solid geometry in the 21st century - ams · 2020. 7. 29. · new horizons in geometry,tom m....

AMS / MAA DOLCIANI MATHEMATICAL EXPOSITIONS VOL 50

A MATHEMATICAL SPACE ODYSSEY

Solid Geometry in the 21st CenturyClaudi Alsina and Roger B. Nelsen

10.1090/dol/050

A Mathematical SpaceOdyssey

Solid Geometry in the21st Century

About the coverJeffrey Stewart Ely created Bucky Madness for the Mathematical ArtExhibition at the 2011 Joint Mathematics Meetings in New Orleans. Jeff, anassociate professor of computer science at Lewis & Clark College, describesthe work: “This is my response to a request to make a ball and stick modelof the buckyball carbon molecule. After deciding that a strict interpretationof the molecule lacked artistic flair, I proceeded to use it as a theme. Here,the overall structure is a 60-node truncated icosahedron (buckyball), but eachnode is itself a buckyball. The center sphere reflects this model in its surfaceand also recursively reflects the whole against a mirror that is behind theobserver.” See Challenge 9.7 on page 190.

c� 2015 byThe Mathematical Association of America (Incorporated)

Library of Congress Catalog Card Number 2015936095

Print Edition ISBN 978-0-88385-358-0

Electronic Edition ISBN 978-1-61444-216-5

Printed in the United States of America

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

The Dolciani Mathematical Expositions

NUMBER FIFTY

A Mathematical SpaceOdyssey

Solid Geometry in the21st Century

Claudi AlsinaUniversitat Politecnica de Catalunya

Roger B. NelsenLewis & Clark College

Published and Distributed by

The Mathematical Association of America

DOLCIANI MATHEMATICAL EXPOSITIONS

Council on Publications and Communications

Jennifer J. Quinn, Chair

Committee on Books

Fernando Gouvea, Chair

Dolciani Mathematical Expositions Editorial Board

Harriet S. Pollatsek, Editor

Elizabeth Denne

Ricardo L. Diaz

Emily H. Moore

Michael J. Mossinghoff

Margaret M. Robinson

Ayse A. Sahin

Dan E. Steffy

Robert W. Vallin

Joseph F. Wagner

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, RobertM. Young

14. The Wohascum County Problem Book, George T. Gilbert, Mark Krusemeyer, andLoren 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

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

18. Which Way Did the Bicycle Go? . . . and Other Intriguing Mathematical Myster-ies, 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 and RogerB. Nelsen

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 B. Nelsen

46. A Guide to Plane Algebraic Curves, (MAA Guides #7), Keith Kendig

47. New Horizons in Geometry, Tom M. Apostol and Mamikon A. Mnatsakanian

48. A Guide to Groups, Rings, and Fields, (MAA Guides #8), Fernando Q. Gouvea

49. A Guide to Functional Analysis, (MAA Guides #9), Steven G. Krantz

50. A Mathematical Space Odyssey: Solid Geometry in the 21st Century, ClaudiAlsina and Roger B. Nelsen

MAA Service CenterP.O. Box 91112

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

Throw your dreams into space like a kite, and you do notknow what it will bring back, a new life, a new friend, a newlove, a new country.

Anaıs Nin

Space is the breath of art.Frank Lloyd Wright

It is often helpful to think of the four coordinates of an eventas specifying its position in a four-dimensional space calledspace-time. It is impossible to imagine a four-dimensionalspace. I personally find it hard enough to visualize three-dimensional space!

Stephen W. HawkingA Brief History of Time (1989)

Preface

Cubes, cones, spheres, cylinders or pyramids are the greatprimary forms which light reveals to advantage; the imageof these is distinct and tangible within us without ambiguity.It is for this reason that these are beautiful forms, the mostbeautiful forms.

Le Corbusier (Charles-Edouard Jeanneret-Gris)

Solid geometry is the traditional name for what we call today the geometry ofthree-dimensional Euclidean space. Courses in solid geometry have largelydisappeared from American high schools and colleges. But we are convincedthat a mathematical exploration of three-dimensional geometry merits someattention in today’s curriculum. This book is devoted to presenting tech-niques for proving a variety of mathematical results in three-dimensionalspace, techniques that may improve one’s ability to think visually. The re-sults and methods employ some of the traditional icons of solid geometry,namely: prisms, pyramids, platonic solids, cylinders, cones, and spheres, andwe devote one chapter to each of the following techniques: enumeration, rep-resentation, dissection, plane sections, intersection, iteration, motion, projec-tion, and folding and unfolding.

Augustus De Morgan (1806–1871) once wrote:

Considerable obstacles generally present themselves to the beginner, instudying the elements of Solid Geometry, from the practice . . . of neversubmitting to the eye of the student, the figures on whose properties heis reasoning.

An analogous claim was later expressed by David Hilbert (1862–1943):

The tendency toward intuitive understanding fosters a more immediategrasp of the objects one studies, a live rapport with them, so to speak,which stresses the concrete meaning of their relations.

These quotes from De Morgan and Hilbert, expressing the desire for avisual and intuitive approach to the geometry of space, have guided our se-lection of topics for this book. A Mathematical Space Odyssey is organized

ix

x Preface

as follows. After this preface we present in the first chapter a number ofexamples of some of the types of questions and problems we address inthe subsequent chapters. We also introduce (or re-introduce) the reader tothe primary objects that inhabit space. We then develop, in nine chapters,the methods in the chapter titles that are fruitful for exploring and provingmany basic facts and theorems using three-dimensional figures. By meansof these methods we review properties concerning figurate numbers, meansand inequalities, volumes and surface areas, and a variety of the celebratedresults in classical solid geometry. In addition to many figures illustratingtheorems and their proofs, we have included a selection of photographs ofthree-dimensional works of art and architecture.

Readers should be familiar with high school algebra, plane and analyticgeometry, and trigonometry. While brief appearances of calculus occur inChapters 5, 9, and 10, no knowledge of calculus is necessary to enjoy thisbook.

Each chapter concludes with a selection of Challenges for the reader to ex-plore further properties and applications of each method. After the chapterswe give hints and solutions to all the Challenges in the book. A MathematicalSpace Odyssey concludes with references and a complete index.

As in our previous books with the MAA, we hope that both secondaryschool and college and university teachers may wish to use the book fora 21st century journey into solid geometry with their students. The bookmay also be used as a supplement for problem-solving sessions or classroomdiscussions, augmented by hands-on materials and modern software.

Special thanks to Harriet Pollatsek and the members of the editorial boardof the Dolciani series for their careful reading of earlier drafts of the bookand their many helpful suggestions. We would also like to thank StephenKennedy, Carol Baxter, Beverly Ruedi, and Samantha Webb of the MAA’sbook publication staff for their expertise in preparing this book for publi-cation. Finally, special thanks to Don Albers, the MAA’s former editorialdirector for books, who as on previous occasions encouraged us to pursuethis project.

Claudi AlsinaUniversitat Politecnica de Catalunya

Barcelona, Spain

Roger B. NelsenLewis & Clark College

Portland, Oregon

Contents

Preface ix

1 Introduction 11.1 Ten examples . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Inhabitants of space . . . . . . . . . . . . . . . . . . . . . . 91.3 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2 Enumeration 272.1 Hex numbers . . . . . . . . . . . . . . . . . . . . . . . . . 272.2 Counting calissons . . . . . . . . . . . . . . . . . . . . . . 282.3 Using cubes to sum integers . . . . . . . . . . . . . . . . . 292.4 Counting cannonballs . . . . . . . . . . . . . . . . . . . . . 352.5 Partitioning space with planes . . . . . . . . . . . . . . . . 372.6 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3 Representation 453.1 Numeric cubes as geometric cubes . . . . . . . . . . . . . 453.2 The inclusion principle and the AM-GM inequality for

three numbers . . . . . . . . . . . . . . . . . . . . . . . . . 493.3 Applications to optimization problems . . . . . . . . . . . . 523.4 Inequalities for rectangular boxes . . . . . . . . . . . . . . 553.5 Means for three numbers . . . . . . . . . . . . . . . . . . . 593.6 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4 Dissection 654.1 Parallelepipeds, prisms, and pyramids . . . . . . . . . . . . 654.2 The regular tetrahedron and octahedron . . . . . . . . . . . 674.3 The regular dodecahedron . . . . . . . . . . . . . . . . . . 714.4 The frustum of a pyramid . . . . . . . . . . . . . . . . . . . 724.5 The rhombic dodecahedron . . . . . . . . . . . . . . . . . . 744.6 The isosceles tetrahedron . . . . . . . . . . . . . . . . . . . 764.7 The Hadwiger problem . . . . . . . . . . . . . . . . . . . . 774.8 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 79

xi

xii Contents

5 Plane sections 835.1 The hexagonal section of a cube . . . . . . . . . . . . . . . 835.2 Prismatoids and the prismoidal formula . . . . . . . . . . . 855.3 Cavalieri’s principle and its consequences . . . . . . . . . . 895.4 The right tetrahedron and de Gua’s theorem . . . . . . . . . 935.5 Inequalities for isosceles tetrahedra . . . . . . . . . . . . . 965.6 Commandino’s theorem . . . . . . . . . . . . . . . . . . . 975.7 Conic sections . . . . . . . . . . . . . . . . . . . . . . . . 995.8 Inscribing the Platonic solids in a sphere . . . . . . . . . . . 1045.9 The radius of a sphere . . . . . . . . . . . . . . . . . . . . 1075.10 The parallelepiped law . . . . . . . . . . . . . . . . . . . . 1085.11 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 110

