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Alexander Soifer

From the Mountains of Colorado to the Peaks of Mathematics

The Colorado Mathematical Olympiad: The Third Decade and Further Explorations

The Colorado Mathematical Olympiad:The Third Decade and Further Explorations

Alexander Soifer

The Colorado MathematicalOlympiad: The Third Decade

and Further Explorations

From the Mountains of Coloradoto the Peaks of Mathematics

Forewords by

Branko GrunbaumPeter D. Johnson, Jr.

Alexander SoiferCollege of Letters, Arts, and SciencesUniversity of Colorado at Colorado Springs1420 Austin Bluffs ParkwayColorado Springs, CO [email protected]

ISBN 978-3-319-52859-5 ISBN 978-3-319-52861-8 (eBook)DOI 10.1007/978-3-319-52861-8

Library of Congress Control Number: 2017934206

Mathematics Subject Classification: 2000

© Alexander Soifer 2017This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part ofthe material is concerned, specifically the rights of translation, reprinting, reuse of illustrations,recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmissionor information storage and retrieval, electronic adaptation, computer software, or by similar ordissimilar methodology now known or hereafter developed.The use of general descriptive names, registered names, trademarks, service marks, etc. in thispublication does not imply, even in the absence of a specific statement, that such names are exemptfrom the relevant protective laws and regulations and therefore free for general use.The publisher, the authors and the editors are safe to assume that the advice and information in thisbook are believed to be true and accurate at the date of publication. Neither the publisher nor theauthors or the editors give a warranty, express or implied, with respect to the material containedherein or for any errors or omissions that may have been made. The publisher remains neutral withregard to jurisdictional claims in published maps and institutional affiliations.

Cover Illustrations: The illustrations on the front cover, from the upper left clockwise, come fromsolutions of problems 28.5; 21.5; Further Exploration E30; and solution of problem 26.3; allpresented in this book.

Printed on acid-free paper

This Springer imprint is published by Springer NatureThe registered company is Springer International Publishing AGThe registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

mailto:10.1007/978-3-319-52861-8

To all those peoplethroughout the world

who create Olympiadsfor new generations of mathematicians.

Forewords to “The Colorado Mathematical

Olympiad, The Third Decade and Further

Explorations: From the Mountains of

Colorado to the Peaks of Mathematics”

Having written enthusiastic forewords to the first and second install-

ments of Alexander Soifer’s series of books about the ColoradoMathematical Olympiads that covered the first two decades of thesetremendously successful and popular events, it is not hard to antici-pate my feelings about the third installment. I need to admit that I amalmost speechless (and will stop speaking when done with this fore-word) facing the ingenuity and inventiveness demonstrated in theproblems proposed in the third decade of these Olympics. However,equally impressive is the drive and persistence of the originator andliving soul of them. It is hard for me to imagine the enthusiasm andcommitment needed to work singlehandedly on such an endeavorover several decades. True, in the more recent past there was helpfrom various quarters (in contrast to the situation Soifer encounteredin some of the past years)—but still the single-minded drive is morethan admirable.

Any mathematician will derive much pleasure reading about theproblems, but so will students (even middle and high school ones)who are interested in challenging puzzles that do not require any“higher mathematics.” As Soifer repeatedly points out, no advancedknowledge is needed to understand the problems, and only

vii

willingness to think logically and with full attention is needed for thesolutions. The full benefit of the Olympiad and the spirit it fosters isevident by the many alumni who went on to become successful inboth mathematics and other fields.

The text in enlivened by Soifer’s comments and anecdotes, whichgreatly enrich the experience provided by the book. Soifer does nothesitate to be explicit about his opinions regarding ethics in mathe-matics (and in general). In particular, he is deeply wounded by thenaming of a medal (for outstanding mathematical research) of theInternational Mathematical Union after Rolf Nevanlinna. DuringWorld War II Nevanlinna lavishly praised Hitler and was active in

recruitment in Finland of the infamous SS troops.The enthusiasm with which students and their teachers endorse

Soifer’s Olympiads points to a regrettable gap in the curricula ofour schools. Painting with a very broad brush, one can say thatthroughout their education the students are not exposed to any kindof visual geometry of the type exemplified in the problems of theOlympiads, nor are they getting any exposure to combinatorial think-ing. Both of these are often helpful in students’ later education,and life.

As of this writing, Soifer continues in organizing the fourth decadeof the Olympics. One can only wish him (and us) many more years ofsuccess.

Branko GrünbaumDepartment of Mathematics

University of Washington, SeattleNovember 24, 2016

viii Foreword

Here is another gem from Alexander Soifer, the third in a series on theColorado Mathematical Olympiad, founded by Professor Soifer in1984 and recreated by him every year since. As in the previous twovolumes, the meat of the feast is in the problems and solutions foreach Olympiad, but these delights are accompanied and seasoned byan account, for each Olympiad, of who won what, which schools thewinners came from, what grade they were in (the Olympiad hascontestants from grades 6 through 12!), who their teachers were,and whether they had achieved distinction in previous Olympiads.Right, all of this is for the record, like a line of hieroglyphics detailingthe accomplishments of a pharaoh, but the record for each year refersto years past, and even if you skim these sections, paying scantattention, the drama builds: Hannah Alpert places third (out of hun-dreds) one year, second the next, and then, in her senior year. . .secondagain. But in that year she and the fellow that placed ahead of her thatyear and the year before were judged to be so many streets ahead ofeverybody else that no third place was awarded—this had neverhappened before and has never happened since. We find that someyears later, after finishing their undergraduate degrees, both contes-tants wound up in the mathematics Ph.D. program at M.I.T. Yes, thatM.I.T.

And then there are the accounts of difficulties in holding theevent—snow storms about every third year, Alex Soifer in and outof the hospital with a kidney stone around the time of the 2006Olympiad. Another strand that I found to be interesting has to dowith Alex’s indefatigable efforts to get politicians to attend the

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Awards Ceremony of the Olympiad. They never come, but many sendletters, every one of which is published in the book—letters from statesenators and the three governors that were in office in Coloradoduring the 10-year span covered by this volume.

Not that the letters are especially interesting to read, but the factthat they were written is of great interest in view of the fact that duringthe first 5 years of the Olympiads, 1984–1988, Professor Soifer had tofight like the dickens to keep the Olympiads alive. A shortsighteddepartment head colluded with misguided administrators at UCCS toarrive at the conclusion that holding an annual math test for grades 6–12 was an unworthy pursuit for a UCCS professor, just not worth the

trouble and expense. It was touch-and-go for a while, but eventuallyAlex was not fired and the Olympiads were not discontinued, thanksto Professor Soifer’s political agility and the presence of UCCSadministrators of a different breed from the ones threatening Olym-piad destruction. What I see in Alex’s pursuit of the attention ofpoliticians and journalists, and his policy of prominently displayingfavorable newspaper coverage and letters from eminent personswhenever possible, is a shrewd and skillful cementing of the statusof the Colorado Olympiad so that it will never be threatened withabolition again, at least while Alexander Soifer is alive andfunctioning.

One consequence of Professor Soifer’s success in this regard is thatthe historical narrative that accompanies the problems and solutionsin this account of the third decade of the Olympiad cannot possibly beas exciting as the corresponding narrative in the accounts of the first20 Olympiads, although some of us still find it quite interesting. But,in compensation, I must tell you that the other two constituents of thebook—the problems and solutions, and the “further explorations”—have, in my opinion, struck a new level.

Alexander Soifer grew up in the Soviet Union where, as in much ofEastern Europe, mathematical exam competitions for secondaryschool students are common, and where the ability to make upsuitable problems for such exams is a prized talent. Alex himselfcredits participation in such competitions for awakening his interestin mathematics, and he has always been an advocate of introducingstudents to mathematics by posing problems—interesting problems,appropriate to their level. If I am not mistaken, his first book wasentitled Mathematics as Problem Solving.

x Foreword

I am of the opinion that if you work at something a long time—say,20 years or more—and really try to get better at it, whatever it is, thenyou will get better at it, even if you were pretty good at it to start with.As I peruse the problems of the third decade of the Colorado Olym-piad, I think I see that this has happened to Alex Soifer. He hasalways, in adulthood, had a great nose for great problems, and hehas always been good at making up problems himself, but now, afterdecades of hunting for Olympiad problems, and struggling to createOlympiad problems, he has become an extraordinary connoisseur andcreator of Olympiad problems. The Olympiad problems were verygood, from the beginning, but in the third decade the problems have

become extraordinarily good. Every brace of five problems is a workof art. The harder individual problems range in quality from brilliantto work-of-genius. I wish that I could find time in my busy schedule towallow in this book for a couple of months. Every time I pick aproblem at random and give it some time—at least half an hour, say—I feel myself getting smarter.

And the same goes for the “Further Explorations” part of the book.Great mathematics and mathematical questions are immersed in asauce of fascinating anecdote and reminiscence. If you could haveonly one book to enjoy while stranded on a desert island, this wouldbe a good choice. If you know a teenager who is even mildlyinterested in mathematics and you wanted to give the teenager a giftof one book, this would be a good choice.

Peter D. Johnson, Jr.Department of Mathematics and Statistics

Auburn UniversityDecember 5, 2016

Foreword xi

The 2011 Forewords for “The Colorado

Mathematical Olympiad and Further

Explorations: From the Mountains of

Colorado to the Peaks of Mathematics”

We live in an age of extreme specialization—in mathematics as well

as in all other sciences, in engineering, in medicine. Hence, to say thatprobably 90% of mathematicians cannot understand 90% of mathe-matics currently published is, most likely, too optimistic. In contrast,even a pessimist would have to agree that at least 90% of the materialin this book is readily accessible to, and understandable by, 90% ofstudents in middle and high schools. However, this does not mean thatthe topics are trivial—they are elementary in the sense that they donot require knowledge of lots of previously studied material, but aresophisticated in requiring attention, concentration, and thinking thatis not fettered by preconceptions. The organization in groups of fiveproblems for each of the “Olympiads,” for which the participantswere allowed four hours, hints at the difficulty of finding completesolutions. I am convinced that most professional mathematicianswould be hard pressed to solve a set of five problems in two hours,or even four.

There are many collections of problems, for “Olympiads” of var-ious levels, as well as problems in a variety of journals. What sets thisbook apart from the “competition” are several aspects that deserve tobe noted.

xiii

• The serenity and enthusiasm with which the problems, and theirsolutions, are presented;

• The absence of prerequisites for understanding the problems andtheir solutions;

• The mixture of geometric and combinatorial ideas that are requiredin almost all cases.

The detailed exposition of the trials and tribulations endured by theauthor, as well as the support he received, shed light on the variety ofinfluences which the administration of a university exerts on thefaculty. As some of the negative actions are very probably a conse-quence of mathophobia, the spirit of this book may cure at least a few

present or future deciders from that affliction.Many mathematicians are certainly able to come up with an inter-

esting elementary problem. But Soifer may be unique in his persis-tence, over the decades, of inventing worthwhile problems, andproviding amusing historical and other comments, all accessible tothe intended pre-college students.

It is my fervent hope that this book will find the wide readership itdeserves, and that its readers will feel motivated to look for enjoy-ment in mathematics.

Branko GrünbaumDepartment of Mathematics

University of Washington, Seattle

xiv Foreword

Here is another wonderful book from Alexander Soifer. This one is amore-than-doubling of an earlier book on the first 10 years of theColorado Mathematical Olympiad, which was founded and nourishedto robust young adulthood by Alexander Soifer.

Like The Mathematical Coloring Book, this book is not so muchmathematical literature as it is literature built around mathematics, ifyou will permit the distinction. Yes, there is plenty of mathematicshere, and of the most delicious kind. In case you were unaware of, orhad forgotten (as I had), the level of skill, nay, art, necessary to posegood olympiad—or Putnam exam-style problems, or the effect thatsuch a problem can have on a young mind, and even on the thoughtsof a jaded sophisticate, then what you have been missing can be foundhere in plenty—at least a year’s supply of great intellectual gustation.If you are a mathematics educator looking for activities for a math

club—your search is over! And with the Further Explorations sec-tions, anyone so inclined could spend a lifetime on the mathematicssprouting from this volume.

But since there will be no shortage of praise for the mathematicaland pedagogical contributions of From the Mountains of Colorado. . ., let me leave that aspect of the work and supply a few words aboutthe historical account that surrounds and binds the mathematicaltrove, and makes a story of it all. The Historical Notes read like awar diary, or an explorer’s letters home: there is a pleasant, mundanerhythm of reportage—who and how many showed up from where,who the sponsors were, which luminaries visited, who won, whattheir prizes were—punctuated by turbulence, events ranging from

xv

A. Soifer’s scolding of a local newspaper for printing only the namesof the top winners, to the difficulties arising from the weather and theshootings at Columbine High School in 1999 (both matters of life anddeath in Colorado), to the inexplicable attempts of university admin-istrators to impede, restructure, banish, or destroy the ColoradoMathematical Olympiad, in 1985, 1986, 2001, and 2003. It is fasci-nating stuff. The very few who have the entrepreneurial spirit toattempt the creation of anything like an Olympiad will be forewarnedand inspired.

The rest of us will be pleasurably horrified and amazed, oursympathies stimulated and our support aroused for the brave ones

who bring new life to the communication of mathematics.

Peter D. Johnson, Jr.Department of Mathematics and Statistics

Auburn University

xvi Foreword

In the common understanding of things, mathematics is dispassionate.This unfortunate notion is reinforced by modern mathematical prose,which gets good marks for logic and poor ones for engagement. Butthe mystery and excitement of mathematical discovery cannot bedenied. These qualities overflow all preset boundaries.

On July 10, 1796, Gauss wrote in his diary

EϒΡΗKΑ! num ¼ Δþ Δþ Δ

He had discovered a proof that every positive integer is the sum ofthree triangular numbers 0, 1, 3, 6, 10, . . . n nþ 1ð Þ=2, . . .f g. Thisresult was something special. It was right to celebrate the momentwith an exclamation of Eureka!

In 1926, an intriguing conjecture was making the rounds ofEuropean universities.

If the set of positive integers is partitioned into two classes, thenat least one of the classes contains an n-term arithmetic progres-sion, no matter how large n is taken to be.

The conjecture had been formulated by the Dutch mathematicianP. J. H. Baudet, who told it to his friend and mentor Frederik Schuh.B. L. van der Waerden learned the problem in Schuh’s seminar at theUniversity of Amsterdam. While in Hamburg, van der Waerden toldthe conjecture to Emil Artin and Otto Schreier as the three hadlunch. After lunch, they adjourned to Artin’s office at the Universityof Hamburg to try to find a proof. They were successful, and the

xvii

result, now known as van der Waerden’s Theorem, is one of the ThreePearls of Number Theory in Khinchine’s book by that name.The story does not end there. In 1971, van der Waerden published aremarkable paper entitled How the Proof of Baudet’s Conjecture wasFound.1 In it, he describes how the three mathematicians searched fora proof by drawing diagrams on the blackboard to represent theclasses, and how each mathematician had Einf€alle (sudden ideas)that were crucial to the proof. In this account, the reader is a fourthperson in Artin’s office, observing with each Einfall the rising antic-ipation that the proof is going to work. Even though unspoken, eachof the three must have had a “Eureka moment” when success was

assured.From Colorado Mountains to the Peaks of Mathematics presents

the 20-year history of the Colorado Mathematical Olympiad. It issymbolic that this Olympiad is held in Colorado. Colorado is knownfor its beauty and spaciousness. In the book there is plenty of spacefor mathematics. There are wonderful problems with ingenious solu-tions, taken from geometry, combinatorics, number theory, and otherareas. But there is much more. There is space to meet the participants,hear their candid comments, learn of their talents, mathematical andotherwise, and in some cases to follow their paths as professionals.There is space for poetry and references to the arts. There is space fora full story of the competition—its dreams and rewards, hard workand conflict. There is space for the author to comment on matters ofgeneral concern. One such comment expresses regret at the limita-tions of currently accepted mathematical prose.

In my historical-mathematical research for The Mathematical Col-oring Book, I read a good number of nineteenth-century Victorianmathematical papers. Clearly, the precision and rigor of mathemat-ical prose has improved since then, but something charming waslost—perhaps, we lost the “taste of time” in our demand for an“objective,” impersonal writing, enforced by journal editors andmany publishers. I decided to give a historical taste to my Olym-pians, and show them that behind Victorian clothing we can findthe pumping heart of the Olympiad spirit. [p. 297]

1Studies in Pure Mathematics (Presented to Richard Rado), L. Mirsky, ed., Academic Press,

London, 1971, pp 251–260.

xviii Foreword

Like Gauss, Alexander Soifer would not hesitate to inject Eureka!at the right moment. Like van der Waerden, he can transform adispassionate exercise in logic into a compelling account of suddeninsights and ultimate triumph.

Cecil RousseauProfessor Emeritus

University of MemphisChair, USA Mathematical Olympiad Committee

Foreword xix

The 1994 Forewords for “Colorado

Mathematical Olympiad: The First

10 Years and Further Explorations”

Love! Passion! Intrigue! Suspense! Who would believe that the

history of a mathematics competition could accurately be describedby words that more typically appear on the back of a popular novel?After all, mathematics is dull; history is dull; school is dull. Isn’t thatthe conventional wisdom?

In describing the history of the Colorado Mathematical Olympiad,Alexander Soifer records the comments of a mathematics teacher whoanonymously supported the Olympiad in each of its first 10 years.When asked why, this unselfish teacher responded “I love my profes-sion. This is my way to give something back to it.” Alexander alsoloves his profession. He is passionate about his profession. And heworks hard to give something back.

The Colorado Mathematical Olympiad is just one way Alexanderdemonstrates his love for mathematics, his love for teaching, his lovefor passing on the incredible joy of discovery. And as you read thehistory of the Olympiad, you cannot help but be taken up yourselfwith his passion.

But where there is passion, there is frequently intrigue. Here itinvolves the efforts of school administrators and others to help—or tohinder—the success of the Olympiad. But Alexander acknowledges

xxi

that we all must have many friends to help us on the journey tosuccess. And the Olympiad has had many friends, as Alexander socarefully and thankfully records.

One of the great results of the Olympiad is the demonstration thatreal mathematics can be exciting and suspenseful. But the Olympiadalso demonstrates the essence of mathematical research, or whatmathematicians really do as they move from problem to example togeneralization to deeper results to new problem. And in doing so itprovides an invaluable lesson to the hundreds of students who partic-ipate each year.

It is appropriate on an anniversary to look back and take stock. It is

also appropriate to look forward. This book does both, for the Colo-rado Mathematical Olympiad is alive and well, thanks to its manyardent supporters. And for that we can all rejoice.

Philip L. EngelPresident, CNA Insurance Companies

Chairman of the Board, MATHCOUNTS FoundationMarch 28, 1994, Chicago, Illinois

xxii Foreword

The author started the Colorado Olympiad in 1984, 10 years ago, andit was a complete success and it is continuing. Several of the winnershave already got their PhDs in Mathematics and Computer Science.

The problems are discussed with their solutions in great detail.A delightful feature of the book is that in the second part more relatedproblems are discussed. Some of them are still unsolved; e.g., theproblem of the chromatic number of the plane—two points of theplane are joined if their distance is 1—what is the chromatic numberof this graph? It is known that it is between 4 and 7. I would guess thatit is greater than 4 but I have no further guess. Just today (March8, 1994) Moshe Rosenfeld asked me—join two points of the plane iftheir distance is an odd integer—is the chromatic number of thisgraph finite? He proved that if four points are given, the distancescannot all be odd integers.

The author states an unsolved problem of his and offers a prize of$100 for it. For a convex figure F in the plane, S(F) denotes theminimal positive integer n, such that among any n points inside or onthe boundary of F there are three points that form a triangle of area14Fj j or less, where |F| is the area of F. Since for any convex figure F,

S(F)¼ 5 or 6, it is natural to ask for a classification of all convexfigures F, such that S(F)¼ 6.

xxiii

I warmly recommend this book to all who are interested in difficultelementary problems.

Paul ErdosMember of the Hungarian Academy of Sciences

Honorary Member of the NationalAcademy of Sciences of the USA

Boca Raton, FloridaMarch 8, 1994

xxiv Foreword

Alexander Soifer, who founded and still runs the famous ColoradoMathematics Olympiad, is one of the world’s top creators of signif-icant problems and conjectures. His latest book covers the Olym-piad’s first 10 years, followed by additional questions that flow fromOlympiad problems.

The book is a gold mine of brilliant reasoning with special empha-sis on the power and beauty of coloring proofs. Stronglyrecommended to both serious and recreational mathematicians onall levels of expertise.

Martin GardnerHendersonville, North Carolina

March 10, 1994

xxv

Many of us wish we could contribute to making mathematics moreattractive and interesting to young people. But few among profes-sional mathematicians find the time and energy to actually do much inthis direction. Even fewer are enterprising enough to start acompletely new project and continue carrying it out for many years,making it succeed against overwhelming odds. This book is anaccount of such a rare endeavor. It details one person’s single-mindedand unwavering effort to organize a mathematics contest meant forand accessible to high school students. Professor Soifer managed tosecure the help of many individuals and organizations; surprisingly,he also had to overcome serious difficulties which should not havebeen expected and which should not have arisen.

The book is interesting in many ways. It presents the history of thestruggle to organize the yearly “Colorado Mathematical Olympiad”;

this should help others who are thinking of organizing similar pro-jects. It details many attractive mathematical questions, of varyingdegrees of difficulty, together with the background for many of themand with well-explained solutions, in a manner that students as well asthose who try to coach them will find helpful. Finally, the “FurtherExplorations” make it clear to the reader that each of these ques-tions—like all of mathematics—can be used as a stepping stone toother investigations and insights.

I finished reading the book in one sitting—I just could not put itdown. Professor Soifer has indebted us all by first making the effort

xxvii

to organize the Colorado Mathematical Olympiads, and then makingthe additional effort to tell us about it in such an engaging anduseful way.

Branko GrünbaumUniversity of Washington

Seattle, WashingtonMarch 18, 1994

xxviii Foreword

If one wants to learn about the problems given at the 1981 Interna-tional Mathematical Olympiad, or to find a statistical summary of theresults of that competition, the required information is contained inMurray Klamkin’s book International Mathematical Olympiads1979–1985. To find out about the members of the 1981 USA team(Benjamin Fisher, David Yuen, Gregg Patruno, Noam Elkies, JeremyPrimer, Richard Stong, James Roche, and Brian Hunt) and what theyhave accomplished in the intervening years, one can read the bookletWho’s Who of U.S.A. Mathematical Olympiad Participants 1972–1986 by Nura Turner. For a view from behind the scenes at the 1981IMO, there is the interesting article by Al Willcox, “Inside the IMO,”in the September–October, 1981, issue of Focus. News accounts ofthe IMO can be found in the Time and Newsweek as well as majornewspapers. However, even if one is willing to seek out these various

sources, it is hard to get a full picture of such a MathematicalOlympiad, for it is much more than a collection of problems and astatistical summary of results. Its full story must be told in terms ofdreams, conflicts, frustration, celebration, and joy.

Now thanks to Alexander Soifer, there is a book about the ColoradoMathematical Olympiad that gives more than just the problems, theirsolutions, and statistical information about the results of the compe-tition. It tells the story of this competition in direct, human terms.Beginning with Soifer’s own experience as a student in Moscow,Colorado Mathematical Olympiad describes the genesis of the math-ematical competition he has created and gives a picture of the workrequired to gain support for such a project. It mentions participants byname and tells of some of their accomplishments. It acknowledges

xxix

those who have contributed problems and it reveals interestingconnections between the contest problems and mathematicalresearch. Of course, it has a collection of mathematical problemsand solutions, very beautiful ones. Some of the problems are fromthe mathematical folklore, while others are striking original contri-butions of Soifer and some of his colleagues. Here’s one of myfavorites, a problem contributed by Paul Zeitz.

Twenty-three people of positive integral weight decide to playfootball. They select one person as referee and then split up intotwo 11-person teams of equal total weight. It turns out that nomatter who is chosen referee this can always be done. Prove that all

23 people have the same weight.

The problems alone would make this book rewarding to read. ButColorado Mathematical Olympiad has more than attractive mathe-matical problems. It has a compelling story involving the lives ofthose who have been part of this competition.

Cecil RousseauMemphis State University

Coach, U.S.A. team for the InternationalMathematical Olympiad

Memphis, TennesseeApril 1, 1994

xxx Foreword

Contents

Forewords to “The Colorado Mathematical Olympiad,The Third Decade and Further Explorations: From theMountains of Colorado to the Peaks of Mathematics” . vii

Foreword by Branko Grünbaum . . . . . . . . . . . . viiForeword by Peter D. Johnson, Jr. . . . . . . . . . . . ix

The 2011 Forewords for “The Colorado MathematicalOlympiad and Further Explorations: From theMountains of Colorado to the Peaks of Mathematics” . xiii

Foreword by Branko Grünbaum . . . . . . . . . . . . xiiiForeword by Peter D. Johnson, Jr. . . . . . . . . . . . xvForeword by Cecil Rousseau . . . . . . . . . . . . . . xvii

The 1994 Forewords for “Colorado MathematicalOlympiad: The First 10 Years and FurtherExplorations” . . . . . . . . . . . . . . . . . . . . . . . . . xxi

Foreword by Philip L. Engel . . . . . . . . . . . . . . xxiForeword by Paul Erdos . . . . . . . . . . . . . . . . . xxiiiForeword by Martin Gardner . . . . . . . . . . . . . . xxvForeword by Branko Grünbaum . . . . . . . . . . . . xxviiForeword by Cecil Rousseau . . . . . . . . . . . . . . xxix

xxxi

Greetings to the Reader—2017 . . . . . . . . . . . . . . . xxxvii

Greetings to the Reader—2011 . . . . . . . . . . . . . . . xliii

Greetings to the Reader—1994 . . . . . . . . . . . . . . . xlix

Part I. The Third Decade

Twenty-First Colorado Mathematical Olympiad

April 16, 2004 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Historical Notes 21 . . . . . . . . . . . . . . . . . . . . . . 3Problems 21 . . . . . . . . . . . . . . . . . . . . . . . . . . 8Solutions 21 . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Twenty-Second Colorado Mathematical Olympiad


Twenty-Third Colorado Mathematical Olympiad


Twenty-Fourth Colorado Mathematical Olympiad


Twenty-Fifth Colorado Mathematical Olympiad

April 18, 2008 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71Historical Notes 25 . . . . . . . . . . . . . . . . . . . . . . . 71Problems 25 . . . . . . . . . . . . . . . . . . . . . . . . . . 78Solutions 25 . . . . . . . . . . . . . . . . . . . . . . . . . . 79

xxxii Contents

Twenty-Sixth Colorado Mathematical Olympiad


Twenty-Seventh Colorado Mathematical Olympiad


Twenty-Eighth Colorado Mathematical Olympiad

April 22, 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111Historical Notes 28 . . . . . . . . . . . . . . . . . . . . . . . 111Is Mathematics an Art? . . . . . . . . . . . . . . . . . . . . 115Problems 28 . . . . . . . . . . . . . . . . . . . . . . . . . . 117Solutions 28 . . . . . . . . . . . . . . . . . . . . . . . . . . 117

Twenty-Ninth Colorado Mathematical Olympiad


Thirtieth Colorado Mathematical Olympiad

April 26, 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135Historical Notes 30 . . . . . . . . . . . . . . . . . . . . . . . 135

Problems 30 . . . . . . . . . . . . . . . . . . . . . . . . . . 143Solutions 30 . . . . . . . . . . . . . . . . . . . . . . . . . . 144

A Round Table Discussion of the Olympiad, or LookingBack from a 30-Year Perspective . . . . . . . . . . . . . 153

Contents xxxiii

Part II. Further Explorations of the Third Decade

Introduction to Part II . . . . . . . . . . . . . . . . . . . . . . 169

E21. Cover-Up with John Conway, Mitya Karabash,and Ron Graham . . . . . . . . . . . . . . . . . . . . . . 171

E22. Deep Roots of Uniqueness . . . . . . . . . . . . . . . . . 183

E23: More about Love and Death . . . . . . . . . . . . . . . . 185

E24: One Amazing Problem and Its Connectionsto Everything—A Conversationin Three Movements . . . . . . . . . . . . . . . . . . . . 189

E25: The Story of One Old Erdos Problem . . . . . . . . . . 199

E26: Mark Heim’s Proof . . . . . . . . . . . . . . . . . . . . . 203

E27: Coloring Integers—Entertainmentof Mathematical Kind . . . . . . . . . . . . . . . . . . . . 205

E28: The Erdos Number and Hamiltonian Mysteries . . . . 213

E29: One Old Erdos–Turan Problem . . . . . . . . . . . . . . 217

E30: Birth of a Problem: The Story of Creationin Seven Stages . . . . . . . . . . . . . . . . . . . . . . . . 221

Part III. Olympic Reminiscences in Four Movements

Movement 1. The Colorado Mathematical OlympiadIs Mathematics; It Is Sport; It Is Art.And It Is Also Community,by Matthew Kahle . . . . . . . . . . . . . . . . . 231

Movement 2. I’ve Begun Paying Off My Debtwith New Kids, by Aaron Parsons . . . . . . . 235

Movement 3: Aesthetic of Personal Mastery,by Hannah Alpert . . . . . . . . . . . . . . . . . 239

xxxiv Contents

Movement 4. Colorado Mathematical Olympiad:Reminiscences by Robert Ewell . . . . . . . . . 243

Farewell to the Reader . . . . . . . . . . . . . . . . . . . . . . 249

References . . . . . . . . . . . . . . . . . . . . . . . . . . . 253

Index of Names . . . . . . . . . . . . . . . . . . . . . . . . . . . 257

Index of Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . 259

Contents xxxv

Greetings to the Reader—2017

As far as the laws of mathematics refer to reality,they are not certain, and as far as they are certain,they do not refer to reality.

—Albert Einstein

The world is a dangerous place to live; notbecause of the people who are evil, but becauseof the people who don’t do anything about it.

—Albert Einstein

The Colorado Mathematical Olympiad has survived for over30 years—long live the Olympiad! It has become a part of our culturallife in Colorado. During the first 30 years, 18,000 students partici-pated in the Olympiad. They wrote 89,000 essays and were awardedover $317,000 in prizes. The Olympiad is a unique joint effort ofschool districts, schools, institutions of higher education, businesscommunity, and local and state governments.

What new can I say here after I gave a lot of thought to the previoustwo 2011 and 1994 forewords reproduced below? The author—I—changed over the past 30 years, as did my views of priorities in life,and consequently my priorities in science and education.

Most of my life I have been a student of Beauty in all her mani-festations, especially nature and the arts. Mathematics is one of the

xxxvii

arts for me. However, my 20-year long 1995–2015 work on the bookThe Scholar and the State: In Search of Van der Warden [Soi10] haschanged me. I now view the highest priority of instruction andresearch in upholding high moral principles. This is important!Many of my colleagues believe that mathematics we create is allthat matters, Mathematik €uber Alles, mathematics above all moralconcerns. In my opinion, there is no good science or good art unless itis built on the foundation of high ethical principles. Often publishersof mathematical books and journals downplay and sometimes evendisregard ethical principles. This is so shortsighted! We have seen inhistory time and again how evil the usage of science could be if it is

not built on high moral foundation. Atrocities of Nazi Germany aloneprovide countless examples of how science, technology, and even artcan be used for ill deeds. Mentioned above my book [Soi10] isdedicated precisely to moral dilemmas of a scholar in the ThirdReich and in the world of today, the world in which Russia invadesand wages wars against its sovereign neighbors Moldova, Georgia,and Ukraine. In order for creative work to be good, it must also servethe good. It ought to be humane. It has to be grounded in morality,empathy, compassion, and kindness. The Great Russian poet Alexan-der Pushkin (1799–1837) poetically expressed this. Let me translatehis lines for you:

And people will be pleased with me for years to come,For I awakened kindness with my lyre,For in my cruel age I Freedom praised and sangAnd urged I mercy for the fallen people.

Those fluent in Russian will appreciate the original verses:

И дoлгo буду тeм любeзeн я нapoду,Чтo чувcтвa дoбpыe я лиpoй пpoбуждaл,Чтo в мoй жecтoкий вeк вoccлaвил я CвoбoдуИ милocть к пaдшим пpизывaл.

I hope that you, my young colleagues, would accept the baton ofmercy and humanity and by your creative work will contribute to thehigh culture of our small endangered planet.

To my disbelief, I encountered not only enthusiastic support of myaddressing ethics of a scholar, but also rare yet vigorous opposition.Are mathematicians entitled to a life in Ivory Tower and disregard the

xxxviii Preface

outside world? This must have been a reason the Executive Commit-tee of the International Mathematics Union (IMU) established in 1981The Rolf Nevanlinna Prize that includes a gold medal with the profileof this fine mathematician, who during the World War II served as theChair of Recruitment Committee of the SS troops in Finland, whopraised Hitler (!) as liberator of Europe. No doubt you know that theSS troops were responsible for the majority of crimes against human-ity as was determined in Nuremberg Trials and other post-World WarII tribunals.

I urged the name change of this prize and the medal in my book[Soi10], in my July 23, 2016, public address to the General Assembly

of the International Commission on Mathematics Instruction (ICMI),and in my letter to the Executive Committee of IMU written on behalfof the Executive Committee of the World Federation of NationalMathematics Competitions (of which I am President). IMU PresidentShigefumi Moro promised me to consider this request at the nextmeeting of the IMU Executive Committee in April 2017. They maydecide not to change the name, but we have done all we can, and thatcounts for something. The 1986 Nobel Peace Laureate Elie Wieselcalled upon us to stand up and be counted:

There may be times when we are powerless to prevent injustice, butthere must never be a time when we fail to protest.

I agree with the great Dutch mathematician Luitzen Egbertus JanBrouwer, who wrote:

It is my opinion that the tiniest moral matter is more important thanall of science, and that one can only maintain the moral quality ofthe world by standing up to any immoral project.

If we, mathematicians of today, took a greater care of our ethics,we would not have lost such a genius mathematician as GrigoryPerelman. Following his great achievements of proving the PoincareConjecture and the Geometrization Conjecture, Grigory walked awayfrom mathematics. He did not want to be “a poster boy” for thescience where the majority tolerates immoral acts of the minority ofits members.

I have always believed that the main goal of mathematics instruc-tion ought to be demonstrating what mathematics is and what math-ematicians do. Technically speaking, we cannot teach science—any

Preface xxxix

science, and any art. We can only create atmosphere for our studentswhere they learn our subject by doing it. As Einstein said,

The only source of knowledge is experience.

Unlike almost any other competition of mathematical kind, ourOlympiad builds bridges between problems of Olympiads and prob-lems of “real” mathematics. These bridges show that the differencebetween the two is quantitative and not qualitative. The same goals ofbeauty, elegance, and surprises for our intuition guide both. And thedifference is mere quantitative: our Olympiad’s problems 3, 4, and5 may require 4–5 hours to solve whereas problems of “real” math-

ematics may require 5000 hours, and sometimes more. As a conse-quence, all my Colorado Mathematical Olympiad books, includingthe present one, feature the bridges between Olympiad problems andresearch mathematics, which I call Further Explorations. This bookcontains ten such Explorations, essays that vividly demonstrate thisconnection, the two-way bridge between the problems we use in ourOlympiad and problems mathematicians ponder in their research,including open problems of mathematics. Yes, two-way, for someof the Olympiad problems pave the way for new mathematicalresearch.

Olympiads inevitably have an element of sport. I try to reduce it byoffering five problems of increasing difficulty that require practicallyno topical knowledge, and 4 hours to solve them and write completeessay-type solutions. The influence of Russian Olympiads of all levelscan easily be noticed. However, there is one more essential element inthe Colorado Mathematical Olympiad. I wish to allow young Olym-pians to compete not only with each other but also with the field ofMathematics. I wish to stop discrimination of high school mathema-ticians based on their tender age, and inspire them to engage inresearch. To this end, our more difficult problems often open horizonsand allow for further explorations, leading young Olympians to theforefront of mathematics.

Authors of the problems are listed next to the problems’ title.However, what could I do when I slightly changed a problem thatcame to me from the (Russian) mathematical folklore? Everyone isthe author, and no one is at the same time. In those cases I did not listany authorship. Every year, the problems have been selected andedited by the Problem Committee, which included Col. Dr. Robert

xl Preface

“Bob” Ewell and me. During some years, the Committee alsoincluded Gary Miller.

I thank the historic Springer for inviting this book—my eighthbook with Springer—in its historic publishing house, founded in1842. No words can fully express the depth of my gratitude to BrankoGrünbaum and Peter D. Johnson, Jr., who were the first readers of thisand all my previous books. Their forewords and suggestions havealways been grounded in their great intellectual capacity, mathe-matical brilliance, perfect aesthetic taste, and unwavering moralcompass.

I thank all the numerous people who joined me in this major

undertaking and made the Olympiad possible, judges, proctors, staffmembers of the University of Colorado Colorado Springs, sponsors,and above all Olympians and their families. A special gratitude goesto Bob Ewell, an Olympiad’s senior judge for a quarter a century, whotranslated my sketches into computer-aided illustrations you see inthis book.

I thank all my mathematics teachers from the grade school to mygrown-up stage in life: Matilda I. Koroleva, Klara A. Dimonstein,Nikolai N. Konstantinov, Tatiana N. Fideli, Ivan V. Morozkin, YuriF. Mett, Leonid Ya. Kulikov, and Paul Erdos. I am grateful to myparents, artist Yuri A. Soifer and actress Frieda M. Hoffman Soifer,for moral guidance and intellectual environment they provided.I thank my kids—Mark, Julia, Isabelle, and Leon—for support andinspiration. Once again, I dedicate my Olympiad book to all thosepeople around the world who create mathematical Olympiads for newgenerations of mathematicians.

At the Colorado Mathematical Olympiad, we have been oftenasked a natural question: how does one create a Mathematical Olym-piad and how does it work? This and other related questions areclarified by the University of Colorado, which produced the 2014film “Thirtieth Colorado Mathematical Olympiad—30 Years ofExcellence.” You can find it on the Olympiad’s homepage http://olympiad.uccs.edu/ and on the YouTube.

I started this foreword with two epigraphs from Albert Einstein. Incase you do not see their relevance to this book, let me clarify. Thefirst one, in a sense, implies that mathematics is not a natural science,it is an art. The second quotation calls on us all to stand up and becounted in the pursuit of high professional ethics. I concur.

Preface xli

http://olympiad.uccs.edu/

http://olympiad.uccs.edu/


Beauty is an instance which plainly shows thatculture is not simply utilitarian in its aims, for thelack of beauty is a thing we cannot tolerate incivilization.

—Sigmund Feud, 1930Civilization and Its Discontents [F]

Talent is not performance; arms and legs are nodance.

—Hugo von Hofmannsthal, 1922Buch der Freunde [Ho]

Imagination is more important than knowledge.For knowledge is limited to all we now know andunderstand, while imagination embraces theentire world, and all there ever will be to knowand understand.

—Albert Einstein

I have been often asked: what are Mathematical Olympiads for? Dothey predict who will go far in mathematics and who will not? In fact,today, in 2010, in Russia, Olympiads are officially used as predictorsof success, for those who placed high enough in important enough

xliii

Olympiads get admitted to important enough universities! Havingspent a lifetime in Olympiads of all levels, from participant to orga-nizer, I have certainly given these kinds of questions a lot of thought.

There is no way to do well by accident, by luck in an Olympiad(i.e., essay-type competition, requiring presentation of complete solu-tions). Therefore, those who even sometimes have done well inOlympiads undoubtedly have talent. Does it mean they will succeedin mathematics? As the great Austrian writer Hugo von Hofmannsthalput it in 1922, Talent is not performance; arms and legs are no dance.With talent, it still takes work, hard work to succeed. I would saytalent imposes an obligation on its owner, a duty not to waste the

talent. I must add, nothing is a guarantee of success; life interferes andthrows barriers on our way. It is critical to know in your gut that inorder to succeed no reason for failure should be acceptable.

The inverse here is not true. Those who have not done well inOlympiads do not necessarily lack talent. They may be late bloo-mers—Einstein comes to mind. Their talent may be in another field oractivity. I believe there are no talentless people—there are peoplewho have not identified their talent and have not developed it.

In addition to allowing talents shine, Olympiads introduce young-sters to the kind of mathematics they may have not seen in school. Itwas certainly a case with me. Olympiads showed me the existence ofmathematics that justifies such adjectives as beautiful, elegant,humorous, defying intuition. Olympiad mathematics inspires andrecruits; it passes the baton to new generation of mathematicians.

And one more thing, which apparently I said long ago (see mynewspaper interview in Historical Notes 14). The students who dowell in Olympiads have freedom of thought. They can look at usualthings in a new way, like great painters and poets do. Olympiadsprovide students with an opportunity to express and appreciate thiscreative freedom.

The Colorado Mathematical Olympiad has survived for over twodecades. Happy anniversary to the Olympiad and all people whomade it possible! The Olympiad has become part of life for manystudents, parents, and schools of Colorado. I know that the Olympiadis approaching when I receive e-mails from eagerly awaiting partic-ipants, their parents, and teachers.

It was reasonable to limit participation to students of the State ofColorado. However, I could not refuse guest participation, and we

xliv Preface

have had Olympians traveling from Long Island, New York, Kansas,and even the national team of the Philippines. For a number of years,winners of the Mobile, Alabama mathematics competition have beenreceiving as a prize a visit to compete in the Colorado MathematicalOlympiad.

I have just now noticed that in preparing for Springer new(expanded) editions of my books Mathematics as Problem Solving,How Does One Cut a Triangle?, and Geometric Etudes in Combina-torial Mathematics, I preserved original books within new editionsand added new parts within new editions. I could not alter thewholeness of the first editions because as works of art, these original

books had to be preserved unaltered. This book dramatically expandsupon the 1994 original Colorado Mathematical Olympiad: The First10 Years and Further Explorations. But the original is preserved as“Book I” (I have added some historical sketches, for after The Math-ematical Coloring Book my writing style has changed), followedby the brand new “Book II” and the new “Part V. Winners Speak:Reminiscences in Eight Parts.”

The original Book I presents the history of the first 10 years of theOlympiad. Book I also presents all 51 problems of the first 10 years ofthe Olympiad with their solutions. History, Problems, and Solutionssections are organized by the year of the Olympiad. Problem numberi.j indicates problem j of the i-th Olympiad. After Historical Notes,Problems, and Solutions come Further Explorations, a unique featureof this book, not found in any of the numerous books reportingRussian, American, Chinese, International, and other Olympiads.Each Exploration takes off from one or more Olympiad problemspresented in this book and builds a bridge to the forefront of mathe-matics, in some explorations to open problems of mathematics. Thisis the feature Paul Erdos liked the most, when he decided to write aforeword for the 1994 edition of the book.

The new Book II presents the history and problems of the seconddecade. I can now see how the problems of the Olympiad havematured over the years. Book II then offers ten new Further Explo-rations, for the total of 20 bridges between Olympiad problems andproblems of real mathematics. I mean here “real” in the sense of realmathematicians working on these kinds of problems, riding thesekinds of trains of thought—and not in a sense of “real” life. In 1921Albert Einstein addressed the correlation between mathematics and

Preface xlv

reality in a most convincing way: As far as the propositions ofmathematics refer to reality, they are not certain; and as far as theyare certain, they do not refer to reality.

Books I and II, each consisting of two parts, are followed by Part V,Winners Speak: Reminiscences in Eight Parts. In this part, severalwinners of the Olympiad evaluate the role of the Olympiad in theirlives and describe their young professional careers, life after theOlympiad.

First of all I thank the many thousands of young mathematicians,Colorado Olympians, for without them, their dedication to the Olym-piad, their efforts, and their demonstrated brilliance all our hard work

would make no sense.I am grateful to my Dean of Letters, Arts and Sciences Tom

Christensen and my Chancellor Pam Shockley-Zalaback forestablishing the Admission Window for the Olympiad’s medalists,the window equal to that for the USA Olympic sportsmen; and toPam for proposing and funding Chancellor’s Scholarships for Olym-piad’s Medalists. I thank the two of them and my chair TomWynn forunderstanding and appreciating the intensity of my ongoing work onthe Olympiad for the past 27+ years.

I thank all those who volunteered their time and talent to serve asjudges and proctors of the Olympiad, especially those judges whohave served the longest: Jerry Klemm, Gary Miller, Bob Ewell, ShaneHolloway, and Matt Kahle. I am grateful to the Olympiad’s managerswho gave the long months every year to organizing the Olympiad andits registration: Andreanna Romero, Kathy Griffith, and MargieTeals-Davis. I thank people of Physical Plant, Media Center, andthe University Center, and first of all Dave Schnabel, ChristianHowells, Rob Doherty, Mark Hallahan, Mark Bell, and Jeff Davis.

The Olympiad has been made possible by dedicated sponsors.I thank them all, and first of all the longest-term sponsorsDr. Stephen Wolfram and Wolfram Research; CASIO; Texas Instru-ments; Colorado Springs School Districts 11, 20, 2, and 3; RangelyHigh School; Chancellor’s and Vice Chancellors’ Offices; and theBookstore of UCCS. I am infinitely indebted to Greg Hoffman, whosefive companies all became major contributors most definitely due toGreg’s trust in and loyalty to the Olympiad.

