AMS / MAA DOLCIANI MATHEMATICAL EXPOSITIONS VOL 50

A MATHEMATICAL SPACE ODYSSEY Solid Geometry in the 21st Century

Claudi Alsina and Roger B. Nelsen 10.1090/dol/050 A Mathematical Space Odyssey Solid Geometry in the 21st Century About the cover Jeffrey Stewart Ely created Bucky Madness for the Mathematical Art Exhibition at the 2011 Joint Mathematics Meetings in New Orleans. Jeff, an associate professor of computer science at Lewis & Clark College, describes the work: “This is my response to a request to make a ball and stick model of the buckyball carbon molecule. After deciding that a strict interpretation of 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 each node is itself a buckyball. The center sphere reflects this model in its surface and also recursively reflects the whole against a mirror that is behind the observer.” See Challenge 9.7 on page 190.

c 2015 by The 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): 10987654321 The Dolciani Mathematical Expositions

NUMBER FIFTY

A Mathematical Space Odyssey Solid Geometry in the 21st Century

Claudi Alsina Universitat Politecnica` de Catalunya Roger B. Nelsen Lewis & 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,ˆ 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 from Mary P. Dolciani, Professor of Mathematics at Hunter College of the City University of New York. In making the gift, Professor Dolciani, herself an exceptionally talented and successful expositor of mathematics, had the purpose of furthering the ideal of excellence 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 with distinction, both as a member of the Committee on Publications and as a member of the Board of Governors. It was with genuine pleasure that the Board chose to name the series in her honor. The books in the series are selected for their lucid expository style and stimulating mathematical content. Typically, they contain an ample supply of exercises, many with accompanying solutions. They are intended to be sufficiently elementary for the undergraduate and even the mathematically inclined high-school student to under- stand and enjoy, but also to be interesting and sometimes challenging to the more advanced 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 and Dale 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 Joseph Silverman 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. Benjamin and 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, Dan Kalman 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 Roger B. 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 and Roger 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. Gouveaˆ 49. A Guide to Functional Analysis, (MAA Guides #9), Steven G. Krantz 50. A Mathematical Space Odyssey: Solid Geometry in the 21st Century, Claudi Alsina and Roger B. Nelsen

MAA Service Center P.O. Box 91112 Washington, DC 20090-1112 1-800-331-1MAA FAX: 1-301-206-9789 Throw your dreams into space like a kite, and you do not know what it will bring back, a new life, a new friend, a new love, a new country. Ana¨ıs Nin

Space is the breath of art. Frank Lloyd Wright

It is often helpful to think of the four coordinates of an event as specifying its position in a four-dimensional space called space-time. It is impossible to imagine a four-dimensional space. I personally find it hard enough to visualize three- dimensional space! Stephen W. Hawking A Brief History of Time (1989) Preface

Cubes, cones, spheres, cylinders or pyramids are the great primary forms which light reveals to advantage; the image of these is distinct and tangible within us without ambiguity. It is for this reason that these are beautiful forms, the most beautiful forms. Le Corbusier (Charles-Edouard´ Jeanneret-Gris)

Solid geometry is the traditional name for what we call today the geometry of three-dimensional Euclidean space. Courses in solid geometry have largely disappeared from American high schools and colleges. But we are convinced that a mathematical exploration of three-dimensional geometry merits some attention in today’s curriculum. This book is devoted to presenting tech- niques for proving a variety of mathematical results in three-dimensional space, 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, and we 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, in studying the elements of Solid Geometry, from the practice ...ofnever submitting to the eye of the student, the figures on whose properties he is reasoning. An analogous claim was later expressed by David Hilbert (1862–1943): The tendency toward intuitive understanding fosters a more immediate grasp 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 a visual 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 of examples of some of the types of questions and problems we address in the subsequent chapters. We also introduce (or re-introduce) the reader to the primary objects that inhabit space. We then develop, in nine chapters, the methods in the chapter titles that are fruitful for exploring and proving many basic facts and theorems using three-dimensional figures. By means of these methods we review properties concerning figurate numbers, means and inequalities, volumes and surface areas, and a variety of the celebrated results in classical solid geometry. In addition to many figures illustrating theorems and their proofs, we have included a selection of photographs of three-dimensional works of art and architecture. Readers should be familiar with high school algebra, plane and analytic geometry, and trigonometry. While brief appearances of calculus occur in Chapters 5, 9, and 10, no knowledge of calculus is necessary to enjoy this book. Each chapter concludes with a selection of Challenges for the reader to ex- plore further properties and applications of each method. After the chapters we give hints and solutions to all the Challenges in the book. A Mathematical Space Odyssey concludes with references and a complete index. As in our previous books with the MAA, we hope that both secondary school and college and university teachers may wish to use the book for a 21st century journey into solid geometry with their students. The book may also be used as a supplement for problem-solving sessions or classroom discussions, augmented by hands-on materials and modern software. Special thanks to Harriet Pollatsek and the members of the editorial board of the Dolciani series for their careful reading of earlier drafts of the book and their many helpful suggestions. We would also like to thank Stephen Kennedy, Carol Baxter, Beverly Ruedi, and Samantha Webb of the MAA’s book publication staff for their expertise in preparing this book for publi- cation. Finally, special thanks to Don Albers, the MAA’s former editorial director for books, who as on previous occasions encouraged us to pursue this project.

Claudi Alsina Universitat Politecnica` de Catalunya Barcelona, Spain Roger B. Nelsen Lewis & Clark College Portland, Oregon Contents

Preface ix 1 Introduction 1 1.1Tenexamples...... 1 1.2 Inhabitants of space ...... 9 1.3 Challenges ...... 22 2 Enumeration 27 2.1Hexnumbers...... 27 2.2Countingcalissons...... 28 2.3 Using cubes to sum integers ...... 29 2.4 Counting cannonballs ...... 35 2.5 Partitioning space with planes ...... 37 2.6 Challenges ...... 40 3 Representation 45 3.1 Numeric cubes as geometric cubes ...... 45 3.2 The inclusion principle and the AM-GM inequality for threenumbers...... 49 3.3Applicationstooptimizationproblems...... 52 3.4 Inequalities for rectangular boxes ...... 55 3.5 Means for three numbers ...... 59 3.6 Challenges ...... 60 4 Dissection 65 4.1 Parallelepipeds, prisms, and pyramids ...... 65 4.2 The regular tetrahedron and octahedron ...... 67 4.3 The regular dodecahedron ...... 71 4.4Thefrustumofapyramid...... 72 4.5 The rhombic dodecahedron ...... 74 4.6 The isosceles tetrahedron ...... 76 4.7TheHadwigerproblem...... 77 4.8 Challenges ...... 79

