Euclidean and Non Euclidean Geometry

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EXPERIENCING
GEOMETRY
Euclidean and Non-Euclidean
w ith History
Third Edition
David W. Henderson
Daina Taimina

C ontents
Preface xv
Changes in this Edition xvii
Useful Supplements xviii
Our Background in Geom etry xix
Acknowledgments for the First Edition xxi
Acknowledgments for the Second Edition xxii
Acknowledgments for This Edition xxiii
How to Use This Book XXV
How W e Use this Book in a Course xxvi
But, Do It Y our Own W ay xxvii
Chapter Sequences xxviii
Chapter 0 Historical Strands
of Geometry 1
A rt/Pattem s Strand 1
Navigation/Stargazing Strand 3
Building Structures Strand 4
M otion/M achines Strand 6
vii

viii Contents
Chapter 1 What Is Straight? 9
History: H ow Can W e Draw a Straight Line? 9
P roblem 1.1 W hen Do You Call a Line Straight? 13
The Symmetries o f a Line 17
Local (and Infinitesim al) Straightness 21
Chapter 2 Straightness on Spheres 25
Early H istory o f Spherical Geometry 25
P roblem 2.1 W hat Is Straight on a Sphere? 28
Symmetries o f Great Circles 32
♦Every Geodesic Is a Great Circle 35
♦Intrinsic Curvature 35
Chapter 3 What Is an Angle? 37
P roblem 3.1 W hat Is an Angle? 37
P roblem 3.2 V ertical Angle Theorem (VAT) 39
Hints for Three D ifferent Proofs 41
Chapter 4 Straightness on Cylinders 43
and Cones
P roblem 4.1 Straightness on Cylinders and Cones 44
Cones with Varying Cone Angles 46
Geodesics on Cylinders 49
Geodesics on Cones 50
♦Problem 4.2 Global Properties o f Geodesics 51
*«-Sheeted Coverings o f a Cylinder 51
*«-Sheeted (Branched) Coverings o f a Cone 53
Locally Isometric 55
Is “ Shortest” Always “Straight”? 56
♦Relations to Differential Geometry 57
Chapter 5 Straightness on 59
Hyperbolic Planes
A Short History o f Hyperbolic Geometry 59
D escription o f A nnular H yperbolic Planes 62
Hyperbolic Planes o f D ifferent Radii (Curvature) 64
P roblem 5.1 W hat Is Straight in a Hyperbolic Plane? 66

Contents IX
♦Problem 5.2 Coordinate System on Annular ^
Hyperbolic Plane
♦Problem 5.3 The Pseudosphere Is Hyperbolic 68
Intrinsic/Extrinsic, Local/Global 71
P roblem 5.4 Rotations and Reflections on Surfaces 71
Chapter 6 Triangles and Congruencies 73
♦Geodesics Are Locally Unique 73
P roblem 6.1 Properties o f Geodesics 74
P roblem 6.2 Isosceles Triangle Theorem (ITT) 75
Circles 76
Triangle Inequality 78
P roblem 6.3 Bisector Constructions 79
P roblem 6.4 Side-Angle-Side (SAS) 80
P roblem 6.5 Angle-Side-Angle (ASA) 85
Chapter 7 Area and Holonomy 89
P roblem 7.1 The Area o f a Triangle on a Sphere 90
P roblem 7.2 Area o f Hyperbolic Triangles 91
P roblem 7.3 Sum o f the Angles o f a Triangle 95
Introducing Parallel Transport 96
Introducing Holonom y 98
P roblem 7.4 The Holonom y o f a Small Triangle 100
The Gauss-Bonnet Formula for Triangles 102
♦Problem 7.5 Gauss-Bonnet Formula for Polygons 103
♦G auss-B onnet Formula for Polygons on Surfaces 106
Chapter 8 Parallel Transport 109
P roblem 8.1 E uclid’s Exterior Angle Theorem (EEAT) 109
P roblem 8.2 Symmetries o f Parallel Transported Lines 111
Problem 8.3 Transversals through a M idpoint 114
P roblem 8.4 W hat Is “Parallel” ? 115
Chapter 9 SSS, ASS, SAA, and AAA 117
Problem 9.1 Side-Side-Side (SSS) 117
P roblem 9.2 Angle-Side-Side (ASS) 119
P roblem 9.3 Side-Angle-Angle (SAA) 121
P roblem 9.4 Angle-Angle-Angle (AAA) 123

X Contents
Chapter 10 Parallel Postulates 125
Parallel Lines on the Plane Are Special 125
P roblem 10.1 Parallel Transport on the Plane 126
P roblem 10.2 Parallel Postulates N ot Involving ^ g
(Non-)Intersecting Lines
Equidistant Curves on Spheres and Hyperbolic Planes 130
P roblem 10.3 Parallel Postulates Involving
(Non-)Intersecting Lines
P roblem 10.4 EFP and HSP on Sphere and ^ 4
Hyperbolic Plane
Com parisons o f Plane, Spheres, and Hyperbolic Planes 136
Parallel Postulates within the Building Structures Strand 138
Non-Euclidean Geom etries within the Historical Strands 140
Chapter 11 Isometries and Patterns 143
P roblem 11.1 Isometries 144
P roblem 11.2 Three Points Determine an Isometry 148
P roblem 11.3 Classification o f Isometries 149
K lein’s Erlangen Program 153
Symmetries and Patterns 154
P roblem 11.4 Examples o f Patterns 158
* P roblem 11.5 Classification o f Discrete Strip Patterns 159
*Problem 11.6 Classification o f Finite Plane Patterns 159
““P roblem 11.7 Regular Tilings with Polygons 160
*Other Periodic (and N on-Periodic) Patterns 161
““Geometric M eaning o f A bstract Group Terminology 163
Chapter 12 Dissection Theory 165
W hat Is Dissection Theory? 165
A Dissection Puzzle from 250 b.c. Solved in 2003 167
History o f Dissections in the Theory o f Area 168
P roblem 12.1 Dissect Plane Triangle and Parallelogram 169
Dissection Theory on Spheres and Hyperbolic Planes 170
P roblem 12.2 Khayyam Quadrilaterals 171
P roblem 12.3 Dissect Spherical and Hyperbolic Triangles ^
and Khayyam Parallelograms
““P roblem 12.4 Spherical Polygons Dissect to Lunes 173

Contents XI
Chapter 13 Square Roots, Pythagoras,
and Similar Triangles
Square Roots 178
P roblem 13.1 A Rectangle Dissects into a Square 179
B audhayana’s Sulbasutram 184
Problem 13.2 Equivalence o f Squares 189
Any Polygon Can Be D issected into a Square 190
History o f Dissections 191
Problem 13.3 More D issection-Related Problems 193
* Three-Dimensional D issections and H ilbert’s Third Problem 194
P roblem 13.4 Similar Triangles 195
Chapter 14 Projections of a Sphere
onto a Plane
P roblem 14.1 Charts M ust Distort 198
Problem 14.2 Gnomic Projection 198
P roblem 14.3 Cylindrical Projection 199
P roblem 14.4 Stereographic Projection 200
History o f Stereographic Projection and A strolabe 202
Chapter 15 Circles 205
P roblem 15.1 Angles and Power o f Points for Circles
in the Plane
*Problem 15.2 Power o f Points for Circles on Spheres 208
Problem 15.3 A pplications o f Power o f a Point 212
P roblem 15.4 Trisecting Angles and Other Constructions 213
Chapter 16 Inversions in Circles 217
Early History o f Inversions 217
Problem 16.1 Inversions in Circles 218
P roblem 16.2 Inversions Preserve Angles and Preserve
Circles (and Lines)
Problem 16.3 U sing Inversions to Draw Straight Lines 224
*Problem 16.4 A pollonius’ Problem 226
Expansions o f the N otion o f Inversions 230

x iv Contents
Cosmic Background Radiation 356
P roblem 24.5 Circle Patterns M ay Show
the Shape o f Space
Latest Evidence on the Shape o f Space 361
Appendix A Euclid’s Definitions, 363
Postulates, and Common Notions
Definitions 363
Postulates 366
Common Notions 366
Appendix B Constructions of 36?
Hyperbolic Planes
The Hyperbolic Plane from Paper A nnuli 367
How to Crochet the Hyperbolic Plane 368
{3,7} and {7,3} Polyhedral Constructions 371
Hyperbolic Soccer Ball Construction 371
“ {3 ,6 Vi} ” Polyhedral Construction 372
Bibliography 375
Index 385

Preface
In mathematics, as in any scientific research, we find
two tendencies present. On the one hand, the tendency toward
abstraction seeks to crystallize the logical relations inherent in
the maze of material that is being studied, and to correlate the
material in a systematic and orderly manner. On the other
hand, the tendency toward intuitive understanding fosters a
more immediate grasp of the objects one studies, a live rap­
port with them, so to speak, which stresses the concrete mean­
ing of their relations.
As to geometry, in particular, the abstract tendency has
here led to the magnificent systematic theories of Algebraic
Geometry, of Riemannian Geometry, and of Topology; these
theories make extensive use of abstract reasoning and sym­
bolic calculation in the sense of algebra. Notwithstanding this,
it is still as true today as it ever was that intuitive understand­
ing plays a major role in geometry. And such concrete intui­
tion is of great value not only for the research worker, but also
for anyone who wishes to study and appreciate the results of
research in geometry.
— David Hilbert [EG: Hilbert, p. iii]
These words were written in 1934 by the “father o f Formalism ” David
H ilbert (1862-1943) in the Preface to Geometry and the Imagination by
Hilbert and S. Cohn-Vossen. H ilbert has emphasized the point we wish
to make in this book;
Meaning is important in mathematics and geometry is an im­
portant source o f that meaning.
x v

x v i Preface
We believe that mathematics is a natural and deep part o f human
experience and that experiences o f m eaning in mathematics are accessi­
ble to everyone. M uch o f m athematics is not accessible through formal
approaches except to those with specialized learning. However, through
the use o f nonform al experience and geometric imagery, many levels o f
m eaning in mathematics can be opened up in a way that m ost humans
can experience and find intellectually challenging and stimulating.
Formalism contains the power o f the m eaning but not the meaning.
It is necessary to bring the power back to the meaning.
A formal p ro o f as we normally conceive o f it is not the goal o f
mathematics — it is a tool — a means to an end. The goal is understand­
ing. W ithout understanding we will never be satisfied — with under­
standing we want to expand that understanding and to communicate it to
others. This book is based on a view o f p ro o f as a convincing communir
nation that answers — Why?
M any formal aspects o f mathematics have now been mechanized
and this mechanization is widely available on personal computers or
even handheld calculators, but the experience o f meaning in m athem at­
ics is still a human enterprise that is necessary for creative work.
In this book we invite the reader to explore the basic ideas o f
geom etry from a more mature standpoint. We will suggest some o f the
deeper meanings, larger contexts, and interrelations o f the ideas. We are
interested in conveying a different approach to m athematics, stimulating
the reader to take a broader and deeper view o f mathematics and to
experience for herself/him self a sense o f mathematizing. Through an
active participation with these ideas, including exploring and writing
about them, people can gain a broader context and experience. This
active participation is vital for anyone who wishes to understand m athe­
m atics at a deeper level, or anyone wishing to understand something in
their experience through the vehicle o f mathematics.
This is particularly true for teachers or prospective teachers who are
approaching related topics in the school curriculum. All too often we
convey to students that mathematics is a closed system, with a single
answer or approach to every problem, and often without a larger context.
We believe that even where there are strict curricular constraints, there is
room to change the m eaning and the experience o f m athem atics in the
classroom.
This book is based on a junior/senior-level course that David started
teaching in 1974 at Cornell for mathematics m ajors, high school

Changes in This Edition XVII
teachers, future high school teachers, and others. M ost o f the chapters
start intuitively so that they are accessible to a general reader with no
particular m athematics background except imagination and a willingness
to struggle with ideas. However, the discussions in the book were written
for mathematics m ajors and m athematics teachers and thus assume o f the
reader a corresponding level o f interest and mathematical sophistication.
The course emphasizes learning geometry using reason, intuitive
understanding, and insightful personal experiences o f meanings in geo­
metry. To accomplish this the students are given a series o f inviting and
challenging problem s and are encouraged to write and speak their rea­
sonings and understandings.
M ost o f the problem s are placed in an appropriate history perspec­
tive and approached both in the context o f the plane and in the context o f
a sphere or hyperbolic plane (and sometimes a geometric m anifold). We
find that by exploring the geometry o f a sphere and a hyperbolic plane,
our students gain a deeper understanding o f the geometry o f the (Euclid­
ean) plane.
We introduce the m odem notion o f “parallel transport along a geo­
desic,” which is a notion o f parallelism that makes sense on the plane but
also on a sphere or hyperbolic plane (in fact, on any surface). W hile
exploring parallel transport on a sphere, students are able to appreciate
more fully that the similarities and differences between the Euclidean
geometry o f the plane and the non-Euclidean geometries o f a sphere or
hyperbolic plane are not adequately described by the usual Parallel
Postulate. W e find that the early interplay betw een the plane and spheres
and hyperbolic planes enriches all the later topics whether on the plane
or on spheres and hyperbolic planes. All o f these benefits will also exist
by only studying the plane and spheres for those instructors that choose
to do so.
C h a n g e s i n T h i s E d i t i o n
This book is an expansion and revision o f the book Experiencing Geo­
metry on Plane and Sphere (1996) and the book Experiencing Geometry
in Euclidean, Spherical, and Hyperbolic Spaces (2001). There are
several important changes: First, there are now co-authors — Daina was
a “contributor” to the second edition. She brings considerable experience
with and knowledge o f the history o f mathematics. W e start in Chapter 0
with an introduction to four strands in the history o f geometry and use

x v iii Preface
the fram ew ork o f these strands to infuse history into (alm ost) every
chapter in the book in ways to enhance the students understanding and to
clear up m any misconceptions. There are two new chapters — the old
Chapter 14 (Circles in the Plane) has been split into two new chapters:
Chapter 15 (on circles with added results on spheres and hyperbolic
plane and about trisecting angles and other constructions) and Chapter
16 (on inversions with added m aterial on applications). There is also a
new Chapter 21, on the geometry o f mechanism s that includes historical
m achines and results in m odem mathematics.
We have included discussions o f four new geometric results
announced in 2003-2004: In Chapter 12 we describe the discovery and
solution o f A rchim edes’ Stomacion Problem. Problem 15.2 is based on
the 2003 generalization o f the notion o f pow er o f a point to spheres. In
Chapter 16 we talk about applications o f a problem o f Apollonius to
m odem pharmacology. In Chapter 18 we discuss the newly announced
solution o f the Poincare Conjecture. In Chapter 22 we bring in a new
result about unfolding linkages. In Chapter 24 we discuss the latest
updates on the shape o f space, including the possibility that the shape o f
the universe is based on a dodecahedron. In addition, we have rearranged
and clarified other chapters from the earlier editions.
Useful S upplements
We will m aintain an Experiencing Geometry W eb page:
‘ w w w .m ath.com ell.edu/~henderson/ExpGeom /
containing any errata that readers find, updates on new results and
discoveries, useful W eb references, and other supplem entary material.
For exploring properties on a sphere it is important that you have a
m odel o f a sphere that you can use. Some people find it helpful to pur­
chase Lenart Sphere® sets — a transparent sphere, a spherical compass,
and a spherical “straightedge” that doubles as a protractor. They work
well for small group explorations in the classroom and are available
from Key Curriculum Press. However, considerably less expensive alter­
natives are available: A beach ball or basketball will w ork for classroom
dem onstrations, particularly i f used with rubber bands large enough to
form great circles on the ball. Students often find it convenient to use
worn tennis balls (“w orn” because the fuzz can get in the way) because
they can be w ritten on and are the right size for ordinary rubber bands to

http://www.math.comell.edu/~henderson/ExpGeom/

Useful Supplements XIX
represent great circles. Also, many craft stores carry inexpensive plastic
spheres that can be used successfully.
If you will be studying the hyperbolic geometry parts o f this book,
we strongly urge that you have a hyperbolic surface such as those
described in Chapter 5. Unfortunately, such hyperbolic surfaces are not
readily available commercially. However, directions for making such
surfaces (out o f paper or by crocheting) are contained in Appendix B,
and you can find patterns o f making paper models on the Experiencing
Geometry Web page; also see Daina Taimina, Crocheting Adventures
with Hyperbolic Planes (AK Peters, 2009).
The use o f commercial dynamic geometry software such as Geome­
ters Sketchpad®, Cabri®, or Cinderella® will enhance any geometry
course. These software packages were originally written for exploring
Euclidean plane geometry, but recent versions allow one to explore also
spherical and hyperbolic geometries. There are also available free on the
Web many dynamic geometry programs with more limited functionality.
We will maintain on the Experiencing Geometry Web page links to
information about these and other software packages and to Web pages
that give examples showing their use for self-learning or in a classroom.
A faculty member may obtain from the publisher the Instructor’s
Manual (containing possible solutions to each problem and discussions
on how to use this book in a course) by calling Faculty Services at
1-800-526-0485.
O ur Background in G eometry
Since an early age David loved to explore geometry in various forms in
art classes, woodcarving, carpentry, exploring nature, or by becoming
involved in photography. But he did not realize that the geometry, which
he experienced, was mathematics or even that it was called “geometry”.
He was not calling it geometry — he was calling it drawing or design or
not calling it anything and just doing it. He did not like mathematics in
school because it seemed very dead to him — just memorizing
techniques for computing things, and he was not very good at memoriz­
ing. He especially did not like his high school geometry course with its
formal two-column proofs. This continued on into the university, where
David was a joint physics and philosophy major and took only those
mathematics courses that were required for physics majors. He became
absorbed in geometry-based aspects o f physics: mechanics, optics,