Every year Colorado Governors Roy Romer, Bill Owens, andBill Ritter, as well as Senator and later Director of Labor Jeffrey

xlvi Preface

M. Welles, have written congratulatory letters to the winners of theOlympiad. Most frequent speakers at the Award Presentation Cere-monies were Chancellors Dwayne Nuzum and Pamela Shockley-Zalaback, Deans James Null and Tom Christensen, sponsor GregHoffman, Senator Jeffrey M. Welles, and Deputy SuperintendantsMaggie Lopes of Air Academy District 12 and Mary Thurman ofColorado Springs District 11. My gratitude goes to all of them and allother speakers of the Award Presentations.

I thank the long-term senior judge of the Olympiad, Dr. Col. RobertEwell, for translating some of my hand-drawn illustrations into sharpcomputer-aided designs.

My late parents, the artist Yuri Soifer and the actress FriedaGofman, gave me life and filled it with art. They rose to the occasion,recognized my enthusiasm toward mathematics, and respected myswitch from the art of music to the art of mathematics. They two andmy kids Mark, Julia, Isabelle, and Leon were my love and inspiration.They all participated in the Olympiad and some won awards. Myveteran judges and I will never forget how for years tiny beautifulIsabelle in a Victorian style dress passed the Olympiad’s buttons tothe winners. I owe you so much!

I admire high talents and professionalism, great taste, and kindattention of the first readers of this manuscript, who are also theauthors of the book’s forewords: Philip L. Engel, Branko Grünbaum,Peter D. Johnson Jr., and Cecil Rousseau. Thank you so very much!

I thank my Springer editor Elizabeth Loew for a constant supportand good cheer, and Ann Kostant for inviting this and my other sevenbooks to the historic Springer.

The Olympiad weathered bad weather postponements and admin-istrators impersonating barricades on the road. The Olympiad alsoencountered enlightened administrators and loyal volunteers. It isalive and well. I may meet with you again soon on the pages of thefuture book inspired by the third decade of the Olympiad!

Preface xlvii


Mathematics, rightly viewed, possesses not onlytruth, but supreme beauty. . . capable of a sternperfection such as only the greatest art can show.

—Bertrand Russell

It has been proved by my own experience thatevery problem carries within itself its ownsolution, a solution to be reached by the intenseinner concentration of a severe devotion to truth.

—Frank Lloyd Wright

The Colorado Mathematical Olympiad has survived for a decade.Happy anniversary to the Olympiad and all people who made itpossible!

What is the Olympiad? Where does one get its problems? How tosolve them? These questions came through the years from everycorner of Colorado and the world. This is the right time to reply: wehave accumulated enough of striking history, intriguing problems,and surprising solutions, and we have not forgotten too much yet.

Part I, The First 10 Years, is my reply to these questions. Itdescribes how and why the Olympiad started and gives a fairlydetailed history of every year. Many, although inevitably far from

xlix

all, people who made the Olympiad possible are recognized: spon-sors, problem creators, judges, proctors, and, of course, contestants.

Some people who tried to make the Olympiad impossible arementioned as well. The history of human events is never a bed ofroses. Future organizers of Olympiads need to know that they willface support and understanding from some and indifference and evenopposition from others.

Part I also presents all 51 problems of the first 10 years of theOlympiad with their solutions. Two or even three solutions arepresented when as many different and beautiful solutions have beenfound. Historical notes, Problems, and Solutions sections are orga-

nized by the Olympiads. Accordingly, problem i.j stands for problemj of the i-th Olympiad.

I did enjoy reliving the 10 years of the Olympiad and revisiting allits problems (nearly all these solutions were written for the first timefor this book). Yet, in working on this book my real inspiration camewhen the idea for Part II: Further Explorations occurred to me. This iswhen I began to feel that I was writing a mathematical book.

Part II in its ten essays demonstrates what happens in a mathemat-ical exploration when a problem at hand is solved. Each essay takesan Olympiad problem (or two, or three of them) and shows how itssolution gives birth to deeper, more exciting, and more generalproblems. Some of these second generation problems are open (i.e.,not solved by anyone!). Others are solved in this book. Severalproblems are left unsolved, even when I know beautiful solutions,to preserve for the reader the pleasure of discovering a solution on hisown. In some essays the reader is led to the third generation problems.Several open problems carry a prize for their first solutions. Forexample, this is the first time that I am offering $100 for the firstsolution of problem E5.8 (this number refers to the eighth problem ofchapter E5 of Further Explorations).

To the best of my knowledge, this is the first Olympiad problembook with a whole special part that bridges problems of mathematicalOlympiads with open problems of mathematics. Yet, it is not a totalsurprise that I am able to offer ten mini-models of mathematicalresearch that originate from Olympiad problems. The famousRussian mathematician Boris N. Delone once said, as AndreiN. Kolmogorov recalls in his introduction to [GT], that in fact, “amajor scientific discovery differs from a good Olympiad problem

l Preface

only by the fact that a solution of an Olympiad problem requires5 hours whereas obtaining a serious scientific result requires 5,000hours.”

There is one more reason why Further Explorations is a key part ofthis book. Olympiads offer high school students an exciting additionand alternative to school mathematics. They show youngsters thebeauty, elegance, and surprises of mathematics. Olympiads celebrateachievements of young mathematicians in their competition witheach other. Yet, I am not a supporter of such competitions foruniversity students. University is a time to compete with the field,not with each other!

I wish to thank here my junior high and high school mathematicsteachers, most of whom are not with us any more: KlaraA. Dimanstein, Tatiana N. Fideli, Boris V. Morozkin, and YuriF. Mett. I just wanted to be like them! I am grateful to NikolaiN. Konstantinov for his fabulous mathematics club that I attendedas an eighth grader, and to the organizers of the Moscow UniversityMathematical Olympiad: they convinced me to become a mathema-tician. I am grateful to the Russian Isaak M. Yaglom and the Amer-ican Martin Gardner for their books that sparked my early interest inmathematics above all other great human endeavors.

The wonderful cover of this book was designed by my lifelongfriend Alexander Okun. Thank you Shurik for sharing your talentwith us. I thank David Turner and Mary Kelley, photographers of TheGazette Telegraph, for the permission to use their photographs in thisbook, and The Gazette editors for the permission to quote the articlesabout the Olympiad.

I am grateful to Philip Engel, Paul Erdos, Martin Gardner, BrankoGrünbaum, and Cecil Rousseau for kindly agreeing to be the firstreaders of the manuscript and providing me with most valuablefeedback. I am truly honored that these distinguished people havewritten introductions for this book.

I applaud my Dean James A. Null for bringing about such a climatein our College of Letters, Arts and Sciences that one is free andencouraged to create. I am grateful to my wife Maya for sustainingan intellectual atmosphere under our roof, proofreading this manu-script, and pushing me to a pen in my less inspired times. I thankSteven Bamberger, himself an award winner in the Fourth Colorado

Preface li

Mathematical Olympiad, for deciphering my manuscript andconverting it into this handsome volume.

The Colorado Mathematical Olympiad has survived for a decade.Happy Anniversary to the Olympiad and all people who made itpossible!

lii Preface

Part I

The Third Decade

Twenty-First Colorado MathematicalOlympiadApril 16, 2004

Historical Notes 21

Mark Heim of Loveland Wins AgainAward Presentation Ceremonies Buried under the Spring Snow

To run this year’s Olympiad, I traveled across the United States, fromPrinceton University, where I lived and worked at the time, to Colo-rado Springs. There was no time to jump-start my regular Coloradolife. I stayed at the magnificent Antler’s Hotel in downtown ColoradoSprings, with an unobstructed view of Pikes Peak, and drove aroundin a tiny rental car.

The Twenty-First Colorado Mathematical Olympiad (CMO-2004)took place on April 16, 2004, 4 days before the fifth anniversary of thetragic Columbine High School Massacre. It brought together 570 mid-dle and high school students from all over Colorado. You may recall,

the national team of the Philippines participated in our Olympiad in2001. They wanted to compete in our Olympiad again, but theAmerican Consulate refused their visa application. Our bureaucratsmust have been afraid that the young, energetic, and talentedPhilippine students would like to remain in the U.S. and contributeto the American science.

On April 23, 2004, the day of the Award Presentation Ceremonies,Colorado dumped on us so much snow that the Ceremony had to bepostponed by a week. I needed to return to Princeton, and for one andonly time missed the Awards. I gave my fully prepared transparenciesof two lectures to Gary Miller, who presented them on my behalf.

© Alexander Soifer 2017A. Soifer, The Colorado Mathematical Olympiad: The Third Decadeand Further Explorations, DOI 10.1007/978-3-319-52861-8_1

3

Our Olympiad is a unique event for young mathematicians, andthey appreciate it. Curtis Larimer, a senior at Palmer High School,wrote on his Olympiad paper, “For Dr. Soifer—thank you for 7 yearsof wonderful competition.” In an interview published in Slice (a partof The Gazette, Colorado Springs’ major newspaper), Troy Driller, afreshman at Saint Mary’s Catholic High School, said:

The problems are not your typical equations so you need a goodimagination and the ability to think outside of the math box.I learned a lot about how to think about and work through mathproblems in different ways. It is a good math experience that willprepare me to compete again next year.

For the second straight year, the judges awarded first prize to MarkHeim, now a junior at Thompson Valley High School. He receiveda gold medal of the Olympiad, a $2000 scholarship to be used atany certified American university or 4-year college, a $1000UCCS Chancellor’s Scholarship for CMO Medalists, CASIO“Super-Calculator” Classpad 300, Wolfram Research’s softwareMathematica, and Geometric Etudes in Combinatorial Mathematics,a book by Vladimir Boltyanski and Alexander Soifer.

Second prize was awarded to Christopher Liebmann, a junior atWasson High School. He received a silver medal of the Olympiad, a$1000 scholarship to be used at any certified American university or4-year college, a $1000 UCCS Chancellor’s Scholarship for CMOMedalists, CASIO “Super-Calculator”, Wolfram Research’s softwareMathematica, and Geometric Etudes in Combinatorial Mathematics.

Third prize was presented to the following four Olympians: JonFox, a sophomore, Palmer High School; Bryce Herdt, a junior,Mitchell High School, who won three straight first prizes in theprevious 3 years; Caleb Packer, senior, Rangely High School; andHelen Schulthorpe, a senior, Arapahoe High School. Each of themreceived a bronze medal of the Olympiad, a $1000 UCCS Chancel-lor’s Scholarship for CMO Medalists, Wolfram Research’s softwareMathematica, CASIO “Super-Calculator”, and Geometric Etudes inCombinatorial Mathematics.

The judges also awarded 28 first honorable mentions and 72 secondhonorable mentions. All of the above 102 winners received Wol-fram’s Mathematica software and CASIO CD-ROM ClassPad300Emulation.

4 Twenty-First Colorado Mathematical Olympiad

Literary Awards were presented to Nick Raynor, a sophomore,Wasson High School; Curtis Larimer, a senior, Palmer High School;and Angelina, a sophomore, James Irwin Charter High School. ArtAward went to Molly Lockhart, a freshman, James Irwin CharterHigh School. Teachers Awards were presented to Joe Scott, JamesIrwin Charter High School; and Mel Oliver, Rangely High School,who raised numerous winners in his rural Northwestern Coloradotown, Rangely, located near the Utah border.

The Prize Fund of the Olympiad was generously donated byIngersoll-Rand Company; Wolfram Research, Inc.; CASIO; TexasInstruments; Air Academy School District 20; Harrison School

District 2; Widefield School District 3; Colorado Springs SchoolDistrict 11; St Mary’s High School; Irving Middle School;Timberview Middle School; Sand Creek High School; Falcon HighSchool; Ft Collins High School; The Colorado College; Vice-Chancellor’s Office, Bookstore, and Chancellor’s Office—all fromthe University of Colorado Colorado Springs.

The Award Presentation Program included “Review of Solutions ofthe Olympiad Problems” and the lecture “Etudes on MathematicalPacking and Covering,” both written by me and read by Gary Miller.

The following guests of honor, hosts and sponsors addressed thewinners and presented awards: Pamela Shockley, Chancellor; RogersRedding, Vice Chancellor for Academic Affairs; Jim Henderson,Vice Chancellor for Student Success—all from the University ofColorado Colorado Springs; Honorable Michael Merrifield, StateRepresentative, The State of Colorado; Norman Ridder, Superinten-dent, Colorado Springs School District 11; Gregory C. Hoffman,Ex-Director, Human Resources, Ingersoll-Rand Company; andAlexander Soifer.

The Olympiad received congratulatory letters from ColoradoGovernor Bill Owens and Executive Director of the Department ofLabor and Employment Jeffrey M. Wells. Let me reproduce for you athoughtful letter by Wells, who started his career with a degree inMathematics from Duke University.

Historical Notes 21 5

I continued my 1996–2005 service on the United States of AmericaMathematical Olympiad (USAMO) Subcommittee, a six-membergroup that created and edited problems and graded papers ofUSAMO. In a number of aspects, I prefer our Colorado MathematicalOlympiad (CMO). For one, we allow anyone to come and compete inthe Olympiad (Olympiad by definition is a competition that requires


complete essay-type solutions). This has never been the case withUSAMO. In the first round of the American national competitions,contestants are offered 90 minutes to answer 30 multiple-choicequestions. In the second round, contestants are expected to write inanswers-only to 15 questions (answers are known to be integersbetween 0 and 999—which makes it in a sense a multiple choicewith 1000 options). And so, one needs to be a master of speedyguessing in order to be admitted to USAMO. To me, this system isakin requiring contestants to sing in the first round, to dance in thesecond one, in order to make it into the final round of pianists’competition. :)

A heavy emphasis on knowledge is another aspect I do not partic-ularly like at USAMO. Of course, having knowledge is a plus, but nota big plus in my book, for knowledge is . . . yes, knowable. I espe-cially dislike that USAMO practically requires an inclusion of hardproblems on standard trigonometry and Euclidean geometry. Byusing such problems, the USAMO tries to select a team that woulddo well at the International Mathematical Olympiad (IMO), whichdue to participation of ca. 100 countries, practically always includesthese kinds of standard trigonometry and geometry problems, thusexhibiting the least common denominator of mathematical taste.What is wrong with trigonometry, I hear you asking? Let me answeryour question with a question: Is trigonometry a living branch ofmathematics? I think not, for I would define a living branch to be afield where, rounding to a whole percentage point, 100 % of problemsare open. (Consequently, I would dramatically reduce in high schooltime allotted to trigonometry.) And what about Euclidean geometry,you wonder? I think it is a pretty but obscure cul-de-sac of mathe-matics, an exercise in constructing an axiomatic theory. Am I pro-posing to abandon geometry in high school? No, I propose to replacemuch of Euclidean geometry by combinatorial geometry. It offers anabundance of problems that sound like a ‘regular’ Euclidean geome-try, but require for their solutions synthesis of ideas from variousbranches of mathematics. See examples of synthesis in [Soi3], [Soi6],and [Soi7]. Moreover, combinatorial geometry offers us open-endedseries of problems, trains of thought resembling mathematical pursuitof best result. And it presents an abundance of classic open problemsthat any student can understand, and yet no professional has yetsolved! Let us stop discrimination of our gifted students based on


their youthful age and allow them to smell the roses—I mean, to smell‘real’ mathematics and touch its unsolved problems. Our studentsmay find partial solution; sometimes they may even settle these openproblems completely. And they will then know the answers to whatshould become the fundamental questions of mathematical education:What is Mathematics? What Do Mathematicians Do?

Shortly after the Olympiad, in July 2004, I gave talks at the 10thInternational Congress on Mathematical Education in Copenhagen,Denmark, that were related to our Olympiad: “One Problem—ThreeLives: P. J. H. Baudet, I. Schur, B. L. van der Waerden and Mono-chromatic Arithmetic Progressions of Integers,” and “One Beautiful

Olympiad Problem: Chess 7 � 7.” I also attended there the NationalExhibit on Mathematical Education in Russia. Sadly—but not sur-prisingly—it opened with the portrait of and a smart quotation fromtheir totalitarian President Vladimir Putin. However, vodka and cav-iar were served, and it was nice to visit with old friends, AndreiEgorov and other people working for the Physics-Mathematical mag-azine “Kvant.”

At a General Meeting of the World Federation of National Math-ematics Competitions (also in Copenhagen), I was reelected to thethird 4-year term as the Secretary and member of the ExecutiveCommittee, the post I occupied since my first 1996 election in Seville,Spain.

Problems 21

21.1. Homeland Security (A. Soifer). Intelligence reports that terror-ists have sailed a 1� 5 frigate completely within our coastline which isa 2004-long strip of 1 � 1 cells. Intelligence does not know exactlywhere the frigate is. What is the smallest number of guided missiles wemust launch to destroy the enemy’s frigate? (A missile destroys a 1� 1cell, and a single hit on the frigate destroys it.)


21.2. 2004 Tails (G. Galperin). You are in an absolutely dark roomwith two tables, A and B. On top of A there is an unknown hugenumber of silver dollars, exactly 2004 of which lie tails up. Nothing ison top of B. You may conduct two operations: you may move anynumber of dollars from one table to another; you may flip over anynumber of dollars on one of the tables. (You cannot tell in any waywhether a coin is tails or heads up.)

(A) Can you obtain an equal number of tails on the tables through aseries of allowed operations?

(B) Can you obtain an equal number of heads on the tables through aseries of allowed operations?

21.3. Crawford Cowboy had a Farm, EE-I-EE-I-O (G. Galperinand A. Soifer). A Crawford cowboy is unable to create a rectangularfence around his ranch using all 2004 straight sections of fence that hehas. He decides to break some of the sections, each section no morethan once, so that the resulting set can do the job.

(A) Prove that there is a collection of 2004 sections such that evenafter breaking exactly one section, the cowboy cannot create arectangular ranch using all the sections.

(B) Find the smallest number n, such that breaking n sections isenough to do the job regardless of the lengths of the original2004 sections.

21.4. To have a Cake (A. Soifer).

(A) We need to protect from the rain a cake that is in the shape of anequilateral triangle of side 2.1. All we have are identical tiles inthe shape of an equilateral triangle of side 1. Find the smallestnumber of tiles needed.

(B) Suppose the cake is in the shape of an equilateral triangle of side3.1. Will 11 tiles be enough to protect it from the rain?

21.5. Chess 7 � 7 (A. Soifer).

(A) Each member of two 7-member chess teams is to play once againsteach member of the opposing team. Prove that as soon as 22 gameshave been played, we can choose four players and seat them at around table so that each pair of neighbors has already played.

(B) Prove that 22 is the best possible; i.e., after 21 games the result ofpart (A) cannot be guaranteed.

Problems 21 9

Solutions 21

21.1. Starting from the left side, we put next to each other 400 frigates,leaving four unit cells on the right (Fig. 21.1.1). We thus need tolaunch at least 400 missiles.

Now color red every fifth cell, 400 in all—they are our targets(Fig. 21.1.2). No matter where the frigate is, it contains a target, thus,400 missiles suffice. ■

21.2.(A). Move 2004 dollars from the table A to the table B and flipthem over. We are done!

Indeed, assume that before the move the 2004 coins, selected forthe move, included x tails. This would leave after the move 2004�x tails on the table A. The table B received 2004 coins, includingx tails. After flipping all coins on the table B, we end up with2004 �x tails on B, which is exactly what we ended up with on thetable A. ■

21.2.(B). First flip over all the coins on the table A; then move 2004coins to the table B; and finally flip over all the coins on the table B.We are done.

Indeed, the first operation produces 2004 heads on the table A, withtails on all other coins. Now we are precisely in the same setting as wewere in problem (A), except “head” and “tail” attributes are switched.Therefore, the two operations we used to solve problem (A), would dothe same to the heads what in problem (A) was done to the tails. ■

Fig. 21.1.1

Fig. 21.1.2


21.3. I dedicated this problem to the United States President GeorgeW. Bush, who considered himself a cowboy and whose farm waslocated in Crawford, Texas. I incorporated my homage into a popularfolk song “Old MacDonald Had a Farm, Ee-i-ee-i-o,“ which you canlisten to here: https://www.youtube.com/watch?v¼nVrDNpFhF5E.

21.3.(A). Choose the fence sections of the following lengths: 2004,1p1, 1p2

, :::, 1p2003

, where p1, p2, . . . , p2003 are distinct prime numbers.

Assume that the Crawford Cowboy was able to fence his farm bybreaking just one section. Then it must have been the large section,for no sum of other sections can equal its length. So let the section oflength 2004 be broken into sections A and B and their lengths are suchthat |A|� |B|. Then the sections A and B may not lie on adjacent sidesof the rectangle, for otherwise the length of A would have to beequaled by a sum of inverse primes, whereas even the sum of all2003 inverses is smaller in length than A. So, A and B lie on oppositesides of the farm’s fence. Now look at the other pair of opposite sides.Their lengths are the sums of some inverses of the primes, and whenwe bring these two sums to common denominators and reduce tolower terms, the two denominators will be the products of differentsets of primes. However, if two reduced fractions have distinctdenominators, they cannot be equal to each other. ■

21.3.(B). n ¼ 2. In view of (A), we only need to show that two breaksare enough. We split one section into two new equal sections and usethem for one pair of opposite sides of the farm. The remaining 2003sections we stand up into one straight fence F, and then break F intotwo sections of equal length for the other pair of opposite sides of thefarm. If it so happened that the place of this break falls on the borderof two original sections, we simply do not need to make the secondbreak. ■

21.4.(A). Mark six points in the equilateral triangle of side 2.1: itsvertices and midpoints of the sides (Fig. 21.4.1). A unit equilateral tilecan cover at most one such point, therefore we need at least six tiles.

Solutions 21 11

https://www.youtube.com/watch?v=nVrDNpFhF5E

https://www.youtube.com/watch?v=nVrDNpFhF5E

On the other hand, six tiles can do the job. Let me show you two

different covering. We can cover the corners (Fig. 21.4.2a), and thenuse three more tiles to cover the remaining hexagon (Fig. 21.4.2b).

Fig. 21.4.1

Fig. 21.4.2A


Alternatively, we can cover the top corner (Fig. 21.4.3), and thenuse five triangles to cover the remaining trapezoid. ■

21.4.(B). Each of the two methods of coverings used in the solution ofproblem 21.4.(A) can be used for covering here. We can push the firstmethod by covering a cake of side up to x ¼ 2.25 with six tiles (seeFig. 21.4.4, where x ¼ 0.25).

Fig. 21.4.2B

Fig. 21.4.3

Solutions 21 13

Let us now cover the cake of side, say, 3.1. We first use thecovering of Fig. 21.4.4 to cover a cake of side 2.2, and put thiscovering in the top corner of the cake of side 3.1, and then take careof the remaining trapezoid with five tiles (Fig. 21.4.5).

Alternatively, we can use four tiles in the top corner, and then useseven tiles for a larger trapezoid (Fig. 21.4.6).

Fig. 21.4.4

Fig. 21.4.5


Read an exciting story of this problem and inspired by it train ofthought in Further Exploration E21. ■

21.5. This problem occurred to me while I was reading a wonderfulunpublished 1989 manuscript of the monograph Aspects of RamseyTheory by two well-known German mathematicians Hans JürgenPr€omel and Bernd Voigt at a mountain lake in Southern Bavaria. Ifound a mistake in a lemma, and constructed a counterexample to thatlemma’s statement. This counterexample provides a solution to prob-lem 21.5.(B) here. Problem 21.5.(A) is a corrected particular case ofthat lemma, translated, of course, into the language of a nice ‘real’story. Enjoy several beautiful solutions of this problem.

21.5.(A). First Solution.Given an array of real numbers x1 , x2 , . . . ,x7, whose arithmetic mean is �x. A well-known inequality (that can beeasily derived from the Cauchy-Schwarz inequality) states that

ffiffiffiffiffiffiffiffiffiffiffiP7i¼1

x2i

7

vuuut� �x;

thus

X7i¼1

x2i � 7�x2:

Fig. 21.4.6

Solutions 21 15

This inequality defines ‘convexity’ of the function f(x)¼ x2, which

readily implies convexity of the binomial function x2

� �¼ 1

2x x� 1ð Þ,

i.e.,

X7i¼1

xi2

� �� 7 �x

2

� �: ð1Þ

Observe that above we defined the binomial function x2

� �for all

real x (not just for positive integers, as is usually done). Also, in acertain informality of notations, for a positive integer x we would use

x2

� �not only as a number, but also as a set of all 2-element subsets of

the set {1, 2, ... , x}.Let us call players on each team by positive integers 1, 2, . . ., 7. A

game between a player i of the first team against a player j of thesecond team can conveniently be denoted by an ordered pair (i, j).Assume that the set G of 22 games has been played.

Denote by S( j) the number of games played by the player j of the

second team: S( j)¼ j{i : (i, j)2G}j. Obviously, P7j¼1

S jð Þ ¼ 22.

For a pair (i1, i2) of the first team players we denote by C(i1, i2) thenumber of the second team players j, who played with both players ofthis pair: C(i1, i2)¼ |{j : (i1, j)2G^ (i2, j)2G}|. Adding all C(i1, i2),we get the sum T ¼ P

i1;i2ð Þ2 7

2

� �C i1; i2ð Þ that counts the triples (i1, i2; j)

such that each of the first team’s players i1 , i2 has played with thesame player j of the second team. This number T can be alternatively

calculated as follows: T ¼ P7j¼1

S jð Þ2

� �. Therefore, we get the equality

Xi1;i2ð Þ2 7

2

� �C i1; i2ð Þ ¼X7j¼1

S jð Þ2

� �:

In view of the convexity inequality (1), we finally get


Xi1;i2ð Þ2 7

2

� �C i1; i2ð Þ ¼X7j¼1

S jð Þ2

� �� 7

X7j¼1

S jð Þ

72

0B@

1CA ¼ 7

22

72

� �> 7 3

2

� �¼ 7

2

� �;

i.e.,

Xi1;i2ð Þ2 7

2

� �C i1; i2ð Þ > 7

2

� �:

We showed that the sum of 7

2

� �non-negative integers is greater

than 7

2

� �, therefore, for at least one of the summands we get the

inequality C(i1, i2)� 2 for some i1 , i2. In our notations this meansprecisely that the pair of the first team players i1 , i2 played with thesame two players j1 , j2 of the second team. Surely, you can seat thesefour players at a round table in accordance with the problem’srequirements! ■

21.5.(B). First Solution. We need to construct an example to dem-onstrate that 21 games do not guarantee the existence of the required

four players. It is convenient to portray the teams of players ashorizontal arrays of seven dots each, for each game that has takenplace to connect the two who played it by a segment. We get aso-called bipartite graph with 14 vertices and 21 edges. Not allsuch graphs are created equal. Figure 21.5.1 shows a graph thatcontains no 4-cycles, i.e., closed walks through four distinct edges,which exactly means that the required choice of four chess playersdoes not exist. ■

Fig. 21.5.1

Solutions 21 17

21.5.(A). Second Solution. In the selection and editing process, Col.Dr. Bob Ewell suggested a nice visual representation of using a 7 � 7table to show the games played, which I am using in my solution. Wenumber the players in both teams. For each player of the first team weallocate a row of the table, and for each player of the second team acolumn. We place a checker in the table in the location (i, j) if theplayer i of the first team played the player j of the second team(Fig. 21.5.2).

If the required four players were found, this would manifest itselfin the table as a rectangle with four checkers in its corners, a check-ered rectangle for short. (Sides of the checkered rectangle arerequired to be parallel to the lines of the grid.) The problem thustranslates into the new language as follows:

A 7 � 7 table with 22 checkers must contain a checkeredrectangle.

Assume that a table has 22 checkers but does not contain a check-

ered rectangle. Since 22 checkers are contained in seven rows, by thePigeonhole Principle, there is a row with at least four checkers init. Observe that interchanging rows or columns does not affect the

Fig. 21.5.2


property of the table to have or have not a checkered rectangle. Byinterchanging rows, we place the row with at least four checkers first.By interchanging columns, we now make all checkers of the first rowappear consecutively starting from the left. We consider two cases.

Case 1. Top row contains exactly four checkers (Fig. 21.5.3).

Draw a bold vertical line L after the first four columns countingfrom the left. To the left of L, the top row contains four checkers, andall other rows contain at most one checker each, for otherwise wewould have had a checkered rectangle (that includes a pair from thetop row). Therefore, to the left of L we have at most 4 + 6 ¼ 10checkers. This leaves at least 12 checkers to the right of L, thus at leastone of the three columns to the right of L contains at least fourcheckers; by interchanging columns and rows we put these fourcheckers in the position shown in Fig. 21.5.3. Then each of the lasttwo columns contains at most one checker total in the rows 2 through5, for otherwise we would have had a checkered rectangle. Thus, wehave at most 4 + 1 + 1 ¼ 6 checkers to the right of L in the rows2 through 5 combined. Therefore, in the lower right 2 � 3 part С of

Fig. 21.5.3

Solutions 21 19

the table we have at least 22 – 10 – 6 ¼ 6 checkers—thus C iscompletely filled with checkers, and we get a checkered rectangle inC in contradiction to our assumption.

Case 2. Top row contains at least five checkers (Fig. 21.5.4).

Draw a bold vertical line L after the first five columns countingfrom the left. To the left of L, the top row contains five checkers, andall other rows contain at most one checker each, for otherwise wewould have had a checkered rectangle (that includes a pair from thetop row). Therefore, to the left of L we have at most 5 + 6 ¼ 11checkers. This leaves at least 11 checkers to the right of L, thus at leastone of the two columns to the right of L contains at least six checkers.By interchanging columns and rows we put five of these six checkersin the position shown in Fig. 21.5.4. The last column now contains atmost one checker total in the rows 2 through 6, for otherwise wewould have had a checkered rectangle. Thus, we have at most5 + 1¼ 6 checkers to the right of L in the rows 2 through 6 combined.

Fig. 21.5.4


Therefore, the upper right 1 � 2 part D of the table plus the lowerright 1 � 2 part C of the table have together at least 22 – 11 – 6 ¼ 5checkers—but they only have four cells, and we thus get a desiredcontradiction. ■

21.5.(A). Third Solution. In this solution, I developed Bob Ewell’sidea of moving checkers from one row to another and noting the netchange in the number of 2-element subsets.

Given a placement P of 22 checkers on the 7 � 7 table. We pickone row; let this row have k checkers total in it. The number of

2-element subsets of a k-element set is denoted by k2

� �and is equal

to k2

� �¼ 1

2k k � 1ð Þ. We now define a function C(P) of a placement

P of 22 checkers on the table as the sum of seven such summandsk2

� �, one per each row of our 7 � 7 table.

If there is a row R in the table with r checkers, where r ¼ 0, 1, or2, then there is a row S with s checkers, where s ¼ 4, 5, 6 or 7 (this is

true because the average number of checkers in a row is 22=7 ). Wenotice that s – r – 1 > 0, and observe that moving one checker fromthe row S to any open cell of the row R produces a placement P1 suchthat C(P) > C(P1), because

r2

� �þ s

2

� �� r þ 1

2

� �� s� 1

2

� �¼ s� r � 1 > 0:

By repeating the above procedure of moving one checker at a timeto a less populated row, we will end up with the final placement Pk,where each row has three checkers except one, which has four. Forthe final placement the function C(Pk) can be easily computed as

6 3

2

� �þ 4

2

� �¼ 24. Thus, for the original placement P, C(P) � 24.

On the other hand, the total number of 2-element subsets in a

7-element set is 7

2

� �¼ 21. Since 24 > 21, there are two identical

2-element subsets (see Fig. 21.5.5) among the 24 pairs counted by thefunction C(P). But the checkers that form these two identical pairsform a desired checkered rectangle!

Solutions 21 21

21.5.(B). Second solution. To show that 22 is the lowest possiblenumber of checkers to guarantee the existence of a checkered rectan-gle, we simply present a checker rectangle-free table with 21 checkers(Fig. 21.5.6). And we are done. ■

By the way, our example in the first solution (Fig. 21.5.1) can betranslated into the language of tables as follows (Fig. 21.5.7):

Fig. 21.5.5

Fig. 21.5.6


Our counter examples in Figs. 21.5.6 and 21.5.7 require an obser-vation that no checkered rectangle is contained in them. Confidentmathematicians play on observations to present “proofs withoutwords” by declaring “Behold.” Unbelievable politicians—DonaldJ. Trump comes to mind—also urge “Believe me! Believe me!” Letme present here a proof that requires no reference to observations orbelief. ■

21.5(B). Third solution.

Example is not the main thing in influencing others. It is the onlything. —Albert Schweitzer

Glue a cylinder (!) out of the 7 � 7 table and put 21 checkers on allsquares of the first, second, and fourth diagonals. Figure 21.5.8 showsour cylinder with one such checkered diagonal. Figure 21.5.9 shows,in a flat representation, the cylinder with all three checkereddiagonals.

Fig. 21.5.7

Solutions 21 23

Assume that our 7 � 7 table contains a checkered rectangle. Sincefour checkers of the rectangle lie on three diagonals, by the Pigeon-hole Principle, two checkers lie on the same (checkers-covered)diagonal D of the cylinder. But this means that on the cylinder ourfour checkers form a square! Two other (opposite) checkers a andb thus must be symmetric to each other with respect to D, which

Fig. 21.5.8

Fig. 21.5.9


implies that the diagonals of the cylinder that contain a and bmust besymmetric to each other with respect to D, but no two checker-covered diagonals in our placement of checkers are symmetric withrespect to D. (To see this, observe Fig. 21.5.10 which shows the toprim of the cylinder with bold dots for the starting squares of thecheckered diagonals. Distances measured in the number of squaresbetween the checkered diagonals are all distinct; clockwise, they are1, 2, and 4, whereas symmetry would require a pair of equal dis-tances). This contradiction implies that there are no checkered rect-angles in our placement. Done! ■

In my lifetime, I created many Olympiad problems. Among themall, this one is my favorite. It allows many solutions and introducesmuch beauty. It originated from a research manuscript, and you willsee another connection to ‘real’ mathematics, to finite projectiveplanes, in the conclusion of this train of thought in Further Explora-tion E22. ■

Fig. 21.5.10

Solutions 21 25

Twenty-Second Colorado MathematicalOlympiadApril 22, 2005

Historical Notes 22

Mark Heim of Loveland Wins for the Third Straight Time!

The Twenty-Second Colorado Mathematical Olympiad (CMO-2005)took place on April 22, 2005. It brought together 650 middle and highschool students from all over Colorado: Aurora, Bennett, Brighton,Calhan, Centennial, Colorado Springs, Denver, Divide, Ellicott,Englewood, Falcon, Fort Collins, Fountain, Littleton, Loveland,Manitou Springs, Monument, Rangely, and Widefield.

The judges once again awarded first prize to Mark Heim, now asenior at Thompson Valley High School. He received a gold medal ofthe Olympiad, a $1500 scholarship to be used at any certified Amer-ican university or 4-year college, a $1000 UCCS Chancellor’s Schol-arship for CMO Medalists, CASIO Cassiopeia Pocket Viewer,

CASIO “Super-Calculator” Classpad 300, and Wolfram Research’ssoftware Mathematica. This is the third straight victory for Mark,who won first prize last year, and tied for first prize 2 years ago withanother three-time winner Bryce Herdt. In the 30 years of the Olym-piad, we had only three 3-time winners: David Hunter, Bryce Herdtand Mark Heim.


27

From the left: Mark Heim, Alexander Soifer, and Bryce Herdt, April 29, 2005

Mark Heim and Hannah Alpert were the only Olympians whosolved problem 22.4.(B). Moreover, they found solutions not knownto me. Mark, in fact, proved a much stronger result than the problemasked for. I am looking at the article “Knowledge plus creativity equalmath contest victory” by Bobbi Sankey published in The Slice onMay4, 2005. It captures my contemporaneous impression of Mark Heim’spaper, and his solution of problem 22.4.(B). Let me quote some of thisarticle here:

Written by Soifer and local mathematician Ming Song, theproblems required knowledge of far more than a few algebraicformulas. Even the judges gathered to listen to Soifer explainsolutions to the five problems before beginning to critique stu-dents’ essays.

The Olympiad is not a speed competition. Many students satin their chairs the full four hours. Problems increase in difficulty,

28 Twenty-Second Colorado Mathematical Olympiad

and students who can think outside the box have the best chanceat solving them.

There are numerous ways to reach conclusions and in judgingprocess, creativity is valued just as much as the correct answer.

This year, one student’s creativity set him apart from thehundreds of other competitors. He discovered a solution evenjudges hadn’t considered for one of the test’s most difficultproblems.

According to Soifer, the high school senior wrote the bestessay in Olympiad history, stunning judges and reaching farbeyond what the problem required him to prove.

“I’ve seen thousands of papers in 22 years and it was justamazing,” he said. “He demonstrated use of group theory, ter-minology of computer science – where did he get all that? It wasabsolutely remarkable work. It was written by a mathematician.”

The mathematician is Mark Heim, a student at ThompsonValley High School in Loveland and the Olympiad’s first placewinner. Heim is a familiar Olympian – he won first prize lastyear and tied for first place 2 years ago with three-time Olympiadwinner Bryce Herdt of Mitchell High School.

“He is an independent thinker,” Soifer said. “He is ready to doresearch.”

Second prize was awarded to Sam Elder, a Freshman from PoudreHigh School. He received a silver medal of the Olympiad, a $1000scholarship to be used at any certified American university or 4-yearcollege, a $1000 UCCS Chancellor’s Scholarship for CMO Medal-ists, CASIO Cassiopeia Pocket Viewer, CASIO “Super-Calculator”,and Wolfram Research’s software Mathematica.

Third prize was presented to Hannah Alpert, a sophomore fromFairview High School. She received a bronze medal of the Olympiad,a $500 scholarship to be used at any certified American university or

4-year college, a $1000 UCCS Chancellor’s Scholarship for CMOMedalists, Wolfram Research’s software Mathematica, CASIO Cas-siopeia Pocket Viewer, and CASIO “Super-Calculator”. RememberHannah: you will meet her several more times in this book!

The judges also awarded 5 fourth prizes, 24 first honorable men-tions, and 101 second honorable mentions.


Literary Awards were presented to Melisa Lizzaraga, a junior fromSt. Mary’s High School; Tim Henrie, a freshman from Sand CreekHigh School; and Crystal Kossow, a sophomore from Falcon HighSchool. Here is one of the winners’ poem:

A Writer in a Crowdby Crystal Kossow

This is me,A writer in a crowd,Geniuses by my sideSolving problems by the secondAs I search my mind for words,Words that will grab at you,To make you see.

I get straight A’s of courseBut for me it’s a struggle,Hours of notes, days of readingTo have an answerBut not understandingThey see me as smartSo they invite me here.

I go blankAs I read the problemSo I writeTo convey my thoughtsSo you may understandI am a writer in a crowdGeniuses by my side.

This year’s Prize Fund of the Olympiad was generously donatedby CASIO, Inc.; Wolfram Research, Inc.; Intermap Technologies,Inc.; Texas Instruments, Inc.; Air Academy School District 20; Har-rison School District 2; Widefield School District 3; Colorado SpringsSchool District 11; St Mary’s High School; Irving Middle School;Timberview Middle School; Sand Creek High School; Falcon HighSchool; Ft. Collins High School; Mrs. Mary Sokol of Long Island,NY; Vice-Chancellor’s Office, Bookstore, and Chancellor’s Office—all from the University of Colorado Colorado Springs.


The Award Presentation Program included “Review of Solutions ofthe Olympiad Problems” by Alexander Soifer and his lecture “Geo-metric Cover-Up.”

The following guests of honor, hosts and sponsors addressed thewinners and presented awards: Pamela Shockley-Zalabak, Chancel-lor; Rogers Redding, Vice Chancellor for Academic Affairs; JimHenderson, Vice Chancellor for Student Success—all from the Uni-versity of Colorado Colorado Springs; Honorable CongressmanMichael Merrifield, State of Colorado; Matt Weiss, Director of Mar-keting and Education, CASIO, Inc.; Maggie Lopez, Assistant Super-intendent, Colorado Springs School District 20; Gregory C. Hoffman,

Director of Human Relations, Intermap Technologies, Inc.; and Alex-ander Soifer.

The Olympiad received congratulatory letters from Colorado Gov-ernor Bill Owens and Executive Director of the Department of Laborand Employment Jeffrey M. Wells. Let me reproduce for you theletter by Governor Owens. (There is a typo in this letter, the governormeant to greet the 22nd Olympiad, not the 23rd.)


On May 28, 2005, I attended Bryce Herdt’s high school graduationparty. It was a delight to celebrate his achievements with his parentsand two brothers.

Right after the Olympiad, on April 26, 2005, our senior judgeDr. Col. Bob Ewell sent me an astute if critical e-mail:

“I realize that technology is a wonderful thing, but watching you dothe proofs using the pre-supplied drawings is like going to a concertand having the conductor put on a record. It feels like you are readinga proof instead of doing a proof as when you drew it out by hand. It’snot quite as satisfying.”


Indeed, the use of technology appears to be a zero-sum game.Fulfilling Bob’s desires, I reverted in part to “doing a proof” infront of our Olympians’ eyes.

Problems 22

22.1. Cover-up (A. Soifer). Can a square of area 2005 be covered by401 squares of area 5 each?

22.2. Tea Time (M. Song). Is it possible to arrange the numbers1–100 in a 10 � 10 grid so that the entries of any T-shaped figureconsisting of four unit squares of the grid, called T-tetromino(depicted below) sum up to an even number?

T-tetromino

22.3. One L of a Grid (A. Soifer). What is the minimum number ofsquares to be colored red in a 10� 10 grid so that any L-shaped figureconsisting of four unit squares of the grid, called L-tetromino, con-tains at least two red squares?

22.4. Red and White (A. Soifer).(A) Each vertex of a regular 11-gon is colored red or white. Prove that

there are two congruent monochromatic triangles of the samecolor.

(B) Each vertex of a regular 2005-gon is colored red or white. Provethat there are two congruent monochromatic 10-gons of the samecolor.

A polygon is called monochromatic if all of its vertices are coloredin the same color. Distinct polygons may have some (but not all)vertices in common.

Problems 22 33

22.5. Love and Death (A. Soifer, inspired by Martin Klazar’sresearch).(A) The DNA of bacterium bacillus anthracis (causing anthrax) is a

sequence, each term of which is one of 2005 genes. How long canthe DNAbe if no two consecutive termsmay be the same gene, andno two distinct genes can reappear in the same order? That is, ifdistinct genes α, β occur in that order (with or without any numberof genes in between), the order α, . . ., β cannot occur again.

(B) The DNA of bacterium bacillus amoris (causing love) is asequence, each term of which is one of 2005 genes. No threeconsecutive terms may include the same gene more than once,

and no three distinct genes can reappear in the same order. Thatis, if distinct genes α, β, and γ occur in that order (with or withoutany number of genes in between), the order α, . . ., β, . . ., γ cannotoccur again. Prove that this DNA is at most 12,032 long.

Solutions 22

22.1. Assume it can be done. Since 401 � 5 ¼ 2005, we have no areato spare, and thus the covering squares must not intersect in insidepoints, and share sides exactly like in a square grid (Fig. 22.1.1).

Look at a side of the large square: it must be partitioned into n sides

of small squares, and we get the equationffiffiffiffiffiffiffiffiffiffi2005

p ¼ ffiffiffi5

pn for an integer

n, i.e., 401 ¼ n2, which is absurd. Thus, the covering cannot bedone. ■

2005

5

Fig. 22.1.1


22.2. Solution by Ming Song.

Consider any five numbers in the cross shape, as shown in Fig. 22.2.1.We are given that a + b + c + d¼ even and a + b + c + e¼ even. Thus,d and e have the same parity. Similarly a, b, d and e all have the sameparity. Because a + b + c + d ¼ even and a, b, and d have the sameparity, c must have the same parity as a, b, and d. Thus, a, b, c, d ande all have the same parity. Since all squares on the board except thecorners are part of a cross shape, 96 entries must all have the sameparity. Since this is not true for the given numbers 1 through 100, weget the desired contradiction. ■22.3. Coloring the 10 � 10 grid in a chessboard fashion red and whitehelps: no matter where an L-tetromino is located on the grid, it willcover exactly two red squares. So, 50 red squares suffice.

Assume now that 49 squares have been colored red and have donethe required job. Partition the grid into 24 L-tetrominoes and one2 � 2 square S in the middle of the board (Fig. 22.3.1).

da

b ce

Fig. 22.2.1

Solutions 22 35

Each L-tetromino contains at least two red squares. Thus, the 24 -L-tetrominos of the tiling contain at least 48 red squares. This leavesat most one red square for the square S. But then an L-tetromino thatshares three white squares with S contains at most one red square,which contradicts our assumption. Thus, 50 red squares is the mini-mum. ■

22.4. (A). First Solution. The shape of such a triangle is uniquelydetermined by the number of sides of the regular 11-gon P betweenthe vertices of the inscribed triangle (Fig. 22.4.1), i.e., by the numberof the integral partitions of 11 into the sum a + b + c ¼ 11;0 < a � b � c.

