrease forever (described as bounds in Figure 9); for example, perhaps it can only take on a finite number of different values.

Figure 9. Monovariants in “practice”

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In the examples we’ve seen so far, the steps are supplied, so we only needed to do tasks 2 through 4. In Part II we’ll tackle more difficult examples where we have to come up with both the steps and the monovariant. There we’ll also see some more diverse applications of monovariants, for . . . china shops, mansions, and swamps are obviously not the only places where monovariants are permitted.

7. Mansion Appendix for the Advanced Reader by Evan O’Dorney6

7.1. “Mission Impossible”. Recall mansion Problem 2 with 123 rooms and 2000 people: every minute, someone moves to a room with at least as many people as the room he started in. Letting ai be the number  of people 2 in room i, we constructed the concentration monovariant f = 123 i=1 ai and showed that eventually everyone gathers in one room. The concentration at this end state, Es , is simply the maximum f (Es ) = 20002 . The goal of the current section is to find the longest time it takes for the process to reach Es , that is, to terminate. Our solution will rely on several other previously proven facts. Problem 4 demonstrated that the minimal concentration occurs in the most “even” distribution, namely, when 91 rooms have 16 people each and the remaining 32 rooms have 17 people each (cf. the Hints section). We call this distribution the beginning state Bs ; its concentration is f (Bs ) = 91·162 +32·172 = 32544. Furthermore, according to Exercise 1, f increases by at least 2 at every step, and the minimal increase of 2 is obtained exactly when a person moves to an equally crowded room, i.e., a move (n, n) → (n − 1, n + 1) is performed. We refer to such an optimal move as a short push, or s–push (the terminology will be justified later). Combining everything, we concluded earlier that the end state Es appears in at most (f (Es ) − f (Bs ))/2 = 1,983,728 minutes. If we could construct a path P from Bs to Es (cf. Fig. 10a) consisting entirely of s–pushes, then this upper bound of almost 2 million would be the answer we are looking for, and there would be no need for this Appendix. As our discussion below will confirm, such a “lucky” long path P does not exist because some other moves (besides s–pushes) are needed to reach Es . But there are so many paths to Es . . . which one is the longest? You could ask a computer to draw the humongous directed graph 7 whose vertices are all possible states and whose edges correspond to all legal moves, find the longest path to Es and calculate its length. Yet, this would give you no clue why the final answer is what it is, nor would it help you in solving analogous problems with slightly changed initial data . . . . 6 Evan O’Dorney is a 9th grader at the Berkeley Math Circle. 7 This sentence requires some basic knowledge in graph theory and can be skipped

without harming your comprehension of the solution.

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It looks like we are on a “mission impossible”; however, a math researcher never backs away from challenges. The main result in this section, which we will show without the help of computers but via various creative math techniques, is as follows:




Theorem 1 (O’Dorney). The process in mansion Problem 2 will terminate in at most 67064 minutes. Moreover, this is the least upper bound on the number of steps since the length of the longest paths reaching Es is 67064. This newly proposed exact upper bound is about 30 times smaller than our old bound of nearly 2 million! So, it is worth verifying Theorem 1, but don’t expect the proof to be easy since the result is the strongest possible. 7.2. Longest paths and dualizing. A longest path Pmax evidently ends in Es , or else we could have extended it by further legal moves until it ends in Es . If the path starts at some state X , we run the process backwards (i.e., people move to less crowded rooms, as in Problem 4) and construct an “anti-path” from X to the beginning state Bs (cf. Fig. 10a); by reversing its steps, we find a path of legal moves from Bs to X . Thus, the original path can be enlarged by prefixing the path from Bs to X . Consequently, any longest path must begin at Bs and end in Es . Es

Es

X

P

Bs

122

X Bs

0 1 2 3 4 5 6

···

2000

109

···

···

13 14 15

··· 0 1 2 3

21 22 91 32 16 17 18

··· 2000

Figure 10. Paths P : Bs → Es What moves can a longest path contain? We already mentioned the s– pushes, but do we need any other “optimal” types of moves? Our discussion of these moves will be much more efficient if we use an alternative interpretation of the mansion game, a sort of a “dual” to it: the ith room becomes pawn pi placed on a cell numbered by ai . For instance, if the 7th room has 5 people, then pawn p7 is placed on cell 5. Thus, consider a 1 × 2001 board numbered from 0 to 2000. At the outset Bs , 91 pawns lie on cell 16 and 32 on cell 17 (in Figure 10b, the cell numbers are below and the numbers of pawns are above the cells). A move consists of selecting two pawns, neither at an end cell, and moving them away from one another by one space.8 The pawns may be on the same cell (a short push), or on different cells – call this move a long push, or l–push. The object is to make as many moves as possible before reaching the end state Es of 122 pawns on cell 0 and 1 on cell 2000. 8 In the original mansion problem, this is the move (a, b) → (a − 1, b + 1) with a ≤ b.

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12. MONOVARIANTS. PART I

Now, not all l–pushes are allowed in constructing maximal paths. If there is a third pawn between or on the pushed pawns’ cells such a move can always be replaced by two moves (cf. Fig. 11). Such l–pushes are “bad” and cannot be used in maximal paths. The “good” l–pushes, for which no third pawn is between or on the pushed pawns’ cells, can be used in maximal paths and will be referred to as ¯l–pushes. = “good” l–pushes, “bad” l–push

or ¯l–pushes =

Figure 11. Some l–pushes 7.3. Our second monovariant. We observed that the concentration f works well with any s–push, as the latter increases f by exactly 2. But ¯l–pushes are a different matter altogether, for f has no control over them: they can increase f by various large amounts (why?), making it impossible to decide if a given path is maximal or not. Thus, we need a second monovariant that “behaves well” under ¯l–pushes. To this end, define the distance d of a state X as the sum of distances9 between any two pawns:  |ai − aj |, d(X ) = 1≤i p, a contradiction. 2 Thus, there are some 0-pawns and the remaining pawns are single occupants of all cells from 1 to k except some cell j, where 1 ≤ j ≤ k ≤ r. Therefore,   −j =p Σ(Ms ) = (1 + 2 + · · · + k) − j = k+1 2 k+1 k k+1 ⇒ 2 >p≥ 2 −k = 2 .     > p, and j = k+1 − p. For So k is the smallest integer such that k+1 2 2 √ a specific formula, k = (1 + 1 + 8p)/2. The construction   allows for no
trapped blank in Ms exactly when j = k, and hence p = k2 . r+1 64 124 In the original problem, p = 2000 < 2 = 2 , and 2 > 2000 > 64 63 2 , so that k = 63, and j = 2 − 2000 = 16 yields the trapped blank.   7.9.2. Case 2: Relatively many people. Let p ≥ r+1 2 . If there are some 0-pawns, the remaining pawns occupy singly cells 1 through r with  at least − 1 < p, one cell missing a pawn, i.e., Σ(Ms ) ≤ (1 + 2 + · · · + r) − 1 = r+1 2 a contradiction. Thus, there are no 0-pawns, and the r pawns spread out on cells t through t + r except some cell j, where 1 ≤ t < j ≤ t + r. Therefore,    −j =p Σ(Ms ) = t(r + 1) + r+1   r+12  − (t + r) ⇒ t(r + 1) + 2 − t > p ≥ t(r + 1) + r+1 2 r  r+1 + tr > p ≥ 2 + tr. ⇔ 2   + tr > p, and So t can be described as the smallest integer such that r+1 2 r    + t(r + 1) − p. For a specific formula, t = (p − j = r+1 2 2 )/r. The exactly when j = t + r, and construction allows for no trapped blanks in M s  r
in this case, t = (p − 2 )/r. As an example in this case, r = 3 rooms and p = 10 people. 3+1consider  + 3 = 9 < 10 but + 2 · 3 = 12 > 10, we have t = 2, t + r = 5, Since 3+1 2 2 j = 4, and Ms has a pawn on each of the cells 2, 3 and 5. One can verify Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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277

this result by brute-force flattening of Bs : Bs = ! 0 0 0 2 1 0 · · · " → ! 0 0 1 0 2 0 0 · · · " → Ms = ! 0 0 1 1 0 1 0 · · · ". We have shown Lemma 4. For a mansion game with p people and r rooms, the middle state Ms can be described as follows:     > p, let k be the smallest integer such that k+1 > p and (a) If r+1 2 k+1 2 j = 2 − p. Then Ms has single pawns on cells 1 through k except j; the remaining pawns are on cell 0.     ≤ p, let t be the smallest integer such that r+1 + tr > p (b) If r+1 2 r+12 and j = 2 + tr + t − p. Then Ms has single pawns on cells t through t + r except j. The borderline case Ms = ! 011...100...0 r+1" has a pawn on each of the cells 1, 2, . . . , r and occurs exactly when p = 2 . The discussion in this Appendix can be summarized by Theorem 2 (O’Dorney–Stankova). For a mansion game with p people and r rooms, the maximum number of steps it takes for the process to terminate is [f (Ms ) − f (Bs ) + d(Es ) − d(Ms )] /2, where the middle state Ms is given in Lemma 4. Two of the four required monovariant values are trivial to compute: d(Es ) = (r − 1)p and f (Bs ) = nm2 + (r − n)(m + 1)2 . The other two, f (Ms ) and d(Ms ), need some calculations.   (as in Lemma 4(a)), show that Exercise 15. When p ≤ r+1 2 (a) f (Ms ) =

k(k+1)(2k+1) − j2; 6 k+1 j  k−j+1 − 2 − 3 2

+ (r − k + 1)p;   k+1 (c) maximum # steps = p(k − m − 1) + r m+1 − 3 . 2

(b) d(Ms ) =

Check that for p = 2000, r = 123, m = 16, and k = 63, the last formula in (c) matches with the previously found least upper bound  of67064. The reader is welcome to work out a similar formula when p > r+1 2 . 7.10. Epilogue on duality. Did we get extremely lucky in finding two seemingly unrelated monovariants, f and d, that nevertheless fit so perfectly to yield the solution? Such a “lucky” coincidence does not occur often and needs a closer inspection. In fact, not only are f and d related, but they can be thought of, in some sense, as dual monovariants. Represent the pawns by stars and the divisions between the cells by bars. For example, a configuration ! 1 2 1 0 0 " would be represented as ∗ | ∗ ∗ | ∗ | |. In a move, one pawn moves left: | ∗ changes to ∗ | , and one pawn moves right: ∗ | changes to | ∗. Thus, an s¯–push is represented by | ∗ ∗ | → ∗ | | ∗. As we saw, this is the only type of move optimal under both f and d. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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12. MONOVARIANTS. PART I

Recall that f increases with increasing distance between the two cells, in fact, by 2 for every cell separating the two pushed pawns (cf. Exercise 1); in other words, f increases by 2 plus twice the number of bars between the stars to be moved. The monovariant d, on the other hand, goes up by 2 for every pawn between or on the pushed pawns’ cells; thus, d increases by 2 plus twice the number of stars between the stars to be moved. One can dualize the entire mansion process by switching ∗ with | and time-reversing it. Thus, if an s –push stands for an s–push on a cell with 3 or more pawns, then an ¯l–push dualizes to an s –push, and vice versa: ¯ l

s


dual

| ∗ | | | ∗ | → ∗ | | | | | ∗  ∗ | ∗ ∗ ∗ | ∗  | ∗ ∗ ∗ ∗ ∗ | . Similarly, if an l –push stands for a non–¯l long push (a “bad” l–push), then an l –push dualizes to an l –push, and an s¯–push dualizes to an s¯–push: l


l


dual

| ∗ | ∗ | | ∗ | → ∗ | | ∗ | | | ∗  ∗ | ∗ | ∗ ∗ | ∗  | ∗ ∗ | ∗ ∗ ∗ | , s¯

s¯

dual

| ∗ ∗ | → ∗ | | ∗  ∗ | | ∗  | ∗ ∗ | . When a game with p bars and r stars is dualized into a game with r bars and p stars10, a maximal path Bs → Ms → Es is dualized to a maximal path Es∗ → M∗s → Bs∗ (cf. Fig. 15a). Along the way, the optimal moves for d are dualized into the optimal moves for f , and vice versa, and thus the monovariants f and d “switch places” in the dual game. The symmetry is even more visible if we consider the (unique) flattening L∗s of Es∗ in the dual game and pull it back to its own dual Ls in the original game (cf. Fig. 15b). As it turns out, Ls has no trapped blanks and is flat, except for one 2. The maximal path is now divided into three parts: Bs → Ls via s – and s¯–pushes, i.e., optimal for monovariant f ; Ls → Ms via s¯–pushes, i.e., optimal for both monovariants f and d; and Ms → Es via ¯l– and s¯– pushes, i.e., optimal for monovariant d. Es d

¯ l,¯ s

Ms

Es∗

 f 

s s
,¯

M∗s

Es ¯ l,¯ s

d

Ms f,d

f

d

s
,¯ s

¯ l,¯ s

f Bs



Bs∗

f 

Bs

d 

s¯

L∗s



s s
,¯

s
,¯ s

M∗s f,d

s¯

Ls

Es∗



¯ l,¯ s

Bs∗

Figure 15. Dual games and dual monovariants f and d 10 The dual game with r bars and p stars and the original game with r bars and p stars have the same rules for movement but different beginning and ending states.

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8. Hints and Solutions to Selected Problems Problem 4. We have to show that f attains its minimal value when the people are most “evenly” distributed throughout the mansion: ?