XX Preface
electricity and magnetism, and relativity. On the other hand, his first
mathematics research paper (on the geometry o f Venn diagrams for more
than four classes) evolved from a course on the philosophy o f logic.
There were no university geometry courses except for one on “Analytic
Geometry and Linear A lgebra,” which only lightly touched on anything
geometric. In his last year as an undergraduate he switched into m athe­
m atics because he finally saw that the geometry he loved really was a
part o f mathematics. In graduate school he studied geometric topology
under his mentor, R. H. Bing, who taught without lectures or textbooks
in a style that is often known as the Moore Method, named after B ing’s
graduate m entor R. L. M oore at the University o f Texas. David joined
the faculty at Cornell in 1966 teaching topology and did not start teach­
ing geometry until 1974. D avid’s teaching o f the geometry course and
the initial writing o f this book evolved from this background.
D aina’s teaching and interests in geometry and history o f m athe­
m atics grew out o f her experiences in Latvia, which has had a strong
tradition o f excellence in mathematics at all educational levels. For three
years in high school D aina’s mathematics teacher was a form er univer­
sity m athematics professor with a passion for geometry — his method o f
teaching was to give the students large num bers o f interesting problems.
This teacher always paid a lot o f attention to how the students drew geo­
metric diagrams; he encouraged Euclidean constructions with compass
and straight edge but also free-hand drawing o f geometric figures, insist­
ing on accurate shapes and proportions. At university, D aina took many
traditional geometry courses; she always enjoyed and excelled at the
drawing aspects o f geometry but did not think it had anything to do with
art or aesthetic sensibilities. During her graduate years at the university
she taught in middle and high school and found that she could raise the
students’ interest in mathematics by telling them stories from the history
o f mathematics. She wanted to learn more history, but there was no
history o f m athematics course at the university. She complained about
this to the chair o f the department in 1979 and he im m ediately assigned
her to develop and teach one. In 1990 her History o f Mathematics:
Textbook fo r University Students (in Latvian) was published. D aina has
taught geometry courses since 1977. Her Ph.D. thesis was in theoretical
com puter science under the direction o f Rusins Freivalds, who still is her
model o f a true mathematician.

Chapter 0
Historical Strands
of G eometry
All people by nature desire knowledge.
— Aristotle (384 b.c -322 b.c.), Metaphysics
History is the witness that testifies to the passing of time; it
illumines reality, vitalizes memory, provides guidance in daily
life and brings us tidings of antiquity.
— Cicero (106 b.c.—43 b.c.), Pro Publio Sestio
Inherited ideas are a curious thing, and interesting to observe
and examine.
— Mark Twain (1835-1910)
A Connecticut Yankee in King Arthur’s Court
We think that the main aspects o f geometry today em erged from fo u |
strands o f early hum an activity that seem to have occurred in most
cultures: art/pattem , building structures, m otion/machines, and naviga-
tion/stargazing. These strands developed more or less independently into
varying studies and practices that eventually from the 19th century on
were woven into what we now call geometry.
A r t / P a t t e r n S t r a n d
To produce decorations for their weaving, pottery, and other objects,
early artists experimented with symmetries and repeating patterns. See
Figure 0.1. Later the study o f symmetries o f patterns led to tilings, group
1

2 Chapter 0 Historical Strands of Geometry
theory, crystallography, and finite geometries. See examples in Chapters
1, 11, 18, and 24.
Figure 0.1 Ancient Greek geometric art with examples of strip patterns
(Chapter 11)
Figure 0.2 Giving o f the Keys to Saint Peter by Pietro Perugina using perspective
(Chapter 20)

Art/Pattem Strand 3
Early artists also explored various methods o f representing physical
objects, and living things. These explorations in Rennaisance (see Figure
0.2) led to the study o f perspective and then later to projective geometry
and descriptive geometry. See examples in Problems 17.6 and 20.6.
In the last two centuries this strand has led to security codes, digital
picture com pression, com puter-aided graphics, the study o f computer
vision in robotics, and computer-generated movies.
For more details on the A rt/Pattem Strand, see [AD: A lbam ], [GC:
Bain], [EM: Devlin], [GC: Eglash], [AD: Field], [GC: Gerdes], [AD:
Ghyka], [AD: Gombrich], [SG: Hargittai], [AD: Ivins], [AD: Kappraf].
N a v i g a t i o n / S t a r g a z i n g S t r a n d
Figure 0.3 Armilary sphere based on Greeks’ view of the universe (Chapter 2)
For astrological, religious, agricultural, and other purposes, ancient
humans attempted to understand the movem ent o f heavenly bodies
(stars, planets, sun, and moon) in the apparently hemispherical sky. See
Figure 0.3. Early humans used the stars and planets as they started
navigating over long distances on the earth and on the sea. They used
this understanding to solve problem s in navigation and in attempts to
understand the shape o f the earth. Ideas o f trigonometry apparently were
first developed by Babylonians in their studies o f the m otions o f

4 Chapter 0 Historical Strands of Geometry
heavenly bodies. Even Euclid wrote an astronom ical work, Phaenomena,
[AT: Berggren], in which he studied properties o f curves on a sphere.
Navigation and large-scale surveying developed over the centuries
around the world and along with it cartography, trigonometry, and
spherical geometry. This strand is represented wherever intrinsic geome­
try o f surfaces is discussed, especially Chapters 4, 7, 8, and 18. Map
making is discussed in Chapter 16 and trigonom etry in Chapter 20.
Examples m ost closely associated with this strand in the last two
centuries are the study o f surfaces and m anifolds, which led to many
m odem spatial theories in physics and cosmology. See Figure 0.4 and
Chapters 2, 5, 18, 22, and 24.
Figure 0.4 Picture of the cosmic background radiation taken by NASA’s
WMAP satellite, released February 11, 2003 (Chapter 24)
For more details on the Navigation/Stargazing Strand, see [CE:
Bagrow], [HI: Burkert], [UN: Ferguson], [UN: Osserman], [SP:
Todhunter].
B u i l d i n g S t r u c t u r e s S t r a n d
As hum ans built shelters, altars, bridges, and other structures, they dis­
covered ways to make circles o f various radii, and various poly-
gonal/polyhedral structures. See Figure 0.5. In the process they devised
systems o f measurem ent and tools for measuring. The (2000-600 b.c .)
Sulbasutram [AT: Baudhayana] is written for altar builders and contains
at the beginning a geometry handbook with proofs o f some theorems and

Building Structures Strand 5
a clear general statement o f the “Pythagorean” Theorem, (see Chapter
13). Building upon geometric knowledge from Babylonian, Egyptian,
and early Greek builders and scholars, Euclid (325-265 b.c.) wrote his
Elements, which became the m ost used m athematics textbook in the
world for the next 2300 years and codified w hat we now call Euclidean
geometry.
Figure 0.5 Step pyramid of Memphis, Egypt (about 3000 b.c.)
Using the Elements as a basis in the period 300 b.c. to about 1000
a.d., Greek and Islamic m athematicians extended its results, refined its
postulates, and developed the study o f conic sections and geometric
algebra. W ithin Euclidean geometry, analytic geometry, vector geometry
(linear algebra and affine geometry), and algebraic geometry developed
later. The Elements also started w hat became known as the axiomatic
method in mathematics. For the next 2000 years m athematicians attem p­
ted to prove Euclid’s Fifth (Parallel) Postulate as a theorem (based on the
other postulates); these attempts culm inated around 1825 with the
discovery o f hyperbolic geometry. See Appendix A and Chapter 10 for a
discussion o f E uclid’s postulate and various other parallel postulates.
Chapters 3, 6, 9, 12, 15, 16, 19, and 23 contain m any o f the basic topics
o f Euclidean geometry.
Further developments with axiomatic methods in geometry led to
the axiomatic theories o f the real numbers and analysis and to elliptic

6 Chapter 0 Historical Strands of Geometry
geometries, axiomatic projective geometry, and other axiom atic geome­
tries. See the three sections at the end o f Chapter 10 for m ore discussion.
And, o f course, Euclidean geometry has continued to be used in the
building o f m odem structures (for example, Figure 0.6).
Figure 0.6 Apartment building in Baltimore
For more detail on the Building Structures Strand, see [AT:
Baudhayana], [AD: Blackwell], [HI: Burkert], [GC: Datta], [ME:
DeCamp], [TX: H artshom e], [HI: Heilbron], [HI: Seidenberg],
M o t i o n / M a c h i n e s S t r a n d
Early hum an societies used the wheel for transportation, for making
pottery, in pulleys, and in pumps.
In ancient Greece, Archimedes (see Figure 0.7), Hero, and other
geometers used linkages and gears to solve geometrical problem s —
such as trisecting an angle, duplicating a cube, and squaring a circle. See
Problem 15.4. These solutions were not accepted in the Building Struc­
tures Strand, and this leads to a common m isconception that these
problems are unsolvable and/or that Greeks did not allow m otion in
geometry. See [TX: M artin] and [HI: Katz],

Motion/Machines Strand 7
Figure 0.7 Endless screw such as used by Archimedes and da Vinci
We do not know much about the influence o f m achines on m athe­
matics after ancient Greece until the advent o f machines in the Renais­
sance. In the 17th century, Descartes and other m athematicians used
linkages to study curves. This study o f curves led to the developm ent o f
analytic geometry. M echanical computing m achines were also developed
in the 17th century.
Figure 0.8 Wankel engine based on a Reuleaux triangle (Chapter 21)

8 Chapter 0 Historical Strands of Geometry
As we will discuss in Chapter 21, there was an interaction between
m athematics and mechanics that led to marvelous machine design
(Figure 0.8) and continues to the m odem m athematics o f rigidity and
robotics.
For more details on the M otion/M achines Strand, see [ME: De-
Camp], [ME: Dyson], [ME: Ferguson 2001], [ME: Kirby], [ME:
M oon], [ME: Ramelli], [ME: W illiams],
t

W hat Is Straight?
_________ Chapter 1
Straight is that of which the middle is in front of both ex­
tremities.
— Plato, Parmenides, 137 E [AT: Plato]
A straight line is a line that lies symmetrically with the points
on itself.
— Euclid, Elements, Definition 4 [Appendix A]
H i s t o r y : H o w C a n W e D r a w a S t r a i g h t L i n e ?
W hen using a compass to draw a circle, we are not starting with a model
o f a circle; instead we are using a fundamental property o f circles that
the points on a circle are a fixed distance from a center. Or we can say
we use Euclid’s definition o f a circle (see Appendix A, D efinition 15).
So now w hat about drawing a straight line: Is there a tool (serving the
role o f a compass) that will draw a straight line? One could say we can
use a straightedge for constructing a straight line. W ell, how do you
know that your straightedge is straight? How can you check that some­
thing is straight? W hat does “straight” mean? Think about it — this is
part o f Problem 1.1 below.
You can try to use Euclid’s definition above. I f you fold a piece o f
paper the crease will be straight — the edges o f the paper needn’t even
be straight. This utilizes m irror symmetry to produce the straight line.
Carpenters also use symmetry to determine straightness — they put two
boards face to face, plane the edges until they look straight, and then
turn one board over so the planed edges are touching. See Figure 1.1.
They then hold the boards up to the light. If the edges are not straight,
9

10 Chapter 1 What Is Straight?
there will be gaps between the boards where light will shine through. In
Problem 1.1 we will explore m ore deeply symmetries o f a straight line.
Planed edges—–
__ _
—– > z
_______________
_____________________ ✓
Figure 1.1 Carpenter’s method for checking straightness
W hen grinding an extremely accurate flat mirror, the following
technique is sometimes used: Take three approxim ately flat pieces o f
glass and put pumice betw een the first and second pieces and grind them
together. Then do the same for the second and the third pieces and then
for the third and first pieces. Repeat many times and all three pieces o f
glass will become very accurately flat. See Figure 1.2. Do you see why
this works? W hat does this have to do with straightness?
Figure 1.2 Grinding flat mirrors
We can also use the usual high school definition, “A straight line is
‘th e shortest distance betw een two points.” This leads to producing a
straight line by stretching a string.
This use o f symmetry, stretching, and folding can also be extended
to other surfaces, as we will see in Chapters 2, 4, and 5. Sometimes we
can get confused when reading in the literature that “straight line” is an
term or that straight lines on the sphere are “defined to be arcs
o f great circles.” We find that putting “ straight” in the context o f the
four historical strands helps clarify this: “Symmetry” comes m ostly from
the A rt/Pattem Strand, “ term ” comes from the axiomatics in
the Building Structures Strand, and “shortest distance” comes m ostly
from the Navigation/Stargazing Strand.

History: How Can We Draw a Straight Line? 11
But there is still left unanswered the question o f whether there is a
mechanism analogous to a compass that will draw an accurate straight
line. We can find answers to this question in the history o f m echanics,
which leads us into the M otion/M achines Strand and to another meaning
o f “straight” .
Turning circular m otion into straight line m otion has been a practi­
cal engineering problem since at least the 13th century. As we can see in
some 13th century drawings o f a sawmill (see Figure 1.3) linkages (rigid
bars constrained to be near a plane and joined at their ends by rivets)
were in use in the 13th century and probably were originated much
earlier. Georgius Agricola’s (1494-1555) geological writings [ME: A gri­
cola] reflect firsthand observations not ju st o f rocks and minerals, but o f
every aspect o f m ining technology and practice o f the time. In the pic­
tures o f his w ork one can see link w ork that was widely used for convert­
ing the continuous rotation o f a w ater wheel into a reciprocating m otion
suited to piston pumps. In 1588, Agostino Ram elli published his book
[ME: Ramelli] on m achines where linkages were widely used. Both o f
these books are readily available; see the Bibliography.
In the late 18th century people started turning to steam engines for
power. James W att (1736-1819), a highly gifted designer o f machines,
worked on im proving the efficiency and power o f steam engines. In a

12 Chapter 1 What Is Straight?
steam engine, the steam pressure pushes a piston down a straight cylin­
der. W att’s problem was how to turn this linear m otion into the circular
m otion o f a wheel (such as on steam locomotives). It took W att several
years to design the straight-line linkage that w ould change straight-line
m otion to circular one. Later, W att told his son,
Though I am not over anxious after fame, yet I am more proud
of the parallel motion than of any other mechanical invention I
have ever made, (quoted in [ME: Ferguson 1962], p 197])
“Parallel motion” is a name W att used for his linkage, w hich was
included in an extensive patent o f 1784. W att’s linkage was a good solu­
tion to the practical engineering problem. See Figure 1.4, where his link­
age is the parallelogram and associated links in the upper left comer.
Figure 1.4 A steam engine with Watt’s “parallel motion” linkage
But W att’s solution did not satisfy mathematicians, who knew that
his linkage can draw only an approximate straight line. M athem aticians
continued to look for a planar straight-line linkage. A linkage that would
draw an exact straight-line was not found until 1864-1871, when a
French army officer, Charles Nicolas Peaucellier (1832-1913), and a
Russian graduate student, Lipmann I. Lipkin (1851-1875), independ­
ently developed a linkage that draws an exact straight line. See Figure
1.5. (There is not m uch known about Lipkin. Some sources mentioned

History: How Can We Draw a Straight Line? 13
that he was bom in Lithuania and was a graduate student o f Chebyshev
in Saint Petersburg but died before completing his doctoral dissertation.)
(For more discussion o f this discovery see also Philip Davis’s delightful
little book The Thread [EM: Davis], Chapter IV.)
Figure 1.5 Peaucellier/Lipkin linkage for drawing a straight line
The linkage in Figure 1.5 works because, as we will show in Prob­
lem 16.3, the point Q will only move along an arc o f a circle o f radius
(s2 – d 2) f / ( g 2 – f ). This allows one to draw an arc o f a large circle with­
out using its center. When the lengths g a n d /a re equal, then Q draws the
arc o f a circle with infinite radius. (See [EG: Hilbert], pp. 272-273, for
another discussion of this linkage.) So we find in the Motion/Machines
Strand another notion o f straight line as a circle o f infinite radius (see the
text near Figure 11.4 for discussion o f circles o f infinite radius).
P roblem 1.1 W hen Do Y ou Call a Line Straight?
In keeping with the spirit o f the approach to geometry discussed in the
Preface, we begin with a question that encourages you to explore deeply
the concept o f straightness. We ask you to build a notion o f straightness
from your experiences rather than accept a certain number o f assump­
tions about straightness. Though it is difficult to formalize, straightness is
a natural human concept.