For a ¼ 1, we can choose b ¼ 1, 2, 3, 4, 5, and so we get fivedistinct partitions;

For a ¼ 2, we can choose b ¼ 2, 3, 4, and we get three distinctpartitions;

For a ¼ 3, we can choose b ¼ 3, 4, and thus get two partitions.

Fig. 22.3.1


Thus, there are total of 10 non-congruent shapes of inscribed tri-angles. Should the vertices of the 11-gon P be colored red and white,we would get at least six vertices of the same color, say red, which

gives us 6

3

� �¼ 20 distinct monochromatic red triangles. Since

20 > 10, we get at least two congruent all-red triangles.Observe, we have proven a bit more: for each shape of a triangle,

there are two monochromatic triangles of this shape, or there are atleast three congruent monochromatic triangles. ■

22.4. (A). Second Solution. As in the previous solution, we have20 distinct all-red triangles. We add to these triangles their imagesunder rotations about the center of the 11-gon P that map P onto P, forthe total of 220 triangles, which we will call rotated triangles. The

total number of the inscribed triangles is 11

3

� �¼ 165. Since

220 > 165, two of the rotated triangles coincide, but this means thatfor these two red monochromatic triangles, one triangle can be rotatedinto another, thus they are congruent.

Observe, we have proven a bit more: one of the two monochro-matic congruent triangles can be rotated to coincide with another. ■

22.4.(B). The two methods have worked to solve problem 22.4(A).However, the first method fails here if one uses it without special care,

Fig. 22.4.1. This inscribed triangle is of type 2 + 3 + 6

Solutions 22 37

for 1003

10

� �< 2004

9

� �. However, Hannah Alpert, a sophomore from

Poudre High School, managed to make it work by reducing the

number of types of triangles down from 2004

9

� �. The second method

works easily, as the following argument demonstrates.Two-color red and white the 2005 vertices of the given polygon P.

At least 1003 vertices must be of the same color, say red, thus we get1003

10

� �red monochromatic 10-gons. We add to these 10-gons their

images under rotations about the center of P that map P onto P, for

the total of 2005� 1003

10

� �10-gons, which we will call rotated

10-gons. The total number of the inscribed 10-gons is 2005

10

� �. Since

2005� 1003

10

� �> 2005

10

� �, two of the rotated 10-gons must coincide,

but this means that for this pair of the red monochromatic 10-gons,one 10-gon can be rotated into another, thus they are congruent.

Observe: we have proven a bit more: one of the two monochromaticcongruent 10-gons can be rotated to coincide with another. ■

In Further Exploration E26 you will find a much stronger resultthan problem 22.4.(B) asked for. It was found during the Olympiad bythe winner Mark Heim!

22.5. (A). First Solution. Let us prove that in a DNA satisfying thetwo given conditions, there is a gene that occurs only once. Indeed, letus assume that each gene appears at least twice. For each gene selectits first two appearances from the left and call them a pair. The firstgene from the left α is in the first pair. This pair ααmust be separated,thus the pair of the second gene β from the left is nestled inside thefirst pair, for otherwise we would get a forbidden sequence

α. . .β. . .α. . .β. The second pair must be separated, and thus the pairof the third gene from the left must be nestled inside the second pair,etc. As there are finitely many genes, we end up with a pair of genes(nestled inside all other pairs) that is not separated, a contradiction.

We will now prove by mathematical induction on the number n ofgenes that the DNA that satisfies the two conditions and uses n genesis at most 2n � 1 gene long. For n ¼ 1 the statement is true, as thelongest DNA is 2 � 1 ¼ 1 gene long.


Assume that a DNA that satisfies the required conditions and usesn genes is at most 2n � 1 gene long. Now let S be a DNA sequencethat satisfies the two conditions and uses n + 1 genes; we need toprove that it is at most 2(n + 1) � 1 ¼ 2n + 1 gene long.

In the first paragraph of our solution, we proved that there a geneg that occurs only once in S; we throw it away. The only violation thatthis throwing may create is making two copies of another geneadjacent—if so, we throw one of them away too. We get the sequenceS0 that uses only n genes. By the inductive assumption, S0 is at most2n� 1 gene long. But S is at most two genes longer than S0, i.e., S is atmost 2n + 1 gene long. The induction is complete.

We need to demonstrate that a DNA of length of 2n� 1 on n genesis attainable. But this is easy: just pick the following sequence 1, 2,. . ., n � 1, n, n � 1, . . ., 2, 1. ■

22.5. (A). Second Solution. We will prove by mathematical induc-tion on n that the DNA that satisfies the problem conditions and usesn genes is at most 2n� 1 gene long. For n¼ 1 the statement is true, aslongest DNA is 2 � 1 ¼ 1 gene long.

Assume that for any positive integer k, k < n, a DNA that satisfiesthe conditions and uses k genes, is at most 2k � 1 gene long. Now letS be the longest DNA sequence that satisfies the problem conditionsand uses n genes. We need to prove that S is at most 2n� 1 gene long.

Let the first gene of S be 1, then the last term must be 1 as well, forotherwise we can make S longer by adding a 1 at the end. Indeed,assume that the added 1 has created a forbidden DNA. This meansthat we now have a subsequence a, . . ., 1, . . ., a, . . ., 1 (with the added1 at the end); but then the original DNA that started with 1, alreadyhad the forbidden subsequence 1 . . . a . . . 1 . . . a.

Let us consider two cases.Case 1. If there are no more 1’s in the DNA, we throw away the

first 1 and the last 1, and we get a sequence S0 that uses n � 1 genes(no more 1’s). By the inductive assumption, S0 is at most 2n� 1 geneslong. But S is two genes longer than S0, i.e., S is at most 2n + 1genes long.

Case 2. Assume now that there is a 1 between the first 1 and the last 1.The DNA then looks as follows: 1, S0, 1, S00, 1. Observe that if a genem appears in the sequence S0, it may not appear in the sequence S00,for this would create the prohibited subsequence 1 . . . m . . . 1 . . . m.

Solutions 22 39

Let the sequence 1, S0, 1 use n0 genes and the sequence 1, S00, 1 use n00

genes. Obviously, n0 + n00 � 1 ¼ n (we subtract 1 in the left sidebecause we counted the middle gene 1 in each of the two subse-quences). By the inductive assumption, the lengths of the sequences1, S0, 1 and 1, S00, 1 are at most 2n0 � 1 and 2n00 � 1 respectively.Therefore, the length of S is (2n0 � 1) + (2n00 � 1) � 1 (we subtract1 because the gene 1 between S0 and S00 has been counted twice). But(2n0 � 1) + (2n00 � 1)� 1¼ 2(n0 + n00)� 3¼ 2(n + 1)� 3¼ 2n� 1 asdesired. The induction is complete.

This proof allows us to find a richer set of examples of DNAs oflength of 2n � 1 (and even describe all such examples if necessary).

For example:

1, 2, . . ., k, k + 1, k, k + 2, k, . . ., k, 2005, k, k� 1, k� 2, . . ., 2, 1.■

22.5.(B). Assume S is the longest DNA string satisfying the problemconditions. Partition S into blocks of three terms starting from the left(the last block may be incomplete and have fewer than three terms, ofcourse). We will call a block extreme if a gene from the given set ofgenes {1, 2, . . ., 2005} appears in this block for the first or the lasttime. There are at most 2 � 2005 extreme blocks.

We claim that there are no complete (i.e., 3-gene) non-extremeblocks.

Indeed, assume the block B, which consists of genes α, β, γ in someorder, is not extreme (in the original DNA string S these three genesdo not have to be consecutive). This means that the genes α, β, γ eachappears at least once before and at least once after appearing in B. Wewill prove that then the DNA would contain the forbidden subse-quence of the type σ, τ, ω, σ, τ, ω. Let A denote the ordered triple ofthe first appearances of α, β, γ (these three genes may very well comefrom distinct 3-blocks). Without loss of generality we can assume thatin A the genes α, β, γ appear in this order. Let C denote the orderedtriple of the last appearances of α, β, γ in some order. Let us look at the9-term subsequence ABC and consider three cases, depending uponwhere α appears in the block B.

Case 1. If α is the first gene in B (Fig. 22.5.1), then we can choose βalso in B and γ in C to form α, β, γ which with α, β, γ from A gives usthe forbidden sequence α, β, γ α, β, γ.


Case 2. Let α be the second gene in B (Fig. 22.5.2). If β follows α inB, then with γ from C we get α, β, γ, which with α, β, γ fromA produces the forbidden sequence α, β, γ α, β, γ. Thus, β mustprecede α in B. If the order of the genes β, γ in C is β, γ, then wecan combine an α from B with this β, γ to form α, β, γ, which withα, β, γ from A gives us the forbidden α, β, γ α, β, γ. Thus, the order inCmust be γ, β. Now we can choose α, γ fromA followed by β, α fromB, followed by γ, β from C to get α, γ, β α, γ, β, which is forbidden.

Case 3. Let α be the third gene in B (Fig. 22.5.3), and is thuspreceded by β in B. If the order in C is β, γ, then we get α, β, γ fromA followed by α from B and β, γ from C to get the forbiddenα, β, γ α, β, γ. Thus, the order in C must be γ, β, and we choose α, γfrom A, followed by β, α from B, and followed by γ, β from C to formthe forbidden α, γ, β, α, γ, β.

a b g

A

a

B C

b g b

Fig. 22.5.3

a b g

A

a

B C

b bg

Fig. 22.5.2

Fig. 22.5.1

Solutions 22 41

We are done, for the DNA sequence consists of at most 2 � 2005extreme 3-blocks plus perhaps an incomplete block of at most twogenes—or 12,032 genes at the most.

For a much stronger bound, obtained by Martin Klazar of Prague,read Further Exploration E23. ■


Twenty-Third Colorado MathematicalOlympiadApril 21, 2006

Historical Notes 23

Sam Elder of Fort Collins solves “The Famous Five”

A few days before the Olympiad, our major newspaper The Gazetteran an article about the Olympiad. Here it is for you.

Prof gets lyrical about math essay contest

By Brian Newsome, The Gazettehttp://gazette.com/prof-gets-lyrical-about-math-essay-contest/arti

cle/18325April 18, 2006—12:00 a.m.


43

http://gazette.com/prof-gets-lyrical-about-math-essay-contest/article/18325

http://gazette.com/prof-gets-lyrical-about-math-essay-contest/article/18325

UCCS professor Alex Soifer began the UCCS Mathematical Olympiad in 1984and has coordinated it annually since then. The essay-style math contest is thelargest of its kind in the nation. Photo by Mark Reis, The Gazette

Alex Soifer describes math in the same words some peoplewould use to describe Romantic literature: beautiful, elegant,exciting and surprising.

The 57-year-old math professor at the University of Coloradoat Colorado Springs fondly recalls his undergraduate days whenstudents and faculty would spend hours after class working onmath problems. That, he says, was when the real learning began.It’s this love of number-riddled proofs that drove Soifer to startthe UCCS Mathematical Olympiad in 1984 for middle and highschool students. Despite the math scholar’s lengthy list of aca-demic accomplishments — Soifer has been a visiting professorat Princeton University and has published four books — theproblem-solving showdown remains the “most important thingin my life,” he said.

This year’s competition will be held Friday. About 900 studentswill spend four hours sweating through five complex problems fora chance to win prizes, scholarships and intellectual braggingrights.

It’s the largest essay-style math contest in the country, Soifersaid, and he is the chief author of the questions. Students earn

44 Twenty-Third Colorado Mathematical Olympiad

more points for the creativity shown in their work than getting acorrect answer.

In a disheveled office brimming with books and papers, theprofessor spouted stories about past Olympiad winners. Therewas Mark Heim of Loveland, who went to study math and com-puters at the Massachusetts Institute of Technology this year.Gideon Yaffe of the Colorado Springs School took first place in1988. He went to Harvard, where he studied math and drama, andnow teaches philosophy at the University of California atBerkeley.

Then there was Matthew Kahle, the C-student from Air Acad-

emy who was denied admission to UCCS. He’s finishing workfor a doctorate at the University of Washington in Seattle.

The Olympiad will begin at 9 a.m. Friday at UCCS. About100 to 150 awards will be presented April 28, also at UCCS. Thisyear’s special guest will be Hideshi Fukaya, the lead inventor ofthe world’s first graphing calculator. He will present winnerswith one of the CASIO calculators that he designed, according toa UCCS news release.

The old e-mail reminded me the particulars of this Olympiad year.Here is a clip from my April 25, 2006, e-mail to the author of thearticle above, Brian Newsome:

Hello Brian,

The Olympiad took place last Friday the 21st. It was quite atrying experience for me, as last Tue my kidney stone started amove. I spent Tuesday at Memorial [Hospital], Wednesday atPenrose [Hospital] for general anesthesia surgery, and Thursdaynight-early Friday another general anesthesia surgery at Penrose.Somehow I showed up at UCCS a few hours after my 2 a.m.release from the hospital, and adrenaline led me through.

Thursday night, before the surgery, I called Bob Ewell and askedhim to run the Olympiad if I am unable to show up by 7 a.m. on theOlympiad day. The night before the Award Presentation CeremoniesI spent yet another night at Memorial Hospital. Released in the earlymorning, I was in time at the Antler’s Hotel to pick up our distin-guished speaker Fykaya Hideshi of CASIO and drive us to theOlympiad Award Presentation. As they say, what does not break us,


makes us stronger. True, but it also made my family weaker, but thatis another story for another book.

Twenty-Third Colorado Mathematical Olympiad (CMO-2006)took place on April 21, 2006. It brought together some 450 middleand high school students from all over Colorado: Denver, Calhan,Grand Junction, Rangely, Aurora, Bailey, Manitou Springs, Cheraw,Parker, Highlands Ranch, Calhan, Boulder, Englewood, Loveland,Falcon, Woodland Park, Monument, United States Air Force Acad-emy, and Colorado Springs.

The judges awarded first prize to Sam Elder, a sophomore atPoudre High School of Fort Collins. He received a gold medal of

the Olympiad, a $1200 scholarship to be used at any certified Amer-ican university or 4-year college, a $1000 UCCS Chancellor’s Schol-arship for CMO Medalists, CASIO “Super-Calculator” Classpad300, CASIO Watch, a complete set of four books by AlexanderSoifer, and a major Wolfram Research’s software. Sam’s mathemat-ics teacher Rita Winski received a Texas Instruments software pack-age. For the first time in 23-year long history of the Olympiad, thewinner solved completely “The Famous Five” problems of the Olym-piad. This was Sam’s first victory but he was second last year behindthe three-time winner Mark Heim.


Hannah Alpert receives Second Prize from UCCS Chancellor Pam Shockley-Zalabak and Alexander Soifer. Photo by Silva Chang, Hannah’s mother

Second prize was awarded to Hannah Alpert, a junior at FairviewHigh School in Boulder. She received a silver medal of the Olympiad,a $600 scholarship to be used at any certified American university or4-year college, a $1000 UCCS Chancellor’s Scholarship for CMOMedalists, CASIO “Super-Calculator”, CASIOWatch, a complete setof four books by Alexander Soifer, and a major Wolfram Research’ssoftware. Hannah was a bronze medalist in the previous year.

Third prize was presented to Leonid Ovanesyan, a junior at SmokyHill High School. He received a bronze medal of the Olympiad, a$300 scholarship to be used at any certified American university or4-year college, a $1000 UCCS Chancellor’s Scholarship for CMOMedalists, CASIO “Super-Calculator”, CASIOWatch, a complete setof four books by Alexander Soifer, and a major Wolfram Research’ssoftware.

Fourth prize was presented to Daniel Pascua, a sophomore atLiberty High School. The judges also awarded 24 first honorablementions and 43 second honorable mentions. Literary Award waspresented to Tyler Henrie, a sophomore at Sand Creek High School.Art Award went to Emily Giles, a junior at Falcon High School.


Chess Tournament, by Emily Giles

The Prize Fund of the Olympiad was generously donated byCASIO, Inc.; Wolfram Research, Inc.; Texas Instruments, Inc.; AirAcademy School District 20; Colorado Springs School District 11;Harrison School District 2; Widefield School District 3; Chancellor’sOffice, Vice-Chancellor for Academic Affairs, Vice-Chancellor forAdministration and Finance, Vice-Chancellor for Student Success,Bookstore—all from the University of Colorado Colorado Springs;City of Colorado Springs; Sanborn Corporation; Jon & Susan Godla;Anthony & Tori Paniogue; Winston & Janelle Howe; and the Judgesof the Olympiad Matt Ewell, Carter Budwell, and Alexander Soifer.

The Award Presentation program featured a lecture Mathematicsand Technology by the celebrated co-inventor of the world’s firstgraphing calculator, Mr. Fukaya Hideshi, Director of Research andDevelopment at CASIO Corporation. Review of Solutions of the


Olympiad Problems and the lecture Coloring Proofs were presentedby Alexander Soifer.

Fukaya Hideshi, co-inventor of the world’s first graphing calculator and Direc-tor of Research & Development at CASIO Corporation. Photo by AlexanderSoifer

The following guests of honor, hosts and sponsors addressed thewinners and presented awards: Pamela Shockley, Chancellor, andThomas Christensen, Dean of the College of Letters, Arts and Sci-ences—both from the University of Colorado Colorado Springs;Maggie Lopez, Assistant Superintendent, Colorado Springs SchoolDistrict 20; and Alexander Soifer.

In early 2006, I invited, for the first time ever, both candidates forthe Governor of Colorado, the Democratic Nominee Bill Ritter,Jr. and the Republican Nominee Bob Bauprez, to address the winnersof the Olympiad. Bauprez did not reply, while Ritter sent a letter,which I am reproducing here. On November 7, 2006, Bill Ritterbecame the new Governor of Colorado. Did Ritter win because hecared to greet our talented Olympians?


Following the Olympiad, in the summer 2006, I gave a related talk“Imagining the Real, Realizing Imaginary” at the Congress of theWorld Federation of National Mathematics Competitions (WFNMC)at the University of Cambridge, England. There I was also presentedwith the Paul Erdos Award by the WFNMC President PetarKenderov. History possess its unique sense of humor: in July 2016,as the current President of WFNMC, I presented the Paul ErdosAward to Petar Kenderov of Bulgaria!

Petar Kenderov presents the Paul Erdos Award to Alexander Soifer, CambridgeUniversity, July 22, 2006


Problems 23

23.1. On a Collision Course (G. Galperin). Ninety three identicalballs move along a line, 59 of them move from left to right with speedv; the remaining 34 balls move from right to left toward the first groupof balls with speed w. When two balls collide, they exchange theirspeeds and direction of motion. What is the total number of collisionsthat will occur?

23.2. A horse! A horse! My kingdom for a horse! (A. Soifer). TheGood, the Bad, and the Ugly divide a pile of gold and a horse. The pileconsists of 2006 gold coins, and they draw in turn 1, 2 or 3 coins fromthe pile. The Good gets the first turn, the Bad draws second, and theUgly takes last. The Ugly does not trust the Bad and never draws thesame number of coins as the Bad has drawn immediately before him.The one who takes the last coin is left behind, while the two otherscross the prairie together on horseback. Who can guarantee himselfthe ride out on the horse regardless of how the others draw coins?How can he do this?

23.3. Chess Tournament. Professional and amateur players, n in all,participated in a round robin chess tournament. Upon its completion,it was observed that each player earned half of his total score in games

against the amateurs. Prove thatffiffiffin

pis an integer.

Round robin is a tournament where every pair of participants playseach other once. A winner of a game earns 1 point; the loser, 0; a drawgives ½ point to each player.

23.4. The Denver Mint. The Denver Mint manufactured 2006 silverdollars. As a quality control procedure, every pair of consecutivelymade coins was weighed against each other, and the difference inweight was less than 1/100 of an ounce. Prove that the coins can bepartitioned into two 1003-coin groups in such a way that the totalgroup weights differ by less than 1/100 of an ounce.

23.5. Math Party (A. Soifer). We say that mathematicians a andb have each other’s number n if there is a coauthor-chain of mathe-maticians a ¼ a0, a1, . . ., an ¼ b such that every consecutive pair inthe chain are coauthors on at least one publication.

A party is attended by at least three mathematicians, and every pairof attendees has a coauthor-chain at the party. Prove that all


mathematicians can be seated at a round table in such a way that everypair of neighbors has each other’s number not exceeding 3.

Solutions 23

23.1. Instead of collisions, we can think of the balls passing througheach other, much like two teams lining up after a game to shakehands1. Thus, the total number of encounters is 34 � 59 ¼ 2006. ■

23.2. Observe that 2006 ¼ 6k + 2 for some integer k. The Good startsby taking 1 coin. Since the Ugly will not draw the same as the Bad,the number of coins drawn by the Bad and the Ugly combined in anyround add up to 3, 4, or 5. Thus, by always drawing the differencebetween 6 and the sum of the previous draws by the Bad and the Ugly,the Good will eventually bring the number of coins to 1 after his turn.Thus, the Bad will pick up the last coin, and the Good and the Uglywill ride off together.

You must have noticed my use of the classics: the title “A horse! Ahorse! My kingdom for a horse!” comes from the drama “Richard III”by William Shakespeare. “The Good, the Bad, and the Ugly” is thetitle of the best “Spaghetti Western” directed by Sergio Leone. ■

23.3. Let the number of professionals be p and the number of ama-teurs a. We have n ¼ p + a. Since every game adds 1 to the totalscore of all participants of a group, the total number of points earned

by the amateurs against the amateurs is a2

� �¼ a2 � að Þ=2. This is

half of the total score of all amateurs, therefore the amateursscored the same total (a2� a)/2 against the professionals. Similarly,the professionals scored against the amateurs the total of

p2

� �¼ p2 � pð Þ=2 points. Thus, the total of the scores in games of

the professionals against the amateurs is (a2� a)/2 + (p2� p)/2.Since the total score in games of the professionals against the

amateurs is equal the number of games (1 point per game), we get

1The handshaking analogy was observed by Dr. Ewell.

Solutions 23 53

the total score of pa. As we have computed the same quantity in twoways, we get the equality

pa ¼ a2 � a� �

=2þ p2 � p� �

=2;

i:e:, p2 � 2paþ a2 ¼ pþ a;

hence, n ¼ p + a ¼ (p � a)2, i.e.,ffiffiffin

p ¼ p� a is an integer. ■

Robert Ewell discovered an interesting classification of solutionsp and a of problem 23.3. Following is his observation.

Theorem (Robert Ewell). If p and a are integers such that (p � a)2 ¼p + a, then p and a are consecutive triangular numbers [in a certainorder, A.S.].

We first prove a required tool.

Tool. Every odd perfect square is 8 � (triangular number) + 1.Let T(n) be the triangular number function:

T(1) ¼ 1

T(2) ¼ 3. . .T(n) ¼ (n2 + n)/2

Let n ¼ 2k + 1 be an odd integer. Then n2 ¼ 4k2 + 4k + 1.That is, every odd perfect square is of the form 4k2 + 4k + 1. But4k2 + 4k + 1 ¼ 8[(k2 + k)/2] + 1 ¼ 8T(k) + 1. ■

Now we are ready to prove the theorem.

If (p � a)2 ¼ p + a, then p2 � 2pa + a2 � p � a ¼ 0. This is aquadratic equation in p:

p2 � 2aþ 1ð Þpþ a2 � a� � ¼ 0:

Solving it, we get

p ¼ ½ 2aþ 1� ffiffiffiffiffiffiffiffiffiffiffiffiffiffi8aþ 1

p� �:


Since a and p are integers, 8a + 1 must be a perfect square. Since8a + 1 is an odd perfect square, due to the tool, it must be of the form8 � (triangular number) + 1. Hence a must be a triangular number:

a ¼ T(n) ¼ (n2 + n)/2 for some integer n. Thenffiffiffiffiffiffiffiffiffiffiffiffiffiffi8aþ 1

p ¼ 2nþ 1,and we get two solutions of the quadratic equation in p:

p1 ¼ ½ n2 þ nþ 1þ 2nþ 1ð Þ� � ¼ ½ n2 þ 3nþ 2� �

¼ ½ nþ 2ð Þ nþ 1ð Þ ¼ T nþ 1ð Þ;p2 ¼ ½ n2 þ nþ 1� 2nþ 1ð Þð Þ ¼ ½ n2 � nð Þ

¼ ½n n� 1ð Þ ¼ T n� 1ð Þ;

whereas a ¼ T(n). ■

23.4. Denote the coins and their weights in the order of manufacture

by the same symbols c1, c2, . . ., c2006. Put all odd-numbered coins intothe first group, G1; its total weight is Cstart ¼ c1 + c3 + . . . + c2005; theweight of the second groupG2 is, of course,Cend¼ c2 + c4 + . . . + c2006.Let D denote the total weight of all 2006 coins. If the difference in thegroup weights is under 1/100 of an ounce, we are done. Otherwise,assume that the difference in the group weights is greater than 1/100,and let the first group be lighter than the second. We start thefollowing substitution process in G1:

1) Replace c1 with c2;2) Replace c3 with c4;. . . . . . . . . . . . . . . . .1003) Replace c2005 with c2006.

At every stage i of this process, the weight Ci of G1 changes by less

than 1/100 (per given quality control information). The weight of G1

starts at Cstart and ends at Cend; in between there is a step at which theweight Ci of G1 increases from being under D/2 to Ci+1 which isabove it. Since Ci+1 � Ci < 1/100, one of Ci, Ci+1 differs from D/2 byless than 1/200 (Fig. 23.4).

Solutions 23 55

This weight, Ci or Ci+1, determines the stage in our substitutionprocess (i or i + 1) that produces the required G1, which differs fromG2 by less than 2(1/200) ¼ 1/100. ■

23.5. Select a vertex (point) on the plane for each mathematician atthe party and connect each pair of coauthors by an edge. We get theCoauthors Graph G. Since for every pair of mathematicians there is acoauthor-chain in G connecting them, G is a connected graph. Wenow create the graph G3 on the same set of vertices as G byconnecting every pair of vertices, corresponding to mathematicianswho have each other’s number not exceeding 3. We need to proveonly that G3 contains a so-calledHamiltonian cycle, i.e., a closed paththat includes each of the vertices of G3 and exactly once. A graphcontaining a Hamiltonian cycle is called Hamiltonian.

A graph is called Hamiltonian-connected if for every two points v,w it has a path starting at v and ending at w that goes through eachvertex of the graph and exactly once.

For our graph G we can easily choose a spanning tree T, i.e., aconnected subgraph with the same vertex set, but without cycles (juststart throwing away one edge at a time from cycles until there will be nocycles left). For this Twe can construct T3 (just like we constructedG3).Surely, if T3 is Hamiltonian-connected, so is G3. Furthermore, if G3 is

Hamiltonian-connected, it is Hamiltonian (just take a pair v, w ofadjacent points; they are connected by a path P starting at v and endingat w and going through each vertex once; now add the edge vw to P,and you get a Hamiltonian cycle). All there is left to prove is thefollowing:

For any tree T, T3 is Hamiltonian-connected.

Let us prove it by mathematical induction. There is just one treeT on 3 vertices, for it T3 is a triangle K3 and as such is Hamiltonian.

<1/100

D/2 Ci+1Ci

Fig. 23.4


Assume now that n> 3 and for any tree R on less than n vertices, R3 isHamiltonian-connected. Let now T be a tree on n vertices and v, w bea pair of vertices in T. We divide the proof of the inductive step intotwo cases.

Case 1. If v and w are adjacent in T, we remove the edge vw. We geta graph T � vwwhich consists of two (not connected with each other)trees, Tv that includes v and Tw that includes w. By the inductiveassumption, Tv

3 and Tw3 are Hamiltonian-connected. Let u be a vertex

in Tv that is adjacent to v, and x be a vertex in Tw adjacent to w, seeFig. 23.5.1. (If Tv or Tw consist of merely one point, we simply setu ¼ v or/and x ¼ w.) Since the edge-distance in T between u andx does not exceed 3, u and x are adjacent in T3.

Let Pv be a Hamiltonian v-u path in Tv3, and Pw be a Hamiltonian

w-x path in Tw3. Then the required Hamiltonian v-w path in T3 can be

obtained by combining Pv, the edge ux, and Pw.Case 2. Let now v and w be not adjacent in T. Since T is a tree, there

is a unique path going from v to w; denote the first vertex of this path(after v) by u—see Fig. 23.5.2. We remove the edge vu. We get agraph T � vu which consists of two (not connected with each other)trees: Tv that includes v and Tu that includes u. By inductive assump-tion, Tv

3 and Tu3 are Hamiltonian-connected. Let x be a vertex in Tv

that is adjacent to v, (If Tv consist of merely one vertex, we simply setx ¼ v). Since the edge-distance between x and u in T does not exceed2, T3 contains an edge xu.

v xwu

Fig. 23.5.1

Solutions 23 57

Let Pv be a Hamiltonian v-x path in Tv3, and Pw a Hamiltonian u-w

path in Tw3. Then the required Hamiltonian v-w path in T3 can be

obtained by combining Pv, the edge xu, and Pw.

Read Further Exploration E28 for a couple of related thoughts. ■

xv

w

u

Fig. 23.5.2


Twenty-Fourth Colorado MathematicalOlympiadApril 20, 2007

Historical Notes 24

Sam Elder and Hannah Alpert Shine Again

To run this year’s Olympiad, I traveled from Princeton, where I livedand worked at the time. It was an enjoyable time, full of meetingoriginal colleagues in mathematics, history, physics, economics, cin-ema, and other disciplines. The 1994 Nobel Laureate John F. Nash,Jr., aka “A Beautiful Mind” and I shared a dear friend, HaroldW. Kuhn. Harold wrote a triple essay “Economics of FrankP. Ramsey, John von Neumann, and John F. Nash” for “The Mathe-matical Coloring Book” I was finishing after 18 years of mathemat-ical and historical research.


59

With Harold W. Kuhn, Princeton, July 2007

With John F. Nash, Jr., Princeton, July 2007

60 Twenty-Fourth Colorado Mathematical Olympiad

Twenty-Fourth Colorado Mathematical Olympiad (CMO-2006)took place on April 20, 2007. It brought together some 420 middleand high school students from all over Colorado: Aurora, Bailey,Boulder, Calhan, Cheraw, Colorado Springs, Denver, Englewood,Falcon, Fort Collins, Florence, Greeley, Highlands Ranch, Littleton,Loveland, Manitou Springs, Monument, Parker, Pagosa Springs,Rangely, and the United States Air Force Academy. We also hadguest participants from the Alabama School of Mathematics andScience, Mobile, Alabama, who earned their trip to our Olympiadby winning their local mathematics competition.

The judges awarded first prize to Sam Elder, a junior from Poudre

High School of Fort Collins. He received a gold medal of the Olym-piad, a $1200 scholarship to be used at any certified Americanuniversity or 4-year college, a $1000 UCCS Chancellor’s Scholarshipfor CMO Medalists, CASIO Graphing Calculator, CASIO Watch,CASIO ClassPad 300, Wolfram Research’s Mathematica 5.2 soft-ware, andWolfram Research’s Crystal Star Necklace. This was Sam’ssecond victory; he was also second behind the winner Mark Heimin 2005.

Second prize was awarded to Hannah Alpert, a senior fromFairview High School of Boulder. She received a silver medal ofthe Olympiad, a $900 scholarship to be used at any certified Americanuniversity or 4-year college, a $1000 UCCS Chancellor’s Scholarshipfor CMO Medalists, CASIO Graphing Calculator, CASIO Watch,CASIO ClassPad 300, Wolfram Research’s Mathematica 5.2 soft-ware, and Wolfram Research’s necklace. Hannah was also a closesecond last year, and third the year before.

In view of a dramatic distance in success between the top twocontestants, Sam Elder and Hannah Alpert, and all other Olympians,the judges decided not to award a third prize this year. This was theonly year when such a decision was made. The judges did award6 fourth prizes, 13 first honorable mentions, and 49 second honorablementions.

The future of our top two winners promises to be spectacular.Following their undergraduate studies, both Hannah and Sam wereadmitted to a Ph.D. Program in Mathematics at the MassachusettsInstitute of Technology. Read more about Hannah and an essay byHannah in Part III of this book, “Olympic Reminiscences in FourMovements.”


The Art Award was presented to Emily Giles, a senior from FalconHigh School. I am reproducing her drawing here for you.

Secrets of Tables, by Emily Giles

This Year’s Prize Fund of the Olympiad was generously donatedby CASIO, Inc.; Wolfram Research, Inc.; Texas Instruments, Inc.;Air Academy School District 20; Colorado Springs School District


11; Harrison School District 2; Cotopaxi Schools; Irving MiddleSchool; Fremont School Dist. RE-2; Rangely High School;Ft. Collins High School; Erie Middle School; Chancellor, Vice-Chancellor for Academic Affairs, Vice-Chancellor for Administra-tion and Finance, Vice-Chancellor for Student Success, and Book-store—all from the University of Colorado Colorado Springs; City ofColorado Springs; and Alexander Soifer.

The Award Presentation program featured lectures One AmazingProblem and its Connections to . . . Everything: A Conversation inThree Parts, and Review of Solutions of the 24th Colorado Mathe-matical Olympiad Problems by Alexander Soifer.

The following guests of honor, hosts and sponsors addressed thewinners and presented awards: Pamela Shockley, Chancellor, andThomas Christensen, Dean of the College of Letters, Arts and Sci-ences—both from the University of Colorado Colorado Springs;Maggie Lopez, Assistant Superintendent, Colorado Springs SchoolDistrict 20; Mary Thurman, Deputy Superintendent, School District11; and Alexander Soifer.

Problems 24

24.1. Stone Age Entertainment Returns (A. Soifer). Fred Flintstoneand Barney Rubble in turn add pebbles to a pile. On each turn, theymust add at least one pebble and may not add more pebbles than thereare already in the pile. The player who makes the pile consist ofexactly 2007 pebbles wins. Find a strategy that allows Fred or Barneyto win regardless of how the other may play. There is originally justone pebble in the pile, and Fred goes first.

24.2. Secrets of Tables (A. Soifer). Each cell of an 8 � 8 chessboardis filled with a 0 or 1. Prove that if we compute sums of numbers ineach row, each column, and each of the two diagonals, we will get atleast three equal sums.

24.3. More Secrets of Tables.(A). Prove that no matter how each cell of a 5� 41 table is filled with

a 0 or 1, one can choose three rows and three columns whichintersect in nine cells filled with identical numbers.

(B). Prove that 41 in part (A) is the lowest possible; i.e., the statementis not true for a 5 � 40 table.

Problems 24 63

24.4. Looking for the Positive (A. Soifer). A number is placed ineach angle of a regular 2007-gon so that that the sum of any tenconsecutive numbers is positive. Prove that one can choose an anglewith the number a in it, such that when we label all 2007 numbersclockwise a ¼ a1, a2, . . ., a2007, each sum a1, a1 + a2, . . ., a1 +a2 + . . . + a2007 will be positive.

24.5. Natural Split (Lord Rayleigh). Prove that if a and b arepositive irrational numbers with 1/a + 1/b ¼ 1, then the two setsA¼ {b1ac, b2ac, . . .bnac, . . .}, and B¼ {b1bc, b2bc, . . . , bnbc, . . .}split the set N of positive integers.

A number a is called irrational if it cannot be presented in a forma ¼ p/q with p, q integers and q 6¼ 0. The symbol bcc stands for thelargest whole number not exceeding c. We say that sets A and B splitN if each positive integer n is in either A or B but not in both A and B.

Solutions 24

24.1. Fred wins by leaving after his move the following numbers ofpebbles in the pile:

2 ! 6 ! 14 ! 30 ! 61 ! 124 ! 250 ! 501 ! 1003 ! 2007.

Fred can do it because the rules allow him to go from leaving a pileof n pebbles after his move to leaving a pile of 2n + 1 or 2n + 2pebbles after his next move, no matter what (allowed) move Barneymakes. Indeed, Barney can change the pile from n pebbles to a pilewith the number of pebbles ranging from n + 1 up to 2n pebbles,which allows Fred to increase the pile to 2n + 1 or 2n + 2 pebbles,whichever of these two quantities Fred desires. ■

24.2. Let a “line” stand for a row, a column or a diagonal. There are18 lines, and nine options for a line’s sum: from 0 to 8. Therefore,there are three lines with equal sums or else each sum from 0 to8 appears as a sum of exactly two lines. Assume the latter. If adiagonal has the sum 8, i.e., consists of all 1s, we would not havetwo lines with the sum 0 (for only the other diagonal could still add upto 0). Therefore, the sum 8 is attained in a row or a column. Withoutloss of generality, let the sum of 8 appear in the column C1 (we arestarting to fill in cells in Fig. 24.1). Since every cell of C1 contains a


1, there is no row or a diagonal with the sum 0; therefore, we have twocolumns, say, columns C2 and C3 with the sums 0, i.e., every cell ofC2 and C3 is filled with a 0. This implies that the second sum of 8 isattained in a column, call it C4. Now neither the sum of 7 nor the sumof 1 can be obtained in a row or a diagonal since each row anddiagonal already has two 0s and two 1s in it. Therefore, the twolines with the sums of 7 must occur in columns, say, columns C5and C6, and the sum of 1 must be attained in columns C7 and C8.

Let the only 0s of columns C5 and C6 be in rows R1 and R2(R1 and R2 are not necessarily distinct); and the only 1s of thecolumns C7 and C8 be in the rows R3 and R4 (R3 and R4 are not

necessarily distinct, and not necessarily different from R1 and R2).We can easily see that the entries in the remaining (at least) four rowsare all identical, and therefore their sums are equal (see Fig. 24.1).This contradicts the assumption that every sum appears exactly twiceand proves that there are at least three lines with equal sums. ■

24.3. (A). There are 5

3

� �¼ 10 ways to place three 0s in a 5-cell

column, as there are ten ways to place three 1s. Thus, there are

20 ways of having three identical numbers in a column. Since weare given 41 columns, by the Pigeonhole Principle, there are threecolumns with the same three numbers identically arranged. Wechoose these three columns and the three rows that contain theidentical numbers. ■

C1 C2 C3 C4 C5 C6 C7 C8R1 1 0 0 1 0 1 0 0R2 1 0 0 1 1 0 0 0R3 1 0 0 1 1 1 1 0R4 1 0 0 1 1 1 0 1R5 1 0 0 1 1 1 0 0R6 1 0 0 1 1 1 0 0R7 1 0 0 1 1 1 0 0R8 1 0 0 1 1 1 0 0

Fig. 24.1

Solutions 24 65

24.3.(B). Fill the 5 � 40 table as follows: ten columns with distinctways of three 0s in each, while filling the remaining two cells in eachrow with 1s. Obtain the next ten columns by interchanging 0s and 1sin the first ten columns. Finally repeat each of the first 20 columns onemore time, for the total of 40 columns. Assume we can choose threerows and three columns which intersect in nine cells filled withidentical numbers. But then we must have had three columns withthe identical three numbers in the same positions, which contradictsour selection of the columns. ■

24.4. First Solution. Proof in two parts.1). Let us prove that the sum S of all 2007 given numbers is positive.

By permuting the summands cyclically, we get:

a1 þ a2 þ :::þ a2007 ¼ Sa2 þ a3 þ :::þ a1 ¼ S::::::::::::::::::::::::::::::::::a10 þ a11 þ :::þ a9 ¼ S

By adding together these ten equalities we get a positive sum inthe left-hand side (as the sum of the 2007 columns, each columnpositive due to being the sum of ten consecutive numbers). On theright we get 10S, which is thus positive, and so is S.

2). Consider the following array of 2007 � 2 ¼ 4014 sums, where weadd one summand at a time moving clockwise:

s1,1 ¼ a1s1,2 ¼ a1 + a2. . .. . .. . .. . .. . .. . .s1,2007 ¼ a1 + a2 + . . . + a2007s1,2008 ¼ a1 + a2 + . . . + a2007 + a1s1,2009 ¼ a1 + a2 + . . . + a2007 + a1 + a2. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .s1,4014 ¼ a1 + a2 + . . . + a2007 + a1 + a2 + . . . + a2007

Choose 1� k� 2007 such that the sum s1,k is the lowest among thefirst 2007 sums; if the same lowest sum appears more than once, wechoose s1,k with the largest k. This sum is in fact the lowest among all4014 sums because the latter 2007 sums are equal the former 2007sums plus S, where S is the sum of all given numbers (S> 0 as shownin part 1 of the proof). We are done because


0 < s1,k+1 � s1,k ¼ ak+10 < s1,k+2 � s1,k ¼ ak+1 + ak+2. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .0 < s1,k+2007 � s1,k ¼ ak+1 + ak+2 + . . . + a2007 + a1 + . . . + ak ■

24.4. Second Solution (by Bob Ewell). Call the sequence a(1),a(1) + a(2), a(1) + a(2) + a(3) + . . . + a(n) partial sums. For m � nlet us also introduce the notation s(m, n) as the partial sum from a(m)to a(n), recognizing that s(m, m) ¼ a(m).

To show that there exists a(1) such that all partial sums from a(1) all the way the around to a(2007) are positive, we need only showthere exists a(1) such that all partial sums from a(1) to a(9) arepositive. Thereafter, the fact that the sum of any ten consecutivenumbers is positive kicks in. For example, s(1, 12) ¼ s(1, 2) + s(3, 12). But s(3, 12) is positive because it is the sum of ten consecutivenumbers. Therefore, if s(1, 2) is positive, then s(1, 12) is positive, ands(1, 22) is positive, and so on.

So how can we show that there exists a sequence of nine consec-utive numbers with all partial sums positive? If the sum of every set often consecutive numbers is positive, then there is at least one positivenumber in every ten. Pick one such positive number and call it a(1).

Now let us start from a(1), moving clockwise computing partialsums along the way. If we can get all the way to s(1, 9) and all thepartial sums along the way are positive, then we are done.

Suppose we can’t. That is, at some point we come to s(1, i) that isnot positive. Now we work counterclockwise from a(1), calling thefirst element a(�1), the next a(�2), and so on until we get to the first a(�j) that makes s(�j, i) positive. We know that j � 10 � i since thesum of every ten consecutive elements is positive.

I contend that all partial sums s(�j, k) for k ¼ �j + 1 to i arepositive.

We know that s(�j, i) > 0. We also know that since a(�j) is thefirst element to make s(�j, i) positive, for all k < j that s(�k, i) � 0.All partial sums s(�j,�k) for 1� k< jmust be positive because s(�j,i) is positive, and s(�j, i)¼ s(�j,�k) + s(�k + 1, i). If s(�j,�k) is notpositive, we have two non-positive numbers adding to a positive. Inparticular s(�j, �1) is positive. For 1 � k � i � 1, we get s(�j, k) ¼s(�j, �1) + s(1, k). As the sum of two positive numbers, s(�j, k) is

Solutions 24 67

positive. We already know that s(�j, i) is positive; therefore, allpartial sums from a(�j) to a(i) are positive.

Furthermore, the string is now longer by at least one element,a(�j). The minimum length of the first string created in this processis three (a(�j), a(1), a(i), where j¼ 1 and i¼ 2). If i + j¼ 9 or 10, weare done. Otherwise, we repeat the process, choosing a(�j) as a(1).Each iteration will extend the string by at least two: a(�j) and a(i). Soafter a maximum of four runs through this process, we will have astring of nine positive partial sums. ■

24.5. This problem was published by the Canadian mathematicianSamuel Beatty (1881–1970) in 1926. The following solution comesfrom the 1954 book [YY] by the Russian twin brothers, mathemati-cians Akiva M. Yaglom and Isak M. Yaglom.

Proof. Given positive irrational numbers a and b such that 1/a + 1/b ¼ 1. Let S be the set of all numbers na and nb, n ¼ 1, 2, . . . If wewere to show that

for any positive integer P, exactly one element of S lies

between P� 1 and P,ð1Þ

the problem would be solved: the integer parts bnac , bnbc of elementsof S would take on values of each positive integer and exactly once.

We will prove (1) by showing that

there are exactly P� 1 elements of S that are < P: ð2ÞIndeed, assume (2) is proven. For P ¼ 1, (2) states that no element

of S is less than 1. For P¼ 2, (2) states that exactly one element of S isless than 2; it must be between 1 and 2. For P ¼ 3, (2) states thatexactly two elements of S are less than 3. Since one element was lessthan 2, the other must be between 2 and 3. Continuing this process

(formally: using mathematical induction) we establish (1), i.e.,exactly one element of S lies between P � 1 and P.

Thus, all we need to prove is (2). Observe that we have na < P for

n ¼ 1, 2, . . . , Pa

� �, and similarly nb < P for n ¼ 1, 2, . . . , P

b

� �. Hence,

since na 6¼mb for any positive integersm, n (for otherwise a¼ (m/n)band plugging (m/n)b in place of a in the given equality 1/a + 1/b ¼ 1forces b ¼ (m + n)/m to be rational), the number of elements of S less


than P is Pa

� �þ Pb

� �. Since the numbers P

a andPb are non-integers (they

are irrational), we observe that

P

a� 1 <

P

a

� �<

P

a;

P

b� 1 <

P

b

� �<

P

b:

By adding these two inequalities together, we get:

P1

aþ 1

b

� � 2 <

P

a

� �þ P

b

� �< P

1

aþ 1

b

� :

Since we are given 1a þ 1

b ¼ 1, the above inequality simplifies to

P� 2 < Pa

� �þ Pb

� �< P. Since there is only one integer between

P � 2 and P, we conclude that Pa

� �þ Pb

� � ¼ P� 1. Thus there are

P � 1 elements of S less than P, and so (2) is proven.See Further Exploration E24 for more mathematics and history!■

Solutions 24 69

Twenty-Fifth Colorado MathematicalOlympiadApril 18, 2008

Historical Notes 25

The Colorado Mathematical Olympiad Celebrates a Quarter aCentury!

