Icons of Mathematics · The DOLCIANI MATHEMATICAL EXPOSITIONS series of the Mathematical...

AMS / MAA DOLCIANI MATHEMATICAL EXPOSITIONS VOL 45

Icons of MathematicsAN EXPLORATION OF TWENTY KEY IMAGES

Claudi Alsina and Roger B. Nelsen

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Icons of Mathematics

An Exploration ofTwenty Key Images

10.1090/dol/045

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c� 2011 byThe Mathematical Association of America (Incorporated)

Library of Congress Catalog Card Number 2011923441

Print ISBN 978-0-88385-352-8

Electronic ISBN 978-0-88385-986-5

Printed in the United States of America

Current Printing (last digit):10 9 8 7 6 5 4 3 2 1

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The Dolciani Mathematical Expositions

NUMBER FORTY-FIVE

Icons of Mathematics

An Exploration ofTwenty Key Images

Claudi AlsinaUniversitat Politecnica de Catalunya

Roger B. NelsenLewis & Clark College

Published and Distributed by

The Mathematical Association of America

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DOLCIANI MATHEMATICAL EXPOSITIONS

Committee on Books

Frank Farris, Chair

Dolciani Mathematical Expositions Editorial Board

Underwood Dudley, Editor

Jeremy S. Case

Rosalie A. Dance

Tevian Dray

Thomas M. Halverson

Patricia B. Humphrey

Michael J. McAsey

Michael J. Mossinghoff

Jonathan Rogness

Thomas Q. Sibley

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The DOLCIANI MATHEMATICAL EXPOSITIONS series of the Mathematical As-sociation of America was established through a generous gift to the Association fromMary P. Dolciani, Professor of Mathematics at Hunter College of the City Universityof New York. In making the gift, Professor Dolciani, herself an exceptionally talentedand successful expositor of mathematics, had the purpose of furthering the ideal ofexcellence in mathematical exposition.

The Association, for its part, was delighted to accept the gracious gesture initiat-ing the revolving fund for this series from one who has served the Association withdistinction, both as a member of the Committee on Publications and as a member ofthe Board of Governors. It was with genuine pleasure that the Board chose to namethe series in her honor.

The books in the series are selected for their lucid expository style and stimulatingmathematical content. Typically, they contain an ample supply of exercises, manywith accompanying solutions. They are intended to be sufficiently elementary for theundergraduate and even the mathematically inclined high-school student to under-stand and enjoy, but also to be interesting and sometimes challenging to the moreadvanced mathematician.

1. Mathematical Gems, Ross Honsberger

2. Mathematical Gems II, Ross Honsberger

3. Mathematical Morsels, Ross Honsberger

4. Mathematical Plums, Ross Honsberger (ed.)

5. Great Moments in Mathematics (Before 1650), Howard Eves

6. Maxima and Minima without Calculus, Ivan Niven

7. Great Moments in Mathematics (After 1650), Howard Eves

8. Map Coloring, Polyhedra, and the Four-Color Problem, David Barnette

9. Mathematical Gems III, Ross Honsberger

10. More Mathematical Morsels, Ross Honsberger

11. Old and New Unsolved Problems in Plane Geometry and Number Theory,Victor Klee and Stan Wagon

12. Problems for Mathematicians, Young and Old, Paul R. Halmos

13. Excursions in Calculus: An Interplay of the Continuous and the Discrete,Robert M. Young

14. The Wohascum County Problem Book, George T. Gilbert, Mark Krusemeyer,and Loren C. Larson

15. Lion Hunting and Other Mathematical Pursuits: A Collection of Mathematics,Verse, and Stories by Ralph P. Boas, Jr., edited by Gerald L. Alexanderson andDale H. Mugler

16. Linear Algebra Problem Book, Paul R. Halmos

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17. From Erdos to Kiev: Problems of Olympiad Caliber, Ross Honsberger

18. Which Way Did the Bicycle Go?: : : and Other Intriguing MathematicalMysteries, Joseph D. E. Konhauser, Dan Velleman, and Stan Wagon

19. In Polya’s Footsteps: Miscellaneous Problems and Essays, Ross Honsberger

20. Diophantus and Diophantine Equations, I. G. Bashmakova (Updated by JosephSilverman and translated by Abe Shenitzer)

21. Logic as Algebra, Paul Halmos and Steven Givant

22. Euler: The Master of Us All, William Dunham

23. The Beginnings and Evolution of Algebra, I. G. Bashmakova and G. S.Smirnova (Translated by Abe Shenitzer)

24. Mathematical Chestnuts from Around the World, Ross Honsberger

25. Counting on Frameworks: Mathematics to Aid the Design of Rigid Structures,Jack E. Graver

26. Mathematical Diamonds, Ross Honsberger

27. Proofs that Really Count: The Art of Combinatorial Proof, Arthur T. Benjaminand Jennifer J. Quinn

28. Mathematical Delights, Ross Honsberger

29. Conics, Keith Kendig

30. Hesiod’s Anvil: falling and spinning through heaven and earth, Andrew J.Simoson

31. A Garden of Integrals, Frank E. Burk

32. A Guide to Complex Variables (MAA Guides #1), Steven G. Krantz

33. Sink or Float? Thought Problems in Math and Physics, Keith Kendig

34. Biscuits of Number Theory, Arthur T. Benjamin and Ezra Brown

35. Uncommon Mathematical Excursions: Polynomia and Related Realms, DanKalman

36. When Less is More: Visualizing Basic Inequalities, Claudi Alsina and Roger B.Nelsen

37. A Guide to Advanced Real Analysis (MAA Guides #2), Gerald B. Folland

38. A Guide to Real Variables (MAA Guides #3), Steven G. Krantz

39. Voltaire’s Riddle: Micromegas and the measure of all things, Andrew J.Simoson

40. A Guide to Topology, (MAA Guides #4), Steven G. Krantz

41. A Guide to Elementary Number Theory, (MAA Guides #5), Underwood Dudley

42. Charming Proofs: A Journey into Elegant Mathematics, Claudi Alsina andRoger B. Nelsen

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43. Mathematics and Sports, edited by Joseph A. Gallian

44. A Guide to Advanced Linear Algebra, (MAA Guides #6), Steven H. Weintraub

45. Icons of Mathematics: An Exploration of Twenty Key Images, Claudi Alsina andRoger Nelsen

MAA Service CenterP.O. Box 91112

Washington, DC 20090-11121-800-331-1MAA FAX: 1-301-206-9789

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Dedicated to the icons of our lives —our heroes, our mentors, and our loved ones.

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Preface

Of all of our inventions for mass communication, picturesstill speak the most universally understood language.

Walt Disney

An icon (from the Greek εικων, “image”) is defined as “a picture that isuniversally recognized to be representative of something.” The world is fullof distinctive icons. Flags and shields represent countries, graphic designsrepresent commercial enterprises; paintings, photographs and even peoplethemselves may evoke concepts, beliefs and epochs. Computer icons are es-sential tools for working with a great variety of electronic devises.

What are the icons of mathematics? Numerals? Symbols? Equations?After many years working with visual proofs (also called “proofs withoutwords”), we believe that certain geometric diagrams play a crucial role invisualizing mathematical proofs. In this book we present twenty of them,which we call icons of mathematics, and explore the mathematics that lieswithin and that can be created. All of our icons are two-dimensional; three-dimensional icons will appear in a subsequent work.

Some of the icons have a long history both inside and outside of math-ematics (yin and yang, star polygons, the Venn diagram, etc.). But most ofthem are essential geometrical figures that enable us to explore an extraor-dinary range of mathematical results (the bride’s chair, the semicircle, therectangular hyperbola, etc.).

Icons of Mathematics is organized as follows. After the Preface we presenta table with our twenty key icons. We then devote a chapter to each, illus-trating its presence in real life, its primary mathematical characteristics andhow it plays a central role in visual proofs of a wide range of mathemati-cal facts. Among these are classical results from plane geometry, propertiesof the integers, means and inequalities, trigonometric identities, theoremsfrom calculus, and puzzles from recreational mathematics. As the American

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x Preface

actor Robert Stack once said (speaking of icons of a different sort), “theseare icons to be treasured.”

Each chapter concludes with a selection of Challenges for the reader toexplore further properties and applications of the icon. After the chapters wegive solutions to all the Challenges in the book. We hope that many readerswill find solutions that are superior to ours. Icons of Mathematics concludeswith references and a complete index.

As with our previous books with the MAA, we hope that both secondaryschool and college and university teachers may wish to use portions of it as asupplement in problem solving sessions, as enrichment material in a courseon proofs and mathematical reasoning, or in a mathematics course for liberalarts students.

Special thanks to Rosa Navarro for her superb work in the preparation ofpreliminary drafts of this manuscript. Thanks too to Underwood Dudley andthe members of the editorial board of the Dolciani series for their carefulreading of an earlier draft of the book and their many helpful suggestions.We would also like to thank Carol Baxter, Beverly Ruedi, and Rebecca Elmoof the MAA’s book publication staff for their expertise in preparing this bookfor publication. Finally, special thanks to Don Albers, the MAA’s editorialdirector for books, who as on previous occasions encouraged us to pursuethis project and guided its final production.

Claudi AlsinaUniversitat Politecnica de Catalunya

Barcelona, Spain

Roger B. NelsenLewis & Clark College

Portland, Oregon

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Twenty Key Iconsof Mathematics

The Bride’s The SemicircleChair

Zhou BiSuan Jing

Garfield’sTrapezoid

CeviansSimilarFigures

The RightTriangle

Napoleon’sTriangles

Two Circles VennArcs andAngles

Polygons withCircles Diagrams

Yin and Yang StarOverlappingFigures

PolygonalLines Polygons

Self-similarFigures

Tatami TilingThe RectangularHyperbola

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Contents

Preface ix

Twenty Key Icons of Mathematics xi

1 The Bride’s Chair 11.1 The Pythagorean theorem—Euclid’s proof and more . . . . 21.2 The Vecten configuration . . . . . . . . . . . . . . . . . . . 41.3 The law of cosines . . . . . . . . . . . . . . . . . . . . . . 71.4 Grebe’s theorem and van Lamoen’s extension . . . . . . . . 81.5 Pythagoras and Vecten in recreational mathematics . . . . . 91.6 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Zhou Bi Suan Jing 152.1 The Pythagorean theorem—a proof from ancient China . . 162.2 Two classical inequalities . . . . . . . . . . . . . . . . . . 172.3 Two trigonometric formulas . . . . . . . . . . . . . . . . . 182.4 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3 Garfield’s Trapezoid 213.1 The Pythagorean theorem—the Presidential proof . . . . . 223.2 Inequalities and Garfield’s trapezoid . . . . . . . . . . . . . 223.3 Trigonometric formulas and identities . . . . . . . . . . . . 233.4 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4 The Semicircle 294.1 Thales’ triangle theorem . . . . . . . . . . . . . . . . . . . 304.2 The right triangle altitude theorem and the geometric mean . 314.3 Queen Dido’s semicircle . . . . . . . . . . . . . . . . . . . 324.4 The semicircles of Archimedes . . . . . . . . . . . . . . . 344.5 Pappus and the harmonic mean . . . . . . . . . . . . . . . 374.6 More trigonometric identities . . . . . . . . . . . . . . . . 384.7 Areas and perimeters of regular polygons . . . . . . . . . . 39

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xiv Contents

4.8 Euclid’s construction of the five Platonic solids . . . . . . . 404.9 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5 Similar Figures 455.1 Thales’ proportionality theorem . . . . . . . . . . . . . . . 465.2 Menelaus’s theorem . . . . . . . . . . . . . . . . . . . . . 525.3 Reptiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535.4 Homothetic functions . . . . . . . . . . . . . . . . . . . . 565.5 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 58