14____ Chapter 1 What Is Straight?
a. How can you check in a practical way i f something is straight?
How do you construct something straight — lay out fence posts
in a straight line, or draw a straight line? Do this without
assuming that you have a ruler, fo r then we will ask, “How can
you check that the ruler is straight? ”
At first, look for examples o f straightness in your experiences. Go out
and actually try walking along a straight line and then along a curved
path; try drawing a straight line and then checking that a line already
xlrawn is straight. As you look for properties o f straight lines that distin­
guish them from non-straight lines, you will probably rem em ber the fol­
lowing statem ent (which is often taken as a definition in high school
geometry): j4 straight line is the shortest distance between two points.
But can you ever m easure the lengths o f all the paths betw een two
points? How do you find the shortest path? I f the shortest path between
two points is in fact a straight line, then is the converse true? Is a straight
line betw een two points always the shortest path? We will return to these
questions in later chapters.
‘ A powerful approach to this problem is to think about lines in terms
o f symmetry. This will become increasingly important as we go on to
other surfaces (spheres, cones, cylinders, and so forth). Two o f the sym­
m etries o f lines are as follows:
♦ ‘Reflection symmetry in the line, also called bilateral symme­
try — reflecting (or m irroring) an object over the line.
Figure 1.6 Reflection symmetry of a straight line
♦ ’Half-turn symmetry — rotating 180° about any point on the
line.
Figure 1.7 Half-turn symmetry of a straight line

Problem 1.1 When Do You Call a Line Straight?_____15
A lthough we are focusing on a symmetry o f the line in each o f these
examples, notice that the symmetry is not a property o f the line by itself
but includes the line and the space around the line. The symmetries pre­
serve the local environm ent o f the line. Notice how in reflection a n d
half-turn symmetry the line and its local environm ent are both part o f the!
symmetry action and how the relationship between them is integral to
the action. In fact, reflection in the line does not move the line at all bub
exhibits a way in which the spaces on the two sides o f the line are the
same*
Definitions. AnHsometry is a transform ation that preserves
distances and angle m easures. A sym metry o f a fig u re is an
isometry o f a region o f space that takes the figure (or the por­
tion o f it in the region) onto itself. You will show in Problem
11.3 that every isometry o f the plane is either a translation, a
rotation, a reflection, or a composition o f them!
b. What symmetries does a straight line havel
Try to think o f other symmetries o f a line as well (there are quite a few).
Some symmetries hold only for straight lines, while some work for other
curves as well. Try to determine which symmetries are specific to •
straight lines and why. Also think o f practical applications o f these sym­
m etries for constructing a straight line or for determining i f a line is
straight.
c. What is in common among the different notions o f straightness ?*
Can you write a definition o f “straight lin e’Kl
Look for things that you call “straight.” W here do you see straight lines?
Why do you say they are straight? Look for both physical lines and non­
physical uses o f the word “straight” . W hat symmetries does a straight
line have? How do they fit with the examples that you have found and
those mentioned above? Can we use any o f the symmetries o f a line to
define straightness? The intersection o f two (flat) planes is a straight liqe
— why does this work? Does it help us understand “straightness”?
Imagine (or actually try!) walking while pulling a long thread w ith4a
small stone attached. When will the stone follow along your path? Why?
This property is used to pick up a fallen w ater skier. The boat travels by
the skier along a straight line and thus the tow rope follows the path o f

16 Chapter 1 What Is Straight?
the boat. Then the boat turns in an arc in front o f the skier. Because the
boat is no longer following a straight path, the tow rope moves in toward
the fallen skier. W hat is happening?
Another idea to keep in m ind is that straightness m ust be thought o f
as a local property. Part o f a line can be straight even though the whole
line may not be. For example, if we agree that this line is straight,
and then we add a squiggly part on the end, like this,
would we now say that the original part o f the line is not straight, even
though it hasn’t changed, only been added to? Also note that we are not
making any distinction here betw een “ line” and “line segment.” The
more generic term “line” generally works well to refer to any and all
lines and line segments, both straight and non-straight.
You are likely to bring up m any ideas o f straightness. It is necessary
then to think about what is common among all o f these straight
phenomena.
Think about and formulate some answers for these questions before
you read any further. Do not take anything for granted unless you see
why it is true. No answers are predeterm ined. You may come up with
something that we have never imagined. Consequently, it is important
that you persist in following your own ideas. Reread the section “How to
Use This Book” starting on page xxv.
j j You should not read further until you have expressed your
thinking and ideas through writing or talking to someone else.

The Symmetries of a Line 17
T h e S y m m e t r i e s o f a L i n e
Reflection-in-the-line symmetry: It is m ost useful to think o f
reflection as a “m irrdr” action with the line as an axN rather than as a
“flip-over” action that involves an action in 3-space. In this way we cart
extend the notion o f reflection symmetry to a spherfl (the flip-over action
is not possible on a sphere). Notice that this symmetry cannot be used as
a definition for straightness because we use straightness to define reflec­
tion symmetry. This same comment applies to most o f the other symme­
tries discussed below.
Figure 1.8 Reflection-in-the-line symmetry
In Figures 1.8-1.14, the light gray triangle is the image o f the dark gray
triangle under the action o f the symmetry on the space around the line.
♦ Practical applicatidn: We can produce a straight line by
folding a piece o f paper because this action forces symmetry
along the crease. Above we showed a carpenter’s example.
Reflection-perpendicular-to-the-line symmetry: A reflection
through any axis perpendicular to the line will take the line onto itself.
Note that circles also have this symmetry about any diameter. See Figure
1.9.
Figure 1.9 Reflection-perpendicular-to-the-line symmetry

18 Chapter 1 What Is Straight?
♦ Practical applications: \ \ ) u call tell i f a straight segment is
perpendicular to a m irror by seeing i f it looks straight with its
reflection. Also, a straight line can be folded onto itself.
‘ Half-turn symmetry: A rotation through h a lf o f a full revolution
about any point P on the line takes the part o f the line before P onto t h |
part o f the line after P and vice versa. Note that some non-straight line%
such as the letter Z, also have half-turn symmetry — but not about every
‘point. See Figure 1.10.
♦ Practical applications: Half-turn symmetry exists for the
slot on a screw and the tip o f the screwdriver (unless you are
using Phillips-head screws and screwdrivers, which also have
quarter-turn symmetry) and thus we can more easily put the
tip o f the screwdriver into the slot. Also, this symmetry is
involved when a door (in a straight wall) opens up flat
against the wall.
»Rigid-motion-along-itself symmetry: For straight lines in the
plane, we call this ‘translation symmetry. Any portion o f a straight line
may be m oved along the line without leaving the line. This property o f
being able to move rigidly along itself is not unique to straight lines; cir­
cles (rotation symmetry) and circular helixes (screw symmetry) have this
property as well. See Figure 1.11.

The Symmetries of a Line 19
♦ Practical applications: Slide joints such as in trombones,
drawers, nuts and bolts, and so forth, all utilize this
symmetry.
>
Figure 1.11 Rigid-motion-along-itself symmetry
3-dimensional-rotation symmetry: In a 3-dimensional space,
rotate the line around itself through any angle using itse lf as an axis.
Figure 1.12 3-dimensional-rotation symmetry
♦ Practical applications: This symmetry can be used to check
the straightness o f any long thin object such as a stick by
tw irling the stick w ith itse lf as the ax i£ I f the stick does not
appear to wobble, then it is straight. This is used for pool
cues, axles, hinge pins, and so forth.
Central symmetry or point symmetry: Central symmetry
through the point P sends any point A to the point on the line determined
by A and P at the same distance from P but on the opposite side from P.
See Figure 1.13. In two dimensions central symmetry does not differ
from half-turn symmetry in its end result, but they do differ in the ways
we imagine them and construct them.

20____ Chapter 1 What Is Straight?
♦ in 3-space, central symmetry produces a result different than
hny single rotation or reflection (though we can check that it
does give the same result as the composition o f three reflec-
• tions through m utually perpendicular planes). To experience
central symmetry in 3-space, hold your hands in front o f you
with the palms facing each other and your left thumb up and
your right thumb down. Your two hands now have approxi­
mate central symmetry about a point midway betw een the
center o f the palms; and this symmetry cannot be produced
by any reflection or rotation.
Figure 1.13 Central symmetry
Similarity or self-similarity “quasi-symmetry”: A ny segment
o f a straight line (and its environs) is sim ilar to (that is, can be m agnified
or shrunk to become the same as) any other segment. See Figure 1.14.
This is not a symmetry because it does not preserve distances but it
could be called a “quasi-symm etry” because it does preserve the measure
o f angles.
Figure 1.14 Similarity “quasi-symmetry”
4 Logarithmic spirals such as the chambered nautilus have self­
sim ilarity as do m any fractals. (See example in Figure 1.15.)

The Symmetries of a Line____ 21
C learly,’other objects besides lines have some o f the symmetries
mentioned here. It is important for you to construct your own such exam ­
ples and attempt to find an object that has all o f the symmetries but is
not a line. This will help you to experience that straightness and the
seven symmetries discussed here are intimately related. You should
come to the conclusion that while other curves and figures have some o f
these symmetries, only straight lines have all o f them.
Local (and I nfinitesimal) Straightness
Previously, you saw how a straight line has reflection-in-the-line symme­
try and half-turn symmetry: One side o f the line is the same as the other.
But, as pointed out above, straightness is a local property in that whether
a segment o f a line is straight depends only on w hat is near the segment
and does not depend on anything happening away from the line. Thus
each o f the symmetries m ust be able to be thought o f (and experienced)
as applying only locally. This will become particularly important later
when we investigate straightness on the cone and cylinder. (See the dis­
cussions in Chapter 4.) For now, it can be experienced in the following
way:
When a piece o f paper is fo ld ed not in the center, the crease
is still straight even though the two sides o f the crease on the

22 Chapter 1 What Is Straight?
paper are not the same. (See Figure 1.16.)
So what is the role o f the sides when we are checking fo r
straightness using reflection symmetry? Think about what is
important near the crease in order to have reflection sym­
metry.
Figure 1.16 Reflection symmetry is local
W hen we talk about straightness as a local property, you m ay bring
oi?t some notions o f scale. For example, i f you see only a small portion
o f a very large circle, it w ill be indistinguishable from a straight line*.
This can be experienced easily on m any o f the m odem graphing pro­
grams for computers. Also, a microscope with a zoom lens will provide
an experience o f zooming. If a curve is smooth (or differentiable), then i f
we “zoom in” on any point o f the curve, eventually the curve will be
indistinguishable from a straight line segment. See Figure 1.17.
We sometimes use the terminology, infinitesimally straight, in place
o f the more standard terminology, differentiable. We say that a curve is
infinitesimally straight at a point p i f there is a straight line / such that if

Local (and Infinitesimal) Straightness____ 23
we zoom in enough on p , the line and the curve become indis­
tinguishable.1 W hen the curve is param etrized by arc length this is
equivalent to the curve having a well-defined velocity vector at each
point.
Figure 1.18 Straightness and smoothness depend on the scale
In contrast, we can say that a curve is locally straight at a point if
that point has a neighborhood that is straight. In the physical world the
usual use o f both smooth and locally straight is dependent on the scale at
which they are viewed. For example, we may look at an arch made out o f
wood — at a distance it appears as a smooth curve; then as we move in
closer we see that the curve is made by many short straight pieces o f fin­
ished (planed) boards, but when we are close enough to touch it, we see

24 Chapter 1 What Is Straight?
that its surface is made up o f smooth waves or ripples, and under a
microscope we see the non-sm oothness o f numerous twisting fibers. See
Figure 1.18. 1
1 This is equivalent to the usual definitions of being differentiable at p . For example, if
t(x) = f i p ) + f ‘ ( p ) ( x – p ) is the equation of the line tangent to the curve (x j{ x) ) at the
point ( p fi p ) ), then, given e > 0 (the distance of indistinguishability), there is a 8 > 0 (the
radius of the zoom window) such that, for |x – p\ < 8 (for x within the zoom window), J/(x) - t(x)\ < e [ f i x ) is indistinguishable from /(jc)] . This last inequality may look more familiar in the form f i x ) - t(x) — f i x ) - f i p ) ~ f ’(p)(x - p ) = { \fix) — fi p ) ] l( x - p ) - f ' i p ) }(x - p ) < e. In general, the value of S might depend on p as well as on s. Often the term smooth is used to mean continuously differentiable, which the interested reader can check is equivalent (on closed finite intervals) to, for each s > 0, there being one 8 > 0 that works
for all p.

_____________ Chapter 2
Straightness on S pheres
… [I]t will readily be seen how much space lies between the two
places themselves on the circumference of the large circle which is
drawn through them around the earth. … [W]e grant that it has been
demonstrated by mathematics that the surface of the land and water is
in its entirety a sphere, … and that any plane which passes through
the center makes at its surface, that is, at the surface of the earth and
of the sky, great circles, and that the angles of the planes, which
angles are at the center, cut the circumferences of the circles which
they intercept proportionately,…
— Ptolemy, Geographia (ca. 150 a.d.) Book One, Chapter II
This chapter asks you to investigate the notion o f straightness on a
sphere, drawing on the understandings about straightness you developed
in Problem 1.1.
Early History of S pherical G eometry
Observations o f heavenly bodies were carried out in ancient Egypt and
Babylon, mainly for astrological purposes and for making a calendar,
which was important for organizing society. Claudius Ptolem y (c.
100-178), in his Almagest, cites Babylonian observations o f eclipses and
stars dating back to the 8th century b.c. The Babylonians originated the
notion o f dividing a circle into 360 degrees — speculations as to why
360 include that it was close to the num ber o f days in a year, it was
convenient to use in their hexadecimal system o f counting, and 360 is
the num ber o f ways that seven points can be placed on a circle without
regard to orientation (for the ancients there were seven “wandering
bodies” — sun, moon, M ercury, Venus, M ars, Saturn, and Jupiter). But,
more important, the Babylonians developed a coordinate system
25

26 Chapter 2 Straightness on Spheres
(essentially the same as what we now call “spherical coordinates”) fork
,th e celestial sphere (the apparent sphere on which the stars, sun, moon,
and planets appear to move) vWth its pole at the north star. Thus it is a
m isconception to think that the use o f coordinates originated with
D escartes in the 17th century.
Figure 2.1 Armillary sphere (1687) showing (from inside out):
earth, celestial sphere, ecliptic, and the horizon
The ancient Greeks became fam iliar with Babylonian astronomy
around 4th century b .c . Eudoxus (408-355 b.c.) developed the “two-
sphere m odel” for astronomy. In this m odel the stars are considered to be
on the celestial sphere (which rotates one revolution a day westward
about its pole, the north star) and the sun is on the sphere o f the ecliptic,
whose equator is the path o f the sun and which is inclined to the equator
o f the celestial sphere at an angle that was about 24° in Eudoxus’ time
and is about 23lA 0 now. The sphere o f the ecliptic is considered to be
attached to the celestial sphere and has an apparent rotation eastward o f

Early History of Spherical Geometry____ 27
one revolution in a year. Both o f these spheres appear to rotate about
their poles. See Figure 2.1.
Autolycus, in On the Rotating Spheres (333-300 b.c.), introduced a
third sphere whose pole is the point directly overhead a particular
observer and whose equator is the visible horizon. Thus the angle
between the horizon and the celestial equator is equal to the angle
(measure at the center o f the earth) between the observer and the north
pole. Autolycus showed that, for a particular observer, some points
(stars) o f the celestial sphere are “always visible,” some are “always
invisible,” and some “rise and set.” See Figure 2.1.
The earliest known mathematical works that m ention spherical
geometry are A utolycus’ book ju st mentioned and Euclid’s Phaenomena
[AT: Berggren] (300 b.c.). Both o f these books use theorems from
spherical geometry to solve astrological problem s such as What is the
length o f daylight on a particular date at a particular latitude? Euclid
used throughout definitions and propositions from spherical geometry.
The definitions include A great circle is the intersection o f the sphere by
a plane through its center and The intersection o f the sphere by a plane
not through the center form s a (small) circle that is parallel to a unique
great circle. The assumed propositions include, for example, Suppose
two circles are parallel to the same great circle C but on opposite sides;
then the two circles are equal i f and only i f they cut o ff from some other
great circle equal arcs on either side o f C. (We will see sim ilar results in
Chapter 10.) There are other more com plicated results assumed, includ­
ing one about the comparison o f angles in a spherical triangle; see [A T :
Berggren], page 25. Thus, it is implied by A utolycus’ and E uclid’s
writings that there were previous works on spherical geometry available
to their readers.
Hipparchus o f Bithynia (190-120 b.c .) took the spherical coordi­
nates o f the Babylonians and applied them to the three spheres (celestial,
ecliptic, and horizon). The solution to navigational and astrological
problem^ (such as W hen will a particular star cross my horizon?) neces­
sitated relating the coordinates on one sphere with the coordinates on the
other spheres. This change o f coordinates necessitates what we now call
spherical trigonometry, and it appears that it was this astronomical
problem with spherical coordinates that initiated the study o f trigonom e­
try. Plane trigonometry, apparently studied system atically first by H ip­
parchus, seems to have been originally developed in order to help with
spherical trigonometry, which we will study in Chapter 20.