6 Intersection 1176.1 Skew lines . . . . . . . . . . . . . . . . . . . . . . . . . . 1186.2 Concurrent lines in the plane . . . . . . . . . . . . . . . . . 1196.3 Three intersecting cylinders . . . . . . . . . . . . . . . . . 1206.4 The area of a spherical triangle . . . . . . . . . . . . . . . . 1216.5 The angles of a tetrahedron . . . . . . . . . . . . . . . . . . 1246.6 The circumsphere of a tetrahedron . . . . . . . . . . . . . . 1266.7 The radius of a sphere, revisited . . . . . . . . . . . . . . . 1276.8 The sphere as a locus of points . . . . . . . . . . . . . . . . 1296.9 Prince Rupert’s cube . . . . . . . . . . . . . . . . . . . . . 1306.10 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 131

7 Iteration 1337.1 Is there a four color theorem in space? . . . . . . . . . . . . 1337.2 Squaring squares and cubing cubes . . . . . . . . . . . . . 1347.3 The Menger sponge and Platonic fractals . . . . . . . . . . 1367.4 Self-similarity and iteration . . . . . . . . . . . . . . . . . 1397.5 The Schwarz lantern and the cylinder area paradox . . . . . 1407.6 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 143

8 Motion 1478.1 A million points in space . . . . . . . . . . . . . . . . . . . 1488.2 Viviani’s theorem for a regular tetrahedron . . . . . . . . . 1498.3 Dissecting a cube into identical smaller cubes . . . . . . . . 1528.4 Fair division of a cake . . . . . . . . . . . . . . . . . . . . 1538.5 From the golden ratio to the plastic number . . . . . . . . . 1538.6 Hinged dissections and rotations . . . . . . . . . . . . . . . 154

Contents xiii

8.7 Euler’s rotation theorem . . . . . . . . . . . . . . . . . . . 1568.8 The conic sections, revisited . . . . . . . . . . . . . . . . . 1578.9 Instant Insanity . . . . . . . . . . . . . . . . . . . . . . . . 1588.10 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 161

9 Projection 1659.1 Classical projections and their applications . . . . . . . . . 1659.2 Mapping the earth . . . . . . . . . . . . . . . . . . . . . . 1699.3 Euler’s polyhedral formula . . . . . . . . . . . . . . . . . . 1779.4 Pythagoras and the sphere . . . . . . . . . . . . . . . . . . 1789.5 Pythagoras and parallelograms in space . . . . . . . . . . . 1809.6 The Loomis-Whitney inequality . . . . . . . . . . . . . . . 1829.7 An upper bound for the volume of a tetrahedron . . . . . . . 1849.8 Projections in reverse . . . . . . . . . . . . . . . . . . . . . 1859.9 Hamiltonian cycles in polyhedra . . . . . . . . . . . . . . . 1879.10 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 189

10 Folding and Unfolding 19310.1 Polyhedral nets . . . . . . . . . . . . . . . . . . . . . . . . 19410.2 Deltahedra . . . . . . . . . . . . . . . . . . . . . . . . . . 19610.3 Folding a regular pentagon . . . . . . . . . . . . . . . . . 20010.4 The Delian problem: duplicating the cube . . . . . . . . . . 20110.5 Surface areas of cylinders, cones, and spheres . . . . . . . 20310.6 Helices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20910.7 Surface areas of the bicylinder and tricylinder . . . . . . . 21110.8 Folding strange and exotic polyhedra . . . . . . . . . . . . 21410.9 The spider and the fly . . . . . . . . . . . . . . . . . . . . 21710.10 The vertex angles of a tetrahedron . . . . . . . . . . . . . 21910.11 Folding paper in half twelve times . . . . . . . . . . . . . . 21910.12 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . 222

Solutions to the Challenges 227Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 227Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 230Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 234Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 237Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 239Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . 243Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . 245Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . 246

xiv Contents

Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . 249Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . 253

References 259

Index 265

About the Authors 272

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) four is the number of vertices and of faces in the tetrahedron.

(b) six is the number of vertices in the octahedron and the number ofedges in the tetrahedron.

(c) six is the number of faces in the cube and the number of verticesin the octahedron, eight is the number of vertices in a cube and thenumber faces in the octahedron, and twelve is the number of edgesin both the cube and the octahedron.

1.2. Tetrahedron: equilateral triangles, squares.Octahedron: squares, regular hexagons.Dodecahedron: equilateral triangles, squares, and regular pentagons,hexagons, and decagons.Icosahedron: regular pentagons and decagons (from planes perpendic-ular to a line joining opposite vertices), and regular nonagons and do-decagons (from planes parallel to a face).The octahedron and icosahedron also have equilateral triangular sec-tions when the plane contains a face of the polyhedron.

1.3. One type of pentahedron is a pyramid whose base is a convex quadrilat-eral and a second type is a prism with triangular bases. See Figure S1.1.

Figure S1.1.

227

228 Solutions to the Challenges

1.4. Yes, all seven are reptiles. The union of three cubes is a rep-8 reptile,the other three in the first row of Figure 1.2.3 are rep-64 reptiles, andthe three in the second row are each rep-8. For the union of three cubes,arrange four copies in each of two layers using the pattern in Figure1.3.2. For the “T” and “L” shaped four-cube polycubes in the first rowof Figure 1.2.3, arrange four copies into a 4 � 4 � 1 block as shown inFigure S1.2a, then four copies of this block form a 4 � 4 � 4 cube, andfour such cubes form a larger version of the tile.

3 3 7 76 7 7 4

6 8 8 48 8 5 5

top layer

1 3 3 41 1 2 4

6 1 2 26 5 5 2

bottom layer

(a) (b)

Figure S1.2.

For the Z shaped polycube in the first row of Figure 1.2.3, arrange eightcopies into a 4� 4� 2 block as shown in Figure S1.2b, then two copiesof this block form a 4 � 4 � 4 cube, and four such cubes form a largeversion of the tile. For each of the polycubes in the second row of Figure1.2.3, two copies of the polycube form a 2 � 2 � 2 cube, and four suchcubes form a larger version of the tile.

1.5. In the box on the left in Figure 1.3.3 the radius of each sphere is a=2mand the total volume of the m3 spheres is m3 � .4=3/�.a=2m/3 D�a3=6, which is independent of m. Hence the total volume of all thespheres is the same for both boxes.

1.6. No. Let the maximum number of edges of a face of such a polyhedronbe n. Then the other numbers of edges of faces are elements of theset f3; 4; � � � ; n � 1g. There are only n � 3 numbers in this set, yetthere must be at least n other faces meeting the face with n edges. Soevery polyhedron must have at least two faces with the same number ofedges.

1.7. The answer is all six! As shown in the Figure S1.3a, the person may beinside a large cube, and from a corner see the six faces (walls, floor, andceiling), or in S1.3b one may look at a cube in front of a mirror, or inS1.3c the cube may have transparent faces, etc.

Solutions to the Challenges 229

(a) (b) (c)

Figure S1.3.

1.8. No. See Figure S1.4, where the vertices are seven of the eight verticesof a cube.

Figure S1.4.

1.9. See Figure S1.5.

xxFigure S1.5.

1.10. Suppose not, so that every diameter of the sphere has two unpaintedends, or one end painted and one unpainted. If we reflect the sphere inits center, then the image of every painted point must be an unpaintedpoint. Thus the image N of the set P of painted points is a subset ofthe set U of unpainted points. Hence N and P are disjoint subsetsof the sphere with equal areas. But this is impossible since the area


of P (and of N , and so also for U/ is more than half the area of thesphere.

1.11. Draw a third diagonal on another face, as shown in Figure S1.6. Sincethe triangle formed by the three diagonals is equilateral, the anglebetween two diagonals is 60ı.

Figure S1.6.

Chapter 22.1. See Figure S2.1a.

(a) (b)

Figure S2.1.

2.2. Count the calissons in Figure S2.1b.

2.3. Each square (except 1) is the sum of two consecutive triangular num-bers, so the sum of the first n odd squares is the same as the sum of thefirst 2n � 1 triangular numbers. Then (2.3) yields

12 C 32 C � � � C .2n � 1/2 D T1 C T2 C T3 C � � � C T2n�1

Dn.2n � 1/.2nC 1/

3:


2.4. Using the hint yields the pile of cubes in Figure S2.2. Now count theunit cubes in vertical slices.

Figure S2.2.


Figure S2.3.

2.6. The number of cubes in (a) is 6.12/ D 1 � 2 � 3, in (b) 6.12 C 22/ D2 � 3 � 5, and in (c) 6.12C 22C 32/ D 3 � 4 � 7. In each step we “wrap” ak � .kC 1/� .2kC 1/ box with six 1� .kC 1/� .kC 1/ boxes. Aftern steps we have 6.12C22C� � �Cn2/ cubes in a n� .nC1/� .2nC1/box, from which (2.1) follows [Kalajdzievski, 2000].

2.7. The elements of the sequence f4qng1nD1 are the odd squares dimin-ished by 1 and the even squares, so fqng1nD1 is the sequence ofsquares divided by 4 (the “quarter-squares”) rounded down. See Fig-ure S2.4 for an illustration of quarter-squares as square and oblongnumbers.


Figure S2.4.


Figure S2.5.

2.9. The nth octahedral number is the sum of the .n � 1/st and nth squarepyramidal numbers, and hence it equals

.n � 1/n.2n � 1/

6Cn.nC 1/.2nC 1/

6Dn.2n2 C 1/

3:

2.10. We count the cubes by considering the location of, say, the uppernortheast corner of the cube. Thus there are n3 1 � 1 � 1 cubes,.n � 1/3 2 � 2 � 2 cubes, .n � 2/3 3 � 3 � 3 cubes, and so on to1 n � n � n cube. Hence the total number of cubes is, by (2.4), thesquare of the nth triangular number.