In February 2008, to my own disbelief I finished The MathematicalColoring Book. 18 years of researching mathematics and history,writing for the first time the history of Ramsey Theory as evolutionof ideas with biographies of the creators of those ideas, running intodead ends, enduring personal losses, undertaking four moves fromColorado Springs to Princeton and back. All of a sudden this sweat-tears-blood had suddenly ended—the book was born and ready to liveher own, grown-up life, quite separately from me. During the finalyears of writing, I refused all invitations for joint work and travel inorder to give all of myself to finally finish this book-of-my-life. I

informed my Greek colleagues that at long last I was willing to accepttheir old offer to bring me to the Land of Pythagoras and Euclid togive lectures in Thessaloniki, Athens, and Corfu Island. What betterplace to celebrate birth of my book than on this Ancient Land thatgave birth to Mathematics herself! As a homage to the Land ofPythagoras, I named the first problem of the 2008 Olympiad TheCorfu Test, for Corfu Island, with its deep blue water caressing thecliffs, became my favorite place in Greece.


71

Thessaloniki Beauty 128–138 A.D.; photo by Alexander Soifer

Upon my return from Greece, I dove into creating problems and

preparing the Olympiad. The Colorado Mathematical Olympiadreached an historic landmark: a Quarter a Century! Twenty-FifthColorado Mathematical Olympiad (CMO-2008) took place on April18, 2008. It brought together some 400 middle and high schoolstudents from all over Colorado: Aspen, Aurora, Bailey, Boulder,Branson, Calhan, Canon City, Colorado Springs, Dacono, Denver,Englewood, Ellicott, Falcon, Fort Collins, Fort Lupton, Littleton,Longmont, Loveland, Monument, Parker, Pueblo, Rangeley, Secu-rity, U.S. Air Force Academy, Woodland Park. We hosted partici-pants from Hudson, Ohio, and Los Altos, California. We also hadguest participants from the Alabama School of Mathematics andScience, Mobile, Alabama, who earned their trip to our Olympiadby winning their local mathematics competition.

The judges awarded first prize to Marshall Carpenter, a senior atFairview High School of Boulder. He received a gold medal of the

72 Twenty-Fifth Colorado Mathematical Olympiad

Olympiad, a $1000 scholarship to be used at any certified Americanuniversity or 4-year college, a $1000 UCCS Chancellor’s Scholarshipfor CMO Medalists, CASIO or Texas Instruments Graphing Calcula-tor, CASIO Watch, Wolfram Research’s Mathematica 5.2 (hardcopy) and the new, not yet released Mathematica 6.0 (soft copy)software, Wolfram Research’s Crystal Star Necklace, and the books”Colorado Mathematical Olympiad: The First Ten Years and FurtherExplorations” by Alexander Soifer and “Geometric Etudes in Com-binatorial Mathematics” by Vladimir G. Boltyanski and Soifer.

Second prize was awarded to Daniel Pascua, a senior at LibertyHigh School. He received a silver medal of the Olympiad, a $1000

scholarship to be used at any certified American university or 4-yearcollege, a $1000 UCCS Chancellor’s Scholarship for CMO Medal-ists, CASIO or Texas Instruments Graphing Calculator, CASIOWatch, Wolfram Research’s Mathematica 5.2 (hard copy) and thenew, not yet released Mathematica 6.0 (soft copy) software, WolframResearch’s necklace, and the books ”Colorado Mathematical Olym-piad: The First Ten Years and Further Explorations” and “GeometricEtudes in Combinatorial Mathematics.”

Third prizewas presented to BenAlpert, a freshman at FairviewHighSchool, and Ryan Bethe, a sophomore at Poudre High School. Each ofthem received a bronzemedal of theOlympiad, a $250 scholarship to beused at any certified American university or 4-year college, a $1000UCCS Chancellor’s Scholarship for CMO Medalists, CASIO or TexasInstruments Graphing Calculator, CASIO Watch, Wolfram Research’sMathematica 5.2 (hard copy) and the new, not yet releasedMathematica6.0 (soft copy) software,WolframResearch’s necklace, and the books ”Colorado Mathematical Olympiad: The First Ten Years and FurtherExplorations” and “Geometric Etudes in Combinatorial Mathematics.”Did you recognize the “Alpert” name?Yes, Benwas a kid brother of ourcelebrated winner Hannah Alpert.


Literary Awards were presented to my daughter Isabelle SoulaySoifer, a junior at Los Altos High School, Los Altos, California, andChelsea Gardner, a junior at Ponderosa High School. Our problem“The Old Glory Returns” inspired Isabelle to capture hardship of ayoung soldier in a war. No doubt, this powerful poem expressedIsabelle’s concern about the American wars in Afghanistan and


Iraq. Rereading now this poem, written in a stroke of inspirationdurinig the Olympiad, in pensil, without any corrections, I am inawe at the masterpiece created the 16-year old Isabelle.

RENEWALby Isabelle Soulay Soifer

Night and day,Dusk till dawn,Young men creeping through the field,Brilliant azure uniforms, their golden pendants bright,Future so dim, no one to hear. . . ..Their screams, their shouts, their gasping breaths,

their mutters of prayer.No mother to nurture their young souls,no father to aid them in their scores. . . ..Mud cakes their rugged boots, sweat drips down

their nervous brows. . . ..This is his first time, engaging in this excuse of a war.

Voices echoing in his mind: stand up for your country! Fight,o young one, prove your worth. Serve this country that hashelped you since birth.Eyes become hazy, the sun seems not near,the world has left him, cold and in fear.Yet then, over the horizon, the image comes clear,a cloth on a stick, what a trick or the light!It is merely a cloth, of red, white, and blue hue.50 stars embracing the tattered sheet.Bullet holes seem to have stricken its surface.He shakes his head in disbelief.“What have you ever done for me! Your worth hasdiminished, my hope is in shards. I’ll never again feelthat cool breeze whip my face, nor the feeling of putting on aclean, fresh new shirt. Off with you, please, I’m letting outmy heart and soul.”Then suddenly, a voice entrances his mind, a cool, soothingvoice, yet of age and of wisdom.

“Remember this flag, this seemingly insignificanttarnish. The glory shines right before you, yet you choosenot to see. Awaken yourself, let imagination take flight.”And with that the soldier gazed up to the heavens, a single


clear teardrop streaming down his worn cheek. He stares at thegray landscape set out before him. A single word is uttered,yet is shaky, hesitant, as if saying it would leave him paralyzed,

“Glory. . . ..”

The Prize Fund of the Olympiad was generously donated byIntermap Technologies, Inc.; CASIO, Inc.; Wolfram Research;Texas Instruments; City of Colorado Springs; Air Academy District20; Colorado Springs School District 11; Harrison School District 2;Rangely High School; Fort Collins High School; Falcon District 49;Chancellor, Vice Chancellor for Academic Affairs, Vice Chancellorfor Student Success, Vice Chancellor for Administration & Finance,Bookstore, College of Letters, Arts, and Sciences, Office of CampusActivities—all from the University of Colorado Colorado Springs;and Alexander Soifer.

The Award Presentation program featured lectures Creators ofMathematical Coloring, and Review of Solutions of the 25th Colo-rado Mathematical Olympiad Problems by Alexander Soifer.

The following guests of honor, hosts and sponsors addressed thewinners and presented the awards: Pamela Shockley, Chancellor, andThomas Christensen, Dean of the College of Letters, Arts and Sci-ences—both from the University of Colorado Colorado Springs; GregHoffman, Director of Human Recourses, Intermap Technologies Inc.,Maggie Lopez, Assistant Superintendent, Colorado Springs SchoolDistrict 20; Mary Thurman, Deputy Superintendent, School District11; and Alexander Soifer.

Colorado Governor Bill Ritter sent us a letter, which I amreproducing here for you. The Governor’s letter was as concise as itwas insightful: “Mathematics is a critical part of our livelihood. Mathis truly the one universal language spoken by all. It is the disciplinethat bridges divides.”


Following the Olympiad, in July 2008, I attended the (quadrennial)International Congress on Mathematical Education in Monterrey,Mexico. I presented there four talks related to the Colorado Mathe-matical Olympiad: Building a Bridge: From Problems of Mathemat-ical Olympiads to Open Problems of Mathematics, Parts I, II, and III;and A Quarter a Century of Discovering & Inspiring Young GiftedMathematicians: All the Best from the Colorado MathematicalOlympiad.

At the General Meeting of the World Federation of NationalMathematics Competitions (also in Monterrey), I was elected to thepost of Senior Vice President and Chair of the Program Committee.


“Building a Bridge” talk in Monterrey, Mexico, 2008

Leaders of National Mathematical Olympiads from five Continents, Monterrey,Mexico, 2008


Problems 25

25.1. The Corfu Test (A. Soifer). The Greek Island of Corfu has2008 students. They all took a test, on which Takis and Ada made fivemistakes each, and everyone else made fewer mistakes. Prove thatthere are 402 students who made the same number of mistakes.

25.2. Indivisible Neighbors (A. Soifer).(A) You are putting each integer 1, 2, . . ., 20082 once into the cells of

a 2008 � 2008 square grid, one integer per cell, so that thedifference between any two neighbors is not divisible by n. Findthe smallest positive integer n for which this is possible todo. Neighbors are integers in cells that have at least one point incommon.

(B) Replace the “difference” by the “sum” and solve the sameproblem.

25.3. The Old Glory Returns (A. Soifer). A 2008 � 2008 flag wasshot in battle 2008 times. Prove that we can cut out of it two250 � 250 squares that have an equal number of bullet holes.

25.4. Map Coloring in the Year 2008 (E. B. Dynkin, S. A.Molchanov and A. L. Rozental’). Maps consisting of 2008 countriesare drawn on both sides of a 2008 � 2008 glass. A map on one side iscolored in 2008 colors, one country per color. We need to color themap on the other side with the same 2008 colors, one country percolor. As a result, some regions of the glass will be colored the samecolor on both sides; call them monochromatic. Can we always makethe combined area of all monochromatic regions at least 2008 byappropriately coloring the second side of the glass?

25.5. One Old Paul Erdos’ Problem (P. Erdos). Prove that theminimum number f(n) of distinct distances determined by n pointsin the plane satisfies the inequality

f nð Þ � n� 3

4

� �12

� 1

2:


Solutions 25

25.1. The 2006 students (2008 minus Takis and Ada), made numbersof mistakes taking on five values: 0, 1, 2, 3, 4. Since2006 ¼ 5 � 401 + 1, by the Pigeonhole Principle, there are at least402 students who made the same number of mistakes. ■

25.2. (A). Since the difference a � b is divisible by n if and only ifa and b produce the same remainders upon division by n, the problemis actually about coloring the 2008 � 2008 square grid into n colors(the remainders): 0, 1, . . ., n� 1 without neighboring squares coloredin the same color. Since at every node of the grid 4 neighboringsquares meet, n must be at least 4. To show that the required coloringis possible for n¼ 4, we just color the 4 unit squares of a 2� 2 squarein four colors, and translate this 2 � 2 square to tile the 2008 � 2008grid, thus determining the coloring of the entire given square grid(Fig. 25.2.1).

One way to achieve this coloring is simply to number the columnsstarting from the left edge of the given grid, and consecutively insertodd numbers into the odd-numbered columns, and even numbersconsecutively into the even-numbered columns. ■

25.2.(B). Since the sum a + b is divisible by n if and only if a andb produce ‘complementary’ remainders upon division by n, the prob-lem is actually about coloring the 2008 � 2008 square grid inton colors (the remainders): 0, 1, . . ., n� 1 without neighboring squarescolored in the complementary pairs of colors. We need to definecomplementary colors/remainders, but this is easy: complementary

1 3

2 4Fig. 25.2.1

Solutions 25 79

colors/remainders are those that add up to n (mod n).1 Of course, 0 iscomplementary to itself (which is fine with us).

Look at a 2 � 2 square in the given grid: its four cells must containat least two even or two odd numbers, which add up to a numberdivisible by n ¼ 2, thus proving that our desired n is at least 3.

Assume now that n ¼ 3 allows the desired coloring. Partition thegiven grid into 2� 2 squares. Each 2� 2 square cannot contain morethan one entry 0 (as it is self-complementary). Therefore, the totalnumber of 0s can be at most 25 % of the given grid. This creates acontradiction since about 1/3 of the original numbers are divisible by3. Thus, n must be at least 4.

For n ¼ 4 the desired coloring exists. We color the given grid intofour colors as follows: the cells of the first column from the left wecolor all in color 1; in the second column we alternate 0 and 2; thethird column is entirely colored 3; the fourth alternates 0 and 2; and soon, cyclically repeating colorings of the first four columns (seeFig. 25.2.2).

Here is one way to achieve this coloring. Number the columns fromthe left to the right. If a column number is congruent to 1 (mod 4), itgets the consecutive numbers congruent to 1 (mod 4), i.e., the firstcolumn would be 1, 5, 9, . . . If a column number is congruent to3 (mod 4), it gets the consecutive numbers congruent to 3 (mod 4),i.e., the third column would be 3, 7, 11, . . . The even-numberedcolumns would contain the consecutive even integers. ■

1We say that a is congruent to b (mod n) if a – b is divisible by n.

1

1

1

1

0 3

3

3

3

0

0 0

2 2

22

Fig. 25.2.2


25.3. Since 250 � 8 ¼ 2000 < 2008, we can partition the flag into8 � 8 ¼ 64 squares of the size 250 � 250 plus a narrow strip leftover(Fig. 25.3.1). If the numbers of bullet holes in the 64 squares were alldistinct, the least possible numbers of bullet holes would be 0, 1, . . .,63. However, 0 + 1 + . . . + 63¼ 64� 63/2¼ 2016, which contradictsthe condition that there are only 2008 in the whole flag. Therefore, thenumbers of bullet holes in our 64 squares are not all distinct. ■

25.4. Denote the colors by integers 1, 2, . . ., 2008, and denote by Rij

the area of the intersection of the regions colored with color i on thefirst side and to be colored with color j on the back side (i, j¼ 1, 2, . . .,2008). Now take a look at the following table:

R1,1 R2, 2 . . . R2008, 2008

R1, 2 R2,3 . . . R2008, 1

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

R1, k R2,k +1 . . . R2008, k�1

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

R1, 2008 R2, 1 . . . R2008, 2007

The sum of all the entries of the table is equal to the area of thewhole glass, which is 2008� 2008. Therefore, there is a row, say k-throw, with the combined area of at least 1/2008 of the total table,which is the area 2008. Observe: in the k-th row, the second subscriptis equal to the first subscript plus k � 1 (mod 2008). This tells us howto permute colors for coloring the back side of the glass to achieve thedesired result: in place of color k use color 1, in place of color k + 1use color 2, . . ., in place of color n use color [n� (k� 1)] (mod 2008).In the new color names, the k-th row will look like

R1,1 R2, 2 . . . R2008, 2008

each square is 250 ´ 250

Fig. 25.3.1

Solutions 25 81

which is exactly the sum of areas of intersecting regions of the samecolors! ■

25.5. Solution by Paul Erdos (I added reasoning at the end of Paul’sproof to clarify its last step). Let S be the set of n points that realizesthe minimum number f(n) of distinct distances among all n-elementsets in the plane. Let P1 be an arbitrary vertex of the convex hull H ofthe set S. The convex hull H of the set S is the minimal polygon thatcontains all points of S inside or on its boundary. (You can think of itas a result of sliding a rubber band around all points of S.) Denote byK the number of distinct distances among the distances P1Pi (i ¼ 2,3, . . ., n). If N is the maximum number of times the same distanceoccurs, then clearly

KN � n� 1:

Since f(n) � K, we get from the above inequality:

f nð Þ � K � n� 1

N: ð1Þ

If r is a distance that occurs N times, then among the n points of theset S there are N points on the circle with the center P1 and radius r.Moreover, these N points must all lie on the same semi-circle since P1is a vertex of the convex polygon H (Fig. 25.5.1). Denoting theseN points Q1, Q2, . . ., QN, we get N� 1 distinct distances Q1Q2, Q1Q3,. . ., Q1QN, and so we have another estimate of f(n):

f nð Þ � N � 1: ð2Þ

Fig. 25.5.1


Inequalities (1) and (2) show that f nð Þ � max N � 1; n�1N

� �, where

the right side reaches its minimum when N � 1 ¼ n�1N (this is because

N� 1 is increasing in N while n�1N is decreasing in N). Paul Erdos says

here “This yields” the result. Let us see how. The last equality impliesN2 � N � (n � 1) ¼ 0. This equality, viewed as a quadratic equation

in N, has a single positive solution N1 ¼ 1þ ffiffiffiffiffiffiffiffi4n�3

p2

¼ n� 34

� �12 þ 1

2.

Therefore, f nð Þ � N1 � 1 ¼ n� 34

� �12 � 1

2as desired.

Do read Further Exploration E25 for the history and mathematicsinspired by this classic Paul Erdos result. ■

Solutions 25 83

Twenty-Sixth Colorado MathematicalOlympiadApril 17, 2009

Historical Notes 26

Colorado Springs sophomore Alan Gardner wins gold

I have never been active in organized politics. I do not really appre-ciate giving up part of my freedom for the good of a party. Anothermatter is a good of my country, and so I published in New York Timesa few letters addressing presidential debates. In 2007, Anthony Lake,Barack Obama Campaign’s chief foreign policy adviser and formerNational Security Adviser for President Bill Clinton, visitedPrinceton University. We shared lunch and conversation. Tony said,“Barack Obama is the right person for this time to lead the country.”This is how I became a member of the Obama Campaign’s EducationPolicy Committee. “Yes-we-can!” was Barack’s inspiring call toaction. There was an incredible aura of hope in the air on November

4, 2008, the election day. The three generations of Soifers: Naomi(age 1), her father and my son Mark, and I knocked on doors, andwere universally met with a smile and excitement. On that day, one ofthe great presidents of the United States was born. The time will tellwhether my assessment is correct. Needless to say, the opinion is allmine, and does not represent the views of my publisher Springer.

With the Twenty-Sixth Colorado Mathematical Olympiad, westarted our second quarter of a century. As the whole Colorado wasbracing for a major snowstorm, I learned that the Olympiad was “on”only at 6 in the morning on the Olympiad’s day, April 17, 2009, whenthe Physical Plant Director Rob Doherty and Police Chief James


85

Spice called me with the good news that the campus of the Universityof Colorado at Colorado Springs would open on schedule after all.However, all Colorado Springs’ and many other schools of the stateclosed down their doors. This created a huge problem for studentswho wanted to come and compete: they were provided no buses andno accompanying teachers. Of course, local Colorado Springs stu-dents had an advantage of being closer to our campus, and yet only asmall number of them braved the weather. On the other hand, studentsfrom snowed-in Monument and such faraway places as Boulder, FortCollins, and even Rangely (the extreme North-Western corner ofColorado) made it to the Olympiad. We also had guest participants

from the Alabama School of Mathematics and Science, Mobile,Alabama, who earned their trip to our Olympiad by winning theirlocal mathematics competition.

Participating schools included Doherty High School, FairviewHigh School, Fort Collins High School, Lewis-Palmer High School,Liberty Common School, Mesa Ridge High School, Murphy HighSchool, Palmer High School, Poudre High School, Rangely HighSchool, Sand Creek High School, Sproul Middle School, St. Mary’sHigh School, Summit Middle Charter School, and the ClassicalAcademy.

As soon as Olympians completed their work by 1 p.m., UniversityVice Chancellor Brian Burnett issued orders to close the campusdown at 1:30 p.m. due to snow storm continuing to pile up whitesubstance on the streets and parking lots. However, Brian allowed theOlympiad judges to continue our work until 3 p.m. And so we workednon-stop until the job was done—at 5 p.m. I thank the judges forchoosing duty over fears of weather: Russel Shafer (1st prize winnerof 1st Olympiad in 1984, who came from Wyoming), Dr. RobertEwell, Dr. Ming Song, Matthew Ewell, Eric Conrad, ShaneHolloway, Jerry Klemm, Bill Hodson, Brian Peterson, Chris Kemp,Matthew Rixman, Lee Overman, Dr. Dale Peterson, Steve Sibert, BillYoung, and Todd Tomlinson. Margie Teals-Davis came from hersnowed in home in the mountains to organize registration and manyother aspects of the Olympiad.

The judges awarded first prize to Allan Gardner, a sophomore inthe mathematics class of Doug James, St. Mary’s Catholic HighSchool in Colorado Springs. In the 26 years of the Olympiad, thiswas the first victory by a student from a private school. Allan received

86 Twenty-Sixth Colorado Mathematical Olympiad

a gold medal of the Olympiad, a $1500 scholarship to be used at anycertified American university or four-year college, a $1000 UCCSChancellor’s Scholarship for CMO Medalists, “The MathematicalColoring Book: Mathematics of Coloring and the Colorful Life ofIts Creators”—the new 2009 book by Alexander Soifer, a CASIOwatch, and software “Mathematica for Students” by WolframResearch.

Second prize was awarded to David Wise, a sophomore at theClassical Academy, Colorado Springs. David received a silvermedal of the Olympiad, a $500 scholarship to be used at any certifiedAmerican university or four-year college, a $1000 UCCS Chancel-

lor’s Scholarship for CMO Medalists, “The Mathematical ColoringBook: Mathematics of Coloring and the Colorful Life of Its Creators,”a CASIO watch, and software by Wolfram Research.

Third prize was presented to Michael Morton, a senior at RangelyHigh School, Rangely, Colorado. He received a bronze medal of theOlympiad, a $500 scholarship to be used at any certified Americanuniversity or four-year college, a $1000 UCCS Chancellor’s Schol-arship for CMOMedalists, “The Mathematical Coloring Book: Math-ematics of Coloring and the Colorful Life of Its Creators”, a CASIOwatch, and software by Wolfram Research.


The Prize Fund of the Olympiad was generously donated byIntermap Technologies, Inc.; Wolfram Research; Colorado SpringsSchool District 11; Air Academy District 20; Fairview High School;Sand Creek High School; Harrison School District 2; Chancellor,Provost, Vice Chancellor for Student Success, Vice Chancellor forAdministration & Finance, College of Letters, Arts, and Sciences,Bookstore—all from the University of Colorado Colorado Springs;City of Colorado Springs; and Alexander Soifer.

The Award Presentation program featured lectures Mysteries ofTwo-Colored Planes, and Review of Solutions of the 26th ColoradoMathematical Olympiad Problems by Alexander Soifer.

The following guests of honor, hosts and sponsors addressed thewinners and presented awards: Jeannie Ritter, The First Lady of theState of Colorado; Colorado State Senator John Morse; ColoradoCongressman Mike Merrifield; Pamela Shockley, Chancellor, andThomas Christensen, Dean of the College of Letters, Arts and


Sciences—both from the University of Colorado Colorado Springs;Greg Hoffman, Director of Human Recourses, Intermap Technolo-gies, Inc.; Maggie Lopez, Assistant Superintendent, Colorado SpringsSchool District 20; Mary Thurman, Deputy Superintendent, SchoolDistrict 11; and Alexander Soifer.

How did we earn the high honor of The First Lady’s presence? Thiscould be a subject of a whole book. OnWednesday, October 29, 2008,Colorado Governor Ritter came to the University of Colorado atColorado Springs (UCCS) to speak in favor of Amendment58, which was on the November–2008 ballot. He was introduced byChancellor Pam Shockley-Zalaback at 1:15 p.m. Among other, he

said that this amendment would bring money to higher education:“We need to create more brain power.”

Upon finishing his talk, the Governor asked for questions. I raisedmy hand, and he invited me to speak:

– Governor Ritter, you have just spoken about the need for creatingmore brain power. We at UCCS have been doing just that. For thepast quarter a century, every year we ran the Colorado Mathemat-ical Olympiad for the best and the brightest middle and high schoolstudents of the State of Colorado. Every year, Governors Romer,Owens and you have sent me letters to read on your behalf to thewinning students. But in a quarter a century a Governor has nevercome to address the winners in person. Our Award PresentationCeremony is on April 24, 2009 or so.

– You are putting me on the spot. I am taking it as an invitation, andwill try to come.

Of course, I was putting the Governor on the spot: that was theidea! I came to the podium and gave Governor Ritter my businesscard. After the questions and answers were over, Governor Ritter andI continued our conversation with a handshake.

– Governor, I am a professor here, and need to go teach in a minute.May I ask you for your direct e-mail, so that I can send you detailsof your talk at the Olympiad’s Award Presentation in April 2008?

– I made a policy of not giving my e-mail to anyone. He (pointing athis Communications Director) will give you his. And I did notpromise to speak, I promised to do my best. Who knows what canhappen in April of next year, hurricane [for the reader: hurricanesdo not happen in Colorado].


I turned to the Communications Director who handed me his card,while the Governor padded me on the back. It was time for my 1:40class (I was, in fact, 6 minutes late). In January 2009, I reminded theGovernor our conversation via his Communications Director EvanDreyer:

“Honorable Governor Ritter,

In short, I am taking the pleasure of inviting you present in personyour vision of math education in our State at the Award PresentationCeremony of the Colorado Mathematical Olympiad, 4:00 p.m. onFriday, April 24, 2009 at the Lodge on UCCS campus.

In long, on Wednesday, October 29, 2008, when you came toUCCS to speak in favor of Amendment 58, you said: “We need tocreate more brain power.” I replied with a public question:

“Governor Ritter, you have just spoken about the need for creatingmore brain power. We at UCCS have been doing just that. For the pastquarter a century, every year we ran the Colorado MathematicalOlympiad for the best and the brightest middle and high school

students of the State of Colorado. Every year, Governors Romer,Owens and you have sent me letters to read on your behalf to thewinning students. But in a quarter a century a Governor has nevercome to address the winners in person. Our Award PresentationCeremony is on April 24, 2009 or so.”

You publicly replied:

“You are putting me on the spot. I am taking it as an invitation, andwill try to come.”

After the questions and answers were over, we continued ourconversation privately, with a handshake.

– Governor, I am a Professor here, and need to go teach in a minute.May I ask you for your direct e-mail, so that I can send you detailsof your talk at the Olympiad’s Award Presentation in April 2008?

– I made a policy of not giving my e-mail to anyone. He (pointing athis Communications Director) will give you his. And I did notpromise to speak, I promised to do my best. Who knows what canhappen in April of next year, hurricane.

Your Communications Director should have recognized my name(his two e-mails to me are attached). Through him, we gave you a


podium to address hundreds of best and brightest Colorado studentsbefore you were elected governor.

Truthfully, Governor: I am not sure we the citizen are able to trustour politicians. In 1990 President G.H.W. Bush and [Colorado] Gov-ernor Romer promised to make the US first in math and scienceeducation by the year 2000. Have they succeeded? No. Have theydone anything to succeed? No. I hope you will pay respect to theyoung & gifted, for they are our future!

Needless to say, I would be happy to provide you with any infor-mation you desire and answer any questions. The only acceptablesubstitution for you would be Mrs. Ritter, who looked beautiful and

sounded eloquent when she introduced [The First Lady] MichelleObama in Colorado Springs during the waning days of the presiden-tial campaign.”

Having received no reply, I reminded the Governor of my existenceon March 18, 2009:

“Honorable Governor Ritter,

Just as an addendum to my recent letter to you, I am sending you aphoto that shows the governor [Arnold Schwarzenegger] of the greatstate of California receiving the organizers and winning studentsof BAMO.

Bay Area Mathematical Olympiad (BAMO) was started by PaulZeitz who was a Colorado Springs teacher and assisted me with theColorado Mathematical Olympiad (CMO). He informed me thatBAMO was inspired by CMO. They even use our system of5-problems 4-hours.

So, the governor of California, in person, recognizes winners of noteven state-wide, but regional Olympiad. Surely, hurricanes should notprevent you from doing the same!”

Then out of the blue came a phone call to me from . . . The FirstLady of Colorado Jeannie Ritter! “I read your correspondence withmy husband. Would you make peace with him if I come and speak atyour Olympiad’s Award Presentation?” I was elated to accept such anunexpected generous offer. Jeannie Ritter came in all her stunningbeauty and gave a thoughtful and delightful address. Our winnerswere surprised and honored. My goal of recognizing young talentswas achieved, for the year.


The First Lady of Colorado Jeannie Lewis Ritter

Following the Olympiad, in May 2009, I ran the internationalworkshop “Ramsey Theory: Yesterday, Today, and Tomorrow” atthe Center for Discrete Mathematics and Theoretical Computer Sci-ence (DIMACS) at Rutgers University in Piscataway, New Jersey. Itwas a joy and honor—and a year of my life—to organize this event.The 30 participants never missed a single lecture. The names of theplenary speakers sounded like ‘Who is Who in Ramsey Theory.’Springer publisher was impressed and published a book of the work-shop’s talks and open problems assembled by delegates in its mostprestigious series “Progress in Mathematical Sciences” [Soi14]. Thetitle of the workshop and the book, conveys its structure: day one,“Yesterday,” was dedicated to talks about the history of RamseyTheory; day two, “Today”—to greatest theorems; and day three,“Tomorrow”—to open problems and conjectures.


Four of the Plenary Speakers at the 2009 DIMACS Ramsey Theory Workshop(from the left): Alexander Soifer, Joel H. Spencer, Jaroslav Nesetril, and RonaldL. Graham


Problems 26

26.1. Termites Cubed (A. Soifer). 2009 termites live in a woodencube Q of volume 2008. Prove that we can cut out of Q a cube ofvolume 251 with at least 252 termites living in it.

26.2. What’s the Difference? (A. Soifer). Numbers 1, 2,. . ., 2009 arewritten in a certain order. Any two neighboring numbers can bereplaced by their difference. This operation is repeated until we endup with just one number. Can this number be 0?

26.3. Red Square (A. Soifer). Is it possible to color red some of theunit squares of a 2009 � 2009 grid so that every unit square shares aside with exactly one red square?

26.4. Stimulus Package (A. Soifer). These are hard economic times.And so, when 2009 University of Colorado alumni got together, theydecided to help each other. A stimulus consists of each alumnussimultaneously retaining half of his money and spreading the otherhalf equally among his 2008 fellow alumni. At the end of the stimulusall totals are rounded up to the nearest dollar, compliments of theUniversity Credit Union. A series of stimuli terminates if furtherapplications of stimuli no longer change the amounts of money heldby alumni. Is there a set of initial funds of the 2009 alumni such thatthe series of stimuli never terminates?

26.5. On Candy and Girls (M. Song). 2009 girls holding a combinedtotal of 2010 pieces of candy sit at a round table. Angela has no candy.A move consists of a girl who has at least two pieces of candy, givingone piece each to her two neighbors. Prove that following a series ofmoves, Angela will get a piece of candy.

Solutions 26

26.1. The cube Q can be partitioned into 23 ¼ 8 cubes of half the sideand of volume 2008/8 ¼ 251 (Fig. 26.1). Since 2009 ¼ 8 � 251 + 1,by the Pigeonhole Principle, one of these smaller cubes must containat least 252 termites living it (we include the surface in each cube).■

Solutions 26 93

26.2. At any stage of the process, the difference created is an alge-braic sum of some of the given numbers (algebraic sum means thatsome of the summands may appear with a plus and other with a minussign). The final number is an algebraic sum of all numbers 1, 2,. . .,2009 in some order (it is easy to prove by induction that the differenceof two algebraic sums with distinct summands is an algebraic sumwith distinct summands). But since the order of summands does notmatter, and among the given numbers we have odd quantity of oddnumbers, this algebraic sum must be odd. Zero is not odd, and thuscannot be equal to the final number. ■

26.3. Pick a square S on the diagonal. A red square, neighboring S is aneighbor to one more diagonal square! (Fig. 26.3).

Fig. 26.1

Fig. 26.3


Therefore, pairs of diagonal squares share a red neighbor. But thetotal number of squares on a diagonal (2009) is odd, therefore therequired coloring does not exist. This problem belongs to my favoritecategory “It is easy to see, especially after someone showed it to you.”Alas, no one solved this problem. ■

26.4. In a stimulus, the maximal amount M an alumnus holds cannotincrease. Similarly, the minimal amount m cannot decrease. More-over, if not all 2009 amounts are equal, the minimal amount m will infact increase, for the amounts transferred in will be greater than theamounts paid out.

You may be wondering, why are the amounts after each stimulusrounded up by the University? I included this provision in the prob-lem in order to guarantee that the difference between the initialmaximum amount M0 and the initial minimum amount m0 will bewiped out in a series of at most M0�m0 stimuli. ■

26.5. Observe that since 2010 candies are held by 2009 girls, by thePigeonhole Principle there is a girl with at least two candies. This

means that we always have a girl who can make a move, until the stopcondition (Angela gets a candy) is satisfied.

Number the girls from 1 to 2009 clockwise starting from Angela.As a result, each girl is assigned a number between 1 and 2009.Angela’s number is 1. Let each of the 2009 candies have a numberat any stage of the game, which is the number of the child who has thecandy at that moment. We call this the candy number.

Let the girl number k, 1 < k < 2009 (i.e., not Angela and not thegirl number 2009) makes a move, i.e., gives out two candies to hertwo neighbors. Then the sum of the squares of the two given candynumbers changes from k2 + k2¼ 2k2 to (k� 1)2 + (k + 1)2¼ 2k2 + 2.

Thus, the total sum of squares of all candy numbers increases by2. However, the total sum of squares cannot keep increasing by2 indefinitely. (The maximum it can reach is 2010 � 20092 whenall the candies belong to the girl 2009).

Therefore, eventually Angela will get a candy, or the girl number2009 will get at least two candies and give two of them away, one toAngela. ■

Solutions 26 95

Twenty-Seventh Colorado MathematicalOlympiadApril 23, 2010

Historical Notes 27

Colorado Springs Allan Gardner wins again!

The XXVII Colorado Mathematical Olympiad opened its doors at0800 hours on April 23, 2010, while many parts of the State wereburied under the snow, and Monument Hill between ColoradoSprings and Denver was as icy as it had been famous to be, with carcrashes and rolled over vehicles. Some school districts closed downall their schools. The 45 Olympians from Rangely did not make it toColorado Springs. At 0900 hours, the starting time of the Olympiad,we had only 27 students present. However new arrivals trickled in,and by 1000 hours we had 174 Olympians, who came from Aurora,Boulder, Colorado Springs, Fort Collins, Manitou Springs, Monu-ment, and Parker. We also had guest participants from the AlabamaSchool of Mathematics and Science, Mobile, Alabama, who earnedtheir trip to our Olympiad by winning their local mathematicscompetition.

A special thank you goes to the Olympiad Judges: Russel Shafer,1st prize winner of 1st Olympiad in 1984, who drove fromWyoming;and Col. Dr. Robert Ewell, who dug himself out from monumentalsnow (he lives in Monument:-) and came to serve as a judge for the20th year.

Simultaneously with our Olympiad, the Chess World Champion-ship started in Bulgaria. Two former world champions, rated in topfour, competed: the challenger Veselin Topalov of Bulgaria and thecurrent champion Viswanathan Anand of India. The best of 12 match


97

was decided only in the very last game on May 11, 2010: Anand wonthe game making the score 6.5 to 5.5.

The judges awarded first prize to Allan Gardner, a junior at AirAcademy High School of Colorado Springs, CO. This was his secondconsecutive victory in the Olympiad. Allan received a gold medal ofthe Olympiad, a $1000 Scholarship to be used at any certified Amer-ican university or four-year college, a $1000 UCCS Chancellor’sScholarship, The Mathematical Coloring Book: Mathematics of Col-oring and the Colorful Life of Its Creators by Alexander Soifer, aCasio Watch, a Wolfram Mathematica online license, and a WolframMathematica for Students download.

Second prize was awarded to Christopher Guthrie, a junior atFairview High School of Boulder. He received a silver medal of theOlympiad, a $1000 Scholarship to be used at any certified Americanuniversity or four-year college, a $1000 UCCS Chancellor’s Schol-arship, The Mathematical Coloring Book: Mathematics of Coloringand the Colorful Life of Its Creators, a Casio Watch, WolframMathematica for Students. Don’t forget Chris—you will meet himagain the next year.

Third prizes were presented to three Olympians: Margaret Koehler,a senior at Lewis-Palmer High School in Monument; David Wise, ajunior at The Classical Academy in Colorado Springs (he was secondprize winner last year); and Ben Alpert, a junior at Fairview HighSchool in Boulder (he was fourth prize winner last year and thirdprize winner in 2008). Each of them received a bronze medal of theOlympiad, a $1000 UCCS Chancellor’s Scholarship, The Mathemat-ical Coloring Book: Mathematics of Coloring and the Colorful Life ofIts Creators, a Casio Watch, Mathematica for Students. Don’t forgetDavid and Ben—you will meet them again the next year.

The judges also awarded 12 first honorable mentions and 10 secondhonorable mentions.

This year, among the judges we had the 2008 winner of the LiteraryAward, Isabelle S. Soifer, now a freshman at the University ofColorado Colorado Springs. She was in charge of selecting the winnerof this year’s Literary Award, which was presented to DawitGebresellassie, 8th grader at Aurora Quest Middle School.

98 Twenty-Seventh Colorado Mathematical Olympiad

Olympiad Judges Isabelle S. Soifer and Matt Rixman. Photo by AlexanderSoifer

The Prize Fund of the Olympiad was generously donated byIntermap Technologies, Inc.; Casio, Inc.; Wolfram Research; Cityof Colorado Springs; Air Academy District 20; Colorado SpringsSchool District 11; Manitou Springs School District 14; Falcon Dis-trict 49; Widefield District 3; Chancellor, Provost, Vice Chancellorfor Student Success, Vice Chancellor for Administration and Finance,College of Letters, Arts, and Sciences, Bookstore—all at the Univer-sity of Colorado Colorado Springs; Silva Chang & Bradley Alpert;and Alexander Soifer.

The Award Presentation program featured a lecture Mysteries ofTwo-Colored Planes, and Review of Solutions of the 27th ColoradoMathematical Olympiad Problems by Alexander Soifer. For the first

time, I gave three commandments to the Olympiad winners:


1. Beware of bureaucracy. We need some of it. But bureaucracy, likea virus, replicates itself.

2. Speak truth to authority, because silent majority creates a breedingground for our ills.

3. And remember: in order to succeed, no reason for failure should beacceptable.

The following guests of honor, hosts and sponsors addressed thewinners and presented the awards: Larry Small, Vice Major of theCity of Colorado Springs; Pamela Shockley, Chancellor, and ThomasChristensen, Dean of the College of Letters, Arts and Sciences—bothfrom the University of Colorado Colorado Springs; Greg Hoffman,

Director of Human Recourses, Intermap Technologies, Inc.; MaggieLopez, Assistant Superintendent, Colorado Springs School District20; and Alexander Soifer.

Warm congratulatory letters came to us from Colorado GovernorBill Ritter, Jr., Colorado Springs Mayor Lionel Rivera, and ColoradoState Senate Majority Leader John P. Morse. Let me share with youhere one of these letters, sent by Senator Morse, who characterizedthe Colorado Mathematical Olympiad as a “fully realized dream.”


Following the Olympiad, in July 2010, I gave related talks,Between the Line and the Plane: Chromatic Etude in 6 Movementsand “In Order to Create a More Perfect Union. . .” at the 6th Con-gress of the World Federation of National Mathematics Competitionsin Riga, Latvia. When the Chief Organizer of the Congress, Professorof Latvian University Agnis Angans fell ill, on his request I composedthe International Program of the Congress, which included talks,


workshops, discussions, and a problem creating session. Two bookswere published based on talks and original Olympiad problemspresented at the Congress.

When during the waning minutes of 1999, Russian President BorisYeltsin passed the ‘crown’ to the former KGB Colonel VladimirPutin, I decided that while Russia is the biggest country, it is notbig enough to contain Putin and me. Yet in Riga I was just one-hourflying time from Moscow, and so I flew to Moscow, for the first andonly time during Vladimir Putin’s reign. I visited my friends, world’sgreat animator Yuri Norstein and feature film director AndreyZvyagintsev. It is nontrivial to visit one of the world’s most sought

after film directors. Here is how it came about. I called AndreyZvyagintsev over the phone.

– Greetings, Andrey! This is Alexander Soifer. I am in Moscow andwould like to visit with you.

– Hello, Alexander. You know, I am editing my new film “Elena”every day from 7 a.m. till 9 p.m.

– What are you trying to say, that you are too busy to see me or youare inviting me to come over after 9 p.m.?

– Come tonight after 9, said Andrey with a laughter.

In 2008, Yuri Norstein traveled from Moscow to Colorado Springsfor us to jointly teach a course “Yuri Norstein and the Great Art ofAnimation.” You may not know, but in 1984 Yuri’s animation “FairyTale of Fairy Tales” won in Los Angeles the title of “The GreatestAnimation of All Time.” All creative people ought to view it! I gotacquainted with Andrey Zvyagintsev in 2007 in a modern way, via aSkype conversation. In 2014 Andrey came to Colorado Springs tojoin me in teaching a course “Great European Film Directors.” It wasa delight to have one of them—a great European film director—nextto me in the classroom! Our conversations from mornings to latenights created an atmosphere of a complete trust, when we discoveredour shared critical view of Russian invasion of parts of Ukraine, andunderstanding of patriotism. Andrey once told me, “Alexander, youare a true American patriot: you criticize America when there aregrounds for it.” Shortly after our visit, Andrey won in Hollywood theGolden Globe Award for his new masterpiece “Leviathan.”

I also visited Dr. Olga Orlova, the Science Editor at the radiostation “Freedom,” funded by the U.S. Congress, the only station


delivering truth and nothing but the truth in Russia. Olga and I gotacquainted the year before, when she found me in the United Statesand conducted a series of three interviews with me, aired over theentire Russian territory, about the early Russian version of my book“The Scholar and the State: In Search of Van der Waerden”[Soi10]. Olga served a traditional Russian dinner in the kitchen, fullof animated conversations, food and drinks, lasting from 8 p.m. till4 a.m.

Problems 27

27.1. Divide and Conquer (M. Song). Is there a way to partitionpositive integers from 1 to 2010 into sets such that in each set there isa number equal to the sum of all other numbers of the same set?

27.2. More Stone Age Entertainment (A. Soifer, based on M.Dufour and S. Heubach’s research). Four piles of 100, 2009,200 and 2010 pebbles are evenly spaced around a circular table.Fred Flintstone and Barney Rubble in turn take off the table anynumber of pebbles from any two consecutive piles, at least one pebbletotal. A pile with zero pebbles continues to count as a pile involved inthe game. The player who takes the last pebble off the table wins.Find a strategy which allows one of the players to win regardless ofthe moves of his opponent. Oh, and Fred goes first!

27.3. Checkers Squared. Find the minimum number of checkers tobe placed on a 4 � 4 checkerboard (at most one checker per cell) sothat no matter which two rows and two columns are removed, theremaining 2 � 2 board will contain a checker.

27.4. “In Order to Form a more Perfect Union. . .” (A. Soifer) TheUnited Nations Organization includes 192 Member States, every pairof which has one disagreement. In order to form a more perfect Union,a negotiation is introduced: if representatives of four Member Statesare seated at a round table so that each pair of representatives seatednext to each other has a disagreement, the negotiation resolves one ofthese four disagreements. A series of consecutive negotiations reducesthe total number of disagreements to n. What is the minimum of n?

Problems 27 103

27.5. Colorful Integers (M. Kahle).(A). What is the minimum number of colors necessary for coloring

the set of positive integers so that any two integers which differby any power of 2 are colored in different colors? (Observe that1 is a power of two: 20 ¼ 1).

(B). What is the minimum number of colors necessary for coloringthe set of positive integers so that any two integers which differby any factorial are colored in different colors?

Solutions 27

27.1. Assume that such a partition exists. Then the sum of thenumbers in each set would be even (the sum of the greatest numbern plus the sum of all the other numbers in the set, which is equal to n).But the sum of the numbers in all the sets is 1 + 2 + . . . + 2010, whichis odd because it includes an odd number—1005—of odd summands.This contradiction proves that the required partition does not exist. ■

27.2. Fred wins. In his first move (Fig. 27.2.1), Fred transforms thestarting position (100, 2009, 200, 2010) into the position (100, 2009,100, 2009):

At this point, numbers of pebbles, symmetric to each other withrespect to the center of the table, are equal. Any move Barney canmake destroys this symmetry, and Fred in his following move can

restore it. Since at every move the total number of pebbles decreases,eventually Fred will leave the position (0, 0, 0, 0) after his move,which means that Fred will take the last pebble. ■

Þ

100 2009 100 2009

2010 200 2009 100

Fig. 27.2.1


27.3. Assume that 6 checkers are placed on a 4 � 4 checkerboard.Then by the Pigeonhole Principle, there is a row with at least 2 check-ers. In fact, there is a row with at least 3 checkers or two rows with atleast 2 checkers each. Therefore, there are two rows containing atleast 4 checkers total. Removing these two rows leaves at most2 checkers that occupy at most two columns; removing these twocolumns leaves no checkers on the remaining board. Thus, 6 checkersare not enough.