28 Chapter 2 Straightness on Spheres
\
The first systematic account o f spherical geometry was Sphaerica
o f Theodosius (around 200 b .c .) It consisted o f three books o f theorems
a^td construction problems. M ost o f the propositions o f Sphaerica were
extrinsic theorems and constructions about a sphere as it sits with its
center in Euclidean 3-space; but there were also propositions form ulated
in terms o f the intrinsic geometry on the surface o f a sphere without
‘ reference to either its center or 3-space. We will discuss the distinction
between intrinsic and extrinsic later in this chapter.
A more advanced treatise on spherical trigonometry was On the
Sphere by M enelaus (about 100 a.d.) There exist only edited Arabic
versions o f this work. In the introduction M enelaus defined a spherical
triangle as part o f a spherical surface bounded by three arcs o f great
circles, each less than a semicircle; and he defined the angles o f these
triangles. M enelaus’ treatise expounds geometry on the surface o f a
sphere in a way analogous to Euclid’s exposition o f plane geometry in
his Elements.
Ptolem y (100-178 a.d.) worked in Alexandria and wrote a book on
geography, Geographia (quoted at the beginning o f this chapter), and
Mathematiki Syntaxis (Mathematical Collections), which was the result
o f centuries o f knowledge from Babylonian astronomers and Greek
geometers. It became the standard W estern work on mathematical
astronomy for the next 1400 years. The Mathematiki Syntaxis in gener­
ally known as the Almagest, which is a Latin distortion o f the book’s
name in Arabic that was derived from one o f its Greek names. The
Almagest is important because it is the earliest existing work containing
a study o f spherical trigonometry, including specific functions, inverse
functions, and the computational study o f continuous phenomena.
M ore aspects o f the history o f spherical geometry will appear later
in this book in the appropriate places. For more readings (and references
to the prim ary literature) on this history, see [HI: Katz], Chapter 4, and
[HI: Rosenfeld], Chapter 1.
Problem 2.1 W hat Is S traight on a S phere?
Drawing on the understandings about straightness you developed in
Problem 1,1, this problem asks you to investigate the notion o f straight­
ness on a sphere. It is important for you to realize that, if you are not
building a notion o f straightness for yourself (for example, i f you are
taking ideas from books w ithout thinking deeply about them ), then you

P r o b l e m 2 .1 What Is Straight o n a Sphere?____ 29
will have difficulty building a concept o f straightness on surfaces other
than a plane. Only by developing a personal m eaning o f straightness for
yourself does it become part o f your active intuition. W e say active intui­
tion to emphasize that intuition is in a process o f constant change and
enrichment, that it is not static.
a. Imagine yo u rself to be a bug crawling around on a sphere. (This
bug can neither fly nor burrow into the sphere.) The b u g ’s
universe is ju st the surface’, it never leaves it. What is “straight”
fo r this bug? What will the bug see or experience as straight?
How can you convince yo u rself o f this? Use the properties o f
straightness (such as symmetries) that you talked about in
Problem 1.1.
b. Show (that is, convince yo u rself and give an argument to
convince others) that the great circles on a sphere are straight
with respect to the sphere, and that no other circles on the
sphere are straight with respect to the sphere.
S uggestions
Great circles are those circles that are the intersection o f the sphere with
a plane through the center o f the sphere. Examples include longitude
lines and the equator on the earth. Any pair o f opposite points can be
considered as the poles, and thus the equator and longitudes with respect
to any pair o f opposite points will be great circles. See Figure 2.2.
Figure 2.2 Great circles

3 0 Chapter 2 Straightness on Spheres
The first step to understanding this problem is to convince yourself
tftat great circles are straight lines on a sphere. Think what it is about the
great circles that would make the bug experience them as straight. To
better visualize w hat is happening on a sphere (or any other surface, for
that matter), you must use models. This is a point we cannot stress
enough. The use o f models will become increasingly important in later
problems, especially those involving more than one line. You must make
lines on a sphere to fully understand w hat is straight and why. An orange
or an old, worn tennis ball work well as spheres, and rubber bands make
good lines. Also, you can use ribbon or strips o f paper. Try placing these
it£ms on the sphere along different curves to see what happens.
Also look at the symmetries from Problem 1.1 to see i f they hold for
straight lines on the sphere. The important thing here is to think in
terms o f the surface o f the sphere, not the solid 3-dimensional ball.
Always try to imagine how things w ould look from the bug’s point o f
view. A good example o f how this type o f thinking works is to look at an
insect called a water strider. The water strider walks on the surface o f a
pond and has a very 2-dimensional perception o f the world around it —
to the water strider, there is no up or down; its whole world consists o f
the 2-dimensional plane o f the water. The w ater strider is very sensitive
to m otion and vibration on the w ater’s surface, but it can be approached
from above or below w ithout its knowledge. Hungry birds and fish take
advantage o f this fact. This is the type o f thinking needed to visualize
adequately properties o f straight lines on the sphere. For more discus­
sion o f w ater striders and other animals with their own varieties o f
intrinsic observations, see the delightful book The View from the Oak, by
Judith and Herbert Kohl [NA: Kohl and Kohl],
Definition. Paths that are intrinsically straight on a sphere
\ ( o r other surfaces) are called geodesics.
This leads us to consider the concept o f intrinsic or geodesic curvature
versus extrinsic curvature* As an outside observer looking at the sphere
in 3-space, all paths on the sphere, even the great circles, are curved -4-
that is, they exhibit extrinsic curvature. But relative to the surface o f the
sphere ( intrinsically), the lines may be straight and thus have intrinsic
curvature zeA). See the last section o f this chapter, Intrinsic Curvature.
B e sure to understand this difference and to see why all symmetries
(such as reflections) m ust be carried out intrinsically, or from the bug’s
V point o f view.

P r o b l e m 2 .1 What Is Straight on a Sphere? 31
It is natural for you to have some difficulty experiencing straight­
ness on surfaces other than the 2-dimensional plane; it is likely that you
will start to look at spheres and the curves on spheres as 3-dimensional
objects. Imagining that you are a 2-dimensional bug walking on a sphere
helps you to shed your limiting extrinsic 3-dimensional vision o f the
curves on a sphere and to experience straightness intrinsically. Ask
yourself the following:
♦ W hat does the bug have to do, when walking on a non-planar
surface, in order to w alk in a straight line?
♦ How can the bug check i f it is going straight?
Experim entation with m odels plays an important role here. W orking
with m odels that you create helps you to experience great circles as, in
fact, the only straight lines on the surface o f a sphere. Convincing
^ourself o f this notion will involve recognizing that straightness on the
plane and straightness on a sphere have common elements. W hen you
are comfortable with “great-circle-straightness,” you will be ready to
transfer the symmetries o f straight lines on the plane to great circles on a
sphere and, later, to geodesics on other surfaces. Here are some activities
that you can try, or visualize, to help you experience great circles and
their intrinsic straightness on a sphere. However, it is better for you to
come up with your own experiences.
♦ Stretch something elastic on a sphere. It will stay in place on
a great circle, but it will not stay on a small circle if the
sphere is slippery. Here, the elastic follows a path that is
approxim ately the shortest because a stretched elastic always
moves so that it will be shorter. This a very useful practical
criterion o f straightness.
♦ Roll a ball on a straight chalk line (or straight on a freshly
painted floor!). The chalk (or paint) will mark the line o f
contact on the sphere, and it will form a great circle.
♦ Take a narrow stiff ribbon or strip o f paper that does not
stretch, and lay it “flat” on a sphere. It will only lie (without
folds and creases) along a great circle. Do you see how this
property is related to local symmetry? This is sometimes
called the Ribbon Test. (For further discussion o f the Ribbon
Test, see Problems 3.4 and 7.6 o f [DG: Henderson].)

32 Chapter 2 Straightness on Spheres
♦ The feeling o f turning and “nontum ing” comes Up. W hy is it
t(iat on a great circle there is no turning and on a latitude line
there is turning? Physically, in order to avoid turning, the
bug has to move its left feet the same distance as its right
feet. O n a non-great circle (for example, a latitude line that is
not th e equator), the bug has to w alk faster with the legs that
lire on the side closer to the equator. This same idea can be
experienced by taking a small toy car with its wheels fixed to
parallel axes so that, on a plane, it rolls along a straight line.
On a sphere, the car will roll around a great circle; but it will
i)ot roll around other curves.
♦ Also notice that* on a sphere, straight lines are intrinsic
\circles (points on the surface a fixed distance along the
surface away from a given point on the surface) — special
circles whose circumferences are straight! N ote that the
equator is a circle with two intrinsic centers: the north pole
and the south pole. In fact, any circle (such as a latitude
circle) on a sphere has two intrinsic centers.
These activities will provide you with an opportunity to investigate
the relationships betw een a sphere and the geodesics o f that sphere.
Along the way, your experiences should help you to discover how great
circles on a sphere have m ost o f the same symmetries as straight lines on
a plane.
—J You should pause and not read further until you have
expressed your thinking and ideas about this problem.
S ymmetries of G reat C ircles
Reflection-through-itself symmetry: We can see this globally
by placing a hemisphere on a flat mirror. The hemisphere together with
the image in the m irror exactly recreates a whole sphere. Figure 2.3
shows a reflection through the great circle g.
Reflection-perpendicular-to-itself symmetry: A reflection
through any great circle will take any great circle (for example, g ‘ in
Figure 2.3) perpendicular to the original great circle onto itself.

Symmetries of Great Circles 33
Half-turn symmetry: A rotation through h a lf o f a full revolution
about any point P on a great circle interchanges the part o f the great
circle on one side o f P with the part on the other side o f P. See Figure
Figure 2.4 Half-turn symmetry
Rigid-motion-along-itself symmetry: For great circles on a
sphere, we call this a translation along the great circle or a rotation
around the poles o f that great circle. This property o f being able to move
rigidly along itself is not unique to great circles because any circle on the
sphere will also have the same symmetry. See Figure 2.5.

34 Chapter 2 Straightness on Spheres
Central symmetry, or point symmetry: Viewed intrinsically
(from the 2-dimensional b ug’s point-of-view), central symmetry through
a point P on the sphere sends any point A to the point at the same great
circle distance from P but on the opposite side. See Figure 2.6.
Intrinsically
Figure 2.6 Central symmetry through P
Extrinsically (viewing the sphere in 3-space) central symmetry through P
would send A to a point o ff the surface o f the sphere as shown in Figure
2.6. The only extrinsic central symmetry o f the sphere (and the only one
for great circles on the sphere) is through the center o f the sphere (which
is not on the sphere). The transform ation that is intrinsically central
symmetry is extrinsically half-turn symmetry (about the diameter
through P). Intrinsically, as on a plane, central symmetry does not differ
from half-turn symmetry with respect to the end result. This distinction
betw een intrinsic and extrinsic is important to experience at this point.

Symmetries of Great Circles____ 35
3 – d im e n s io n a l- r o ta tio n s y m m e tr y : This symmetry does not
hold for great circles in 3-space; however, it does hold for great circles
in a 3-sphere. See Problem 22.5.
You will probably notice that other objects on the sphere, besides
great circles, have some o f the symmetries mentioned here. It is impor­
tant for you to construct such examples. This will help you to realize that
straightness and the symmetries discussed here are intimately related.
* E v e r y G e o d e s i c I s a G r e a t C i r c l e
Notice that you were not asked to prove that every geodesic (intrinsic
straight line) on the sphere is a great circle. This is true but more diffi­
cult to prove. M any texts simply define the great circles to be the
“ straight lines” (geodesics) on the sphere. W e have not taken that
approach. We have shown that the great circles are intrinscially straight
(geodesics), and it is clear that two points on the sphere are always
joined by a great circle arc, which shows that there are sufficient great-
circle geodesics to do the geometry we wish.
To show that great circles are the only geodesics involves some
notions from differential geometry. In Problem 3.2b o f [DG: Henderson]
this is proved using special properties o f plane curves. M ore generally, a
geodesic satisfies a differential equation with the initial condition being
a point on the geodesic and the direction o f the geodesic at that point
(see Problem 8.4b o f [DG: Henderson]). Thus it follows from the analy­
sis theorem on the existence and uniqueness o f solutions to differential
equations that
Theorem 2 .1 . A t every point and in every direction on a smooth
surface there is a unique geodesic going from that point in that
direction.
From this it follows that all geodesics on a sphere are great circles. Do
you see why?
♦ I n t r i n s i c C u r v a t u r e
You have tried wrapping the sphere with a ribbon and noticed that the
ribbon will only lie flat along a great circle. (If you haven’t experienced
this yet, then do it now before you go on.) Arcs o f great circles are the
only paths on a sphere’s surface that are tangent to a straight line on a
piece o f paper wrapped around the sphere.

36____ Chapter 2 Straightness on Spheres
I f you wrap a piece o f paper tangent to the sphere around a latitude
circle (see Figure 2.7), then, extrinsically, the paper will form a portion
o f a cone and the curve on the paper will be an arc o f a circle when the
paper is flattened. The intrinsic curvature o f a path on the surface o f a
sphere can be defined as the curvature ( 1/radius) that one gets when one
“unw raps” the path onto a plane. For more details, see Chapter 3 o f
[DG: Henderson],
D ifferential geometers often talk about intrinsically straight paths
(geodesics) in terms o f the velocity vector o f the m otion as one travels at
a constant speed along that path. (The velocity vector is tangent to the
curve along which the bug walks.) For example, as you w alk along a
great circle, the velocity vector to the circle changes direction, extrinsi­
cally, in 3-space where the change in direction is toward the center o f the
sphere. “Tow ard the center” is not a direction that makes sense to a
2-dimensional bug whose whole universe is the surface o f the sphere.
Thus, the bug does not experience the velocity vectors as changing direc­
tion at points along the great circle; however, along non-great circles the
velocity vector will be experienced as changing in the direction o f the
closest center o f the circle. In differential geometry, the rate o f change,
from the bug’s point o f view, is called the covariant (or intrinsic)
derivative. As the bug traverses a geodesic, the covariant derivative o f
the velocity vector is zero. This can also be expressed in terms o f paral­
lel transport, which is discussed in Chapters 7, 8, and 10 o f this text. See
[DG: Henderson] for discussions o f these ideas in differential geometry.