xi xii Contents

5 Plane sections 83 5.1 The hexagonal section of a cube ...... 83 5.2Prismatoidsandtheprismoidalformula...... 85 5.3 Cavalieri’s principle and its consequences ...... 89 5.4 The right tetrahedron and de Gua’s theorem ...... 93 5.5 Inequalities for isosceles tetrahedra ...... 96 5.6 Commandino’s theorem ...... 97 5.7Conicsections...... 99 5.8 Inscribing the Platonic solids in a sphere ...... 104 5.9 The radius of a sphere ...... 107 5.10Theparallelepipedlaw...... 108 5.11 Challenges ...... 110 6 Intersection 117 6.1Skewlines...... 118 6.2 Concurrent lines in the plane ...... 119 6.3 Three intersecting cylinders ...... 120 6.4 The area of a spherical triangle ...... 121 6.5 The angles of a tetrahedron ...... 124 6.6 The circumsphere of a tetrahedron ...... 126 6.7 The radius of a sphere, revisited ...... 127 6.8 The sphere as a locus of points ...... 129 6.9 Prince Rupert’s cube ...... 130 6.10 Challenges ...... 131 7 Iteration 133 7.1 Is there a four color theorem in space? ...... 133 7.2 Squaring squares and cubing cubes ...... 134 7.3 The Menger sponge and Platonic fractals ...... 136 7.4Self-similarityanditeration...... 139 7.5 The Schwarz lantern and the cylinder area paradox . . . . . 140 7.6 Challenges ...... 143 8Motion 147 8.1 A million points in space ...... 148 8.2 Viviani’s theorem for a regular tetrahedron ...... 149 8.3 Dissecting a cube into identical smaller cubes ...... 152 8.4Fairdivisionofacake...... 153 8.5Fromthegoldenratiototheplasticnumber...... 153 8.6 Hinged dissections and rotations ...... 154 Contents xiii

8.7 Euler’s rotation theorem ...... 156 8.8 The conic sections, revisited ...... 157 8.9InstantInsanity...... 158 8.10 Challenges ...... 161 9 Projection 165 9.1Classicalprojectionsandtheirapplications...... 165 9.2 Mapping the earth ...... 169 9.3 Euler’s polyhedral formula ...... 177 9.4 Pythagoras and the sphere ...... 178 9.5 Pythagoras and parallelograms in space ...... 180 9.6 The Loomis-Whitney inequality ...... 182 9.7 An upper bound for the volume of a tetrahedron ...... 184 9.8Projectionsinreverse...... 185 9.9 Hamiltonian cycles in polyhedra ...... 187 9.10 Challenges ...... 189 10 Folding and Unfolding 193 10.1 Polyhedral nets ...... 194 10.2 Deltahedra ...... 196 10.3 Folding a regular pentagon ...... 200 10.4 The Delian problem: duplicating the cube ...... 201 10.5 Surface areas of cylinders, cones, and spheres ...... 203 10.6Helices...... 209 10.7 Surface areas of the bicylinder and tricylinder ...... 211 10.8 Folding strange and exotic polyhedra ...... 214 10.9Thespiderandthefly...... 217 10.10 The vertex angles of a tetrahedron ...... 219 10.11 Folding paper in half twelve times ...... 219 10.12 Challenges ...... 222 Solutions to the Challenges 227 Chapter 1 ...... 227 Chapter 2 ...... 230 Chapter 3 ...... 234 Chapter 4 ...... 237 Chapter 5 ...... 239 Chapter 6 ...... 243 Chapter 7 ...... 245 Chapter 8 ...... 246 xiv Contents

Chapter 9 ...... 249 Chapter 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 1 1.1. (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 of edges in the tetrahedron. (c) six is the number of faces in the cube and the number of vertices in the octahedron, eight is the number of vertices in a cube and the number faces in the octahedron, and twelve is the number of edges in 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, and the 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 Figure 1.3.2. For the “T” and “L” shaped four-cube polycubes in the first row of Figure 1.2.3, arrange four copies into a 4 4 1 block as shown in Figure S1.2a, then four copies of this block form a 4 4 4 cube, and four such cubes form a larger version of the tile.

(a) (b) 1334 3377 1124 6774 6122 6884 6552 8855 bottom layer top layer

Figure S1.2.

For the Z shaped polycube in the first row of Figure 1.2.3, arrange eight copies into a 4 4 2 block as shown in Figure S1.2b, then two copies of this block form a 4 4 4 cube, and four such cubes form a large version of the tile. For each of the polycubes in the second row of Figure 1.2.3, two copies of the polycube form a 2 2 2 cube, and four such cubes 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=2m and 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 the spheres is the same for both boxes. 1.6. No. Let the maximum number of edges of a face of such a polyhedron be n. Then the other numbers of edges of faces are elements of the set f3; 4; ;n 1g. There are only n 3 numbers in this set, yet there must be at least n other faces meeting the face with n edges. So every polyhedron must have at least two faces with the same number of edges. 1.7. The answer is all six! As shown in the Figure S1.3a, the person may be inside a large cube, and from a corner see the six faces (walls, floor, and ceiling), or in S1.3b one may look at a cube in front of a mirror, or in S1.3c 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 vertices of a cube.

Figure S1.4.

1.9. See Figure S1.5.

x x

Figure S1.5.

1.10. Suppose not, so that every diameter of the sphere has two unpainted ends, or one end painted and one unpainted. If we reflect the sphere in its center, then the image of every painted point must be an unpainted point. Thus the image N of the set P of painted points is a subset of the set U of unpainted points. Hence N and P are disjoint subsets of the sphere with equal areas. But this is impossible since the area 230 Solutions to the Challenges

of P (and of N , and so also for U/ is more than half the area of the sphere. 1.11. Draw a third diagonal on another face, as shown in Figure S1.6. Since the triangle formed by the three diagonals is equilateral, the angle between two diagonals is 60ı.

Figure S1.6.

Chapter 2 2.1. 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 the first 2n 1 triangular numbers. Then (2.3) yields

2 2 2 1 C 3 CC.2n 1/ D T1 C T2 C T3 CCT2n1 n.2n 1/.2n C 1/ D : 3 Solutions to the Challenges 231

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

Figure S2.2.

2.5. See Figure S2.3.

Figure S2.3.

2.6. The number of cubes in (a) is 6.12/ D 1 2 3,in(b)6.12 C 22/ D 2 3 5, and in (c) 6.12 C 22 C 32/ D 3 4 7. In each step we “wrap” a k .k C 1/ .2k C 1/ box with six 1 .k C 1/ .k C 1/ boxes. After n steps we have 6.12 C 22 CCn2/ cubes in a n .n C 1/ .2n C 1/ box, from which (2.1) follows [Kalajdzievski, 2000].

1 2.7. The elements of the sequence f4qngnD1 are the odd squares dimin- 1 ished by 1 and the even squares, so fqngnD1 is the sequence of squares divided by 4 (the “quarter-squares”) rounded down. See Fig- ure S2.4 for an illustration of quarter-squares as square and oblong numbers. 232 Solutions to the Challenges

Figure S2.4.

2.8. See Figure S2.5.

Figure S2.5.

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

.n 1/n.2n 1/ n.n C 1/.2n C 1/ n.2n2 C 1/ C D : 6 6 3

2.10. We count the cubes by considering the location of, say, the upper northeast 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 to 1 n n n cube. Hence the total number of cubes is, by (2.4), the square 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 n circles in the plane. We obtain a maximum when each pair of circles intersect in two points and no three are concurrent. Clearly C.0/ D 1, C.1/ D 2, and C.2/ D 4. Suppose k1 circles partition the plane into C.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 into 2.k 1/ arcs. Each of these arcs divides in two one of the 2-regions formed by the first k 1 circles, so that C.k/ D C.k 1/ C 2.k 1/. Solutions to the Challenges 233

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 1 C 2Tn1

where Tk denotes the kth triangular number, and consequently 2 C.n/ D 2 C 2Tn1 D n n C 2.