6 Cevians 616.1 The theorems of Ceva and Stewart . . . . . . . . . . . . . . 626.2 Medians and the centroid . . . . . . . . . . . . . . . . . . . 656.3 Altitudes and the orthocenter . . . . . . . . . . . . . . . . . 666.4 Angle-bisectors and the incenter . . . . . . . . . . . . . . . 686.5 Circumcircle and circumcenter . . . . . . . . . . . . . . . . 706.6 Non-concurrent cevians . . . . . . . . . . . . . . . . . . . 726.7 Ceva’s theorem for circles . . . . . . . . . . . . . . . . . . 736.8 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 74

7 The Right Triangle 777.1 Right triangles and inequalities . . . . . . . . . . . . . . . 787.2 The incircle, circumcircle, and excircles . . . . . . . . . . . 797.3 Right triangle cevians . . . . . . . . . . . . . . . . . . . . 847.4 A characterization of Pythagorean triples . . . . . . . . . . 857.5 Some trigonometric identities and inequalities . . . . . . . 867.6 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 87

8 Napoleon’s Triangles 918.1 Napoleon’s theorem . . . . . . . . . . . . . . . . . . . . . 928.2 Fermat’s triangle problem . . . . . . . . . . . . . . . . . . 938.3 Area relationships among Napoleon’s triangles . . . . . . . 958.4 Escher’s theorem . . . . . . . . . . . . . . . . . . . . . . . 988.5 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 99

9 Arcs and Angles 1039.1 Angles and angle measurement . . . . . . . . . . . . . . . 1049.2 Angles intersecting circles . . . . . . . . . . . . . . . . . . 1079.3 The power of a point . . . . . . . . . . . . . . . . . . . . . 1099.4 Euler’s triangle theorem . . . . . . . . . . . . . . . . . . . 1119.5 The Taylor circle . . . . . . . . . . . . . . . . . . . . . . . 1129.6 The Monge circle of an ellipse . . . . . . . . . . . . . . . . 1139.7 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 114

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Contents xv

10 Polygons with Circles 11710.1 Cyclic quadrilaterals . . . . . . . . . . . . . . . . . . . . . 11810.2 Sangaku and Carnot’s theorem . . . . . . . . . . . . . . . 12110.3 Tangential and bicentric quadrilaterals . . . . . . . . . . . 12510.4 Fuss’s theorem . . . . . . . . . . . . . . . . . . . . . . . . 12610.5 The butterfly theorem . . . . . . . . . . . . . . . . . . . . 12710.6 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 128

11 Two Circles 13111.1 The eyeball theorem . . . . . . . . . . . . . . . . . . . . . 13211.2 Generating the conics with circles . . . . . . . . . . . . . 13311.3 Common chords . . . . . . . . . . . . . . . . . . . . . . . 13511.4 Vesica piscis . . . . . . . . . . . . . . . . . . . . . . . . . 13711.5 The vesica piscis and the golden ratio . . . . . . . . . . . . 13811.6 Lunes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13911.7 The crescent puzzle . . . . . . . . . . . . . . . . . . . . . 14111.8 Mrs. Miniver’s problem . . . . . . . . . . . . . . . . . . . 14111.9 Concentric circles . . . . . . . . . . . . . . . . . . . . . . 14311.10 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 144

12 Venn Diagrams 14912.1 Three-circle theorems . . . . . . . . . . . . . . . . . . . . 15012.2 Triangles and intersecting circles . . . . . . . . . . . . . . 15312.3 Reuleaux polygons . . . . . . . . . . . . . . . . . . . . . 15512.4 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 158

13 Overlapping Figures 16313.1 The carpets theorem . . . . . . . . . . . . . . . . . . . . . 16413.2 The irrationality of

p2 and

p3 . . . . . . . . . . . . . . . 165

13.3 Another characterization of Pythagoreantriples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

13.4 Inequalities between means . . . . . . . . . . . . . . . . . 16713.5 Chebyshev’s inequality . . . . . . . . . . . . . . . . . . . 16913.6 Sums of cubes . . . . . . . . . . . . . . . . . . . . . . . . 16913.7 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 170

14 Yin and Yang 17314.1 The great monad . . . . . . . . . . . . . . . . . . . . . . . 17414.2 Combinatorial yin and yang . . . . . . . . . . . . . . . . . 17614.3 Integration via the symmetry of yin and yang . . . . . . . . 17814.4 Recreational yin and yang . . . . . . . . . . . . . . . . . . 17914.5 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 181

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xvi Contents

15 Polygonal Lines 18315.1 Lines and line segments . . . . . . . . . . . . . . . . . . . 18415.2 Polygonal numbers . . . . . . . . . . . . . . . . . . . . . . 18615.3 Polygonal lines in calculus . . . . . . . . . . . . . . . . . . 18815.4 Convex polygons . . . . . . . . . . . . . . . . . . . . . . . 18915.5 Polygonal cycloids . . . . . . . . . . . . . . . . . . . . . . 19215.6 Polygonal cardioids . . . . . . . . . . . . . . . . . . . . . 19615.7 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 198

16 Star Polygons 20116.1 The geometry of star polygons . . . . . . . . . . . . . . . . 20216.2 The pentagram . . . . . . . . . . . . . . . . . . . . . . . . 20616.3 The star of David . . . . . . . . . . . . . . . . . . . . . . . 20816.4 The star of Lakshmi and the octagram . . . . . . . . . . . . 21116.5 Star polygons in recreational mathematics . . . . . . . . . . 21416.5 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 217

17 Self-similar Figures 22117.1 Geometric series . . . . . . . . . . . . . . . . . . . . . . . 22217.2 Growing figures iteratively . . . . . . . . . . . . . . . . . . 22417.3 Folding paper in half twelve times . . . . . . . . . . . . . . 22717.4 The spira mirabilis . . . . . . . . . . . . . . . . . . . . . . 22817.5 The Menger sponge and the Sierpinski carpet . . . . . . . . 23017.6 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 231

18 Tatami 23318.1 The Pythagorean theorem—BhNaskara’s proof . . . . . . . . 23418.2 Tatami mats and Fibonacci numbers . . . . . . . . . . . . . 23518.3 Tatami mats and representations of squares . . . . . . . . . 23718.4 Tatami inequalities . . . . . . . . . . . . . . . . . . . . . . 23818.5 Generalized tatami mats . . . . . . . . . . . . . . . . . . . 23918.6 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 240

19 The Rectangular Hyperbola 24319.1 One curve, many definitions . . . . . . . . . . . . . . . . . 24519.2 The rectangular hyperbola and its tangent lines . . . . . . . 24519.3 Inequalities for natural logarithms . . . . . . . . . . . . . . 24719.4 The hyperbolic sine and cosine . . . . . . . . . . . . . . . 24919.5 The series of reciprocals of triangular numbers . . . . . . . 25019.6 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 251

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Contents xvii

20 Tiling 25320.1 Lattice multiplication . . . . . . . . . . . . . . . . . . . . . 25420.2 Tiling as a proof technique . . . . . . . . . . . . . . . . . . 25520.3 Tiling a rectangle with rectangles . . . . . . . . . . . . . . 25620.4 The Pythagorean theorem—infinitely many proofs . . . . . 25720.5 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 258

Solutions to the Challenges 261Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287Chapter 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290Chapter 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293Chapter 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295Chapter 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297Chapter 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298Chapter 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302Chapter 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303Chapter 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305Chapter 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306

References 309

Index 321

About the Authors 327

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Solutions to the Challenges

Many of the Challenges have multiple solutions. Here we give but one solu-tion to each Challenge, and encourage readers to search for others.

Chapter 11.1. (a) Yes, but they are all similar to the 3-4-5 right triangle. If we set b D

aCd and c D aC2d , then a2Cb2 D c2 implies a2�2ad�3d2 D 0,so that a D 3d , b D 4d , and c D 5d . (b) Yes, and they are all similarto the right triangle with sides 1,

p2, and

p3. (c) If we set b2 D a2r

and c2 D a2r2, then r2 D r C 1 so that r must be the golden ratio� D .1C

p5/=2.

1.2. Let x; y; z denote the sides of the outer squares, and let A;B;C denotethe measures of the angles in the central triangle opposite sides a; b; c,respectively. For (a), the law of cosines yields a2 D b2Cc2�2bc cosAand x2 D b2C c2� 2bc cos.180ı�A/ D b2C c2C 2bc cosA, hencex2 D 2b2C2c2�a2. Similarly y2 D 2a2C2c2�b2 and z2 D 2a2C2b2�c2, from which the result follows. In (b), we have x2Cy2Cz2 D6c2 and z2 D c2, hence x2 C y2 D 5z2.

1.3. The line segmentsPbPc andAPa are perpendicular and equal in length,as are the line segments PbPc and APx , hence A is the midpoint ofPaPx . The other two results are established similarly.

1.4. Label the vertices of the squares and triangles as shown in Figure S1.1,let the sides of the square AEDP be a; and the sides of the squareCHIQ be b. Triangles ABE and BCH are congruent with legs a andb, and the corollary in the paragraph preceding Figure 1.7 tells us thattriangles ABE andDEK have the same area, as do triangles BCH andKHI , and triangles KMN and DKI . Hence triangles ABE, DEK,KHI , and BCH all have the same area (ab=2/, and furthermore,jPQj D 2.aC b/.

261

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262 Solutions to the Challenges

DE

A B C

H I

M

N

K

P Q

Figure S1.1.

Thus

area.KMN/ D area.DKI/ D area.PDKIQ/ � area.PDIQ/

D .a2 C b2 C 4 � ab=2C area.BEKH//

�.1=2/ � 2.aC b/ � .aC b/

D .aC b/2 C area.BEKH/ � .aC b/2

D area.BEKH/

as claimed. (Area.BEKH/ D a2 C b2.)

1.5. The areas of triangle ABH and quadrilateral HIJC are equal. Leta D jBC j and b D jAC j. Since triangles ACI and ADE aresimilar, jCI j=b D a=.aC b/ so that jCI j D ab=.aC b/ andarea(ACI/ D ab2=2.aC b/. Similarly area.BJC/ D a2b=2.aC b/,and thus area.ACI/C area.BJC/ D ab=2 D area.ABC/. Hence

area.AJH/ C 2area.HIJC/C area.BIH/

D area.ACI/C area.BJC/

D area.ABC/

D area.AJH/C area.HIJC/

C area.BIH/C area.ABH/;

and thus area(HIJC ) = area(ABH ) [Konhauser et al., 1996].

1.6. Construct congruent parallelograms ABED0 and ADFB 0, as shown inFigure S1.2 and observe that Q and S are the centers of ABED0 andADFB 0. A 90ı clockwise rotation about R takes ABED0 to ADFB 0

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Solutions to the Challenges 263

and the segment RQ to RS . HenceQR and RS are equal in length andperpendicular. A similar 90ı counterclockwise rotation about T showsthat QT and TS are equal in length and perpendicular, from which itfollows that QRST is a square.

A

B

C

D

Q

R

S

T

E

F

B

C

D

Figure S1.2.

1.7. Triangles ACF and BCE are congruent, and a 90ı turn takes one tothe other. Hence AF and BE are perpendicular at P . Since †APEis a right angle, it lies on the circumcircle of square ACED. Hence†DPE D 45ı as it subtends a quarter circle. Similarly †FPG D 45ı,so that †DPG D 45ı C 90ı C 45ı D 180ı, i.e., D;P , and G arecollinear [Honsberger, 2001].