Let us show that 7 checkers, placed as in Fig. 27.3.1, deliver therequired position. Assume not, i.e., there is a pair of rows R1, R2 and apair of columns C1, C2 containing all checkers. The lower right corner

checker is contained only in the last row and the last column, one ofthese lines has to belong to the removed 4 lines. Due to the symmetrywith respect to the main diagonal, we can assume that the removedcolumn C1 is the last column.

Then R1, R2 are two of the first three rows. Removing them leavescheckers in three columns (counting C1), which are impossible toremove by removing only two columns.

Note: I conjecture that this placement is unique up to a series ofinterchanges of rows and columns. Can you prove or disprove it? ■

27.4. In creating this problem, I consulted Wikipedia, which informs:“The United Nations Organization (UNO) or simply UnitedNations (UN) is an international organization whose stated aims arefacilitating cooperation in international law, international security,economic development, social progress, human rights, and the

Fig. 27.3.1

Solutions 27 105

achieving of world peace. . . There are currently 192 member states,including nearly every sovereign state in the world.”

(1) Let each Member Country be represented by a vertex of a graph,

in which we connect two vertices by an edge if and only if thecorresponding two countries have a disagreement. Then the InitialDisagreements Graph is the complete graph K192 on 192 vertices(a set of 192 vertices, every two of which are connected by anedge). A negotiation selects a 4-cycle subgraph C4 of K192 (“rep-resentatives of four countries are seated at a round table so thateach pair of neighbors has a disagreement”) and removes oneedge from it. The problem, thus translated into the language ofgraphs, asks to find the minimum number of edges in a Disagree-ments Graph obtained from the initial K192 by a series of consec-utive removals of an edge from a 4-cycle.

(2) Observe first that the removal of an edge in a C4 subgraph pre-serves connectivity of the Disagreement Graph (i.e., ability totravel between any pair of vertices through a series of edges).If the series of consecutive negotiations were to eliminate allcycles, we would get a connected cycle-free graph, called a tree,on 192 vertices. Such a tree has exactly 191 edges (proof by aneasy mathematical induction). Observe that for any two points of atree we have a unique path through the edges (for otherwise wewould have created a cycle in the union of two distinct paths).This observation allows us to prove that any tree is 2-colorable(so that vertices of the same color are not adjacent). Indeed, colora point A in color 0, and any other point B in color 0 or 1 dependingupon the parity of the edge distance from A to B (edge distance issimply the number of edges in the path). Observe now that theproperty of 2-colorability is preserved under the removal of anedge from a 4-cycle, and under the reverse operation of complet-ing a 4-path to a 4-cycle. The Initial Disagreements Graph K192 isnot 2-colorable (it requires 192 colors!), therefore we will neverget to a tree as a result of a series of negotiations! We thus provedthat a 191-edge Disagreement Graph is unreachable.

(3) On the other hand, we can ‘fly a kite’ and in the process get aDisagreements Graph with 192 edges.


Through the series of negotiations, we can get from the Kite-0 graph, which is K192 (it is depicted in Fig. 27.4.1 with not all ofits edges drawn), to the Kite-1 graph, which consists of K191 with anattached 1-edge “tail.” Indeed (see Fig. 27.4.1), from the 4-cycle {1,3, 4, 5} we remove the edge {1, 3}; from {1, 4, 5, 6} remove {1, 4};. . ., from {1, 190, 191, 192} remove {1, 190}; from {1, 191, 192, 2}remove {1, 191}. Finally, from the 4-cycle {1, 2, 3, 192}, we remove{1, 192}, getting the desired graph Kite-1 (Fig. 27.4.2).

Continuing this process (you can formalize it by a simple mathe-matical induction), we will get Kite-189 graph, which consists of a

Fig. 27.4.1 Kite-0

Fig. 27.4.2 Kite-1

Fig. 27.4.3 Kite-189

Solutions 27 107

triangle K3 with a tail of length 189 (Fig. 27.4.3), which has exactly192 edges as desired. ■

HOMEWORK. Determine which of the graphs in Figs. 27.4.4 and

27.4.5 can be obtained from the Initial Disagreements Graph K192

through a series of negotiations.

27.5. (A) Clearly 3 colors are necessary, since the numbers 1, 2,3 pairwise differ by powers of 2 and thus require three distinct colors.But coloring the positive integers cyclically modulo 3 does the trick,because under this coloring the difference between two numbers ofthe same color is a multiple of 3, which is never equal to a power of2. So 3 colors are also sufficient. ■

27.5. (B) The author of this wonderful problem Matthew Kahlesubmitted the problem to me in 2008 with a solution by AdamHesterberg, who was a high school student at the time, and solved

Fig. 27.4.4 Cycle C192

Fig. 27.4.5 Cycle C191 with a 1-tale


the problem in one day. In order to prove the existence of a mysteri-ous “r”, Adam used nesting intervals. We often prove an existence inmathematics without discovering the value of the existing number.Robert Ewell calculated the value of Adam’s ”r.” However, onOctober 11, 2016, Matthew Kahle sent me a solution that explicitlydetermines the value of one such mysterious “r” as a part of thesolution. I choose to present this latest solution here. Do read FurtherExploration E27 for much more of related history and mathematics.

Assume 3 colors suffice. Since 1! ¼ 1 and 2! ¼ 2, any threeconsecutive integers must be colored in 3 distinct colors a, b, c.Numbers 1 through 6 must be colored a, b, c, a, b, c. Accordingly,number 7 must be colored a, but this is not allowed because 7 – 1¼ 3!– a contradiction. Thus, at least four colors are needed.

Suppose for a moment that there exists a number r, (necessarilyirrational), such that n!r is in the interval [1, 3] (mod 4), for everypositive integer n. Then we can determine which of the 4 colors to useon the integer k by looking at kr (mod 4): the color-defining-intervals[0, 1) [1, 2) [2, 3) [3, 4) (mod 4) determine the 4-coloring of the set ofintegers Z.

Thus defined 4-coloring satisfies the conditions of the problem.Indeed, suppose |i – j| ¼ n! for some n. By multiplying through by r,we get |ri – rj| ¼ rn!, which is between 1 and 3 (mod 4). In particular,ri and rj belong to different color-defining-intervals modulo 4, andthus i and j received different colors.

All that is left to prove is the existence of the desired r. Thefollowing lemma is useful.

Lemma. For k � 1, we have

1

4kð Þ!þ1

4 k þ 1ð Þð Þ!þ1

4 k þ 2ð Þð Þ!þ � � � � 1

4k � 1� 1

4k � 1ð Þ!

Proof of Lemma. Provided that �1 < R < 1, the formula for the sumof an infinite geometric series is

aþ aRþ aR2 þ aR3 þ � � � ¼ a

1� R

Solutions 27 109

Setting a¼ 1/(4k)!, R¼ 1/4k, and comparing term-by-term thefactorial series in the statement of the lemma with the geometricseries, the result immediately follows. ■

We claim that the following number r,

r¼ 1þ 1

4!þ 1

8!þ 1

12!þ � � �

¼ 1:0416914703416917479394211141 . . .

has the desired property.Let n � 1 be an integer. Suppose that k is the smallest integer such

that n < 4k. Then

n!r ¼ n! 1þ 1

4!þ 1

8!þ 1

12!þ � � �

� �

¼ n!þ n!

4!þ n!

8!þ n!

12!þ � � �

¼ n!þ n!

4!þ � � � n!

4 k � 1ð Þð Þ!� �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

An

þ n!

4kð Þ!þn!

4 k þ 1ð Þð Þ!� �|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

Bn

Observe that An is a sum of integers. All but the last of theseintegers is a multiple of 4. The last integer summand n ! /(4(k� 1))!is either 1 or 2 (mod 4), depending on n (mod 4). If n� 0 or 1 (mod 4),then n ! /(4(k� 1))!� 1 (mod 4), and if n� 2 or 3 (mod 4), then n ! /(4(k� 1))! � 2 (mod 4). Therefore, An � 1 or 2 (mod 4) for every n.

The Lemma implies that 0 < Bn < 1. Indeed, since n � 4k � 1 wehave

Bn � 4k � 1ð Þ! 1

4kð Þ!þ1

4 k þ 1ð Þð Þ!þ � � ��

� 4k � 1ð Þ! � 1

4k � 1� 1

4k � 1ð Þ!¼ 1

4k � 1

< 1:

Summing up, we conclude that n!r 2 [1, 3] (mod 4) for every n.Read more about this train of thought in Further Exploration E27.■


Twenty-Eighth Colorado MathematicalOlympiadApril 22, 2011

Historical Notes 28

Three New Olympians Win, while the Second Olympiad Book isborn

A few days before the Olympiad, on April 19, 2011, Barry Noreenwrote a lovely essay about our Olympiad for the major city newspa-per, The Gazette. Let me share it with you.

Deep-rooted math tradition bears fruit

By: BARRY NOREENApril 19, 2011http://gazette.com/noreen-deep-rooted-math-tradition-bears-fruit/

article/116522

Go figure.

That’s been what Alexander Soifer has been encouraging youngstudents to do for the last 28 years. On Friday, the University ofColorado at Colorado Springs will welcome about 400 studentsto the 28th annual Colorado Math Olympiad (CMO), where theywill be given four hours to complete five tough essay-style mathproblems. It’s hardly about trivia.

The Soifer-inspired event has drawn 15,000 participants to date.Simple arithmetic, but the contributions Math Olympiad winnershave made to the world have long since crossed over into thegeometric realm.


111

http://gazette.com/noreen-deep-rooted-math-tradition-bears-fruit/article/116522

http://gazette.com/noreen-deep-rooted-math-tradition-bears-fruit/article/116522

Soifer, a Russian emigre (Russia’s loss, America’s gain) who is atenured mathematics professor at UCCS, has watched as Olym-piad winners (they get medals, scholarships and other prizes)have gone on to impressive colleges and careers. Some areprofessors now, others started their own companies. More recentparticipants are still attending college.

Cal-Berkeley. Stanford. Harvard. Princeton. The MassachusettsInstitute of Technology. There are too many personal stories toinclude in a single column, so check out my blog to read aboutthe Olympiad’s brilliant minds.

Exhibit A: Matthew Kahle, who participated in his first CMO asan eighth-grader in 1987. Despite his exceptional mind, Kahlestruggled in high school, getting a C grade in geometry andgraduating from Air Academy High School with a 1.9 gradepoint average.

UCCS would not admit Kahle, who attended Pikes PeakCommunity College, then received a full-ride scholarship toColorado State University, where he received his bachelor’sand master’s degrees. He earned a doctorate in math from theUniversity of Washington.

There’s more. Kahle won a post-doctoral position at Stanfordand in 2010 received a fellowship to study at the Institute forAdvanced Study in Princeton—the institution made famous byanother brilliant mind who was bored with school as a youngster:Albert Einstein.

“He personifies the Olympiad ideals more than anyone,” saidSoifer, whose book, “Colorado Mathematics Olympiad and Fur-ther Explorations” will soon be coming out.

Soifer recalled another local student, Bryce Herdt, who hasAsperger’s syndrome. “He won first place as an eighth-grader.For him, it was important that he was recognized.”

It’s all too true that often, the most brilliant kids are not the oneswho live in the fast lane in junior high or high school. Time andcircumstances have a way of leveling that out eventually.

It’s nice that long before “eventually” rolls around, there is anevent such as the Olympiad, and that the passionate and brainySoifer keeps it going strong.

112 Twenty-Eighth Colorado Mathematical Olympiad

http://barrysblog.freedomblogging.com/

A few days before the 2011 Olympiad, my new Olympiad book[Soi9] was born. It included the history, problems, and solutions ofthe first 20 years of the Colorado Mathematical Olympiad, as well as20 Further Explorations, bridges between some Olympiad problemspresented in this book and the forefront of mathematics.

This time, on April 22, 2011, Colorado treated the Twenty-EighthColorado Mathematical Olympiad to a Colorado-perfect blue-skiedsunny day. The 196 Olympians came from Aurora, Boulder, CherryCreek, Colorado Springs, Denver, Fort Collins, Holyoke, ManitouSprings, Monument, Parker, Rangely, Widefield, and Yoder. We alsohad guest participants from the Alabama School of Mathematics and

Science, Mobile, Alabama, who earned their trip to our Olympiad bywinning in their local mathematics competition.

First prizes were awarded to the following three Olympians: BenAlpert and Christopher Guthrie, both seniors at Fairview High Schoolof Boulder; and David Wise, a senior, homeschooled in Monument.Each of them received a gold medal of the Olympiad, a $500 Schol-arship to be used at any certified American university or four-yearcollege, a $1000 UCCS Chancellor’s Scholarship, Casio fx-9750G11Graphing Calculator; The Mathematical Coloring Book: Mathematicsof Coloring and the Colorful Life of Its Creators by Alexander Soiferor a biography of Paul Erdos My Brain is Open by Bruce Schechter;and Wolfram software Mathematica for Students download. DavidWise became the first homeschooled Olympian to win first prize.

Ben Alpert also received a Creativity Award for his solution ofproblem 28.3. (B).

Second prize was awarded to Albert Soh, a freshman at FairviewHigh School of Boulder. He received a silver medal of the Olympiad,a $300 Scholarship to be used at any certified American university orfour-year college, a $1000 UCCS Chancellor’s Scholarship, Casiographing calculator; The Mathematical Coloring Book: Mathematicsof Coloring and the Colorful Life of Its Creators orMy Brain is Open;and Wolfram Mathematica for Students.

Third prizes were awarded to Charles Xu, a senior at Fairview HighSchool in Boulder; and Allan Gardner, a senior at Air Academy HighSchool in Colorado Springs. Each of them received a bronze medal ofthe Olympiad, a $1000 UCCS Chancellor’s Scholarship; Casiographing calculator; My Brain is Open by Bruce Schechter; andWolfram Mathematica for Students.


Forth prize was awarded to Artem Bolshakov, a sophomore atRidgeview Classical Schools of Fort Collins. The Judges alsoawarded 10 first honorable mentions and 20 second honorablementions.

A Literary Award was presented to Andrew Purvis, a senior,Manitou Springs High School. An Art Award was presented to AnaSeini Fifita, an eighth grader, Aurora Quest K-8 School.

Olympiad’s Judges at work: Bob Ewell (left) and Shane Holloway. Photo byAlexander Soifer

The Prize Fund of the Olympiad was generously donated by Casio,Inc.; Wolfram Research; City of Colorado Springs; BroadmoorTowne Center Sears Store; Air Academy School District 20; Colo-rado Springs School District 11; Manitou Springs School District 14;Falcon School District 49; Widefield High School; Cherry CreekWest Middle School; Robert and June Ewell; Chancellor, Provost,Vice Chancellor for Student Success, Vice Chancellor for Adminis-tration and Finance, College of Letters, Arts, and Sciences, Book-store—all from the University of Colorado Colorado Springs; andAlexander Soifer.


The Award Presentation Program featured lectures A Proof fromthe Book, and Review of Solutions of the 28th Colorado MathematicalOlympiad Problems by Alexander Soifer.

The following guests of honor, hosts and sponsors addressed thewinners and presented awards: Chancellor Pamela Shockley-Zalabakand Tom Christensen, Dean, College of Letters, Arts, and Sciences—both from the University of Colorado Colorado Springs (UCCS);Mr. Richard Skorman, Candidate for Mayor of Colorado Springs;Larry Small, Vice Mayor, City of Colorado Springs; Greg Hoffman,Personnel Manager, Intermap Technologies, Inc.; Karin Reynolds,Asst. Superintendent, Dist. 20; Brenda LeBrasse, Executive Director,

School District 11; and Alexander Soifer.At the time of this year’s Olympiad, I put on paper a few of my

thoughts. Let me share them with you here.

Is Mathematics an Art?

You and I met a myriad of people, who are exemplary specialists intheir (narrow) field, and do not know much outside of it. Worse yet,often they do not want to know. I agree with Maya Soifer who told meon occasion that I would have contributed to mathematics more if Icut my interests in other arts. What is wrong with this pragmaticdesire to learn only one’s field that puts bread on the table? Let merespond with my own “Shovel Aphorism”:

In Order to Dig Deep, One Ought to Dig Wide

I feel that my studies of visual arts, art of cinema, literary art, andmusic, forged what is ‘me’ today. And without those days and years incaptivity of the arts, I would not have created the books that I havewritten, and taught the courses I have created.

I see the difference between Art and Science in the subject matterthey study. (Natural) Science reflects what is outside of us, while Artreflects what is within. And so, let me ask you a natural question:

Is Mathematics an Art? 115

Is Mathematics an Art or a Science?

It depends upon your view. The great majority of mathematicians arePlatonists (after the great Greek philosopher Plato), i.e., they believethat mathematics discovers what already exists in Nature, outsideof us. And so for Platonists mathematics is a science. For me,mathematics is an Art. Mathematics can reflect what is out there,but it can also invent Beauty inspired from within that challenges theBeauty of Nature.

On September 2011, the University of Colorado held a grandopening of the exhibit “Art of the Fang People from Alexander SoiferCollection.” If mathematics is an art, it has a connection with the artof Africa!


Problems 28

28.1. Plane Drawings (A. Soifer). Is it possible to draw 2011 linesegments on the plane so that no three have a point in common andeach of them intersects exactly three other segments?

28.2. Gridlock (A. Soifer). Each U.S. senator likes 50 other senators.In order to sponsor a bill, two senators are needed who like each other.Will there necessarily be such a pair of sponsors? (There are100 U.S. senators.)

28.3. Squaring Triangles (A. Soifer).(A) Can a square be cut into triangles, all congruent to each other,

whose sides are in the ratio 3:4:5?(B) Can a square be cut into triangles, all congruent to each other,

whose angles are in the ratio 1:2:3?

28.4. The Moscow Metro (A. Soifer). The Moscow metro systemincludes 2011 stations. Among any three stations there are two thatare at most 1 mile apart. Prove that there is a circular disk of radius1 mile that contains at least 1006 metro stations inside or on itsboundary.

28.5. Won’t you be my neighbor! (A. Soifer). Each unit square of a2011 � 2011 square grid is colored in one of 2011 colors so that eachcolor is used. A pair of distinct colors is called a neighbor pair if theyappear as colors of a pair of unit squares sharing a side.

(A) Find the maximum M of the number of neighbor pairs.(B) Find the minimum m of the number of neighbor pairs.

Solutions 28

28.1. Assume that such a drawing is possible. Then the total numberof intersections is I ¼ 2011 � 3. In I, we counted each intersectiontwice, once for each of the intersecting segments, therefore I mustbe even, which it is not, a contradiction. Such a drawing does notexist. ■

Solutions 28 117

28.2. The total number of “likes” is L ¼ 100 � 50 ¼ 5000; they areour pigeons. The total number of (unordered) pairs of senators is

100

2

� �< L; they are our pigeonholes. Place “a likes b” in the pigeon-

hole {a, b}. By the Pigeonhole Principle, there is a hole with at leasttwo pigeons on it; i.e., there are two senators who like each other. ■

28.3. (A) Two right (“Egyptian”) triangles of side lengths 3, 4,and 5, can be arranged to form a 3 � 4 rectangle (Fig. 28.3.1).A 12 � 12 rectangle can be easily cut into rectangles of side lengths3 and 4 (Fig. 28.3.2), and we are done. ■

28.3. (B) The angles of the tiling triangle T are 30�, 60�, and 90�; letits sides be 1,

ffiffiffi3

p, and 2; then the area of T is

ffiffiffi3

p/2. Assume that there

is a square S that can be cut into n triangles, all congruent to T. Then

Fig. 28.3.1

Fig. 28.3.2


the area of S is nffiffiffi3

p/2. The side s of S is a sum of sides of T, therefore

s ¼ a + bffiffiffi3

p, where a and b are integers; and thus the area of S is

aþ bffiffiffi3

p� �2. We have found two expressions for the area of S,

therefore we get the equality

aþ bffiffiffi3

p� �2

¼ nffiffiffi3

p=2;

i:e:, a2 þ 3b2� �þ 2ab

ffiffiffi3

p¼ n

2

ffiffiffi3

p:

Sinceffiffiffi3

pis irrational, we get a2 + 3b2¼ 0, and thus a ¼ b ¼ 0, i.e.,

the side length of S is zero. So, only the ‘zero-square’ can be tiled—byzero copies of the given triangle (or any other tile!). If we do not countthe zero-square as allowed, then the desired square S does not exist.

Tiling is an exciting topic in mathematics. You can read moreabout it in [Soi7]. There is a fabulous classic text on this subject:Tiling and Patterns by Branko Grünbaum and G.C. Shephard [GS].■

28.4. Construct a graph G with vertices in metro stations, and twovertices adjacent if and only if the stations are at most distance1 apart. The given statement guarantees that any three vertices a,b and c contain an adjacent pair.

A degree of a vertex v in a graph is defined as the number of edgesemanating from v and denoted by degv. Let a be a vertex of maximumdegree dega (Fig. 28.4.1). If dega �1005, then the desired unit diskhas center at the station corresponding to a.

Fig. 28.4.1

Solutions 28 119

Assume dega �1004; then there are at least 1006 vertices notadjacent to a; call one of them b. Since the degb � dega � 1004,among these 1006 vertices there is one, call it c, not adjacent to b.Thus, we have found vertices a, b and c no pair of which is adjacent, acontradiction. Therefore, the desired disk exists. ■

28.5. Form a graph G with 2011 vertices, one per color, and twovertices adjacent if and only if the corresponding colors form aneighbor pair somewhere on the colored grid. The graph G isconnected, for there is always a rook’s path on the grid connectingunit squares of any pair of colors (Fig. 28.5.1). Where the color alongthe path changes, we get an edge in G. Thus, every two vertices ofG are connected by a path along its edges, and hence, G is connected.

Any tree on 2011 vertices features the minimum number 2010 ofedges in a connected graph (an easy proof by mathematical induction);the complete graph K2011 (i.e., the graph where every pair of vertices is

adjacent) sports the maximum number 2011

2

� �of edges. All there is left

to demonstrate is a coloring of the grid that induces a tree graph T, anda coloring of the grid that induces the complete graph K2011.

In order to achieve a tree, we first color the 2011�2011 square gridin a chessboard fashion in color 1 and a temporary color 2012. Wethen replace color 2012 with colors 2, 3,. . ., 2011 with each of thesecolors used on at least one unit square. We get a tree with the root atthe vertex corresponding to color 1, and 2010 edges connecting thisvertex to all other 2010 vertices of the graph (Fig. 28.5.2).

Fig. 28.5.1


Let us now induce the complete graph K2011.The row 1 we color in two alternating colors, 1 and the temporary

color 2012. We then replace the color 2012 with one unit square ofeach of the colors 2, 3,. . ., 1006.

The row 2 we color in two alternating colors, 2 and the temporarycolor 2012. We then replace the color 2012 with one unit square ofeach of the colors 3, 4,. . ., 1007; and so on.

The row 2011 we color in two alternating colors, 2011 and thetemporary color 2012. We then replace the color 2012 with one unitsquare of each of the colors 1, 2,. . ., 1005.

Due to a cyclic nature of coloring, it suffices to verify that color1 neighbors on each of the other colors. It neighbors on colors 2, 3,. . .,1006 in the first row. For 1007� i� 2011, color i neighbors on color1 in the row i , where color i neighbors on colors i+1, i+2,. . ., i+1006(we calculate these sums modulo 2011, i.e., subtract 2011 from thecolor number as soon as it exceeds 2011). Since i + 1 � 2012 and2013 � i+1006, color 1 will appear as a neighbor of color i in therow i. ■

Fig. 28.5.2

Solutions 28 121

Twenty-Ninth Colorado MathematicalOlympiadApril 20, 2012

Historical Notes 29

And the Winner is Albert Soh of Boulder

On April 20, 2012, Colorado treated the Twenty-Ninth ColoradoMathematical Olympiad to a Colorado-perfect blue skies. The223 Olympians came from Aurora, Boulder, Castle Rock, CherryCreek, Colorado Springs, Denver, Fort Collins, Holyoke, HighlandsRanch, Manitou Springs, Lakewood, Monument, Parker, Rangely,Pueblo, Woodland Park, and Yoder. We also had a guest Olympianfrom Sugarland, Texas. Once again we welcomed participants fromthe Alabama School of Mathematics and Science, Mobile, Alabama,who earned their trip to our Olympiad by winning their local math-ematics competition.


123

Albert Soh is receiving the first prize, with Chancellor Pam Shockley-Zalabak,Greg Hoffman, and Alexander Soifer

First prize was awarded to Albert Soh, a sophomore at FairviewHigh School of Boulder. He received a gold medal of the Olympiad, a$1000 Scholarship to be used at any certified American university orfour-year college, a $1000 UCCS Chancellor’s Scholarship, high endCasio Graphing Calculator; The Mathematical Coloring Book: Math-ematics of Coloring and the Colorful Life of Its Creators by Alexan-der Soifer or A New Kind of Science by Dr. Stephen Wolfram; andWolfram Mathematica for Students 8.0 software Download. Albertalso received Creativity Award for being the only one to solveproblem 29.3(B). This award included a 4-day 3-night stay right onthe Atlantic Ocean in the sunny Florida, and was sponsored byMr. Julian Felder, General Manager of Wyndham Deerfield BeachResort Hotel. Last year, Albert won second prize of the Olympiad.

Second prize was awarded to Artem Bolshakov, a junior atRidgeview Classical Schools of Fort Collins. He received a silvermedal of the Olympiad, a $500 Scholarship to be used at any certifiedAmerican university or four-year college, a $1000 UCCS Chancel-lor’s Scholarship, Casio Prizm fx-CG 10 Color Graphing Calculator;The Mathematical Coloring Book: Mathematics of Coloring and theColorful Life of Its Creators or A New Kind of Science; and WolframMathematica software. Last year, Artem won a fourth prize.

124 Twenty-Ninth Colorado Mathematical Olympiad

Third prize was awarded to Jesse Zhang, a freshman at FairviewHigh School in Boulder. He received a bronze medal of the Olym-piad, a $1000 UCCS Chancellor’s Scholarship; Casio Prizm fx-CG10 Color Graphing Calculator; The Mathematical Coloring Book:Mathematics of Coloring and the Colorful Life of Its Creators orA New Kind of Science; and Wolfram Mathematica software.


The Prize Fund of the Olympiad was generously donated by Casio,Inc.; Wolfram Research; Julian Felder, General Manager WyndhamDeerfield Beach Resort, Florida; Air Academy District 20; Colorado

Springs School District 11; Fairview High School; Rangely Schools;Robert and June Ewell; Chancellor, Provost, Vice Chancellor forStudent Success, Vice Chancellor for Administration and Finance,College of Letters, Arts, and Sciences, Bookstore—all from theUniversity of Colorado Colorado Springs (UCCS); and AlexanderSoifer.

The Award Presentation Program featured lectures Birth of aProblem: The Story of Creation in Seven Stages, and Review ofSolutions of the 29th Colorado Mathematical Olympiad Problemsby Alexander Soifer.

The following guests of honor, hosts and sponsors addressed thewinners and presented the awards: Chancellor Pamela Shockley-Zalabak and Tom Christensen, Dean, College of Letters, Arts, andSciences—both from the University of Colorado Colorado Springs;Greg Hoffman, Personnel Manager, Intermap Technologies, Inc.;Todd Morse, Executive Director Learning Services, School 20;Dave Sawtelle, K-12 Mathematics Facilitator, School District 11;and Alexander Soifer.

Following the Olympiad, on July 8–15, 2012, I attended the 12thInternational Congress on Mathematical Education in Seoul, SouthKorea. There I gave five talks relevant to the Olympiad:

1. What “Problem Solving” Ought to Mean and How CombinatorialGeometry Answers this Question: Divertismento in Nine Move-ments (a plenary talk).

2. Combinatorial Geometry Offers Gifted Students a Field for DoingResearch: Concerto in Four Movements.


3. The Goal of Mathematics Education, Including Competitions, Is toLet Student Touch “Real” Mathematics: We Ought to Build thatBridge.

4. Some of My Favorite Problems of Combinatorial Geometry,Solved and Unsolved: What Can Be Offered in Classroomand How.

5. Creating Problems for the Colorado Mathematical Olympiad: AnEtude in Four Movements. Mini-Conference of the World Feder-ation of National Mathematics Competitions, Seoul, South Korea,July 7, 2012.

Giving one of my five talks at the Congress in Seoul, July 2012


At a General Meeting of the World Federation of NationalMathematics Competitions (in Seoul), I was elected to the post ofPresident, thus becoming the first president to come from the UnitedStates.

Problems 29

29.1. Summing Up (A. Soifer).(A) Suppose three integers are added pairwise, and the results are

403, 704, and 905. Find the integers if they exist.(B) Solve the same problem if the pairwise sums are 403, 704, and

906.

29.2. Triangulation (A. Soifer). The convex 2012-gon P ispartitioned into finitely many triangles so that each side ofP belongs to one triangle, and no vertex of any triangle lies in aninner point of a side of another triangle. Can the map thus formed becolored in two colors, red and blue, so that no triangles which share aside are the same color and all the triangles along the perimeter ofP are blue?

29.3. Just Your Average Cube (A. Soifer). Each corner of the initialcube contains a number, and these numbers include 2012 and 2013.An averaging operation creates the second cube by replacing eachcorner number by the mean of the three numbers one edge away fromit. We then repeat the averaging operation obtaining the thirdcube, etc.(A) Can it so happen that all the numbers of the 2012th cube coincide

with the corresponding numbers of the initial cube? If yes, find allpossible arrays of the corner numbers.

(B) Can it so happen that all the numbers of the 2013th cube coincidewith the corresponding numbers of the initial cube? If yes, find allpossible arrays of the corner numbers.

Problems 29 127

29.4. Beyond the Finite I (A. Soifer).(A) Given 11 real numbers in the form of infinite decimal represen-

tations, prove that there are two of them which coincide ininfinitely many decimal places.

(B) Show that 11 is the lowest possible quantity to guarantee theresult of part (A).

29.5. Beyond the Finite II (A. Soifer). Infinitely many circular disksof radius 1 are given inside a bounded figure in the plane. Prove thatthere is a circular disk of radius 0.9 that is contained in infinitely manyof the given disks. A circular disk consists of all points on and insideof a circle.

Solutions 29

29.1. (A) Arrange the three given integers in non-decreasing ordera1 � a2 � a3. In the sum of all three pairwise sums, each givennumber will appear two times, i.e.,

2 a1 þ a2 þ a3ð Þ ¼ 403þ 704þ 905 ¼ 2012:

And thus a1 + a2 + a3 ¼ 1006. Now we can easily find all the givennumbers: a1¼ 1006 – 905¼ 101; a3¼ 1006 – 403¼ 603; a2 ¼ 403 –101 ¼ 302. ■

29.1. (B) Similarly to (A), we get 2(a1 + a2 + a3)¼ 403 + 704 + 906¼2013, which is absurd since 2013 is not divisible by 2, and thereforesuch numbers do not exist. ■

29.2. Figure 29.2 shows a coloring for a 12-gon that satisfies all theconditions of the problem. Assume that such a coloring exists for the2012-gon. Let R and B be the numbers of sides of all red and bluetriangles respectively, with B� R for definiteness. Surely, both R andB are divisible by 3, and so does their difference. On the other hand,each side is shared by one red and one blue triangle, except the sidesof the given 2012-gon; therefore, B – R¼ 2012. However, 2012 is notdivisible by 3. This contradiction proves that the required coloringdoes not exist for a 2012-gon. ■


29.3. (A) & (B).Denote the initial cube by C1; the second cube by C2;etc. Let the maximum and the minimum of the corner numbers ofcube Ci beMi and mi respectively, where i ¼ 1, 2,. . ., 2013. Since weaverage the numbers, we get the following inequalities:

M1 � M2 � � � � � M2013;

m1 � m2 � � � � � m2013:

Assuming that following a series of averaging we end up with theoriginal set of the corner numbers, we get the equalities

M1 ¼ M2 ¼ � � � ¼ M2013 ¼ M;

m1 ¼ m2 ¼ � � � ¼ m2013 ¼ m:

Look at the last, 2013th cube C2013, and at its maximum cornernumber M, whose position is depicted by a symbol ■ (Fig. 29.4.1).

Fig. 29.2

Solutions 29 129

Since M is the mean of the three numbers from the cube C2012,these three numbers must all be equal to M; their positions aredepicted by a symbol ■ in Fig. 29.4.2.

This implies that the four numbers of the preceding cube, C2011,must all be equal M; their positions are depicted by a symbol ■ inFig. 29.4.3.

Fig. 29.4.1

Fig. 29.4.2


Similarly, the minimum number m of the cube C2013 appears inthree corners of the cube C2012 and consequently four corners of thecube C2011. We have now uniquely accounted for all eight cornernumbers of the cube C2011: fourM and four m. Moreover, eachM hasall its neighbors (corners one edge away) to be equal to m and viceversa. Since the initial corner numbers include 2013 and 2012,M 6¼m. In fact, M ¼ 2013 and m ¼ 2012. And so the cubes in oddplaces C1, C3, . . ., C2013 must have the same numbers 2012 and 2013in the same positions, while even numbered cubes C2, C4,. . .,C2012

must coincide with each other, and can be obtained from C1 byinterchanging numbers 2012 and 2013.

And so the answer is “no” to problem 29.4. (A) and “yes” toproblem 29.4. (B). There are two arrays of corner numbers for theinitial cube that satisfy the conditions of problem 29.4. (B): one hasentries 2013 in the positions shown in Fig. 29.4.3 and the remainingentries 2012, and another set can be obtained from the first one byinterchanging numbers 2013 and 2012. ■

29.4. (A) Assume that among the given 11 numbers in their infinitedecimal representation there is no pair of numbers that coincides ininfinitely many decimal places. Then the set S of all decimal placeswhere at least one pair of the given numbers coincides is finite. Let uslook at the decimal place d, which is further to the right than anydecimal places from the set S. There are only 10 possible digits 0, 1,. . ., 9 for the position d, but we have 11 given decimal fractions;

Fig. 29.4.3

Solutions 29 131

therefore, by the Pigeonhole Principle, there is a pair of the givenfractions that coincides in the place d, which contradicts the choice ofthe set S. ■

29.4. (B) Just look at the following 10 infinite decimal fractions thathave no coincidences in any decimal position at all:

0:00 . . . 0 . . .0:11 . . . 1 . . .. . . . . . . . . . . . :0:99 . . . 9 . . . ■

29.5. The idea of looking at the centers of disks in this solution wascontributed by Robert Ewell. A visualization of the problem setting isdepicted in Fig. 29.5.1. Since the given figure F is bounded, it can besurrounded by a square S. Create on S a square grid (Fig. 29.5.2) of a‘small’ unit side a (the value of a to be determined later).

Fig. 29.5.1


There are infinitely many centers of the given disks and finitelymany cells in the grid in Fig. 29.5.2. By the Infinite PigeonholePrinciple, there is a cell c1 that contains infinitely many centers(Fig. 29.5.3). Denote by S this infinite set of the disks whose centerslie in c1.

Fig. 29.5.2

Fig. 29.5.3

Solutions 29 133

Let us now calculate how small the side a of the cell c1 should be sothat we can find a disk D1 of radius 0.9 that is contained in each ofdisks of the family S. In Fig. 29.5.4, I depict the cell c1 with center O;A is the center of the shown disk from the family S, and B a point onthe boundary of that disk.

According to the triangle inequality,

OBj j � ABj j � OAj j ¼ 1� OAj j � 1� a=ffiffiffi

2p

,where vertical bars denote the length of a segment, and the equality

is attained when the point B is in the position B1 and the triangle AOBis degenerate. Therefore, for any position of B on the circle, we get

OBj j � 1� a=ffiffiffi

2p

.

Now we can choose a to satisfy the condition 1� a=ffiffiffi

2p � 0:9. We

can thus choose a ¼ 0:1ffiffiffi

2p

. We are done, for the circular disk of

radius 0.9 and center in O, is contained in all disks from the infinitefamily S.

Of course, in the statement of this problem, we can replace 0.9 byany positive number less than 1.

Read much more about this train of thought in Further ExplorationE30. ■

Fig. 29.5.4


Thirtieth Colorado MathematicalOlympiadApril 26, 2013

Historical Notes 30

Celebrating the Centenary of Paul Erdos’ Birth(March 26, 1913–September 20, 1996)

Paul Erdos and Isabelle Soifer, Kalamazoo, MI, June 4, 1992. Photo by Alex-ander Soifer


135

During our anniversary 30th year, The Gazette featured a story byDebbie Kelley about the Olympiad. Let me reproduce it for you.

This year’s math Olympiad ‘brain-wrecking’

April 30, 2013 Updated: April 30, 2013 at 6:35 pmhttp://gazette.com/this-years-math-olympiad-brain-wrecking/arti

cle/1500102

Students take part in The Math Olympiad Friday, April 26, 2013 at UCCS.Photo courtesy of Jeff Foster, UCCS

No matter who you ask about this year’s Colorado MathematicalOlympiad, everyone agrees: It was hard.

136 Thirtieth Colorado Mathematical Olympiad

http://gazette.com/this-years-math-olympiad-brain-wrecking/article/1500102

http://gazette.com/this-years-math-olympiad-brain-wrecking/article/1500102

‘Much harder than last year’s,’ said Pranit Nanda, a seventh-grader from Aurora Quest K-8, near Denver, who won secondhonorable mention in the 2012 Olympiad. ‘It was a lot of visualthinking, rather than just using numbers to find the answers.’

‘Brain-wrecking’ is how Varun Roy, a sixth-grader from Moun-tain Ridge Middle School in Colorado Springs described thechallenge, held April 26 at the University of Colorado at Colo-rado Springs.

‘Those were the five hardest questions I’ve ever seen,’ he said.

‘It was very long and confusing,’ piped in North Middle Schoolstudent Regan Ogilvie. ‘I answered all of them but just guessedat one.’Now in its 30th year, the competition gives participants up tofour hours to solve five essay-style questions. This year, 255mid-dle- and high-school students from around the state entered.Winners will be announced and receive medals at a publicceremony at UCCS on Friday. To commemorate the 30th anni-versary, the event also will feature an Olympiad documentarywith the late mathematical genius Paul Erdos and a roundtablediscussion with former winners and other math whizzes.The complexity of this year’s problems isn’t surprising, saidDavid Hunter, professor and head of the department of statisticsat Pennsylvania State University.

‘The problems are always pretty much the same: easy to explain,yet moderately to very difficult to solve,’ said Hunter, the1986–1988 Olympiad champion and 1988 Palmer High Schoolvaledictorian. ‘The part many people don’t understand is thatthere is no such thing as ‘the answer’ to an Olympiad problem. Asolution is an explanation of why such-and-such must be true,and, often, there is more than one way to explain it.’A panel of judges found that out, as they listened for hours toOlympiad founder Alexander Soifer review the numerous waysin which the problems could be solved.

‘The problems are in the form of stories and because they are thesame for sixth through 12th-graders, they don’t require much

knowledge, but rather talent, creativity and original thinking,’said Soifer, a UCCS professor who teaches math, art history and


European cinema. ‘In the Olympiad of 2001, for example, aneighth-grader won first prize.’The Olympiad isn’t like other math contests for kids. There’s noreward for speed or regurgitating memorized math facts.

‘Somehow nervousness was less of a factor than with othertests,’ Hunter said. ‘You felt that you really had time to sit andthink, and that you wouldn’t be penalized for day dreaming alittle—which in fact often led to interesting solution ideas.’Matt Kahle, who graduated in the bottom 8% of his class at AirAcademy High School, excelled at the Olympiad. He took firstplace in 1990 and 1991, and today holds a Ph.D. in mathematicsfrom the University of Washington and is an assistant professorof mathematics at Ohio State University.He’s also been an Olympiad judge for the past decade.

‘The Olympiad helped me grow as a problem-solver and helpedme learn to write more clearly,’ he said. ‘It would be hard tooverestimate the impact the Olympiad has had on bright youngminds in Colorado for the past 30 years.’Soifer created the Olympiad based on the USSR National Math-ematical Olympiad, in which he competed and then served onthe committee.Mark Heim, a three-time Olympiad champion from 2003 to2005, judged entries for the first time this year to ‘experiencethe Olympiad from the opposite point of view.’ Instead offocusing on how he tackled tough problems as a contestant,Heim as a judge could see the variety of approaches studentsuse and evaluate their merit and accuracy.

‘I appreciate the Olympiad for its praise and encouragement oforiginal thinkers to a wide audience—something not often donebeneath the college level,’ said Heim, now a senior at ColoradoState University.


Alexander Soifer with two three-time Olympiad winners David Hunter (left)and Mark Heim, interviewed above

Of course, everyone appreciated a round number 30: our Olympiadsurvived for three decades, long live the Olympiad! But there was amore important anniversary to observe, to which I dedicated the 30thOlympiad: the great Paul Erdos’ Centenary!

On April 26, 2013, Colorado treated the Thirtieth Colorado Math-ematical Olympiad to the vibrant Colorado mountain sun. The257 Olympians came from all over Colorado: Aurora, Boulder, CastleRock, Cherry Creek, Colorado Springs, Denver, Fort Collins, Hol-yoke, Highlands Ranch, Manitou Springs, Lakewood, Monument,Parker, Rangely, Pueblo, Woodland Park, and Yoder. Once againwe had participants from the Alabama School of Mathematics andScience, Mobile, Alabama, who earned their trip to our Olympiad bywinning their local mathematics competition.


Past Olympiad winners serving as judges: Matthew Kahle (left) and RusselShafer, April 26, 2013. Photos by Alexander Soifer

First prize of the Olympiad was awarded to Albert Soh, a junior,and Jesse Zhang, a sophomore, both from Fairview High School ofBoulder. Each received a gold medal of the Olympiad, a $750 Schol-arship to be used at any certified American university or four-yearcollege, a $1000 UCCS Chancellor’s Scholarship, Casio Prism cal-culator; Wolfram Mathematica for Students download, The Mathe-matical Coloring Book: Mathematics of Coloring and the ColorfulLife of Its Creators by Alexander Soifer, and an Olympiad-relatedbook published by Springer. This was the second victory forAlbert Soh.

Second prize was awarded to Shawn Ong, a junior at DouglasCounty High School, in Castle Rock. He received a silver medal ofthe Olympiad, a $1000 UCCS Chancellor’s Scholarship, Wyndham


Deerfield Beach Resort Hotel 3-day 2-nights gift certificate; CasioPrism calculator; Wolfram Mathematica for Students, The ColoradoMathematical Olympiad: From the Mountains of Colorado to thePeaks of Mathematics by Alexander Soifer; and an Olympiad-relatedbook published by Springer.

Judging the Olympiad proved hereditary: Bob Ewell (right) and his son Mat-thew Ewell at work, April 26, 2013. Photo by Alexander Soifer

Third prizes were awarded to Jeremy Schiff, a freshman at CherryCreek High School in Denver; and Sisi Peng, a sophomore at RampartHigh School in Colorado Springs. Each of them received a bronzemedal of the Olympiad, a $1000 UCCS Chancellor’s Scholarship;Casio Prism calculator; Wolfram Mathematica for Students, TheColorado Mathematical Olympiad: From the Mountains of Coloradoto the Peaks of Mathematics, and an Olympiad-related bookpublished by Springer.



On May 3, 2013, starting at 1:15 p.m. in the prestigious BergerHall, we conducted unique Award Presentation Ceremonies. TheUniversity of Colorado Colorado Springs Chancellor PamelaShockley-Zalaback moderated The Round Table that included thepast Olympiad first prize winners Professor David Hunter, ProfessorMatthew Kahle, and Mr. Mark Heim, the host of the first 1984 andthird 1986 Olympiads and Columbia University Professor DennisMithaug, a sponsor Greg Hoffman, Olympiad’s veteran judges Col.Dr. Robert Ewell and Gary Miller, and I. This Round Table merits aseparate chapter, which is what you will find next in this book.

The Award Presentation Program also featured three talks by

Alexander Soifer: Review of Solutions of the 30th Colorado Mathe-matical Olympiad Problems; Paul Erdos at 100; and Charge to theWinners: The Ten Commandments. The last talk was concise; let mereproduce my Commandments for you here, although without illus-trations and rationale I presented in that talk.

Charge to the Winners: The Ten Commandmentsby Alexander Soifer

1. Choose a Creative Endeavor2. Make Your Life Meaningful


3. Love What You Do to in Order to Be Free4. Give All of Yourself to Your Work5. Broaden Your Knowledge6. Value Imagination above Knowledge7. Believe in Miracles8. View Talent as a Duty9. Do Not Give Up too Easily

10. Build on High Moral Ground

The Prize Fund of the Olympiad was generously donated by Casio,Inc.; Wolfram Research; Springer publisher; City of ColoradoSprings; Air Academy School District 20; Colorado Springs School

District 11; Garfield County School District; Rangely High School;Robert and June Ewell; Julian Felder, General Manager WyndhamDeerfield Beach Resort; Chancellor, Provost, Vice Chancellor forStudent Success, Vice Chancellor for Administration and Finance,College of Letters, Arts, and Sciences—all from the University ofColorado Colorado Springs; and Alexander Soifer.