We now find the maximum number S.n/ of “3-regions” (three- dimensional regions) in space determined by n spheres. We obtain the maximum number when each pair of spheres intersect in a circle, and those circles on spheres have the same properties as the circles in planes described above. Clearly S.0/ D 1, S.1/ D 2, and S.2/ D 4. Suppose k1 spheres partition space into S.k1/ 3-regions. Thus the surface 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), and each of those 2-regions divides in two the 3-regions into which the first k1 spheres had partitioned space. Thus S.k/ D S.k1/CC.k1/, or S.k/ S.k 1/ D 2 C 2Tk2. Summing from k D 1 to k D n yields

S.n/ S.n 1/ D 2 C 2Tn2

S.n 1/ S.n 2/ D 2 C 2Tn3

S.n 2/ S.n 3/ D 2 C 2Tn4 : :

S.3/ S.2/ D 2 C 2T1 S.2/ S.1/ D 2

.n 2/.n 1/n S.n/ 2 D 2.n 1/ C 2 6 where we have used (2.3) to sum the first n 2 triangular numbers. Thus S.n/ D 2n C .n 2/.n 1/n=3 D n.n2 3n C 8/=3. 234 Solutions to the Challenges

2.12. Within the tetrahedral pile of cannonballs in Figure 2.4.3, we can find six cannonballs touching a given cannonball in one layer, plus another three 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 n.n C 1/.2n C 1/ .2n/.2n C 1/.2n C 2/ 4 D : 6 6 You can illustrate this result by re-stacking the tetrahedral pile of can- nonballs into four square pyramidal piles, as indicated here for the eighth tetrahedral number 120: 120 D .1 C 3/ C .6 C 10/ C .15 C 21/ C .28 C 36/ D 4 C 16 C 36 C 64 D 4.1 C 4 C 9 C 16/:

Chapter 3 3.1. The result follows directly from (3.12) and the AM-GM inequality 2 2 2 (3.5), both applied to the three numbers dxy, dyz, and dxz. 3.2. Yes. If the package is in the shape of a right circular cylinder with height h in and base radius r in, then the volume is V D r2h in3 and the length + girth is S D h C 2r in, so that p p 3 V D 3 r r h .2r C h/=3 D S=3 D 36: Hence the largest acceptable cylindrical package has volume V D 363= 14851in3. 3.3. The number of cubes in one set of slices begins with n2 and ends with n.n C 1/, and in the other set begins with n.n C 1/ C 1 and ends with .n C 1/2 1. Solutions to the Challenges 235

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 in Figure 3.9.2. Then the length S of string is S D 2x C 4y C 6z D 12.If V denotes the volume, then 48V =48xyz = 2x 4y 6z. From (3.4) we have p p 3 48V D 3 2x 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 D 3z. Hence 3z 3z=2 z D 4=3, z D 2=3 ft, y = 1 ft, and x =2ft. 3.6. Label the dimensions of the box so that the area of the top and bottom is xy, of the front and back is xz, and of the other two sides is yz. Then V = xyz and the cost C is given by C D 2cxy C 2bxz C 2ayz.Bythe AM-GM inequality (3.5) we have C 3 8abcV 2 D .2cxy/.2bxz/.2ayz/ 3 with equality if and only if cxy D bxz D ayz. Thus the minimum cost is 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 h2 V 2 D 2r4h2 D 2 2r2 2r2 D 2 2r2 2r2 .s2 4r2/ 4 3 3 2r2 C 2r2 C .s2 4r2/ s2 2 D 2 ; 3 3 p 3 2 2 2 2 or V s =3p 3, with equality if and only if 2r D s 4r D h =4, i.e., h D 2r 2p. So the maximum volume is obtained when the height of the barrel is 2 times its diameter. 3.8. When the perimeter is given, so is s.Using(3.5)wehave p p p .s a/ C .s b/ C .s c/ 3=2 AD s .s a/.s b/.s c/ s 3 s2 D p 3 3 with equality if and only if a D b D c. Hence the triangle with maxi- mum area is an equilateral triangle. 236 Solutions to the Challenges

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 a2x2 C b2y2 C c2z2 to both sides yields (3.15). 3.10. Yes, it is possible. See Figure S3.1 for one solution, exhibiting the three 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 42s6=243. Since s2 D r2 Ch2 we have r2h 2 2r4h2 42 r2 r2 V 2 D D D .s2 r2/ 3 9 9 2 2 42 1 r2 r2 3 42s6 C C .s2 r2/ D ; 9 3 2 2 243 p with equality if r2=2 D s2 r2 D h2, i.e., when r D h 2. 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 two 1 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 and radius 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-eighth of a sphere of radius 1, volume 4=3. Hence the volume of B is abc C 2.ab C ac C bc/ C .a C b C c/ C 4=3. Solutions to the Challenges 237

Chapter 4 4.1. 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 diagonal direction.”

Figure S4.1.

4.2. Yes, see Figure S4.2 where the three 1 1 1 cubes are white and the six 122 boxes are different shades of gray depending on orientation. The complementary problem (constructing a 3 3 3 cube from the nine smaller pieces) is known as the Slothouber-Graatsma puzzle.

Figure S4.2. p 4.3. The side lengthp of the enclosing cube in Figure 4.8.2b is s 2 so its vol- 3 ume is 2s p2, and thep volume of each of the eight pyramids removed 3 3 is .1=6/.s=p2/ D ps 2=24. Hencep the volume of the cuboctahedron is V D 2s3 2 s3 2=3 D 5s3 2=3.

4.4. Using the notationp from Section 4.2 yields V D volO .3s/ 3 6volRSP .s/ D 8s 2. 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 whichp is shown in Figurep S4.3. Sincep the edge of the half-cube has length s 2, V D 8 1=2 .s 2/3 = 8s3 2. 238 Solutions to the Challenges

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 edges be a, a0; b, b0; and c, c0; and assume a C b C c D a C b0 C c0 D a0 C b C c0 D a0 C b0 C c. Then b C c D b0 C c0 and b C c0 D b0 C c,or equivalently 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 the tetrahedron is isosceles. 4.6. Using the notation in Figure 4.6.1, we have b2 C c2 a2 D 2z2 >0so that b2 C c2 >a2, which implies that the angle opposite side a in the face is acute. Similarly the angles opposite sides b and c are acute, so each 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/ C 32.13/ D 43, 38: 1.33/ C 37.13/ D 43, 39: 1.63/ C 18.33/ C 1.23/ C 19.13/ 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 and 16 16 16 as subcubes. Similarly 1.33/ C 1.43/ C 1.53/ D 63 but this does not make 3 3-admissible. 4.9. Figure 4.2.2 shows that a regular tetrahedron with edge length 2s can be dissected into four regular tetrahedra with edge length s and a regular octahedron 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 can be 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/. Solutions to the Challenges 239

4.10. See Figure S4.4.

Figure S4.4.