2 f (n1 , n2 , . . . , n123 ) ≥ 162 + · · · + 162 + 17 · · + 172 = 32544.  + · 91

32

To prove this, we’ll do something seemingly “illegal”: we’ll allow people to move freely to other, possibly less crowded, rooms. Since any configuration of people yields a valid value for f , it doesn’t matter how the people move around – we just keep on recording the resulting values of f . Moving people from more crowded to less crowded rooms will decrease the concentration by at least 2 (why?), unless we do something silly like the move (n, n + 1) → (n + 1, n), where the difference of the population in the two rooms is 1. Obviously, such a move will preserve the concentration. We conclude that, as long as there are two rooms with a difference of 2 or more people, we can decrease the concentration by moving a person from the more crowded to the less crowded room. Therefore, the smallest concentrations occur only in configurations where all values ni differ by at most 1, i.e., some rooms have a people in them, and the remaining rooms have (a + 1) people. Since 2000 is a sum of 123 numbers, of which some are a’s and some are (a + 1)’s, we have 123a < 2000 < 123(a + 1), and the only integer satisfying these inequalities is a = 16. If k is the number of rooms with 16 people each, then 16k + 17(123 − k) = 2000 gives k = 91, from which the minimal concentration f is 32544, attained at the desired configuration of people.
Exercise 4. Consider the situation where someone from a room with a 1 1 + b+1 with people moves to a room with b people, b ≥ a. Compare a+1 1 1 + . Another related measure of concentration is the number of possible a b+2 handshakes by the people in a room (suggested by Joshua Zucker after the math circle demonstration at MathFest ’07). ♦ Exercise 5. Suppose the following move occurs: (5, 6, 0) → (3, 7, 1), i.e., two people from a 5-person room move to a 6-person room and an empty room, respectively, and all other rooms stay the same. The concentration defined as sum of squares decreases: 52 +62 +02 = 61 > 59 = 32 +72 +12 . Ditto with the concentration defined via reciprocals ni1+1 : 16 + 17 + 11 > 1 > 14 + 18 + 12 . Hence the original single-monovariant definitions do not work on the generalized mansion problem.
Problem 6. Let a person move from room X with a people to room Y with b people, where b ≥ a, and let no other moves occur (just as in the original mansion). The concentration then increases: (2b+1 + 2a−1 ) − (2b + 2a ) = 2b − 2a−1 ≥ 2a − 2a−1 = 2a−1 > 0. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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12. MONOVARIANTS. PART I

Now suppose that, in addition to the move above, several other people from room X file into some other rooms. The concentration in X can decrease at most to 20 , leaving no people in X , while the concentration in the rest of the mansion can only increase: in Y, for instance, the concentration at least doubles from 2b to 2b+1 , and there will be additional increases caused by the other people. Therefore, the net change in concentration is at least (2b+1 − 2b ) − (2a − 20 ) = 2b − 2a + 1 ≥ 2a − 2a + 1 = 1 > 0.       room Y

room X

Check that replacing 2 by a larger number will simply make the (positive) net changes in concentration even larger. However, if we replace 2 by a smaller number, say, 1.9, things will go very wrong. If two rooms X and Y have k + 1 people each and k rooms are empty, suppose one person from X goes to Y, while the remaining k people from X spread to the k empty rooms, one per room: (k + 1, k + 1, 0, 0, . . . , 0) → (0, k + 2, 1, 1, . . . , 1).       k

k

The concentration then changes by     k+2 + 1.9k + 1 − 2 · 1.9k+1 + k = · · · = 0.1(9k + 10 − 1.9k+1 ). 1.9 But the exponential 1.9k grows faster than the linear 9k + 10; hence for large enough k the net change will be negative, e.g., for k = 6: 0.1(64 − 1.97 ) ≈ −2.54 < 0. Similar calculations can be carried out for any other small base instead of 1.9. ♦ Exercise 6. If w > m in room X and a woman leaves X , then the new imbalance |w − 1 − m| in X will be smaller than the old imbalance |w − m|: 0 ≤ |w − 1 − m| = (w − 1) − m < w − m = |w − m|. Verify that the imbalance of X will also decrease if a man walks into X .
Exercise 7. Due to the symmetry of the rules we need only consider the move of a woman from a room with wj > mj to a room with wk < mk . Check that imbalance in each of the two rooms decreases by 1: |wj − 1 − mj | − |wj − mj | = (wj − 1 − mj ) − (wj − mj ) = −1, |wk + 1 − mk | − |wk − mk | = (mk − (wk + 1)) − (mk − wk ) = −1, while all other room imbalances remain the same. The total imbalance, therefore, decreases by exactly 2. The bound of 1000 cannot be improved. Indeed, let 1000 men be in one room X and 1000 women be in another room Y. All men from X move to Y, one a time, thus requiring 1000 steps to terminate the process.
Exercise 8. After we distribute all the women, there is only one way to distribute the men so that each room is gender-balanced. So, we can confine ourselves to counting the ways to distribute (without any restrictions) the  1122 = women only. If the rooms are distinguishable, there are 1000+123−1 123−1 122 Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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ways (cf. the Balls in Urns formula on p. 43.) If the rooms are indistinguishable, the problem is equivalent to finding the number of ways to partition 1000 (the women) into k parts (the non-empty rooms), where k ≤ 123.
Problem 8. You can use the two monovariants in a tag team: show that the total imbalance always either decreases or stays the same at each step; and if it stays the same, the concentration has to increase. ♦ Exercise 10. The process stops when each of the 1998 middle lily pads (lily pads not at the ends) has one frog or none. We assume that the frogs are indistinguishable, but the lily pads are distinguishable. If exactly k of the   ways of choosing these middle lily pads have frogs on them, there are 1998 k k middle lily pads, for k = 0, 1, . . . , min{1998, n}. The remaining (n − k) frogs can be distributed in any way we like on the two end lily pads, i.e., in n − k + 1 ways (why?). Adding up over k yields min{1998,n}



1998 k

(n − k + 1)

k=0

for the number of final frog configurations. Since n ≤ S ≤ 20002 n and S increases by exactly 2 at each step, the number of steps needed to end the process is at most (20002 − 1)n/2. This can be improved since the only way for S = n is when all n frogs start on the first lily pad, but then the process has already terminated in 0 steps.
Try to find a better (or the best?) upper bound on the number of steps. Problem 10. First we show that every sector will eventually have a frog in it. Indeed, if some sector E never gets visited by a frog, then we can reduce to Problem 9 (as in Fig. 7) and conclude that the frogs must eventually stop jumping. But the frogs keep jumping forever by PHP, as we concluded in the text, a contradiction. Next, pick any two adjacent sectors S1 and S2 . We know that S1 will be visited by a frog at some point. But then there will always be some frog in at least one of these two sectors. Indeed, if the number of frogs in one of S1 or S2 decreases, the number of frogs in the other sector increases (why?), so thereafter we’ll never have 0 frogs on both S1 and S2 . Finally, split the swamp into the adjacent pairs of sectors {S1 , S2 }, {S3 , S4 }, . . . , {S1999 , S2000 }. We know that after a while, each pair of sectors will be populated by at least one frog, which yields 1000 different sectors containing a frog each. How do we get the extra 1001st frog-populated sector? Suppose exactly 1000 different sectors are frog-populated at this moment. Since no two adjacent sectors can be frog-vacant simultaneously, WLOG we can assume that each of the sectors S1 , S3 , S5 , . . . , S1999 is frog-populated, while the even indexed sectors have no frogs (why?). Recall again that the jumping continues forever: say, two frogs from sector S3 now jump into sectors S2 and S4 . But then here are our 1001 frog-populated sectors: S1 , S2 ,
S4 , S5 , S7 , . . . , S1999 . Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Note: The solution can be made to work for a total of 2000 frogs; however, since the jumping does not need to continue forever (why?), some modifications are in order. Exercise 11. (Carroll’s original solution) (a) By definition, the height h is the product of all k’s in twin pairs (k − 1, k), and hence h is always a positive integer. We claim that h increases with every operation. Indeed, consider the operation in which switches k, k+1, k+2 are turned. Before this operation, switches k − 1 and k may (or may not) have pointed in the same direction, as may switches k +2 and k +3; no other such pairs can be broken, so the height is divided by at most k(k + 3). However, the pairs (k, k + 1) and (k + 1, k + 2) are created, multiplying the height by (k + 1)(k + 2). Thus, if h is the height of the configuration, then the new height h after one move satisfies k 2 + 3k + 2 (k + 1)(k + 2) =h· > h. h ≥ h · k(k + 3) k 2 + 3k So, as claimed, the height does increase at every step. Since the height is an increasing integer and it cannot exceed 1 · 2 · · · n = n!, it can only increase finitely many times, and the result follows.
√ (b) If h is defined as the sum of k for all twin pairs (k − 1, k), the same argument works: you need only show that the largest possible √ √ √ √ loss of k + k + 3 is less than the smallest possible gain of k + 1 + k + 2, so that the new h always increases, i.e., prove that √ √ √ ? √ k + k + 3 < k + 1 + k + 2 for all k ≥ 1. One subtlety here is that the values of h are not necessarily integers! However, since there are only finitely many possibilities of the k’s to plug into the sum h, there are only finitely many values of h. Thus, h cannot increase forever and must eventually stop. ♦ Exercise 12. The value of d(Ck ) for Ck = ! 0 11 · · · 1 0 · · · " with k 1’s equals       1 + 2 + · · · + (k − 1) + 1 + 2 + · · · + (k − 2) + · · · + 1 + 2 + 1         + · · · + 32 + 22 = k+1 ♦ = k2 + k−1 2 3 . The last equality is known as the “hockey stick theorem” from its visual appearance in Pascal’s Triangle (cf. The Art and Craft of Problem Solving [101]). Exercise 13. The distance |ai −aj | between the two pushed pawns increases by 2 regardless of what move is applied. If pawn pk is not between or on the pushed pawns’ cells, then the sum |ai − ak | + |aj − ak | does not change, since one summand decreases by 1 and the other increases by 1. However, if pk is an “interfering” pawn (i.e., between or on a pushed pawn’s cell), then both of these summands increase by 1, causing the total monovariant to increase by at least 4. As the only moves with no “interfering” pawns are s¯– and ¯l–pushes, they are the only ones causing an increase of 2 to d. ♦ Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Lemma 2. To prove the claimed flattening Y → Yf , we use induction on the sum a + b. If a + b = 0, i.e., a = 0 = b, then Y = ! 2 " → ! 101 " = Yf . For the induction step, assume that Y flattens to Yf for any sum a + b ≤ n and that the interval of Y is extended in Yf by one cell in both directions. Let Y have a 1’s on the left and b 1’s on the right, with a + b = n + 1. WLOG, let b ≥ 1. Removing the rightmost 1, we apply IH to the resulting Y  : . . 1 2 1 . . 1 " → Yf = !11 . . . 1 0 1 .  . . 11 " Y  = !1  .  .   a

b−1

a+1

b



. . . 1 0 1 . . . 1 2 " = !11 . . . 1 0 Y  ", ⇒ Y = !Y 1 " → !11     b

a

b

where Y  has a 1’s on the left and no 1’s on the right of the 2. Thus, IH . . . 1 ", so that flattens Y  to ! 1 0 11   a+1

. . . 1 " = !11 . . . 1 0 11 . . . 1 ". Y → !11 . . . 1 1 0 11         b

a+1

b+1

a+1

Note that the interval of Yf extended the interval of Y by one cell in both directions. As for Z → Zf , it now follows directly from the case for Y. Still, an explanation is necessary for why running into the 0-cell will not change the result of the lemma. ♦ Exercise 15. Since the 0-pawns have no effect on f ,  − j2. f (Ms ) = ki=1 i2 − j 2 = k(k+1)(2k+1) 6 The 0-pawns, however, do influence the value of d(Ms ). Recall Ck from Exercise 12. The middle state Ms in Case 1 is obtained from Ck by removing the pawn on cell j and adding (r − k + 1) 0-pawns. Thus, d(Ms ) = d(Ck ) − (j-cell’s contribution) + (0-pawns’ contribution)  k    j−1  − i=1 i − k−j = k+1 i=1 i − j i=1 i + (r − k + 1) 3  k+1    j  k−j+1 − − + (r − k + 1) − j . = k+1 3 2 2  2   p

For the maximal number of steps, substitute the above findings in the formula from Theorem 2 and simplify, using several relations between the variables k+1 ♦ involved such as j = 2 − p and n = (m + 1)r − p.

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Epilogue 1. What Comes from Within It is the 1980’s. A sunny 5th grade classroom in Bulgaria. The math teacher opens the class register, calls two girls to the board, and gives each a problem. Soon enough, one of the girls writes a correct solution, receives an A, and goes back to her seat. The other girl is stuck; she tries one approach, then another; but the boat and ship in her problem go up and down the river and refuse to meet in simple mathematical equations . . . . Meanwhile, the other students “tame” the vessels in their notebooks, and the teacher moves on with the new lesson. The girl remains at the board for the rest of the period, her tears making it even harder to think about the problem. The bell rings. The teacher beckons the girl and asks: “You know what grade you deserve, yes?” A nod. “Well, I will not give it to you if you explain the correct solution to me by the next math class.” The still sobbing girl goes home in a miserable mood, yet with a big hope. Her father (a shipbuilding engineer, speaking of coincidences) helps her derive a system of two linear equations in two variables. From here on the solution is easy, and so the girl explains it to the teacher the next day. Having avoided the poor grade, she doesn’t stop there: “May I come to your math circle?” she inquires. Three months later, to her classmates’, parents’, and her own amazement, that girl wins the local Math Olympiad with a perfect score. Her fate is sealed right then: math will be her future. Sure enough, she will continue for years with her ballet, piano, and guitar lessons; she will attend a poetry circle and compete at science and literature olympiads; but her passion . . . her passion will always be for math problem solving. Later that year she would devise her own way of conquering the last row of the Rubik’s cube (having learned to solve the first two at her math circle); in a couple of years she would represent Bulgaria at the International Mathematical Olympiads (IMO); then go onto a math major at Bryn Mawr and a doctorate at Harvard; train the USA math team for the IMO’s . . . and come full circle by founding the Berkeley Math Circle in 1998. 285

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That girl is me – not angry at my middle school math teacher for putting me on the spot in front of the whole class, rather, grateful to her for giving me a second chance, for seeing the seed of talent in me, for accepting me and nurturing my mathematical curiosity at her math circle, and for propelling me forward with the belief that “what comes from within will take you far.”

2. The Culture of Circles 2.1. All you need is love. There is more than one way to fall in love with mathematics. Many Eastern European mathematicians have come along the path of math circles, where they have learned for the first time that the world of math is larger than one could imagine, more interesting, and more diverse. The math circle culture is ingrained in the societies in these countries. During the communist era, established mathematicians and pre-college teachers considered it their duty to expose the younger generation to the wonders of mathematics. And so they teamed together to found and run math circles. In my hometown of Rousse (
), for example, the math circles used to meet twice a week in the afternoons or after dinner for 1.5-hour sessions. The elementary/middle school math circle started in 3rd grade and included about 25 kids of the same age from my school. The high school math circle started in 8th grade, held its sessions at the local science youth center, and involved about 15 students from several schools, about half of whom made the circle’s core and competed at local, national, and international olympiads. Concurrently, there were identically organized circles at all grade levels 3–11. The material covered ranged from basic algebra and geometry to advanced olympiad problem solving, to lower- and upper-division college topics. 2.2. Worthy of a circle. Mathematics was not the only subject “worthy of a circle”. Starting in late middle or early high school, there were circles in chemistry, physics, and biology; in English, poetry, and literature. I participated in just about all of them at one time or another. I tried many fields because the opportunities were there for me to explore. The math circles were only part of a large net of pre-college circles created to draw children and discover their talents. It was no more prestigious or “cool” to attend a soccer club or take music lessons than to be a member of, say, a high school physics circle. In fact, parents knew how important the advanced knowledge gained in circles would be for their children’s future and hence enthusiastically supported circle participation. 2.3. And they said higher math wasn’t practical? It was to my advantage to attend math circles in particular. The type of thinking and specific knowledge I mastered there helped me win science olympiads, e.g., devise systems of equations to balance chemical elements or solve a quadratic equation in physics in 7th grade. I was heavily courted by my high school teacher to participate in biology olympiads, for they often involved combinatorial gene-counting or probability theory: a piece of cake for math circlers. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Even composing poetry and critiquing literature apparently benefited from my “math-set” of mind. My favorite story here (which, incidentally, landed in the Philadelphia Inquirer in the early 1990’s) goes back to the mandatory two-semester freshman English course at Bryn Mawr. As the only non-native speaker of English in my class, I put a tremendous effort into the weekly essay assignments, practically sleeping with the dictionary under my pillow every weekend before the homework was due. Still awaiting my final grade in January, I was sitting one day on the floor in my dormitory and assembling my spring schedule. The phone rang, and, to my surprise, my English instructor spoke at the other end of the line: “Are you a math major, by any chance?” I answered affirmatively and steeled myself for the worst. The instructor exclaimed: “It figures! You write so clearly and in such a structured way, yet your personality shows through your words! Even though I disagree with half of the arguments in your final essay [‘How to Read the Beatles’] you wrote them so convincingly, like a true mathematician . . . . I exempt you from the second semester of English. You should take a higher-level course: I can teach you nothing more in writing in this course.”