W hat Is an A ngle?
_________ Chapter 3
A (plane) angle is the inclination to one another of two lines in
a plane which meet one another and do not lie in a straight
line. — Euclid, Elements, Definition 8 [Appendix A]
In this chapter you will be thinking about angles. In 3.1 we will investi­
gate various notions and definitions o f angles and what it means for
them to be considered to be the same (congruent). In 3.2 we will prove
the important Vertical Angle Theorem (VAT). It is not necessary to do
these parts in order — you may find it easier to do Problem 3.2 before
3.1 because it may help you think about angles. In a sense, you should be
working on 3.1 and 3.2 at the same time because they are so closely
intertwined. This provides a valuable opportunity to apply and reflect on
what you have learned about straightness in Chapters 1 and 2. This will
also be helpful in the further study o f straightness in Chapters 4 and 5;
but, if you wish, you may study this chapter after Chapters 4 and 5.
P r o b l e m 3.1 W h a t I s a n A n g l e ?
Give some possible definitions o f the term “angle.” Do all o f
these definitions apply to the plane as well as to spheres’? What
are the advantages and disadvantages o f each?
For each definition, what does it mean fo r two angles to be
congruent? How can we check?
Suggestions
Etymologically, “angle” comes through Old English, Old French, Old
German, Latin, and Greek words for “hook.” Textbooks usually give
37

38 Chapter 3 What Is an Angle?
some variant o f the definition: An angle is the union o f two rays (or
segments) with a common endpoint.
I f we start with two straight line segments with a common endpoint
and then add squiggly parts onto the ends o f each one, would we say that
the angle has changed as a result? Likewise, look at the angle form ed at
the lower-left-hand com er o f this piece o f paper. Even first grade
students will recognize this as an example o f an angle. Now, tear o ff the
com er (at least in your imagination). Is the angle still there, on the piece
you tore off? Now tear away more o f the sides o f the angles, being
careful not to tear through the com er. The angle is still there at the
com er, isn’t it? See Figure 3.1.
\S o what part o f the angle determines how large the angle is, or i f it
is an angle at all? W hat is the angle? It seems it cannot be merely a
\mion o f two rays. Here is one o f the m any cases where children seem to
know more than we do. Paying attention to these insights, can we get
better definitions o f “angle”? Do not expect to find one formal definition
that is completely satisfactory; it seems likely that no formal definition
cap capture all aspects o f our experience o f w hat an angle is.
There are at least three different perspectives from which we can
define “angle,” as follows:
♦ a dynamic notion o f angle — angle as movement;
♦ angle as measure; and,
♦ angle as geometric shape.
A dynamic notion o f angle involves an action: a rotation, a turning
point, or a change in direction between two lines. Angle as measure may
be thought o f as the length o f a circular arcs or the ratio betw een areas o f

Problem 3.1 What Is an Angle? 39
circular sector^. Thought o f as a geometric shape, an angle may be seen
as the delineation o f space by two intersecting lines. Each o f these
perspectives carries with it methods for checking angle congruency. You
can check the congruency o f two dynamic angles by verifying that the
actions involved in creating or replicating them are the same. If you feel
that an angle is a measure, then you must verify that both angles have the
same measurfe. I f you describe angles as geometric shapes, then you
describe how one angle can be made to coincide with the other using
isometries. W hich o f the above definitions has the m ost m eaning for
you? Are there any other useful ways o f describing angles?
N ote that we sometimes talk about directed angles, or angles with
direction. W hen considered as directed angles, we say that the angles a
and P in Figure 3.2 are not the same but have equal m agnitude and
opposite directions (or sense). Note the sim ilarity to the relationship
between line segments and vectors.
Figure 3.2 Directed angles
P roblem 3.2 V ertical A ngle T heorem (VAT)
a =P
Prove: Opposite angles fo rm ed by two intersecting straight lines are
congruent. [Note: Angles such as a and /? are called vertical

40____ Chapter 3 What Is an Angle?
angles.] What properties o f straight lines and/or the plane are
you using in your proof} Does your p r o o f also work on d
sphere? Why? Which definitions from Problem 3.2 are you using
Hn your proof?
Show how you would “m ove” a to make it coincide with p. We do not
have in mind a formal two-column proof that used to be in American
high school geometry. M athem aticians in actual practice usually use
V ‘proof” to mean “a convincing comm unication that answers — W hy?”
This is the notion o f p ro o f we ask you to use. There are three features o f
\ a proof:
• It m ust communicate (the words and drawings need to clearly
express what it is that you want to say — and they m ust be
understandable to your reader and/or listener.)
• It m ust be convincing (to yourself, to your fellow students, and to
your teacher; preferably it should be convincing to someone who
was originally skeptical).
• It m ust answer — Why? (W hy is it true? W hat does it m ean? Where
\ did it come from?)
The goal is understanding. W ithout understanding we will never be
fully satisfied. With understanding we want to expand that understand­
ing and to communicate it to others.
\ Symmetries were an important elem ent o f your solutions for
Problems 1.1 and 2.1. They will be very useful for this problem as well.
It is perfectly valid to think about m easuring angles in this problem , but
proofs utilizing line symmetries are generally simpler. If often helps to
think o f the vertical angles as whole geometric figures. Also, keep in
mind that there are many different ways o f looking at angles, so there are
many ways o f proving the vertical angle theorem. Make sure that your
notions o f angle and angle congruency in Problem 3.1 are consistent
with your proofs in Problem 3.2, and vice versa. Any o f the definitions
from 3.1 can, separately or together, help you prove the Vertical Angle
Theorem.
■33 You should pause and not read further until you have expressed
ybur own thinking and ideas about Problems 3.1 and 3.2.

P roblem 3 .2 Vertical Angle Theorem (VAT)____ 41
H i n t s f o r T h r e e D i f f e r e n t P r o o f s
In the following section, we will give hints for three different proofs o f
the Vertical Angle Theorem. N ote that a particular notion o f angle is
assumed in each proof. Pick one o f the proofs, or find your own different
proof that is consistent with a notion o f angle and angle congruence that
is most meaningful to you.
1st proof:
Figure 3.4 VAT using angle as measure
Each line creates a 180° angle. Thus, a + y – /3+ y. See Figure 3.4.
Therefore, we can conclude that a = /?. But why is this so? Is it
always true that i f we subtract a given angle from two 180° angles then
the rem aining angles are congruent? See Figure 3.5.
Figure 3.5 Subtracting angles and measures
Num erically, it does not make any difference how we subtract an
angle, but geometrically it makes a big difference. Behold Figure 3.6!
Here, s really cannot be considered the same as 8. Thus, m easure does
not completely express what we see in the geometry o f this situation. If
you wish to salvage this notion o f angle as measure, then you must

42 Chapter 3 What Is an Angle?
explain why it is that in this p ro o f o f the Vertical Angle Theorem y e a n
be subtracted from both sides o f the equation a + y= P + ?.
Figure 3.6 5 is not the same as e
2 n d p ro o f:* Consider two overlapping lines and choose any point
on them. Rotate one o f the lines, m aintaining the point o f intersection
and making sure that the other line rem ains fixed. See Figure 3.7.
\ W hat happens? W hat notion o f angle and angle congruency is at
\w o r k here?
3 rd p ro o f: W hat symmetries will take or onto p i See Figure 3.3 or,
3.4. Use the properties o f straight lines you investigated in Chapters 1»
\ and 2.

Chapter 4
Straightness on
Cylinders and Cones
If a cut were made through a cone parallel to its base, how
should we conceive of the two opposing surfaces which the cut
has produced — as equal or as unequal? If they are unequal,
that would imply that a cone is composed of many breaks and
protrusions like steps. On the other hand if they are equal, that
would imply that two adjacent intersection planes are equal,
which would mean that the cone, being made up of equal
rather than unequal circles, must have the same appearance as
a cylinder; which is utterly absurd.
— Democritus of Abdera (-460 — 380 b.c.)
This quote shows that cylinders and cones were the subject o f m athe­
m atical inquiry before Euclid (-3 6 5 — 300 b.c.). In this chapter we in­
vestigate straightness on cones and cylinders. You should be
comfortable with straightness as a local intrinsic notion — this is the
bug’s view. This notion o f straightness is also the basis for the notion o f
geodesics in differential geometry. Chapters 4 and 5 can be covered in
either order, but we think that the experience with cylinders and cones in
4.1 will help the reader to understand the hyperbolic plane in 5.1. If the
reader is comfortable with straightness as a local intrinsic notion, then it
is also possible to skip Chapter 4 i f Chapters 18 and 24 on geometric
m anifolds are not going to be covered. However, we suggest that you
read the sections at the end o f this chapter — Is “ Shortest” Always
“ Straight”? and Relations to Differential Geometry — at least enough to
find out what E uclid’s Fourth Postulate (see Appendix A) has to do with
cones and cylinders.
43

44 Chapter 4 Straightness on Cylinders and Cones
W hen looking at great circles on the surface o f a sphere, we were
able (except in the case o f central symmetry) to see all the symmetries o f
straight lines from global extrinsic points o f view. For example, a great
circle extrinsically divides a sphere into two hem ispheres that are mirror
images o f each other. Thus on a sphere, it is a natural tendency to use the
more fam iliar and comfortable extrinsic lens instead o f taking the bug’s
local and intrinsic point o f view. However, on a cone and cylinder you
must use the local, intrinsic point o f view because there is no extrinsic
view that will work except in special cases.
Problem 4.1 Straightness on C ylinders and C ones
a. What lines are straight with respect to the surface o f a cylinder
or a cone? Why? Why not?
b. Examine:
♦ Can geodesics intersect themselves on cylinders and
cones?
♦ Can there be more than one geodesic joining two
points on cylinders and cones?
♦ What happens on cones with varying cone angles,
including cone angles greater than 360°? These are
discussed starting in the next section.
Suggestions
Problem 4.1 is sim ilar to Problem 2.1, but this time the surfaces are
cylinders and cones.
Make paper models, but consider the cone or cylinder as continuing
indefinitely with no top or bottom (except, o f course, at the cone point).
Again, imagine yourself as a bug whose whole universe is a cone or
cylinder. As the bug crawls around on one o f these surfaces, what will
the bug experience as straight? As before, paths that are straight with
respect to a surface are often called the “geodesics” for the surface.
As you begin to explore these questions, it is likely that many other
related geometric ideas will arise. Do not let seemingly irrelevant excess
geometric baggage worry you. Often, you will find yourself getting lost
in a tangential idea, and th at’s understandable. Ultimately, however, the

Problem 4.1 Straightness on Cylinders and Cones____ 45
exploration o f related ideas will give you a richer understanding o f the
scope and depth o f the problem. In order to w ork through possible
confusion on this problem, try some o f the following suggestions others
have found helpful. Each suggestion involves constructing or using
models o f cones and cylinders.
♦ You may find it helpful to explore cylinders first before
beginning to explore cones. This problem has many aspects,
but focusing at first on the cylinder will sim plify some things.
♦ I f we make a cone or cylinder by rolling up a sheet o f paper,
will “straight” stay the same for the bug when we unroll it?
Conversely, if we have a straight line drawn on a sheet o f
paper and roll it up, will it continue to be experienced as
straight for the bug crawling on the paper? We are assuming
here that the paper will not stretch and its thickness is
negligible.
♦ Lay a stiff ribbon or straight strip o f paper on a cylinder or
cone. Convince yourself that it will follow a straight line with
respect to the surface. Also, convince yourself that straight
lines on the cylinder or cone, when looked at locally and
intrinsically, have the same symmetries as on the plane.
♦ I f you intersect a cylinder by a flat plane and unroll it, what
kind o f curve do you get? Is it ever straight? (One way to see
this curve is to dip a paper cylinder into water.)
♦ On a cylinder or cone, can a geodesic ever intersect itself?
How many tim es? This question is explored in more detail in
Problem 4.2, which the interested reader may turn to now.
♦ Can there be more than one geodesic joining two points on a
cylinder or cone? How many? Is there always at least one?
Again this question is explored in more detail in Problem 4.2.
There are several important things to keep in mind while working
on this problem. First, you absolutely must make models. I f you
attempt to visualize lines on a cone or cylinder, you are bound to make
claims that you would easily see are mistaken i f you investigated them
on an actual cone or cylinder. M any students find it helpful to make
models using transparencies.

46 Chapter 4 Straightness on Cylinders and Cones
Second, as with the sphere, you m ust think about lines and triangles
on the cone and cylinder in an intrinsic way — always looking at things
from a bug’s point o f view. We are not interested in w hat’s happening in
3-space, only what you would see and experience if you were restricted
to the surface o f a cone or cylinder.
And last, but certainly not least, you m ust look at cones o f different
shapes, that is, cones with varying cone angles.
C o n e s w i t h V a r y i n g C o n e A n g l e s
Geodesics behave differently on differently shaped cones. So an impor­
tant variable is the cone angle. The cone angle is generally defined as
the angle measured around the point o f the cone on the surface. Notice
that this is an intrinsic description o f angle. The bug could measure a
cone angle (in radians) by determining the circumference o f an intrinsic
circle with center at the cone point and then dividing that circumference
by the radius o f the circle. We can determine the cone angle extrinsically
in the following way: Cut the cone along a generator (a line on the cone
through the cone point) and flatten the cone. The m easure o f the cone
angle is then the angle m easure o f the flattened planar sector.
Attach
180°
Figure 4.1 Making a 180° cone
For example, if we take a piece o f paper and bend it so that h a lf o f
one side meets up with the other h a lf o f the same side, we will have a
180-degree cone (Figure 4.1). A 90° cone is also easy to make — just
use the com er o f a paper sheet and bring one side around to meet the
adjacent side.
Also be sure to look at larger cones. One convenient way to do this
is to make a cone with a variable cone angle. This can be accomplished
by taking a sheet o f paper and cutting (or tearing) a slit from one edge to
the center. (See Figure 4.2.) A rectangular sheet will work but a circular

Cones with Varying Cone Angles____ 47
sheet is easier to picture. Note that it is not necessary that the slit be
straight!
Figure 4.2 A cone with variable cone angle (0 -360°)
You have already looked at a 360° cone — it’s ju st a plane. The
cone angle can also be larger than 360°. A common larger cone is the
450° cone. You probably have a cone like this somewhere on the walls,
floor, and ceiling o f your room. You can easily make one by cutting a
slit in a piece o f paper and inserting a 90° slice (360° + 90° = 450°) as in
Figure 4.3.
Figure 4.3 How to make a 450° cone
You may have trouble believing that this is a cone, but rem ember
that ju st because it cannot hold ice cream does not mean it is not a cone.
If the folds and creases bother you, they can be taken out — the cone
will look ruffled instead. It is important to realize that when you change
the shape o f the cone like this (that is, by ruffling), you are only chang­
ing its extrinsic appearance. Intrinsically (from the bug’s point o f view)
there is no difference. You can even ruffle the cone so that it will hold
ice cream i f you like, although changing the extrinsic shape in this way
is not useful to a study o f its intrinsic behavior.
It may be helpful for you to discuss some definitions o f a cone, such
as the following: Take any simple (non-intersecting) closed curve a on a

48 Chapter 4 Straightness on Cylinders and Cones
sphere and the center P o f the sphere. A cone is the union o f the rays
that start at P and go through each point on a. The cone angle is then
equal to (length o f a)/(radius o f sphere), in radians. Do you see why?
Y ou can also make a cone with variable angle o f more than 180°:
Take two sheets o f paper and slit them together to their centers as in
Figure 4.4. Tape the right side o f the top slit to the left side o f the bottom
slit as pictured. Now slide the other sides o f the slits. Try it!
Experim ent by making paper examples o f cones like those shown in
Figure 4.4. W hat happens to the triangles and lines on a 450° cone? Is
the shortest path always straight? Does every pair o f points determine a
straight line?
Finally, also consider line symmetries on the cone and cylinder.
Check to see if the symmetries you found on the plane will w ork on
these surfaces, and rem em ber to think intrinsically and locally. A special
class o f geodesics on the cone and cylinder are the generators. These are
the straight lines that go through the cone point on the cone or go paral­
lel to the axis o f the cylinder. These lines have some extrinsic symme­
tries (can you see which onesl), but in general, geodesics have only
local, intrinsic symmetries. Also, on the cone, think about the region
near the cone point — w hat is happening there that makes it different
from the rest o f the cone?
It is best if you experiment with paper models to find out
what geodesics look like on the cone and cylinder before reading
further.
Figure 4.4 Variable cone angle larger than 360°