Chapter 5 5.1. Approximately 7.6 m.

5.2. As noted in Section 5.2 the regular octahedron is a right triangularp 2 antiprism.p If we let s denote itsp edge length, then A1 D A2 D s 3=4, 2 Am D 3s 3=8, and h D s 6=3. Hence 1 s p s2 p 3s2 p s3 p vol .s/ D 6 2 3 C 4 3 D 2: O 6 3 4 8 3

5.3. At a height xp2 Œr; r the cross-sectionp of the torus is anp annulus with 2 2 2 2 2 2 2 2 area Œ.RC r x / .R r x / D 4R r p x , and 2 2 the area of thep rectangular cross-section of the cylinder is 2 r x 2R D 4R r2 x2. 5.4. At a height x in Œh=2; h=2 the cross-section of the bead is an annulus with area .r2 x2/ a2 D Œ.h=2/2 x2 (since a2 C .h=2/2 D r2/, which is the same as the area of the circular disk cross-section of the 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 determined by the axes of the two cylinders are squares, and each square has area 4= times the area of the corresponding circular section of a sphere. Hence the volume of the bicylinder is .4=/ 4r3=3 D 16r3=3. 5.6. Following the hint, the volume of the spherical segment is the same as the volume of the corresponding segment of the tetrahedron. But this segment is a prismatoid, so we can use the prismoidal formula 240 Solutions to the Challenges

to find its volume. Since the corresponding sections of the spherical and tetrahedral segments have the same area, we need only apply the 2 prismoidal formula to the spherical segment. Clearly A1 D a and 2 A2 D b , so we need only find Am.Todoso,letr denote the radius of the sphere, insert a y-axis through the poles of the sphere with its origin at the center, let the upper plane cut the axis at s and the lower plane at t, so that a2 C s2 D r2, b2 C t 2 D r2, and 2 2 2 2 h D s t. Then Am D .r Œ.s C t/=2 / D .=4/Œ4r .s C t/ . Thus

h h V D .A C 4A C A / D .a2 C b2 C 4r2 s2 2st t 2/ 6 1 m 2 6 h h D .3a2 C 3b2 C s2 2st C t 2/ D .3a2 C 3b2 C h2/: 6 6

5.7. The spherical cap is a spherical segment, as seen in Figure 5.11.4 with a D 0, so its volume V is V D .h=6/.3b2 C h2/.But r2 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 the volume of the cone. 5.8. Yes. The solids we are using for comparison with Cavalieri’s principle are spheres, and the prismoidal formula gives the correct volume for spheres. 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 area of the triangle two ways yields ch=2 D ab=2, so that .1=h/2 D c2=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 Figure 5.4.1. 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 half that of the corresponding sections of a right circular cylinder with base radius 1 in and height 2 in. Thus the volume of the plug is 1=2 12 2 D in3. Solutions to the Challenges 241

(b) The plug in Figure 5.10.6c is a right circular cylinder with two cylindrical wedges removed. The volume of the cylinder is 2 in3 and the volume of each wedge is, by Example 5.7, 4=3 in3, thus the volume of the plug is 2 8=3 in3. (c) There are infinitely many different shapes for the plug. Perhaps the simplest is just the union of the three shaded sections in Figure 5.10.6b. 5.12. No. Consider the cube and prism in Figure S5.1, where the bases of the prism are the same as for the cube, two of the lateral faces (front and 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 the rectangular 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 areap of PQR, then ŒABC D 4, ŒABD D 3, and ŒACD = ŒBCD D 7=2; hence ŒABC 2 D ŒABD2 C ŒACD2 C ŒBCD2. 5.14. Let V , B; and H be the volume, base area, and altitude of the tetrahedron 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 and hence 1 1 1 2 2 1 2 v D bh D B H D BH D V: 3 3 9 3 27 3 27 5.15. (a) Consider the triangle with vertices .0;0;0/, .a;b;c/, and .a C x;b C y;c C z/. (b) Since square roots are nonnegative we need only prove that

.a C x/2 C .b C y/2 C .c C z/2 p p 2 a2 C b2 C c2 C x2 C y2 C z2 (5.5) 242 Solutions to the Challenges

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

.a C 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 p p C 2 a2 C b2 C c2 x2 C y2 C z2 p p 2 D a2 C b2 C c2 C x2 C y2 C z2 :

To show that (5.5) implies (5.3), consider (5.5) for the absolute values jaj, jbj, jcj, jxj, jyj, jzj, expand the squares and use the fact that jax C by C czj jajjxj C jbjjyj C jcjjzj. 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 the icosahedron will kiss a unit sphere whose center coincides with the center of the circumsphere. But in Section 5.7 we learned that the p edge length s of the icosahedron satisfies s D d= 2 C '. Thus p s D 4= 2 C ' 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 7' C 4 2 3 Vdodec D p 2:785 2 ' 3 5'2 2 3 and Vicos D p 2:536: 6 2 C '

5.18. The proof uses the following easily verified facts from plane geometry and trigonometry: the side length of a regular n-gon inscribed in a circle of radius r is 2r sin.=n/, cos.=5/ D '=2, and sin.=10/ D 1=.2'/, where ' denotes the golden ratio. Then p D 2r sin.=5/ and d 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. Solutions to the Challenges 243

Chapter 6

6.1. No. Let each line be parallel to the other two, but with l3 not in the plane determined by l1 and l2. Then it is impossible for any line intersecting both l1 and l2 to also intersect l3. 6.2. Since each edge of the tetrahedron connects two vertices, summing SA D ˛ Cˇ C and three similar expressions for the other vertices yields T D 2D 4.

6.3. At most two. Spheres †1 and †2 intersect in a circle, and that circle intersects †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) is along the z-axis while the altitude from (1,1,0) is the line parallel to the y-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 isosceles triangles with edges 4, 4, and 7, and so all the edges in the faces satisfy the triangle inequality. Figure S6.1 shows that for a tetrahedron to exist whenp five sides have length 4, the sixth side must have length less than 4 3 6:93.

4 4 4 43 4 4

Figure S6.1.

6.6. Denote thePn fixed points by .ai ;bi ;ci / for i D 1;P 2; ;n, and let n N n 2 aN D .1=n/ iPD1 ai and similarly for b and cN. SinceP iD1 .x ai / D 2 n 2 n 2 n.xP Na/ C iD1 .ai Na/ (and similarly for iD1 .y bi / and n 2 iD1 .z ci / /, it follows that

Xn Xn Xn 2 2 2 2 .x ai / C .y bi / C .z ci / D k iD1 iD1 iD1 244 Solutions to the Challenges

is equivalent to

.x Na/2 C .y b/N 2 C .z Nc/2 Xn 2 1 2 2 2 D k Œ.ai Na/ C .bi b/N C .ci Nc/ ; n iD1

and the locus is a sphere centered at .a;N b;N c/N for sufficiently large k. 6.7. Similar spherical triangles (i.e., triangles with identical angles) will have 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 intersect in three parallel lines or intersect at a common point P . Consequently each of the pairwise intersection lines (for the planes of the light gray and medium gray faces, and the plane of the hidden face in the back of 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 not planar). Hence the figure does not represent a polyhedron [Gardner, 1992].