2.11. We proceed as we did in Section 2.5, first finding the maximum num-ber C.n/ of “2-regions” (two-dimensional regions) determined by ncircles in the plane. We obtain a maximum when each pair of circlesintersect in two points and no three are concurrent. Clearly C.0/ D 1,C.1/ D 2, and C.2/ D 4. Suppose k�1 circles partition the plane intoC.k�1/ 2-regions. The kth circle intersects each of the first k�1 cir-cles in two points, and these 2.k � 1/ points divide the kth circle into2.k � 1/ arcs. Each of these arcs divides in two one of the 2-regionsformed by the first k � 1 circles, so that C.k/ D C.k � 1/C 2.k � 1/.


Summing from k D 1 to k D n yields

C.n/ � C.n � 1/ D 2.n � 1/

C.n � 1/ � C.n � 2/ D 2.n � 2/

C.n � 2/ � C.n � 3/ D 2.n � 3/

:::

C.2/ � C.1/ D 2 � 1

C.1/ � C.0/ D 1

C.n/ � 1 D 1C 2Tn�1

where Tk denotes the kth triangular number, and consequentlyC.n/ D 2C 2Tn�1 D n

2 � nC 2.

We now find the maximum number S.n/ of “3-regions” (three-dimensional regions) in space determined by n spheres. We obtainthe maximum number when each pair of spheres intersect in a circle,and those circles on spheres have the same properties as the circles inplanes described above. Clearly S.0/ D 1, S.1/ D 2, and S.2/ D 4.Suppose k�1 spheres partition space into S.k�1/ 3-regions. Thus thesurface of the kth sphere is divided into C.k � 1/ 2-regions (the argu-ment for circles on spheres is the same as for circles in the plane), andeach of those 2-regions divides in two the 3-regions into which the firstk�1 spheres had partitioned space. Thus S.k/ D S.k�1/CC.k�1/,or S.k/ � S.k � 1/ D 2 C 2Tk�2. Summing from k D 1 to k D n

yields

S.n/ � S.n � 1/ D 2C 2Tn�2

S.n � 1/ � S.n � 2/ D 2C 2Tn�3

S.n � 2/ � S.n � 3/ D 2C 2Tn�4:::

S.3/ � S.2/ D 2C 2T1

S.2/ � S.1/ D 2

S.n/ � 2 D 2.n � 1/C 2.n � 2/.n � 1/n

6

where we have used (2.3) to sum the first n � 2 triangular numbers.Thus S.n/ D 2nC .n � 2/.n � 1/n=3 D n.n2 � 3nC 8/=3.


2.12. Within the tetrahedral pile of cannonballs in Figure 2.4.3, we can findsix cannonballs touching a given cannonball in one layer, plus anotherthree in the layers above and below, as shown in Figure S2.6.

Figure S2.6.

2.13. The theorem is: Four times the nth square pyramidal number is the(2n/th tetrahedral number, and a proof is

4n.nC 1/.2nC 1/

6D.2n/.2nC 1/.2nC 2/

6:

You can illustrate this result by re-stacking the tetrahedral pile of can-nonballs into four square pyramidal piles, as indicated here for theeighth tetrahedral number 120:

120 D .1C 3/C .6C 10/C .15C 21/C .28C 36/

D 4C 16C 36C 64 D 4.1C 4C 9C 16/:

Chapter 33.1. The result follows directly from (3.12) and the AM-GM inequality

(3.5), both applied to the three numbers d2xy , d2yz , and d2xz .

3.2. Yes. If the package is in the shape of a right circular cylinder with heighth in and base radius r in, then the volume is V D �r2h in3 and thelength + girth is S D hC 2�r in, so that

3p�V D

3p�r � �r � h � .2�r C h/=3 D S=3 D 36:

Hence the largest acceptable cylindrical package has volume V D363=� � 14851in3.

3.3. The number of cubes in one set of slices begins with n2 and ends withn.nC 1/, and in the other set begins with n.nC 1/C 1 and ends with.nC 1/2 � 1.


3.4. Set .a; b; c/ D .x2; y2; z2/ in (3.8).

3.5. Let x = length, y = width, and z = height of the rectangular box inFigure 3.9.2. Then the length S of string is S D 2xC 4yC 6z D 12. IfV denotes the volume, then 48V = 48xyz = 2x � 4y � 6z. From (3.4) wehave

3p48V D 3

p2x � 4y � 6z � .2x C 4y C 6z/=3 D S=3 D 4;

and so the maximum volume is V D 43=48 D 4=3 ft3 with x D 2y D3z. Hence 3z � 3z=2 � z D 4=3, z D 2=3 ft, y = 1 ft, and x = 2 ft.

3.6. Label the dimensions of the box so that the area of the top and bottomis xy, of the front and back is xz, and of the other two sides is yz. ThenV = xyz and the cost C is given by C D 2cxy C 2bxzC 2ayz. By theAM-GM inequality (3.5) we have

8abcV 2 D .2cxy/.2bxz/.2ayz/ �

�C

3

�3with equality if and only if cxy D bxz D ayz. Thus the minimum costis C D $3.8abcV 2/1=3.

3.7. Let r and h denote the base radius and height, respectively, of the barrel.Then V D �r2h and s2 D 4r2 C h2=4 so that (3.5) yields

V 2 D �2r4h2 D �2 � 2r2 � 2r2 �h2

4D �2 � 2r2 � 2r2 � .s2 � 4r2/

� �2�2r2 C 2r2 C .s2 � 4r2/

3

�3D �2

�s2

3

�3;

or V � �s3=3p3, with equality if and only if 2r2 D s2�4r2 D h2=4,

i.e., h D 2rp2. So the maximum volume is obtained when the height

of the barrel isp2 times its diameter.

3.8. When the perimeter is given, so is s. Using (3.5) we have

ADpsp.s� a/.s� b/.s� c/ �

ps

�.s� a/C .s� b/C .s� c/

3

�3=2

Ds2

3p3

with equality if and only if a D b D c. Hence the triangle with maxi-mum area is an equilateral triangle.


3.9. Using the hints yields 2axby � 2 jaxbyj � a2y2 C b2x2, 2axcz �2 jaxczj � a2z2 C c2x2, and 2bxcz � 2 jbxczj � c2y2 C b2z2.Adding the three inequalities and then adding a2x2C b2y2C c2z2 toboth sides yields (3.15).

3.10. Yes, it is possible. See Figure S3.1 for one solution, exhibiting thethree layers in the box with nine bricks in each [Hoffman, 1981].

Top layer Middle layer Bottom layer

Figure S3.1.

3.11. It suffices to show that V 2 � 4�2s6=243. Since s2 D r2Ch2 we have

V 2 D

��r2h

3

�2D�2r4h2

9D4�2

9�r2

2�r2

2� .s2 � r2/

�4�2

9�

�1

3

�r2

2Cr2

2C .s2 � r2/

��3D4�2s6

243;

with equality if r2=2 D s2 � r2 D h2, i.e., when r D hp2.

3.12. Since '2 D ' C 1 we have ' D 1 C .1='/, so that if we set a D 1

and b D 1=' in the identity associated with Figure 3.1.3 we have'3 D 13 C .1='/3 C 3 � 1 � .1='/ � ', hence '3 � .1='/3 = 4.

3.13. The set B can be partitioned into the following sets: (i) A itself, vol-ume abc, (ii) two a � b � 1 bricks, two a � 1 � c bricks, and two1 � b � c bricks, volume 2ab C 2ac C 2bc, (iii) four quarter cylin-ders of length a and radius 1, four quarter cylinders of length b andradius 1, and four quarter cylinders of length c and radius 1, vol-ume .a C b C c/� , and (iv) eight spherical sectors, each one-eighthof a sphere of radius 1, volume 4�=3. Hence the volume of B isabc C 2.ab C ac C bc/C �.aC b C c/C 4�=3.


Chapter 44.1. Dudeney’s solution: “The mystery is made clear by the illustration. It

will be seen at once how the two pieces slide together in a diagonaldirection.”

Figure S4.1.

4.2. Yes, see Figure S4.2 where the three 1 � 1 � 1 cubes are white and thesix 1�2�2 boxes are different shades of gray depending on orientation.The complementary problem (constructing a 3 � 3 � 3 cube from thenine smaller pieces) is known as the Slothouber-Graatsma puzzle.

Figure S4.2.

4.3. The side length of the enclosing cube in Figure 4.8.2b is sp2 so its vol-

ume is 2s3p2, and the volume of each of the eight pyramids removed

is .1=6/.s=p2/3 D s3

p2=24. Hence the volume of the cuboctahedron

is V D 2s3p2 � s3

p2=3 D 5s3

p2=3.

4.4. Using the notation from Section 4.2 yields V D volO.3s/ �6volRSP .s/ D 8s3

p2. For an alternate solution, slice the truncated

octahedron through its center along the planes of the edges of the orig-inal octahedron, cutting the truncated octahedron into eight half-cubes,one of which is shown in Figure S4.3. Since the edge of the half-cubehas length s

p2, V D 8 � 1=2 � .s

p2/3 = 8s3

p2.


Figure S4.3.4.5. If the tetrahedron is isosceles, then the faces are congruent and have the

same perimeter. In the other direction, let the pairs of opposite edgesbe a, a0; b, b0; and c, c0; and assume a C b C c D a C b0 C c0 D

a0CbCc0 D a0Cb0Cc. Then bCc D b0Cc0 and bCc0 D b0Cc, orequivalently b � b0 D c0 � c and b � b0 D c � c0. Thus c0 � c D c � c0

so that c D c0, and it now follows that a D a0 and b D b0, so that thetetrahedron is isosceles.

4.6. Using the notation in Figure 4.6.1, we have b2C c2� a2 D 2z2 > 0 sothat b2 C c2 > a2, which implies that the angle opposite side a in theface is acute. Similarly the angles opposite sides b and c are acute, soeach face is an acute triangle.