1.8. Yes, but only in a very special case. Let the original triangle have sidesa; b; c and angles ˛, ˇ, � . Since each vertex angle of a regular polygonwith n sides measures .n � 2/180ı=n we need

360ı �2.n � 2/180ı

n�maxf˛; ˇ; �g > 0;

or maxf˛; ˇ; �g < 720ı=n. Hence n � 12. If each flank triangle hasthe same area as the original triangle, then bc sin˛=2 D

bc sin.720ı=n � ˛/=2, i.e., sin˛ D sin.720ı=n � ˛/ (and similarly forˇ and � ). Hence n = 4, the classical configuration with squares, or ˛ Dˇ D � D 360ı=n and n D 6, i.e., regular hexagons around an equi-lateral triangle.

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Chapter 22.1. We have equality in (2.1) if and only if sin � D 1, i.e., � D 90ı, in

Figure 2.5. This implies that the shaded triangles are isosceles righttriangles and thus

pa Dpb, so that a D b. We have equality in (2.2)

if and only if � D 90ı in Figure 2.6 and jax C byj D jaxj C jbyj, i.e.,ax and by have the same sign. This implies that the shaded triangles aresimilar, so that jxj=jyj D jaj=jbj, or jayj D jbxj. If ax and by have thesame sign, we have ay D bx.

2.2. The area of the white parallelogram in Figure S2.1a is sin.�=2�˛Cˇ/and sin.�=2 � ˛ C ˇ/ D cos.˛ � ˇ/ [Webber and Bode, 2002].

(a) (b)

cos α

sin β

sin α cos β

β

α

1

1 – α + βπ

2

Figure S2.1.

2.3. Set x D sin t and y D cos t in Figure 2.5.

2.4. Using (2.2) we have

jax C by C czj � jax C byj C jczj

�pa2 C b2

px2 C y2 C jczj

�p.a2 C b2/C c2

p.x2 C y2/C z2:

We can similarly prove by induction the n-variable Cauchy-Schwarzinequality

ja1x1 C a2x2 C � � � C anxnj

�

qa21 C a

22 C � � � C a

2n

qx21 C x

22 C � � � C x

2n:

2.5. Set x and y equal to 1=2 in (2.2):

aC b

2D

ˇˇa � 12 C b �

1

2

ˇˇ �

pa2 C b2

r1

2D

ra2 C b2

2:

2.6. No. Consider any isosceles triangle with †A D †B , †C ¤ 90ı.

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Chapter 33.1. Use (3.1) with a D jsin � j and b D jcos � j.

3.2. See Figure S3.1. The right triangles with legs a,pab and

pab, b

are similar, and thus the shaded triangle is a right triangle. Its hy-potenuse a C b is at least as long as the base 2

pab of the trapezoid,

thus .aC b/=2 �pab.

ab

a

b

a+b

ab

Figure S3.1.

3.3. See Figure S3.2.

a

ab

b

b–a

b+a

Figure S3.2.

3.4. Let p D F2n and q D F2n�1 in (3.4). Then pCq D F2nC1 and p2Cpq C 1 D F 22n C F2nF2n�1 C 1. By Cassini’s identity (Fk�1FkC1 �F 2kD .�1/k for k � 2/, F 22nC1 D F2n�1F2nC1, and thus p2CpqC

1 D F2n�1.F2n C F2nC1/ D q � F2nC2. Hence q=.p2 C pq C 1/ D1=F2nC2.


cos α cos β

sin

α c

os β

sin α sin β

cos

α s

in β

αβ

α

sin β cos

β

1

α

Figure S3.3.

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(a) (b)

sec α tan β

tan

α ta

n β

αβ

α

sec

α 1

tan αtan β tan α tan β

αβ

α

1

αsec α tan β se

c α tan

α

1

tan

β

Figure S3.4.

3.7. See Figure S3.5 [Burk, 1996].

ab ac bcba

c

a 2

c 2

b 2

ab

Figure S3.5.

3.8. Compute the lengths of the hypotenuse of each right triangle, as shownin Figure S3.6. Then in the shaded right triangle we have sin � D2z=.1C z2/ and cos � D .1 � z2/=.1C z2/[Kung, 2001].

θ/2

θ/2θ/2

1

z

z2

z

1 +

z2

1 + z2

1 + z 2

z

Figure S3.6.

3.9. If we place the trapezoid in the first quadrant, as shown in Figure S3.7,then the coordinates of Pk are

Pk D ..b C ak/=2; .aC bk/=2/ D .b=2; a=2/C k.a=2; b=2/;

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so Pk lies on the ray from .b=2; a=2/ with slope b=a.

a

akb

bk

Figure S3.7.

3.10. In the Garfield trapezoid in Figure S3.8 we havep2.x C 1=x/ �

px C

p1=x, from which the first inequality in (3.5) follows. The

second follows from the AM-GM inequality.

x

1/xx

1/x2/x

2x

2(x+1/x)

Figure S3.8.

Chapter 44.1. Draw AD and BD parallel to BC and AC , respectively, intersecting in

D. Since C is a right angle, ACBD is a rectangle. Hence its diagonalsAB and CD are equal in length and bisect each other at O . Since Ois equidistant from A;B; and C , it is the center of the circumcircle oftriangle ABC , with AB as diameter. See Figure S4.1.

BA

C

D

O

Figure S4.1.

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4.2. Let jPQj D a, jQRj D b, and jQS j D h, so that h2 D ab followsfrom the right triangle altitude theorem. If A denotes the area of thearbelos and C the area of the circle with diameter QS , then

A D�

2

�aC b

2

�2��

2

�a2

�2��

2

�b

2

�2D �

�ab

4

�D �

�h

2

�2D C:

4.3. (a) Set � D arcsin x, sin � D x, and cos � Dp1 � x2 in Figure

4.14 and in the two half-angle tangent formulas. (b) Set � D arccos x,cos � D x, and sin � D

p1 � x2 in Figure 4.14 and in the two half-

angle tangent formulas. (c) Set � D arctan x, sin � D x=p1C x2, and

cos � D 1=p1C x2 in Figure 4.14 and in the two half-angle tangent

formulas.

4.4. With the solid line as the altitude, the area of the triangle is .1=2/ sin 2� .With the dashed line as the altitude, its area is .1=2/.2 sin �/ cos � ,and hence sin 2� D 2 sin � cos � . Using the law of cosines, the length2 sin � of the chord satisfies .2 sin �/2 D 12C 12 � 2 cos 2� , and hencecos 2� D 1 � 2 sin2 � .


1xx

arcsinxarctanx

Figure S4.2.

4.6. The area of the shaded region is �.R2 � r2/=2. However, a is the geo-metric mean ofRCr andR�r , hence a2 D .RCr/.R�r/ D R2�r2.

4.7. Let T denote the area of the right triangle, L1 and L2 the areas of thelunes, and S1 and S2 the areas of the circular segments (in white) inFigure S4.3. Then the version of the Pythagorean theorem employed inthe proof of Proposition 4 from Archimedes’ Book of Lemmas yieldsT C S1 C S2 D .S1 C L1/C .S2 C L2/, so T D L1 C L2.

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S1S

2

L1

L2

T

Figure S4.3.

4.8. We first compute the lengths of the segments in Figure 4.18 using tri-angle geometry: Let jABj D d , then

jAF j D dp6=3; jBF j D d

p3=3;

jAEj D dp2=2; and jBN j D d=�

p3:

(a) For a tetrahedron with edge s, consider the cross-section passingthrough one edge and the altitudes of the opposite faces, as shown inFigure S4.4.

ss

sss 3 2

s 3 3s 3 6

s 63

d2

d/2

sd/2

Figure S4.4.

Since the two altitudes have length sp3=2, s and d are related as shown

in the gray right triangle, we have sp6

3�d

2

!2C

sp3

3

!2D

�d

2

�2

and thus s D dp6=3 D jAF j.

(b) For a cube with edge s, the diameter d of the sphere is also thediagonal of an s-by-s

p2 rectangle, so that d2 D s2 C 2s2 and hence

s D dp3=3 D jBF j.

(c) For an octahedron with edge s, d is the diagonal of a square withside s and hence s D d

p2=2 D jAEj.

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(d) For a dodecahedron with edge s, consider a cross-section passingthrough a pair of opposite edges, as shown in Figure S4.5.

d

s

th

Figure S4.5.

The cross-section is a hexagon whose remaining four edges are altitudesof the pentagonal faces, of length h D .s tan 72ı/=2. The angle betweentwo such edges is a dihedral angle arccos.�1=

p5/ of the dodecahedron.

If we let t be the length of the indicated segment in the figure, then thelaw of cosines along with identities such as sec 72ı D 2� and � C.1=�/ D

p5 yields t2 D s2.1C�/2. Hence we have d2 D s2Cs2.1C

�/2 D 3s2.1C �/ D 3s2�2, or s D d=�p3 D jBN j.

4.9. Assume without loss of generality that the radius of the small semicir-cles is 1 and the radius of the large semicircle is 2. Consequently theperimeter of the cardioid is 4� . It suffices to consider lines through thecusp that make an angle of � in Œ0; �=2� with the common diameter ofthe semicircles, as shown in Figure S4.6. The length of the portion ofperimeter to the right of the line is 2� C 2.� � �/ D 2� , half the totalperimeter.

θ 2θ

π – θ

Figure S4.6.

Chapter 55.1. Since (a; b; c/ and .a0; b0; c0/ are similar, a0 D ka, b0 D kb, and c0 D

kc for some positive k. Hence

aa0 C bb0 D ka2 C kb2 D kc2 D cc0:

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5.2. Let s and t be the lengths of the segments indicated in Figure S5.1.Since each of the shaded triangles is similar to ABC , b0=b D s=c anda0=a D t=c. Hence

a0

aCb0

bCc0

cDt

cCs

cCc0

cDc

cD 1

as claimed [Konhauser et al., 1996].

BA

C

ab

c

Pts

ts

Figure S5.1.

5.3. Draw the diameter PS and the segment QS as shown in Figure S5.2.Since†QPS D †PQR, right triangle PQS is similar to right trianglePQR. Thus jQRj=jPQj D jPQj=jPS j, so that jPQj is the geometricmean of jQRj and jPS j.

P

Q

R

S

Figure S5.2.

5.4. See Figure S5.3. Draw KE parallel to CD. Triangles AKE andAFC are similar, thus jFC j=jKEj D jAC j=jAEj D 2, so that

BA

CD

E

F

G

xx

K

Figure S5.3.

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jFC j D 2 jKEj. Since x = 22.5ı, †DFG D 67:5ı D †DGF . Tri-angles DFG and EGK are similar, and since DFG is isosceles, so isEKG. Thus jGEj D jKEj, so that jFC j D 2 jGEj.

5.5. Since the triangles have the angle at D in common, it is sufficient (andnecessary) to have the angles marked x and y equal. Hence AD mustbe the bisector of the angle at A. See Figure S5.4.

A

BC

D

Ex

y x

Figure S5.4.

5.6. In Figure S5.5ab we see that the I and P pentominoes are rep-4 reptiles.Since the L pentomino tiles a 2�5 rectangle (see Figure S5.5c) andthe Y pentomino tiles a 5�10 rectangle (see Figure S5.5d), each tiles a10�10 square and is thus a rep-100 reptile.

(a) (b) (d)(c)

Figure S5.5.