Figure S6.2.

6.9.p The side length of the regular hexagonal section of a unit cube is 2=2. If a square with side greater than 1 is inscribed in the hexagon, then it liesp within its circumcircle. But the radius of the circumcir- pcle is alsop 2=2, and every square within this circle has side at most 2=2 2 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 its area will be strictly less than 1. Solutions to the Challenges 245

6.10. The locus of points on a sphere equidistant from two given points is a great circle. Given the spherical triangle ABC,letK be the great circle bisecting AB, and let K0 be the great circle bisecting AC. K and K0 meet at two antipodal points; let P be the one inside ABC equidistant from A, B, and C . The great circle determined by P , its antipodal point, 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 volume n Vn equals V0.1=2/ . Hence the Sierpinski´ tetrahedron has finite surface area and zero volume. (b) If the generating octahedron has surface area A0 and volume V0, then after n iterations the surface area An equals n n A0.3=2/ and the volume Vn equals V0.3=4/ . Hence the Sierpinski´ octahedron has infinite surface area and zero volume. 7.2. The surface area of the fractal cube is the same as that of the original cube, 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 inscribed in each band is .h=6m/Œ2An C 4A2n, where Ak is the area of a regular k-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 take 2 the limit as n !1. Since An and A2n both approach r ,thevolume of 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 step in the iteration, a star-shaped hole and ring of six hexagons and six triangles replace each hexagonal region of the section, and a hexagon and three triangles replace each triangular region of the section. 7.7. (a) ln 4=ln 2 D 2,(b)ln6=ln 2 2:585,(c)ln4=ln 2 D 2,(d)ln9=ln 3 D 2.

2 7.8. Let Sn D 1 C 4 CCn . Then the number Tn of cubes in the nth 2 aggregation is Tn D Sn C2Sn1 CSn2 D .2n 1/.2n 2n C 3/=3. 246 Solutions to the Challenges

Figure S7.1. 7.9. See Figure S7.2.

Figure S7.2.

Chapter 8 8.1. a) In Section 4.6 we learned that the faces of an isosceles tetrahedron are 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

1 1 1 1 V D Kd C Kd C Kd C Kd 3 1 3 2 3 3 3 4

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 Platonic solids, 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 cubes have one red face, 12.n2/ cubes have two red faces, and 8 cubes have three red faces.

2 8.3. Let Vk D pkpkC1 denote the volume of the kth Padovan brick. It is easy to prove by mathematical induction that V0 C V1 CCVn D pnpnC1pnC2, so the total volume of the bricks equals the volume of the box. Figure S8.1, which illustrates the induction step in the proof, shows that the bricks actually fit into the box. Solutions to the Challenges 247

pn+3 p n+2

p n+1

pn pn+1

Figure S8.1.

8.4. Replacing cube IV by cube V changes the codes in Table 8.1 to the following table: Cube (1) (2) (3) I 1615 II 2 10 15 III256 V 3425 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 similar U-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 248 Solutions to the Challenges

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

Figure S8.3.

8.7. (a) The area of the top is 12 D and the lateral area is 2 11=2 D . (b) More generally we show how to “square” an a b rectangle: A semicirclep with diameter a C b yields a line segment with length ab, the side of the required square. With a D 1=2 and b D 2 p p we have ab D . See Figure S8.4.

ab ab

Figure S8.4.

(c) No. The argument is literally “circular,” since you need to be able to 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 in Section 8.8 yields x2 C .yN cos Nz sin /2 D r2 as the equation of the cylinder. Setting zN D 0 and replacing yN by y yields x2 C .cos2 /y2 D r2 as the equation of the intersection. This is a circle, an ellipse, or two parallel 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 p . x2 C .yN cos Nz sin /2 R/2 C .yN sin CNz cos /2 D r2

as the equation of the torus. Setting zN D 0 and replacing yN by y yields p x2 C y2 2R x2 C y2 cos2 C R2 D r2: Solutions to the Challenges 249

After rearranging terms and squaring both sides we have

.x2 Cy2/2 C2.x2 Cy2/.R2 r2/C.R2 r2/2 D 4R2.x2 Cy2 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;

or Œ.x2 C y2/ .R2 r2/2 .2rx/2 D 0; which factors to yield

Œ.x r/2 C y2 R2Œ.x C r/2 C y2 R2 D 0:

Hence the intersection is two circles with radius R whose centers are 2r units apart.

Chapter 9 9.1. See Figure S9.1, where the segments and arc labeled a) through e) have the required lengths in Table 9.1. a) b) λ d) e)

1 c) 2 λ π λ 42 λ

Figure S9.1.

9.2. No. If an ideal map did exist, it would project a spherical triangle onto a plane 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 edge joins two vertices, so the average number of edges per vertex is 2E=V , and this average is at least 3. (b) Three non-colinear vertices can only form a triangle in a plane, so V 4. Hence 2E 3V 12. 250 Solutions to the Challenges

(c) If E D 7 then 3V 14 so that V D 4. But since each edge joins a distinct pair of vertices, the maximum number of edges for 4 a polyhedron with four vertices is D 6. 2 (d) A pyramid whose base is a regular n-gon for n 3 has 2n edges, so all even numbers 6 and greater are possible values for E (see Figure S9.2a for a projection of the skeleton for the case n D 5/. Cutting off a base corner of the pyramid as shown in Figure S9.2b adds three edges, so the resulting polyhedron has 2n C 3 edges. Hence all odd numbers 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 since 1=E > 0 we have .1=n/ C .1=d/ > .1=2/. (c) Clearly n 3 and d 3,butn and d cannot both be greater than 3, 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 yield the five regular polyhedra studied earlier. 9.5. (a) The average number of edges per vertex is 2E=V ,so2E 3V .So by Euler’s formula, 3V D 6 C 3E 3F ,so6 C 3E 3F 2E and hence 6 C E 3F ,or2E 6F 12. (b) If 2E 6F 12, then 2E < 6F , so that 2E=F < 6, i.e., the average number of sides per face is less than 6. 9.6. When F D 5 Euler’s polyhedron formula reduces to E D V C 3. Since F D 5, all the faces must be triangles or quadrilaterals, and the degree of each vertex must be 3 or 4. Let Fn be the number of faces that are n-gons, and Vk the number of vertices of degree k.NowF3 C F4 D 5 and 3F3 C 4F4 D 2E, and the positive integer solutions to these equations are .F3;F4;E/= (4,1,8), (2,3,9), and (0,5,10). Similarly V3C Solutions to the Challenges 251

V4 D V and 3V3 C 4V4 D 2E D 2V C 6, from which it follows that V3 C 2V4 D 6. The positive integer solutions to this equation (and the corresponding 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 D 2E, hence 6F 2E D F5. Property (ii) implies 2E D 3V , multi- plying Euler’s formula by 6 yields 6V 6E C 6F D 12, thus 6F 2E D 12 and hence F5 D 12. A third polyhedron satisfying the two properties is the hexagonal truncated trapezohedron, a polyhedron with 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 the spherical triangle. Then the measures of the angles and sides (in radians) are approximately C D 2:1760485, a D :8499911, and b D :7751140, so that (9.1) yields cos c D :1723279. Hence c D 1:3976039 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 the area of ABC is less than or equal to the sum of the areas of the other three faces follows from (1) the area of a each of the other faces is at least 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 superset of ABC. 252 Solutions to the Challenges

9.10. Let C 0 and D0 be the projections of C and D, respectively, onto the 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 areas of ABC 0 and ABD0.ButtheareaofABC 0 is less than or equal to the area of ABC, and the area of ABD0 is less than or equal to theareaofABD. 9.11. Yes for both solids. See Figure S9.4a for the cuboctahedron and S9.4b for Durer’s¨ solid.