Problems 30

30.1. Cover-up (A. Soifer). Is it possible to cover the plane with 2013straight lines?

30.2. Games Mathematicians Play (A. Soifer). Nine checkersoccupy the 3 � 3 lower left square of a 30 � 30 checkerboard. In amove you are allowed to have one checker jump over another onehorizontally, vertically, or diagonally onto an unoccupied square. Isthere a series of moves that transfers the nine checkers to

(A) The lower right 3 � 3 square of the checkerboard?(B) The upper right 3 � 3 square of the checkerboard?

Problems 30 143

30.3. Cross-Town Friendships (A. Soifer). Three schools have 2013students total, and each student has the same total number n, n> 0, offriends from among students in the other two schools. Prove that ineach school there is a student with a friend in each of the otherschools.

30.4. Watching the Super Bowl (A. Soifer). In order to watch theSuper Bowl together, all families in Castle Rock, Colorado, go to thenearest neighbor house. There are 2013 houses in Castle Rock, anddistances between the houses are all distinct.

(A) Prove that there will be a house where nobody will watch thegame. Prove that on the other hand, there will be a house with atleast two families watching.

(B) Prove that there will be over 333 houses with equally manyfamilies watching the game.

(C) Prove that there will be at least 671 houses with equally manyfamilies watching the game.

(D) Construct an example where the maximum number of houseshosting equally many families does not exceed 671.

30.5. One Old Erdos Problem (P. Erdos). Given a set of 22013 +1 distinct positive integers, prove that all its two-term sums cannot becomposed of the same 2013 prime numbers.

An integer m is composed of distinct primes p1, p2, . . ., pn if everyprime factor ofm is one of these primes. In two-term sums we excludesums of two copies of the same integer.

Solutions 30

30.1. First Solution. No. Pick a point O and draw through O linesparallel to the given 2013 lines. Now choose a line L through O of anunused slope. L shares at most one point with each given line. ThusL has infinitely many points not shared with the 2013 given lines. ■

30.1. Second Solution (by Robert Ewell). No. Draw a circle on theplane. Each line can intersect the circle at most twice so at most 4026points of the circle are on one of the lines. There are an infinitenumber of points on the circle left over. ■


30.1. Third Solution. Draw a regular 2014-gon on the plane. It takes2014 lines to cover its sides. ■

30.2. (A). At any allowed move, checkers remain on the same color of

the board colored in a chessboard fashion (Fig. 30.2.1). The startingposition has 5 black squares whereas the lower right corner positiononly has 4. Therefore, the answer is no. ■

30.2. (B). The chessboard coloring is powerless in this case, as bothstarting and ending positions have equal numbers of black and whitesquares. No problem, we can use a column type coloring, under whichthe initial colors of the checkers are preserved under the allowedmoves as well (Fig. 30.2.2). But now the initial position has 6 blacksquares, whereas the upper right position has only 3. Therefore, theanswer is no again. Observe: this coloring solves simultaneously bothproblems 30.2. (A) and 30.2.(B). ■

Fig. 30.2.1

Solutions 30 145

30.3. Let us name the three schools by the numbers of students in them:U, V, W respectively. Observe:W¼ 2013–U–V.Assume that in schoolU there is no student with friends in both schools V and W. Then theU students of that school are naturally partitioned into U1 students withfriends only in the school V and U–U1 students with friends only in theschoolW. Thus, the number of friendships between the students of theschools U and V is nU1, and the number of friendships between thestudents of the schools U and W is n(U–U1). We can now count thenumber of friendships between the students of the schools V, W in twodifferent ways. From V the number of friendships going into W is nV–nU1. FromW the number of friendships going into V is n(2013–U–V)–n(U–U1). We get the equality:

nV � nU1 ¼ n 2013� U � Vð Þ � n U � U1ð Þ:

Therefore, V � U1 ¼ 2013 � U � V � (U � U1), or

2U þ 2V � 2U1 ¼ 2013,

Fig. 30.2.2


which is absurd because 2013 is odd. A similar argument works forschools V and W. ■

30.4. Translation of the Problem into the Language of Digraphs.Let us create a directed graph, or digraph, a diagram with nodesrepresenting houses. Two houses A, B are connected by a directededge A!B if the family from A goes to B to watch the Super Bowl.Some edges could be bi-directional. Let the symbol |AB| denote thedistance between the houses A and B.

30.4. (A). Since the number of vertices is odd, there is at least oneone-directional edge, say, B!A. If no edge goes into B, we are done,for B will be empty. Otherwise there is a house C differentfrom A with an edge C!B. While continuing this process ofextending a directed path, we will never create a cycle. Indeed,assume we got a cycle D1!Dn!Dn�1 . . .!D3!D2!D1!Dn.Since families go to the nearest home, we get the following seriesof inequalities: |D1Dn|> |DnDn� 1|> . . . > |D3D2|> |D2D1|> |D1Dn|,thus |D1Dn|> |D1Dn|, which is absurd. We will stop when we addto our path an edge W!V with no edge going into W.

Now that we know that the house W will host no families, we areleft with 2013 families going to 2012 houses, and, by the PigeonholePrinciple, at least one house will host at least two families for theSuper Bowl. ■

30.4. (B). Let us first prove that no house will end up with six or morefamilies watching the game. Assume that the families from six houses

A, B, C, D, E, and F all went to the house O (Fig. 30.4.1).There is an angle α at O that is at most 60� (360�/6). Since all the

distances between the houses are distinct, all angles of the triangleOCD are distinct, and α is not the largest of this triangle’s angles. Letthe largest angle of that triangle be β. Then the largest side of thistriangle is OD, and so the family from D will not go to O, a contra-diction. Therefore, numbers of families watching the Super Bowltogether range from 0 to 5 (the family from O will go somewhereelse), six numbers in all. Since 2013 > 6 � 333, there will be at least334 houses hosting equal number of families for the Super Bowl. ■

Solutions 30 147

30.4. (C) (My short solution inspired by a long one by Robert Ewell).Let Xi denote the number of houses hosting i families, i ¼ 0, 1,. . .,5.We get two equalities induced by out-degrees and in-degrees of ourdigraph respectively:

X0 þ X1 þ X2 þ X3 þ X4 þ X5 ¼ 2013;

X1 þ 2X2 þ 3X3 þ 4X4 þ 5X5 ¼ 2013:

By subtracting the second equality from double the first one, we get

2 X0 � 671ð Þ þ X1 � 671ð Þ ¼ X3 þ 2X4 þ 3X5:

Thus, X0 or X1 is at least 671, for otherwise the left side is negativewhereas the right side is not. ■

30.4. (D). Consider a map in which the houses are distance-partitioned into 671 three-house groups, each group forming a scalenetriangle of unique side lengths, and three-house hamlets far awayfrom each other. This configuration produces 671 houses with

Fig. 30.4.1


0 families, 671 with one family, and 671 with two families watchingthe Super Bowl. ■

30.4. (E). Extra Problem, posed and solved by Robert Ewell (I only

cleaned-up the solution).

• Find max min (number of houses with equal numbers of familieswatching).

Denote byM the minimum number of houses hosting equal numberof families. If M� 135, we get from the in-degree equality from thesolution of problem 30.4. (C): 2013¼ B + 2C + 3D + 4E + 5F�2025,

which is absurd. Thus, M�134. On the other hand, there is a config-uration with M ¼ 134. We need just one example.

Form 134 7-house groups with one 5-family house, one 2-familyhouse, and five 0-family houses in each group (Fig. 30.4.2): let housesb, c, d, e, f go to house a, and houses a and g go to house b.

Form 134 7-house groups with one 4-family house, one 3-familyhouse, and five 0-family houses in each group (Fig. 30.4.3): let housesb, c, d, e go to house a; let houses a, f, and g go to house b.

Fig. 30.4.2

Solutions 30 149

We now have 10 � 134 ¼ 1340 0-family houses, 134 5-familyhouses, 134 4-family houses, 134 3-family houses, and 134 2-familyhouses with 137 houses left over. Take 134 of those and form 67 pairsso that we have 134 1-family houses.

Take the three houses that are left over and form them so that thereis one 2-family house, one 1-family house, and one 0-family house. Ascalene triangle delivers this configuration. The totals are:

Family per house Houses

0 1341

1 135

2 135

3 134

4 134

5 134

30.5. To commemorate Paul Erdos’s 100th birthday, I dove into hisold papers. I extracted and adopted this problem from the paper On aproblem in the elementary theory of numbers by Paul Erdos and PaulTuran [ET]. In this paper Erdos and Turan used a larger upper boundof 3� 22012, which can be reduced to 22013 + 1 without altering ideasof the proof. We need the following tool:

Tool 30.5 A. Let a1, a2,. . ., a2n+1 be a set of positive integers and p>2a prime number. It is always possible to select out of this set at leastn + 1 integers b1, b2,. . ., bn+1 such that no two-term sum bi + bj isdivisible by a greater power of p than bi and bj are.

Fig. 30.4.3


Remark. If bi, bj are divisible by pm, pn respectively, then of coursebi + bj is divisible by pmin(m, n). The crux of the tool is the existenceof such bi, i ¼ 1, 2,. . .,n +1, that no bi + bj is divisible by a greaterpower of p than pmin(m, n).

Proof of Tool 30.5A. Divide each of the given numbers a1, a2,..., a2n+1by the highest possible power of p to obtain the quotients c1, c2,. . ., c2n+1(some of them being possibly equal), such that no member of this newsetC is divisible by p. Now partitionC into two classes depending upontheir smallest positive residue mod p (i.e., remainders upon division byp), being less than or greater than p/2. By the Pigeonhole Principle, atleast one of these two classes, C1, must contain n + 1 numbers. Weretain only C1. The two-term sums in C1 are not divisible by p. Theintegers ai corresponding to these ci satisfy the requirement of the tool.

Now we are ready to solve problem 30.5.

30.5. Let a1, a2,. . .,a22013þ1 be distinct positive integers such that all

their two-term sums are composed of 2013 primes p1, p2, . . ., p2013.Our set of primes must contain 2, for otherwise for any three givennumbers a1, a2, a3, at least one of their two-sums is even and thuscannot be composed without 2. Let p1 ¼ 2.

Applying tool 30.5A with p ¼ p2013, we obtain 22012 + 1 integerswith the property stated in the Tool 30.5A. Repeating the processwith p ¼ p2012, we obtain 22011 + 1 integers with the propertystated in the tool 30.5A with respect to both p2013 and p2012, andso on. We end up with the three numbers a1, a2, a3 that satisfy thestatement of the Tool 30.5A with respect to all primes p2, p3, . . .,p2013, i.e.:

a1 þ a2 ¼ 2m1,1pm1,22 � � � pm1,2013

2013 ;

a2 þ a3 ¼ 2m2,1pm2,22 � � � pm2,2013

2013 ;

a3 þ a1 ¼ 2m3,1pm3,22 � � � pm3,2013

2013 :

Due to the Tool 30.5A, the first of these equalities implies that each

a1, a2 must be divisible by pm1,22 � � � pm1,2013

2013 . Neither of a1, a2 can alsobe divisible by 2m1,1 , for otherwise a1 or a2 would be divisible by

2m1,1 � pm1,22 � � � pm1,2013

2013 , which is equal to their sum a1 + a2, and this isabsurd. But when can the sum a1 + a2 be divisible by a greater powerof 2 than each summand? Only when a1 and a2 are divisible by the

Solutions 30 151

same maximum power of 2. Similarly, a2, a3 must be divisible by thesame maximum power of 2. Thus, a1, a2, a3 are divisible by the samemaximum power of 2, denote it by 2u. By dividing both sides of theabove three equalities by 2u, we get

b1 þ b2 ¼ 2αw1;

b2 þ b3 ¼ 2βw2;

b3 þ b1 ¼ 2γw3;

where b1, b2, b3 are distinct odd integers and wi ¼ pmi,22 � � � pmi,2013

2013 ;

i ¼ 1,2,3. Look at the first equality: by the Tool 30.5A, both b1 andb2 are divisible by w1. The quotients b1/w1, b2/w1 are distinctodd integers and as such add up to at least 1 + 3 ¼ 4; thus2α¼ b1/w1 + b2/w1� 4 and α� 2. Similarly β� 2 and γ� 2.

Now add together the last three equalities. The right side is divis-ible by 4, whereas the left side 2(b1 + b2 + b3) is not. This contradic-tion proves the desired result.

If you enjoyed this Paul Erdos problem 30.5 you ought to now readFurther Exploration E29. ■


A Round Table Discussionof the Olympiad, or Looking Backfrom a 30-Year Perspective

It is impossible to translate into a written word the excitement of the30-Year Anniversary Award Presentation and the Round Table Panelthat took place. Yet, let me attempt to give you a glimpse of thisevent.

I opened the celebration brochure with the following address.Dear Friends, Olympians, Parents, Teachers and Guests!

Welcome to the Awards Presentation of the 30th Colorado Math-ematical Olympiad. Yes, we have reached 30 years. And this Pro-gram will be unique in celebrating this milestone. We dedicate thisOlympiad to the great Paul Erdos, who would have turned 100 onMarch 26, 2013. In 1989 Paul talked about our Olympiad in thedocumentary which we will play for you.

Our Olympiad is listed in the World Compendium of MathematicsCompetitions by Peter O’Halloran. In 2011, Springer published mybook The ColoradoMathematical Olympiad and Further Explorations:


153

From the Mountains of Colorado to the Peaks of Mathematics. This450-page book covers the first 20 years of the Olympiad’s history,problems, solutions, and builds 20 bridges from the problems of theOlympiad to the forefront of mathematics. It also includes essays byseveral past winners about the Olympiad’s influence on them and theirfuture careers.

In fact, several of the Olympiad’s past first prize winners will bewith us to share their memories and present aspirations: DavidHunter, Matthew Kahle, and Mark Heim. Aaron Parsons sent us hisgreetings. The early host of the Olympiad Dean Dennis Mithaug willbe with us for the first time since his appearances in 1984 and 1986.

Thanks to the generosity of local and national sponsors, listed inthis program, winners will receive fine awards. However, today’sprogram is not only for the winners. It is for all Colorado kids andtheir friends, who enjoy fun of the mathematical kind. I hope all ofyou, Olympians, your parents and teachers will enjoy and benefitfrom discussing with me Solutions of the Olympiad Problems, attend-ing my lecture Paul Erdos and His Problems at 100, and a receptionwhere you can meet and befriend fellow talented students from allcorners of Colorado.

Roundtable Moderator and UCCS Chancellor Pamela Shockley-Zalabak

154 A Round Table Discussion of the Olympiad, or Looking Back from a 30-Year. . .

The Panel was sponsored and moderated by the Chancellor of myUniversity Pamela Shockley-Zalabak. Here is a complete list of thepanelists with their short biographies at the time of the celebration.

Professor David Hunter was First Prize Winner in the ColoradoMathematical Olympiads of 1986, 1987, and 1988. In 1988 he was theValedictorian of Palmer High School. Upon graduation fromPrinceton University in Mathematics (where he was on the volleyballand soccer teams), Dr. Hunter fulfilled his high school plans “to payback [to public education]” by working for two years as a high schoolteacher. He then earned his Ph.D. degree from the University ofMichigan, Ann Arbor. He is presently Professor and Head of Statis-tics Department at the Pennsylvania State University.

Professor Matthew Kahle was First Prize Winner in the ColoradoMathematical Olympiads of 1990 and 1991. Having graduated withtwo degrees from Colorado State University, Fort Collins. He earnedhis Ph.D. degree from the University of Washington. Dr. Kahle wasthen awarded prestigious Postdoctoral Fellowships, first by theStanford University and then by the Institute for Advanced StudyPrinceton. He is presently a Professor at Ohio State University.Dr. Kahle competed in 5 Colorado Mathematical Olympiads andhas served as a judge in 9 Olympiads.

Mark Heim was First Prize Winner the Colorado MathematicalOlympiads of 2003, 2004, and 2005. He is presently a senior,majoring in Mathematics and Computer Science, at Colorado StateUniversity, Fort Collins.

Gary Miller, BA in Mathematics and Physics, Western State Col-lege, Gunnison, CO, 1966. MAT in Secondary Mathematics Educa-tion, Harvard University, 1968. He served as a Computer Coordinatorof Colorado Springs School District, and Mathematics teacher atCoronado High School from 1970—1997. From 1987 to present,

Mr. Miller has served as a Judge for the Colorado MathematicsOlympiad, and since 1989 also a Member of the Problem Committee.A Tandy Corporation’s national award winner for Innovation andCreativity in Science and Mathematics Teaching.

Robert N. (Bob) Ewell, Ed.D., Lt Col, USAF, Ret. Dr. Ewellgraduated from Clemson University with a B.S. in Mathematics in

A Round Table Discussion of the Olympiad, or Looking Back from a 30-Year. . . 155

1968. He earned his doctorate in education from Auburn University in1984. He has judged every Olympiad but one since 1989. He served inthe Air Force from 1970–1990, had his own statistical consultingbusiness from 1990–2001, and now serves with The Navigators.

Professor Dennis E. Mithaug, B.A., Dartmouth College; M.A., M.Ed., Ph.D., University of Washington. Dr Mithaug was Professor andDean of the School of Education at the University of Colorado atColorado Springs when he served as a host of first (1984) and third(1986) Colorado Mathematical Olympiads. In 1991 he became Pro-fessor of the Teachers College of Columbia University and Chair ofthe Department of Special Education (1991–1996) and later its Direc-tor when it became a program in the Department of Health andHuman Behavior.

Greg Hoffman is presently Director of Human Resources forIntermap Technologies, Inc. Prior to joining Intermap, Mr. Hoffmanserved in progressively responsible HR leadership roles in the high-tech industry with Digital Equipment Corp., Quantum Corp., AppleComputer, Ingersoll Rand, and The Sanborn Map Company. He hasbeen the most dedicated sponsor of the Olympiad, ever since he andDr. Soifer met in 1995 at the Apple Computer Plant in Fountain,Colorado. In 1998 Mr. Hoffman was recognized by the Medal of theColorado Mathematical Olympiad.

Professor Alexander Soifer has founded and led the Colorado Math-ematical Olympiad for 30 years. He served on USSR National Math-ematical Olympiad (1970–1973) and USA Mathematical Olympiad(1996–2005). He is the elected President (2012–2018) of the WorldFederation of National Mathematics Competitions, where during thepreceding 16 years he served as Secretary and Senior Vice President.He has been a Professor at the University of Colorado ColoradoSprings for 34 years. During 2002–2004 and 2006–2007 he servedas a Visiting Fellow at Princeton University and Long Term VisitingScholar at Rutgers University.

The following memorable photographs depict the AnniversaryRound Table.


Discussion Panel, from, the left: Alexander Soifer, Gary Miller, Robert Ewell,Matthew Kahle, Mark Heim, David Hunter, Dennis Mithaug, Greg Hoffman,and the moderator Pamela Shockley-Zalabak

Panelists, from, the left: Alexander Soifer, Gary Miller, and Robert Ewell


Panelists, from, the left: Robert Ewell, Matthew Kahle, and Mark Heim

Panel, from, the left: David Hunter, Dennis Mithaug, and Greg Hoffman

What did the panelists share? Here are but a few quotations to giveyou a flavor of the Round Table.

Pamela Shockley-Zalabak: We have a panel from all over thecountry, who have a connection to this Olympiad.

David Hunter: One of the questions still seems to me like one of thearchetypes of Olympiad question. Paint the plane in two colors, provethat you can find two points that are exactly one inch apart that are of


the same color. For most people this problem does not even sound likemath. You can explain it to somebody, you can explain what planemeans, what points mean, and you can explain that we are talkingabout abstract concepts and not things that you can actually draw.And so it really does not jive with what most people consider math-ematics, and jet it is exactly what is on the Olympiad, and it is realmathematics. . . It takes creativity to come up with a solution andexplain why it is a solution.

Matthew Kahle: It is hard to imagine the Olympiad without him. Myhope now is that it continues indefinitely, and someone eventuallyrises to that, and continues to organize, and continues to composeproblems, so that the tradition can continue.

Mark Heim: I think it is enlightening to see all the different people’sideas and interpretations of the problems that are not all the same.

Gary Miller: So many things like this are tied to a personality, thatthe personality gets old, goes away, and the idea dies. I am hoping . . .

Robert Ewell: These problems have a unique character for the mostpart that even if you can’t do it, when somebody shows you how to doit, you say oh, oh, I see that.

Dennis E. Mithaug: In thirty years he e-mailed me and asked to comeback and share with him all of his successes. This is a guy who strikesvery favorable deals. He is my friend. He is a very charming man.

Greg Hoffman: We need that kind of thought process in the U.S. andthis is one way, and it is happening in our hometown and we need tosupport it.

Alexander Soifer:We reward arms and legs of Olympic sportsmen, Iapplaud them, and it is a delight to have Olympians in your classes,but why not reward the brains too and apply the same admissionwindow to [math] Olympiad medalists? [such an admission windowfor Olympiad’s medalists was eventually approved by the College ofLetters, Arts, and Sciences, but never approved by the College ofEngineering and Applied Science.]

The University produced a 30-minute film about the 30 years of theOlympiad that included interviews with the Round Table panelists.


Enjoy it on your laptop or smartphone: https://www.youtube.com/watch?v¼EOTiA4YWU-k.

We invited the Olympiad winner Aaron Parsons, who was unableto leave the University of California Berkeley. Instead he sent me thefollowing e-mail.

From: [email protected] [mailto:[email protected]] On Behalf of Aaron ParsonsSent: Monday, April 1, 2013 11:25 p.m.To: Alexander Soifer, [email protected]: Re: XXX Olympiad

Hi Alexander,

I am so sorry to be so slow in responding. I was hoping to find away to be able to respond affirmatively to you invitation, andthat caused this long stall. Unfortunately, there just doesn’t seemto be any way to do it for me this time.

Berkeley is treating me very well. I love the astronomy depart-ment here, and am really enjoying my teaching and research. It’sa great privilege. I’m spending a lot of time building a radiotelescope to discover the first stars that formed in the universe13 billion years ago, and it’s a blast.

Thank you for thinking of me for the 30th anniversary. CMO hasbeen such an institution for math in Colorado. I’m proud to havebeen a part of it, and am very very appreciative of all of the timeand effort you have put in to keep this running for so long. AsI’m sure you know, CMO really helped open my eyes to adifferent side of academia than I had previously experienced,and in the end, I find that my first experiences with the kinds ofproblems you posed us have been more representative of thekinds of problems I find in research than anything I found in anyclassroom.

I hope you have a rollicking good time celebrating this anniver-sary. I’m sorry I will miss it.

All the best,

Aaron


https://www.youtube.com/watch?v=EOTiA4YWU-k



The Olympiad touches not only participants, but also their families.The following card I received from Patty Rusinger, Matt Kahle’smother, right after the 30th Olympiad:


April 29, 2013

Greetings:

On behalf of the State of Colorado, it is my pleasure to congratulateall the student participants and award winners in the 30th AnnualColorado Mathematical Olympiad.

Your participation in this competition is a testament to your hardwork and academic dedication in the area of mathematics. We areconfident that your continued commitment to your studies will bringyou bright and successful futures. Keep up the good work!

You have my best wishes, both now and in the years to come.

Sincerely,

John W. HickenlooperGovernor


We received a letter from the President of the Colorado State SenateJohn P. Morse:


Greetings also came from the Mayor of the City of ColoradoSprings Steve Bach:


Part II

Further Explorations of the ThirdDecade

Introduction to Part II

The more you know, the more you knowyou don’t know.

—Aristotle

Science is always wrong. It never solves aproblem without creating ten more.

—George Bernard Shaw

What does the Aristotle mean by this paradoxically sounding state-ment: “The more you know, the more you know you don’t know”? Ofcourse, the more you know, the more you know! However, the moreyou know, the better you understand how much more there is toknow! It is like climbing a mountain: the higher your rise, the widerhorizon you see.

The playwright Bernard Shaw observes that a problem is neversolved without giving birth to a number of new problems. This isexactly what Part II of this book will present. The fifty problems ofthe Colorado Mathematical Olympiad are solved in Part I of thisbook. Their solutions give birth to a number of more exciting, deeper,sometimes open problems, that we are going to discuss here inFurther Explorations. Voltaire believed that “The secret of being abore is to tell everything.” I agree. In this part of the book I will leaveout much of what I know (and everything I do not know:), so that youhave your space to enjoy and discover on your own!


169

E21. Cover-Up with John Conway, MityaKarabash, and Ron Graham

Inspired by Problem 21.4: “To Have a Cake”

John Horton Conway and Alexander Soifer, Fine Hall, Princeton University,July 2007

The problem title comes from a famous proverb “To have a cake andeat it too,” which in my early American years I could not understand.Surely you have to “have a cake” in order “to eat it”! A betterformulation of this folk wisdom would have been “To keep a cakeand eat it too,” which is obviously impossible, hence a moral of the


171

proverb. But that is not what I would like to share with you here.A version of what follows first appeared in the second, 2009 Springeredition of my book “How Does One Cut a Triangle?” [Soi6]. But itfits here so well that I am including its new updated and expanded2016 version.

During the years 2002–2004 and 2006–2007, I was a “VisitingFellow” at Princeton University. In American translation from theBritish, a “fellow” stands for a researcher.1 A fine MathematicsDepartment was appropriately located in Fine Hall. My first office(2003–2004) was on the third floor near the office of the celebratedmathematician John Horton Conway, the John von Neumann Profes-

sor of Mathematics. We soon became friends. John would come inand litter my blackboard with virtuoso calculations of polyhedralinvariants. In the photo, you can see an example of John’s work,with “Please leave” message to the custodians. I have never seen Johnwith a calculator—he did not need one. John brilliantly simplifiedcalculations to the point they became trivial—it was nothing short ofa show watching John calculate.

My Princeton office blackboard beautifully ‘decorated’ by John H. Conway

During one of the breaks in our work, John took me to the PrincetonCemetery to pay homage to Kurt G€odel. On other breaks, we walked

1At the same time, I was a “Long Term Visiting Scholar” at Rutgers University, Piscataway,

but there I was involved in totally different problems, jointly with the genius Saharon Shelah.

Read about it in my “The Mathematical Coloring Book: Mathematics of Coloring and the

Colorful Life of Its Creators” [Soi3].

172 E21. Cover-Up with John Conway, Mitya Karabash, and Ron Graham

to a Chinese restaurant, where John taught me to use chopsticks. Myfinal chop-exam was to pick ice cubes with chopsticks. On one of thewalks to pick up John’s youngest son Gareth from a kindergarten,John shared with me his ideas on the foundations of set theory.

The contrast between two British ‘imports’ to Princeton-Math,Andrew Wiles and John Conway, could not be more dramatic. Asan intellectual, Conway knew much about much, and was curiousacross many human endeavors. Wiles’ interests were rather narrow.Having conquered—jointly with Richard Taylor—the most famousproblem in the history of mathematics, Fermat’s Last Theorem (FLT),Andrew believed that Fermat did not have a proof of his FLT, while

romantic Conway—and I—trusted Fermat had a proof, of course,different from the one found by Taylor and Wiles. During2006–2007 academic year, Princeton-Math Department ManagerScott Kenney suddenly got rid of the Fine Hall Commons Room’slibrary. These books were donated over a half a century by Princetonmathematicians. The library contained rare first editions of mathe-matical books. AndrewWiles, the Chair of the Department, reacted asfollows to the inquiry of the Vice-Chair Simon B. (Si) Kochen: “I wasout of the country. But they were obsolete anyway.” This act upset theintellectuals of the department, such as Conway, Kochen, and JosephKohn. After all, a first edition is a treasure to those who value books,and “obsolete” to those who don’t. The now empty wall-long bookcase was thrown away and replaced by one more blackboard.

Perhaps, a contrast with AndrewWiles and other successful narrowspecialists, prompted a chip on John Conway’s shoulder, as he oncetold me, “I haven’t done anything major in mathematics.” I replied indisagreement. John was perhaps the most ingenious mathematician Imet, with an unparalleled breadth, insight, and irresistible taste forbeauty—in mathematics, philosophy, literature, and life.

During my second Princeton stay, 2006–2007, John suffered astroke. However, even that did not stop him. I recall visiting John ashe was recovering at his house, when he most energetically demandedfrom Robert MacPherson, an editor of Annals of Mathematics, topublish Thomas Hales’ computer-aided proof on the nearly 400-yearold Johannes Kepler’s optimal ball packing conjecture. In a sense, JohnConway won, for today I read on the journal’s Internet page that

Computer-assisted proofs of exceptionally important mathemat-ical theorems will be considered by the Annals.

Inspired by Problem 21.4: “To Have a Cake” 173

We lived close to each other, and I witnessed John’s determinationto get back in shape after a triple-bypass heart surgery, as he dailystubbornly walked around our neighborhood, leaning on a cane.

One of my main projects at Princeton was writing a book “TheScholar and the State: In Search of Van der Waerden” [Soi10]. It wasboth enjoyable as it was profitable to share my newly uncoveredarchival documents with and to get feedback from my colleagues atPrinceton-Math, especially from my friends John Conway and HaroldW. Kuhn.

Princeton-Math maintained an historic tradition of a daily 3 to4 p.m. coffee hour in the Commons Room, on the third floor of the

Fine Hall, attended by everyone, from undergraduate students to the“Beautiful Mind” (John F. Nash, Jr.). For one such coffee hour, I camethinking about a natural partition of an equilateral triangle into n2

congruent equilateral triangles. Of course, n2 unit equilateral trianglescan cover an equilateral triangle of side n. I asked myself a question,where does the continuous clashes with the discrete? What if we wereto enlarge the side length of the large triangle from n to n +ε, where εis a ‘however small’ positive value, how many unit triangles will weneed to cover it? This comprised my new open problem:

Cover-Up Problem E21.1. Find the minimum number of unit equi-lateral triangles required to cover an equilateral triangle of side n +ε.

During the next coffee hour, I posed the problem to a few Princetoncolleagues. The problem immediately excited John Conway. From theCommons Room he went to the airport, to fly to a conference. Onboard of the airplane, John found a way to do the job with just n2 +2 unit triangles. (Area considerations alone show the need for at leastn2 + 1 of them.) Conway shared his cover-up with me upon hisreturn—at a coffee hour, of course. Now it was my turn to travel to aconference. I usually have a quality time on an airplane: being amongthe strangers is akin to a solitude. What I found onboard was a totallydifferent cover-up with the same number, n2 + 2 unit triangles.

Upon my return, at a coffee hour, I shared my cover-up with JohnConway. We decided to publish our results together. John suggestedsetting a new world record in the number of words in a paper, andsubmitting it to the American Mathematical Monthly. On April28, 2004, at 11:50 a.m. (computers record the exact time!),I submitted our paper that included just two words, “n2 + 2 can”


and our two drawings. I am compelled to reproduce our submissionhere in its entirety (Figs. E21.1 and E21.2).

Can n2 + 1 unit equilateral triangles cover an equilateral triangleof side > n, say n + ε?

John H. Conway & Alexander SoiferPrinceton University, Mathematics, Fine Hall, Princeton, NJ

08544, [email protected] & [email protected]

n2 + 2 can:

Fig. E21.1

Fig. E21.2


The American Mathematical Monthly was surprised, and did notknow what to do about our new world record of a 2-word article. Twodays later, on April 30, 2004, the Editorial Assistant Mrs. MargaretCombs acknowledged the receipt of the paper, and continued:

The Monthly publishes exposition of mathematics at manylevels, and it contains articles both long and short. Your article,however, is a bit too short to be a good [sic] Monthly article. . .A line or two of explanation would really help.

The same day, at the coffee hour, I showed John The Monthlye-mail and asked, “What do you think?” His answer was concise,

“Do not give up too easily.” Accordingly, I replied The Monthly thesame day:

I respectfully disagree that a short paper in general—and thispaper in particular—merely due to its size must be “a bit tooshort to be a good Monthly article.” Is there a connectionbetween quantity and quality? ... We have posed a fine (in ouropinion) open problem and reported two distinct ‘behold-style’proofs of our advance on this problem. What else is there toexplain?

The Monthly, apparently felt outgunned, for on May 4, 2004, thereply came from The Monthly’s top gun, Editor-in-Chief Bruce Palka:

The Monthly publishes two types of papers: “articles,” which aresubstantive expository papers ranging in length from about six totwenty-five pages, and “notes,” which are shorter, frequentlysomewhat more technical pieces (typically in the one-to-fivepage range). I can send your paper to the notes editor if youwish, but I expect that he’ll not be interested in it either becauseof its length and lack of any substantial accompanying text. . .The standard way in which we use such short papers these daysis as “boxed filler” on pages that would otherwise contain a lot ofthe blank space that publishers abhor. . . If you’d allow us to useyour paper in that way, I’d be happy to publish it.

John Conway and I accepted the “filler,” and in January 2005 issueour paper [CS1] was published. The Monthly, however, invented thetitle without any consultation with the authors, and added our title tothe body of the article!


We also published a short article [CS2] in Geombinatorics, wherewe additionally observed that the ‘equilaterality’ is essential, forotherwise n2 + 1 triangles, similar to the large triangle T and withthe ratio of sizes 1: (n + ε) can cover T (Fig. E21.3).

Mitya Karabash receiving the book from Alexander Soifer, Princeton, 2007

Then Mitya Karabash, whom I first met when he was a highschooler in New York and now a brilliant student at ColumbiaUniversity, joined me for further explorations of this problem. Firstof all, we observed the following result, which is better than its simpleproof:

Fig. E21.3


Non-Equilateral Cover-Up E21.2 [KS1]. For every non-equilateraltriangle T, n2 + 1 triangles similar to T and with the ratio of linearsizes 1: (n + ε), can cover T.

Proof. An appropriate affine transformation maps an equilateraltriangle and its covering shown in Fig. E21.2 onto T. This transfor-mation produces a covering of T with n2 + 2 triangles similar to T. Wecan now cover the top triangle with only 2 tiling triangles instead of3 as shown in Fig. E21.4, thus reducing the total number of coveringtriangles to n2 + 1. ■

Mitya and I then generalized the problem from covering a triangle

to covering more complex figures we named trigons.n-Trigon Tn is the union of n triangles from the standard triangu-

lation of the plane such that a triangular rook can find a path betweenany two triangles of Tn , i.e., the union of n edge-connected triangles.If the triangulation is equilateral, then we say that the n-trigon isequilateral. You can see an example of an equilateral 9-trigon inFig. E21.5.

Fig. E21.4

Fig. E21.5


In our cover-up games, we assume that the ratio of thecorresponding sides of the triangles forming the trigon and the titlingtriangles is (1 + ε)/1. We proved the following result:

Karabash-Soifer’s Trigon Theorem E21.3 [KS1]. An n-trigon Tncan be covered with

1. n + 2 triangles if the trigon is equilateral;2. n + 1 triangles if the trigon is non-equilateral.

In spite of all the progress, however, one ‘little’ question remainsopen. I will formulate it here as a conjecture: in 2004 John Conwayand I thought we knew the answer – we just had no idea how to proveit!

Conway-Soifer Cover-Up Conjecture E21.4. Equilateral triangle ofside n + ε cannot be covered by n2 + 1 unit equilateral triangles.

If proven, this conjecture would chaaracterize the equilateral trian-gle as the only triangle requiring n2 + 2 unit covering triangles. Thisproperty could then be used as a definition of an equilateral triangle!

Right after the Cover-Up Problem E21.1, I created the Cover-UpSquared Problem. Naturally, a square of side n can be covered by n2

unit squares. When, however, we let the side length increase merelyto n + ε, we get a new open problem:

Cover-Up Squared Problem E21.5 [Soi2]. Find the smallest number

P(n) of unit squares that can cover a square of side length n + ε.

I devised a covering approach illustrated in Fig. E21.6.


My results were followed by the joint ones by Mitya Karabash andI. The best Mitya and I were able to do in the Cover-Up Squared, wasto match Paul Erdos and Ronald L. Graham’s dual 1975 result [EG1]on packing squares in a square:

Karabash-Soifer’s Theorem E21.6 [KS2]. P(n) ¼ n2 + O(n7/11).Let me explain the “big O” notation. We write f(n) ¼ O(g(n)) if

asymptotically the function f(n) grows not faster than a constantmultiple of g(n).

Immediately upon the publication of this result, on July 18, 2008,I received an e-mail from Ron Graham:

Hi Sasha,I received the latest issue of Geombinatorics with your nicearticle on square covering.

I am mentioning this e-mail because, like in movie series made fortelevision, there was a continuation to Ron’s e-mail. Ameremonth later,he sentme amanuscript, jointlywrittenwith his wife and a distinguishedmathematician in her own rights Fan Chug, entitled Packing equalsquares into a large square. In it, they improved the 33-year oldGraham-Erdos result [GE], and also Mitya’s and my Theorem E21.6:

A B

C

D

E

n – k k + e

Fig. E21.6


Chung-Graham’s Theorem E21.7 [CG].

P(n) ¼ n2 + O n 3þ ffiffi

2pð Þ=7logn

� �

.

Do you see why Chung-Graham result is slightly better thanKarabash-Soifer? Here is why:

7=11 ¼ 0:636363 . . .

3þffiffiffi

2p� �

=7 ¼ 0:630601 . . .

Would you like to hear “tttha rrrest of the story,” as Paul Harveyused to say on the radio? On November 5, 2008, I was asked by theEditor-in-Chief Ole Warnaar to referee Chung-Graham manuscriptsubmitted to the Journal of Combinatorial Theory, Series A! OnDecember 12, 2008, I wrote my report:

REVIEW OF CHUNG-GRAHAM MS

Dear Professor Warnaar,I have read the manuscript with great interest and pleasure.The results represent a small but important improvement,attained by clever improvements in the approach of Erdos-Gra-ham 1975. . .I enthusiastically recommend this article for publicationin JCTA.

The Chung-Graham paper appeared shortly after [CG].The Cover-Up Squared Problem remains open, both in search for

the asymptotically lowest possible answer and for exact values forsmall n. Mitya and I conjecture:

Karabash-Soifer Cover-Up Square Conjecture E21.8 [KS2].P(n) ¼ n2 + Ω(n1/2).

We write f(n) ¼ Ω(g(n)) if asymptotically the function f(n) growsnot slower than a constant multiple of g(n).

If you haven’t started working on these exciting problems yet, youare wasting your time. :) Do share with me any and all advances youachieve!


Ronald L. Graham and Alexander Soifer, March 2009, Boca Raton, FL


E22. Deep Roots of Uniqueness

Inspired by Problem 21.5: “Chess 7�7”

Obviously, any solution of problem 21.5.(B) can be presented in aform of 21 checkers on a 7� 7 board (see left 7� 7 part with 21 blackcheckers in Fig. 21.5.9). It is not obvious that the solution is unique,i.e., by a series of interchanges of rows and columns, any solution ofthis problem can be brought to match precisely the one I presented in

Fig. 21.5.9! Of course, such interchanges mean merely renumberingof players of the same teams.

The uniqueness of the solution of problem 21.5.(B) is preciselyanother way of stating the uniqueness of what is known in mathemat-ics as the Projective Plane of Order 2, the so called “Fano Plane,”denoted by PG(2,2). It was named after Gino Fano (1871–1952), theItalian geometer who pioneered the study of finite projectivegeometries.

The Fano Plane is an abstract construction, with symmetry (dual-ity) between points and lines: it consists of seven points and sevenlines. You can think of rows and columns of our 7 � 7 table as linesand points respectively, with three points on every line and three linesthrough every point. In Fig. E22.1 you can see a traditional depictionof the Fano Plane, where one of the lines is represented by a circle.


183

http://dx.doi.org/10.1007/978-3-319-52861-8_1

http://dx.doi.org/10.1007/978-3-319-52861-8_1

Observe that if on our 7� 7 board we replace each of the 21 check-ers by 1 and the rest of the squares fill with zeroes, we get theso-called Incidence Matrix of the Fano Plane.

In general, a Finite Projective Plane of Order n is defined as a setof n2 + n+ 1 points with the properties that:

1. Any two points determine a line,2. Any two lines determine a point,3. Every point has n + 1 lines through it,4. Every line contains n + 1 points on it.

Fig. E22.1 Depiction of the Projective Plane of Order 2

184 E22. Deep Roots of Uniqueness

E23: More about Love and Death

Inspired by Problem 22.5: “Love and Death”

I hope you did not take the DNAs featured in my problem 22.5 tofaithfully reflect reality. Remember, we are in the Illusory World ofMathematics! To whet your appetite for the problem, I invented thebacterium bacillus anthracis, causing anthrax (death), in problem22.5.(A). In problem 22.5.(B), I went even further by imagining thebacterium bacillus amoris, causing love. I was inspired by a talk by aPh.D. student Martin Klazar that I attended in 1996 during my longterm visit of Charles University in beautiful Prague, Czech Republic.Now Martin is a professor at that same university. The notes I tookin 1996 during Klazar’s talk, contained at the end the followingremark:

By overlapping the 3-gene blocks by their end terms and usingthe same argument, Martin showed that the upper bound can bereduced from 6n +2 (n is here the number of available genes) to4n + 2, and with clever observation of the starting and endingtriples to even 4n – 4. It is possible to achieve the bound of 4n – 7,proof of which would require further cleverness.

These bounds, of course, are stronger than the ones I asked for inproblem 22.5.(B). Their proofs were not presented during Klazar’stalk. Now, Twenty Years After, as Alexandre Dumas named his sequelto The Three Musketeers, I asked Martin Klazar to enlighten us. Hereis his reply, containing the proof of a much stronger upper bound,with my minimal stylistic editing.


185

From: Martin Klazar [mailto:[email protected]]Sent: Monday, September 12, 2016 12:53 p.m.To: Alexander Soifer [[email protected]]Subject: Re: Hello, Martin!

Dear Sasha,

Here is my proof that a 3-sparse word u [i.e., no three con-secutive terms in u may include the same gene more than once]over n-element alphabet avoiding the pattern abcabc as a subse-quence has length at most 4n – 4 (for n > 1).

We denote by F the first occurrences (of a letter) in u, by L thelast occurrences, and by S the intersection F\ L. The intersec-tion consists of exactly the letters that appear in u just once.

We may assume that u has the length |u| of at least six (else thebound holds) and split u into three words u ¼ u’vu” where |u’| ¼|u”|¼ 3. Note that each of the three terms of u’ lies in F and thoseof u” lie in L.

We look now for an upper bound of the length |v| of the middlepart of u. We cover v by k intervals I1, . . ., Ik of length 3 each andby at most one residual term at the end, so that Ii and Ii+1 sharetheir endpoints (thus if v ¼ abcadeca then I1 ¼ abc, I2 ¼ cad,I3 ¼ dec plus the residual term a). If k¼ 0 then there may be tworesidual terms. Hence |v| is at most 3 + 2(k–1) + 1 ¼ 2k + 2.

Consider one of these intervals I ¼ Ii ¼ xyz. By the sparsenesscondition for u, the x, y, z are of course distinct. If x is not inL, y is not in F[L, and z is not in F, then u has an abcabcsubsequence (for then y, z are forced to appear before I and x,y after I). Thus at least one of the following statements is true:(x is in L) or (y is in F[L) or (z is in F). I select one of these threeelements of I (i.e., one for which the clause holds) and call itgood (so all three terms in I may be good, or two of them, butcertainly at least one term of I is good). I hope now it is clearwhat I mean by “good” elements.

Let G be the set of good terms in v.

186 E23: More about Love and Death

We bound k by the number |G| of good terms in v. Since G is asubset of F[L, we have that |G| is at most 2n. Since the Ii are notdisjoint, we may have chosen some g inG for two (but not more)intervals Ii. But if this happens then g is the last term in Ii, thefirst term in Ii+1, and is in S. Thus k is at most |F[L|’ where theapostrophe means that each element of the subset S of F[L iscounted with the weight 2. But we still have that |F[L|’ is at most2n (it is < 2n only if some of the n letters do not appear in u atall), and so k is at most 2n. But k is in fact at most 2n – 6 becausethe 6 terms in u’ and u” lie in F[L and not in S, but not in v andare not used in any Ii.

Summarizing, |u| ¼ |u’| + |v| + |u”| ¼ 6 + |v|, which is at most6 + 2k + 2, which is at most 6 + 2(2n – 6 )+ 2 ¼ 4n – 4.

Best,

Martin

P.S.: I do not know [a] better bound. I think I have somewherestated and proved some lower bound and posed a problem to deter-

mine the extremal function Ex(abcabc, n) exactly, which should bedoable, but as far as I know, has not been done.

Let us formulate the results we in fact proved in problems 22.5.(A) and 22.5.(B) in the notations of Martin Klazar’s post scriptum.

Problem 22.5. (A). Ex(abab, n) ¼ 2n – 1.

Problem 22.5. (B). Ex(abcabc, n) � 6n + 2.

In this notation, Martin proved above the following result:

Upper Bound (Martin Klazar). Ex(abcabc, n) � 4n – 4.

Open Problem (Martin Klazar). Find the exact value forEx(abcabc, n).