5.7. No. Condition (5.3) yields ebx D keax=k so that at x D 0 we havek D 1, i.e., a D b.

Chapter 66.1. From Figure S6.1 we have 2ma � bC c, 2mb � cCa, 2mc � aCb,

so that ma C mb C mc � 2s. The shaded triangle has sides 2ma,2mb , 2mc and medians 3a=2, 3b=2, 3c=2, so by the above inequal-ity we have 3.aC b C c/=2 � 2.ma C mb C mc/, or 3s=2 � maC

mb Cmc .

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2a

2b

2ca

b

c

mb

mb

mc

mc

ma

ma

Figure S6.1.

6.2. See Figure S6.2a, where a > b > c. Since the sides are in arithmeticprogression, aC c D 2b so the semiperimeter is s D 3b=2. Hence thearea K satisfies K D rs D 3br=2 and K D bh=2, thus h D 3r sothat the incenter I lies on the dashed line parallel to AC one-third ofthe way from AC to B (see Figure S6.2b). The centroid G lies on themedian BMb one-third of the way from AC to B , hence G also lieson the dashed line [Honsberger, 1978].

ac

b

rr

rA

B

Ch

Ir

A

B

Ch

I G

Mb

(a) (b)

Figure S6.2.

6.3. Draw an altitude to one side, cut along two medians from the foot ofthe altitude in the resulting small triangles, and rotate the pieces asshown in Figure S6.3. Then rotate the entire new triangle 180ı [Kon-hauser et al., 1996].

Figure S6.3.

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6.4. Our solution is from [Erdos, 1940]. Since a cevian is shorter thaneither of the sides adjacent to the vertex from which it is drawn, itis shorter than the longest side of the triangle. Draw PR and PS par-allel, respectively, to AC and BC as shown in Figure S6.4, so that�PRS is similar to �ABC . Since PZ is a cevian in �PRS and RSis its longest side, jPZj < jRS j.

BA

C

XY

Z

PQ

SR

T

Figure S6.4.

Draw QT through P parallel to AB . �QPY is similar to �ABY ,and since AB is the longest side of �ABY , QP is the longest side of�QPY . Since AQPR is a parallelogram, we have jPY j < jQP j DjARj. Similarly, jPX j < jPT j D jSBj, and hence

jPX j C jPY j C jPZj < jSBj C jARj C jRS j D jABj :

6.5. Triangles AHY and BHX in Figure 6.7 are similar, and hencejAH j=jHY j D jBH j=jHX j, so that jAH j � jHX j D jBH j � jHY j. Inthe same way jBH j � jHY j D jCH j � jHZj.

6.6. Since the sides of �XYZ are parallel to the sides of �ABC , the alti-tudes of �XYZ are perpendicular to the sides of �ABC . See FigureS6.5, and letO be the orthocenter of�XYZ. Thus�AOZ is congru-ent to �BOZ, and jAOj D jBOj. Similarly jBOj D jCOj, whenceO is equidistant from A;B , and C .

O

BA

C

XY

Z

Figure S6.5.

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6.7. See Figure S6.6, and let M be the midpoint of CD. Thus ME is par-allel to BC . Since BC is perpendicular to AC , so is ME, hence itlies on the altitude to AC in�ACE. In�ACE, CD is an altitude, soM is the orthocenter of �ACE. Thus AM lies on the third altitude,so AM is perpendicular to CE. Since A andM are midpoints of sidesof�CDF , AM is parallel to FD, and so FD is also perpendicular toCE [Honsberger, 2001].

BA

C

ED

F

M

Figure S6.6.

6.8. (a) Draw PQ and CR perpendicular to AB , as shown in Figure S6.7.

BA

C

XY

Z

P

RQ

Figure S6.7.

Then right triangles PQZ and CRZ are similar, and so

jPZj

jCZjDjPQj

jCRjDjPQj � jABj=2

jCRj � jABj=2DŒABP �

ŒABC �:

Similarly,jPX j

jAX jDŒBCP �

ŒABC �andjPY j

jBY jDŒCAP �

ŒABC �:

Hence

jPX j

jAX jCjPY j

jBY jCjPZj

jCZjDŒABP �C ŒBCP �C ŒCAP �

ŒABC �DŒABC �

ŒABC �D 1:

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(b) Since jPAj=jAX j D 1 � jPX j=jAX j, etc., we have

jPAj

jAX jCjPBj

jBY jCjPC j

jCZjD 3�

�jPX j

jAX jCjPY j

jBY jCjPZj

jCZj

�D 3�1 D 2:

6.9. See Figure S7.8. Since X and Y are midpoints of BC and AC , XYis parallel to AB and half its length, or jAZj D jXY j D jBZj. SincePQ is parallel to AB and XY , it follows that PR, RS , and SQ areeach half the length of AZ, XY , and BZ, respectively, hence jPRj DjRS j D jSQj [Honsberger, 2001].

BA

C

XY

QPSR

Z

Figure S6.8.

6.10. We need only show that the semiperimeter s of a triangle of area 1=�exceeds 1. If r denotes the inradius of the triangle, we have rs D 1=� ,or s D 1=�r . The area �r2 of the incircle is less than the area ofthe triangle, so �r2 < 1=� , so 1 < 1=.�r/2, or 1 < 1=�r . Hences D 1=�r > 1 [Honsberger, 2001].

Chapter 77.1. See Figure S7.1. Equality holds in (a) and (b) if and only if ay�bx D

0. i.e., a=b D x=y, and in (c) if and only if a=b D x=y or �y=x.

a +b22

x +y22

|ax+by|

|ay–bx|

a –b22 x –y 22

|ax–by||ay–bx|

(a) (b)

Figure S7.1.


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3x +y22

3y +x22

2y(x–y) 2x(x–y)x+y

x+y

Figure S7.2.

7.3. (a) Two nonintersecting circles have two internal tangents that inter-sect at a point between the circles on the center line, the line joiningthe centers. Hence vertex A, the intersection of the lines containingside b and side c, lies on the line joining Ib and Ic ; and similarly forvertices B and C .

(b) Two nonintersecting circles also have two external tangents thatintersect at a point to one side of both circles on the center line. Fur-thermore, the configuration of two circles and four tangent lines issymmetric with respect to the center line. So if one internal and oneexternal tangent are perpendicular, so are the other internal and exter-nal tangents. The excircles on sides a and c have the perpendicularlines of the legs as internal and external tangents, and hence the left-most dashed line is perpendicular to the line of the hypotenuse. Similarconsideration of the a excircle and the incircle, the b excircle and theincircle, and the b and c excircles shows that the other three dashedlines are also perpendicular to the line of the hypotenuse and thus par-allel to one another.

7.4. Yes. We have .z C 2/2 C Œ.2=z/C 2�2 D Œz C 2C .2=z/�2 and if z isrational, so are all three sides.

7.5. (a) The length of the overlap is aC b � c D 2r .

(b) With sides a; b; c and altitude d , we have, applying part (a) of theproblem to each of the three right triangles,

2r C 2r1 C 2r2 D .aC b � c/C .jADj C h � b/C .jBDj C h � a/

D 2hC jADj C jBDj � c D 2h;

and hence r C r1 C r2 D h [Honsberger, 1978].

7.6. (a) Since c D 2R D aCb�2r ,RCr D .aC b/=2 �pab D

p2K.

(b) Since rs D K D ch=2 � .2R �R/=2 D R2, R2 � rs � 0: Buts D 2R C r implies rs D 2rR C r2, hence R2 � 2Rr � r2 � 0, or.R=r/2 � 2.R=r/ � 1 � 0. Because R=r > 0, R=r is at least as large

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as the positive root 1Cp2 of x2 � 2x � 1 D 0. Equality holds when

the triangle is isosceles.

7.7. If we let c D 1 �pxy and b D

p.1 � x/.1 � y/, then a D

pc2 � b2 D

px C y � 2

pxy. By the arithmetic-geometric mean

inequality a is real so a; b, and c form a right triangle with b � c;a D 0 if and only if x D y.

7.8. Introduce a coordinate system with the origin at C , AC on the pos-itive x-axis, and BC on the positive y-axis, so that the coordinatesof the points of interest are M.b=2; 0/, N.0; ra/, and I.r; r/. Writingthe equation for the line MN and using (7.2), it follows that I lies onMN .

7.9. Let a; b; c be the sides of the triangle, and s the side of the square, asshown in Figure S7.3.

a

b

c

s

tx

y

s

Figure S7.3.

Using similar triangles, s D tc=b, s=.b � t / D a=c, and hences D abc=.ab C c2/. If (a; b; c/ is a primitive Pythagorean triple,then abc and ab C c2 are relatively prime and s is not an integer.If (a; b; c/ is replaced by (ka, kb, kc) where k is a positive integer,then s D kabc=.ab C c2/ which will be an integer if and only if k isa multiple of ab C c2. Thus the smallest such triangle similar to theprimitive triangle is obtained by setting k D abC c2. Since the small-est (in terms of area and perimeter) primitive Pythagorean triple is(3, 4, 5), and since it minimizes abC c2, the desired triple is obtainedby multiplying by 3 �4C52 D 37 to get (111,148,185) [Yocum, 1990].

7.10. The bowtie lacing (d) is the shortest and the zigzag lacing (c) is thelongest. We will show that the length order of the lacings is (d)< (a)<(b) < (c) by comparing edge lengths in right triangles whose verticesare the eyelets. Removing lace segments of equal length from the (d)and (a) lacings yields Figure S7.4. Comparing a leg and hypotenuse ofa right triangle yields (d) < (a).

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(a)(d)

Figure S7.4.

Removing lace segments of equal length from the (a) and (b) lacingsyields the patterns on the left in Figure S7.5, which are four copiesof the simpler patterns in the center of the figure. On the right werearrange the lace segments and use the fact that a straight line is theshortest path between two points to conclude (a) < (b).

(b)(a) (b)(a) 4 4

Figure S7.5.

Repeating the above procedure with lacings (b) and (c) yields FigureS7.6, and hence (b) < (c).

(c)(b)

Figure S7.6.

7.11. See Figure S7.7, and compute sin � and cos � using the shadedtriangle.

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θ

2θ

cos 2θ

cos θ

1

θ

2θ

cos θ

11

sin

2θ

Figure S7.7.


cos β

(α – β)/2

sin α

cos α

sin β

β

απ – (α – β)

1

1

sin β

cos α

1 + 1 –

2cos(π – (α

– β))

22

Figure S7.8.

7.13. The mean xyp.x C y/=.x3 C y3/ is smaller than the harmonic

mean, and the meanp.x3 C y3/=.x C y/ is larger than the contra-

harmonic mean. Thus we can add two rows to Table 7.1:

c b a

0)2xy

x C yxy

rx C y

x3 C y3

p3xy jx � yjp

.x3 C y3/.x C y/

5)

sx3 C y3

x C y

x2 C y2

x C y

jx � yjpxy

x C y

Chapter 88.1. It suffices to show that †PGR D 60ı and jPGj D 2 jGRj. Let T be

the midpoint of BC . We first show that triangles PCG and RTG aresimilar. See Figure S8.1.

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A B

C

P

Q

G

T

R

Figure S8.1.

Since RT is parallel to AC , jPC j D jAC j D 2 jRT j. Since G isalso the circumcenter of equilateral �BQC , GT is perpendicular toBC , �CGT is a 30ı-60ı-90ı triangle, so †GCT D 30ı and jCGj D2 jGT j. But†RTB D †ACB , so that†RTG D †ACBC90ı. How-ever, †PCG D 60ıC†ACBC 30ı, hence triangles PCG and RTGare similar, and since jCGj D 2 jGT jwe have jPGj D 2 jGRj. Finally,†PGR D †PGT C†TGR D †PGT C†PGC D 60ı, so �PGRis a 30ı-60ı-90ı triangle, as claimed [Honsberger, 2001].