4.7. 29: 5.23/ C 24.13/ D 43, 34: 26.23/ C 8.13/ D 63, 36: 4.23/ C32.13/ D 43, 38: 1.33/C 37.13/ D 43, 39: 1.63/C 18.33/C 1.23/C19.13/ D 93, 41: 25.23/C 16.13/ D 63, 43: 3.23/C 40.13/ D 43, 45:1.63/C 36.23/C 8.13/ D 83, and 46: 1.63/C 18.33/C 27.13/ D 93.

4.8. No, since no dissection of 41� 41� 41 can have both 33� 33� 33 and16 � 16 � 16 as subcubes. Similarly 1.33/ C 1.43/ C 1.53/ D 63 butthis does not make 3 3-admissible.

4.9. Figure 4.2.2 shows that a regular tetrahedron with edge length 2s can bedissected into four regular tetrahedra with edge length s and a regularoctahedron with edge length s. Hence

4volT .s/C volO.s/ D volT .2s/ D 8volT .s/;

and thus volO.s/ D 4volT .s/.(b) Figure 4.2.3 shows that a regular octahedron with edge length s canbe dissected into two regular square pyramids with edge length s. Hence

2volRSP .s/ D volO.s/ D 4volT .s/;

and thus volRSP .s/ D 2volT .s/.



Figure S4.4.

Chapter 55.1. Approximately 7.6 m.

5.2. As noted in Section 5.2 the regular octahedron is a right triangularantiprism. If we let s denote its edge length, thenA1 D A2 D s2

p3=4,

Am D 3s2p3=8, and h D s

p6=3. Hence

volO.s/ D1

6�s

3

p6

�2 �s2

4

p3C 4 �

3s2

8

p3

�Ds3

3

p2:

5.3. At a height x 2 Œ�r; r� the cross-section of the torus is an annulus witharea �Œ.RC

pr2 � x2/2 � .R�

pr2 � x2/2� D 4�R

pr2 � x2, and

the area of the rectangular cross-section of the cylinder is 2pr2 � x2 �

2�R D 4�Rpr2 � x2.

5.4. At a height x in Œ�h=2; h=2� the cross-section of the bead is an annuluswith area �.r2 � x2/� �a2 D �Œ.h=2/2 � x2� (since a2C .h=2/2 Dr2/, which is the same as the area of the circular disk cross-section ofthe sphere. Hence the volume of the bead is .4=3/�.h=2/3.

5.5. The cross-sections of the bicylinder parallel to the plane determinedby the axes of the two cylinders are squares, and each square has area4=� times the area of the corresponding circular section of a sphere.Hence the volume of the bicylinder is .4=�/ � 4�r3=3 D 16r3=3.

5.6. Following the hint, the volume of the spherical segment is the sameas the volume of the corresponding segment of the tetrahedron. Butthis segment is a prismatoid, so we can use the prismoidal formula


to find its volume. Since the corresponding sections of the sphericaland tetrahedral segments have the same area, we need only apply theprismoidal formula to the spherical segment. Clearly A1 D �a2 andA2 D �b2, so we need only find Am. To do so, let r denote theradius of the sphere, insert a y-axis through the poles of the spherewith its origin at the center, let the upper plane cut the axis at s andthe lower plane at t , so that a2 C s2 D r2, b2 C t2 D r2, andh D s� t . Then Am D �.r2� Œ.s C t /=2�2/ D .�=4/Œ4r2� .sC t /2�.Thus

V Dh

6.A1 C 4Am C A2/ D

�h

6.a2 C b2 C 4r2 � s2 � 2st � t2/

D�h

6.3a2 C 3b2 C s2 � 2st C t2/ D

�h

6.3a2 C 3b2 C h2/:

5.7. The spherical cap is a spherical segment, as seen in Figure 5.11.4with a D 0, so its volume V is V D .�h=6/.3b2 C h2/. Butr2 D .r � h/2 C b2 so that b2 D 2rh � h2, and hence V D.�h=6/.6rh � 2h2/ D .�h2=3/.3r � h/, which is the same as thevolume of the cone.

5.8. Yes. The solids we are using for comparison with Cavalieri’s principleare spheres, and the prismoidal formula gives the correct volume forspheres.

5.9. Use de Gua’s theorem from Section 5.4 and apply Guba’s inequality(3.11). Equality holds when the hypotenuse triangle is equilateral.

5.10. (a) Let c denote the length of the hypotenuse. Computing the areaof the triangle two ways yields ch=2 D ab=2, so that .1=h/2 Dc2=a2b2 D .a2 C b2/=a2b2 D .1=a/2 C .1=b/2.

(b) Let K denote the area of the hypotenuse triangle ABC in Figure5.4.1. From de Gua’s theorem we have 4K2 D b2c2 C a2c2 C

a2b2, or equivalently, .2K=abc/2 D .1=a/2 C .1=b/2 C .1=c/2.But the volume of the right tetrahedron is both abc=6 and hK=3,hence 2K=abc D 1=h.

5.11. (a) The triangular sections of the plug in Figure 5.10.6b have area halfthat of the corresponding sections of a right circular cylinder withbase radius 1 in and height 2 in. Thus the volume of the plug is1=2 � �12 � 2 D � in3.


(b) The plug in Figure 5.10.6c is a right circular cylinder with twocylindrical wedges removed. The volume of the cylinder is 2� in3

and the volume of each wedge is, by Example 5.7, 4=3 in3, thusthe volume of the plug is 2� � 8=3 in3.

(c) There are infinitely many different shapes for the plug. Perhapsthe simplest is just the union of the three shaded sections in Figure5.10.6b.

5.12. No. Consider the cube and prism in Figure S5.1, where the bases ofthe prism are the same as for the cube, two of the lateral faces (frontand back) are parallelograms, and two (left and right) are rectangles.The cross-sections of both solids are congruent squares. The parallel-ogram faces have the same area as the square faces of the cube, but therectangular faces have greater area.

(a) (b)

Figure S5.1.

5.13. Only †ACB is a right angle, so ABCD is not a right tetrahedron. If[PQR] denotes the area of �PQR, then ŒABC � D 4, ŒABD� D 3, andŒACD� = ŒBCD� D

p7=2; hence ŒABC �2 D ŒABD�2 C ŒACD�2 C

ŒBCD�2.

5.14. Let V , B; and H be the volume, base area, and altitude of thetetrahedron ABCD in Figure 5.11.8, and v, b; and h the volume,base area, and altitude of the shaded tetrahedron. Then V D

.1=3/BH , b D .2=3/2.1=4/B D .1=9/B , and h D .2=3/H andhence

v D1

3bh D

1

3�1

9B �

2

3H D

2

27�1

3BH D

2

27V:

5.15. (a) Consider the triangle with vertices .0; 0; 0/, .a; b; c/, and .a Cx; b C y; c C z/.

(b) Since square roots are nonnegative we need only prove that

.aC x/2 C .b C y/2 C .c C z/2

��p

a2 C b2 C c2 Cpx2 C y2 C z2

�2(5.5)


is equivalent to the Cauchy-Schwarz inequality (5.3). To show that(5.3) implies (5.5), we have:

.aC x/2 C .b C y/2 C .c C z/2

D a2 C b2 C c2 C x2 C y2 C z2 C 2.ax C by C cz/

� a2 C b2 C c2 C x2 C y2 C z2

C 2pa2 C b2 C c2

px2 C y2 C z2

D�p

a2 C b2 C c2 Cpx2 C y2 C z2

�2:

To show that (5.5) implies (5.3), consider (5.5) for the absolutevalues jaj, jbj, jcj, jxj, jyj, jzj, expand the squares and use the factthat jax C by C czj � jaj jxj C jbj jyj C jcj jzj.

5.16. Consider a regular icosahedron inscribed in a sphere whose diame-ter d is 4. Then twelve unit spheres centered at the vertices of theicosahedron will kiss a unit sphere whose center coincides with thecenter of the circumsphere. But in Section 5.7 we learned that theedge length s of the icosahedron satisfies s D d=

p2C '. Thus

s D 4=p2C ' � 2:103, so that no two of the unit spheres centered

at adjacent vertices of the icosahedron touch one another.

5.17. The regular dodecahedron has the greater volume, since

Vdodec D7' C 4

2

�2

'p3

�3� 2:785

and Vicos D5'2

6

�2

p2C '

�3� 2:536:

5.18. The proof uses the following easily verified facts from plane geometryand trigonometry: the side length of a regular n-gon inscribed in acircle of radius r is 2r sin.�=n/, cos.�=5/ D '=2, and sin.�=10/ D1=.2'/, where ' denotes the golden ratio. Then p D 2r sin.�=5/ andd D 2r sin.�=10/ D r=', and thus

a2 D p2 � r2 D 4r2 sin2.�=5/ � r2 D 4r2Œ1 � cos2.�=5/� � r2

D 3r2 � 4r2 cos2.�=5/ D r2.3 � '2/:

But 3 � '2 D 1='2 (since Œ' � .1='/�2 D 1/ so that a D r=' D d .


Chapter 6

6.1. No. Let each line be parallel to the other two, but with l3 not in the planedetermined by l1 and l2. Then it is impossible for any line intersectingboth l1 and l2 to also intersect l3.

6.2. Since each edge of the tetrahedron connects two vertices, summingSA D ˛CˇC�� and three similar expressions for the other verticesyields T D 2D � 4� .

6.3. At most two. Spheres †1 and †2 intersect in a circle, and that circleintersects †3 in at most two points.

6.4. No. Consider the tetrahedron whose vertices have xyz-coordinates(0,0,0), (–1,0,0), (1,1,0), and (0,0,1). Then the altitude from (0,0,1) isalong the z-axis while the altitude from (1,1,0) is the line parallel to they-axis through the point (1,1,0), which does not intersect the z-axis.