I didn’t end up taking another English course (probably a mistake on my part); but needless to say, my math (and literature) circle training was responsible for the above remarkable exemption.

3. Eastern European vs. USA Math Circles 3.1. He loves me; he loves me not! On the larger scale, the math circles shaped my future by drawing me like a powerful magnet to the world of mathematical problem solving. Since that 5th grade dramatic experience, I knew within me that no discipline but math would complete me, and no profession other than one in mathematics would be satisfactory to me. It is because I loved math at school that I went to the math circle to get more of it. Unfortunately, students in the U.S. by-and-large do not like their math classes. And let us not deceive ourselves: generally, the talented middle and high school students are bored by the low-level math, the relentless repetition, and the lack of advanced ideas or challenging problems. And it is because they don’t like math in school that they come to the Berkeley Math Circle. Ironic, isn’t it? 3.2. Frequently asked questions. Here are some more differences between Eastern European and U.S. math circles. Keep in mind that not all U.S. circles follow the BMC model, and neither are my hometown math circles (HMC) identical twins of the other Eastern European math circles. 3.2.1. Age of circlers. While in HMC all students were about the same age, U.S. math circles may incorporate students of a variety of ages, e.g., BMC engages students in grades 4–12, all sitting and learning in the same room. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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3.2.2. Logistics. HMC met twice a week for 1.5 (or more) hours. The HMC were numerous and organized in such a way that students ordinarily could go there and get home without parents’ assistance. U.S. math circles, due to transportation issues and conflict with other established school and out-ofschool activities (e.g., volleyball team, music lessons, chorus, etc.), may meet only once a week for 2-hour sessions. The large area covered by the one BMC (from Davis to San Jose, from Palo Alto to Orinda and Danville) calls for parents to drive their kids across the long distances and forces the evening BMC time (6–8 pm) during the week, or alternative weekend sessions whose timing presents other obstacles to families and organizers. 3.2.3. Home base. While HMC were either based at a school or at a local math/science center, their U.S. counterparts are usually university-based. A sufficient number of teachers in Eastern Europe were qualified to lead math circles on their own, with some occasional support of materials and instructors from a nearby university. Alas, this is not the case in the U.S. 3.2.4. Topics in HMC were organized in modules, providing continuity and gradual increase of difficulty and depth of the material. This was possible mostly because the students had very similar math background, level of knowledge, and mathematical maturity and because circlers attended all sessions: transportation issues did not exist and other activities were deprioritized by the math circles. In the U.S., the circlers may vary from beginners to seasoned members of the national USA math team, and hence single powerful sessions incorporating the various levels and backgrounds are more practical than long sequences of linked sessions. Besides, the sparsity of U.S. math circles and competing activities means regular weekly attendance is not always possible; hence missing one session should not preclude understanding the following one. For example, the BMC sessions are usually singletons, with occasional series of 2 and rarely 3 sessions. 3.2.5. Session leaders in HMC were only one or two teachers who organized the specific math circle. Occasionally we had guest speakers from the local university, and once in a while we were visited by professors from Sofia University or the National Youth Science/Math Center who trained the Bulgarian national team. In contrast with HMC, each BMC instructor leads an average of 1–2 sessions per year, accounting for approximately 18 instructors every year. They are mathematicians from nearby universities and colleges, some specially trained high school teachers, some professionals working in related fields, and even some alumni and current advanced circlers. 3.2.6. Popularity. Everyone in Eastern Europe knew about the math circles; children and parents alike were well aware of the opportunity to enroll and of the possibilities which successful participation might open in the students’ future. What portion of the U.S. population has an inkling that math circles exist? Negligible. What status do math circles have in U.S. society and its educational system? Unclear. Can they compare in popularity to membership of a high school football or debate team? No, they can’t. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Figure 1. Football or Math Circle? 3.2.7. Government support. The overall organization and funding in the socialist model math circle was entirely secured by the state; a math circle was an extracurricular activity roughly equivalent to one course each semester and was thus correspondingly compensated by the Ministry of Education. To the contrary, SF Bay Area math circles, for instance, are partially funded (if at all) by private sources; the remaining “funds” are donated by volunteers’ time, effort, professionalism, and enthusiasm. Undoubtedly, the reader has more questions, and the comparison list can go on and on. But this Epilogue is not intended as an exhaustive study of the math circle phenomenon. For more details on U.S. Math Circles, see Sam Vandervelde’s “Circle in a Box” [95]. 3.3. Get to the point. One way to resolve most of the problems associated with math circles in the U.S. is . . . (OK, start dreaming!) . . . to have a math circle at every college and university. (1) The professor organizing and running the math circle will receive a one- or two-course release from the math department, depending on the frequency, length, and intensity of the circle sessions. This will compensate for the huge effort involved in directing a math circle and will hopefully encourage more mathematicians to get involved in educating the talented youth of the U.S. (2) The math circle can be formally organized as a math course and, thus, be open also to undergraduates. (3) Undergraduate and graduate students, as well as interested postdocs and tenured faculty, can be vertically integrated in this model. (4) A reasonable modest semester fee for non-university participants (pre-college students, teachers and others) will provide honoraria to the session leaders. (5) The math department can provide secretarial and computing support and office supplies, as well as a work-study student assistant and web administrator. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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The math circle will be an invaluable math program offered to the local community and can be viewed as part of the math department’s outreach activities. This network model will resolve transportation problems at least for the urban and suburban areas (i.e., areas with an institution of higher education), will mobilize previously disinterested math faculty, and will give some tangible and formal recognition to the work of math circle leaders. An NSF VIGRE grant for the University of Utah ensured the above model for their math circle, led by Peter Trapa and Dan Ciubotaru (cf. [93]). Other university-based circles approaching this model were founded at San Jose State University [80], University of California at Davis [20], Stanford University [88], University of California at Los Angeles [57], and others. Below we’ll examine more closely the model of the Berkeley Math Circle [8].

4. History and Power Despite the shortfalls of U.S. math circles’ set-up, don’t get me wrong: I founded and ran one such circle for a decade and plan on doing so for at least another decade. If I had to describe the Berkeley Math Circle in one phrase, it would simply be a “high-power version of my hometown math circle”. But let’s start from the beginning. 4.1. To marvel and to be appalled. By my last year of graduate studies at Harvard, I had taught enough math courses to question the quality and depth of pre-college math education in the U.S. The few strong (very strong!) undergraduates never took calculus or linear algebra (apparently having taken them at some university while in high school) but jumped directly to upper-division courses like real analysis, abstract algebra, or number theory, to name a few. The cream of the crop, former USAMO winners and IMO medalists, even ventured into graduate courses like algebraic geometry or topology, or Lie algebras (why not?). Each and every such top student had beaten his/her own path out of the jungle of U.S. secondary math education by hiring tutors, by escaping to a nearby university, or, if extremely talented in problem solving, by qualifying for the 30-student one month Mathematical Olympiad Summer Program (MOSP), in preparation for the IMO’s. As I marveled at the super-advanced math knowledge and skills those relatively rare students had acquired through very special personal circumstances, I was appalled at the general math level of the remaining huge bulk of undergraduates. We are talking here about problems in dealing with fractions and simple algebraic manipulations, with which, I am sure, a 6th grader in Bulgaria would have felt perfectly comfortable! 4.2. The missing link. In addition to the outrageous discrepancy between the “top” and the “generic” math student, the link between secondary and college math education – the math circles – was nonexistent as a system. It Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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seemed to me there was no statewide system in the U.S. to meet the needs of talented math students, to discover and train them, to inspire them to continue on with advanced mathematics. And so, I decided it was high time to get acquainted, first-hand, with the secondary education in the US: I enrolled in the Massachusetts teachers certification program. The two high schools for my practicum, Newton North and Chelmsford, offered me an interesting mixture of classes from almost remedial algebra to a problem-solving course of my design. I saw the mathematical potential in a number of students, the desire to go beyond the regular school curriculum. But I realized too that the math teachers were overburdened with courses, never-ending administrative chores and extracurricular activities; the additional load of running math circles (assuming some teachers were qualified and willing) was inconceivable unless the school supported the enterprise financially and administratively. And these were two of the good and prosperous schools in the Boston area. 4.3. The chicken or the egg. I didn’t have time to think about the situation in the bad schools, as I graduated from Harvard and moved in 1997 to Berkeley to take up a postdoctoral position at the Mathematical Sciences Research Institute (MSRI). It wasn’t a month into my new job, when I got an e-mail from Hugo Rossi (then the Deputy Director of MSRI) asking MSRI members for suggestions on possible outreach activities to the community. About 10 minutes later, Hugo and I were in agreement that a regional Math Olympiad for pre-college students would be the right thing to do: an Olympiad different from the numerous fast-type calculational contests, an Olympiad consisting of a few hard essay-proof problems for several hours, in the true fashion of Eastern Europe. I met Paul Zeitz (University of San Francisco) a week later, and definite plans to start the Bay Area Mathematical Olympiad (BAMO) were set in motion. To publicize the plan, in the late fall of 1997 MSRI asked me to give a talk to an audience of 400 people at a bi-annual public event. Sandwiched between two spectacular lectures on the mathematics behind “Brain Waves” and “Toy Story”, was my modest presentation “The High School Olympiads - Excitement, Talent, and Determination” (see MSRI streaming video [89]). Years afterward, people still remember it by a single picture: that of a chicken and an egg. The idea was that BAMO would get its participants mainly through newly founded school-based math circles around the SF Bay Area and would serve as an annual focal event for their activities. The Olympiad and the Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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math circles would complete and strengthen each other and would be founded at the same time: neither would exist without the other. The mathematical community would support the math circles with materials and occasional session leaders; but the circles would be run by teachers at their schools. In the audience were Tom Davis (Silicon Graphics), Tom Rike (Oakland High School), Quan Lam (UC Berkeley President’s Office), Brian Conrey (Director of the American Institute of Mathematics in Palo Alto (AIM)), and Donald Knuth (Stanford), who all expressed desire to help with the new circle and Olympiad movement. MSRI and AIM then launched a series of events with local teachers and the media to publicize BAMO and to encourage the start-up of many math circles. Alexander Givental and Bjorn Poonen (UC Berkeley), John McCuan (MSRI), Dmitry Fuchs (UC Davis), Tatiana Shubin (SJSU), Joshua Zucker (then at Henry Gunn High School), and others were attracted through these events and pledged their support. 4.4. The “temporary” is the most permanent. One of these public events stands out in my mind as the conception of the Berkeley Math Circle. It was half a year later, in April 1998. Thirty or so local teachers had gathered at MSRI to learn about BAMO and to experience a math circle mock-session. Everyone was elated after the presentations; people were talking excitedly to one another. But when a poll was taken of how many teachers were interested in starting a math circle at their own school, guess what? There was not a single hand up in the air! This was a wake-up call for all of us . . . more precisely, a bucket of icy water on my hot head. I remember sitting in my chair and puzzling it over: “What shall we do? BAMO can’t survive without math circles . . . . But the teachers are obviously not ready to undertake the enterprise on their own. Is this the end of it?” Still reeling with the thought, I started circulating among my colleagueprofessors asking if they were willing to deliver several sessions a year at a temporary math circle, to serve as an example to teachers, so that they would learn how it is done and would then start their own math circles. I got affirmative answers from seven and undertook the task of organizing a 1–2 year trial math circle in Berkeley. I must have been out of my mind, not realizing at the time what an enormous responsibility, both academic and administrative, I was willingly adding to my full-time job. But that’s what a new baby requires: sacrifice and effort and devotion. I had more than enough of each, as I was carrying a lifelong gratitude for my own childhood math circles and wanted to convey the wonders of mathematics to the young generation of the United States, my new home. What I didn’t know was that this project was far from temporary: that it would go on year after year, until we would be celebrating now a decade of the Berkeley Math Circle and the present book would be our new baby. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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There must have been more “crazy” people in the SF Bay Area at that time. A twin to BMC was born: the San Jose Math Circle [80] came into existence the same week as BMC, mid-September 1998, under the tender care and never-ending enthusiasm of Tatiana Shubin and Tom Davis, and is still operational. For a few years Tom Rike, Joshua Zucker, and John Howe led their own school-based circles in Oakland, Henry Gunn, and Presentation High Schools, respectively. Sam Vandervelde had a circle for two years at Stanford [88] (now led by parents). With MSRI’s guidance and support, Paul Zeitz and Brandy Wiegers launched a different type of math circles in San Francisco [79] and Oakland [69]. Sharon Madison opened the Sudbury Math Circle (Canada) as a chapter of BMC, and Olga Radko also fashioned the LA Math Circle [57] after BMC. The SF Bay Area network has expanded now to a number of math circles across the U.S.: very few school-based and not nearly as many as needed, but certainly way more than a decade ago. 4.5. Mapping out the future. Zooming back in on the Berkeley Math Circle, the services it offers begin with the weekly sessions and the monthly contests, but certainly do no end there. BMC has become a center for communications between students, parents, instructors, teachers, educators, and university administrators, where the circlers’ present and future mathematical education is mapped out. This kind of mentoring is possible only in the presence of both “sides”: high quality instructors and students. The more than 50 BMC instructors range from teachers and students to university faculty and real world tycoons. Among them are mathematicians: Alexander Givental, Alexandre Chorin, Bernd Sturmfels, Bjorn Poonen, Dmitry Fuchs, Elwyn Berlekamp, Federico Ardilla, Joe Buhler, Kiran Kedlaya, Olga Holtz, Ravi Vakil, Robin Hartshorne, Serge Lang, Vera Serganova, and many more. Some famous alumni have also contributed sessions to the circle: Gabriel Carroll, Maxim Maydanskiy, Inna Zakharevich, Neil Herriot, Andrew Dudzik, Austin Shapiro, and Oaz Nir, all of whom have chosen career paths in or related to mathematics. The accomplishments of the BMCers are stellar. For example, half of the BAMO grand prizes and brilliancy awards have been captured by BMCers, including the only brilliancy award won by a girl, Hoan Ngo (Oakland High School), as well as half a dozen gold and silver medals at the IMO’s and a dozen USAMO wins. In 2007, Evan O’Dorney, as an 8th grader, scored perfectly at BAMO and won the National Spelling Bee; next year he scored highest at the USAMO and received the Clay Olympiad Scholar Award [16] for one of his solutions. Several multiple-time Putnam Fellows1 are among our students. But most importantly, original mathematical research has been conducted by several circlers, including Gabriel Carroll, Tiankai Liu, Maksim Maydanskiy, Evan O’Dorney, and others. 1William Lowell Putnam Mathematical Competition [100] is the premier Mathematical Olympiad for college students in the world. A Putnam Fellow is among the top 5 scorers.