8.2. (a) Since X and Z are the midpoints of AC and CD, XZ is parallelto AD and jXZj D .1=2/ jADj. Similarly YZ is parallel to BC andjYZj D .1=2/ jBC j. Since jADj D jBC j, we have jXZj D jYZj.Because of parallels, †YXZ C †XYZ D 120ı and thus †XZY D60ı. Hence �XYZ is equilateral.

(b) Since †AC†B D 120ı, then †C C†D D 240ı. Thus

†PCB D 360ı � †C � 60ı D 300ı � .240ı � †D/

D 60ı C†D D †ADP:

So triangles ADP and BCP are congruent and hence the angle be-tween AP and BP is 60ı. Thus �APB is equilateral [Honsberger,1985].

8.3. The triangle with dashed sides in Figure 8.14 is the Napoleon triangle,since the outer vertex of a small gray triangle is the centroid of theequilateral triangle erected on the entire side of the given triangle, asillustrated in Figure S8.2.

Figure S8.2.

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8.4. Yes. There are two cases to consider in computing the area of a flanktriangle. We use the formula .xy sin �/=2 for the area of a triangle withsides x and y and included angle � . When ˛ > 60ı equality of areasof ABC and the flank triangle requires sin˛ D sin.240ı � ˛/, or ˛ D120ı, as shown in Figure S8.3a.

b

cα

120 + α

(a) (b) (c)240 – α

b cα

Figure S8.3.

When ˛ < 60ı equality of areas of ABC and the flank triangle requiressin˛ D sin.120ıC˛/, or ˛ D 30ı, as shown in Figure S8.3b. The onlytriangle all of whose angles are 30ı or 120ı is a 30ı�30ı�120ı triangle,as illustrated in Figure S8.3c.

8.5. Let a and b be the sides of ABCD, as shown in Figure S8.4. The graytriangles ABE and BCF are congruent, and hence jBEj D jBF j.Since the obtuse angles in the gray triangles are 150ı, the two grayangles at vertex B sum to 30ı, hence †EBF D 60ı and triangleBEF is equilateral, with side length jBEj D jBF j D jEF j Dpa2 C b2 C ab

p3 from the law of cosines. Let G and H denote

the centroids of ADE and CDF and O the center of ABCD. SincejGOj D .3aC b

p3/=6 and jHOj D .3b C a

p3/=6, jGH j2 D

jGOj2 C jOH j2 D .a2 C b2 C abp3/=3, and thus jGH j D jEF j �p

3=3 as claimed.

A

B C

D

E

Fa

bH

G

O

Figure S8.4.

8.6. Since triangles BC 0C , BA00C , and BB 0C have the same base BCand equal angles C 0, A00, and B 0, they have the same circumcircle, as

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shown in Figure S8.5. Thus †CA00B 0 D †B (in triangle ABC ). Since†BCA00 D †B C†C (in triangle ABC ), †C 0CA00 D †C and henceline segments A00B 0 and C 0C are parallel. Similarly A00C 0 and B 0B areparallel, and hence AC 0A00B 0 is a parallelogram as claimed.

B C

B´C´

A´

A

Figure S8.5.

8.7. See Figure S8.6 [Bradley, 1930].

c–a

a

b

c

Figure S8.6.

Chapter 99.1. Computing the power of A with respect to the circle with b D jAC j

and c D jABj yields .c � a/.c C a/ D b2, or c2 D a2 C b2.

9.2. LetAB ,CD, andEF be the three chords, and set a D jPAj, b D jPBj,c D jPC j, d D jPDj, e D jPEj, and f D jPF j. Then a C b DcCd D eCf and computing the power of P with respect to the circleyields ab D cd D ef . If we let s denote the common value of the sumsand p the common value of the products, then each set fa; bg, fc; dg,fe; f g are the roots of x2 � sx C p D 0 and thus the sets are the same.Without loss of generality set a D c D e. Then the points A, C , and

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E lie on a circle with center P and also on the given circle, so the twocircles coincide and P is the center of the given circle [Andreescu andGelca, 2000].

9.3. fA;X;Zb; Ycg are concyclic since AXZb and AXYc are right triangleswith a common hypotenuse AX , the diameter of the common circum-circle; and similarly for fB; Y;Xc ; Zag and fC;Z;Xb; Yag. See FigureS9.1a.

BA

C

BA

C(a) (b)

Figure S9.1.

9.4. fX; Y;Xc ; Ycg are concyclic since XYXc and XY Yc are right trian-gles with a common hypotenuse XY , the diameter of the common cir-cumcircle. Similarly for fY;Z; Ya; Zag and fZ;X;Zb; Xbg. See FigureS9.1b.

9.5. The distance between the observatory and the center of the earth is6382.2 km, and thus the power is 53,592.84 km2. The distance to thehorizon is the square root of the power, or about 231.5 km.

9.6. As usual, let a; b; c denote the sides opposite angles A;B;C , respec-tively. The power ofAwith respect to the circumcircle ofCBZY yieldsc jAZj D b jAY j and the power ofB with respect to the circumcircle ofACXZ yields c jBZj D b jBX j. Adding yields c2 D a jBX jCb jAY j.In both cases jBX j D a � b cosC and jAY j D b � a cosC , andhence

c2 D a.a � b cosC/C b.b � a cosC/ D a2 C b2 � 2ab cosC:

[Everitt, 1950].

9.7. See Figure S9.2, where two additional angles have measure y. Hence†DOB D 2y and †B D 90ı � y since �DOB is isosceles. SinceABDC is a cyclic quadrilateral, xC.90ı�y/ D 180ı, or x�y D 90ı.

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A B

DC x y

y

y

2yO

Figure S9.2.

Chapter 1010.1. Rotate the inner square 45ı, as shown in Figure S10.1.

Figure S10.1.

10.2. Since Q is cyclic, Ptolemy’s theorem pq D acC bd holds, and sinceQ is tangential, aCc D bCd . Hence two applications of the AM-GMinequality yield

8pq D 2.4acC4bd/ � 2Œ.aC c/2C .bCd/2� D .aCbC cCd/2:

10.3. The centers of the two circles in Figure 10.20 lie on the bisector OCof †AOB , which is also the altitude from O in �AOB (see FigureS10.2). If we set jOAj D 1, then jOAj D 1 and jODj D

p2=2.

AO

BC

D

Figure S10.2.

Hence the diameter jCDj D .2 �p2/=2. But from Challenge 7.5a

the diameter of the incircle of �AOB is jODj D 2 �p2 D 2 jCDj

[Honsberger, 2001].

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10.4. Let segments jOAj, etc. be labeled as a, etc., and let K1, etc. de-note the areas of the shaded triangles in Figure S10.3. Since verticalangles at O are equal, as well as at A and C , and at Band D, wehave

K1

K2Dae

cg;K3

K4Ddf

bh;K1

K4Dax

by; and

K3

K2Ddx

cy:

QP

AC

X O Y

D

B

K1 K2

K3 K4

a

b

c

d

e

f

g

h

Figure S10.3.

HenceK1K3

K2K4Dadef

bcghDadx2

bcy2;

so that computing the powers of the points X and Y with respect tothe circle yields

x2

y2Def

ghD.p � x/.q C x/

.p C y/.q � y/Dpq � x.q � p/ � x2

pq C y.q � p/ � y2:

Simplifying, pq.y � x/ D xy.q � p/, from which the desired resultfollows [Bankoff, 1987].

10.5. See Figure S10.4, where we have drawnABCD, and also the diameterDF and chord CF . Because each angle intercepts the same arc of thecircle †DAE D †DFC , and �DCF is a right triangle.

A

B

C

D

E

F

Figure S10.4.

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Hence †ADE D †FDC , and thus jABj D jCF j as chords in equalarcs. Therefore

jAEj2 C jBEj2 C jCEj2 C jDEj2 D jABj2 C jCDj2

D jCF j2 C jCDj2 D .2R/2:

10.6. If ABCD is a bicentric trapezoid, as shown in Figure S10.5a, then†AC†C D †B C†D and †AC†D D †B C†C.D 180ı), andhence †A D †B and ABCD is isosceles. Then arc AC D arc BDand hence arc AD D arc BC . Thus jADj D jBC j D u (say), and letx and y denote the bases as shown in Figure S10.5b.

BA

CD

h

y

x

uu

(y–x)/2

(a) (b)

Figure S10.5.

Then x C y D 2u so that u D .x C y/=2. Let h denote the altitude ofABCD. Applying the Pythagorean theorem to the gray triangle yieldsh Dpxy.

Chapter 1111.1. Draw line segments PQ, BP , and CQ, as shown in Figure S11.1.

Since�APB and�AQC are isosceles andBP andCQ are parallel,we have

†BAC D 180ı � .†PAB C†QAC/

D 180ı � .1=2/.180ı � †APB C 180ı � †AQC/

D .1=2/.†APB C†AQC/ D .1=2/.180ı/ D 90ı:

A

B

C

QP

Figure S11.1.

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11.2. Using angle measurement methods from Section 9.2, we have†QPR D x=2 D .z � y/=2, and hence xCy D z. It is not necessaryfor the circles to be tangent.

11.3. Draw AO intersecting the given circle at P as shown in FigureS11.2. Since AO bisects †BAC , we need only show that BP bi-sects †ABC to conclude that P is the incenter of �ABC . Because�OPB is isosceles, we have

†CBO C†PBC D †PBO D †OPB D †PAB C†APB:

But †BPO D †PAB since corresponding sides are perpendicular,hence †PBC D †APB and BP bisects †ABC .

A

B

C

OP

Figure S11.2.

11.4. Orient the lamina as shown in Figure S11.3. Computing its momentabout the vertical axis of the crescent as the difference of the mo-ments of the two disks yields

2x � .� � 12 � �x2/ D 1 � .� � 12/ � x � .�x2/:

1 2x 2x

Figure S11.3.

Simplifying yields x3 � 2x C 1 D 0, which has three real roots:1, –�, and 1=�, with only the last one being meaningful [Glaister,1996].

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11.5. Two roots (p1 and

p4/ are the radius and diameter of each of the

circles. The others can be seen in Figure S11.4.

2

35

Figure S11.4.

11.6. Since the circular window lies between two arcs of radii (say) r and2r , its radius is r=2 and its center lies at the intersection of two arcswith radii 3r=2. See Figure S11.5.

Figure S11.5.

11.7. LetRQPST be a chord of the outer circle, as shown in Figure S11.6.The power of P with respect to each circle is constant, say jPQj �jPS j D c and jPRj � jPT j D k. Since jST j D jQRj we have

kD.jPQjCjQRj/.jPS jCjST j/D .jPQjCjQRj/.jPS jCjQRj/

DjQRj�.jQRjCjPQjCjPS j/CcDjQRj � .jQRj C jQS j/C c:

P

Q RST

Figure S11.6.

Thus jQRj � .jQRj C jQS j/ is constant, so jQRj will be a maximumwhenever jQS j is a minimum. The arithmetic mean-geometric mean

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inequality applied to jPQj and jPS j with jPQj � jPS j D c tellsus that jQS j is a minimum when P is the midpoint of QS , and PbisectsQS if and only if PR is perpendicular to the segment joiningP to the common center of the two circles [Konhauser et al., 1996].

11.8. Since †OMP is a right angle, the locus is the arc of the circle withdiameter OP inside C .

11.9. Ratios of corresponding sides in similar triangles yields

jVP j

rDjVP j C r

RDjVP j C r C r 0

r 0;

from which it follows that R D 2rr 0=.r C r 0/.