Geodesics on Cylinders 49
G e o d e s i c s o n C y l i n d e r s
Let us first look at the three classes o f straight lines on a cylinder. W hen
walking on the surface o f a cylinder, a bug might w alk along a vertical
generator. See Figure 4.5.
Figure 4.5 Vertical generators are straight
It might walk along an intersection o f a horizontal plane with the
cylinder, what we will call a great circle. See Figure 4.6
Figure 4.6 Great circles are intrinsically straight
Or, the bug might walk along a spiral or helix o f constant slope
around the cylinder. See Figure 4.7.
Figure 4.7 Helixes are intrinsically straight

50 Chapter 4 Straightness on Cylinders and Cones
W hy are these geodesics? How can you convince yourself? And
why are these the only geodesics?
G e o d e s i c s o n C o n e s
Now let us look at the classes o f straight lines on a cone.
Walking along a generator: W hen looking at straight paths on a
cone, you will be forced to consider straightness at the cone point. You
might decide that there is no way the bug can go straight once it reaches
the cone point, and thus a straight path leading up to the cone point ends
there. Or you might decide that the bug can find a continuing path that
has at least some o f the symmetries o f a straight line. Do you see which
path this is? O r you might decide that the straight continuing path(s?) is
the limit o f geodesics that ju st miss the cone point. See Figure 4.8.
Walking straight and around: If you use a ribbon on a 90° cone,
then you can see that this cone has a geodesic like the one depicted in
Figure 4.9. This particular geodesic intersects itself. However, check to
see that this property depends on the cone angle. In particular, if the
cone angle is more than 180°, then geodesics do not intersect them ­
selves. And i f the cone angle is less than 90°, then geodesics (except for
generators) intersect at least two times. Try it out! Later, in Problem 4.2,
we will describe a tool that will help you determine how the num ber o f
self-intersections depends on the cone angle.
Figure 4.8 Bug walking straight over the cone point

Geodesics on Cones 51
Figure 4.9 A geodesic intersecting itself on a 90° cone
* P r o b l e m 4.2 G l o b a l P r o p e r t i e s o f G e o d e s i c s
Now we will look more closely at long geodesics that wrap around
on a cylinder or cone. Several questions have arisen.
a. How do we determine the different geodesics connecting two
points’? How many are there! How does it depend on the cone
angle! Is there always at least one geodesic joining each pair o f
points! How can we justify our conjectures’?
b. How many times can a geodesic on a cylinder or cone intersect
itself! How are the self-intersections related to the cone angle!
A t what angle does the geodesic intersect itself! How can we
justify these relationships!
S uggestions
Here we offer the tool o f covering spaces, which may help you explore
these questions. The method o f coverings is so named because it utilizes
layers (or sheets) that each cover the surface. We will first start with a
cylinder because it is easier and then move on to a cone. *
* /7-Sh e e t e d C o v e r i n g s o f a C y l i n d e r
To understand how the method o f coverings works, imagine taking a
paper cylinder and cutting it axially (along a vertical generator) so that it
unrolls into a plane. This is probably the w ay you constructed cylinders
to study this problem before. The unrolled sheet (a portion o f the plane)
is said to be a 1-sheeted covering o f the cylinder. See Figure 4.10. I f you
marked two points on the cylinder, A and B, as indicated in the figure,
when the cylinder is cut and unrolled into the covering, these two points

52 Chapter 4 Straightness on Cylinders and Cones
become two points on the covering (which are labeled by the same
letters in the figure). The two points on the covering are said to be lifts
o f the points on the cylinder.
■>
Figure 4.10 A 1-sheeted covering of a cylinder
Now imagine attaching several o f these “sheets” together, end to
end. W hen rolled up, each sheet will go around the cylinder exactly once
— they will each cover the cylinder. (Rolls o f toilet paper or paper
towels give a rough idea o f coverings o f a cylinder.) Also, each sheet o f
the covering will have the points A and B in identical locations. You can
see this (assuming the paper thickness is negligible) by rolling up the
coverings and making points by sticking a sharp object through the
cylinder. This means that all the A ’s are coverings o f the same point on
the cylinder and all the B ‘s are coverings o f the same point on the cylin­
der. We ju st have on the covering several representations, or lifts, o f
each point on the cylinder. Figure 4.11 depicts a 3-sheeted covering
space for a cylinder and six geodesics joining A to B. (One o f them is the
most direct path from A to B and the others spiral once, twice, or three
times around the cylinder in one o f two directions.)
Figure 4.11 A 3-sheeted covering space for a cylinder

*«-Sheeted Coverings of a Cylinder 53
We could also have added more sheets to the covering on either the
right or left side. You can now roll these sheets back into a cylinder and
see what the geodesics look like. Rem ember to roll sheets up so that
each sheet o f the covering covers the cylinder exactly once — all o f the
vertical lines between the coverings should lie on the same generator o f
the cylinder. Note that i f you do this with ordinary paper, part or all o f
some geodesics will be hidden, even though they are all there. It may be
easier to see w hat’s happening if you use transparencies.
This method works because straightness is a local intrinsic
property. Thus, lines that are straight when the coverings are laid out in a
plane will still be straight when rolled into a cylinder. Rem ember that
bending the paper does not change the intrinsic nature o f the surface.
Bending only changes the curvature that we see extrinsically. It is im por­
tant always to look at the geodesics from the bug’s point o f view. The
cylinder and its covering are locally isometric.
Use coverings to investigate Problem 4.2 on the cylinder. The
global behavior o f straight lines may be easier to see on the covering.
* /7-Sh e e t e d ( B r a n c h e d ) C o v e r i n g s o f a C o n e
Figure 4.12 1-sheeted covering of a 270° cone
Figure 4.12 shows a 1-sheeted covering o f a cone. The sheet o f paper
and the cone are locally isometric except at the cone point. The cone
point is called a branch poin t o f the covering. We talk about lifts o f
points on the cone in the same way as on the cylinder. In Figure 4.12 we

54____ Chapter 4 Straightness on Cylinders and Cones
depict a 1-sheeted covering o f a 270° cone and label two points and their
lifts.
A 4-sheeted covering space for a cone is depicted in Figure 4.13.
Each o f the rays drawn from the center o f the covering is a lift o f a
single ray on the cone. Similarly, the points marked on the covering are
the lifts o f the points A and B on the cone. In the covering there are four
segments joining a lift o f A to different lifts o f B. Each o f these segments
is the lift o f a different geodesic segment joining A to B.
Think about ways that the bug can use coverings as a tool to expand
its exploration o f surface geodesics. Also, think about ways you can use
coverings to justify your observations in an intrinsic way. It is important
to be precise; you don’t want the bug to get lost! Count the num ber o f
ways in which you can connect two points with a straight line and relate
those countings with the cone angle. Does the num ber o f straight paths
only depend on the cone angle? Look at the 450° cone and see if it is
always possible to connect any two points with a straight line. Make
paper models! It is not possible to get an equation that relates the cone
angle to the number o f geodesics joining every pair o f points. However,
it is possible to find a form ula that works for most pairs. M ake covering
spaces for cones o f different size angles and refine the guesses you have
already made about the numbers o f self-intersections.

w-Sheeted (Branched) Coverings of a Cone 55
In studying the self-intersections o f a geodesic / on a cone, it may
be helpful for you to consider the ray R that is perpendicular to the line /.
(See Figure 4.14.) Now study one lift o f the geodesic l and its relation­
ship to the lifts o f the ray R. N ote that the seams between individual
wedges are lifts o f R.
L o c a l l y I s o m e t r i c
By now you should realize that when a piece o f paper is rolled or bent
into a cylinder or cone, the bug’s local and intrinsic experience o f the
surface does not change except at the cone point. Extrinsically, the piece
o f paper and the cone are different, but in terms o f the local geometry
intrinsic to the surface they differ only at the cone point.
Two geometric spaces, G and H , are said to be locally isometric at
the points G in G and H in H i f the local intrinsic experience at G is the
same as the experience at H. That is, there are neighborhoods o f G
and H that are identical in terms o f their intrinsic geometric properties. A
cylinder and the plane are locally isometric (at each point) and the plane
and a cone are locally isometric except at the cone point. Two cones are
locally isometric at the cone points only i f their cone angles are the
same.
Because cones and cylinders are locally isometric with the plane,
locally they have the same geometric properties. Later, we will show that
a sphere is not locally isometric with the plane — be on the lookout fo r a
result that will imply this.

56 Chapter 4 Straightness on Cylinders and Cones
Is ” S h o r t e s t ” A l w a y s ” S t r a i g h t ” ?
We are often told that “a straight line is the shortest distance between
two points,” but is this really true? As we have already seen on a sphere,
two points not opposite each other are connected by two straight paths
(one going one way around a great circle and one going the other way).
Only one o f these paths is shortest. The other is also straight, but not the
shortest straight path.
Consider a model o f a cone with angle 450°. Notice that such cones
appear comm only in buildings as so-called “outside com ers” (see Figure
4.3). It is best, however, to have a paper model that can be flattened.
Figure 4.15 There is no straight (symmetric) path from A to B
Use your model to investigate which points on the cone can be
joined by straight lines (in the sense o f having reflection-in-the-line
symmetry). In particular, look at points such as those labeled A and B in
Figure 4.15. Convince yourself that there is no path from A to B that is
straight (in the sense o f having reflection-in-the-line symmetry), and for
these points the shortest path goes through the cone point and thus is not
straight (in the sense o f having symmetry).
Figure 4.16 The shortest path is not straight (in the sense of symmetry)

Is “Shortest” Always “Straight”? 57
Here is another example: Think o f a bug crawling on a plane with a
tall box sitting on that plane (refer to Figure 4.16). This combination
surface — the plane with the box sticking out o f it — has eight cone
points. The four at the top o f the box have 270° cone angles, and the four
at the bottom o f the box have 450° cone angles (180° on the box and
270° on the plane). W hat is the shortest path betw een points X and Y,
points on opposite sides o f the box? Is the straight path the shortest? Is
the shortest path straight? To check that the shortest path is not straight,
try to see that at the bottom com ers o f the box the two sides o f the path
have different angular measures. (In particular, i f X and Y are close to
the box, then the angle on the box side o f the path measures a little more
than 180° and the angle on the other side measures alm ost 270°.)
♦ R e l a t i o n s t o D i f f e r e n t i a l G e o m e t r y
We see that sometimes a straight path is not shortest and the shortest
path is not straight. Does it then makes sense to say (as most books do)
that in Euclidean geometry a straight line is the shortest distance
between two points? In differential geometry, on “smooth” surfaces,
“straight” and “shortest” are m ore nearly the same. A smooth surface is
essentially w hat it sounds like. More precisely, a surface is smooth at a
point if, when you zoom in on the point, the surface becomes indistin­
guishable from a flat plane. (For details o f this definition, see Problem
4.1 in [DG: Henderson]. See also the last section and especially the
endnote in Chapter 1.) Mote that a cone is not smooth at the cone point,
but a sphere and a cylinder are both smooth at every point. The follow ­
ing is a theorem from differential geometry:
T heorem 4.1: I f a surface is smooth, then an intrinsically
straight line (geodesic) on the surface is always the shortest
path between “nearby” points. I f the surface is also complete
(every geodesic on it can be extended indefinitely), then any
two points can be jo in ed by a geodesic that is the shortest path
between them. See [DG: Henderson], Problems 7.4b and 7.4d.
Consider a planar surface with a hole removed. Check that for points
near opposite sides o f the hole, the shortest path (on the planar surface
with hole rem oved) is not straight because the shortest path must go
around the hole. We encourage the reader to discuss how each o f the
previous examples and problems is in harmony with this theorem.

58____ Chapter 4 Straightness on Cylinders and Cones
Note that the statement “every geodesic on the surface can be
extended indefinitely” is a reasonable interpretation o f E uclid’s Second
Postulate: Every limited straight line can be extended indefinitely to a
(unique) straight line [Appendix A, Postulate 2], Note that the Second
Postulate does not hold on a cone unless you consider geodesics to
continue through the cone point.
Also, Euclid defines a right angle as follows: When a straight line
intersects another straight line such that the adjacent angles are equal
to one another, then the equal angles are called right angles [Appendix
A, Definition 10]. N ote that if you consider geodesics to continue
through the cone point, then right angles at a cone point are not equal to
right angles at points where the cone is locally isometric to the plane.
And Euclid goes on to state as his Fourth Postulate: All right angles
are equal [Appendix A, Postulate 4], Thus, Euclid’s Second Postulate or
Fourth Postulate rules out cones and any surface with isolated cone
points. W hat is further ruled out by E uclid’s Fourth Postulate would
depend on formulating more precisely ju st what it says. It is not clear (at
least to the authors!) whether there is som ething we w ould want to call a
surface that could be said to satisfy E uclid’s Fourth Postulate and not be
a smooth surface. However, it is clear that Euclid’s postulate at least
gives part o f the m eaning o f “smooth surface,” because it rules out
isolated cone points.
W hen we were in high school geometry class we could not under­
stand why Euclid w ould have made such a postulate as his Postulate 4 —
how could they possibly not be equal? In this chapter we have discov­
ered that on cones right angles are not all equal.

Chapter 5
Straightness on
Hyperbolic Planes
[To son Janos:] For God’s sake, please give it [work on
hyperbolic geometry] up. Fear it no less than the sensual
passion, because it, too, may take up all your time and deprive
you of your health, peace of mind and happiness in life.
— Wolfgang Bolyai (1775-1856)
[EM: Davis and Hersh], page 220
We now study hyperbolic geometry. This chapter may be skipped i f the
reader will not be covering Chapter 17 and if, in the rem ainder o f this
book, the reader leaves out all mentions o f hyperbolic planes. However,
to skip studying hyperbolic planes w ould be to skip an important notion
in the history o f geometry and also to skip the geometry that m ay be the
basis o f our physical universe.
As with the cone and cylinder, we must use an intrinsic point o f
view on hyperbolic planes. This is especially true because, as we will
see, there is no standard embedding o f a complete hyperbolic plane into
3-space.
A S h o r t H i s t o r y o f H y p e r b o l i c G e o m e t r y
Hyperbolic geometry initially grew out o f the Building Structures Strand
through the work o f Janos Bolyai (1802-1860, Hungarian), and N. I.
Lobachevsky (1792-1856, Russian). Hyperbolic geometry is special
from a formal axiomatic point o f view because it satisfies all the postu­
lates (axioms) o f Euclidean geometry except for the parallel postulate. In
hyperbolic geometry straight lines can converge toward each other
59

60 Chapter 5 Straightness on Hyperbolic Planes
without intersecting (violating Euclid’s Fifth Postulate, see Appendix
A), and there is more than one straight line through a point that does not
intersect a given line (violating the usual high school parallel postulate,
which states that through any point P not on a given line / there is one
and only one line through P not intersecting /). See Figure 5.1.
Figure 5.1 Two geodesics through a point not intersecting a given geodesic
The reader can explore more details o f the axiomatic nature o f
hyperbolic geometry in Chapter 10. N ote that the 450° cone also violates
the two parallel postulates mentioned above. Thus the 450° cone has
some o f the properties o f the hyperbolic plane.
Hyperbolic geometry has turned out to be useful in various
branches o f higher mathematics. Also, the geometry o f binocular visual
space appears experimentally to be best represented by hyperbolic
geometry (see [HY: Zage]). In addition, hyperbolic geometry is one o f
the possible geometries for our three-dimensional physical universe —
we will explore this connection more in Chapters 18 and 24.
Hyperbolic geometry and non-Euclidean geometry are considered in
many books as being synonymous, but as we have seen there are other
non-Euclidean geometries, especially spherical geometry. It is also not