(a) (b)

Figure S9.4.

9.12. Color the vertices as shown in Figure S9.5, and observe that each edge 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 Hamiltonian cycle.

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 diagonal segment joining the vertices. 9.14. No. To have equality in the inequality of Section 9.6, the faces ABC and BCD must be equilateral triangles, and the dihedral angle at Solutions to the Challenges 253

edgep BC must be =2 (see Figure 9.6.1a). But in this case jADj D M 6=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 CCanbnj AB ja j jb j ja j jb j ja j jb j 1 1 C 2 2 CC n n A B A B A B ! 1 ja j2 jb j2 ja j2 jb j2 ja j2 jb j2 1 C 1 C 2 C 2 CC n C n D 1; 2 A2 B2 A2 B2 A2 B2

and hence ja1b1 C a2b2 CCanbnj AB.

2=3 9.16. The proof is by contradiction. Assume maxfAx;Ay ;Azg

(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 regular tetrahedra glued to two of its faces. 254 Solutions to the Challenges

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 E C F D 2, 4F D 2E, and so 2V E D 4. But since the polyhedron is convex, V D V3, so that 3V D 2E and hence 2V 3V =2 D 4 and V D 8, E D 12, and F D 6. Thus the only solution is the cube. 10.4. See Figure S10.2.

Figure S10.2.

10.5. Here are two solutions, there are infinitely many more [Hunter and Madachy, 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 tetrahedron ABCD is 180ı. If we unfold the tetrahedron by cutting along the three edges emanating from D, we obtain a triangle as shown in Figure S10.4, 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 jDCj, that is, opposite edges AB and DC have the same length. Similarly, Solutions to the Challenges 255

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

D 1

AB

D D 2 C 3

Figure S10.4.

Conversely, if the tetrahedron is isosceles then the three angles at a vertex are the same as the three angles in the opposite face, and hence sum 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’s surface that is tropical is 2rh=4r2 D h=2r D sin.23:4378ı/ 39:8%. 10.8. The two graphs have exactly the same length. Set r D 1 in Figure 10.7.2 and the discussion that follows. 10.9. 30 ft. There are three feasible ways to unfold the room yielding paths of length 30 ft and approximately 31.3 ft and 33.5 ft. p 10.10. The shortest path has length 3 2. See Figure S10.5.

Figure S10.5.

10.11. As seen in the unfolding of the box illustratedp in Figure S10.6a, the length of the shortest path from A to X is 8. Let B be the point one- fourth of the way down the diagonal, as shown in the unfolding of 256 Solutions to the Challenges

the box illustrated in Figures S10.6b and S10.6c. In either ofp the two routes shown, the length of the shortest path from A to B is 8:125. By symmetry there are two other routes along hidden sides of the box. Hence the point B where the shortest path from A to B is longest is not X [Gardner, 1996].

(a)X (b) (c) B B A A A

Figure S10.6.

10.12. Cut the cone along the dashed line in Figure 10.12.5 directly opposite the line from the vertex through the given point, and open the cone as shown in Figure S10.7. The cone is now a sector of a circle, and the shortest path consists of the two lines perpendicular to the linesp resulting from the cut [Steinhaus, 1969]. The condition h>r 3 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 volume of the tetrahedron in Figure 10.12.6ap is 1=6 while the volume of the tetrahedron in Figure 10.12.6b is 5=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 each triangle form tetrahedral nets. S folds into a regular tetrahedron with positive volume, T folds into a degenerate tetrahedron (a square) with zero volume, but both tetrahedra have all faces with area 1. Solutions to the Challenges 257

10.15. The leftmost nets in the two rows of Figure S10.2 can be modified to 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 in Figure S10.8b [Gardner, 2001].

(a) (b)

Figure S10.8.

10.16. Yes. The triangular prism with edges of unit length whose net is shown in Figure S10.9 has the property. For a proof that this is the only 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 2rh, we have, using the AM-GM inequality (3.5) for three numbers,

V 2 2h.3r h/2 Œ.2h C .3r h/ C .3r h//=33 1 D D : S 3 144r3 144r3 18 We have equality if and only if 2h D 3r h,orh D r, i.e., when the spherical 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 the icosahedron, the numbers are 4 and 3, respectively, as seen in their nets in Figure S10.10. 258 Solutions to the Challenges

Figure S10.10.

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

S T r y R

r–h h

Figure S10.11.

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

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Admissible number 77, 78, 81, 238 Cartography 169–176 Algebraic identities and boxes Cauchy-Schwarz inequality 62, 95, 45–49 96, 115, 182, 183, 192, 241, Angles between diagonals of a box 242 25, 58 Cavalieri’s principle 89–92, 111, in lunes 122 112, 114, 240 of a tetrahedron 124–126, 219 Chromatic number 133 Antiprism 18, 19, 87, 88, 141, 199 Circular helix 209–211 Antiprisms in Platonic solids 87, 88 Circumcenters in the faces of a Archimedes’ screw 211 tetrahedron 126, 127 Arithmetic mean 8, 31, 50, 51, 59, Circumsphere of a tetrahedron 126, 60, 128, 191, 252 127 Arithmetic mean-geometric mean Coil 209 inequality for three numbers 50, Colors in space 6, 133, 134, 216, 51 225, 257 Arithmetic mean-geometric mean Commandino’s theorem 97–99 inequality for two numbers 8 Completely incongruent boxes 135, Arithmetic mean-root mean square 136 inequality 59, 60 Concurrent lines 118, 119 Azimuthal projection 170 Cone 20, 21, 63, 79, 92, 99–103, equidistant projection 172, 189 157, 158, 169, 170, 175, 176, 205, 206, 224, 240, 256 Barrel 62, 235 Conformal projection 171, 174, 175, Bicylinder 5, 111, 112, 120, 121, 189 211–214, 239 Conic sections 99–103 Binomial coefficient 38, 39 Conical canvas tent 52, 53 Blivet 184 projection 169, 175, 176 Box inequalities 55–59 Convex polyhedron 10, 125, 177, Buckminsterfullerene 190 178, 190–192, Cosine law and gnomic projection Calissons 28, 29, 40, 230 178, 179 Camera obscura 168 Counting calissons 28, 29, 40, Cannonballs 35, 36, 41, 43, 234 230 Cantor dust 144 cannonballs 35, 36, 41, 43, 234 Cantor principle 27 Csaszar´ polyhedron 216, 217