6.5. Two of the faces are equilateral triangles, and the other two are isoscelestriangles with edges 4, 4, and 7, and so all the edges in the faces satisfythe triangle inequality. Figure S6.1 shows that for a tetrahedron to existwhen five sides have length 4, the sixth side must have length less than4p3 � 6:93.

4

4 3

4

444

Figure S6.1.

6.6. Denote the n fixed points by .ai ; bi ; ci / for i D 1; 2; � � � ; n, and letNa D .1=n/

PniD1 ai and similarly for Nb and Nc. Since

PniD1 .x � ai /

2 D

n.x � Na/2 CPniD1 .ai � Na/

2 (and similarly forPniD1 .y � bi /

2 andPniD1 .z � ci /

2/, it follows that

Xn

iD1.x � ai /

2 CXn

iD1.y � bi /

2 CXn

iD1.z � ci /

2 D k2


is equivalent to

.x � Na/2 C .y � Nb/2 C .z � Nc/2

D k2 �1

n

Xn

iD1Œ.ai � Na/

2 C .bi � Nb/2 C .ci � Nc/

2�;

and the locus is a sphere centered at . Na; Nb; Nc/ for sufficiently large k.

6.7. Similar spherical triangles (i.e., triangles with identical angles) willhave the same area by Girard’s theorem, and hence they must be con-gruent.

6.8. No. Three planes, no two of which are parallel, will either intersectin three parallel lines or intersect at a common point P . Consequentlyeach of the pairwise intersection lines (for the planes of the light grayand medium gray faces, and the plane of the hidden face in the backof the “solid”) must pass through P . As the dashed extensions in Fig-ure S6.2 indicate, they do not (also note that the light gray face is notplanar). Hence the figure does not represent a polyhedron [Gardner,1992].

Figure S6.2.

6.9. The side length of the regular hexagonal section of a unit cube isp2=2. If a square with side greater than 1 is inscribed in the hexagon,

then it lies within its circumcircle. But the radius of the circumcir-cle is also

p2=2, and every square within this circle has side at mostp

2=2 �p2 D 1, so a square with side greater than 1 cannot be inscribed

in the hexagon. Furthermore, any square inscribed in the regular hexag-onal section cannot have all four vertices on the circumcircle, and so itsarea will be strictly less than 1.


6.10. The locus of points on a sphere equidistant from two given points is agreat circle. Given the spherical triangle ABC, letK be the great circlebisecting AB, and let K 0 be the great circle bisecting AC. K and K 0

meet at two antipodal points; let P be the one inside ABC equidistantfrom A, B , and C . The great circle determined by P , its antipodalpoint, and the midpoint of BC is the third bisector.

Chapter 7

7.1. (a) If the generating tetrahedron has surface area A0 and volume V0,then after n iterations the surface area An equals A0 and the volumeVn equals V0.1=2/n. Hence the Sierpinski tetrahedron has finite surfacearea and zero volume. (b) If the generating octahedron has surface areaA0 and volume V0, then after n iterations the surface area An equalsA0.3=2/

n and the volume Vn equals V0.3=4/n. Hence the Sierpinskioctahedron has infinite surface area and zero volume.

7.2. The surface area of the fractal cube is the same as that of the originalcube, while the volume of the fractal is zero.

7.3. E D 60, A D 40=3, and V D 5=7.

7.4. The volume is zero, but the surface area is the same as the original cube.

7.5. The limit of the volume of the lantern equals the volume of the cylinder.Using the result of Example 5.2, the volume of the antiprism inscribedin each band is .h=6m/Œ2AnC 4A2n�, where Ak is the area of a regulark-gon inscribed in a circle of radius r . Thus the volume of the lantern is.h=6/Œ2An C 4A2n�. Since this is independent of m, we need only takethe limit as n!1. Since An and A2n both approach �r2, the volumeof the lantern approaches �r2h, the volume of the cylinder.

7.6. See Figure S7.1 for the first three iterations of the section. At each stepin the iteration, a star-shaped hole and ring of six hexagons and sixtriangles replace each hexagonal region of the section, and a hexagonand three triangles replace each triangular region of the section.

7.7. (a) ln 4=ln 2 D 2, (b) ln 6=ln 2 � 2:585, (c) ln 4=ln 2 D 2, (d) ln 9=ln 3D 2.

7.8. Let Sn D 1 C 4 C � � � C n2. Then the number Tn of cubes in the nthaggregation is Tn D SnC2Sn�1CSn�2 D .2n � 1/.2n2 � 2nC 3/=3.


Figure S7.1.


Figure S7.2.

Chapter 8

8.1. a) In Section 4.6 we learned that the faces of an isosceles tetrahedronare congruent triangles, hence they all have the same area K. Pro-ceeding as we did in the algebraic proof in Section 8.2, we have

V D1

3Kd1 C

1

3Kd2 C

1

3Kd3 C

1

3Kd4

and hence d1 C d2 C d3 C d4 is the constant 3V =K.b) Yes, it holds for any convex polyhedron with the property that all

the faces have the same area. This includes the other four Platonicsolids, the rhombic dodecahedron, bipyramids with congruent trian-gular faces, etc.

8.2. The interior .n � 2/3 small cubes have no red faces, 6.n � 2/2 cubeshave one red face, 12.n�2/ cubes have two red faces, and 8 cubes havethree red faces.

8.3. Let Vk D p2kpkC1 denote the volume of the kth Padovan brick. It is

easy to prove by mathematical induction that V0 C V1 C � � � C Vn DpnpnC1pnC2, so the total volume of the bricks equals the volume ofthe box. Figure S8.1, which illustrates the induction step in the proof,shows that the bricks actually fit into the box.


pn+3

pn+2

pn+1

pn pn+1

Figure S8.1.

8.4. Replacing cube IV by cube V changes the codes in Table 8.1 to thefollowing table:

Cube (1) (2) (3)I 1 6 15II 2 10 15III 2 5 6V 3 4 25

Now there is only one set of codes—(6,10,5,3)—whose product is 900,and so we can only have all four colors on “top-bottom” or “front-back,”not both.

8.5. (a) Cut the torus horizontally and then vertically to produce four similarU-shaped pieces. Then stack the four pieces and cut as indicated in Fig-ure S8.2a to produce four large and eight small pieces. (b) Nine pieces.See Figure S8.2b.

(a) (b)

Figure S8.2.

8.6. Making a slice around the torus in the form of a full-twist Mobius strip(so that it has two sides) by rotating the knife 360ı will result in two


interlocked rings. See Figure S8.3 (images from the Stack ExchangeNetwork).

Figure S8.3.

8.7. (a) The area of the top is � �12 D � and the lateral area is 2� �1 �1=2 D� .

(b) More generally we show how to “square” an a � b rectangle: Asemicircle with diameter a C b yields a line segment with lengthpab, the side of the required square. With a D 1=2 and b D 2�

we havepab D

p� . See Figure S8.4.

a bab

Figure S8.4.

(c) No. The argument is literally “circular,” since you need to be ableto construct a 2� � 1=2 rectangle (equivalent to squaring the circle)to construct the cylinder!

8.8. Rotating the y- and z-axes through an angle � (0 � � � �=2/ as inSection 8.8 yields x2C . Ny cos � � Nz sin �/2 D r2 as the equation of thecylinder. Setting Nz D 0 and replacing Ny by y yields x2C .cos2 �/y2 Dr2 as the equation of the intersection. This is a circle, an ellipse, or twoparallel lines when � D 0, 0 < � < �=2, or � D �=2, respectively.

8.9. Rotating the y- and z-axes through an angle � as in Section 8.8 yields

.px2 C . Ny cos � � Nz sin �/2 �R/2 C . Ny sin � C Nz cos �/2 D r2

as the equation of the torus. Setting Nz D 0 and replacing Ny by y yields

x2 C y2 � 2Rpx2 C y2 cos2 � CR2 D r2:


After rearranging terms and squaring both sides we have

.x2Cy2/2C2.x2Cy2/.R2�r2/C.R2�r2/2 D 4R2.x2Cy2 cos2 �/:

Using sin � D r=R and further simplifying yields

.x2 C y2/2 � 2.x2 C y2/.R2 � r2/C .R2 � r2/2 D 4r2x2;

orŒ.x2 C y2/ � .R2 � r2/�2 � .2rx/2 D 0;

which factors to yield

Œ.x � r/2 C y2 �R2�Œ.x C r/2 C y2 �R2� D 0:

Hence the intersection is two circles with radius R whose centers are2r units apart.

Chapter 99.1. See Figure S9.1, where the segments and arc labeled a) through e) have

the required lengths in Table 9.1.

λ

a)

c)

d) λ

1 λ

π4 2

λ

b)

e)

2

Figure S9.1.

9.2. No. If an ideal map did exist, it would project a spherical triangle onto aplane triangle whose angle sum is � , but by Girard’s theorem (see Sec-tion 6.4) the angle sum of a spherical triangle is always greater than � .

9.3. (a) From the hint, 2E=F � 3 so that 2E � 3F . Similarly, each edgejoins two vertices, so the average number of edges per vertex is2E=V , and this average is at least 3.

(b) Three non-colinear vertices can only form a triangle in a plane, soV � 4. Hence 2E � 3V � 12.


(c) If E D 7 then 3V � 14 so that V D 4. But since each edgejoins a distinct pair of vertices, the maximum number of edges for

a polyhedron with four vertices is

�4

2

�D 6.