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5. Does the U.S. Need Top-Tier Math Circles?2 “I wish to state in no uncertain terms how important programs for our talented young people are to the future of this country. The best place to develop the highest end mathematical talent is in groups where young people can feed off each others’ excitement, guided by the best minds in the field. The model of top-tier math circles has been honed over decades in other countries. An American version has been in place for a decade and has shown measurable and almost unbelievable results. Now is the time to make these programs a permanent feature of our educational landscape. The community is ready to assist in any way possible. Universities are happy to provide facilities. Professors are happy to volunteer their time. Parents are happy to spend countless hours. And the reason we do this is that when you see these kids catch fire, it takes your breath away.” Ravi Vakil Four-time Putnam Fellow Professor of Mathematics Stanford University

5.1. Early birds. Creative people start at a very young age to think “outside-of-the-box” and to make significant contributions to the world. Some noticeable examples are Bill Gates, who at age 20 dropped out of Harvard to run Microsoft full-time; Steve Jobs founded Apple at age 19; and recently Mark Zuckerberg created Facebook, a social graph platform, also at age 19. The best young minds in the U.S. deserve our support. The Top-Tier Math Circles are venues for such support: they nurture individuals who are capable of significant accomplishments by giving them advanced training in problem solving tools that are found in no other U.S. educational institution. As another example, a month before Evan O’Dorney [58] qualified for his first IMO in Spain ’08, the 9th grader was exempted from his final in a linear algebra class at UC Berkeley. The reason: he solved an open problem posed in an article by Professor William Kahan [48]; more precisely, Evan found out how small one can make the Cayley transform of a real orthogonal matrix by reversing the signs on selected columns. “BMC has taught me a number of useful mathematical concepts and theories and exposed me to challenging problems. Writing problems for the Monthly Contests provided an outlet for my creative mind. BMC also introduced me to the top local, national, and international mathematical contests. The mentorship I receive through BMC is invaluable.” Evan O’Dorney, BMCer Perfect BAMO ’07 and top USAMO ’08 scores Clay Olympiad Scholar Award ’08 National Spelling Bee Champion ’07 2 Excerpts from [98].

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BMCer Gabriel Carroll was a high school junior when he took time off from IMO participation to work at the Research Science Institute at MIT. Without any prior experience in algebraic topology, he studied the link between posets and geometric figures, and his paper “Homology of Narrow Posets” [73] won the third prize at the Intel Science Talent Search ’01. Gabriel went on to win two gold and one silver medal at the IMO’s, achieving one of only four perfect scores at IMO ’01. He conquered the Putnam four times, two of which while in high school. A quote by him appears in the beginning of the Introduction. After winning a grand BAMO prize and the Regents’ and Chancellor’s Scholarship to UC Berkeley, BMC alumnus Maksim Maydanskiy attended two top undergraduate research programs: the Penn State REU and the REU in Duluth, Minnesota. His first project was inspired by Monsky’s Theorem on triangulations of the square and resulted in the paper “Triangles Gone Wild” [49]. His Duluth work “The Incidence Coloring Conjecture for Graphs of Maximum Degree 3” [61] extended the previously known result that all Hamiltonian cubic graphs have incidence 5-coloring to all cubic graphs. “The impact of the math circle program on my personal mathematical development is hard to overestimate. It was, and continues to be, the single most vibrant source of mathematical activity for high school students in the Bay Area. The lectures introduced me to many areas of mathematics, a number of which came up again in my later studies. The opportunity to meet a variety of people from fellow students to professors, the college campus setting, the overall atmosphere – all of that made the Berkeley Math Circle unique. The program helped me to shape my plans for undergraduate education. It was an experience no other sources could provide. The program has a great effect on mathematical youth in the Bay Area. It provides an interaction media and stimulating environment, both encouraging further involvement from students already interested in mathematics and promoting mathematics to a wider audience.” Maxim Maydanskiy, BMC alumnus BAMO ’00 grand prize Ph.D. student in mathematics, MIT

5.2. The ultimate measure: more testimonials. An important contribution that top-tier math circles make is to challenge the exceptional students and by doing so to keep them interested in science and mathematics. “The math circle was so crucial to my education and interest in math; I can hardly imagine studying math at Harvard if it weren’t for it.” Tiankai Liu, SJMC and BMC alumnus Three-time IMO gold medalist Two-time Putnam Fellow Ph.D. student in mathematics, MIT Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Over and over again, our circlers write about the impact of the program on their understanding of mathematics and their future; about a “different side of math” which they can acquire at the math circle but not at school; about “mind-bending” and “constantly challenging” sessions; about “gaining confidence” and finding a place where they “feel accepted”. Every year we conduct anonymous evaluations of the Berkeley Math Circle. Here is a short selection of some common remarks by students and parents. • “I really liked BMC talks because they were very diverse in nature. One day we might learn about an unordinary geometry, where the concept of parallel lines is different from the ordinary; next time we will learn how to fold a piano into a Möbius strip. This Math Circle really broadened my horizons about what math is really about.” • “BMC made me see a different side of math. The math I experience at math circle is totally different from the math at school. At school, we do derivatives, integrals, infinite series, etc. However, at math circle we don’t do anything like that. It made me realize that there’s a lot more to math.” • “BMC makes me understand how large things interact with math in the world and how mathematics relates to society. Thank you for the Berkeley Math Circle. Because of this circle, I discovered that mathematics is very important to our lives, and I learned a great deal about mathematics that I never learned in school.” • “This is the first year my sons did any kind of mathematics activity outside of the schools and it really opened their eyes! Tuesday nights at BMC is the night they never miss if they don’t have to.” • “It has provided interesting topics with scientific and practical applications to learn about and helped solidify my goals in math and science.” • “I was impressed by the amount of mathematical material covered. There were many great speakers!” • “BMC taught me how to write proofs, which has helped me tremendously in my university math classes . . . .” • “BMC rocks! Longer sessions!!” • “I’ve made a lot of friends through BMC, and I really enjoy going to the lectures when possible. BMC helped me gain self-confidence. It also provided a comfortable peer group and a place where I felt accepted.” • “The Berkeley Math Circle provides me with a period of fun, intense, and mind-bending sessions of mathematical obstacle-courses. Constantly challenging and altogether educating, I really feel the Math Circle has taken a positive toll on my view of mathematics.”

5.3. The gathering storm. There are a number of studies of the deteriorating situation in U.S. math and science education and its impact on the scientific and technological presence of the U.S. in the world. To describe just how critical the situation is, we refer below to three such reports.

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“The United States is losing its edge in innovation and is watching the erosion of its capacity to create new scientific and technological breakthroughs. Increased global competition, lackluster performance in mathematics and science education, and a lack of national focus on renewing its science and technology infrastructure have created a new economic and technological vulnerability as serious as any military or terrorist threat.” A Commitment to America’s Future, 2005 [13]

Recently the National Academy of Sciences has also called to our attention the need for the U.S. to raise its capabilities in mathematics, science and engineering, in a report entitled “Rising Above the Gathering Storm: Energizing and Employing America for a Brighter Economic Future” [64]. According to it: • The U.S. has long depended on foreign-born and -trained mathematicians, engineers and scientists to help maintain its intellectual lead. • The global competition for these talented individuals has greatly intensified in recent years and will continue to do so, as the rest of the world increases its technical capabilities and living standards. • To remain competitive, the U.S. needs to devote considerably more effort and resources to foster excellence in mathematics, science and engineering.

The majority of talented individuals in these fields recruited by U.S. universities and technology companies are from China, Europe, India, and the former Soviet Union. A 2006 report on Science, Technology, Engineering, and Mathematics Education (STEM, [68]) brought forward related troubling trends and numbers: • In 2004, China graduated approximately 500,000 engineers; India graduated 200,000 engineers; and the U.S. graduated 70,000 engineers. On the other hand, South Korea graduates as many engineers as the U.S. even though it has only one sixth of the U.S. population. • More than half of all engineering doctorates awarded in the U.S. go to foreign-born students. In 2003, 25% of all college-educated workers and 40% of all doctorate holders were foreign-born. Over half of the doctorate holders in several fields who resided in the U.S. were foreign-born: computer science; electrical, civil, and mechanical engineering. • From 1994 to 2004, there has been a steady increase in the percentage of U.S. patents granted with a foreign origin, including foreign-owned companies and foreign inventors. In one decade this number has increased from 18% in 1994 to 48% in 2004!

What do these foreign countries do differently from the U.S.? There are many differences and each country is unique. India and China value technical education as a path to prosperity; admission to technical schools there is based on rank in national exams. In the former Soviet Union and Eastern Europe, mathematically talented individuals are identified very early and are provided with the resources needed to reach their full potential. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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5.4. Raising the ceiling. What can be done in the U.S.? Hung-Hsi Wu, Professor of Mathematics at UC Berkeley, has been involved in the education of U.S. mathematics teachers for the last decade. He was on the Task Group on Teachers in the National Mathematics Advisory Panel appointed by President Bush and is currently serving on the National Research Council Panel on the Study of Teacher Preparation Programs. “A main purpose of both panels is to address the crisis in teacher quality among math teachers so as to insure the production of a large enough pool of mathematically literate students to fill our technological needs. However, to insure that we also produce first rate scientists and mathematicians, a different kind of approach would be necessary. This is where the Math Circles come in. It is programs like the Math Circles that can provide the needed guidance and stimulation for the cream of the crop of this pool. While the work done by the abovementioned panels is designed to raise the floor to make our nation competitive in the global market, what the Math Circles do is to raise the ceiling in order to maintain our worldwide leadership position in science and technology. At a time of need in our nation’s mathematics education, the work done in top-tier math circles such as the Berkeley Math Circle and the San Jose Math Circle is of vital importance.” Hung-Hsi Wu Professor of Mathematics University of California at Berkeley

While it is unlikely that math circles will have a large impact on the value system of the American public, the top-tier math circles in the U.S. do play a significant role in meeting the challenges described above by preparing our best young minds for their future role as mathematics, science, and technology leaders. With your help, we can establish a dense network of math circles across the U.S. With hope, Zvezdelina Stankova Berkeley Math Circle Director Berkeley, June 2008

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Symbols and Notation N Z R C α, β, γ I(A, r) I(A) [ABC] AB |AB| AB − − → AB ABC ∠ABC ∼ = ∠A = ∠B ∼ ⊥




⊂ (⊂) ∈ (∈)  () A∩B aA |A| Σ (A) (a, b) [a, b] x x min{a, b} lcm(a, b) a | b (a
b) a ≡ b (mod c)

set of natural numbers set of integers set of real numbers set of complex numbers alpha, beta, gamma: letters from the Greek alphabet inversion with center A and radius r inversion with center A and unspecified radius area of triangle ABC segment AB or its length depending on context distance from A to B; used if AB is ambiguous arc AB ray AB triangle ABC angle ABC geometric congruence congruence of angles written also as ∠A ∼ = ∠B geometric similarity is perpendicular to is parallel to is a subset of (is not a subset of) is an element of (is not an element of) passing (not passing) through intersection of set A and set B mass point (a, A) number of elements in set A sum of elements in set A open interval from a to b: all x ∈ R such that a < x < b closed interval from a to b: all x ∈ R such that a ≤ x ≤ b floor of x: greatest integer ≤ x ceiling of x: least integer ≥ x minimum of a and b least common multiple of a and b a divides b without remainder (a does not divide b) a is congruent to b modulo c 299

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 ϕ(n) ⇒ ⇐ ⇔, iff
♦

repeated operation number of integers between 1 and n relatively prime to n implies, only if if, is implied by if and only if end of proof end of hint or partial solution ? questionable proof n! n factorial, 1 · 2 · 3 · · · (n − 1) · n P (n, k) number of permutations of n objects taken k at a time n binomial coefficient n choose k, n!/(r!(n − r)!) k Re{z} real part of complex number z Im{z} imaginary part of complex number z z conjugate of complex number z √ i imaginary unit, −1 e base of natural log, ≈ 2.71828 π ratio of circumference √ to diameter of a circle,√≈ 3.14159 φ; φ golden ratio, (1 + 5)/2; its conjugate, (1 − 5)/2

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Abbreviations AA AHSME AIM AIME AMC AMS ARML BAMM BAMO BMC GPHP HMC IH, IHs IMO LA LHS MAA MASS MC MI, MIs MIT MOSP MSRI NSF NYCML PHP PST REU RHS SAS SF SJMC SJSU SOS

angle-angle criterion for similarity of triangles American High School Mathematics Examination American Institute of Mathematics American Invitational Mathematics Examination American Mathematics Competition American Mathematical Society American Regional Mathematics League Bay Area Mathematics Meet Bay Area Mathematical Olympiad Berkeley Math Circle Generalized Pigeonhole Principle hometown math circles Inductive Hypothesis, Strong Inductive Hypothesis International Mathematical Olympiad Los Angeles left-hand side Mathematical Association of America Mathematics Advanced Study Semesters Monthly Contest at the Berkeley Math Circle mathematical induction, strong form of mathematical induction Massachusetts Institute of Technology Mathematical Olympiad Summer Program Mathematical Sciences Research Institute National Science Foundation New York City Mathematical League Pigeonhole Principle problem solving technique Research Experience for Undergraduates right-hand side side-angle-side criterion for similarity/congruence of triangles San Francisco San Jose Math Circle San Jose State University sum of squares 301

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302 SSS TLC USAMO VIGRE WLOG

ABBREVIATIONS side-side-side criterion for similarity/congruence of triangles tangent line-chord USA Mathematical Olympiad Vertical Integration of Research and Education without loss of generality

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Biographical Data Mira Bernstein was born in Russia. She completed her undergraduate studies at Yale and wrote her Ph.D. thesis at Harvard on moduli spaces of curves, in the field of algebraic geometry. As a postdoctoral fellow at UC Berkeley, Mira led a number of sessions at the Berkeley Math Circle. One of her first contributions to BMC was a paper on proofs, which is still widely used by the circlers and is the basis for Session 5 in this book. For twelve summers, her passion for problem solving and working with talented young mathematicians has been channeled towards organizing and running the Canada/USA Mathcamp [14]. Mira joined Wellesley College in 2002. Her research interests also include combinatorics and information theory; her paper “Some Canonical Sequences of Integers” [9] investigates “eigen-sequences” associated with transformations of integer sequences. Lately, she has been interested in applied mathematics as well, working on a project in mathematical population genetics in collaboration with a laboratory at the Harvard Medical School. Gabriel Carroll was a student at Oakland Technical High School when he attended the Berkeley Math Circle for three years. He won three consecutive BAMO grand prizes and three ARML top individual prizes, received two gold and one silver IMO medal (including a perfect score at IMO ’01 in Washington, D.C.), and was among the top five-ranked Putnam scorers from 2000–2003, becoming one of only seven four-time Putnam Fellows. Gabriel co-coordinated the BMC Monthly Contest for two years and has enjoyed writing problems for BAMO and USAMO. While still a circler, he presented a number of topics to the more advanced BMC students; one of these sessions became the basis for Monovariants in this book. His paper “Homology of Narrow Posets” [73] placed him third at the Intel Science Talent Search ’01 [46]. Gabriel has a variety of interests outside mathematics: he is an outstanding piano player, loves drawing cartoons (six of which are in this book), edited the Harvard-based humor magazine Swift [36], and taught English in Hunan 303

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province in China. In 2006–2007, he was employed by the National Bureau of Economic Research [65], where he worked for a group studying procrastination in retirement savings. Gabriel is currently a graduate student in economics at MIT.