11.10. From Figure S11.7, we have

jABj2 D .r1 C r2/2 � .r1 � r2/

2 D 4r1r2

and the result follows.

PQ

BA

r1 r

2

Figure S11.7.

11.11. Using the solution to Challenge 11.10, we have jAC j D 2pr1r3,

jCBj D 2pr2r3, and jAC j C jCBj D jABj D 2

pr1r2. Hence

pr2r3 C

pr1r3 D

pr1r2, and division by

pr1r2r3 yields the de-

sired result.

Chapter 1212.1. Using the Hint and the notation in Figure 12.5b, we have AC B C

C CD D T1 C T2 D T .

12.2. The right angles at Q;R, and S enable us to draw circles throughP , R, B , Q and through P , R, S , C as shown in Figure S12.1.Hence †PRS C †PCS D 180ı. But †PCS D 180ı � †PBA D

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A

B

C

P

Q

R

S

Figure S12.1.

†PBQ D †PRQ, and thus †PRS C †PRQ D 180ı. Hence Q,R, and S lie on the same line.

12.3. View the intersecting circles as a graph with V vertices (points ofintersection of the circles), E edges (arcs on the circles) and F faces(regions in the plane), and use Euler’s formula: V � E C F D 2.Each circle intersects the others, so it will have 2.n� 1/ vertices and2.n� 1/ edges. Hence E D 2n.n� 1/. Since each vertex belongs totwo circles, V D n.n�1/, and thus F D 2C2n.n�1/�n.n�1/ Dn2 � nC 2.

12.4. Two. One is AB . The other is determined by PAQ where the circleC3 is symmetric to C2 with respect to the tangent line to C2 at A. SeeFigure S12.2.

A

BC1 C2

C3

P

Q

Figure S12.2.

12.5. Let K denote the area of the shaded region in Figure 12.15. ThenK D 2L C 4M � 1 where L is the area of the lens-shaped regioninside two quarter circles whose centers are at the endpoints of oneof the diagonals of the square, and M is the area of one of the fourregions adjacent to a side of the square. Hence L D 2.�=4/ � 1 D

�=2� 1 and sinceM is the difference between the area of the squareand the sum of the areas of two circular sectors (with angle �=6,i.e., area �=12/ and an equilateral triangles with side 1, M D 1 �

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2.�=12/�p3=4 D 1��=6�

p3=4. HenceK D 1C�=3�

p3 �

0.315.

12.6. The solution of the bun puzzle comes from the fact that, with the rel-ative dimensions of the circles as given, the three diameters will forma right-angled triangle, as shown in Figure S12.3a. It follows that thetwo smaller buns are exactly equal in area to the large bun. There-fore, if we give David and Edgar the two halves of the largest bun,they will have their fair shares—one quarter of the confectioneryeach (Figure S12.3b). Then if we place the small bun on the top ofthe remaining one and trace its circumference in the manner shownin Figure S12.3c, Fred’s piece (in gray) will exactly equal Harry’ssmall bun with the addition of the white piece—half the rim of theother. Thus each boy gets an equal share, and only five pieces arenecessary.

(a) (b) (c)

Figure S12.3.

12.7. Join the centers of the circles to form an equilateral triangle. Thenthe area of the shaded region in Figure 12.17 is the area

p3r2 of

the triangle minus the sum �r2=2 of the areas of three sectors, i.e.,.p3 � �=2/r2.

12.8. Yes. See Figure S12.4.

21

3

45

6

7

Figure S12.4.

12.9. Triangle PAB will be isosceles if jPAj D jPBj, jPAj D jABj, orjPBj D jABj. Since �ABC is equilateral, Figure S12.5 shows thatthere are ten positions for the point P [Honsberger, 2004].

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BA

C P

Figure S12.5.

12.10. Since the three angles at F in Figure 12.18 are each 120ı, andthe angles at A0, B 0, and C 0 are each 60ı, the circumcircles of thethree shaded triangles all pass though F . Hence the line segmentjoining the centers of the circumcircles of �ABC 0 and �AB 0C isperpendicular to the common chord AF , and similarly for the chordsBF and CF .

Chapter 1313.1. The result follows from the carpets theorem if the sum of the areas of

quadrilaterals AMCP and BNDQ equals the area of ABCD. SeeFigure S13.1. The area of quadrilateral AMCP equals the sum ofthe areas of triangles ACM and ACP . The area of triangle ACMis one-half the area of triangle ABC and the area of triangle ACPis one-half the area of triangle ACD, so the area of quadrilateralAMCP is one-half of the area of quadrilateral ABCD. A similarresult holds for quadrilateral BNDQ, and hence the sum of the ar-eas of quadrilaterals AMCP and BNDQ equals the area of ABCD[Andreescu and Enescu, 2004].

BA

C

D

M

N

P

Q

Figure S13.1.

13.2. To reach the conclusion we need only show that the sum of the areasof PMQN and PRQS equals the area of ABCD. Label the vertices

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of the three rectangles as shown in Figure S13.2, and draw the diag-onals of PMQN and PRQS (the dashed lines), and let O be theircommon point of intersection. SinceO is the midpoint of PQ, it lieson the vertical axis of symmetry of ABCD, and sinceO lies ofMNit lies on the horizontal axis of symmetry of ABCD, and hence O isthe center of ABCD. It now follows that jDS j D jCN j D jAM j,and thus SM is parallel to AD.

BA

CD

M

N

PQ

R

S

O

Figure S13.2.

The shaded quadrilateral PMQS is the union of two triangles PMSandQSM . The area of PMS is one-half the area of AMSD and thearea of QSM is one-half the area of MBCS , so the area of PMQSis one-half the area of ABCD. Similarly, PMQS is the union of thetwo triangles PQM and PQS , and thus it is equal in area to one-half of PMQN plus one-half of PRQS , and thus the area ofABCDequals the sum of the areas of PMQN and PRQS , as claimed [Kon-hauser et al., 1996].

13.3. In (a) we have a=2 C b=2 �pab, the arithmetic mean-geometric

mean inequality [Kobayashi, 2002]. In (b) we have a2=2 C b2=2 �Œ.aC b/=2�2, which on taking square roots yields the arithmeticmean-root mean square inequality.

13.4. Three sets of four cocircular points are shown as black dots in FigureS13.3: the four corners of the isolated rectangle, the four points onthe small white disk, and the points labeled A, B , C , and D. To seethis draw BD and consider it as the diameter of a circle. Since theangles at A and C are right angles, they lie on the circle of whichBD is the diameter.

There is a fifth set of four cocircular points: A, the unmarkedintersection immediately below A on the right, B , and the unmarkedcorner just above B . The points all lie on the circle whose diameteris the line segment joining B to the point below A.

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A

B

CD

Figure S13.3.

13.5. A covers more than half the area ofB—consider the area of the trian-gle whose vertices are the common corner ofA andB , the right-mostcorner of B , and the left-most corner of A.

Chapter 1414.1. Assume the radius of the monad is 1. In each case we need only show

yin is bisected. (a) Yin is bisected since the area of a small circle is�=4. (b) The area inside the circular cut is �=2, and by symmetry thearea of the part of yin inside the cut is �=4. (c) With the radii given inthe Hint, the part of yin adjacent to the center of the monad has area

�

8�2C��2

8��

8D.2 � �/�

8C.1C �/�

8��

8D�

4

[Trigg, 1960]. (d) Use the procedure shown in Figure 14.3, but withfour regions rather than seven.

14.2. See Figure S14.1 [Duval, 2007].

Figure S14.1.

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14.3. See Figure S14.2 [Larson, 1985].

123

nn+1

...

Figure S14.2.

14.4. Let N D 2n.2kC 1/ where n � 0 and k � 1, and express 2N as theproduct of m D minf2nC1; 2K C 1g and M D maxf2nC1; 2K C 1g.Since .M �mC 1/=2 and .M Cm � 1/=2 are positive integerswhose sum is M , we can represent 2N by the array of balls shownin Figure S14.3:

m

M

M–m+12

M+m–12

Figure S14.3.

Thus N D M�mC12C M�mC3

2C � � � C MCm�1

2[Frenzen, 1997].

14.5. Arrange three copies of 12 C 22 C 32 C � � � C n2 cubes as shown inFigure S14.4a into the object shown in Figure 14.4b. Two copies itform a rectangular box with dimensions n � .n C 1/ � .2n C 1/ asshown in Figure S14.4c, and the formula follows from division by 6.

(a) (b) (c)

Figure S14.4.

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14.6. Use symmetry to obtain (a) �=2, (b) .� ln 2/=8, (c) 1, (d) 0, (e) 1=2,(f) � .

14.7. Draw a line from A internally tangent to the boundary of yin at T ,and draw a line from the end of the diameter opposite A through T tointersect the boundary of yin at S . Then B can be located anywhereon the portion of the boundary between T and S passing through A.See Figure S14.5a.

AAA

BC

(a) (b) (c) (d)

B

C

AB

C

TS

Figure S14.5.

In Figures 14.5b, c, and d we see three cases for the right angle at Band the location of vertex C .

Chapter 1515.1. Let n be trapezoidal, specifically the sum of m consecutive positive

integers beginning with k C 1. Then (with T0 D 0/

n D .k C 1/C .k C 2/C � � � C .k Cm/ D TkCm � Tk

D.k Cm/.k CmC 1/

2�.k/.k C 1/

2Dm.2k CmC 1/

2:

One of m or 2k CmC 1 is odd and the other is even, and hence if nis trapezoidal, it is not a power of 2 [Gamer et al., 1985].

15.2. Both recursions follow immediately from (15.1).

15.3. See Figure S15.1 [Nelsen, 2006].

b

c

an

Figure S15.1.

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15.4. In Section 15.5 we showed that the sum of the squares of the diag-onals from a vertex and the two adjacent sides is 2nR2. Repeatingfor each vertex counts the squares of all the diagonals and all thesides twice. Hence the sum is .1=2/n.2nR2/ D n2R2 [Ouellette andBennett, 1979].

15.5. The two acute angles at the base each measure �=2 � �=n, and eachof the other n�2 obtuse angles measures ��=n. See Figure 15.17b,and observe that at each vertex of the polygonal cycloid the angle inthe shaded triangle measures �=n and each of the two angles in thewhite triangle measures �=2 � �=n.

15.6. Let Lk denote the length of the segment of the polygonal cardioidin the isosceles triangle with equal sides dk in Figure 15.18b. Sincedk D 2R sin.k�=n/, we have

Lk D 4R sink�

nsin

2�

nD 2R

�cos

.k � 2/�

n� cos

.k C 2/�

n

�

and hence the length of the polygonal cardioid is

Xn�1

kD1Lk D 2R

Xn�1

kD1

�cos

.k � 2/�

n� cos

.k C 2/�

n

�

D 4R

�1C 2 cos

�

nC cos

2�

n

�D 8R cos2

�

nC 8r

(r D R cos.�=n//.

15.7. If the angle of inclination of the chord is �0, then the length of thechord is

2a.1C cos �0/C 2a.1C cos.�0 C �// D 4a:

Chapter 1616.1. See Figure S16.1, and apply the law of sines to the gray tri-

angle to obtain .a2 C b3/=.c3 C d1/ D sinD=sinA. Similarly,we have .b2 C c3/=.d3 C e1/ D sinE=sinB , .c2 C d3/=.e3 Ca1/ D sinA=sinC , .d2 C e3/=.a3 C b1/ D sinB=sinD, and.e2 C a3/=.b3Cc1/ D sinC=sinE. Multiplying the equations yields(16.2) [Lee, 1998].