Did you get hooked on these sequences and would like to learn andsolve more? Let me quote a paragraph from a relevant page ofWikipedia, so that you will know what to search for:

E23: More about Love and Death 187

In combinatorics, a Davenport–Schinzel sequence is asequence of symbols in which the number of times any twosymbols may appear in alternation is limited. The maximumpossible length of a Davenport–Schinzel sequence is boundedby the number of its distinct symbols multiplied by a small butnon-constant factor that depends on the number of alternationsthat are allowed. Davenport–Schinzel sequences were firstdefined in 1965 by Harold Davenport and Andrzej Schinzel.

188 E23: More about Love and Death

E24: One Amazing Problem and ItsConnections to Everything—A Conversation in Three Movements

Inspired by problem 24.5: “Natural Split”

Some proofs command assent. Others woo and charm the intellect.They evoke delight and an overpowering desire to say, “Amen,Amen”.

—Lord Rayleigh

This is an unusual story, which merits an unusual exposition: ourtrain of thought will be moving backward and forward in time! Nofully chronological presentation seems to fit all of the discoveries wewill touch upon in this exploration.

Part I. 1926: Samuel Beatty

On Friday, April 20, 2007, during the 24th Colorado MathematicalOlympiad, contestants worked on the following problem, known inmathematics as Beatty’s Theorem:

Beatty’s Theorem. Prove that if a and b are positive irrationalnumbers with 1/a + 1/b ¼ 1, then the two sets A¼ {b1ac, b2ac, . . .bnac, . . .}, and B¼ {b1bc, b2bc, . . . , bnbc, . . .} split the set N of pos-itive integers.

A number a is called irrational if it cannot be presented in a forma ¼ p/q with p, q integers and q 6¼ 0. The symbol bcc stands for thelargest whole number not exceeding c. We say that sets A and B splitN if each positive integer n is in either A or B but not in both A and B.


189

Samuel Beatty (1881–1970) published this problem in March 1926in The American Mathematical Monthly [Bea1]. Two solutions werepublished a year later, a joint solution by the Russian-born Germanmathematician Alexander Ostrowski of G€ottingen University andJames Hyslop of University of Glasgow, and another solution byAlexander Craig Aitken of University of Edinburgh. Beatty likelysubmitted the problem with a solution, but I do not know what kindthat solution was.

Clark Kimberling writes [Kim] that Samuel Beatty was the firstperson to receive a Ph.D. degree in mathematics from a Canadianuniversity. He was a colorful teacher and problemist who became the

chairman of the mathematics department, and later, the 21st Chan-cellor of the University of Toronto.

The Russian—and later American—mathematician James VictorUspensky [Usp] proved that a similar problem for more than twoirrational numbers does not have a solution. He appears to be first togive the problem its current formulation, which is simpler than theone used by Beatty.

The Ostrowski-Hyslop type of solution was more elegantlypresented by the Russian twin brothers, mathematicians Akiva andIsaak Yaglom in 1954, and it is their solution of problem 24.5 that Ipresented to you in the present book. It comes from their fabulous1954 book Non-Elementary Problems in Elementary Exposition[YY].

I knew Isaak Moiseevich Yaglom very well; we shared a passionnot only for mathematics, but also for Russian Avant-Garde Art of theearly XX century. In fact, in his Moscow apartment I saw an oilpainting by the famous Russian avant-garde artist Robert Falk. Idedicated to Isaak Yaglom two of my books, How Does One Cut aTriangle? and Geometric Etudes in Combinatorial Mathematics(a joint book with Yaglom’s frequent coauthor VladimirG. Boltyanski). I met Akiva Yaglom only once, in his Researchinstitute of Crystallography of the Soviet National Academy ofSciences.

Pairs of sequences that split the positive integers are now called theBeatty Sequences. They have been the subject of hundreds of papersby now.

190 E24: One Amazing Problem and Its Connections to Everything. . .

Samuel Beatty (MacTutor History of Mathematics Archive, University of StAndrews, Scotland)

Part II. 1907: Willem Abraham Wythoff

Let us move back in time on our train of thought, to the year 1907.That was the year when the Dutch number theorist Willem AbrahamWythoff (1865–1939) published his “A Modification of the Game ofNim” [Wyt]. Wythoff writes:

The game is played by two persons. Two piles of counters areplaced on a table, the number of each pile being arbitrary. Theplayers play alternately and either take from one of the piles anarbitrary number of counters or from both piles an equal number.The player who takes up the last counter or counters, wins.

This game has become quite popular. In literature it is called theWythoff Game, or the Wythoff Nim.

Part II. 1907: Willem Abraham Wythoff 191

The state of the game can be represented by the ordered pair ofnon-negative integers (m, n) with m, n being numbers of counters inthe piles after a move. The position (m, m) is losing (for the playerwho left it after his move), as are positions (n, 0) and (0, n). Withoutloss of generality we can assume now that m < n. The first winningposition is (1, 2),with thedifferencebetween coordinatesn¼2�1¼1.The second winning position is such that cannot be brought to (1, 2) inone move, but must be brought to (1, 2) on the second move regard-less of what the first move is. Clearly, in the second winning positionwe cannot use coordinates 1 and 2, and cannot use the same difference2 � 1 ¼ 1 between coordinates as has been used before. Hence the

second winning position is (3, 5) with difference between coordinatesn ¼ 5 � 3 ¼ 2. Similarly the third winning position, n ¼ 3, is (4, 7).Soon we compile the following table:

n an bn

1 1 2

2 3 5

3 4 7

4 6 10

5 8 13

6 9 15

7 11 18

8 12 20

9 14 23

10 16 26

Observe that

1. an is the least positive integer which is not equal to a1, b1, a2, b2,. . ., an�1, bn�1.

2. bn ¼ an + n.

It is not hard to show that these properties uniquely determine pairsof entries of our table above, and these pairs are all the winningpositions of the Wythoff Nim (see proof in [YY], from which I ampresenting a summary here.)

I hear you wondering, OK, this is nice, but how is it relevant to theBeatty’s Theorem?

Just take a ¼ τ ¼ffiffi

5p þ1

2(this number is the famous Golden Ratio!) in

the Beatty’s Theorem, and consequently calculate b from the equality


1/a + 1/b ¼ 1—you will easily get b ¼ τ2. The split of the positiveintegers N gives us precisely the pairs of winning positions: an ¼ bnτc,bn ¼ bnτ2c.

You would agree with me that this is a remarkable fact, as is thefact that in the solution of his gameWythoff pulled the Golden Ratio τout of a thin air, 19 years before Beatty published his problem!

Willem Abraham Wythoff was born in Amsterdam in 1865,the son of an operator of a sugar refinery. He received a Ph.D. inmathematics from the University of Amsterdam in 1898. From 1899to 1929, Dr. Wythoff was a collaborator of Revue Semestrielle desPublications Mathematiques, a forerunner of Mathematical Reviews[Kim].

Willem Wythoff, reproduced with a kind permission of Clark Kimberling

Yaglom and Yaglom [YY] probably did not know of Wythoff’spaper. They present the same game and refer to it as an old Chinesefolk game tsyan-shidzi, which may mean “choosing stones” in one ofthe languages of China. The Google’s Questions and Answers [Goo]suggest that the proper translation would be jian shi zi, which inChinese characters looks like

潯坒赽

Quite possibly the Chinese were first, ages before Wythoff, but I donot know. If you succeed in establishing the Chinese origin of thisgame, do share your findings with me!

Part II. 1907: Willem Abraham Wythoff 193

Part III. 1894: John William Strutt, Lord Rayleigh

The Romanian-American mathematician Isaac Jacob Schoenberg[Sch] truly amazed me: he somehow discovered that the BeattyTheorem is in reality not Beatty’s! Schoenberg showed that thefamed physicist John William Strutt, Lord Rayleigh stated it32 years before Beatty, in 1894, in his celebrated 2-volume mono-graph The Theory of Sound, and not in its first 1877–1878 edition, butin the second 1894 edition [Ray]! Being at Princeton University washelpful: I went to the Fine Library in the basement of Fine Hall, andverified Schoenberg’s historical discovery: both editions were on theshelf.

Before we look at Rayleigh’s result, I propose an excursion intothe Age of Enlightenment.

Father Marin Mersenne (1588–1648), was an extraordinary man:mathematician for whom Mersenne Primes were named, theologian,philosopher, and music theorist. He was the first ‘walking’ researchjournal: Mersenne circulated problems and ideas through his corre-spondence with such giants as Rene Descartes, Pierre de Fermat,Etienne and Blaise Pascal, Girard Desargues, Galileo Galilei, andChristiaan Huygens. It was said that “To inform Mersenne of adiscovery, meant to publish it throughout the whole of Europe.”Mersenne held meetings at his home with the likes of Fermat, Pascal,and Roberval, and in 1635 founded something like the world’s firstacademy of sciences, the informal private Academie Parisienne theprecursor to the Academie Royal des Sciences.


Father Marin Mersenne (MacTutor History of Mathematics Archive, Universityof St Andrews, Scotland)

You probably know about theMersenne Primes, i.e., primes of theform 2p � 1 for a prime p. Mersenne also dedicated much of hisenergy to studies of music instruments and published his results as thebook Traite de l’harmonie universelle. He discovered the frequencyformula: the frequency f of vibrations of a uniform string AB oflength ¼ l under tension is

f ¼ c=l,

where c is a constant depending on the tension and linear density ofthe string. If all units are chosen such that c¼ 1, then for a unit distantstring (l ¼ 1) we get f ¼ 1.

Let us now leave the Age Enlightenment and move forward intime, to Issac Schoenberg, who summarizes physical information weneed to know in order to understand Rayleigh’s book:

“The string AB may vibrate so that the n � 1 points, which divideAB into n equal parts, are stationary (Fig. E24.1). These points are thenodes, and frequency of the vibrations becomes fn¼ n (n¼ 1, 2, . . .).”

Part III. 1894: John William Strutt, Lord Rayleigh 195

Now we are ready to move back in time to Lord Rayleigh himself[Ray, pp. 122–123]. He considers holding at rest a point (point X inFig. E24.1), looks at the example where it divides the string in theproportion 3:2, and observes that between any consecutive positiveintegers (frequencies of the string without a fixed point) there will be“a number to be found” either in the left side of the string AX or in theright side of the string XB. He observes that the frequencies on the leftand on the right may in some instances coincide and continues:

The coincidences may be avoided by dividing the stringinconmeasurably. Thus, if x be an inconmeasurable [irrational]number less than unity, one of the series of quantities m/x,m/(1 � x), where m is a whole number, can be found whichshall lie between any consecutive integers, and but one suchquantity can be found.

You can recognize in Rayleigh’s statement your problem 24.5: justchoose a ¼ 1/x. Thus, the result ought to be called Rayleigh’sTheorem!

I would like to share with you here excerpts from J. J. O’Connorand E. F. Robertson’s biography of Lord Rayleigh [Mac].

John William Strutt, Lord Rayleigh was born 1842 in LangfordGrove (near Maldon), Essex, England. . .. He entered TrinityCollege, Cambridge, in October 1861. . .. In 1866 Rayleighwas elected a Fellow of Trinity College, Cambridge. . .. He didnot need an academic post to earn his living. Rather when hereturned from the trip to the United States he purchased equip-ment for undertaking scientific experiments and set it up on thefamily estate at Terling.

Rayleigh’s theory of scattering, published in 1871, was thefirst correct explanation of why the sky is blue. . .. The firstvolume of The Theory of Sound, on the mechanics of a vibrating

0 113

A BX1

223

Fig. E24.1 After I. J. Schoenberg, drawn by Robert Ewell


medium which produces sound, was published in 1877, while thesecond volume on acoustic wave propagation was published thefollowing year. . .. From 1879 to 1884 Rayleigh was the secondCavendish professor of experimental physics at Cambridge. . ..The laboratory had been opened five years earlier and Maxwellhad been the first Cavendish professor. . .. In 1884 he [Rayleigh]resigned his Chair at Cambridge to return to his research on hisown estate at Terling. . ..

Rayleigh had been elected as a Fellow of the Royal Society in1873. He received the Royal Medal from the Society in 1882,and became secretary of the Society in 1885, being awarded the

Society’s Copley Medal in 1899. He gave the Society’s BakerianLecture in 1902 and he was elected President of the Society in1905, holding the position until 1908. Rayleigh served as Pres-ident of the London Mathematical Society in 1876–78 and hewas awarded the Society’s Medal in 1890. He also had connec-tions with the Royal Institution, becoming professor of naturalphilosophy there in 1887. He became chancellor of CambridgeUniversity in 1908. . ..

Rayleigh is perhaps most famous for his discovery the inertgas argon in 1895, work which earned him a Nobel Prize in1904. . ..

Remarkably there are 446 items in the list [of Rayleigh’spublications]. They cover an incredible range of topics inapplied mathematics and physics. . .. In addition to the moreusual topics of applied mathematics and physics . . . he wroteon more unusual topics such as Insects and the colour of flowers(1874), On the irregular flight of a tennis ball (1877), Thesoaring of birds (1883), The sailing flight of the albatross(1889), and The problem of the Whispering Gallery (1910).

Let me finish this Exploration with Rayleigh’s words (quoted inH. E. Hunter The Divine Proportion, New York, 1970), which I usedas an epigraph for this exploration, for they are so fitting when wediscuss his remarkable Theorem and its beautiful proof:

Some proofs command assent. Others woo and charm the intel-lect. They evoke delight and an overpowering desire to say,“Amen, Amen.”

Part III. 1894: John William Strutt, Lord Rayleigh 197

Lord Rayleigh (MacTutor History of Mathematics Archive, University of StAndrews, Scotland)


E25: The Story of One Old ErdosProblem

Inspired by problem 25.5: “One Old Paul Erdos’ Problem”

Paul Erdos and Alexander Soifer at the University of Colorado at ColoradoSprings, March 19, 1989, Photo by Tom Kimmell

Paul Erdos’ 3-page 1946 paper [E4603] was so rich on beauty, thatwe already used its “Theorem 3” as problem 14.4 in 1997 at the 14thColorado Mathematical Olympiad (CMO) in a slightly stronger ver-sion of the upper bound [Soi9]:

The Erdos Problem E25.1 (CMO Problem 14.4; P. Erdos, 1946,[E4603]). Given a positive integer n. Let the maximum and minimumdistances determined by n points in the plane be denoted by R andr respectively.


199

(a) Prove that in any set of n points in the plane r can occur at most 3ntimes.

(b) Prove that in any set of n points in the plane R can occur at mostn times.

In 2008, I returned to this Erdos classic paper to use the lowerbound of his “Theorem 1” as problem 25.5 in the 25th ColoradoMathematical Olympiad. Recall, in problem 25.5, Olympians wereasked to prove that the minimum number f(n) of distinct distancesdetermined by n points in the plane satisfies the inequality

f nð Þ � n� 3

4

� �12

� 1

2:

In fact, Paul Erdos proved [E4603] both lower and upped bounds:

n� 3

4

� �12

� 1

2� f nð Þ � cn= lognð Þ1=2:

Paul believed that his upper bound was best possible, and formu-lated a conjecture accordingly.

Paul Erdos’ Conjecture E25.2. f(n)¼ cn/(logn)1/2.

He liked this problem and his conjecture and repeated them in manyof his problem papers, for example [E7525], [E7525], and [E8609].

The first improvement of the lower bound came from Leo Moser,who in 1952 proved [Mos] that

f nð Þ � cn2=3:

The problem then fell in a coma for 32 years. Fan Chung awakenedit in 1984 by proving [Chu] that

f nð Þ � cn5=7:

After Fan, many fine mathematicians achieved gradual improve-ments of the lower bound of f(n). Finally, Larry Guth and Nets HawkKatz, using an ingenious setup created by Gy€orgy Elekes and MichaSharir, surprised the world by proving that

200 E25: The Story of One Old Erdos Problem

f nð Þ � cn=logn:

Thus, Guth and Katz almost completely solved this classic Erdosproblem, for we now know that

c1n=logn � f nð Þ � c2n=ffiffiffiffiffiffiffiffiffilogn

p:

It would be a great achievement and a hard problem for you tocomplete the solution of these problems, and see the lower and upperbounds coincide. Just don’t spend your entire life on it! ;)

Guth–Katz proof first appeared on November 17, 2010 in the openaccess arXiv: 1011.4105. As a result of communications between theauthors and referees, the second and third versions followed in arXivin 2011, and finally a printed version [GK] came out in 2015 inPrinceton’s Annals of Mathematics. On September 5, 2016, LarryGuth advised me by an e-mail that a “slower version” (read: moredetailed one) has just been published as a book by the AmericanMathematical Society [Gut].

E25: The Story of One Old Erdos Problem 201

https://en.wikipedia.org/wiki/ArXiv

https://arxiv.org/abs/1011.4105

E26: Mark Heim’s Proof

Inspired by problem 22.4(B): “Red and White”

In my solution of problem 22.4.(B) I used rotations. The winner of theOlympiad Mark Heim, a senior from Thompson Valley High School,implemented line symmetries instead. He proved that there are twocongruent monochromatic 501-gons, which is a much stronger resultthan the one required in the problem! Let us see how Mark did it;I edited and shortened his presentation.

Mark Heim’s Solution of Problem 22.4.(B). The regular 2005-gonP has 2005 mirror reflections, each with one vertex stationary and 1002pairs transposed, with the total of 2005 � 1002 ¼ 2,009,010 individualtransposition components. Since the total number of transpositions is

2005

2

� �¼ 2,009,010, each must have been counted exactly once in

reflections (corresponding to the geometric fact that for every pair ofdistinct vertices in P there is a unique line, the perpendicular bisector,transposing the given pair). Among two-colored vertices of the polygonP, there are at least 1003 vertices of the same color, say blue, which

guarantee at least 1003

2

� �¼ 502,503 transpositions of blue pairs. Since

502,503/2005> 250.6, there is a vertex v ofPwith at least 251 reflectionscontainingmonochromatic transpositions of blue pairs, all with respect toa line through v. These 251 blue pairs contain 502 blue vertices. There-fore, we have a monochromatic 502-gon with a mirror symmetry. Thisimmediately allows us to choose two distinct monochromatic 501-gons,each a mirror image of the other with respect to a line through v.■


203

Is 501-gon the best we can get?

Open Problem E26.1. Find the largest integer f(2005), such that nomatter how the vertices of a regular 2005-gon are two-colored, there

will be two monochromatic n-gons of the same color, congruent toeach other.

Mark Heim gave us a start: f(2005) � 501.More generally:

General Open Problem E26.2. Given a positive integer m � 7. Findthe largest integer n ¼ f(m), such that no matter how the vertices ofa regular m-gon are two-colored, there will be two monochromaticn-gons of the same color, congruent to each other.

As you noticed, Mark Heim proved even more: his congruent251-gons are mirror images of each other.

Perhaps, such a requirement would affect the answer to the GeneralOpen Problem E26.2:

Open Problem E26.3. Given a positive integer m � 7. Find thelargest integer n ¼ fM(m), such that no matter how the vertices of aregular m-gon are two-colored, there will be two monochromaticn-gons of the same color that are mirror images of each other.

You may recall, in our initial solution of problem 22.4.(B), weobtained a pair of congruent 10-gons, such that one was an image ofanother under a rotation. How would such an add-on requirementaffect the outcome?

Open Problem E26.4. Given a positive integer m � 7. Find thelargest integer n ¼ fR(m), such that no matter how the vertices of aregular m-gon are two-colored, there will be two monochromaticn-gons of the same color that are images of each other under rotations.

204 E26: Mark Heim’s Proof

E27: Coloring Integers—Entertainmentof Mathematical Kind

Inspired by problem 27.5: “Colorful Integers”

In 1987, Paul Erdos posed the following problem to the well-knownIsraeli mathematician Yitzhak Katznelson, a Stanford professor, whorecollects 14 years later [Kat]:

In 1987 Paul Erdos asked me if the Cayley graph defined onZ [the set of integers] by a lacunary sequence has necessarily afinite chromatic number. Below is my answer [in the positive],delivered to him on the spot but never published [until 2001].

As usual in my writings, I am naming this result after both con-tributors, the author of the conjecture and the prover.

The 1987 Erdos–Katznelson Theorem. Let ε> 0 be fixed andsuppose that S¼ {n1, n2, ... , nj, ...} is a sequence of positive integerssuch that nj+ 1> (1 + ε)nj for all j � 1.1 Define a graph G ¼ G(S)with vertex set Z by letting the pair (n, m) be an edge if and only ifjn � mj 2 S. Is the chromatic number χ(G) finite?

Katznelson presented the Erdos conjecture and his proof at a 1991seminar attended by then young Israeli mathematician Yuval Peres.

Peres, currently a professor at the University of California Berkleyand a researcher at Microsoft, jointly with Wilhelm Schlag, presentlyat the University of Chicago, [PS] recently improved Katznelson’supper bound for the chromatic number in the Erdos problem. From

1Such a sequence is called lacunary.


205

here, I let Matthew Kahle, Professor at Ohio State University and theColorado Mathematical Olympiad 1990 and 1991 winner, tell the restof the story. Matt writes to me in the October 12, 2016, e-mail:

Dear Sasha,Here is a brief history of this problem. Feel free to extract

whatever is interesting for your own creative purposes. :)I saw an interesting seminar talk by Yuval Peres (about his

joint work with Schlag) when I was a graduate student at Uni-versity of Washington. He addressed some questions of Erdosabout coloring graphs on integers like this. As long as thesequence of distances is increasing at least exponentially fast,

the chromatic number is finite, and they can even get an explicitupper bound on the chromatic number. So I asked at the end ofhis talk about the factorial graph.

Not sure why this was my first question, but I guess it was thefirst sequence I thought of that grew super-exponentially fast.But the factorial graph is also particularly appealing since noperiodic coloring will work, and periodic colorings are what youwant to try.

Yuval said he didn’t know, but he guessed that their proofwould show that it was probably less than 10.

So I asked my office mate at the time, Tristram Bogart, whowas also studying combinatorics, what is the chromatic numberof the factorial graph. We quickly established that you need atleast 4 colors, and applying what we remembered from Peres’sproof, we were able to prove that the chromatic number was atmost 5. So we knew that chromatic number was either 4 or 5. Webet a beer on the outcome: I bet 4 and Tristram bet 5.

There it stood for a few years, until I asked the question tosome bright [high school] students at Canada/USA Mathcamp. Ijust defined the graph and offered $20 for figuring out what itschromatic number is. But I did not give them any hint. I did notexplain the idea of Peres’s proof, or what Tristram and I knewso far.

Amazingly, Adam Hesterberg came back the next day andclaimed the $20, showing that the chromatic number is exactly4. His proof of the existence of such a number was very similarto, or the same to Peres’s proof, but he was a high school student,rediscovering the methods of the professionals! And he

206 E27: Coloring Integers—Entertainment of Mathematical Kind

improved on whatever Tristram and I had been doing, becausewe were only able to find a 5-coloring so he must have been alittle more efficient!

I was happy to pay Adam the $20. I did collect on the beerfrom Tristram sometime later, and joked that $20 was expensivefor a beer (the net result for me), but I didn’t mind because I washappy that my guess was right.

When I sat down to remember the proof, for your book, Iremember being a little bit dissatisfied with the nested intervals.It is a subtle fact of analysis that an infinite sequence of nestedclosed intervals must have a point of intersection. After all, itfails if the intervals are open instead of closed! So I wondered if Icould instead find an explicit r that works, perhaps as an infiniteseries. Playing around for a few minutes, I found the r in theproof I gave you. Somehow this is more satisfying to me, since itis slightly more elementary, and because the coloring is moreexplicit.

Adam Hesterberg is currently a Ph.D. student in mathematicsat MIT. His research interests include graph theory, computa-

tional geometry, and theoretical computer science.Warm regards,Matt

The high school student in this story, Adam Hesterberg, soon afterwon the 2007 USA Mathematical Olympiad. He then graduated fromPrinceton University, and is now a graduate student at the Massachu-setts Institute of Technology (MIT).

Matt submitted problems 27.5.(A) and 27.5.(B) to me for ourOlympiad. At that time he wrote up Adam Hesterberg’s solution of27.5.(B). Adam’s main idea of the existence of an “r” was nothingshort of brilliant; he used a sequence of nesting segments for it. Thesolution of problem 27.5.(B) that you saw in the chapter dedicated tothe 27th Olympiad, was written especially for this book by MatthewKahle.

Imagine, 2 days after the Olympiad, I received a remarkable e-mailfrom the Olympiad judge and the first prize winner of the FirstColorado Mathematical Olympiad in 1984 Russel Schaffer.

Monday, April 26, 2010 3:22 p.m.To: Alexander Soifer <[email protected]>

E27: Coloring Integers—Entertainment of Mathematical Kind 207

Cc: [email protected]: Alternate Solution to 5b

Alexander,On the drive back to Wyoming on Saturday afternoon, I

thought a bit about problem 5b and came up with an alternatesolution.

Today, I formalized it and wrote it up.This isn’t the simple solution that you asked for on Friday. It is

less elegant than the solution that you presented. It is, however, aworkmanlike solution based on more straightforward intuition.No flashes of daring brilliance required. The intuition is such thata smart high school student could reasonably come up with it in

the allotted time.Russel

Let me now reproduce for you Russel’s e-mail attachment.

Russel Schaffer’ Traveling Solution of Problem 27.5.(B). Definethe color for each integer x � 0 to be:

c xð Þ ¼X1

i¼0

xþPi�1

j¼1

4j� 1ð Þ!

4ið Þ!

26664

37775

0BBB@

1CCCA mod 4ð Þ

where we use square brackets [ ] to indicate the integer part of a realnumber. Clearly this defines a coloring with four colors. To demon-strate that no two integers have the same color if they are separated byn! for some integer n, consider (c(d + n!) � c(d)) (mod 4) for someintegers d� 0 and n� 0. As a notational convenience, we let k¼ [n/4].


c d þ n!ð Þ � c dð Þð Þ �Xk�1

i¼0

d þ n!þXi�1

j¼1

4j� 1ð Þ!

4ið Þ!

266664

377775�

d þXi�1

j¼1

4j� 1ð Þ!

4ið Þ!

266664

377775

0BBBB@

1CCCCA

þd þ n!þ

Xk�1

j¼1

4j� 1ð Þ!

4kð Þ!

266664

377775�

d þXk�1

j¼1

4j� 1ð Þ!

4kð Þ!

266664

377775

þX1

i¼kþ1

d þ n!þXi�1

j¼1

4j� 1ð Þ!

4ið Þ!

266664

377775�

d þXi�1

j¼1

4j� 1ð Þ!

4ið Þ!

266664

377775

0BBBB@

1CCCCA

The summand on the first line will be congruent to 0 modulo4 because

4k�� n!

4ið Þ! for all 0 � i � k � 1:

Because n!4k! is an integer, the summand on the second line equals:

1 � 1 mod4ð Þ if n � 0 mod4ð Þn � 1 mod4ð Þ if n � 1 mod4ð Þn n� 1ð Þ � 2 mod4ð Þ if n � 2 mod4ð Þn n� 1ð Þ n� 2ð Þ � 2 mod4ð Þ if n � 3 mod4ð Þ

Because 0 < n!4ið Þ! < 1, for all i > k, we know that each term under

the summation in the third line must be 0 or 1. The entire summand onthe third line is thus 0 or 1 because at most one term under thesummation can be non-zero. Assume to the contrary, that there areintegers a and b, k < a < b, such that:

d þ n!þXa�1

j¼1

4j� 1ð Þ! � xa 4að Þ! > d þXa�1

j¼1

4j� 1ð Þ!

d þ n!þXb�1

j¼1

4j� 1ð Þ! � xb 4bð Þ! > d þXb�1

j¼1

4j� 1ð Þ!


for some integers xa and xb. Then we would have:

1þd þ

Xa�1

j¼1

4j� 1ð Þ!

n!� xa 4að Þ!

n!>

d þXa�1

j¼1

4j� 1ð Þ!

n!

1þd þ

Xa�1

j¼1

4j� 1ð Þ!

n!�

xb 4bð Þ!�Xb�1

j¼a

4j� 1ð Þ!

n!>

d þXa�1

j¼1

4j� 1ð Þ!

n!

Now, the center quantities in both inequalities are both integers,and they are both bounded by an identical pair of values which differby 1, therefore they must be equal. But we cannot have

xa 4að Þ! ¼ xb 4bð Þ!�Xb�1

j¼a

4j� 1ð Þ!

because (4a)! divides the left hand side and all terms of the right handside except for (4a � 1)!. We have thus reached a contradiction andcan conclude that at most one term under the summation can benon-zero.

Therefore, in the above expansion of (c(d + n!)� c(d)) (mod 4) wesee that the first summand will always be 0, the second 1 or 2, and thethird 0 or 1. We conclude that (c(d + n!) � c(d)) (mod 4) is alwaysnonzero. ■

Having finished his solution, Russel continues:

Believe it or not, there is some intuition behind this solution.Consider the coloring where each integer x is assigned the colorx (mod 4). This works just fine for pairs of numbers whosedifference is 1, 2, or 6. In fact, not only does it work for 1, 2,and 6, it also works for 1 + 1, 2 + 1, and 6 + 1.

This gives us room to squeeze in an adjustment to make thingswork for pairs of numbers that differ by 24. We add 1, modulo4, to all numbers in the block of 24 integers contained in [24, 47].We add 2, modulo 4, to all numbers in the next block of24 integers; add 3, modulo 4, to the following block of 24 inte-gers, and so on.As observed above, the adjustment did not cause problems withthe coloring for pairs whose difference is 1, 2, or 6. And the


adjusted coloring also works for pairs whose difference is24, 120, 720, and 5040. We run into trouble again only whenwe need to compare pairs of numbers that differ by 8!.

We resolve the problems with 8! by doing another adjustment,adding increasing increments to successive blocks of 8! integers.We need only be careful that no two block boundaries get tooclose to the block boundaries from the first adjustment. To thisend, we begin our adjustment 3! before the end of the first blockof 8! integers.

As before, the adjusted coloring works until we reach the nextdifference of the form 4i!. This is 12!. Again, we perform an

adjustment whose block boundaries are guaranteed not to be tooclose to any previous block boundaries. We continue in thismanner, making adjustments for all differences of the form 4i!.The coloring given at the head of the first page formalizes aninfinite sequence of these adjustments.

Let me repeat one lyrical line from Russel’s solution:

Believe it or not, there is some intuition behind this solution.

Indeed, Russel possesses some intuition!In fact, I believe that if Russel were not a senior when in 1984 I

started the Olympiad, he would have won as many Olympiads as hewere to enter. My fault: I started the Colorado Mathematical Olym-piad too late. :)

When a fabulous problem 27.5.(B) gets solved, we are inspired tosee better, look further, aspire a higher ground. Recall the famousChromatic Number of the Plane Problem:

Find the minimum number of colors χ(E2) required for coloringthe Euclidian plane E2 in such a way that no two points of thesame color are at a distance 1 apart.

In 1950, it was established (easy) that 4� χ(E2)� 7, and that is stillall we know today in general case! Read much more about thisproblem in my The Mathematical Coloring Book: Mathematics ofColoring and the Colorful Life of Its Creators [Soi3]. Inspired by hisproblem 27.5.(B), Matthew Kahle proposes to increase the set offorbidden monochromatic distances in the plane from a singleton{1} to all factorials {1!, 2!, . . ., n!, . . .}.


Open Factorial Coloring Problem on the Plane (M. Kahle). Findthe minimum number of colors χF(E

2) required for coloring theEuclidian plane E2 in such a way that no two points of the samecolor are at a factorial distance apart.

We do not even know whether χF(E2) is finite, so you have plenty

of enjoyable research to undertake!Of course, the dimension in this problem can be raised, and thus we

find ourselves in space, in the Euclidean n-dimensional space En.

Open Factorial Coloring Problem in n-Space. Find the mini-mum number of colors χF(E

n) required for coloring the Euclidian n-space En in such a way that no two points of the same color are at afactorial distance apart.

The answers to both open problems could be not finite but ratherinfinite cardinal numbers. Reading on basic Set Theory could aid yourengagement in this infinite fun!


E28: The Erdos Numberand Hamiltonian Mysteries

Inspired by problem 23.5 “Math Party”

The problem 23.5 opens with a definition: We say that mathemati-cians a and b have each other’s number n if there is a coauthor-chainof mathematicians a¼ a0 , a1 , . . . , an¼ b such that every consecu-tive pair in the chain are coauthors on at least one publication.

I was inspired to create this definition by the so-called ErdosNumber of a mathematician b, which is equal to the length of ashortest coauthor-chain where the mathematician a is Paul Erdos.Paul Erdos’ Erdos number is defined as 0. The Erdos number 1 isassigned to coauthors of Erdos (I am happy to be one of them). TheErdos number 2 is assigned to those who had a joint paper with a PaulErdos’ coauthor, etc.

Wikipedia provides relevant information (https://en.wikipedia.org/wiki/Erd%C5%91s_number):

Erdos wrote around 1,500 mathematical articles in his lifetime,mostly co-written. He had 511 direct collaborators; these are thepeople with Erdos number 1. The people who have collaboratedwith them (but not with Erdos himself) have an Erdos number of2 (9267 people as of 2010, those who have collaborated withpeople who have an Erdos number of 2 (but not with Erdos oranyone with an Erdos number of 1) have an Erdos number of

3, and so forth. A person with no such coauthorship chainconnecting to Erdos has an Erdos number of infinity (or anundefined one). Since the death of Paul Erdos, the lowestErdos number that a new researcher can obtain is 2.


213

https://en.wikipedia.org/wiki/Erd%C5%91s_number

https://en.wikipedia.org/wiki/Erd%C5%91s_number

https://en.wikipedia.org/wiki/Defined_and_undefined

The Erdos Number became so popular that Wikipedia lists peopleby their Erdos number for Erdos numbers 1, 2, and 3: https://en.wikipedia.org/wiki/List_of_people_by_Erd%C5%91s_number

These lists probably came from the Internet site The Erdos NumberProject that Jerry Grossman created at his Oakland University:

http://wwwp.oakland.edu/enp/

Among a wealth of trivia information, there is something I wouldlike to share with you here:

There are about 1.9 million authored items in the Math Reviewsdatabase, by a total of about 401,000 different authors.” This

allows to create a graph! “The collaboration graph C has theroughly 401,000 authors as its vertices, with an edge betweenevery pair of people who have a joint publication (with orwithout other coauthors).

Another original Internet site is called Collaboration Distance:

http://www.ams.org/mathscinet/collaborationDistance.html

You can enter names of two people and see their collaborationdistance, i.e., the length of a shortest coauthor-chain, to use theterminology of our problem 23.5. The collaboration distance betweenAlbert Einstein and Paul Erdos is 2: they both were coauthors withErnst Gabor Straus. This implies that the collaboration distancebetween Albert Einstein and I is 3. :)

Let us now return to problems of mathematical kind. In the solutionof problem 23.5, we introduced a definition. Let me repeat it here foryour convenience.

A graph is called Hamiltonian-connected if for every two verticesv, w there is a path starting at v and ending at w that goes through allother vertices of the graph exactly once.

In the solution of problem 23.5, we proved that for any tree T, thegraph T3 is Hamiltonian-connected. It is time to pose a fundamentalgeneral problem:

Open Problem E28.1. Find a necessary and sufficient condition for agraph to be Hamiltonian-connected.

The problem is not just open—no serious conjecture of a necessaryand sufficient condition has been formulated for it.

214 E28: The Erdos Number and Hamiltonian Mysteries

https://en.wikipedia.org/wiki/List_of_people_by_Erd%C5%91s_number

https://en.wikipedia.org/wiki/List_of_people_by_Erd%C5%91s_number

http://wwwp.oakland.edu/enp/

http://www.ams.org/mathscinet/collaborationDistance.html

A Hamiltonian cycle is a path starting and ending at the samevertex that goes through all other vertices of the graph exactly once.A graph that contains a Hamiltonian cycle is called a Hamiltoniangraph. We arrive at another fundamental general problem:

Open Problem E28.2. Find a necessary and sufficient condition for agraph to be Hamiltonian.

This problem too is not just open—no serious conjecture a neces-sary and sufficient condition has been formulated for it. I know anumber of mathematicians who dedicated much of their lives toworking on these two problem. Many sufficient conditions havebeen discovered. However, we have no conjecture for a necessaryand sufficient condition. In fact, if your dream is to be mentioned onthe front page of the New York Times, proving a necessary andsufficient condition for one of these problems would suffice! :)

It is fascinating that a similar-sounding problem of finding a nec-essary and sufficient condition for a graph to be Eulerian is very easyto find. In fact, the great Leonard Euler formulated it in 1736.

A graph is called connected if there is a path between any twovertices of the graph along a series of its edges. A graph is calledEulerian if there is a circuit (i.e., a path that ends at the vertex itstarted from) that contains every edge of the graph exactly once.

Euler’s Criterion E28.3 (L. Euler [Eul], 1736). A connected graph isEulerian if and only if the degree of each its vertex is even.

Euler formulated this necessary and sufficient conditioncompletely, but supplied a proof only in the necessary direction.Did he have a proof of the sufficient condition? I think so, for Eulerproved many much harder results. Perhaps, he omitted this proofbecause it seemed too easy for him.

E28: The Erdos Number and Hamiltonian Mysteries 215

E29: One Old Erdos–Turan Problem

Inspired by problem 30.5: “One Old Erdos Problem”

Paul Erdos and Alexander Soifer, The Last Meeting, Baton Rouge, LA,February 22, 1996

In the early 1930s Budapest, a group of young Jewish studentsregularly met in a park or took excursions to the countryside todiscuss mathematics. On May 28, 2000, in Sydney, Australia, I hadthe pleasure of sharing dinner with two members of this group,George Szekeres and Esther Klein, who recalled for me the regularmembers of this remarkable group: Paul (Pal) Erdos, Tibor Grünwald


217

(later Gallai), Gergor (Geza) Grünwald, Esther Klein, Gy€orgy(George) Szekeres, Lily Szekely (later Sag), Paul (Pal) Turan,Endre Vazsonyi, and Marta Wachsberger (later Sved). The twoPauls became coauthors when still in high school, even before theyactually met: the solution they obtained independently was publishedin the physics-mathematical magazine K€oMaL.

To commemorate Paul Erdos’ 100th birthday, I looked through hisold papers. I extracted and adopted this problem 30.5 from the 1934article written by the 20-year-old Paul Erdos and the 23-year-old PaulTuran [ET]. The authors used a larger upper bound 3� 22012, whichcan be reduced to 22013 + 1 without altering the ideas of the proof.

We say that a positive integer m is composed of distinct primesp1, p2, . . ., pn if every prime factor of m is one of these primes.

The 1934Erdos–TuranTheorem. Two-term sums of a set of 2n+ 1distinct positive integers require for their composition at least n + 1primes.

Let Ƒ(n) be the largest number such that for any set of Ƒ(n) distinctpositive integers, two-term sums are composed of a set of n primes. Inthis notation, the Erdos–Turan Theorem can be reformulated asfollows:

The 1934 Erdos–Turan Upper Bound. Ƒ(n) � 2n.

Erdos and Turan realized that in their theorem, they proved theupper bound of convenience. We mathematicians first prove what wecan! They then came up with a bold conjecture.

The 1934 Erdos–Turan Conjecture. Ƒ(n) ¼ O(n1 + ε) for anyε > 0.

We have already used the “big O” notation in Exploration E21. Wewrite f(n) ¼ O(g(n)) if asymptotically the function f(n) grows notfaster than a constant multiple of g(n).

Having formulated this obviously hard conjecture, the authorsadmit, “but actually we cannot prove this relation.” Not only is theconjecture still open today, the Erdos–Turan Upper Bound appears tonot have been improved in over 80 years and counting.

The Erdos–Turan Theorem has the following corollary:

Corollary. π(n)> log2 x, where π(n) denotes the number of primes<n.

218 E29: One Old Erdos–Turan Problem

This inequality is not very strong, compared to the definitive result:

The Prime Number Theorem. π(x) ~ x/log x.

The symbol ~ means that the left and right side functions are equalasymptotically.

This theorem was first conjectured by Adrien-Marie Legendre in1798, and proved almost a century later, in 1896, independently byJacques Salomon Hadamard [H1], [H2] and Charles-Jean EtienneGustave Nicolas de la Vallee Poussin [VP].

In 1949 Atle Selberg [Sel] and Paul Erdos [E4902] independentlygave the first ‘elementary’ proofs of the Prime Number Theorem(PNT). ‘Elementary’ means without the use of complex analysis,and not at all that the proofs were easy—they were hard.

Then there came a famous scandal. Selberg proved his fundamentalinequality; Erdos used it to prove the generalized Chebyshev Theo-rem; Selberg used Erdos’ result to find his first proof of PNT; Erdosproved PNT at the same time. What were they to do? Erdos, alwaysgenerous, proposed to publish a joint paper. Selberg wanted each ofthem to publish his fragment of the proof separately. The idea ofpublishing individual articles side by side was, according to ErnstG. Straus, defeated by Hermann Weyl, a Permanent Professor at theInstitute for Advance Study from its inception, who brought youngSelberg in the Institute and mentored him. Straus recalls:

It was Weyl who caused the Annals [of Mathematics] to rejectErdos’ article and publish only a version of Selberg thatcircumvented Erdos’ contribution, without mentioning the vitalpart played by Erdos in the first elementary proof, or even in thediscovery of the fact that such a proof was possible.

The situation was politicized to a point that the Bulletin of theAmerican Mathematical Society informed Erdos that the refereerecommended against publication. For more detailed historical dis-course, I refer you to the fine 2009 article The Elementary Proof of thePrime Number Theorem by Joel Spencer and Ronald Graham [SG].

The crux of the matter is very educational. Selberg wanted tocompete with Erdos and other mathematicians as if it were an Olym-pic sprint dash, and coming first meant everything. Erdos, as always,alone and with numerous coauthors, competed with the field, the field

E29: One Old Erdos–Turan Problem 219

of mathematics, for the glimpse of The Book. I heard Paul’s clarifi-cation of his notion of The Book several times. Here is one headdressed to the Colorado Mathematical Olympiad in March 1989:

God has a transfinite Book, which contains all theorems and theirbest proofs, and if He is well intentioned toward those, He showsthem the Book for a moment. And you don’t even have tobelieve in God, but you have to believe that The Book exists.

220 E29: One Old Erdos–Turan Problem

E30: Birth of a Problem: The Storyof Creation in Seven Stages

Inspired by problem 29.5: “Beyond the Finite II”

Problem solving requires the existence of problems. Somebody has tocreate them. But how?

Let me share here with you how creating problems for the Olym-piad usually works for me. Every early March I attend the Southeast-ern International Conference on Combinatorics, Graph Theory, andComputing at Florida Atlantic University. Listening to talks thereaccelerates the gears of my mind. Ideas of research mathematicsinspire my problem creating. Of course, research mathematics oftendeals with a sophisticated topic matter, often in a break-through-the-wall by all available means solutions, and so the first task is to extract‘naked’ beautiful ideas out of research mathematics. The second taskis to dress up these ideas and present them to the Olympians in a formof an enjoyable story. Consequently, some of our problems havespecial titles, such as “Stimulus Package”, “Chess 7 � 7”, “OldGlory”, “Stone Age Entertainment”, “King Arthur and the Knightsof the Round Table”, “Crawford Cowboy Had a Farm, Ee i ee i oh!”,etc. I would like to illustrate here how a problem is created on theexample of the Olympiad problem 29.5 Beyond the Finite II that Icreated over the month of March 2012. You will witness the con-struction of a two-way bridge between problems of mathematicalOlympiads and research problems of mathematics. Let us begin, thetrain of thought is ready to depart from the first station.


221

I wanted to let my Olympians ‘touch’ the infinity, to feel howdifferent it is from finite magnitudes. And so, I created the Stage1 Problem.

Stage 1 Problem. Infinitely many circular disks of diameter 1/700 aregiven inside a unit square. Prove that there is a circular disk ofdiameter 1/2012 that is contained in infinitely many of the givendisks.

A circular disk consists of all points on and inside of a circle.

Solution. My solution was based on inventing a 2012 � 2012square grid on the given unit square and observing that every givendisk contains at least one of the unit cells of the grid. This dictates the

numbers appearing in the problem:700� 2ffiffiffi

2p

< 2012. (We can evenreplace 700 by 711, but not by 712).

Assume that no cell is contained in infinitely many of the givendisks. For each of the finitely many unit cells c of the grid G we getfinitely many disks f(c) containing this cell c, so the sum Σ of allsummands f(c) is finite. This sum includes all the given disks (somepossibly counted more than once), for we have shown that each diskcontains at least one unit cell. We have arrived at contradiction, forwe were given infinitely many disks. Therefore, there is a cell c1 thatis contained in infinitely many of the given disks.

All that is left is to inscribe the required in the problem circular diskin the unit cell c1. ■

I then thought: infinity is a rather large number; why should thesize (1 � 1) of the given square matter? It really does not, we can getrid of the size of the square and move to the Stage 2 Problem.

Stage 2 Problem. Infinitely many circular disks of diameter 1/700 aregiven inside a square. Prove that there is a circular disk of diameter1/2012 that is contained in infinitely many of the given disks.