(d) A pyramid whose base is a regular n-gon for n � 3 has 2n edges, soall even numbers 6 and greater are possible values forE (see FigureS9.2a for a projection of the skeleton for the case n D 5/. Cuttingoff a base corner of the pyramid as shown in Figure S9.2b adds threeedges, so the resulting polyhedron has 2nC 3 edges. Hence all oddnumbers 9 and greater are possible values for E.

(a) (b)

Figure S9.2.

9.4. (a) Both nF and dV count the number of edges twice.(b) Setting F D 2E=n and V D 2E=d in Euler’s polyhedral formula

and simplifying yields .1=n/C .1=d/ D .1=2/C .1=E/, and since1=E > 0 we have .1=n/C .1=d/ > .1=2/.

(c) Clearly n � 3 and d � 3, but n and d cannot both be greater than3, since then .1=n/ C .1=d/ � .1=2/. So either n D 3 or d D 3,which leads to the five pairs in the solution. Observe that these yieldthe five regular polyhedra studied earlier.

9.5. (a) The average number of edges per vertex is 2E=V , so 2E � 3V . Soby Euler’s formula, 3V D 6 C 3E � 3F , so 6 C 3E � 3F � 2Eand hence 6CE � 3F , or 2E � 6F � 12.

(b) If 2E � 6F � 12, then 2E < 6F , so that 2E=F < 6, i.e., theaverage number of sides per face is less than 6.

9.6. When F D 5 Euler’s polyhedron formula reduces to E D V C3. SinceF D 5, all the faces must be triangles or quadrilaterals, and the degreeof each vertex must be 3 or 4. Let Fn be the number of faces that aren-gons, and Vk the number of vertices of degree k. Now F3 C F4 D

5 and 3F3 C 4F4 D 2E, and the positive integer solutions to theseequations are .F3; F4; E/ = (4,1,8), (2,3,9), and (0,5,10). Similarly V3C


V4 D V and 3V3 C 4V4 D 2E D 2V C 6, from which it follows thatV3 C 2V4 D 6. The positive integer solutions to this equation (and thecorresponding values of E/ are .V3; V4; E/ = (6,0,9), (4,1,8), (2,2,7),and (0,3,6). Hence the only two pentahedra satisfy .F3; F4; E/ = (4,1,8)and (2,3,9), the two described in the solution to Challenge 1.3.

9.7. (a) The regular dodecahedron.(b) Let F5 and F6 denote the number of pentagonal and hexagonal

faces, respectively and V , E, and F the total number of vertices,edges, and faces. Then (i) implies F5 C F6 D F and 5F5 C 6F6 D2E, hence 6F � 2E D F5. Property (ii) implies 2E D 3V , multi-plying Euler’s formula by 6 yields 6V �6EC6F D 12, thus 6F �2E D 12 and hence F5 D 12. A third polyhedron satisfying the twoproperties is the hexagonal truncated trapezohedron, a polyhedronwith twelve regular pentagonal faces, two regular hexagonal faces,thirty-six edges, and twenty-four vertices. See Figure S9.3.

Figure S9.3.

9.8. Let A (PDX), B (BCN), and C (north pole) be the vertices of thespherical triangle. Then the measures of the angles and sides (inradians) are approximately C D 2:1760485, a D :8499911, andb D :7751140, so that (9.1) yields cos c D :1723279. Hence c D1:3976039 and thus the distance between BCN and PDX is approxi-mately 8904 km or 5533 mi.

9.9. Consider the face �ABC of the tetrahedron ABCD. The proof that thearea of �ABC is less than or equal to the sum of the areas of the otherthree faces follows from (1) the area of a each of the other faces is atleast as great as the area of its projection in the plane of �ABC, and(2) the union of the three projections is equal to�ABC or is a supersetof �ABC.


9.10. Let C 0 and D0 be the projections of C and D, respectively, ontothe plane of �ABM. Then M is also the midpoint of C 0D0. Since�ABC 0, �ABD0, and �ABM have a common base AB and the al-titude of �ABM is the arithmetic mean of the altitudes of �ABC 0

and �ABD0, the area of �ABM is the arithmetic mean of the areasof�ABC 0 and�ABD0. But the area of�ABC 0 is less than or equalto the area of �ABC , and the area of �ABD0 is less than or equal tothe area of �ABD.

9.11. Yes for both solids. See Figure S9.4a for the cuboctahedron and S9.4bfor Durer’s solid.

(a) (b)

Figure S9.4.

9.12. Color the vertices as shown in Figure S9.5, and observe that eachedge joins a white and a black vertex. So if a Hamiltonian cycle ex-ists, it must pass through an even number of vertices. But the en-neahedron has eleven vertices, so it does not possess a Hamiltoniancycle.

Figure S9.5.

9.13. No. Two quadrilateral faces, the one in the center and the one corre-sponding to the exterior of the diagram, share a pair of opposite ver-tices. Since the faces are planar, they would also share the diagonalsegment joining the vertices.

9.14. No. To have equality in the inequality of Section 9.6, the faces ABCand BCD must be equilateral triangles, and the dihedral angle at


edge BC must be �=2 (see Figure 9.6.1a). But in this case jADj DMp6=2 > M , contradicting the fact that M is the length of the

longest edge.

9.15. The inequality xy � .x2 C y2/=2 for nonnegative x and y yields

ja1b1 C a2b2 C � � � C anbnj

AB

�ja1j

A

jb1j

BCja2j

A

jb2j

BC � � � C

janj

A

jbnj

B

�1

2

ja1j

2

A2Cjb1j

2

B2Cja2j

2

A2Cjb2j

2

B2C � � � C

janj2

A2Cjbnj

2

B2

!D 1;

and hence ja1b1 C a2b2 C � � � C anbnj � AB .

9.16. The proof is by contradiction. Assume maxfAx ; Ay ; Azg < V 2=3,then all three projections have area less than V 2=3. ConsequentlypAxAyAz < V , contradicting the Loomis-Whitney inequality (9.2).

9.17. If there were such a formula, then solving it simultaneously withEuler’s polyhedral formula would yield an equation in just two vari-ables. Hence one of the variables uniquely determines the other. Thethree projections of polyhedral skeletons in Figure S9.6 show that thiscannot happen. Parts (a) and (b) show that V does not uniquely de-termine E or F , parts (b) and (c) show that E does not uniquely de-termine V or F , and parts (a) and (c) show that F does not uniquelydetermine V or E [Barnette, 1983].

(a) (b) (c)

Figure S9.6.

Chapter 10

10.1. The net in Figure S10.1 can be folded to form a (nonconvex) deltahe-dron that consists of a regular tetrahedron with two additional regulartetrahedra glued to two of its faces.


Figure S10.1.

10.2. The net in Figure S10.1 can also be folded to yield a regular octahe-dron.

10.3. If all the faces of a convex polyhedron are congruent squares, then V �ECF D 2, 4F D 2E, and so 2V �E D 4. But since the polyhedronis convex, V D V3, so that 3V D 2E and hence 2V � 3V =2 D 4 andV D 8, E D 12, and F D 6. Thus the only solution is the cube.


Figure S10.2.

10.5. Here are two solutions, there are infinitely many more [Hunter andMadachy, 1975]. Cut on the solid lines and fold on the dashed lines:

Figure S10.3.

10.6. Assume that the sum of the angles at each vertex of a tetrahedronABCD is 180ı. If we unfold the tetrahedron by cutting along the threeedges emanating from D, we obtain a triangle as shown in FigureS10.4, since the three angles at A, B , and C sum to 180ı.

The result is then a triangle D1D2D3 with A, B , and C as mid-points of the sides. Hence jABj D 1=2 � jD2D3j D jD2C j D jDC j,that is, opposite edges AB and DC have the same length. Similarly,


the other two pairs of opposite edges have the same length, and thetetrahedron is isosceles.

A B

C

D 1

D 2 D 3

Figure S10.4.

Conversely, if the tetrahedron is isosceles then the three angles at avertex are the same as the three angles in the opposite face, and hencesum to 180ı.

10.7. Let r denote the radius of the earth. Then the height of the tropi-cal zone is h D 2r sin.23:4378ı/, so that the fraction of the earth’ssurface that is tropical is 2�rh=4�r2 D h=2r D sin.23:4378ı/ �39:8%.

10.8. The two graphs have exactly the same length. Set r D 1 in Figure10.7.2 and the discussion that follows.

10.9. 30 ft. There are three feasible ways to unfold the room yielding pathsof length 30 ft and approximately 31.3 ft and 33.5 ft.

10.10. The shortest path has length 3p2. See Figure S10.5.

Figure S10.5.

10.11. As seen in the unfolding of the box illustrated in Figure S10.6a, thelength of the shortest path from A to X is

p8. Let B be the point one-

fourth of the way down the diagonal, as shown in the unfolding of


the box illustrated in Figures S10.6b and S10.6c. In either of the tworoutes shown, the length of the shortest path from A to B is

p8:125.

By symmetry there are two other routes along hidden sides of thebox. Hence the point B where the shortest path from A to B is longestis not X [Gardner, 1996].

(a) (b) (c)

A

B

X

A

A

B

Figure S10.6.

10.12. Cut the cone along the dashed line in Figure 10.12.5 directly oppositethe line from the vertex through the given point, and open the coneas shown in Figure S10.7. The cone is now a sector of a circle, andthe shortest path consists of the two lines perpendicular to the linesresulting from the cut [Steinhaus, 1969]. The condition h > r

p3

insures that the sector is less than half of a circular disk.

Figure S10.7.

10.13. No. The two tetrahedra have the same edge lengths, but the volumeof the tetrahedron in Figure 10.12.6a is 1=6 while the volume of thetetrahedron in Figure 10.12.6b is

p5=12.