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a1

a2 b

1

b2

c1

c2d1

d2

e1

e2

A

B

CD

E

a3b

3

c3d3

e3

Figure S16.1.

16.2. Since area(CDE/ D area.DEA), the line segment AC is paral-lel to DE. Similarly, each diagonal of the pentagon is parallel toa side. Thus ABCF (see Figure S16.2a) is a parallelogram, andarea(ACF / D area.ABC )D 1.

F

A

B

C

DE

CA

DE

F1

qp

q

(a) (b)

Figure S16.2.

Let p D area.DEF ) and q D area(CDF ) D area(AEF ), with p Cq D 1. Since triangles ACF and AEF have the same base AF andtriangles CDF and DEF have the same base DF , 1=q D q=p, orp D q2 (see Figure S16.2b). Thus q2 C q D 1 so that q D � � 1,p D 2� �, and the area of the pentagon is 1C 1C pC 2q D �C 2[Konhauser et al., 1996].

16.3. (a) Erase some of the lines in Figure 16.23b and draw some others tocreate the grid in Figure S16.3a. Hence the triangle is a right trianglewhose legs are in the ratio 3:4, and the result follows.

(a) (b) (c)

Figure S16.3.

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(b) We need only show that the dark gray portion of the shaded righttriangle in Figure S16.3b has 1=6 the area of the entire shaded righttriangle, and that is done in Figure S16.3c.

16.4. Yes, see Figure S16.4.

4 10 9 1

5

612

23

8

Figure S16.4.

16.5. We need to show that aC b C c D x C y C z (See Figure S16.5).

a

z p q y

b t s c

u r

x

Figure S16.5.

LetN denote the magic constant. Computing the six line sums yields

aC p C uC b DN D x C r C s C y

b C t C s C c DN D y C p C q C z

c C r C q C a DN D z C uC t C x

from which the result follows by adding the three equations and can-celing common terms (p, q, r , s, t , u) on each side. Similarly, in anymagic octagram the sum of the four numbers at the corners of eachlarge square must be the same.


The figure illustrates r.9; 3/ D 10, and 9 is the smallest value of nfor which r.n; 3/ > n.

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Figure S16.6.

16.7. Yes. See Figure S16.7, where � represents a standing tree and a harvested tree.

(a) (b)

Figure S16.7.

16.8. In Figure S16.8 AB is a side of the decagon and FD a side of thef10=3g star. Draw diametersBE andGC through the centerO . Sincethe decagon is regular, the chords AB , GC , and FD are parallel, asare the chordsAF ,BE, andCD. Hence the two shaded quadrilateralsare parallelograms, so that

jFDj D jHC j D jHOj C jOC j D jABj C jOC j

as claimed [Honsberger, 2001].

BA

C

D

E

F

G OH

Figure S16.8.


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(a) (b)

Figure S16.9.

16.10. Let the dimensions of the rectangle be x-by-1 as indicated in FigureS16.10. Since the entire rectangle is similar to the portion shadedgray, we have x=1 D 1=.x=2/ and hence x D

p2.

1

x

Figure S16.10.

Chapter 1717.1. 1

8C 1

16C 1

32C � � � D 1

4and 2

9C 2

27C 2

81C � � � D 1

3.

17.2. See Figure S17.1 for the first iteration, placing 25 non-attackingqueens on a 25-by-25 chessboard [Clark and Shisha, 1988].

Figure S17.1.

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17.3. Let the original triangle have edge length 1, and letAn andLn denotethe area and length of the boundary after n iterations. ThenAn D.3=4/An�1 and Ln D .3=2/Ln�1 with A0D

p3=4 and L0 D 3, so

that An D .3=4/n � .p3=4/, Ln D 3.3=2/n, and the result follows.

17.4. Let An and Pn denote the area and the perimeter of the holes inthe Sierpinski carpet after n iterations. It is easy to show that An D.8=9/n and Pn D .4=5/Œ.8=3/n � 1�, so the result follows.

17.5. See Figure S17.2, where parts a, b, c, d illustrate n D 2; 3; 4; 5; re-spectively. In Figure S17.2a, the right triangle is isosceles, in S17.2bthe acute angles are 30ı and 60ı, in S17.2c the right triangle is arbi-trary, and in S17.2d the legs are 1 and 2.

(a) (b)

(c) (d)

Figure S17.2.

17.6. Since ��1 D 1=�, 1�1=� D 1=�2, and so on, the areas of the whitesquares in Figure 17.22c are 1, 1=�2, 1=�4, 1=�6, etc., which sum tothe area � of the original rectangle.

Chapter 1818.1. 2

9C 2

81C 2

729C � � � D 1

4.

18.2. (a) Let a D Fn and b D FnC1 so that a C b D FnC2 and b � a DFn�1.

(b) The identity in (a) is equivalent to F 2nC1 � FnFnC2 D

Fn�1FnC1 � F2n . Thus the terms of the sequence fFn�1FnC1 �

F 2n g1nD2 have the same magnitude and alternate in sign. So we

need only evaluate the base case n D 2: F1F3 � F 22 D 1 soFn�1FnC1 � F

2n is C1 when n is even and �1 when n is odd, i.e.,

Fn�1FnC1 � F2n D .�1/

n.

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18.3. Figure 18.12 suffices, with ˛ replaced by its complement in the lightgray tatamis [Webber and Bode, 2002].


Figure S18.1.

18.5. In Figure S18.2a we see that 1 � 4pq, so that .1=p/ C .1=q/ D

1=pq � 4. In Figure S18.2b we see that

2

�pC

1

p

�2C2

�qC

1

q

�2�

�pC

1

pC q C

1

q

�2� .1C4/2 D 25:

p q

1

p

q

1

p +1 p q +

1 q

p + 1 p

q + 1 q

(a) (b)

Figure S18.2.

18.6. Using the hint, we have

T8Tn D8Tn.8Tn C 1/

2D 4Tn.2nC 1/

2;

and so if Tn is square, so is T8Tn . Since T1 D 1 is square, this relationgenerates an infinite sequence of square triangular numbers.


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Figure S18.3.

18.8. See Figure S18.4, where in each case the side of the large square isFnC1 D Fn C Fn�1, and in Figures S18.4bc the side of the smallcentral square is Fn�2.

(a) (b) (c)

Figure S18.4.

Chapter 1919.1. If A D .a; 1=a/ and B D .b; 1=b/, then M D ..aC b/=2;

..1=a/C .1=b//=2/ and if P D .p; 1=p/, then

1=p

pD..1=a/C .1=b//=2

.aC b/=2D

1

ab:

Hence p Dpab and 1=p D

p1=ab as required [Burn, 2000].

19.2. The proof follows immediately from the observation that the twoshaded triangles in Figure S19.1 have the same area.

A

B

O

A

B

O

=

Figure S19.1.

19.3. (a) For x > 0, (19.2) implies 1=.1C x/ < Œln.1C x/�=x < 1 and forx in (�1,0), (19.2) implies 1 < Œln.1C x/�=x < 1=.1C x/. (b) Use(19.3) with fa; bg D f1; 1C xg.

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19.4. Replacing x by x1=n � 1 in part (a) of Challenge 19.3 yields 1 �x�1=n < .1=n/ ln x < x1=n�1, or n.1�x�1=n/ < ln x < n.x1=n�1/,which is equivalent to ln x < n.x1=n � 1/ < x1=n ln x. Take the limitas n!1.

19.5. Set fa; bg D f1 � x; 1C xg in (19.3).

19.6. Replace a bypa and b by

pb in (19.3) to obtain

4pab �

2.pb �pa/

ln b � ln a�

paCpb

2

and multiply by .paCpb/=2 to obtain the two inner inequalities in

(19.4). The outer inequalities are equivalent to the AM-GM inequal-ity [Carlson, 1972].

Chapter 2020.1. Yes. See Figure S20.1 for examples from the Real Alcazar in Seville.

Figure S20.1.


Q:

P:

P = 2Q

Figure S20.2.

20.3. This is BhNaskara’s proof. See Section18.1.

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20.4. Using the Hint, we see that the area of the smaller square is 4 and thearea of the larger square is 10, hence the smaller square has 2=5 thearea of the larger. See Figure S20.3.

Figure S20.3.

20.5. Using the Hint (see Figure S20.4), we need only find the area ofthe central triangle, since all four triangles in the Vecten configura-tion have the same area. Enclosing the central triangle in the 20-acredashed rectangle yields its area: 20 � .5=2 C 9=2 C 4/ D 9 acres.Hence the area of the estate is 26C 20C 18C 4 � 9 D 100 acres.

Figure S20.4.

20.6. In Figure S20.5 we see that the sum jajjxj C jbjjyj of the area of thetwo rectangles is the same as the area of a parallelogram with edgelengths

pa2 C b2 and

px2 C y2.

|a|

|b|

|x|

|y|

+ =

a +b2 2x +y2 2

Figure S20.5.

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The area of a parallelogram with given edges lengths is less that thearea of a rectangle with the same edge lengths, and thus

jax C byj � jajjxj C jbjjyj �pa2 C b2

px2 C y2:

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Index

Aczel’s inequality 88Aliquot triangle 72Altitudes of a triangle 66–68Angle 103–115Angle-bisectors 68–70Angle-bisector theorem 68, 69Angle measurement 104–106Angles intersecting circles 107–109Anticenter 118, 120Annulus 131Apollonius’s theorem 65, 127Arbelos 34,35Arc length 189Arcs 103–115, 189Archimedean spiral 230Archimedes’ propositions 34–37, 41,

51Arcs and angles 103–115Arctangent function 23, 24, 26, 42Area of a cardioid 196,197

of a cycloid 193–195of a quadrilateral 255of a regular polygon 39of Napoleon’s triangle 95–98

Arithmetic mean 17, 18, 20, 23, 37,72, 78, 79, 168, 171, 238

Arithmetic mean-geometric meaninequality 17,18, 72, 78, 79,168, 171, 238

Arithmetic mean-root mean squareinequality 20, 23, 78, 79, 168

Arithmetic progression 74, 176, 177

Ballantine rings 157Barbier’s theorem 156

Bells 244Bertrand’s paradox 144BhNaskara’s proof 234Bicentric quadrilateral 118, 125,

126Bimedians 120Blaschke-Lebesgue theorem 156Borromean rings 157, 158Bradwardine’s formula 203, 204Bramagupta’s formula 121, 126Bride’s chair 1–14Bull’s eye illusion 143Butterfly theorem 118, 127, 128

Cardioid 184, 196, 197Cardioid of Boscovich 43Carnot’s theorem 122, 124Carpets theorem 163–165Cassini’s identity 27, 240Catalan numbers 192Cauchy-Schwarz inequality 17, 18,

87, 260Center of gravity 61, 65, 66, 71, 213Central angle 107Centroid 61, 65, 66, 71, 213Ceva’s theorem 62–73Cevians 61–75Chebyshev’s inequality 169Chinese checkers 210Chord 135, 136, 199Circles 117–130, 131–147Circles and polygons 117–130Circular segment 142Circumcenter 70, 71, 80Circumcircle 70, 71, 79–84, 154, 155

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322 Index

Circumradius 70, 71, 79, 80, 194,195, 212

Cocircular points 171, 172Combinatorial yin and yang 176, 177Common chords 135, 136Compass rose 213, 214Complex quadrilateral 127, 128, 130Concentric circles 131, 143, 144Conics 133–135Contraharmonic mean 78, 79Convex polygons 189–192Cordoba proportion 212Cosine 19, 25, 27, 38, 39, 86, 87, 90,