T om Davis competed on the Caltech Putnam team as an undergraduate and earned his Ph.D. in probability and partial differential equations at Stanford. He was a founder and the Principal Scientist of Silicon Graphics. For ten years he has been a freelance mathematician, pursuing his passion: to work on challenging problems with talented students in the setting of math circles. Tom has been involved in the Berkeley Math Circle from its inception. He co-founded and co-directs the San Jose Math Circle [80] and regularly leads sessions at all SF Bay Area math circles. He has also coorganized Teachers’ Math Circles at AIM and MSRI. According to Tom, calculus is not enough to do computer graphics: “People who are interested in making computer-generated dinosaurs for ‘Jurassic Park’ or a liquid metal man in ‘Terminator II’ or who want to have Forrest Gump shake hands with Richard Nixon had better have a solid grounding in advanced calculus and in differential and projective geometry.” Tom’s web site [19] contains free dynamic geometry software, the Rubik’s cube software discussed in his article, and an extensive collection of math circle talks. Tom is also an avid fan of long distances. Recently, he completed the Boston and Rome marathons and won in his age group the first two triathlons in which he participated, Wildflower and San Jose International.

Quan Lam is a native of Macau. After spending his undergraduate years at Caltech, he received his Ph.D. in mathematics from UC Berkeley and an MBA in Finance and Management Science from the Haas School of Business at UC Berkeley. Quan works at the UC Office of the President as a manager for the IM Division of Information Resources and Communication. He has led a number of Berkeley Math Circle sessions from the beginning, with a special emphasis on proof techniques; his article in this book, for example, treats in detail the method of mathematical induction. In 1998–1999, he was the first web administrator for the Berkeley Math Circle. For twelve years, Quan has been the coordinator of ARML [2] practices for the students in the SF East Bay. During this time, the SF Bay Area Ateam placed first 3 times and second 3 times, and two circlers, Gabriel Carroll and Tiankai Liu, scooped the top ARML individual prizes from 1998–2001. Quan has also worked hard in Asia on recruiting new countries for the ARML and thus expanding the contest scope from national to international. Taiwan and Hong Kong joined ARML because of his involvement; his next challenge is to recruit teams from Macau and China and possibly from Thailand, Singapore, and Malaysia. Quan will be assisting in organizing a “Junior” ARML for elementary and middle school students in Hong Kong, and he hopes to “export” the math circles to that region some day. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Ivan Matić earned a silver medal on the Serbian IMO team as a high school student. Upon graduation from Belgrade University, he was accepted to the Ph.D. program in mathematics at UC Berkeley, where he now does research in the intersection of probability and partial differential equations (PDE) on large deviations of Hamilton-Jacobi PDE. Ivan likes to work with young math students around the world. His article on Circle Geometry in this book has fused together several of his engaging olympiad preparation sessions at BMC. He has also served as the BMC monthly coordinator for three years and the Associate Director of the Circle for one year as well. Ivan frequently visits the Canada/USA Mathcamp [14], where he helps in many aspects of the program. When he goes back home, he coaches the Serbian IMO team. Ivan devotes a lot of time and effort to math popularization. He spends many weekends programming and editing the contents of a web site [23] dedicated to math competitions around the world and containing a vast supply of aids and materials for anyone who loves the world of olympiads. He has co-authored the book The IMO Compendium [22], containing the IMO short-listed problems and their (in many cases) multiple solutions.

Evan O’Dorney has attended the Berkeley Math Circle for two years and

co-coordinated the BMC Monthly Contest in 2007–2008. As an 8th grader, he captured the BAMO ’07 grand prize with a perfect score and became the 2007 Scripps National Spelling Bee Champion. The next year he tied for the top score on the USAMO ’08, won the Clay Olympiad Scholar Award [16] for the most creative solution to a USAMO problem, and qualified for the USA team for the IMO ’08 in Spain. Still a 9th grader, he discovered how small one can make the Cayley transform of a real orthogonal matrix by reversing the signs on selected columns and thereby solved an open linear algebra problem posed in an article by Professor William Kahan [48]. The Appendix in Session 12 is based on his original work on a game/graph theory problem from Gabriel Carroll’s Monovariants session. Evan enjoys playing the piano and composing music. He also likes computer programming, juggling, and reading the dictionary and has earned a first degree black belt in tae kwon do.

T om Rike graduated from San Francisco State College in 1968. After spending the next two years taking graduate courses and getting a teaching credential, he taught six years at Westlake Junior High School. In 1974, he went back to school in the evenings and received his M.S. in mathematics from Holy Names College in 1976. Moving to Oakland High School, he taught until he retired in 2003 and now volunteers there three days a week. Tom has served as the high school liaison for BMC and BAMO from the beginning. He has been fascinated by giants of mathematical thought such as Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Archimedes, Euler, and Gauss and has shared their works at BMC sessions on a number of occasions: using the arbelos from the Book of Lemmas by Archimedes [43]; demonstrating Euler’s solution to the Basel Problem [24]; and giving Gauss’s proof from Disquisitiones Arithmeticae that a 17-gon is constructible [31]. It is Archimedes’ lever, on which he “balanced” his Mass Point article in this book. For a number of years, Tom ran a math circle at Oakland High, incorporating into it students and teachers from nearby Skyline and Bishop O’Dowd High Schools. He now also gives talks at the San Jose Math Circle and the new Oakland/East Bay Teachers’ Math Circle. Tom has been deeply involved as a coach with East Bay Mathletes, a monthly competition among local high schools, which began in 1978. In 1983, he helped found in Oakland a middle school mathematics competition which takes place four times a year. Among his interests outside mathematics are the San Jose Sharks, the SF Giants, and opera. He has been playing Go and studying the Japanese language with devotion for over 40 years.

T atiana Shubin went as a high school student to the Special Mathematics and Physics Boarding School of the Academy of Sciences in Novosibirsk. She did her undergraduate work in the USSR at the Kazakh and Moscow State Universities. In 1983 she completed her Ph.D. thesis at UC Santa Barbara on existence of neofields, a cross between combinatorics, abstract algebra, and number theory. After a couple of years at UC Davis, she joined San Jose State University in 1985 and has been teaching there ever since. At the outset, Tatiana was a strong proponent of math circles. She has often said that she owes her life to math circles and is now paying back her debt of gratitude by making math circles available to others. In 1998, Tatiana co-founded the San Jose Math Circle [80] and the highly successful Bay Area Math Adventures (BAMA) talks. The latter have been preserved in Mathematical Adventures for Students and Amateurs [41], co-edited by Tatiana and David Hayes; a second volume is upcoming. Tatiana has also contributed sessions to BMC since 2001, with emphasis on group theory and a variety of hybrid geometry topics. Her very article in this book unites geometry with complex numbers. She is an organizer of math circles for middle school teachers at AIM, MSRI, and Washington, D.C.; has helped start a math circle in Santa Fe and gave talks at another one in Dallas; and is now in the process of expanding this concept nationally: a Special Interest Group of MAA on Math Circles for Students and Teachers will be officially formed at the AMS/MAA Joint Meeting in January 2009. Tatiana’s outstanding teaching was recognized in 2006 via MAA’s Distinguished College or University Teaching of Mathematics Award, of the Northern California, Nevada, and Hawaii section. With all these activities, Tatiana still finds opportunities to practice her favorite pastime, rock hounding, resulting in an expansive rock collection at her home. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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Zvezdelina Stankova was drawn into the world of mathematics when as

a 5th grader she joined the math circle at her school in Bulgaria and won, three months later, the Regional Math Olympiad. She represented her home country at two IMO’s, earning silver medals. Her article on Inversion in the Plane in this book is inspired by one of the very first lectures she heard during the training of the Bulgarian IMO team. As a freshwoman at Sofia University, Zvezda won a competition to study in the U.S. and completed her undergraduate degree at Bryn Mawr College in 1993. She did her first math research in enumerative combinatorics at two summer REU’s in Duluth, Minnesota. The resulting papers contributed to her Alice T. Schafer Prize for Excellence in Mathematics by an Undergraduate Woman, awarded by the Association for Women in Mathematics. In 1997, Zvezda received a Ph.D. from Harvard University, with a thesis on moduli spaces of curves, in the field of algebraic geometry. Meanwhile, she earned a high school teaching certificate in the state of Massachusetts and later in California. As a postdoctoral fellow at MSRI and UC Berkeley in 1997–1999, Zvezda co-founded BAMO [7] and started the Berkeley Math Circle [8]. She trained the USA national team for the IMO’s for six years, including the memorable year 2001 when three of the six team members were BMCers. Since 1999, she has been at Mills College. Her current research interests include classification of restricted patterns in the area of enumerative and algebraic combinatorics. Zvezda’s inspiring style and passion to teach have been recognized by the MAA: in 2004 she was selected as a recipient of the first Henry L. Alder Award for Distinguished Teaching by a Beginning College or University Mathematics Faculty Member. Zvezda enjoys Latin and ballroom dancing, playing the piano and the guitar, jogging, and spending time with her young girl and boy.

Sam Vandervelde is probably best known for the The Mandelbrot Competition [59], a math contest he helped create with Sandor Lehoczky and Richard Rusczyk in 1990, which greatly increased American high school students’ exposure to written mathematical arguments. For a compilation of all problems from the first twelve years of the contest, we direct the reader to two of Sam’s books [96, 97]. In 1989, as a senior in high school, Sam won the ARML and received a silver IMO medal. He later taught for five years at The Roxbury Latin School, a private boys school in Boston for grades 7–12. In 2004, he went back to graduate school and earned his doctorate at the University of Chicago with a thesis on Mahler measure, in the field of number theory. As a visitor at Stanford University, he founded and ran the Stanford Math Circle [88] for two years and wrote a handbook on math circles called Circle in a Box [95]. His wide range of experiences coupled with a natural charisma have made his BMC sessions quite memorable. His Stomp article in this book is full of “exotic” imagery that will engage the reader’s imagination. Purchased from American Mathematical Society for the exclusive use of Sora Kanosue (knsrxo) Copyright no copyright American Mathematical Society. Duplication prohibited. Please report unauthorized use to [email protected]. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.

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In 2007 Sam joined the Mathematics Department at St. Lawrence University. His recent research interests include graph theory and number theory. He contributes problems to the USAMO and is currently on the USAMO Problems and Grading Committees. His other interests include playing soccer, spending time with his two boys, and performing choral music.

Paul Zeitz won the USAMO in 1974 and was on the first USA team to participate at the IMO. He majored in history at Harvard and then taught for several years in high schools in Colorado Springs and San Francisco. In 1992 he received his Ph.D. in ergotic theory from UC Berkeley and is presently a professor at the University of San Francisco. For his teaching, he was honored with the MAA’s national Deborah and Franklin Tepper Haimo Award in 2003. Every spring from 1994–2005, Paul ran a yearly math festival, the Bay Area Math Meet, which was the highlight of the year for hundreds of SF Bay Area students. He coached the USA IMO team for several years, most notably in 1994, when the team attained the only perfect score in the history of the competition. In 1999 he co-founded the Bay Area Mathematical Olympiad [7]. For a decade, Paul has delivered inspiring sessions at BMC: in combinatorics and probability, graph and game theory, number theory and statistics, generating functions and the golden ratio, and other topics. His Combinatorics article in this book demonstrates, among other things, the richness of his mathematical language and the ease with which he delves into mathematical “folklore”. Not surprisingly, his book The Art and Craft of Problem Solving [101], now in its second edition, is known throughout the world. Paul is the director of another model of math circles, the San Francisco Math Circle [79], which is in its third year. When not doing math, Paul likes to tinker with electronics and go on outdoor adventures with his wife (a former national park ranger) and two children.