A Short History of Hyperbolic Geometry 61
accurate to say (as many books do) that non-Euclidean geometry was
discovered about 170 years ago. As we discussed in Chapter 2, spherical
geometry (which is clearly not Euclidean) was in existence and studied
(within the Navigation/Stargazing) by at least the ancient Babylonians,
Indians, and Greeks more than 2000 years ago.
M ost texts and popular books introduce hyperbolic geometry either
axiom atically or via “m odels” o f the hyperbolic geometry in the Euclid­
ean plane. These models are like our fam iliar map projections o f the
surface o f the earth. Like these maps o f the earth’s surface, intrinsic
straight lines on the hyperbolic plane are not, in general, straight in the
model (map) and the model, in general, distorts distances and angles. We
will return to the subject o f projection and m odels in Chapter 17. These
“models” grew out o f the A rt/Pattem Strand.
In this chapter we will introduce the geometry o f the hyperbolic
plane as the intrinsic geometry o f a particular surface in 3-space, in
much the same way that we introduced spherical geometry by looking at
the intrinsic geometry o f the sphere in 3-space. This is more in the flavor
o f the Navigation/Stargazing Strand. Such a surface is called an isomet­
ric embedding o f the hyperbolic plane into 3-space. We will construct
such a surface in the next section. Nevertheless, m any texts and popular
books say that David Hilbert (1862-1943, German) proved in 1901 that
it is not possible to have an isometric embedding o f the hyperbolic plane
onto a closed subset o f Euclidean 3-space. These authors miss what
Hilbert actually proved. In fact, Hilbert [HY: Hilbert] proved that there
is no real analytic isometry (that is, no isometry defined by real-valued
functions that have convergent pow er series). In 1964, N. V. Efimov
[HY: Efimov] extended H ilbert’s result by proving that there is no iso­
metric embedding defined by functions whose first and second deriva­
tives are continuous. W ithout giving an explicit construction, N. Kuiper
[HY: Kuiper] showed in 1955 that there is a differentiable isometric
embedding onto a closed subset o f 3-space.
The construction used here was shown to David by W illiam
Thurston (b.1946, Am erican) in 1978; and it is not defined by equations
at all, because it has no definite embedding in Euclidean space. The idea
for this construction is also included in [DG: Thurston], pages 49 and
50, and is discussed in [DG: Henderson], page 31. In Problem 5.3 we
will show that our isometric model is locally isometric to a certain
smooth surface o f revolution called the pseudosphere, which is well
known to locally have hyperbolic geometry. Later, in Chapter 17, we

62____ Chapter 5 Straightness on Hyperbolic Planes
will explore the various (non-isometric) models o f the hyperbolic plane
(these models are the way that hyperbolic geometry is presented in most
texts) and prove that these models and the isometric constructions here
produce the same geometry.
D e s c r i p t i o n o f A n n u l a r H y p e r b o l i c P l a n e s
In Appendix B we describe the details for five different isometric
constructions o f hyperbolic planes (or approximations to hyperbolic
planes) as surfaces in 3-space. It is very important that you actually
perform at least one o f these constructions. The act o f constructing the
surface will give you a feel for hyperbolic planes that is difficult to get
any other way. We will focus our discussions in the text on the descrip­
tion o f the hyperbolic plane from annuli that was proposed by Thurston.
Figure 5.2 Annular strips for making an annular hyperbolic plane

Description o f Annular Hyperbolic Planes 63
A paper model o f the hyperbolic plane may be constructed as
follows: Cut out many identical annular (“annulus” is the region between
two concentric circles) strips as in Figure 5.2. Attach the strips together
by taping the inner circle o f one to the outer circle o f the other. It is
crucial that all the annular strips have the same inner radius and the same
outer radius, but the lengths o f the annular strips do not matter. You can
also cut an annular strip shorter or extend an annular strip by taping two
strips together along their straight ends. The resulting surface is o f course
only an approximation o f the desired surface. The actual hyperbolic
plane is obtained by letting S —> 0 while holding the radius p fixed. Note
that since the surface is constructed (as S-> 0) the same everywhere it is
homogeneous (that is, intrinsically and geometrically, every point has a
neighborhood that is isometric to a neighborhood o f any other point). We
will call the results o f this construction the annular hyperbolic plane.
We strongly suggest that the reader take the time to cu t o u t carefully
several such a n n u li a n d ta p e th em to g e th e r as indicated.
Daina discovered a process for crocheting the annular hyperbolic
plane as described in Appendix B. The result is pictured in Figures 5.1
and 5.3 and other photos in this chapter.
Figure 5.3 A crocheted annular hyperbolic plane
There is also a polyhedral construction o f the hyperbolic plane that
is not directly related to the annular constructions but is easier for

64 Chapter 5 Straightness on Hyperbolic Planes
students (and teachers!) to construct. This construction (invented by
D avid’s son Keith Henderson) is called the hyperbolic soccer ball. See
Appendix B for the details o f the constructions (and tem plates) and
Figure 5.4 for a picture. It also has a nice appearance if you make the
heptagons a different color from the hexagons. As with any polyhedral
construction we cannot get closer and closer approxim ations to the
hyperbolic plane. There is also no apparent way to see the annuli.
Figure 5.4 The hyperbolic soccer ball
H y p e r b o l i c P l a n e s o f D i f f e r e n t R a d i i ( C u r v a t u r e )
Note that the construction o f a hyperbolic plane is dependent on p (the
radius o f the annuli), which we will call the radius o f the hyperbolic
plane. As in the case o f spheres, we get different hyperbolic planes
depending on the value o f p. In Figures 5.5-5.7 there are crocheted
hyperbolic planes with radii approxim ately 4 cm, 8 cm, and 16 cm. The
pictures were all taken from approxim ately the same perspective and in
each picture there is a centim eter rule to indicate the scale.

– • B3
F ig u r e 5 . 5 H y p e r b o lic p la n e w ith p a 4 c m
Figure 5.6 Hyperbolic plane with p s 8 c m
Figure 5.7 Hyperbolic plane with p » 16 cm
65

66 Chapter 5 Straightness on Hyperbolic Planes
N ote that as p increases, a hyperbolic plane becomes flatter and
flatter (has less and less curvature). Both spheres and hyperbolic planes,
as p goes to infinity, become indistinguishable from the ordinary flat
(Euclidean) plane. Thus, the plane can be called a sphere (or hyperbolic
plane) with infinite radius. In Chapter 7, we will define the Gaussian
Curvature and show that it is equal to U p 2 for a sphere and – U p 2 for a
hyperbolic plane.
P r o b l e m 5 . 1 W h a t I s S t r a i g h t i n a
H y p e r b o l i c P l a n e ?
a. On a hyperbolic plane, consider the curves that run radially
across each annular strip. Argue that these curves are intrinsi­
cally straight. Also, show that any two o f them are asymptotic, in
the sense that they converge toward each other but do not
intersect.
Look for the local intrinsic symmetries o f each annular strip and then
global symmetries in the whole hyperbolic plane. M ake sure you give a
convincing argument w hy the symmetry holds in the limit as S-> 0.
W e shall say that two geodesics that converge in this way are
asymptotic geodesics. Note that there are no geodesics (straight lines) on
the plane that are asymptotic.
b. Find other geodesics on your physical hyperbolic surface. Use
the properties o f straightness (such as symmetries) you talked
about in Problems 1.1, 2.1, and 4.1.
Try holding two points betw een the index fingers and thumbs on your
two hands. Now pull gently — a geodesic segment with its reflection
symmetry should appear betw een the two points. I f your surface is
durable enough, try folding the surface along a geodesic. Also, you may
use a ribbon to test for geodesics.
c. What properties do you notice fo r geodesics on a hyperbolic
plane? How are they the same as geodesics on the plane or
spheres, and how are they different from geodesics on the plane
and spheres?

P roblem 5.1 What Is Straight in a Hyperbolic Plane? 67
Explore properties o f geodesics involving intersecting, uniqueness, and
symmetries. Convince yourself as m uch as possible using your m odel —
full proofs for some o f the properties will have to w ait until Chapter 17.
* P r o b l e m 5.2 C o o r d i n a t e S y s t e m o n A n n u l a r
H y p e r b o l i c P l a n e
First, we will define coordinates on the annular hyperbolic plane that
will help us to study it in Chapter 17. Let p be the fixed inner radius o f
the annuli and let Hs be the approxim ation o f the annular hyperbolic
plane constructed from annuli o f radius p and thickness 8. On Hs pick
the inner curve o f any annulus, calling it the base curve’, and on this
curve pick any point as the origin O and pick a positive direction on this
curve. We can now construct an (intrinsic) coordinate system x s \ R2 -»
H s by defining x s ( 0 ,0) = O, xs(w , 0) to be the point on the base curve at
a distance w from O, and xg(w, 0) to be the point at a distance s from xs
(w, 0) along the radial geodesic through x s (w, 0), where the positive
direction is chosen to be in the direction from outer to inner curve o f
each annulus. Such coordinates are often called geodesic rectangular
coordinates. See Figure 5.8.
Figure 5.8 Geodesic rectangular coordinates on annular hyperbolic plane
a. Show that the coordinate map x is one-to-one and onto from the
whole o f R2 onto the whole o f the annular hyperbolic plane.
What maps to the annular strips, and what maps to the radial
geodesics?

68 Chapter 5 Straightness on Hyperbolic Planes
b. Let X and p be two o f the radial geodesics described in part a. I f
the distance between X and p along the base curve is w, then
show that the distance between them at a distance s = nSfrom
the base curve is, on the paper hyperbolic model,
s/S
Now take the limit as 8 —> 0 to show that the distance between X
and p o n the annular hyperbolic plane is
w exp {-s/p).
Thus, the coordinate chart jc preserves (does not distort) distances
along the (vertical) second coordinate curves but at x(a, b) the distances
along the first coordinate curve are distorted by the factor o f exp(-blp)
when compared to the distances in R2.
* P r o b l e m 5.3 T h e P s e u o o s p h e r e Is H y p e r b o l i c
Show that locally the annular hyperbolic plane is isometric to
portions o f a (smooth) surface defined by revolving the graph o f
a continuously differentiable function o f z about the z-axis. This
is the surface usually called the pseudosphere.
O utline of proof
1. Argue that each point on the annular hyperbolic plane is like any
other point. (Think o f the annular construction. A bout a point
consider a neighborhood that keeps its size as the width o f the
annular strips, S, shrinks to zero.)
2. Start with one o f the annular strips and complete it to a full annulus
in a plane. Then construct a surface o f revolution by attaching to the
inside edge o f this annulus other annular strips as described in the
construction o f the annular hyperbolic plane. (See Figure 5.9.) Note
that the second and subsequent annuli form truncated cones.
Finally, imagine the width o f the annular strips, 8, shrinking to zero.
3. Derive a differential equation representing the coordinates o f a
point on the surface using the geometry inherent in Figure 5.9. If
f r ) is the height (z-coordinate) o f the surface at a distance o f r from

*Problem 5.3 The Pseudosphere Is Hyperbolic 69
the z-axis, then the differential equation should be (rem em ber that p
is a constant)
Why?
dr —r
dz ~ J p 2 – r i ‘
4. Solve (using tables or com puter algebra systems) the differential
equation for z = fir) as a function o f r. N ote that you are not getting
r as a function o f z. This curve is usually called the tractrix.
5. Then argue (using a theorem from first-sem ester calculus) that r is a
continuously differentiable function o f z.
We can also crochet a pseudosphere by starting with 5 or 6 chain stitches
and continuing in a spiral fashion, increasing as when crocheting the
hyperbolic plane. See Figure 5.10. Note that, when you crochet beyond

70 Chapter 5 Straightness on Hyperbolic Planes
the annular strip that lays flat and forms a complete annulus, the surface
forms ruffles and is no longer a surface o f revolution (nor smooth).
Figure 5.10 Crocheted pseudosphere with ruffles
The term “pseudosphere” seems to have originated with Hermann
von H elm holtz (1821-1894, German), who was contrasting spherical
space with what he called pseudospherical space. However, Helmholtz
did not actually find a surface with this geometry. Eugenio Beltrami
(1835-1900, Italian) actually constructed the surface that is called the
pseudosphere and showed that its geometry is locally the same as
(locally isometric to) the hyperbolic geometry constructed by Loba-
tchevsky. (For more historical discussion, see [H I: Katz], pages
781-783.) M athem aticians searched further for a surface (in those days
“surface” meant “real analytic surface”) that would be the whole o f the
hyperbolic plane (as opposed to only being locally isometric to it). This
search was halted when Hilbert proved that such a surface was im possi­
ble (in his theorem that we discussed above at the end o f the first section
in this chapter, A Short History o f Hyperbolic Geometry).

Intrinsic/Extrinsic, Local/Global_____71
I n t r i n s i c / E x t r i n s i c , L o c a l / G l o b a l
On the plane or on spheres, rotations and reflections are both intrinsic in
the sense that they are experienced by a 2-dimensional bug as rotations
and reflections. These intrinsic rotations and reflections are also extrin­
sic in the sense that they can also be viewed as isometries o f 3-space.
(For example, the reflection o f a sphere through a great circle can also be
viewed as a reflection o f 3-space through the plane containing the great
circle.) Thus rotations and reflections are particularly easy to see on
planes and spheres. In addition, on the plane and sphere all rotations and
reflections are global in the sense that they take the whole plane to itself
or whole sphere to itself. (For example, any intrinsic rotation about a
point on a sphere is always a rotation o f the whole sphere.) On cylinders
and cones, intrinsic rotations and reflections exist locally because cones
and cylinders are locally isometric with the plane. However, some intrin­
sic rotations on cones and cylinders are extrinsic and global: for
example, rotations about the cone point on a circular cone with cone
angle <360°, or h a lf turns about any point on a cylinder. Rotations about the cone point on (>360°)-cones are global but not extrinsic. Rigid-
m otion-along-geodesic symmetries are extrinsic and global on cylinders
but are neither on any cone. (Do you see why?) Reflections, in general,
are neither extrinsic nor global (Can you see the exceptions on cones and
cylinders?).
P r o b l e m 5 . 4 R o t a t i o n s a n d R e f l e c t i o n s
o n S u r f a c e s
We can see from our physical hyperbolic planes that geodesics exist
joining every pair o f points and that these geodesics each have
reflection-in-them selves symmetry. (If you did not see this in Problem
5.1c, then go back and explore some more with your physical model. In
Chapter 17 we will prove rigorously that this is in fact true by using the
upper half-plane model.) In Chapter 17 we will show that these reflec­
tions are global reflections o f the whole hyperbolic space. N ote that
there do not exist extrinsic reflections o f the hyperbolic plane (em bed­
ded in Euclidean 3-space). M oreover, given all this, it is not clear that
there exist intrinsic rotations, nor is it necessarily clear what exactly
intrinsic rotations are.

72____ Chapter 5 Straightness on Hyperbolic Planes
a. Let l and m be two geodesics on the hyperbolic plane that inter­
sect at the point P. Look at the composition o f the reflection R/
through l with the reflection Rm through m. Show that this
composition RmR/ deserves to be called a rotation about P. What
is the angle o f the rotation?
Figure 5.11 Composition of two reflections is a rotation
Let A be a point on m and B be a point on /, and let Q be an arbitrary
point (not on m or /). Investigate where A, B, and Q are sent by R/ and
then by RmR,. See Figure 5.11. W hy are all points (except P) rotated
through the same angle and in the same direction.
W e will study symmetries and isometries in more detail in Chapter
11. In that chapter we will show that every isometry (on the plane,
spheres, and hyperbolic planes) is a composition o f one, two, or three
reflections.
*b. Show that Problem 3.2 (VAT) holds on cylinders, cones ( includ­
ing the cone points), and hyperbolic planes.
If you check your proof(s) o f 3.2 and m odify them ( if necessary) to
involve only symmetries, then you will be able to see that they hold also
on the other surfaces.
c. Define “rotation o f a figure about P through an angle 0 ”
without mentioning reflections in your definition.
W hat does a rotation do to a point not at P I
d. A popular high school text defines a rotation as the composition
o f two reflections. Is this a good definition? Why or why not?