234Crescent 131, 141Cube 40, 41Curve of constant width 156Cyclic quadrilateral 107, 110,

118–121, 125–127, 155Cyclogon 193–195Cycloid 184, 192, 193

Degree 104–106Diamond 55Dido’s semicircle 32, 33, 34

problem 32, 33theorem 33

Dodecahedron 40, 41, 207, 208Dominoes 54Double angle formulas 38, 39Dudeney’s puzzles 141, 159, 160,

174–176, 241, 259

Ellipse 133, 134, 138, 144Equiangular spiral 228, 229Equilateral arch 138Equilateral hyperbola 244Escher’s theorem 91, 98, 99Euclid’s construction of the five

Platonic solids 40, 41proof 2–4

Euler’s arctangent identities 24inequality 112line 71triangle inequality 72triangle theorem 111, 112

Excenter 80Excircle 79–81Exradii 80, 81Exterior angle 108, 109Eyeball theorem 132, 133

Fagnano’s problem 67Fermat point 61, 93, 94Fermat’s problem 93, 94Fibonacci identities 236, 240, 242Fibonacci numbers 26, 27, 235, 236,

240, 242Fibonacci spiral 229Figurate numbers 186–188Finsler-Hadwiger theorem 5, 13Flank triangles 4–7Fuss’s theorem 118, 126, 127

Game of Daisy 180, 181Garfield’s proof 22

trapezoid 21–28Gelosia multiplication 254Generalized tatami 239, 240Geodesic domes 92Geometric mean 17,18, 31, 37,72,

78, 79, 168,171, 238, 239progression 48, 222, 223, 225,226

series 222–224Geometric–harmonic mean

inequality 37, 78, 79, 168, 238,239

Gergonne point 70Golden bee 56

earring 145ellipse 144

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Index 323

ratio 41, 56, 138, 139, 144, 145,206, 207, 229, 232

rectangle 41, 229, 232spiral 229

Grad 106Gradian 106Great monad 174–176Grebe’s point 8Grebe’s theorem 8Growing figures iteratively 224–226Growth spiral 228

Hadwiger-Finsler inequality 98Harmonic mean 37, 78, 79, 238,

239Haruki’s theorem 151Helen of geometers 193Heron’s formula 67, 69, 81Hexagonal numbers 187Hexagram 202, 216Hippocrates’ lunes 42, 43Homothetic figures 8, 45–60Homothetic functions 56–58Hyperbola 134, 135, 243, 247–249Hyperbolic

cosine 249, 250sine 249, 250

Icon ix, xiIcosahedron 40, 41Incenter 61, 69, 80, 81Incircle 69, 79–84Inequalities 17, 18, 22, 23, 28, 78,

79, 87, 88, 167, 168, 241,246–249, 252, 260

Inequalities by paper folding 171Inradius 69, 80, 81, 121, 126Inscribed angle 107

triangle 67Integration results 178, 179, 182Interior angle 108

Irrationality ofp2 163, 165

ofp3 165, 166

Iterative figures 221–232

Japanese theorem 118, 121, 122, 124Johnson’s theorem 152

Kiepert’s theorem 95

Lacing shoes 89, 90Lattice multiplication 254Law of cosines 7, 8Law of sines 67, 71Lemoine point 8, 61Lens 131, 137, 142Length of a cycloid 195Linkages 184, 185Logarithm 247–249Logarithmic mean 248

spiral 228Loyd’s puzzles 9, 10, 11, 180, 219,

240, 241Lune 42, 43, 131, 139, 140

Magic constant 214Magic heptagram 215

hexagram 215octagram 215pentagram 214stars 214–216

Mandelbrot cardioid 197, 198Mathematical icons ix, xiMedial triangle 74Median triangle 66Medians 65, 66, 74Menelaus’s theorem 52, 53Menger sponge 230Miquel’s theorem 154, 155Monge circle of an ellipse 113,114Monge’s theorem 113, 114, 152,

153

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324 Index

Moniamond 44Mrs. Miniver’s problem 141, 142

Napier’s bones 254inequality 247rods 254

Napoleon triangle 91–101, 160, 161Napoleon’s theorem 91–93Natural logarithm 247, 248Non-concurrent cevians 72, 73Number e 248

� 117

Octagonal numbers 187Octagram 202, 211, 214Octahedron 40, 41Octant 105, 106Orchard planting problem 216Orthic triangle 67, 68Orthocenter 61, 66–68, 71, 74Overlapping figures 163–172

Pantograph 52Paper folding 227, 228Pappus result 37Parabola 134Parahexagon 93Peaucellier-Lipkin linkage 184, 185Pentagonal numbers 186, 187Pentagram 202, 205–208Pentiamond 55Perfect tiling 55, 56Perimeter of a regular polygon 39Piecewise circular curves 175Platonic solids 40, 41Point of homothety 8Polite number 198Polygonal cardioid 196, 197

cycloids 192–195lines 183–199numbers 184, 186–188

Polygons with circles 117–130Polyiamonds 55Polygram 202Polyominoes 54, 55Power of a point 109–111, 127, 153Primitive Pythagorean triples 85Ptolemy’s inequality 119Ptolemy’s theorem 118, 119Putnam Mathematical Competition

1980 Problem A3, 1781987 Problem B1, 179

Pythagoras tree 222Pythagorean theorem 2–4, 9, 15–17,

22, 47, 234, 257, 258tiling 257–259triples 85, 86, 163, 166, 167

Quadrant 105, 106Quadratic mean 19Quadrilaterals 117–130

Radian 106Radical axis 113, 150, 151Reciprocal Pythagorean theorem 47,

48Reciprocals of triangular numbers

250Recreational mathematics 9, 10, 11,

141, 159, 160, 174–176,214–216, 219, 240, 241,259

Rectangular hyperbola 243–252Reduction compass 50, 51Regular polygons 184, 193–195

polyhedra 40, 41Reptiles 53–55Reuleaux polygons 155–157triangle 155Right triangle 77–90

altitude theorem 31cevians 84, 85

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Index 325

Roof trusses 62Root mean square 20, 23, 78, 79,

168

Salinon 34, 36, 37Sam Loyd’s puzzles 9, 10, 11, 180,

219, 240, 241Sangaku 13,121Self-homothetic function 57Self-similar figures 221–232Semicircle 29–43Semicircles of Archimedes 34–37Semi-inscribed angle 108Sextant 105, 106Shoe lacings 89, 90Sierpinski carpet 230, 231

triangle 231, 232Silver ratio 212

rectangle 212Similar figures 45–60Sine 19, 25, 27, 38, 39, 86, 87, 90,

142, 239Small stellated dodecahedron 207,

208Solomon’s seal 202Sphinx hexiamond 55Spira mirabilis 228–230Square numbers 186Squared figure 139, 140Star numbers 210

of David 202, 208–211of Lakshmi 202, 211, 212polygons 201–219

Stellated polygon 202Stellation 202Stewart’s theorem 64Straight line 184–186Subadditive function 79Sum of arithmetic progression 176,

177of cubes 169, 170, 238

of geometric progression 48, 222,223, 225, 226

of odd numbers 181, 238of squares 182of triangular numbers 226

Symmetry 178

Tangent 25, 26, 27, 38, 39, 90Tangential quadrilateral 118, 125,

126Tatami 233–242

inequalities 238, 239mats 237, 238

Taylor circle 112, 113Tessellation 253Tetrahedron 40, 41Tetriamonds 55Tetrominoes 54Thales’ proportionality theorem 46,

47triangle theorem 30, 107

Three-circle theorems 150–153Tiles 92, 93Tiling 93, 253–260

a rectangle 256, 257Torricelli’s configuration 94Trapezoidal rule 188, 189

numbers 198Tree planting problem 216, 218Triamond 55Triangles and intersecting circles

153–155Triangular numbers 176, 181, 187,

188, 198, 199, 226, 237, 250Triangulations of a convex polygon

121–124, 190–192Trigonometric formulas 18–20,

23–27identities 23–25, 38, 39, 42, 86,87, 90

inequality 87

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326 Index

Triple angle formulas 36, 86–87Trisection of an angle 35Trominoes 54Two-circle theorems 131–147

Uncial calligraphy 106Unicursal hexagram 202, 210, 211

van Lamoen’s theorem 9Varignon parallelogram 213Vecten configuration 4–7, 9–12, 259

point 7, 61, 95

Venn diagrams 149–161Vesica piscis 131, 137–139

Weierstrass substitution 27Weitzenbock’s inequality 91, 96–98Width 156Windmill 1Wittenbauer parallelogram 213

Yin and yang 173–182

Zhou bi suan jing 15–20

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About the Authors

Claudi Alsina was born on 30 January 1952 in Barcelona, Spain. He re-ceived his BA and PhD in mathematics from the University of Barcelona.His post-doctoral studies were at the University of Massachusetts, Amherst,Claudi, Professor of Mathematics at the Technical University of Cataloniahas developed a wide range of international activities, research papers, pub-lications and hundreds of lectures on mathematics and mathematics educa-tion. His latest books include Associative Functions: Triangular Norms andCopulas (with M.J. Frank and B. Schweizer) WSP, 2006; Math Made Vi-sual. Creating Images for Understanding Mathematics (with Roger Nelsen)MAA, 2006; Vitaminas Matematicas and El Club de la Hipotenusa,Ariel, 2008, Geometria para Turistas, Ariel, 2009, When Less Is More:Visualizing Basic Inequalities (with Roger Nelsen) MAA, 2009; AsesinatosMatematicos, Ariel, 2010 and Charming Proofs: A Journey Into ElegantMathematics (with Roger Nelsen) MAA, 2010.

Roger B. Nelsen was born on 20 December 1942 in Chicago, Illinois. Hereceived his B.A. in mathematics from DePauw University in 1964 and hisPh.D. in mathematics from Duke University in 1969. Roger was electedto Phi Beta Kappa and Sigma Xi, and taught mathematics and statistics atLewis & Clark College for forty years before his retirement in 2009. Hisprevious books include Proofs Without Words: Exercises in Visual Thinking,MAA 1993; An Introduction to Copulas, Springer, 1999 (2nd. ed. 2006);Proofs Without Words II: More Exercises in Visual Thinking, MAA, 2000;Math Made Visual: Creating Images for Understanding Mathematics (withClaudi Alsina), MAA, 2006; When Less Is More: Visualizing Basic Inequal-ities (with Claudi Alsina), MAA, 2009; Charming Proofs: A Journey IntoElegant Mathematics (with Claudi Alsina), MAA, 2010; and The CalculusCollection: A Resource for AP and Beyond (with Caren Diefenderfer, edi-tors), MAA, 2010.

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AMS / MAA DOLCIANI MATHEMATICAL EXPOSITIONS

The authors present twenty icons of mathematics, that is, geometrical shapes such as the right triangle, the Venn diagram, and the yang and yin symbol and explore math-ematical results associated with them. As with their previous books (Charming Proofs, When Less is More, Math Made Visual) proofs are visual whenever possible. The results require no more than high-school mathematics to appreciate and many of them will be new even to experienced readers. Besides theo-rems and proofs, the book contains many illustrations and it gives connections of the icons to the world outside of math-ematics. There are also problems at the end of each chapter, with solutions provided in an appendix. The book could be used by students in courses in problem solving, mathematical reasoning, or mathematics for the liberal arts. It could also be read with pleasure by professional mathematicians, as it was by the members of the Dolciani editorial board, who unani-mously recommend its publication.

Icons of MathematicsAN EXPLORATION OF TWENTY KEY IMAGES

Claudi Alsina and Roger B. Nelsen

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