10.14. No. Let S be an equilateral triangle and T an isosceles right triangle,both of area 4. The lines joining the midpoints of the sides of eachtriangle form tetrahedral nets. S folds into a regular tetrahedron withpositive volume, T folds into a degenerate tetrahedron (a square) withzero volume, but both tetrahedra have all faces with area 1.


10.15. The leftmost nets in the two rows of Figure S10.2 can be modifiedto yield strips that wrap around the cube, as shown in Figure S10.8a.Each strip can be then cut into n congruent strips, as illustrated inFigure S10.8b [Gardner, 2001].

(a) (b)

Figure S10.8.

10.16. Yes. The triangular prism with edges of unit length whose net isshown in Figure S10.9 has the property. For a proof that this is theonly other polyhedron with the property in the problem, see [Kruse-meyer et al., 2012].

Figure S10.9.

10.17. Since V D .�h2=3/.3r � h/ and S D 2�rh, we have, using theAM-GM inequality (3.5) for three numbers,

V 2

S3D2h.3r � h/2

144�r3�Œ.2hC .3r � h/C .3r � h//=3�3

144�r3D

1

18�:

We have equality if and only if 2h D 3r � h, or h D r , i.e., when thespherical cap is a hemisphere.

10.18. It is easy to see that the numbers for the tetrahedron, octahedron,and cube are 4, 2, and 3, respectively. For the dodecahedron and theicosahedron, the numbers are 4 and 3, respectively, as seen in theirnets in Figure S10.10.


Figure S10.10.

10.19. Draw radii as shown in Figure S10.11, let y be the radius of the circleof intersection of S and T , and let h be the height of the spherical capof S lying inside T .

Rr

h

y

r–h

ST

Figure S10.11.

Hence y2Ch2 D R2 and y2C.r�h/2 D r2. Solving these equationsfor h yields h D R2=2r . Thus the area of the spherical cap is 2�rh D�R2, independent of r and equal to the area enclosed by a great circleof T .

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Index

Admissible number 77, 78, 81, 238Algebraic identities and boxes

45–49Angles between diagonals of a box

25, 58in lunes 122of a tetrahedron 124–126, 219

Antiprism 18, 19, 87, 88, 141, 199Antiprisms in Platonic solids 87, 88Archimedes’ screw 211Arithmetic mean 8, 31, 50, 51, 59,

60, 128, 191, 252Arithmetic mean-geometric mean

inequality for three numbers 50,51

Arithmetic mean-geometric meaninequality for two numbers 8

Arithmetic mean-root mean squareinequality 59, 60

Azimuthal projection 170equidistant projection 172, 189

Barrel 62, 235Bicylinder 5, 111, 112, 120, 121,

211–214, 239Binomial coefficient 38, 39Blivet 184Box inequalities 55–59Buckminsterfullerene 190

Calissons 28, 29, 40, 230Camera obscura 168Cannonballs 35, 36, 41, 43, 234Cantor dust 144Cantor principle 27

Cartography 169–176Cauchy-Schwarz inequality 62, 95,

96, 115, 182, 183, 192, 241,242

Cavalieri’s principle 89–92, 111,112, 114, 240

Chromatic number 133Circular helix 209–211Circumcenters in the faces of a

tetrahedron 126, 127Circumsphere of a tetrahedron 126,

127Coil 209Colors in space 6, 133, 134, 216,

225, 257Commandino’s theorem 97–99Completely incongruent boxes 135,

136Concurrent lines 118, 119Cone 20, 21, 63, 79, 92, 99–103,

157, 158, 169, 170, 175, 176,205, 206, 224, 240, 256

Conformal projection 171, 174, 175,189

Conic sections 99–103Conical canvas tent 52, 53

projection 169, 175, 176Convex polyhedron 10, 125, 177,

178, 190–192,Cosine law and gnomic projection

178, 179Counting calissons 28, 29, 40,

230cannonballs 35, 36, 41, 43, 234

Csaszar polyhedron 216, 217

265

266 Index

Cube 3, 4, 8, 10, 11, 16, 18, 23–25,29–35, 41, 42, 45–50, 57–61,67, 68, 71, 75, 77, 78, 80,83–85, 87, 104, 105, 120,130–132, 135, 137, 143–145,152, 155, 156, 159–161, 182,183, 186, 195, 201–203, 222,225

Cube algebra 45–47Cubic numbers 3, 45–49Cubing a cube 135

the sphere 93Cubism 30Cuboctahedron 22, 80, 191, 237, 252Cylinder 5, 20, 21, 62, 79, 89, 90, 92,

102, 103, 120, 121, 127, 128,140–143, 162, 169–193, 204,205, 209–214

Cylindrical projection 173–175wedge 92, 93, 241

Dandelin-Quetelet theorem 101–102de Gua’s theorem 93–98, 114, 182,

240Degree 3 power mean 59, 60Degree 3 power mean-root mean

square inequality 59, 60Degree of a polyhedral vertex

188Delian problem 201–203Deltahedron 12, 196–200Developable surface 169, 204Devil’s tuning fork 184Dihedral angle 117, 122–125, 131,

171, 177–179, 252Directrix of a conic 101–103, 157,

238Dissections 4, 11, 12, 65–81,

154–156Dodecahedron 14–16, 71, 72, 87,

104–107, 115, 187, 190, 194,195, 227, 242, 251, 257

Doughnut 111, 162Drawing a regular icosahedron 15,

16Dudeney’s fly and spider 217, 218Dudeney’s puzzles 61, 79, 217, 218,

237Duplication of the cube 201–203

principle 27, 30, 32, 135Durer’s solid 177

Eccentricity of a conic 102Element of a cone, 102

of a cylinder, 102, 209, 210Elements of Euclid 10, 17, 19, 67,

79, 104–107, 115, 126, 129Ellipse 99, 100, 101, 102, 103, 158,

212, 213, 248Enneahedron 191, 252Enumeration 27–43Equirectangular projection 173Euler bricks 9Euler’s rotation theorem 156, 157

polyhedral formula 177, 178, 250,251, 253

Face angles 124, 222Fair division of a cake 153Fat elephant inequality 192False perspective 168, 169Fibonacci numbers 46, 154Figurate numbers 27, 28, 34–36Flexible polyhedron 217Focus of a conic 100, 102, 103,

157Foldable objects 206Folding 193–226

a regular pentagon 200, 201paper in half twelve times

219–221Four color conjecture and theorem 6,

133Fractal 136, 140, 143–145, 245, 246

Index 267

Frustum of a cone 175, 204–208Frustum of a pyramid 72–74, 87,

88Fubini principle 27, 29, 30

General position of lines 118, 119,131

of spheres 131Geometric mean 8, 31, 50, 51, 59,

74Geometric mean-arithmetic mean

inequality 8, 50, 51Geometric mean-harmonic mean

inequality 59, 60Girard’s formula and theorem 122,

123, 244Gnomonic projection 170, 178, 180,

189God’s algorithm and number 160,

161Golden ratio 15, 20, 63, 70, 71, 104,

153, 154, 242Great circles 117, 122, 170, 171,

177, 178, 189Greenwich meridian 99, 169, 173Groin vault 5Guba’s inequality 57, 58, 61, 240Gyrobifastigium 18Gyroelongated square bipyramid

199

Hadwiger problem 77, 78Hamiltonian cycle or circuit 187,

188, 191, 252Harmonic mean 59Harmonic mean-geometric mean

inequality 59, 60Hausdorff dimension 145Helices 209–211Helicoidal ramps 209, 210Heronian mean 74Heron’s formula 62, 96, 97

Hex numbers 27, 28, 40Hexagonal section of a cube 83–85Hexagonal truncated trapezohedron

251Hexahedron 10Hilbert’s third problem 79Hinged dissection 154–156Human poverty index 59Huzita-Hatori axioms 194Hyperbola 99–102, 159Hyperbolic paraboloid 203, 204Hyperboloid of one sheet 203, 204

Icons of solid geometry ixIcosahedron 10, 15, 16, 87, 104, 106,

115, 190, 195, 197, 199, 227,242, 257, 258

Impossible cubed box 135object 184

Inclusion principle 49–52Incongruent boxes 135, 136Inequalities for rectangular boxes

55–59for isosceles tetrahedra 96, 97

Inequality between two means 59,60

Inscribing the Platonic solids in asphere 104–107

Instant Insanity puzzle 158–161,247

Intersecting cylinders 5, 11, 112,117, 120, 121, 212

spheres 186Intersection 117–132Isometry 147Isoperimetric inequality in space 90Isosceles tetrahedron 76, 77, 81, 90,

91, 96, 97, 81, 161, 222, 238,243, 246, 255, 256

Iteration 133–145, 245, 246

Kissing number problem 42, 43, 115

268 Index

Lambert’s equal-area azimuthalprojection 172, 173

equal-area cylindrical projection173, 174

Lateral area of a cone 52Latitude 99, 169–175, 189Law of cosines, spherical 180Longitude 169–175Loomis-Whitney inequality

182–184, 253Loyd’s puzzles 48, 49

Map 5, 6, 133, 134Mapping the earth 169–176Means for three numbers 59, 60Median in a tetrahedron 97–99Mechanical drawing 176Menger sponge 136–139, 144, 145Mercator’s projection 174, 175Method of exhaustion 79

of indivisibles 89Midsection 86–88Million points in space problem 148,

149Minimal paths on boxes 224, 255,

256on cones 224, 256on cylinders 210

Miura fold 221, 222Mobius band or strip 214, 215

polyhedron 214Moscow papyrus 72Motion 147–163Mozartkugel 208

Nappes of a cone 100, 102Net 18, 194–197, 199, 214–217, 219,

222, 225, 253, 254, 256, 257Nondevelopable surface 203

Oblong numbers 31, 32, 41, 231Octagonal prism 17

Octahedron 10, 13, 14, 16, 19, 69,70, 80, 81, 87, 104, 105, 110,131, 138, 143, 145, 199, 225,238, 239, 254, 257