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Bibliography 1. American Mathematics Competitions (AMC), http://www.unl.edu/amc/. 2. American Regions Mathematics League, http://www.arml.com/index.php. 3. T. Andreescu and Z. Feng, 103 Trigonometry Problems from the Training of the USA IMO Team, Birkhäuser, 2004. 4. F. Ardila and R. Stanley, Tilings, http://math.sfsu.edu/federico/Articles/ tilings.pdf, 2004. 5. R. Artino, A. Gaglione, and N. Shell, The Contest Problem Book IV, 1973–1982, Math. Assoc. of America, 1983. 6. A. Benjamin and J. Quinn, Proofs that Really Count: The Art of Combinatorial Proof, Math. Assoc. of America, 2003. 7. Bay Area Mathematical Olympiad, http://www.bamo.org/. 8. Berkeley Math Circle (BMC), http://mathcircle.berkeley.edu/. 9. M. Bernstein and N. Sloane, Some Canonical Sequences of Integers, http://arxiv. org/abs/math/0205301v1, 2002. 10. G. Berzsenyi and S. Maurer, The Contest Problem Book V, 1983–1988, Math. Assoc. of America, 1997. 11. A. Bogomolny, Cut-the-Knot: Circles and Spheres, http://www.cut-the-knot.org/ proofs/circlesAndSpheres.shtml. 12. J. Boyd and P. Raychowdhury, An Application of Convex Coordinates, Two Year College Math. J. (1983), 348–349. 13. Business-Higher Education Forum, A Commitment to America’s Future, http:// www.bhef.com/publications/MathEduReport-press.pdf, 2005. 14. Canada/USA Mathcamp, http://www.mathcamp.org/. 15. F. Chung, R. Graham, J. Morrison, and A. Odlyzko, Pebbling a Chessboard, Amer. Math. Monthly 102 (1995), no. 2, 113–123. 16. Clay Olympiad Scholar Award, http://www.claymath.org/olympiad/. 17. J. Conway and R. Guy, The Book of Numbers, Springer-Verlag, New York, 1996. 18. H. Coxeter, Introduction to Geometry, pp. 216–221, John Wiley & Sons, 1969. 19. T. Davis, http://www.geometer.org/. 20. Davis Math Circle, http://www.math.ucdavis.edu/∼ exploration/mathcircle/. 21. F. Diamond and J. Shurman, A First Course in Modular Forms, vol. 228, SpringerVerlag, New York, 2005. 22. D. Djukić, V. Janković, I. Matić, and N. Petrović, The IMO Compendium: A Collection of Problems Suggested for The International Mathematical Olympiads: 1959– 2004, Problem Books in Mathematics, Springer, 2006. 23. D. Djukić and I. Matić, http://www.imomath.com/. 24. W. Dunham, Euler, The Master of Us All, pp. 43–48, Math. Assoc. of America, 1999. 309

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310

BIBLIOGRAPHY

25. H. Edwards, Fermat’s Last Theorem: A Genetic Introduction to Algebraic Number Theory, vol. 50, Springer, 1977. 26. A. Engel, Problem Solving Strategies, Springer, 2006. 27. EPGY, Education Program for Gifted Youth at Stanford University: EPGY Summer Institutes, http://epgy.stanford.edu/summer/middleschoolprogram.html. 28. D. Fomin, S. Genkin, and I. Itenberg, Mathematical Circles (Russian Experience), Amer. Math. Soc., 1996. 29. J. Gallian, Contemporary Abstract Algebra, Houghton Mifflin, 2005. 30. M. Gardner, Knotted Doughnuts: Look-See Proofs, pp. 192–204, W. H. Freeman and Company, 1986. 31. C. Gauss, Disquisitiones Arithmeticae, English ed., Springer, 1985. 32. I. Gelfand, E. Glagoleva, and A. Kirilov, The Method of Coordinates, Birkhäuser, 1990. 33. I. Gelfand, E. Glagoleva, and E. Shnol, Functions and Graphs, Birkhäuser, 1990. 34. I. Gelfand and M. Saul, Trigonometry, Birkhäuser, 2001. 35. I. Gelfand and A. Shen, Algebra, Birkhäuser, 1993. 36. K. Gewertz, Not Just Numbers: Math Whiz Carroll Lives Inside – and Outside – His Mind, http://www.hno.harvard.edu/gazette/2005/06.09/34-carroll.html,2005. 37. A. Givental, The Pythagorean Theorem: What Is It About?, Amer. Math. Monthly 113 (2006), no. 3, 261–265. 38. P. Halmos, Problems for Mathematicians Young and Old, Dolciani Math. Expositions, no. 12, Math. Assoc. of America, 1991. 39. M. Hausner, The Center of Mass and Affine Geometry, Amer. Math. Monthly (1962), 724–737. , A Vector Space Approach to Geometry, Dover, 1998. 40. 41. D. Hayes and T. Shubin (eds.), Mathematical Adventures for Students and Amateurs, Spectrum Series, Math. Assoc. of America, 2004. 42. Sir T. Heath, The Thirteen Books of Euclid’s Elements, vol. I, Dover, 1956. , The Works of Archimedes, Dover, reissue of the 1897 edition with the 1912 43. Supplement on The Method, 2002. 44. R. Honsberger, Mathematical Gems 5: Geometry via Physics, Two Year College Math. J. (1979), 271–276. 45. T. Hungerford, Abstract Algebra, An Introduction, Saunders College, 1997. 46. Intel Science Talent Search, http://www.societyforscience.org/sts/. 47. H. Jacobs, Geometry, W. H. Freeman and Company, 1974. 48. W. Kahan, Is There a Small Skew Cayley Transform with Zero Diagonal?, Linear Algebra and Its Applications (2006), 335–341. 49. J. Kantor and M. Maydanskiy, Triangles Gone Wild, MASS selecta (2003), 277–288. 50. A. Khodulev, Pebble Spreading, Kvant (1982), 28–31, 55. 51. A. Kiselev, Kiselev’s Geometry: Book 1. Planimetry, adapted from Russian by Alexander Givental, Sumizdat, 2006. 52. M. Klamkin and A. Liu, Three More Proofs of Routh’s Theorem, Crux Mathematicorum 7 (1981), no. 6, 199–203. 53. V. Klee and S. Wagon, Old and New Unsolved Problems, Dolciani Math. Expositions, no. 11, Math. Assoc. of America, 1991. 54. M. Kontsevich, Problem M715, Kvant (1981), 21. 55. B. Kordemsky, The Moscow Puzzles: 359 Mathematical Recreations, Dover, 1992. 56. L. Larson, Problem Solving Through Problems, pp. 304–305, Springer-Verlag, New York, 1983. 57. Los Angeles Math Circle, http://www.math.ucla.edu/∼ radko/circles/. 58. MAA Online, Evan O’Dorney: Spelling Champ and Math Whiz, http://www.maa. org/news/060408odorney.html. 59. Mandelbrot Competition, http://www.mandelbrot.org/.

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311

60. K. Marling, Mickey Mouse at the Andy Warhol Museum, http://www. carnegiemuseums.org/cmag/bk_issue/1999/mayjun/feat1.html. 61. M. Maydanskiy, The Incidence Coloring Conjecture for Graphs of Maximum Degree 3, Discrete Mathematics 292 (2005), 131–141. 62. Mechanics Magazine, Archimedes Moves the Earth, http://www.math.nyu.edu/ ∼crorres/Archimedes/Lever/LeverIntro.html. 63. B. Medigovich, Mass Point Geometry: How to Paint a Triangle, unpublished, 1985. 64. National Academy of Sciences et al., Rising Above the Gathering Storm: Energizing and Employing America for a Brighter Economic Future, http://books.nap.edu/ catalog.php?record_id=11463, 2007. 65. National Bureau of Economic Research, http://www.nber.org/. 66. R. Nelsen, Proofs Without Words, Math. Assoc. of America, 1993. 67. I. Niven, A New Proof of Routh’s Theorem, Amer. Math. Monthly (1976), 25–27. 68. Northern Illinois University, Illinois Status Report onScience, Technology, Engineering, and Mathematics Education, http://www.keepingillinoiscompetitive.niu.edu/ ilstem/pdfs/STEM_ed_Trends_1-4.pdf, 2006. 69. Oakland/East Bay Math Circle, http://oebmc.mathcircles.org/. 70. A. Odlyzko and H. Riele, Disproof of the Mertens Conjecture, J. für die reine und angewandte Mathematik 357 (1985), 138–160, http://www.dtc.umn.edu/∼ odlyzko/ doc/arch/mertens.disproof.pdf. 71. D. Pedoe, Thinking Geometrically, Amer. Math. Monthly (1970), 711–721. , Notes on the History of Geometrical Ideas I: Homogeneous Coordinates, 72. Math. Magazine (1975), 215–217. 73. I. Peterson, Prized Geometric Logic, http://www.maa.org/mathland/mathtrek_4_ 2_01.html, 2001. 74. A. Poorten, Notes on Fermat’s Last Theorem, Wiley-Interscience, 1996. 75. P. Ribenboim, Fermat’s Last Theorem for Amateurs, Springer, 1999. 76. L. Riddle, Sophie Germain and Fermat’s Last Theorem, http://www.agnesscott. edu/lriddle/women/germain-FLT/SGandFLT.htm. 77. J. Robinson, Julia Robinson Math Festival ’07 at Google, Mountain View, CA, http: //www.msri.org/specials/festival, May 2007. 78. C. Salkind, The MAA Problem Book II, 1961–1965, Math. Assoc. of America, 1966. 79. San Francisco Math Circle, http://www.sfmathcircle.org/index.html. 80. San Jose Math Circle, http://www.sanjosemathcircle.org/. 81. M. Saul, G. Kessler, S. Krilov, and L. Zimmerman, The New York City Contest Problem Book, 1975–1984, Dale Seymour, 1986. 82. D. Shklarsky, N. Chentzov, and I. Yaglom, The USSR Olympiad Problem Book, Dover, 1994. 83. S. Singh, Fermat’s Enigma: The Epic Quest to Solve the World’s Greatest Mathematical Problem, Anchor, 1998. 84. S. Singh and J. Lynch, The Proof, http://www.pbs.org/wgbh/nova/proof/, 1997. 85. H. Sitomer and S. Conrad, Mass Points, Eureka 2 (1976), no. 4, 55–62. 86. A. Soifer, Mathematics as Problem Solving, pp. 84–88, Center for Excellence in Math. Education, 1987. 87. I. Sominsky, The Method of Mathematical Induction, Mir, 1975. 88. Stanford Math Circle, http://www.stanfordmathcircle.org/. 89. Z. Stankova, The High School Olympiads: Excitement, Talent, and Determination, http://www.msri.org/publications/ln/msri/1997/bamo/sf/1/index.html, 1997. 90. S. Stein, Archimedes: What Did He Do Besides Cry Eureka?, pp. 7–25, Math. Assoc. of America, 1999. 91. R. Taylor and A. Wiles, Ring-theoretic Properties of Certain Hecke Algebras, Ann. Math. 141 (1995), no. 3, 553–572. 92. E. Uspensky, Cheburashka (
), http://chebur.hobby.ru/chebur.html.

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93. Utah Math Circle, http://www.math.utah.edu/mathcircle/. 94. R. Vakil, A Mathematical Mosaic, Brendan Kelly, 1997. 95. S. Vandervelde, Circle in a Box, to be published by the American Mathematical Society. , The First Five Years: A compilation of problems from the Mandelbrot Com96. petition, 1990–1995, Greater Testing Concepts, 1996. , The Mandelbrot Problem Book: A compilation of problems from the Man97. delbrot Competition, 1995–2002, Greater Testing Concepts, 2003. 98. M. Whitlow, M. Breen, Z. Stankova, and T. Shubin, Sustainable Funding of Top Tier Math Circles, Proposal, 2007. 99. A. Wiles, Modular Elliptic Curves and Fermat’s Last Theorem, Ann. Math. 141 (1995), no. 3, 443–551. 100. William Lowell Putnam Mathematical Competition, http://math.scu.edu/putnam/. 101. P. Zeitz, The Art and Craft of Problem Solving, 2nd ed., John Wiley & Sons, 2006. 102. L. Zimmerman and G. Kessler, ARML-NYSML Contests 1989–1994, Contests in Mathematics, vol. 2, MathPro Press, 1995.

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Credits The American Mathematical Society gratefully acknowledges the kindness of the following institutions and individuals in granting the following permissions: Business-Higher Education Forum The quotation “A Commitment to America’s Future: Responding to the Crisis in Mathematics and Science Education” in the Epilogue, http://www.bhef.com/publications/MathEduReport-press.pdf, Business-Higher Education Forum, 2005. Gabriel Carroll The cartoons “A grime number”, “A lettuce point”, “Factorial”, “n chews “k”, “An unordered pear”, and “Monotone Increasing”. Annette W. Emerson Photograph of Andrew Wiles. The Mathematical Association of America, American Mathematics Competitions Problems from USAMO ’98, AIME ’85, AIME ’88, AIME ’89, AIME ’92, AHSME ’64, AHSME ’65, AHSME ’75, AHSME ’80 used with permission. Rare Book and Manuscript Library, University of Pennsylvania “Give Me a Place to Stand and I Will Move the Earth”, an engraving of Archimedes moving the earth using a lever. Mechanic’s Magazine (cover of bound Volume II, Knight & Lacey, London, 1824).

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Index (3, 3, 7)-configuration, 239 3-dimensional, 124, 128, 140, 247, 251

right, 137, 162, 164, 190, 230 supplementary, 236, 243 tangent line-chord, TLC, 233, 234 vertex, 11 vertical, 237 angle bisector, 136, 148, 152, 153, 246 Angle Bisector Theorem, 136, 146, 152 proof, 136 ant, 129 applied mathematics, 296, 303 arc, 137, 227 circular, 8, 226 intercepted, 227 Archer Design Inc., xviii Archimedes, 123, 128, 129, 151, 306 lever, 128 Ardilla, Federico, 293 area, 100, 128, 135, 139, 146, 148 addition, 135, 148 Argand, Jean-Robert, 182 Aristotle, 92 arithmetic, 52, 128 arithmetic progression/series, 169 ARML, 146, 303, 307 problem, 146 solution, 152, 153 art, xvi, 164 Asia, 304 associativity law, 187 asymptotic manipulations, 219 axiomatic approach of Hausner, 130

absolute value, 67, 185, 198 abstract algebra, xii, 70, 116, 187, 290 AHSME, xviii, 146 problem, 146, 149 solution, 147, 149, 152, 153 AIM, 292, 304, 306 AIME, xviii, 146 problem, 147, 149, 150 solution, 147, 150, 153 Aladdin’s lamp, 3 algebra, xv, 128, 253, 286 algebraic geometry, x, 187, 290, 303, 307 algebraic topology, 295 Alper, Ted, xviii Alternative Centroid Theorem, 21, 24 Alternative Parallelogram Theorem, 24 altitude, 135–138, 148, 149, 152, 234 AMC, xviii, 146 American Mathematical Society, 306 analysis, 187 analytic geometry, 180, 225 angle acute, 149 alternate interior, xii, 11, 247 central, 227 complement, 234 congruent (equal), 17, 22 corresponding, 12 exterior, 162, 227 inscribed, 7, 227, 228 interior, 162 internal, 12 opposite, 12 remote interior, 227

balancing point, 128, 131, 133, 134, 137 Balls in Urns formula, 43, 281 BAMM, 34, 308 BAMO, 225, 242, 291–295, 305, 308 problem, 225, 230, 231, 266 315