Of course, with the size of the square gone, we can increase all theremaining numbers to get to the Stage 3 Problem.

Stage 3 Problem. Infinitely many circular disks of radius 1 are given

inside a square. Prove that there is a circular disk of diameter 1=ffiffiffi

2p

that is contained in infinitely many given disks.

222 E30: Birth of a Problem: The Story of Creation in Seven Stages

My next thought was about the shape. Must we embed our flock ofcircular disks in a square? Does the shape matter? It really does not.We can get rid of the square shape and thus get to the next stage.

Stage 4 Problem. Infinitely many circular disks of radius 1 are giveninside a bounded figure in the plane. Prove that there is a circular disk

of diameter 1=ffiffiffi

2p

that is contained in infinitely many of the givendisks.

The next question I posed to myself was inspired by a typical for

mathematicians search for the best possible result: is 1=ffiffiffi

2p

the largestpossible diameter we can guarantee to exist inside infinitely many ofthe given disks? Here came in handy the idea Bob Ewell used to solveof Stage 1 Problem. Even though Bob used it to solve Stage 1, thisapproach contained a possibility of the next stage.

Stage 5 Problem. Infinitely many circular disks of radius 1 are giveninside a bounded figure in the plane (Fig. E30.1). Prove that there is acircular disk of radius 0.9 that is contained in infinitely many of thegiven disks.

In the April 20, 2012, Colorado Mathematical Olympiad wedecided to use Stage 5 Problem as the most difficult problem 29.5.(B) of that year: Beyond the Finite II. I trust you have read the

Fig. E30.1

E30: Birth of a Problem: The Story of Creation in Seven Stages 223

solution in chapter “Twenty-Ninth Colorado Mathematical Olym-piad” and so there is no need to repeat it here.

A lovely Olympiad problem was born, what else can an Olympicproblem creator desire? Perhaps, a cup of espresso, but a mathema-tician always strives for the next step. Is there the next station for ourtrain of thought? We were embedding circular disks, is the ‘circular-ity’ essential to our games? Not at all! To proceed further, we need adefinition.

Translates F1, F2, . . . of a convex figure F, are just that: figuresobtained from F by translations (Fig. E30.2).

Now we are ready for Stage 6, where circularity of disks disap-pears, and we hope to get the same result as in the previous stage, butnow for all convex figures.

Stage 6 Conjecture. Given a convex figure F. Infinitely many trans-lates of F of area 1 are given inside a bounded figure in the plane(Fig. E30.3). Prove that for any ε> 0 there is a translate of F of area1� ε that is contained in infinitely many of the given translates.

Can we relax the requirement of packing in only translates? AsPresident Barak Obama used to famously say, Yes we can!

Fig. E30.2


Stage 7 Open Problem. Given a convex figure F. Infinitely manyfigures of area 1 congruent to F are embedded inside a bounded figureA in the plane (Fig. E30.4). Find the maximum area ϕ(F) such thatthere is a figure similar to F of area ϕ(F) that is contained in infinitelymany of the given figures, regardless of their embedding in A.

Fig. E30.3

Fig. E30.4


This is a very exciting and very difficult problem. I believe that theresult depends upon the shape of F, hence my introduction of thefunction ϕ(F) defined on convex figures F.

Having created this train of thought, the Stage 6 Conjecture and theStage 7 Open Problem, on March 23, 2012, I asked one of world greatgeometers, the Sage of Convexity and Polyhedra, Branko Grünbaumfor comment. He replied (long live e-mail!) the same day:

Dear Sasha,Your question for arbitrary convex figure is interesting, and I

think it is new. It seems to me that the proof [of Stage 6Conjecture] can be completed by using for the ‘center’ of F the(area) centroid of F. Then it is well known (Minkowski’s mea-sure of asymmetry) that every chord of F through this center isdivided by the center in a ratio that is between 1/2 and 2.1 Thisshould be enough to show that there is a homothetic copy of F,for any given ε> 0. Notice that the whole thing generalizes to alldimensions d, by the same argument, because then Minkowski’sratio is between 1/d and d.

I hope this is satisfactory.Best wishes,Branko

1The Centroid of a plane figure F is the intersection of all straight lines that divide F into two

parts of equal moment about the line. The centroid of a plane figure is the point on which it

would balance when placed on a needle.

Minkowski’s Measure of Asymmetry Theorem (1897) states that every chord of F through

its centroid is divided by the centroid in a ratio that is between 1/2 and 2.


Branko Grünbaum

Branko Grünbaum does not only contribute an idea of an assault onthe Stage 6 Conjecture, but also suggests an n-dimensional general-izations of the Conjecture! For further reading on the topic, let merefer you to the classics of the last two centuries, Minkowski [Min]and Grünbaum [Grü].

There are two reasons I am leaving these continuations of my trainof thought to you, my reader. First of all, you will enjoy it. Secondly, Iwish my book to remain alive: as Pablo Picasso said in 1961,

Unfinished, a picture remains alive, dangerous. A finished workis a dead work, killed.


Part III

Olympic Reminiscences in FourMovements

Movement 1. The ColoradoMathematical Olympiad Is Mathematics;It Is Sport; It Is Art. And It Is AlsoCommunity, by Matthew Kahle

Professor Soifer asked me to share some reminiscences for his newbook. I am happy to! First, I will reprint parts of an essay I wrote in2008, for an earlier Olympiad book.

Professor Alexander “Sasha” Soifer told me that for the newedition of the Colorado Mathematical Olympiad book, hewanted to include a chapter written by past winners, about therole of the Olympiad in their lives, their view of math, and theirfuture careers. I am very happy to contribute an essay for this.

I want to emphasize that there are many features of the CMO[the Colorado Mathematical Olympiad] that distinguish it in mymind from the other contests which I competed in. First, theformat itself was different from most competitions—five ques-tions and four hours! It is true that there are other Olympiadstyle contests that I would come across years later (USAMO,Putnam, . . .), but I first sat for the CMO as an 8th grader, when

our teacher and coach Betty Daniels took a few of us from themiddle school to UCCS [University of Colorado at ColoradoSprings] for the Olympiad.

None of us had ever seen an exam like this before! I rememberSasha coming around to the rooms at the beginning of the day,asking if anyone had questions. I think I asked him if we coulduse calculators. He smiled impishly, shrugged his shoulders andsaid, “Sure, why not?” (Perhaps, needless to say, calculatorsweren’t much help.) I am pretty sure that I spent the wholefour hours every year I took it, even in eighth grade, and Inever solved all of the problems. But it was not an ordeal for


231

me to take a four-hour math test; I could leave any time I wanted.I stayed until the end because I was having fun. This might beone of the most important qualities that separate the CMO in mymind—the spirit of fun, and even the sense of humor.

Not to say that the Olympiad was not serious business, not atall. In fact, I would say it was some of the hardest tests that I evertook, and I am more proud of my performances in the CMO thanin any other math contest. But it was always a pleasure to workon the problems. And I would continue thinking about the onesthat I hadn’t gotten all week, and then come back for the awardceremony to see the answers revealed. In hindsight, Soifer’spresentations at the award ceremony feel like some of the firstreal math lectures I ever saw. They were so far beyond what theywere teaching us in school, so different in style, yet somehowaccessible at the same time. It felt like he was giving us a peek ofan entire other world that we had never been exposed to.

I grew up to become a mathematician, but my road was a littlebit of a long and winding one. In short, I was always much moreexcited about mathematics itself than I ever was about school, somuch so that I barely made it out of high school and dropped outof college twice. But after a few false starts I set my sights ongetting a math Ph.D., which I finished at the University ofWashington in 2007. Since then, I’ve been a postdoctoral fellowat Stanford University. I greatly enjoy both research and teach-ing, and I hope to find a tenure track job as a professor that willallow me to continue both.

I think the Olympiad influenced me in many ways. It intro-duced me to the idea that there are many more math problemsunsolved than there are solved. (I remember Professor Soifersaying, “To nearest percent, 0% of all math problems aresolved.”). It helped me find out that I am capable of obsessingabout math problems for hours, or days, or longer. (Needless tosay, this is an important trait for a math researcher.) And thatwhich problems we work on, and how we work on them, is notonly a matter of ability, but of aesthetics and taste. (The CMOawarded special prizes for creative solutions, as well as literaryprizes for clever poems and stories.) I think it also helped merealize how much I enjoy just talking with people about math—some of the first “serious” math conversations I remember hav-ing with Soifer.

232 Movement 1. The Colorado Mathematical Olympiad Is Mathematics. . .

I appreciate that Sasha treated me as a friend and a peer evenwhen he first met me as a 14-year-old, and I have stayed in touchwith him and connected to the Olympiad since then. I havecontributed three research articles so far to his journalGeombinatorics, and helped judge the CMO several times.I was honored to be able to help judge for the 25th annualCMO last year. I think Sasha and I are just kindred spirits. Wemay naturally have some similar tastes mathematically, but Ialso think we recognized and appreciated the mischief andhumor in each other’s eyes from when we first met. For somereason, right now I am remembering winning the Olympiad, and

shaking hands with Professor Soifer and one of the deans, andSoifer turning to the dean and saying, “See! This is why weshould admit C–students!”

The C student eventually grew up. He received his Ph.D. fromUniversity of Washington in 2007. He completed postdoctoral fel-lowships at Stanford and Institute for Advanced Study [Princeton],and then proceeded to a tenure-track position at Ohio State Univer-sity. Since then, he was awarded an Alfred P. Sloan Research Fel-lowship and an NSF CAREER award, and was awarded tenurein 2015.

Now, back to CMO reminiscences.It was exciting to visit the 30th CMO in 2013. It was wonderful to

spend a week back in my childhood home of Colorado, visiting withmy mom and stepdad in Colorado Springs, seeing old friends aroundthe state, and driving around in the mountains. The week began andended with the Olympiad.

First, it was a day and a half of grading exams with Sasha and apanel of about 20 judges. I am surprised hearing myself to say that itwas fun, because grading calculus exams is usually boring if notdepressing. In the CMO you sometimes have to read for a while tofind the diamond in the rough, but it is always a pleasure beingsurprised by essays with unexpected proofs, or even papers with noproofs but funny poetry or well-drawn art.

The week ended with the award ceremony. The presentation ofawards was preceded by a panel discussion with Sasha and severalformer judges and participants from the Olympiad including myself.It was wonderful to hear many shared memories, and to see some of

Movement 1. The Colorado Mathematical Olympiad Is Mathematics. . . 233

the former Olympians grown up. David Hunter, who came a fewyears ahead of me, is now a statistics professor at Penn State. Itseemed parallel to my own career trajectory as a math professor atOhio State. It was impossible not to speculate about the impact thatthe Olympiad had had on us. The CMO gave encouragement to bothof us growing up, and I am sure helped us both to find our confidenceand passion.

I found what I enjoyed most about that week in Colorado andreconnecting with the Olympiad: it was the people. It was wonderfulto reminisce and reflect with several former winners of the Olympiadincluding David Hunter, Mark Heim, and Russel Shaffer. I enjoyed

catching up with Bob Ewell and Jerry Klemm, who have probablyjudged more than fifty CMO’s between the two of them. This leads tomy final observation. The Colorado Mathematical Olympiad is math-ematics; it is sport; it is art. And it is also community. There is a spiritof friendship and inclusion—people with shared vision and inspira-tion, enjoying each other’s company and quirkiness. I believe we areslowly creating a culture where young people can discover theirinherent intelligence and creativity. It is much too soon to know thelong-term impacts of the CMO.

234 Movement 1. The Colorado Mathematical Olympiad Is Mathematics. . .

Movement 2. I’ve Begun Paying Off MyDebt with New Kids, by Aaron Parsons

Permit me first reproduce a section about Aaron Parsons from my2011 Olympiad book [Soi9].

Best teachers are usually found in major metropolitan centers,where schools are well funded and cared for by enlightened populous.The small town Rangely in the very North-Western corner of thestate of Colorado near Colorado-Utah border did not hold promiseof mathematical inspiration for their small student body—unless amiracle were to happen. It did! The miracle’s name was MelvinOliver, mathematics teacher at Rangely High School.

Mel must have practiced magic, for nearly all students he broughtto the Colorado Mathematical Olympiad through many years, wonvarious awards. (We typically presented awards to no more than 20%of contestants.) In most years, Melvin Oliver and his students traveledtwice to Colorado Springs (a 13-hour round trip by car), to attend boththe Olympiad and a week later the Award Presentation. Among the

many of Mel’s fine Olympians, one stood out: Aaron Parsons, whowon a Silver Medal as a junior in 1997, and a Gold Medal in 1998. Allthis Aaron achieved while successfully competing in track and field—sprint to be precise—on the state’ highest level, placing 3rd inColorado. And Aaron was rewarded by an improbable journey,from the town of Rangely to Harvard University as mathematicsmajor!


235

When invited to Harvard, Aaron shared the good news with me inhis July 25, 1998 letter:

Dear Professor Soifer,I would first like to say that I have enjoyed your contest through-out my high school years. It is one of the few contests inColorado that places all students on an equal level and allowseach to test his/her merit through problem-solving skills andcritical thinking. This contest, to me, comes closer to capturingthe essence of mathematics than any other contest I have evertaken. Thank you very much for putting together such a remark-able contest for so many years.

I would also like to thank you for helping provide formy collegeeducation by offering scholarships to the winners of the contest. Ihave been accepted toHarvardUniversity, andwill pursue a degreein mathematics and physics there this coming year. . .

Thank you again for your wonderful math contest. I am sureyou will be seeing a lot more of my favorite teacher, Mel Oliver,and the Rangely Math Club.

Success had no effect on this modest talented cheerful young man.Ten years passed. I contacted Aaron in December 2008 and asked toshare his experiences during and after the Olympiad. On December11, 2008 Aaron replied:

Dear Prof. Soifer,How wonderful to hear from you! Sorry that I’ve been slow toreturn the message you left on my guestbook (I only check itinfrequently). I would be more than happy to provide a shortnote about myself and CMO:

I participated in the Colorado Math Olympiad [CMO] from1994 to 1998, representing Rangely High School [RHS] andcoached by the generous and committed Melvin Oliver, whosingle-handedly developed and supported the math program atRHS. At a time when most math contests focused on speed,numbers, and arithmetic tricks, CMO stood out as somethingcompletely different.

The first time I took a CMO test, I was flabbergasted—I wasso tuned to the “other” type of contest, I felt I could hardly solvea single problem! No one on our team qualified for the awards

236 Movement 2. I’ve Begun Paying Off My Debt with New Kids, by Aaron Parsons

ceremony that year, and one of the problems just drove menuts—one about polygons of unit area on a grid. Resolving toqualify for the “answer session” next year, I set to work. I didmanage an “Honorable Mention” the next year, and after visitingUCCS and meeting Prof. Soifer for the annual recapitulation ofthe contest, both Mel and I came away with a new understandingof a broader, more abstract, and altogether much more fun sideof mathematics than we had previously seen.

I distinctly remember the following year, when, having qual-ified for the awards ceremony, I found myself unfortunatelyneeding to skip the ceremony in order to participate in a state

track meet (held just an hour away). I met with Prof. Soifer toexcuse myself and to apologize for necessity of my departure.“Oh, that’s quite alright,” he said. “You know, I was a sprintermyself in high school.” And a good one at that, as I found out. Ibegan to understand his bounding energy in front of his students,to guess at a joie d’ vivre that could be expressed both academ-ically and athletically.

After graduating from high school, I studied physics andmathematics (and ran track) at Harvard. There, I discoveredthat math—the real math that mathematicians do—was reallymuch more like the bounding, gleeful CMO math than any othermath I had been exposed to. I grew much better versed inmathematical reasoning, but it still wasn’t until my secondyear away at college that the solution to that demonic polygonproblem finally came to me. Liberated at long last, I moved on toastrophysics. I am currently finishing my doctorate at [the Uni-versity of California] Berkeley, working to discover the firststars that formed in the universe 10 billion years ago.—Aaron Parsons

I asked Aaron to update us on his life’s events that happened since2010. On August 27, 2016, Aaron writes:

Dear Alexander,A decade later, I must again beg your forgiveness for being slowto respond. I suppose this reveals one of my fundamental char-acter flaws. None of us are perfect, but please accept myapologies.

Movement 2. I’ve Begun Paying Off My Debt with New Kids, by Aaron Parsons 237

Many things have happened since we last talked! I’m still atBerkeley, but now as a professor (tenured just last month) withthree wonderful children. Teaching, research, and parenting suitme very well. I tell my students that if they are lucky, they willfind a job where they enjoy the idea of what they are working on(the big picture) and the actual work (the details). I love themath, engineering, and social interaction that comes with run-ning a scientific experiment, and what could be more interestingthan mapping the universe as it appeared 10 billion years ago?One of my great pleasures is sharing with students and childrenthe fun, intricate, and often surprising worlds of math and sci-

ence. Inspired by the CMO and by Rangely’s math heroes Meland Lorraine Oliver, I started up a Family Math Night at ourlocal elementary school. This last year, we learned about fractalsand functions, Mobius strips and curved space, spirals and tes-sellations. Parents are skeptical that these subjects are “math.”Little do they know that this math is more math than the maththey think math is!

Thinking back to our high school trips to participate in theCMO, I’m indebted to you and Mel for teaching me “real” math.As I’ve begun paying off my debt with new kids, I’m finallybeginning to understand why Mel was willing to drive a van withridiculously loud teenagers 12 hours across the state of Colo-rado, just to participate in Dr. Soifer’s Colorado Math Olympiad.

All the best,

Aaron

238 Movement 2. I’ve Begun Paying Off My Debt with New Kids, by Aaron Parsons

Movement 3: Aesthetic of PersonalMastery, by Hannah Alpert

Hannah Alpert was the first girl to win back to back second prizes inthe Colorado Mathematical Olympiad in 2006 and 2007. Moreover,she was a solo second prize winner behind only Sam Elders, who wasa solo first prize winner. Before that, as a sophomore in 2005, Hannahwon third prize, and was, together with the winner Mark Heim, theonly Olympian to solve problem 22.4, in her own way, unknownto me.

As an undergraduate mathematics major at the prestigious Univer-sity of Chicago, Hannah Alpert received a number of awards, includ-ing “the 2010 Alice T. Schafer Prize for excellence in mathematics byan undergraduate woman.” The following year, she won a NationalScience Foundation Research Fellowship, entered a Ph.D. program inmathematics at the Massachusetts Institute of Technology (MIT), andin April 2016 defended her doctoral thesis entitled “Special gradienttrajectories counted by simplex straightening,”

I have been meeting, with great pleasure, with Hannah’s talentedfamily at the Award Presentations for many years. Her father,Dr. Bradley Alpert, has been a scientist at the National Institute ofStandards and Technology in Boulder. Hannah’s mother, SilvaChang, has ran a mathematical circle in Boulder, which producedmany of our Olympiad’s winners. Hanna’s younger brother, BenAlpert, won fourth prize in 2009, third prizes in 2008 and 2010, andfinally first prize in 2011.

OnMay 23, 2016, Hannah proved a theorem about smallness of ourworld, by sharing with me her mathematical interaction with our 1990and 1991 Olympiad winner Matthew Kahle:


239

It turns out that one of the main people I expect to work with nextis Matt Kahle. I will be a postdoc at ICERM [MathematicsResearch Institute at Brown University] during the Topologyin Motion program that he is helping to organize—ICERM issponsoring me for the whole year—and then I have a 3-yearposition at Ohio State. It was funny one day when my mom andI realized that we had each mentioned him on that same day (shehad read about him in your books) but hadn’t realized we weretalking about the same person. He is a friend of my [MIT]advisor Larry Guth so Larry had given me a project to work onthat he’d gotten from Matt.

Hannah has an original style, all her own, and includes imagery inmathematics. Hannah’s success was easy to predict, but she exceededall expectations. On August 25, 2016, Hannah wrote the followingessay especially for this book.

The first time I went to the CMO, I went with a group of studentsfrom my high school. I didn’t know them, so I assumed that theywere the real math experts. I didn’t realize that, having attendedan intensive summer math program, I had more math trainingthan they did. Out of the five problems on the test, I thoughtI had done about one and a half, plus the first one which waseasy. It seemed like an appropriate level of performance forsomeone who didn’t normally do this kind of thing. The otherstudents were saying they’d solved three or four problems,which I assumed was more typical.

We were all invited to the award ceremony, but there had beena big snowstorm, so the drive that would normally already takemultiple hours was going to be much slower. My family decidedwe wouldn’t go, because presumably I wasn’t winning a verysubstantial award anyway. We were quite surprised to hearafterward that I had won third place, better than those “realmath experts” from my school!

I remember other things from other years. It was always funnyto me that there would be so many contestants that we filled upseveral rooms all sitting right next to each other, but then mostpeople would leave after half an hour. You might think thiswould work out fine and the serious contestants would get plenty

240 Movement 3: Aesthetic of Personal Mastery, by Hannah Alpert

of space, except that I knew most of them by this time, so wewould arrive together and sit together, spending the whole testall clumped up and trying not to look at each other’s papers. Wewere too awkward to just stand up and move to an empty tablewhile the test was in progress—what if a proctor wouldn’t let usfinish, or a friend was offended? So I remember that one year inthe last hour of the test, my friend next to me must have beengetting stuck on the math, because she took out a hairbrush andstarted brushing her hair. I was laughing to myself while tryingto make sure that we didn’t look like we were communicating sothat we didn’t get kicked out for cheating.

I also remember that one of my solutions showed up in theaward ceremony. It was a graph theory problem—I just looked itup and it’s number 5, “Math Party”, from 2006—and the problemwas stated in terms of mathematicians sitting around a tableaccording to their collaboration distance. My solution was statedin terms of a completely different story, something about an antcrawling around in the desert and needing to stop every so often toeat or drink or refuel or something. I think the graders found thisamusing. At the time I thought that explaining a proofmeant that Iwas supposed to describe everything that was happening in myhead, so if I was seeing all kinds of gestures and colors, then that’swhat should be communicated. It took me a few years to under-stand that each person has to build a separate understanding, andyou can’t just import it all in one piece from one mind to another.

I like to refer to the books of old CMO problems wheneverI want some stand-alone problems to get people talking.It’s impressive that Professor Soifer and his writing teamshave collected such a large number of problems that areprerequisite-free, immediately engaging, and at just the rightlevel of requiring some thought but still being solvable withoutobscure trickery. These problems are so convenient when I amteaching and need to help a group of students to get to know eachother mathematically, to hear each person’s thinking style andfeel out the shape of a collaboration.

I think what resonates with me most about the CMO is whatI’ll call the “aesthetic of personal mastery.” After high school Iwent to college at the University of Chicago, where they trainedme in a large body of standard mathematical knowledge. Then I

Movement 3: Aesthetic of Personal Mastery, by Hannah Alpert 241

went to grad school at MIT, where I continued to try to learn allthe words and basic concepts so that I might understand whateverybody was talking about. It’s been in the past year, finishingmy Ph.D., that I’ve found my way back to this aesthetic ofpersonal mastery and realized that I had it in high school andthen lost it in college. In the CMO problems, you aren’t judgedon how completely you have assimilated the wisdom of the ages.You can just think about the problem yourself. You can do it inthe way that makes sense to you, using the tools you alreadyknow. This is what I mean by the aesthetic of personal mastery.

There are six published papers on research I did before enter-

ing grad school. I’ve just submitted the fourth paper on research Idid during grad school. I think that what has distinguished eachproject is my ability to move away from community expecta-tions for what I ought to understand—in undergrad this meantleaving campus to go to REU programs—and spend time withwhat I could understand by myself. I guess with all thoseresearch papers it sounds like I kept the aesthetic of personalmastery all along. I sort of did without realizing it. I thoughtI was really supposed to be learning all this other stuff—I’m notsaying it hasn’t been useful, though!—and these research pro-jects were side amusements while I learned to do things prop-erly. Now I’m more intentional about making space to do thingsmy way.

Just like how the CMO focuses on helping students developthe skills of exploring mathematics rather than on specific math-ematical content, I remain most interested in how to work, learn,and think, regardless of the topic of study. I’m not set on being aprofessor, but I do plan to have academic jobs for the next fewyears. I expect that my orientation toward mental processes willhelp me in my teaching. In research, I intend to take on anyprojects where I can find something to understand thoroughly,whether or not it all adds up to a cohesive mathematical researchprogram.

242 Movement 3: Aesthetic of Personal Mastery, by Hannah Alpert

Movement 4. Colorado MathematicalOlympiad: Reminiscences by RobertEwell

Robert N. (Bob) Ewell is a graduate of Clemson University (B.S. inMathematics, 1968). He did graduate study in mathematics atClemson and the University of Nebraska. He went on to pick up amaster’s in Education from Troy State University in 1982, and aDoctor of Education from Auburn University in 1984. An AFROTCcadet at Clemson University, he served in the US Air Force for20 years, retiring as a lieutenant colonel in 1990. After the AirForce, he did statistical consulting for 10 years before entering Chris-tian ministry. He now serves with The Navigators as a leadershipdevelopment coach. In addition to being a fine amateur mathemati-cian, Bob plays piano and enjoys playing and watching sports. Here isBob’s story.

I met Dr. Alexander Soifer in 1989 right after the famous mathe-matician Dr. Paul Erdos had come to town. There was an article afterthe fact in the Gazette, and I called Dr. Soifer to ask if there was a way

we could know about such things in advance. During the conversa-tion, Alex said, what are you doing next Friday? Would you like tojudge the Colorado Mathematical Olympiad?

I’ve missed only one competition since then, right after I moved toAlabama in 2001. The remaining 5 years I was in Alabama, I cameback to judge the Olympiad because I missed it. Other judges havetraveled back to Colorado Springs for the Olympiad after movingaway. Matt Kahle, about whom you’ll read more later, came backfrom Colorado State in Fort Collins, from Washington State, andfrom Ohio State. Shane Holloway moved to Seattle for a few years


243

to work at Amazon and came back for the competition every year hewas gone.

My son, Matthew, who competed in many Olympiads, earning ashigh as first honorable mention, came back to judge from Northwest-ern University in the Chicago area. The competition conflicted withsomething important at school, but I wrote a letter to his mathprofessor, making the case that judging the Olympiad was not onlya valuable public service, he would learn a lot of mathematics whiledoing it. I don’t have the letter anymore, but his professor said it wasvery compelling, and he allowed Matt to return to Colorado Springsto judge. Unfortunately, that year was the only time (I think) that the

competition was postponed by 1 week because of snow.One good thing came from my time in Alabama. I connected with

the Alabama School of Math and Science in Mobile, traveling thethree hours from Montgomery to do a presentation of Olympiadproblems for their math club. Several professors were there, too.Since then, the school has sent one or two students to Coloradoevery year for the competition.

In the early days, we worked very hard. Students were all over thecampus, multiple rooms in multiple buildings, sometimes up to 900!The judges helped count the students and send them with a proctor tofind their room. After I had judged a few years, Alex and I walked therooms together to answer their questions. One year, Alex was in thehospital with kidney stones and was delivered to the campus about30 minutes after the competition started! I was prepared to brief thesolutions to the judges and carry on without him if necessary. Ofcourse, he didn’t walk the rooms that year, starting the currenttradition of my answering most of the students’ questions.

In more recent years, UCCS staff people have taken over thelogistics part of the Olympiad, and things have been much better forthe judges! And more recently, all the students take the test in oneroom, which makes it much easier on me as I answer their questions.

With respect to answering questions, we try to make the problemsclear, but as is the habit of mathematicians, we don’t use more wordsthan we need to. And many of the students just don’t seem able to readcarefully enough to understand what’s being asked. The student canread a sentence that says something like, “Is there a series of allowedmoves to . . . [cause something to happen]?” The student will ask me,“What is an allowed move?” I politely, of course, refer him to the

244 Movement 4. Colorado Mathematical Olympiad: Reminiscences by Robert Ewell

previous sentence in which allowed moves are defined! Sometimesstudents don’t know even the most basic mathematical terms such as“integer.” Some of the young students know the term “whole num-ber” instead.

I was on active duty in the Air Force back in 1989, and even today,when calls from Alex come, the first word is usually, “Colonel. . .”The next words are usually, “Here’s what you don’t understand. . .”Our relationship has been ‘stormy’ from the beginning. I wasn’t afraidto challenge him back in 1989, usually on the elusive question of howmuch “partial credit” to give. He has always welcomed such chal-lenges provided, of course, that one knows “when to hold ‘em and

when to fold ‘em.”I have brought a bit more order, structure, and efficiency to the

judging process. Judging used to take well into Saturday. Now we’refinished by day’s end Friday. One practice we implemented at mysuggestion was that of identifying “senior judges”—those withenough experience not to give credit when none is warranted andnot to miss a solution that might be different from the one Alexpresented. And those with enough experience to know when theyare over their head and pass the paper to Alex himself. Alex, who hasa strong sense of equality and democracy, was resistant to the idea ofsenior judges. But when we had to regrade all the papers because twonovice judges sat together and poorly graded hundreds of papers, herelented. Now every paper is seen by at least one senior judge.

Shane Holloway added to our efficiency by computerizing thereporting of scores. Jerry Kemp, the longest tenured judge, has alwaysorganized papers several hours into the judging so that by the end ofthe day we have all the papers organized by descending scores. In theearly days, Jerry wrote the scores of each student’s paper on theblackboard, a long, tedious process. Shane started bringing his laptopand, using a program he developed to accommodate the Math Olym-piad unorthodox scoring system (+, +/�, ½, �/+, �), we can noweasily record the scores and produce lists of winners with the touch ofa button.

Part of our process used to include selecting a place for the after-competition lunch on Saturday. So Alex would solicit ideas and havethe judges vote on where they wanted to go. One year, someonenominated the Olive Garden. Alex said, “Where is that?” clearlyindicating he had never been there. From then on the voting process

Movement 4. Colorado Mathematical Olympiad: Reminiscences by Robert Ewell 245

was entertaining. I sat in the back and watched because I knew that ifAlex had wanted to go to the Olive Garden, he would have been therealready. Sure enough, Olive Garden got the most votes, but that’s notwhere we went. I don’t remember exactly how he engineered it, butwe kept voting until another restaurant came out on top! In recentyears, we have a “tradition” of certain restaurants. One for Fridaynight for a small group of senior judges and one on Saturday for all.Judges no longer select the restaurant, if, indeed, they ever did.

I have brought efficiency; Dr. Soifer brings the mathematics. WhileI love mathematics and can do it at some level, I will never be nearlyas proficient as our Olympiad winners.

The first years I would have about an hour or so to look at theproblems, and I was usually able to work problems 1–3 during thosetimes. Occasionally, I would solve or at least get started on number4. I never had a clue about any of the number 5s. Then I became amember of the problem committee and got to see the problems inadvance. Given enough time I could usually work problems 1–4, anda couple years, I actually solved all five, given 2 weeks and a bit ofcoaching from Dr. Soifer.

I’m always intrigued by the number of different approaches toproblems. Often Dr. Soifer takes a geometric approach, and I takean algebraic one (when such diversity is possible). One year, how-ever, my approach to a difficult number 5 was simple and geometric,and his was complex and algebraic. When he refined my geometricapproach to a particularly elegant solution, that number 5 has becomeone of his favorite problems. Another year, when he kept modifyingnumber 5 (resulting in the lecture, “7 Stages of Problem Creation”,I actually solved number 5 before he did! He, long-time judge GaryMiller, and I were discussing the problems, and this number 5, whichhe had changed. Just before lunch Alex presented a solution (off thetop of his head) that wasn’t valid. We told him so, and off we went tolunch in separate cars. On the way to lunch, the solution came to me,and I triumphantly announced it when I arrived. As we used to say inthe Air Force, “Even a blind hog finds an acorn sometimes!”

As I say, I will never be as proficient as the Olympiad winners andthere have been some special ones along the way.

One of my oldest son’s high school classmates, Matt Kahle, wonthe Olympiad twice even though neither he nor my son thought muchof high school. Matt made Cs in mathematics and less than that in


everything else, so it was the Olympiad that showed him what he wascapable of. Turned down for Dr. Soifer’s school, UCCS, because hisgrades were too low, he started at the local community college. Hewent on to complete college at a school different from UCCS, got amaster’s, then a doctorate, then post-doctoral work at Stanford. Henow teaches math at Ohio State. Matt is a Colorado Math Olympiadsuccess story!

Another interesting student was Bryce Herdt, who first showed upwhen he was in the sixth grade. Bryce has Asperger’s Syndrome, andI remember his mother accompanying him to that first competition.Even at such a young age, he asked good questions. The next year,

when he was in the 7th grade, he earned first honorable mention.Beginning in the 8th grade, 2001, he won 3 years in a row! Brycecame back after he graduated from high school to help judge.Although he couldn’t really focus well enough to grade papers, heoften helped us with valuable insights. I remember talking with afellow judge with Bryce sitting nearby looking away, appearing not tobe paying attention. Then suddenly, after one of us said something,Bryce turned around and said, “No. That’s not right.” And then wenton to help us understand what we were talking about.

One of the years when I solved all five problems over a 2-weekperiod, a student turned in his paper 90 minutes into the 4-hourcompetition. I said, “Did you finish?” He replied, “Yes. I solved allfive in the first 45 minutes and took another 45 minutes to write themup. I was going to write something for the Literary Award, but since Igot all the problems right, I decided not to.” He was right. He hadsolved all five and won first prize.

In 2009, problem 5 was one I would not have solved no matter howmuch time I had. The approach I was using didn’t work, and I wouldnever have thought of the approach that was required. Olympiad daydawned snowy. I live in Monument, 20 miles north of UCCS, and Ithought we might postpone the competition (something that hashappened only once in 33 years). Not so in 2009. I arrived late, justin time to hear Dr. Soifer finish explaining the solution to the elusiveproblem 5. As I sat down next to a new judge, I was surprised to hearhim say, “I solved it a different way.” When a new judge announcesthat he thinks he’s solved problem 5 “in a different way,” it normallymeans he doesn’t know what he’s doing. So I asked if I could see hissolution. He passed me his paper, and it didn’t take 30 seconds for me

Movement 4. Colorado Mathematical Olympiad: Reminiscences by Robert Ewell 247

to realize that I was in the presence of genius. Not only had he writtena beautiful solution to problem 5, different from Dr. Soifer’s, but healso had perfect solutions to problems 1, 2, and 4—neatly written, inink, on one side of one sheet of paper, in about one hour!

Then I looked up and noticed his name tag: Russel Shaffer, a nameI recognized as the winner of the first Colorado Math Olympiad wayback in 1984. As I introduced myself, I thought, “Some people can dothis better than others!”


Farewell to the Reader

I can see the sun, but even if I cannot see thesun, I know that it exists. And to know that thesun is there—that is living.—Fyodor M. Dostoyevsky, “The BrothersKaramazov”

The world will be saved by beauty!—Fyodor M. Dostoyevsky, “The Idiot”

Pure mathematics is, in its way, the poetry oflogical ideas.—Albert Einstein

Does Dostoyevsky refer to mathematics in the epigraphs above?Probably not, but his lines are so applicable to mathematics, with theirreferences to proving existence and importance of beauty! Einstein, aconnoisseur of Beauty, observes poetic qualities of pure mathematics.Mathematics could be viewed as a science whose truths exist inNature independently of our mind. It is also used as a tool in a varietyof other sciences. In a sense, mathematics is a language, used by manydisciplines to make themselves more rigorous. And mathematicscould be viewed as an art, which not only reflects Nature, but alsocreates Beauty that can compete with the Beauty of Nature. TheColorado Mathematical Olympiad exists to spread this Beauty, andto pass this connoisseurship to young minds.

Our Olympiad is now on its 34th year. It continues to be a lifelongcommitment for me. Even when I lived and worked at Princeton


249

University (2002–2004 and 2006–2007), I flew back to ColoradoSprings to run the Olympiad and its Award Presentations. The Olym-piad has been my way to pay back to America for excepting me as arefugee from tyranny, allowing to live an American dream, and topass the mathematical baton to the kids of Colorado and the nextgenerations of creative thinkers. As you have seen, our winners havebecome professors of mathematics and astronomy, law and philoso-phy, inventors and computer scientists at such brilliant institutions asUniversity of California Berkeley, University of Southern California,Pennsylvania State University, Ohio State University, etc.

I have been getting help from many people from within the

University of Colorado Colorado Springs and outside. And yet, theOlympiad has been in a sense a one inspired person drive, including adrive through all adversities (which is surely a form of problemsolving). In the 30th Anniversary film about the Colorado Mathemat-ical Olympiad, winners and judges ponder what would happen withthe Olympiad in the future. In a pure Paul Erdos style, I wish theOlympiad to survive for centuries, but we shall see!

Thank you for holding my book in your hands. I welcome yourideas, comments, solutions of problems presented here and newproblems you may create and submit to our Olympiad. In due time,I hope to offer your attention a continuation of this book entitledsomething like The Colorado Mathematical Olympiad: The FourthDecade and Further Explorations.

As Paul Erdos used to say at the end of his lectures, “everythingcomes to an end, and so has this” book. However, if you are inclinedto continue your explorations of mathematics with me, I have goodnews for you. This book is one of my 11 books that Springer has or issoon going to publish.

If you enjoy a visual appeal of geometry, you may wish to read theexpanded edition of How Does One Cut a Triangle? [Soi6]. I self-published its first edition in 1990. This book offers a glimpse of ‘real’mathematics, and demonstrates synthesis of ideas from variousbranches of mathematics working together to achieve a purely geo-metric result.

The expanded edition ofGeometric Etudes in Combinatorial Math-ematics [Soi7] shows how geometric insight does wonders in serviceto combinatorics. It also shows how beautifully geometry provessome celebrated theorems of mathematical analysis. Its first edition,

250 Farewell to the Reader

written jointly by the distinguished geometer Vladimir Bolyanskiand I, came out in 1991.

The Election Day, November 4, 2008 (the “yes-we-can” day) sawthe release of the book I dreamed of and worked on for 18 years,The Mathematical Coloring Book: Mathematics of Coloring and theColorful Life of Its Creators [Soi3]. This voluminous book presents abeautiful mathematics of coloring (the so-called Ramsey Theory) asevolution of ideas. It also includes archival historical investigationsinto the lives of mathematicians who created these ideas, from thetime of World War I to the Nazi period in Germany, from thedevastated by World War II Netherlands to the present times. The

book demonstrates aesthetics of mathematics, addresses philosophyof its foundations, and psychology of mathematical and historicaldiscovery. It also demonstrates that mathematics could become agenre of literature—if the author is willing to play the role of SherlockHolmes and treat you, the reader, as if you are Dr. Watson. The NobelLaureate Boris Pasternak expressed my goals in this book moreprecisely and concisely than I could—great poets do magic withwords:

I bring here all: what have I lived thru,And that what keeps my soul alive,My rectitude and aspirations,And what have seen my own eyes.

Here is how it sounds in the original Russian:

Здecь будeт вce: пepeжитoe,И тo, чeм я eщe живу,Moи cтpeмлeнья и уcтoи,И видeннoe нaяву.1

My newest 2015 book, includes practically no mathematics. How-ever, great twentieth century mathematician Van der Waerden andphysicist Werner Heisenberg are the main personages of the book,which is entitled The Scholar and the State: In Search of Van derWaerden [Soi10]. This archivally researched over 20-year period

1[Pas], Translated especially for The Mathematical Coloring Book [Soi11] by Ilya Hoffman.

Farewell to the Reader 251

book is my contribution to the issue that has become for me evenmore important than mathematics and the arts: the issue of profes-sional ethics of a creator.

I ponder questions that have not lost its urgency today, such as therole of a scholar in a state, particularly in a totalitarian state. I found atroubling commonality between the creative class joining Hitler thenand supporting Putin’s war on Ukraine now. Fred Rodell dedicatedhis 1936 book 55 Men: The Story of the Constitution “To the SchoolChildren and the Politicians—for the same reason.” My book waswritten likewise

To the School Children and the Scholars—for the same reason.

What is next for me? Two contracts remain with Springer. A bookof favorite open problems of the legendary mathematician PaulErdos, my friend, mentor, and coauthor: Problems of pgom Erdos[ES]. I would not have attempted to write it, but in 1990 Paul askedme to help him in this endeavor, and so we had been working on thisbook together for years, from 1990 to his passing in 1996. This will beour joint book. . . Paul Erdos (1913–1996) was the greatest problemcreator of all time. You will be able to work on his problems becauseno knowledge is required for understanding many of them. Moreover,often Erdos’ problems allow you, young mathematicians, to advanceand find at least partial solutions.

The second remaining book is Memory in Flashback: A Mathema-tician’s Adventures on Both Sides of the Atlantic [Soi13]. It will be acollection of humorous and noteworthy vignettes sketching colorfulmoments of my life, remarkable people I had the pleasure to interactwith, and world affairs of particular significance. I suspect you willenjoy it, but we will see. :)

Having read this book, you have become my alumnus. This titlecarries a responsibility to stay in touch, to send me your most enjoy-able solutions, your new problems, suggestions and ideas. Be restassured: I will always be delighted to hear back from you!

252 Farewell to the Reader

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[Soi2] Soifer, A. (2005). Cover-up squared. Geombinatorics, XIV(4), 221–226.[Soi3] Soifer, A. (2009). The mathematical coloring book: Mathematics of coloring and the

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[Soi5] Soifer, A. (2009). Mathematics as problem solving (2nd expanded edition). New York:

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[Soi8] Soifer, A. (Ed.). (2011). Ramsey theory yesterday, today and tomorrow. Progress inMathematics. New York: Birkhauser, Springer.

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References 255

Index of Names

AAlpert, B., 73, 98, 113, 239

Alpert, H., 28, 29, 38, 47, 59, 61, 73, 239–242

CCarpenter, M., 72

Christensen, T., 115, 125

Chung, F., 200

Conway, J.H., 172–174, 176, 178–181

DDostoyevsky, D.M., 249

Dynkin, E., 78

EEinstein, A., 112, 214, 249

Elder, S., 29, 43, 46, 59–63, 239

Erdos, P., 51, 78, 83, 113, 135–144, 150,

152–154, 180, 199–201, 213–215, 218,

220, 243, 250, 252

Ewell, M., 86, 141

Ewell, R., 18, 21, 32, 45, 54, 67, 86, 97, 109,

114, 125, 132, 141–144, 148, 149, 155,

157–159, 196, 223, 234, 243–248

GGardner, A., 85–87, 89–91, 93, 95, 97–104,

106–110, 113

Graham, R.L., 92, 171–182, 219

Grünbaum, B., 119, 226, 227

Guthrie, C., 98, 113

HHeim, M., 3–7, 27–29, 31–35, 37–40, 42, 45,

46, 61, 138, 139, 142, 154, 155,

157–159, 203–204, 234, 239

Herdt, B., 4, 27–29, 32, 112, 247

Hesterberg, A., 108

Hoffman, G., 5, 31, 75, 88, 100, 115, 124, 125,

142, 156–159

Holloway, S., 86, 114, 243, 245

Hunter, D., 27, 137, 139, 142, 154, 155, 157,

158, 234

KKahle, M., 45, 104, 108, 112, 138, 140, 142,

154, 155, 157–159, 161, 231–234, 239,

243, 246

Karabash, M., 172–174, 176, 178, 179,

181, 182

Katznelson, Y., 205

Klemm, J., 86, 234

Kuhn, H.W., 59, 60, 174

MMiller, G., 3, 5, 142, 155, 157, 159, 246

Moser, L., 200

NNash, J.F. Jr., 59, 60, 174

OOwens, B., 5, 31, 88


257

PParsons, A., 154, 160, 235–238

Peres, Y., 205, 206

RRitter, B., 49, 75, 88–90, 100

Ritter, J.L., 87, 90, 91

Romer, R., 88–90

SSchaffer, R., 207, 208

Schlag, W., 205, 206

Shokley-Zalabak, P., 5, 31, 49, 63, 75, 87, 100,

115, 125, 142, 154, 157, 158

Soh, A., 113, 123–125, 127–129,

131–134, 140

Song, M., 28, 33, 35, 86, 93, 103

TTuran, P., 150, 217–220

WWaerden, Bartel van der, 8, 103, 174, 251

Wells, J.M., 5, 31

YYaglom, A.M., 68, 190, 193

Yaglom, I.M., 68, 190, 193

ZZhang, J., 125, 140

258 Index of Names

Index of Terms

B“Big O” notation, 180, 218

CChromatic number of the plane, 211

Chromatic number of a graph, 206

Coloring a graph, 120

Coloring a map, 78

Congruent (numbers), 80

Convex figure, 224–226

Convex hull, 82

Convexity of a function, 16

Cycle, 17, 56, 106–108, 147, 215

DDavenport–Schinzel sequence, 188

FFano Plane, 183, 184

Finite Projective Plane, 25, 184

GGraph, 17, 45, 48, 49, 56, 57, 61, 73, 106, 107,

113, 119–121, 124, 147, 148, 214, 215,

221, 241

Graph bipartite, 17

Graph complete, 106, 120, 121

Graph directed/digraph, 147

Graph Eulerian, 215

Graph Hamiltonian, 56, 57, 214

MMersenne Primes, 194, 195

TTree, 56, 57, 106, 120, 214

Tree spanning, 56

Trigon, 178, 179

WWythoff Nim, 191–193


259

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