Optimization problems 52–55Origami 193, 194, 221Orthogonal projection 171, 176,

189Orthographic projection 15,

180–185Outer angles of a tetrahedron 125,

126Outer trihedral angle 125, 126

Padovan’s sequence 154, 161Paper folding 193–222Parabola 99–102Parallelepiped 65–67, 76

law 108, 109Parallelogram law 109Partitioning space with planes

37–39with spheres 42, 232, 233

Penrose triangle 184Pentagon-hexagon-decagon identity

115Pentagonal bipyramid 199

numbers 33, 34Pentahedron 23, 190, 227Perfect cube 135Perfect rectangles 134, 135, 136Perspective 17, 165–169Planar projection 169–175Plane section 83–116Planes in space 37–39Plastic number 153, 154Platonic fractals 136–138Platonic solids 9–16, 18, 23, 88, 138,

187, 189, 190, 196, 225, 246Platonic solids and antiprisms 87,

88Platonic solids in a sphere 104–107

Index 269

Plato’s cubes 48, 49Polycube 11, 12, 23, 24, 134,

228Polyhedral nets 14, 195, 197–199,

222, 254Polyhedron, definition 9, 10Printing 3D objects 110Prism 4, 17–19, 66, 67, 85–87, 257Prismatoid 85–88, 113, 239Prismoid 85Prismoidal formula 85–88, 113Projections 165–192Pronic number 31Puzzles 11, 12, 48, 49, 61, 79, 80,

158–160, 217, 218, 237Pyramid 1, 2, 4, 6, 7, 12, 19, 20,

31–33, 51, 65–70, 72, 73, 80,81, 86, 87, 106, 115, 150, 155,227, 250

Pythagorean theoremfor parallelograms in space

180–182in space 93–96, 180–182on a sphere 178–180

Pythagorean triple 9

Quarter-square numbers 41, 231, 232

Radius of a sphere 4, 5, 71, 107, 108,121–123, 127, 128

Rectangular number 31Recycling symbol 215Regular

polyhedron 10tiling 14

Represention ofcubic numbers 3, 45–48, 50hex numbers 27, 28oblong numbers 31, 32pentagonal numbers 32, 33square numbers 3, 30triangular numbers 34, 35

Reptiles 23, 24, 228Rhombic dodecahedron 74–76, 155,

187, 188Right tetrahedron 94–96Roman dodecahedra 14, 15Root mean square 59Root mean square-arithmetic mean

inequality 59, 60Root mean square-degree 3 power

mean inequality 59, 60Rotation 147–151, 154–157Rubik’s cube 160Ruled surface 203–208

Sanford’s fair division 153Schwarz lantern 140–144

paradox in cylinders 140–144Sections of a torus 117, 118Self-similar solids 133, 139, 140,

143Shadows 166, 147, 170,Shephard’s conjecture 195, 196Sinusoid 209, 212, 213,Skeleton of a polyhedron 177, 187,

188, 189, 191, 194, 195, 250,253

Skew lines in space 118, 119Sierpinski

octahedron 138tetrahedron 138

Slant height of a right circular cone63

Slicing 83Slothouber-Graatsma puzzle 237Snub disphenoid 199Solid angle 125Solid geometry ix,xSoma cube 11, 12, 24Speedcubing 161Spider and the fly problem 217, 218,

223

270 Index

Sphere 21, 24, 25, 42, 43, 101, 102,104–107, 115, 131, 156, 157,178–180, 186, 232, 233, 243

as a locus of points 129, 130radius of a 107, 108, 121–123,

127, 128surface area of a 207, 208volume of a 79, 89–91

Spherical cap 112, 113, 121, 240,258

law of cosines 179lunes 122Pythagorean theorem 180segment 112, 239, 240zone 207, 208

Spherometer 127, 128Square numbers 3, 8, 27, 30, 31, 33,

34, 40, 41, 57, 61, 94–97, 132,181, 182, 230, 231, 234

pyramid 6, 69, 70, 73, 81, 87, 92,238

pyramidal number 36, 43, 234Squaring a circle 93, 201

a square 134, 135Steffen’s flexible polyhedron 217Steinmetz solid 111Stella octangula 70, 71Stellated octahedron 70Stereographic projection 170, 171,

174, 189Steradian 125Sums of cubes 34, 35

of hex numbers 27, 28of oblong numbers 31, 32of outer angles of a tetrahedron

125, 126of pentagonal numbers 33, 34of squares 30of sequences of integers 29–35of triangular numbers 32, 33

Sundial 12, 166Surface area

of bicylinders and tricylinders211–214

of cylinders, cones and spheres203–209

of fractals 136, 139, 140, 144, 245of a frustum of a cone 205, 206of spherical triangles 121–123of the Schwarz lantern 141–143

Szilassi polyhedron 215–217

Tetrahedral kites 68, 69number 36, 41, 43, 234

Tetrahedron, 6, 10–13, 16, 67–70,81, 84, 87, 91–98, 104, 105,114, 124–127, 131, 138–140,149–151, 155, 184, 185, 188,191, 197, 199, 200, 215, 219,225

inequality 191, 251isosceles 76, 77, 81, 90, 91, 96,

97, 161, 222, 238, 246, 247, 255right 93–96

Torus 111, 117, 118, 162, 163, 239,247, 248

Triangular bipyramid 199Triangular numbers 32–34, 36, 38,

41, 230, 232, 233Triaugmented triangular prism 199Tricylinder 5, 120, 121, 211–214Trihedral angle 117, 124, 125, 131Trinomial 47Trirectangular tetrahedron 93Tropical zone 222, 223, 255Truncated cone 100

icosahedron 22, 190octahedron 22, 80, 81, 155, 156,

237, 238pyramid 72–74, 87, 88

Uniform polyhedron 19Unfolding figures 193–226Universal curve 138

Index 271

Upper bound for the volume of atetrahedron 184, 185

Using cubes to sum integers 29–33

Van der Laan’s plastic number 153,154

Vertex angles of a tetrahedron 219Villarceau circles 118Viviani’s theorem for a regular

tetrahedron 149–151, 162, 246Voicu’s inequality 58, 59Volume of a barrel 62

of a bead 111, 239of a bicylinder 5, 112of a cone 52, 63, 79, 90, 92, 113,

240of a cuboctahedron 80of a cylinder 55, 79, 92of a cylindrical wedge 92, 93of a frustum of a square pyramid

72–74, 87of a parallelepiped 65, 66of a plug 113, 240, 241of a prism 4, 66of a prismatoid 86of a pyramid 4, 32, 33, 51, 66–70,

72, 75, 77, 86, 87, 150, 154

of a regular dodecahedron 71, 72of a regular icosahedron 106of a regular octahedron 67, 69, 70,

81, 110of a regular tetrahedron 67, 68, 69,

91, 114of a rhombic dodecahedron 75of a sphere 24, 70, 89, 90of a spherical cap 113, 121, 240of a spherical segment 112of a stella octangula 70of a tetrahedron 184, 185of a tricylinder 120, 12of a torus 111, 239of a truncated octahedron 80, 155,

156, 237of a wedge 72, 93of an antiprism 88of an isosceles tetrahedron 77, 91,

96, 97of fractals 136, 139of the Schwarz lantern 144, 245

Wallace-Bolyai-Gerwien theorem79

Wedge 72, 92, 93, 113, 241Wrapping a sphere 208

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 Catalonia,has 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 (with Roger Nelsen) MAA, 2006; Vitaminas Matematicas and El Clubde la Hipotenusa, Ariel, 2008, Geometria para Turistas, Ariel, 2009; WhenLess is More (with Roger Nelsen) MAA, 2009; Asesinatos Matematicos,Ariel, 2010; Charming Proofs (with Roger Nelsen) MAA, 2010; and Iconsof Mathematics (with Roger Nelsen) MAA, 2011.

Roger B. Nelsen was born in Chicago, Illinois. He received his B.A. inmathematics from DePauw University in 1964 and his Ph.D. in mathematicsfrom Duke University in 1969. Roger was elected to Phi Beta Kappa andSigma Xi, and taught mathematics and statistics at Lewis & Clark Collegefor forty years before his retirement in 2009. His previous books includeProofs Without Words, MAA 1993; An Introduction to Copulas, Springer,1999 (2nd. ed. 2006); Proofs Without Words II, MAA, 2000; Math MadeVisual (with Claudi Alsina), MAA, 2006; When Less Is More (with ClaudiAlsina), MAA, 2009; Charming Proofs (with Claudi Alsina), MAA, 2010;The Calculus Collection (with Caren Diefenderfer), MAA, 2010; Icons ofMathematics (with Claudi Alsina), MAA, 2011, and College Calculus (withMichael Boardman), MAA, 2015.

272

AMS / MAA DOLCIANI MATHEMATICAL EXPOSITIONS

Solid geometry is the traditional name for what we call today

the geometry of three-dimensional Euclidean space. This book

presents techniques for proving a variety of geometric results

in three dimensions. Special attention is given to prisms, pyra-

mids, platonic solids, cones, cylinders and spheres, as well as

many new and classical results. A chapter is devoted to each

of the following basic techniques for exploring space and

proving theorems: enumeration, representation, dissection,

plane sections, intersection, iteration, motion, projection, and

folding and unfolding.

The book includes a selection of challenges for each chapter

with solutions, references and a complete index. The text is

aimed at secondary school and college and university teachers

as an introduction to solid geometry, as a supplement in

problem solving sessions, as enrichment material in a course

on proofs and mathematical reasoning, or in a mathematics

course for liberal arts students.

solid geometry in the 21st century - ams · 2020. 7. 29. · new horizons in geometry,tom m....

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