265 266 Index

Cube 3, 4, 8, 10, 11, 16, 18, 23–25, Doughnut 111, 162 29–35, 41, 42, 45–50, 57–61, Drawing a regular icosahedron 15, 67, 68, 71, 75, 77, 78, 80, 16 83–85, 87, 104, 105, 120, Dudeney’s fly and spider 217, 218 130–132, 135, 137, 143–145, Dudeney’s puzzles 61, 79, 217, 218, 152, 155, 156, 159–161, 182, 237 183, 186, 195, 201–203, 222, Duplication of the cube 201–203 225 principle 27, 30, 32, 135 Cube algebra 45–47 Durer’s¨ solid 177 Cubic numbers 3, 45–49 Cubing a cube 135 Eccentricity of a conic 102 the sphere 93 Element of a cone, 102 Cubism 30 of a cylinder, 102, 209, 210 Cuboctahedron 22, 80, 191, 237, 252 Elements of Euclid 10, 17, 19, 67, Cylinder 5, 20, 21, 62, 79, 89, 90, 92, 79, 104–107, 115, 126, 129 102, 103, 120, 121, 127, 128, Ellipse 99, 100, 101, 102, 103, 158, 140–143, 162, 169–193, 204, 212, 213, 248 205, 209–214 Enneahedron 191, 252 Cylindrical projection 173–175 Enumeration 27–43 wedge 92, 93, 241 Equirectangular projection 173 Euler bricks 9 Dandelin-Quetelet theorem 101–102 Euler’s rotation theorem 156, 157 de Gua’s theorem 93–98, 114, 182, polyhedral formula 177, 178, 250, 240 251, 253 Degree 3 power mean 59, 60 Degree 3 power mean-root mean Face angles 124, 222 square inequality 59, 60 Fair division of a cake 153 Degree of a polyhedral vertex Fat elephant inequality 192 188 False perspective 168, 169 Delian problem 201–203 Fibonacci numbers 46, 154 Deltahedron 12, 196–200 Figurate numbers 27, 28, 34–36 Developable surface 169, 204 Flexible polyhedron 217 Devil’s tuning fork 184 Focus of a conic 100, 102, 103, Dihedral angle 117, 122–125, 131, 157 171, 177–179, 252 Foldable objects 206 Directrix of a conic 101–103, 157, Folding 193–226 238 a regular pentagon 200, 201 Dissections 4, 11, 12, 65–81, paper in half twelve times 154–156 219–221 Dodecahedron 14–16, 71, 72, 87, Four color conjecture and theorem 6, 104–107, 115, 187, 190, 194, 133 195, 227, 242, 251, 257 Fractal 136, 140, 143–145, 245, 246 Index 267

Frustum of a cone 175, 204–208 Hex numbers 27, 28, 40 Frustum of a pyramid 72–74, 87, Hexagonal section of a cube 83–85 88 Hexagonal truncated trapezohedron Fubini principle 27, 29, 30 251 Hexahedron 10 General position of lines 118, 119, Hilbert’s third problem 79 131 Hinged dissection 154–156 of spheres 131 Human poverty index 59 Geometric mean 8, 31, 50, 51, 59, Huzita-Hatori axioms 194 74 Hyperbola 99–102, 159 Geometric mean-arithmetic mean Hyperbolic paraboloid 203, 204 inequality 8, 50, 51 Hyperboloid of one sheet 203, 204 Geometric mean-harmonic mean inequality 59, 60 Icons of solid geometry ix Girard’s formula and theorem 122, Icosahedron 10, 15, 16, 87, 104, 106, 123, 244 115, 190, 195, 197, 199, 227, Gnomonic projection 170, 178, 180, 242, 257, 258 189 Impossible cubed box 135 God’s algorithm and number 160, object 184 161 Inclusion principle 49–52 Golden ratio 15, 20, 63, 70, 71, 104, Incongruent boxes 135, 136 153, 154, 242 Inequalities for rectangular boxes Great circles 117, 122, 170, 171, 55–59 177, 178, 189 for isosceles tetrahedra 96, 97 Greenwich meridian 99, 169, 173 Inequality between two means 59, Groin vault 5 60 Guba’s inequality 57, 58, 61, 240 Inscribing the Platonic solids in a Gyrobifastigium 18 sphere 104–107 Gyroelongated square bipyramid Instant Insanity puzzle 158–161, 199 247 Intersecting cylinders 5, 11, 112, Hadwiger problem 77, 78 117, 120, 121, 212 Hamiltonian cycle or circuit 187, spheres 186 188, 191, 252 Intersection 117–132 Harmonic mean 59 Isometry 147 Harmonic mean-geometric mean Isoperimetric inequality in space 90 inequality 59, 60 Isosceles tetrahedron 76, 77, 81, 90, Hausdorff dimension 145 91, 96, 97, 81, 161, 222, 238, Helices 209–211 243, 246, 255, 256 Helicoidal ramps 209, 210 Iteration 133–145, 245, 246 Heronian mean 74 Heron’s formula 62, 96, 97 Kissing number problem 42, 43, 115 268 Index

Lambert’s equal-area azimuthal Octahedron 10, 13, 14, 16, 19, 69, projection 172, 173 70, 80, 81, 87, 104, 105, 110, equal-area cylindrical projection 131, 138, 143, 145, 199, 225, 173, 174 238, 239, 254, 257 Lateral area of a cone 52 Optimization problems 52–55 Latitude 99, 169–175, 189 Origami 193, 194, 221 Law of cosines, spherical 180 Orthogonal projection 171, 176, Longitude 169–175 189 Loomis-Whitney inequality Orthographic projection 15, 182–184, 253 180–185 Loyd’s puzzles 48, 49 Outer angles of a tetrahedron 125, 126 Map 5, 6, 133, 134 Outer trihedral angle 125, 126 Mapping the earth 169–176 Means for three numbers 59, 60 Padovan’s sequence 154, 161 Median in a tetrahedron 97–99 Paper folding 193–222 Mechanical drawing 176 Parabola 99–102 Menger sponge 136–139, 144, 145 Parallelepiped 65–67, 76 Mercator’s projection 174, 175 law 108, 109 Method of exhaustion 79 Parallelogram law 109 of indivisibles 89 Partitioning space with planes Midsection 86–88 37–39 Million points in space problem 148, with spheres 42, 232, 233 149 Penrose triangle 184 Minimal paths on boxes 224, 255, Pentagon-hexagon-decagon identity 256 115 on cones 224, 256 Pentagonal bipyramid 199 on cylinders 210 numbers 33, 34 Miura fold 221, 222 Pentahedron 23, 190, 227 Mobius¨ band or strip 214, 215 Perfect cube 135 polyhedron 214 Perfect rectangles 134, 135, 136 Moscow papyrus 72 Perspective 17, 165–169 Motion 147–163 Planar projection 169–175 Mozartkugel 208 Plane section 83–116 Planes in space 37–39 Nappes of a cone 100, 102 Plastic number 153, 154 Net 18, 194–197, 199, 214–217, 219, Platonic fractals 136–138 222, 225, 253, 254, 256, 257 Platonic solids 9–16, 18, 23, 88, 138, Nondevelopable surface 203 187, 189, 190, 196, 225, 246 Platonic solids and antiprisms 87, Oblong numbers 31, 32, 41, 231 88 Octagonal prism 17 Platonic solids in a sphere 104–107 Index 269