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316 solution, 231, 242, 243, 245, 266, 282 barycentric coordinates, 140 Bay Area Math Adventures, 306 Bay Area math circles, xiii, 264, 289, 291, 293, 304 Beatles, 287 befuddler notation, 48 beginning state, Bs , 268 Belgrade, 305 Berkeley Math Circle, x, xi, xvii, xviii, 285, 287, 290, 292, 293, 298, 307 Berlekamp, Elwyn, 293 Bernstein, Mira, 87, 155, 303 Binet’s formula, 119 binomial, 38 binomial coefficient, 36, 41 `n´ , n choose k, 36 k interpretation, 36, 37 properties, 37 Binomial Theorem, 39 Alternate (humor), 39 biology, 286 Bombelli, Rafael, 182 Boston, 291, 304, 307 bound, 267 best, 257, 262 least upper, 269, 271 Breen, Mike, ixn, xviii Brown, Tom, xviii Bryant, Robert, xvii Bryn Mawr College, 285, 287, 307 Buhler, Joe, xvii, 293 Bulgaria, 285, 290, 307 Burks, Michael, 260 calculus, xii, 112, 196, 218, 223, 290 Caltech, 304 Canada/USA Mathcamp, 203, 303, 305 Canadian Mathematical Society, 151 Cardano, Girolamo, 181 Cardano-Tartaglia formula, 181, 197 Carroll, Gabriel, ix, xviii, 65, 249, 293, 295, 303–305 Cartesian plane, 216 Cauchy, Augustin-Louis, 151 Cayley transform, 294, 305 ceiling function, x, 163, 165 center of circle, 2, 11, 23 of gravity, 151 of inversion, 4 of mass, 20, 131, 134–136, 141, 151 centroid, 20, 127, 128, 132, 138, 140

INDEX Centroid Theorem, 20, 21, 128, 140 Ceva, Giovanni, 127 Ceva’s Theorem, 142, 145 proof, 142 cevian, 127, 129, 134, 142, 148–150, 152 chameleon, 207, 219 Cheburashka (
), 226 chemistry, 286 Chen, William, xviii China, 297, 304 chocolate bar problem, 250 chords, 14, 19, 21 concurrent, 238 equal, 228 Chorin, Alexandre, 293 chromatic number of the plane, 173 circle, xii, xv, 25, 41, 149, 225, 226 about rectangle, 14, 23 about trapezoid, 23 center of, 226 inscribed, 149 phantom, 236 radius of, 226 under inversion, 7 unit, 179, 185, 194, 198 circle geometry, xii, 225 circumcenter, 13, 14, 23, 136, 137, 238 circumcircle, 12, 13, 14, 24, 137 circumradius, 136 Ciubotaru, Dan, 290 claim, 107 Clay Mathematics Institute, xvii Clay Olympiad Scholar Award, 293, 305 collinear, 10–14, 18–20, 22, 23, 142–144 coloring invariant, 212 combination, 36, 166 combinatorial gene-counting, 286 combinatorial geometry, 41, 120 combinatorics, x, xii, xiv, 25, 251, 303 algebraic, 307 enumerative, 25, 307 common denominator, 153 divisor, 104 factor, 92 tangent, 235 commutativity law, 52, 187 compactify, 5 complement, 33, 34 complex number, 2, 92, 180, 182, 183 addition, 183, 199 argument, arg, 192 Cartesian form, 184, 191, 193

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INDEX conjugate, 191, 198 conjugation, 186, 193, 199 Euler form, 194 Euler’s formula, 194 imaginary part, 183, 184 modulus, 185, 189, 191, 198 multiplication, 188, 191 polar coordinates, 192 polar form, 192, 193 purely imaginary, 186, 198, 200 real part, 183 scalar multiplication, 184 set, C, 184 subtraction, 185 composition, 195 of inversion, 5, 6 computer graphics, 304 computer science, ix, x, 297 computer simulation (Rubik), 47 concentration as monovariant, 252, 268 Concepts bag, 9 concurrent, 128, 134–137, 140, 149, 231 concyclic, 10, 11–14, 18–20, 22, 226, 229, 231, 237 conditions for, 229 configuration of numbers, 207 congruence, 70, 113, 130, 188, 220 as equivalence relation, 73 auxiliary modulus in, 77 exponents in, 76 exponents off-limits, 79 in geometry, 205, 215, 219 modulo d, 70, 208 negative numbers in, 75 powers in, 76, 77 properties, 72, 73 representatives, 80 table-mod approach, 79, 113 world modulo d, 71 conjecture, xv, 103, 130 Conrey, Brian, 292 constant, 258 contest problems, 145 continued fractions, 219 contradiction, 7, 8, 91, 97–101, 218, 244 coordinate, 130, 210, 220 areal, 140 geometry, 2, 130 system, 2, 10, 180 corollary, xv counterexample, 94, 96, 112, 123, 254 counting addition, 32

317 complement, 33 division, 35 encoding, 29, 40 menu process, 26 multi-stage process, 27, 31 multiplication, 26, 31 overcounting, 33, 35 partition, 32 subtraction, 33 criterion AA for similarity, 15 SAS for similarity, 17, 190 SSS for congruence, 243 cubie, 48 center, 48 corner, 48 edge, 48 cycle canonical notation, 58 disjoint, 57 permutation, 57 structure, 58 cyclic quadrilateral, 12, 13, 17, 19 trapezoid, 239 Cyclicity Theorem, 12, 229 Czech and Slovak National Olympiad problem, 26 solution, 44 Dallas, 306 Davis Math Circle, 290 Davis, Tom, xviii, 47, 292, 293, 304 de Moivre’s formula, 197 de Souza, Paulo, xviii de Vera, Wycee, xviii deductive reasoning, xvi definition, xv, 130 degenerate case, 144 denominator, 18, 219 derivative, 296 DeRose, Tony, xviii Descartes, René, 68 diagonal, xii, 41, 119, 147, 190, 199, 221 diameter, 137, 240 difference, 100, 208, 210, 220 differential geometry, 187, 304 dinosaur, 304 distance, 2, 129, 179, 185, 198 as monovariant, 270 between remainders, 70 formula, 15, 16, 18, 21, 23, 186, 198 formula backwards, 21

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318 inversely proportional, 16 under inversion, 15, 16 distributivity law, 187 dividend, 67 divisible, xii, 44, 104, 112, 208, 210, 220 divisor, 68, 210 common, 104, 116 domino, 95, 211, 215 dragon, 99 dual, 142, 269 game, 278 monovariants, 278 duality, 277 Dudzik, Andrew, xviii, 293 Duluth REU, 295, 307 Dunne, Edward, xviii Earth, 129 East Bay Mathletes, 306 problem, 135 solution, 151 Eastern Europe, x, 286, 287, 291, 297 economics, ix, x, 304 Eisenbud, David, xvii elephant, 3, 129 elliptic curves, x end state, Es , 268 engineering, 297 English, 286, 287, 303 equation, 204 congruence, 72 cubic, 180 linear, 130, 285 quadratic, 2, 180, 188, 286 system, 286 vector, 150 equidistant, 148, 226 equivalence relation, 73 reflexivity, 73 symmetry, 73 transitivity, 73 equivalent, 145 Erdős, Paul, 28, 35 Escape of the Clones, 216 etymology, 38 Euclid, 66, 68, 91, 227 Euclidean algorithm, 75 geometry, 129, 133 Euler’s Theorem, 78 Euler, Leonhard, 114, 119, 306 Europe, 297 example, xiii, 93

INDEX exponent, 51 Exterior Angle Theorem, 227, 230, 237 extra construction, 2, 137, 138, 239 face twists, 49 facelet, 48 factorial, 28 factorization, 104, 113, 114 far away at infinity, 16 Fermat, Pierre de, 65 Fermat’s Little Theorem, 78, 79 Fermat-Wiles Theorem, 64, 66 Ferro, Scipione del, 181 Fibonacci identity, 119 sequence, 117 fine china, 249 finite, 254 finitely many, 91 fixing pointwise, 6 flat state, 272 unique, 272 floor function, x, 158 football, 289 Foundation Merriam-Webster, xvii Mosse, xvii National Science, xviii Packard, xvii Toyota, xvii fraction, xii, 92 frog problem circular, 265 linear, 263 Fuchs, Dmitry, xviii, 292, 293 fulcrum, 129 function, xii, 256, 266 convex, 267 exponential, xii, 112, 280 linear, xii, 280 quadratic, xii trigonometric, xii Fund’l Theorem of Arithmetic, 116 Galileo, xiii, xv game theory, xii, 251 Gates, Bill, 294 Gauss, Carl, 35, 109, 151, 182, 306 general position, 239 genetics, 303 geometry, 203, 286 Germain, Sophie, 66 Givental, Alexander, xviii, 292, 293

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INDEX Goldbach, Christian, 114 golden ratio, φ, 119 gopher, 211 Gopher Gun, 214 graph, 295 cubic, 295 directed, 268 Hamiltonian, 295 of function, xii graph theory, x, 308 Greece, 91 Greek mathematics, 128 group theory, x Gump, Forrest, 304 Haas School of Business, 304 Hadamard, Jacques, 182 Harris, Joe, xvii Hartshorne, Robin, 293 Harvard, 285, 290, 294, 303, 307 Harvard Medical School, 303 Hausner, Melvin, 151 Hayes, David, 306 Heron’s formula, 139, 148 Herriot, Neil, xviii, 293 Hesterberg, Adam, 214 High School Bishop O’Dowd, 306 Chelmsford, 291 Henry Gunn, 292, 293 Newton North, 291 Oakland, xiii, 292, 293, 305 Oakland Technical, 303 Presentation, 293 Skyline, 306 high school geometry, 130, 136 Hippasus, 92 Holtz, Olga, 293 Holy Names College, 305 homology, 303 homomorphism, 187 Hong Kong, 304 Howe, John, 293 Hunts, Julian, 264 hypotenuse, 137, 193, 234, 238 SOS, 191 identity, 196 map, 6 move, 51 if and only if, iff, 97, 228 imbalance, 261, 270

319 IMO, ix, xi, 63, 285, 290, 293–295, 303, 305, 307 problem, 13, 14, 64 solution, 14, 23 incenter, 13, 23, 136, 137, 148 incidence coloring, 295 incircle, 12, 13, 14, 23 Incircle Theorem, 23 increasing, 258 independent proof, 145 India, 297 induct on, 119 inequality, 104, 111, 163, 204, 223 strict, 163 to restrict monovariant, 257 triangle, 18, 261 infinite, 254 series, 216 infinitely long, 16 many, 91 infinity, 255 information theory, 303 inradius, 148 inscribed angles, 7 circle, 19 quadrilateral, 1, 21, 24 rectangles, trapezoids, 23 Inscribed Angle Theorem, 227, 244 integer, xii, 64 Z, 66 consecutive, 168 negative, 67 positive, 64 square of, 64 integral, 296 Intel Science Talent Search, 295, 303 interval, xii of a state, 273 invariant, 95, 96, 98, 100, 101, 206, 252 coloring, 211, 212, 221 summation, 219 inverse, 50 moves, 49 inversion, xii–xiv, 1 center, 4, 5, 8, 11, 19 choose center, 10, 13, 17, 19 choose radius, 10 circle under, 6–8 composition, 5 definition, 4, 8–10, 16 distance under, 15, 16

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320 identity map, 6 image under, 4 injective (one-to-one), 9 line under, 6, 7 parallel lines, 8, 11, 13, 14, 19, 22, 23 properties, 5–9, 11, 12, 15, 16 radius, 4, 10, 12, 15 ray under, 7, 8 ray used in, 4, 5 tangent circles under, 8 inverted circle, 7, 23 configuration, 22, 23 elephant, 3, 6, 8, 15 figure, object, 9, 10, 13 image, 7, 9, 16 line, 7 picture, 10–13, 21, 22 point, 4, 15, 17, 18 Ishikawa, Yuki, xviii Ivanov, Valentin, 114 Japan, 306 Jiangsu Province Math Olympiad, 265 Jobs, Steve, 294 Jurassic Park, 304 Kahan, William, 294, 305 Kazakh State Uniiversity, 306 Kedlaya, Kiran, 293 knight, 99 Knuth, Donald, 292 Kontsevich, Maksim, 216 Lam, Quan, xviii, 103, 292, 304 Lang, Serge, 293 lattice, 117 law, ix Law of Sines, 136, 152, 180 proof, 137 least common multiple, lcm, 59 leg, 193 Lehoczky, Sandor, 307 lemma, xv Lie algebras, 290 lily pads, 263 limits, 223 linear change of variables, 181 function, 201, 280 polynomial, 196 linear algebra, x, 119, 251, 290, 294 eigen-spaces, 303 Jordan form, 119

INDEX matrix diagonalization, 119 real orthogonal matrix, 294, 305 lines, xii, 130 concurrent, 239 parallel, xii, 8, 11, 13, 14, 19, 22, 23, 96, 136, 138, 144, 180, 196, 242, 296 tangent, 234 under inversion, 7 literature, 286 Liu, Tiankai, 65, 293, 295, 304 logic, xv long ¯ l–push, 270 long l–push, 269 Los Angeles Math Circle, xi, 290, 293 lowest terms, 92 MAA, 306 Macau, 304 macro, 54 Madison, Sharon, 293 Malaysia, 304 Manin,Yuri, 88 mansion problem, 251 alternate interpretation, 269 alternative solution, 258 gender balanced, 261 generalized, 275 hybrid, 262 infinite, 255 least upper bound, 268 random, 260 refined, 256 solution, 254 Maple, 114 mass points, xii, 127 assigning masses, 128 associativity, 132, 142, 144, 145 balancing point, 131 closure, 132 coincide, 130 commutativity, 132, 141 definition, 130 distributivity, 132 in space, 140 operations, 131 properties, 132 scalar multiplication, 131 splitting masses, 141, 142, 150, 153 subtraction, 133, 150 sum, 131, 153 transversal, 130, 146, 150, 153 using altitudes, 137 math circle, ix, 289, 298, 304

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INDEX Mathematica, 114, 125 mathematical event, 107 mathematical induction, xiii, xv, 103, 155, 158, 162, 164, 191, 200 basic form, 107 basis case, 107 basis step, 107, 157 choose, 159, 161 double, 160 inductive conclusion, 108 inductive hypothesis, 108 inductive rule, 107 inductive step, 108, 157 strong basis step, 115 strong form, 115, 159, 160 strong induction hypothesis, 115 strong inductive step, 115 mathematical model, 204 mathematical research, x, 294, 295, 307 mathematics education, 297 Matić, Ivan, xvii, xviii, 225, 305 Maydanskiy, Maksim, xviii, 293, 295 McCuan, John, 292 median, 20, 21, 128, 134, 135, 148, 153 Medigovich, Bill, 151 Megginson, Bob, xvii Menelaus’s Theorem, 132, 142, 143, 148 proof, 143, 145 menu process, 256 Mertens Conjecture, 114 Mickey Mouse, 226 middle state, Ms , 271 midpoint, 20, 24, 135, 141, 150, 152, 232 midsegment, 150 Mills College, 307 Mironov, Dmitri, xviii Mississippi formula, 35, 39, 40, 43 MIT, ix, 295, 304 Möbius, Augustus Ferdinand, 151 Möbius strip, 296 modular forms, 65 moduli spaces of curves, 303, 307 monomial, 39 monotone increasing, 249 monotonicity, 267 monovariant, 249, 250, 252, 253, 267 concentration, 253, 268 distance, 270 dual, 277, 278 imbalance, 261, 270 in “practice”, 267 non-numerical, 263 tag teams, 257, 258, 264