Plato’s cubes 48, 49 Reptiles 23, 24, 228 Polycube 11, 12, 23, 24, 134, Rhombic dodecahedron 74–76, 155, 228 187, 188 Polyhedral nets 14, 195, 197–199, Right tetrahedron 94–96 222, 254 Roman dodecahedra 14, 15 Polyhedron, definition 9, 10 Root mean square 59 Printing 3D objects 110 Root mean square-arithmetic mean Prism 4, 17–19, 66, 67, 85–87, 257 inequality 59, 60 Prismatoid 85–88, 113, 239 Root mean square-degree 3 power Prismoid 85 mean inequality 59, 60 Prismoidal formula 85–88, 113 Rotation 147–151, 154–157 Projections 165–192 Rubik’s cube 160 Pronic number 31 Ruled surface 203–208 Puzzles 11, 12, 48, 49, 61, 79, 80, 158–160, 217, 218, 237 Pyramid 1, 2, 4, 6, 7, 12, 19, 20, Sanford’s fair division 153 31–33, 51, 65–70, 72, 73, 80, Schwarz lantern 140–144 81, 86, 87, 106, 115, 150, 155, paradox in cylinders 140–144 227, 250 Sections of a torus 117, 118 Pythagorean theorem Self-similar solids 133, 139, 140, for parallelograms in space 143 180–182 Shadows 166, 147, 170, in space 93–96, 180–182 Shephard’s conjecture 195, 196 on a sphere 178–180 Sinusoid 209, 212, 213, Pythagorean triple 9 Skeleton of a polyhedron 177, 187, 188, 189, 191, 194, 195, 250, Quarter-square numbers 41, 231, 232 253 Skew lines in space 118, 119 Radius of a sphere 4, 5, 71, 107, 108, Sierpinski´ 121–123, 127, 128 octahedron 138 Rectangular number 31 tetrahedron 138 Recycling symbol 215 Slant height of a right circular cone Regular 63 polyhedron 10 Slicing 83 tiling 14 Slothouber-Graatsma puzzle 237 Represention of Snub disphenoid 199 cubic numbers 3, 45–48, 50 Solid angle 125 hex numbers 27, 28 Solid geometry ix,x oblong numbers 31, 32 Soma cube 11, 12, 24 pentagonal numbers 32, 33 Speedcubing 161 square numbers 3, 30 Spider and the fly problem 217, 218, triangular numbers 34, 35 223 270 Index

Sphere 21, 24, 25, 42, 43, 101, 102, of bicylinders and tricylinders 104–107, 115, 131, 156, 157, 211–214 178–180, 186, 232, 233, 243 of cylinders, cones and spheres as a locus of points 129, 130 203–209 radius of a 107, 108, 121–123, of fractals 136, 139, 140, 144, 245 127, 128 of a frustum of a cone 205, 206 surface area of a 207, 208 of spherical triangles 121–123 volume of a 79, 89–91 of the Schwarz lantern 141–143 Spherical cap 112, 113, 121, 240, Szilassi polyhedron 215–217 258 law of cosines 179 Tetrahedral kites 68, 69 lunes 122 number 36, 41, 43, 234 Pythagorean theorem 180 Tetrahedron, 6, 10–13, 16, 67–70, segment 112, 239, 240 81, 84, 87, 91–98, 104, 105, zone 207, 208 114, 124–127, 131, 138–140, Spherometer 127, 128 149–151, 155, 184, 185, 188, Square numbers 3, 8, 27, 30, 31, 33, 191, 197, 199, 200, 215, 219, 34, 40, 41, 57, 61, 94–97, 132, 225 181, 182, 230, 231, 234 inequality 191, 251 pyramid 6, 69, 70, 73, 81, 87, 92, isosceles 76, 77, 81, 90, 91, 96, 238 97, 161, 222, 238, 246, 247, 255 pyramidal number 36, 43, 234 right 93–96 Squaring a circle 93, 201 Torus 111, 117, 118, 162, 163, 239, a square 134, 135 247, 248 Steffen’s flexible polyhedron 217 Triangular bipyramid 199 Steinmetz solid 111 Triangular numbers 32–34, 36, 38, Stella octangula 70, 71 41, 230, 232, 233 Stellated octahedron 70 Triaugmented triangular prism 199 Stereographic projection 170, 171, Tricylinder 5, 120, 121, 211–214 174, 189 Trihedral angle 117, 124, 125, 131 Steradian 125 Trinomial 47 Sums of cubes 34, 35 Trirectangular tetrahedron 93 of hex numbers 27, 28 Tropical zone 222, 223, 255 of oblong numbers 31, 32 Truncated cone 100 of outer angles of a tetrahedron icosahedron 22, 190 125, 126 octahedron 22, 80, 81, 155, 156, of pentagonal numbers 33, 34 237, 238 of squares 30 pyramid 72–74, 87, 88 of sequences of integers 29–35 of triangular numbers 32, 33 Uniform polyhedron 19 Sundial 12, 166 Unfolding figures 193–226 Surface area Universal curve 138 Index 271

Upper bound for the volume of a of a regular dodecahedron 71, 72 tetrahedron 184, 185 of a regular icosahedron 106 Using cubes to sum integers 29–33 of a regular octahedron 67, 69, 70, 81, 110 Van der Laan’s plastic number 153, of a regular tetrahedron 67, 68, 69, 154 91, 114 Vertex angles of a tetrahedron 219 of a rhombic dodecahedron 75 Villarceau circles 118 of a sphere 24, 70, 89, 90 Viviani’s theorem for a regular of a spherical cap 113, 121, 240 tetrahedron 149–151, 162, 246 of a spherical segment 112 Voicu’s inequality 58, 59 of a stella octangula 70 Volume of a barrel 62 of a tetrahedron 184, 185 of a bead 111, 239 of a tricylinder 120, 12 of a bicylinder 5, 112 of a torus 111, 239 of a cone 52, 63, 79, 90, 92, 113, of a truncated octahedron 80, 155, 240 156, 237 of a cuboctahedron 80 of a wedge 72, 93 of a cylinder 55, 79, 92 of an antiprism 88 of a cylindrical wedge 92, 93 of an isosceles tetrahedron 77, 91, of a frustum of a square pyramid 96, 97 72–74, 87 of fractals 136, 139 of a parallelepiped 65, 66 of the Schwarz lantern 144, 245 of a plug 113, 240, 241 of a prism 4, 66 Wallace-Bolyai-Gerwien theorem of a prismatoid 86 79 of a pyramid 4, 32, 33, 51, 66–70, Wedge 72, 92, 93, 113, 241 72, 75, 77, 86, 87, 150, 154 Wrapping 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 and Copulas with M. J. Frank and B. Schweizer, WSP, 2006; Math Made Vi- sual (with Roger Nelsen) MAA, 2006; Vitaminas Matematicas´ and El Club de la Hipotenusa, Ariel, 2008, Geometria para Turistas, Ariel, 2009; When Less is More (with Roger Nelsen) MAA, 2009; Asesinatos Matematicos´ , Ariel, 2010; Charming Proofs (with Roger Nelsen) MAA, 2010; and Icons of Mathematics (with Roger Nelsen) MAA, 2011.

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