321 Monsky’s Theorem, 295 Monthly Contest, x, xviii, 225, 294, 303 problem, 234, 235, 240 solution, 234, 235, 237, 240 Moscow State University, 306 MOSP, 290 MSRI, xvii, 291–293, 304, 306, 307 multinomial, 39 coefficient, 35 multiple, 208 multiplication, 51 music, 286, 305, 308 mutually exclusive cases, 32 n-dragon, 99, 101 National Academy of Sciences, 297 natural sciences, x Ngo, Hoan, 293 Nir, Oaz, 65, 293 Nixon, Richard, 304 non-commutativity, 51, 52 non-decreasing, 258 non-proof, xvii, 88 nonagon, 179 Novosibirsk, 306 number, 95 π, 192, 194 e, 194 i, 183 1, as identity, 51 binary, 29 complex, xii, 2, 92, 182 composite, 116 consecutive, 103, 116 cube of, 104, 108, 123 even, 71, 92, 95, 97, 105, 122 imaginary, 182 integer, 92 irrational, 92, 93, 119 natural, xiii, 104, 204 negative, 75 odd, 64, 71, 94, 97, 103, 105, 122 positive, 70 power of, 104 prime, 63, 91 rational, 92, 93, 100 real, xii, 184 square of, 108, 123 sum of squares, 200 transcendental, 92 whole, 92 number theory, x, 63, 203, 251, 290, 308 NYC Senior ‘A’ Mathletes, 151

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322 NYCML, 145 problem, 146, 148, 149 solution, 148, 149, 152, 153 O’Dorney, Evan, xviii, 268, 294, 305 O’Dorney, Jennifer, xviii O’Hara, Kathy, xvii Oakland/East Bay Math Circles, 293 Odlyzko, Andrew, 114 Olsson, Martin, xviii one-to-one correspondence, 31, 34 order finite, 54 natural, 56 of operation, 51 of permutation, 59 of Rubik’s cube move, 53 orthocenter, 137 Orthocenter Theorem, 137 proof, 138 outreach activities, 290, 291 pairwise disjoint, 32 parallel, 150 line, 170 projection, 144 translate, 183 parallelogram, xii, 19, 135, 138, 151, 184, 190, 199 Parallelogram Theorem, 20, 24 parity, 95, 160, 161, 206 invariant, 211, 212, 215 partial derivatives, 219 partial differential equations, 304, 305 partition, 104, 105, 212 Pascal’s Triangle, 38 pattern, 95 Peano axioms, xiii Peavy, Barbara, xviii Penn State REU, 295 perfect square, 64 permutation, 27, 49, 55 cycle notation, 57 identity, 58 number of, P (n, k), 28 two-row notation, 56 perpendicular, 4, 5, 11, 14, 15, 23, 164, 199, 233, 245 bisector, 14, 136–138 foot of, 4, 193, 240 philosophy, ix, xv physical model, 131 physics, 286

INDEX piano, 285, 303, 305, 307 Pigeonhole Principle, xiii, 165, 254 basic (PHP), 53, 166 generalized (GPHP), 44, 166 pigeonhole, 167 proof, 175 plane, 128 poetry, 286 point, 130 fat, 130 trisection, 135, 139, 148, 150, 153 Pointless Machine, 209 policy analysis, ix polygon, 41, 162, 163, 179, 226 convex, 162 cyclic, 10 hexagon, 162, 164, 173 nonagon, 179 octagon, 164 pentagon, 162, 174 self-intersecting, 162, 164 polyhedron, 128 polynomial, xii, 112 polyomino, 212 Poonen, Bjorn, xviii, 292, 293 poset, 295, 303 postulate, 130 power of a point, 247 successive, 53 Power of a Point Theorem, 22 pre-calculus, xii preservation of angle measure, 2 of shape, 3 prime, xiii, 63, 66, 104, 106, 116 Sophie Germain, 66 prime factorization, 91, 104, 106 existence, 116 uniqueness, 116 Prime Number Theorem, 182 principle of the lever, 128, 151 probability, 89, 286, 304 problem solving techniques (PST), xii, xv, xvi, 9 apply case-chasing, 161 apply linear changes, 181 apply monovariant to terminate, 267 apply repeatedly macro moves, 55 assign correctly 3 masses, 147 assign masses via Law of Sines, 137 assign numbers for an invariant, 217 assign numbers to things, 263

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INDEX assume the claim is false, 93 bound the monovariant, 255 find best bound, 262 with inequality, 257 change a single state, 213 change circles to lines, 10 check small cases, 204 consider parity, 207 count how many, not how large, 44 count the complement, 34 create a weighted measure, 252 create mutually exclusive cases, 32 devise extra constructions, 240 devise two-proof solutions, 97 do not be afraid of intuition, 252 do not depend just on pictures, 228 do not miss the “obvious”, 236 draw approximate figure, 134 draw general type, 10 draw inverted picture away, 10 eliminate an undesirable power, 181 equate real and imaginary parts, 194 exploit similar triangles, 12 find a strategy, 160 find an invariant on a sub-board, 212 fix one variable, vary the other, 119 focus on the end result, 90 force convenient center of mass, 134 force given expressions, 74 generalize the problem, 76 induct on increasing quantity, 161 keep in perspective, 17 kill solutions via inequalities, 204 label concepts, 82 list all configurations, 228 look at smaller simpler cases, 49 look for an invariant, 206 look for sequences of moves, 214 look for sets of objects, 169 make connections, 213 make sure tag teams really work, 259 manipulate algebraically, 16 match data with PHP, 167 move from n to n + 1 in solution, 158 multiply choices for each stage, 27 name unknown quantity, 90 note equal altitudes, 135 notice and prove cyclicity, 226 one direction proves the other, 143 order all objects/people, 30 pair up summands, 81 partition and add outcomes, 32 partition and color the board, 212

323 reason backwards in P1 ⇔ P2 , 237 reduce number of holes, 167 reduce to previously solved, 17 regroup similar terms, 69 relate Pn+1 to Pn , 108 require invariant to change, 95 shortcut traveling proofs, 99 show 4 points are concyclic, 229 show existence, 68 show figures are concurrent, 231 show uniqueness, 69 split into prime and composite, 116 split into simpler moduli, 80 start a PHP problem, 167 substitute remainders, 79 switch from numbers to letters, 189 translate into simpler, 10 try different viewpoints, 138 turn equalities into inequalities, 163 turn negative into positive, 172 use area addition or vectors, 128 use formulas backwards, 21 use M I for finitely many claims, 272 use mod 10 for last digit, 78 use parity, 83 use Pigeonhole Principle, 53 use sandwich to prove A < B, 111 use “small” representatives, 80 verify 1-1 correspondence, 31 walk backwards along staircase, 261 walk on staircase of events, 250 watch out for irrelevant info, 232 product, 203 projective geometry, 304 proof, x, xii, xiii, xiv, xvi, 87, 296 by contradiction, 91, 95, 96, 98 by induction, 155 constructive, 93, 96 of existence, 69, 96 of impossibility, 93, 95, 97 of possibility, 93, 97 of uniqueness, 69 “two-headed”, 162 via example, 93 via invariant, 96 property, xv proposition, xv, 107 PST bag, 9 Ptolemy’s Theorem, 1 generalized, 18 inversion equivalent, 1 proof, 2, 3, 17, 18 Putnam, ix, xi, 293–295, 303, 304

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324 pyramid, 128 Pythagorean society, 92 Pythagorean Theorem, xii, 2, 138, 148, 153, 180, 185, 234 quadratic formula, 181, 182, 197 term, 181 quadrilateral, xii, 135, 139, 147, 162 cyclic, 12, 13, 17, 19, 22–24, 237 diagonals, 1, 17 inscribed, 1, 21, 24 nice (convex), 12 opposite angles, 12 opposite sides, 1, 21, 24 sides, 17 quotient, 67 exists, 69 unique, 69 radian, 192 radical axis, 238 radius, 2, 226, 233 of inversion, 4, 10, 12, 15 Radko, Olga, 293 Ramanujan, Srinivasa, 35, 37, 219 Ramsey’s Theorem, 174 Raphael, Brother, 151 ratio, 92, 128, 138, 140–142, 146, 148 after inversion, 21, 24 in medians, 20, 21 of sides, 12, 17 ray under inversion, 7, 8 used in inversion, 4, 5 real analysis, xii, 290 real number system, xiii rectangle, 14, 23, 98, 188, 190, 199, 241 refine the result, 256 relatively prime, 75, 219 remainder, xii, 67, 208, 220 case-chasing, 78 cycling, 76 exists, 69 of exponent, 77 representatives, 80 table-mod approach, 79, 113 unique, 69 research mathematics, 145 Research Science Institute at MIT, 295 respect an operation, 187, 191, 193, 199 restricted patterns, 307 Riele, Herman, 114

INDEX Riemann, Bernhard, 182 Riemann Hypothesis, 114, 182 Rike, Tom, xiii, xviii, 127, 292, 293, 305 Rome, 304 root, 114, 119 of unity, 180 Rossi, Hugo, xvii, 291 Rousse (
), 286 Routh’s Theorem, 139, 140 Roxbury Latin School, The, 307 Rubik (computer simulation), 47 Rubik’s cube, xii, xiv, 47 Rusczyk, Richard, 307 same points, 130 samovar, 90 San Francisco Math Circle, 293, 308 San Francisco State College, 305 San Jose Math Circle, xi, 264, 293, 298, 304, 306 San Jose State University, 292, 306 sandwich technique, 111 Santa Fe, 306 science, 285, 286, 295–297 Scripps Spelling Bee, xviii, 293, 305 seesaw, 129 balance, 129 fulcrum, 129 principle, 134 segment, 2, 15, 18, 21, 22, 24, 127 line, 8 shortest, 245 tangent, 14 under inversion, 7 semicircle, 245 semiperimeter, 139, 148 sentence, 107 sequence, 218 integer, 303 of problems, 106 recursive, x, 116, 174, 178, 219 Serbia, 305 Serganova, Vera, xviii, 293 series, x, 110 geometric, 218 infinite, 296 set, xii, 27 set theory, x Shapiro, Austin, 293 short s¯–push, 270 short s–push, 268 Shubin, Tatiana, ixn, xviii, 179, 293, 306

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INDEX Silicon Graphics, 304 Singapore, 304 Singer, Michael, xvii Sizemore, Steve, xviii Snow, Marsha, xviii soccer, 286, 308 Sofia University, 307 South Korea, 297 Soviet Union, USSR, xiii, 297, 306 space, 128 Spain, 305 sphere, 5 square, 205, 230, 295 of a number, 252 St. Lawrence University, 308 Stanford, xviii, 290, 292, 294, 304, 307 Stanford Math Circle, 260, 293, 307 Stankova, Zvezdelina, ixn, xviii, 1, 63, 155, 307 stars-and-bars dual, 278 statement, 107 Stomp, 211 strengthen the conclusion, 256 Sturmfels, Bernd, 293 subset, 27 Sudbury Math Circle, 293 sum, 93, 100, 103, 108, 203, 206 summation techniques, 219 symmetry, 5, 14, 19, 23, 205, 215, 278 axis, 241 line, 241 mirror, 205 tag teams, 257, 258, 263, 264 Taiwan, 304 tangency, 149 tangent, 2, 4, 5, 15, 19, 233 circles, 8, 10, 11, 13, 14, 22, 24 circles externally, 14 circles internally, 19, 22 circles pairwise, 2 figures, 9, 13, 14, 19, 23 internal, 19 segments, 14 Tartaglia, Niccolo, 181 technology, 297 telescoping method, 110, 123 Terminator II, 304 tetrahedron, 127, 128, 140 opposite edges, 140 tetromino, 95, 98, 101, 211, 215, 216 Thailand, 304 theorem, xv, 130

325 Theorems bag, 9 tiling, 98, 101, 215, 216, 221 time-reversing, 278 TLC Angle Theorem, 234 top-tier math circle, ix, x, xiv, 294, 298 topology, 187, 290 Towers of Hanoi, 121, 162 transformation, x, 3, 179 dilation (rescaling), 3, 195, 201 inversion, 3 reflection, 3, 186, 195, 198, 201 rotation, 3, 188, 195, 196, 201 translation, 3, 23, 195, 201, 242 transversal, 129, 130, 141, 143, 144 Trapa, Peter, 290 trapezoid, xii cyclic, isosceles, inscribed, 19, 23, 239, 241, 242, 247 trapped blank, 273 triangle, xii, xv, 100, 128, 130, 295 30◦ –60◦ –90◦ , 148 3–4–5, 149 acute, 137 congruent, 16, 140 isosceles, 11, 136, 148, 225, 228 right, 2, 137, 148, 150 similar, 2, 7, 11, 12, 15, 16, 234 Triangle Inequality, 18, 261 triangulation, 295 trigonometry, 2, 136, 137, 193, 225 arctangent, arctan, 200 cosine, 194 sine, 194 tromino, 95, 211, 216 Twin Prime Conjectures, 66 UC Berkeley, xvii, 292, 294, 295, 298, 303–305, 307 UC Davis, 292, 306 UC Santa Barbara, 306 undefined, 5 unicolor configuration, 207 figure, 170n, 173 segment, 171 triangle, 174 University of Chicago, 307 University of San Francisco, 291 unordered, 259 US, xiv, 287, 288, 290, 293, 296, 297 USAMO, xviii, 293, 294, 303, 305, 308 problem, 82 solution, 82

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326

INDEX

Utah Math Circle, 290 Vakil, Ravi, xviii, 293, 294 Vallée-Poussin, Charles, 182 Vandervelde, Sam, 203, 293, 307 Varignon’s Theorem, 135, 140 vector, 128, 185 space, 185 vertex, 17, 20, 21 Washington, D.C., 303, 306 weaken the hypothesis, 256 Wellesley College, 303 Wertheimer, David, xviii Wessel, Caspar, 182 Westlake Junior High School, 305 Whitlow, Marc, ixn, xviii Wiegers, Brandy, 293 Wikipedia, xviii Wiles, Andrew, xvi, 63, 64, 65 winning strategy, 93 Wu, Hung-Hsi, 298 Yale, 303 Yeung, Joyce, xviii Zakharevich, Inna, xviii, 293 Zeitz, Paul, xviii, 25, 291, 293, 308 Zucker, Joshua, xviii, 279, 292, 293 Zuckerberg, Mark, 294

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