_____________ Chapter 6
T riangles and Congruencies
Polygons are those figures whose boundaries are made
of straight lines: triangles being those contained by three,…
Things which coincide with one another are equal to
one another.
— Euclid, Elements, Definition 19 &
Common Notion 4 [Appendix A]
At this point, you should be thinking intrinsically about the surfaces o f
spheres, cylinders, cones, and hyperbolic planes. In the problem s to
come you will have opportunities to apply your intrinsic thinking when
you make your own definitions for triangle on these surfaces and investi­
gate congruence properties o f triangles.
In this chapter we will begin our study o f triangles and their congru­
encies on all the surfaces that you have studied: plane, spheres, cones,
cylinders, and hyperbolic spaces. (If you skipped any o f these surfaces,
you should find that this and the succeeding chapters w ill still make
sense, but you will want to limit your investigations to triangles on the
surfaces you studied.)
Before starting with triangles we m ust first discuss a little more
general inform ation about geodesics. *
* G e o d e s i c s A r e L o c a l l y U n i q u e
In previous chapters we have studied geodesics, intrinsically straight
paths. Our m ain criterion has been (in Chapter 2, 4, and 5) that a path is
intrinsically straight (and thus, a geodesic) if it has local intrinsic
reflection-through-itself symmetry. Using this notion, we found that
joining any pair o f points there is a geodesic that, on a sphere, is a great
73

circle and, on a hyperbolic space, has reflection-through-itself symmetry.
However, on more general surfaces, which may have no (even local)
reflections, it is necessary to have a deeper definition o f geodesic in
terms o f intrinsic curvature. (See for example, Chapter 3 o f [DG:
Henderson].) Then, to be precise, we m ust prove that the geodesics we
found on spheres and hyperbolic planes are the only geodesics on these
surfaces. It is easy to see that these geodesics that we have found are
enough to give, for every point and every direction from that point, one
geodesic proceeding from that point in that direction. To prove that these
are the only geodesics, it is necessary (as we have m entioned before) to
involve some notions from differential geometry. In particular, we must
first define a notion o f geodesic that will work on general surfaces that
have no (even local) intrinsic reflections. Then we show that a geodesic
satisfies a second-order (nonlinear) differential equation (see Problem
8.4b o f [DG: Henderson]). Thus, it follows from the analysis theorem on
the existence and uniqueness o f differential equations, with the initial
conditions being a point on the geodesic and the direction o f the
geodesic at that point, that
T heorem 6.0. For any given point and any direction at that
point on a smooth surface there is a unique geodesic starting
at that point and going in the given direction.
From this it follows that the geodesics with local intrinsic reflection-in-
itself symmetry, which we found in Problems 2.1, 4.1, and 5.1, are all
the geodesics on spheres, cylinders, cones, and hyperbolic planes.
P r o b l e m 6.1 P r o p e r t i e s o f G e o d e s i c s
In this problem we ask you to pull together a summary o f the properties
o f geodesics on the plane, spheres, and hyperbolic planes. M ostly, you
have already argued that these are true, but we summarize the results
here to rem ind us what we have seen and so that you can reflect again
about why these are true. Rem ember that cylinders and cones (not at the
cone point) are locally the same geometrically as (locally isometric to)
the plane; thus, geodesics on the cone and cylinder are locally (but not
globally) the same as straight lines on the plane.
a. For every geodesic on the plane, sphere, and hyperbolic plane
there is a reflection o f the whole space through the geodesic.
74____ Chapter 6 Triangles and Congruencies

Problem 6.1 Properties of Geodesics 75
b. Every geodesic on the plane, sphere, and hyperbolic plane can
be extended indefinitely (in the sense that the bug can walk
straight ahead indefinitely along any geodesic).
c. For every pair o f distinct points on the plane, sphere, and hyper­
bolic plane there is a (not necessarily unique) geodesic contain­
ing them.
d. Every pair o f distinct points on the plane or hyperbolic plane
determines a unique geodesic segment joining them. On the
sphere there are always at least two such segments.
e. On the plane or on a hyperbolic plane, two geodesics either
coincide or are disjoint or they intersect in one point. On a
sphere, two geodesics either coincide or intersect exactly twice.
Note that for the plane and hyperbolic plane, parts d and e are equivalent
in the sense that they each imply the other.
Notice that these properties distinguish a sphere from both the
Euclidean plane and from a hyperbolic plane; however, these properties
do not distinguish the plane from a hyperbolic plane.
P r o b l e m 6.2 I s o s c e l e s T r i a n g l e T h e o r e m ( I T T )
In order to start out with some common ground, let us agree on some
terminology: A triangle is a geometric figure form ed o f three points
( vertices) joined by three straight line (geodesic) segments (sides). A
triangle divides the surface into two regions (the interior and exterior).
The (interior) angles o f the triangle are the angles betw een the sides in
the interior o f the triangle. (As we will discuss below, on a sphere you
m ust decide which region you are going to call the interior — often the
choice is arbitrary.)
We will find the Isosceles Triangle Theorem very useful in studying
circles and the other congruence properties o f triangles because the two
congruent sides can be considered to be radii o f a circle.
a. (IT T ) Given a triangle with two o f its sides congruent, then are
the two angles opposite those sides also congruent? See Figure
6 .1. Look at this on all five o f the surfaces we are studying.

76 Chapter 6 Triangles and Congruencies
=> a = P
Figure 6.1 ITT
Use symmetries to solve this problem. First, look at this on the plane and
note what properties o f the plane you use. Then look on other surfaces.
Look for counterexam ples — i f there were counterexam ples, w hat could
they look like? If you think that ITT is not true for all triangles on a
particular surface, then describe a counterexam ple a n d look for a
smaller class o f triangles that do satisfy ITT on that surface. In the
process o f these investigations you will need to use properties o f geodes­
ics on the various surfaces (see Problem 6.1). State explicitly what
properties you are using. Hint: On a sphere two given points do not
determine a unique geodesic segment but two given points plus a third
point collinear to the given two d o determine a unique geodesic segment.
In your p ro o f o f Part a, try to see that you have also proved the fol­
lowing very useful result:
b. Corollary. The bisector o f the top angle o f an isosceles triangle
is also the perpendicular bisector o f the base o f that triangle.
You may also w ant to prove a converse o f ITT, but we will use it in
this book only in Problem 14.4:
c. Converse of ITT. I f two angles o f a triangle are congruent, then
are the sides opposite these angles also congruent?
Use symmetry and look out for counterexam ples — they do exist for the
converse.
C i r c l e s
To study congruencies o f triangles we will need to know something
about circles and constructions o f bisectors and perpendicular bisectors.
We define a c irc le intrinsically: A circle with center P and radius
PQ is the collection o f all points X which are connected to P by a
segment P X which is congruent to PQ.

Circles 77
N ote that on a sphere every circle has two (intrinsic) centers that are
antipodal (and, in general, two different radii). See Figure 6.2.
Now ITT can be used to prove theorems about circles. For example,
T heorem 6 .2 . On the plane, spheres, hyperbolic planes, and
locally on cylinders and cones, i f the centers o f two circles are
disjoint {and not antipodal), then the circles intersect in either
0, 1, or 2 points. I f the centers o f the two circles coincide (or
are antipodal), then the circles either coincide or are disjoint.
P ro o f: Because cylinders and cones are locally isometric to the plane,
locally and intrinsically circles will behave the same as on the plane.
Thus we limit the rem ainder o f this p ro o f to the plane, spheres, and
hyperbolic planes. Let C and C’ denote the centers o f the circles. See
Figure 6.3.
If C and C are antipodal on a sphere and the two circles intersect at
P, then a (extrinsic) plane through P (perpendicular to the extrinsic
diam eter CCr) will intersect the sphere in a circle that m ust coincide with
the two given circles. I f C and C’ coincide on a sphere and the circles
intersect, then pick the antipodal point to C as the center o f the first
circle, which reduces this to the case we ju st considered. I f C and C’
coincide on the plane and hyperbolic planes and the circles intersect,
then the circles have the same radii because two points are joined by
only one line segment. Thus, i f the centers coincide or are antipodal, the
circles coincide or are disjoint.

78____ Chapter 6 Triangles and Congruencies
Thus, we can now assume that C and C’ are disjoint and not antipo­
dal, so there is a unique geodesic joining the centers. I f A and B are two
points o f intersection o f the circles, then AACB and AAC’B are isosceles
triangles.
But given that AACB and AAC’B are isosceles, the Corollary to ITT
asserts that the bisectors o f ZAC B and ZAC ‘B must be perpendicular
bisectors o f their common base. Thus, the union o f the two angle bisec­
tors is straight and joins C and C’. So the union m ust be contained in the
unique geodesic determined by C and C’. Therefore, any pair o f intersec­
tions o f the two circles, such as A and B, m ust lie on opposite sides o f
this unique geodesic. Immediately, it follows that there cannot be more
than two intersections.
T r i a n g l e I n e q u a l i t y
On the plane, we have the following well-known result:
Triangle Inequality on Plane or a Hyperbolic Plane: The combined
length o f any two sides o f a triangle is greater than the length o f
the third side.
Do you see how this follows from our discussion o f circles? See
Figure 6.3.

Triangle Inequality 79
The Triangle Inequality is a partial expression o f the statement that
“a straight line is the shortest distance between two points” and because
o f that we might expect that the triangle inequality is false on spheres
(and cylinders and cones). Can you fin d a counterexample? But we can
make a simple change that makes it work on the sphere:
Triangle Inequality on a Sphere: The combined lengths o f any two
sides is not less than the (shortest) distance between the end
points o f the third side.
This leads to a definition that will be useful in some later chapters:
D e fin itio n . For any line segment, l, with endpoints, A, B, we
define the (special) absolute value o f l (in symbols, |/ |s ) to be
the shortest distance from A to B.
Note that on the plane (or in a vector space) this is the same as the usual
“absolute value” (or “norm”).
P r o b l e m 6.3 B i s e c t o r C o n s t r u c t i o n s
a. Show how to use a compass and straightedge to construct the
perpendicular bisector o f a straight line segment. How do you
know it is actually the perpendicular bisector? How does it work
on the sphere and hyperbolic plane?
Use ITT and Theorem 6.2. Be sure that you have considered all segment
lengths on the sphere. Hint: Use Figures 6.3 and the arguments in the
section Circles.
b. Show how to use a compass and straightedge to fin d the bisector
o f any planar angle. How do you know it actually is the angle
bisector? How does it work on the sphere and hyperbolic plane?
Use ITT and part a. Be sure that you have considered all sizes o f angles.
It is a part o f m athematical folklore that it is impossible to trisect an
angle with compass and straightedge; however, you will show in
Problem 15.4 that, in fact, it is possible. In addition, we will discuss
what is a correct statement o f the impossibility o f trisecting angles.

80 Chapter 6 Triangles and Congruencies
P r o b l e m 6.4 S i d e – A n g l e – S i d e (SAS)
We now investigate properties that will allow us to say that two triangles
are “the same” . Let us clarify some term inology that we have found to be
helpful for discussing SAS and other theorems. Two triangles are said to
be congruent if, through a combination o f translations, rotations, and
reflections, one o f them can be made to coincide with the other. In fact
(as we will prove in Chapter 11), we only need to use reflections. If an
even num ber o f reflections are needed, then the triangles are said to be
directly congruent, because in this case (as we show in Problem 11.3)
the reflections can be replaced pairwise by rotations and translations. In
this text we will focus on congruence and not specifically on direct
congruence’, however, some readers m ay wish to keep track o f the
distinction as we go along.
Figure 6.4 Direct congruence and congruence
In Figure 6.4, AABC is directly congruent to AA’B’C’ but AABC is
not directly congruent to AA”B”C”. However, AABC is congruent to both
A A B ‘C ‘ and A A “B”C” and we write: AABC = A A B ‘C ‘ = AA ‘B”C”.
Figure 6.5 SAS
Are two triangles congruent i f two sides and the included angle
o f one are congruent to two sides and the included angle o f the
other? See Figure 6.5.
In some textbooks SAS is listed as an axiom; in others it is listed as the
definition o f congruency o f triangles, and in others as a theorem to be

Problem 6.4 Side-Angle-Side (SAS) 81
proved. But no m atter how one considers SAS, it still makes sense and is
important to ask, W hy is SAS true on the plane?
Is SAS true on spheres, cylinders, cones, and hyperbolic planes’?
I f you fin d that SAS is not true fo r all triangles on a sphere or
another surface, is it true fo r sufficiently small triangles? Come
up with a definition fo r “small triangles ” fo r which SAS does
hold.
S uggestions
Be as precise as possible, but u se y o u r in tu itio n . In trying to prove SAS
on a sphere you will realize that SAS does not hold unless some restric­
tions are made on the triangles. Keep in m ind that everyone sees things
differently, so there are m any possible definitions o f “small.” Some may
be more restrictive than others (that is, they don’t allow as m any trian­
gles as other definitions). Use whatever definition makes sense for you.
Rem ember that it is not enough to simply state what a small triangle
is; you m ust also prove that SAS is true for the small triangles under
your definition — explain why the counterexam ples you found before
are now ruled out and explain why the condition(s) you list is (are) suffi­
cient to prove SAS. Also, try to come up with a basic, general p ro o f that
can be applied to all surfaces.
And rem em ber what we said before: By “p r o o f ’ we mean what
m ost m athematicians use in their everyday practice, that is, a convincing
communication that answers — Why? We do not ask for the two-column
proofs that used to be common in North Am erican high schools (unless,
o f course, you find the two-colum n p ro o f is sufficiently convincing and
answers — Why?). Y our p ro o f should convey the m eaning you are
experiencing in the situation. Think about why SAS is true on the plane
— think about what it means for actual physical triangles — then try to
translate these ideas to the other surfaces.
So why is SAS true on the plane? We will now illustrate one way o f
looking at this question. Referring to Figure 6.6, suppose that AABC
and AA ‘B ‘C are two triangles such that ABAC = ZB’A’C’, AB = A ‘B’ and
AC = A’C’. Reflect A4 ‘B’C’ about the perpendicular bisector (Problem
6.3) o f AA ‘ so that A ‘ coincides with A. Because the sides AC and A’C ‘
are congruent, we can now reflect about the angle bisector o f ZC’AC.
Now C’ coincides with C. (Why?) I f after this reflection B and B ‘ are not

82 Chapter 6 Triangles and Congruencies
coincident, then a reflection (about AC = A’ C r) will complete the process
and all three vertices, the two given sides, and the included angle o f the
two triangles will coincide. So why is it that, on the plane, the third sides
(BC and B ‘C ) must now be the sam el
Step 1: Two triangles with SAS Step 2: Reflect about the
perpendicular bisector of AA’
Figure 6.6 SAS on plane
Because the third sides (BC and B ‘C ) coincide, AABC is congruent
to SA’B’C’. (In the case that only two reflections are needed, the two
triangles are directly congruent.)
The p ro o f o f SAS on the plane is not directly applicable to the other
surfaces because properties o f geodesics differ on the various surfaces.
In particular, the num ber o f geodesics joining two points varies from
surface to surface and is also relative to the location o f the points on the
surface. On a sphere, for example, there are always at least two straight
paths jo ining any two points. As we saw in Chapter 4, the num ber o f
geodesics joining two points on a cylinder is infinite. On a cone the
number o f geodesics is dependent on the cone angle, but for cones with

Problem 6.4 Side-Angle-Side (SAS) 83
angles less than 180° there is more than one geodesic joining two points.
It follows that the argument made for SAS on the plane is not valid on
cylinders, cones, or spheres. The question then arises: Is SAS ever true
on those surfaces?
Look for triangles for which SAS is not true. Some o f the properties
that you found for geodesics on spheres, cones, cylinders, and hyperbolic
planes will come into play. As you look closely at the features o f trian­
gles on those surfaces, you may find that they challenge your notions o f
triangle. Your intuitive notion o f triangle may go beyond w hat can be
put into a traditional definition o f triangle. W hen you look for a defini­
tion o f sm all triangle for which SAS will hold on these surfaces, you
should try to stay close to your intuitive notion. In the process o f explor­
ing different triangles you m ay come up with examples o f triangles that
seem very strange. Let us look at some unusual triangles.
Figure 6.7 Two counterexamples for SAS on sphere
For instance, keep in m ind the examples in Figure 6.7. All the lines
shown in Figure 6.7 are geodesic segments o f the sphere. The two sides
and their included angle for SAS are marked. As you can see, there are
two possible geodesics that can be drawn for the third side — the short
one in front and the long one that goes around the back o f the sphere.
Remember that on a sphere, any two points define at least two geodesics
(an infinite num ber i f the points are at opposite poles).
Look for sim ilar examples on a cone and cylinder. You may decide
to accept the smaller triangle into your definition o f “small triangle” but
to exclude the large triangle from your definition. But what is a large

84 Chapter 6 Triangles and Congruencies
triangle? To answer this, let us go back to the plane. What is a triangle
on the plane? W hat do we choose as a triangle on the plane?
Figure 6.8 We choose the interior of a plane triangle to have finite area
On the plane, a figure that we want to call a triangle has all o f its
angles on the “inside.” Also, there is a clear choice for inside on the
plane; it is the side that has finite area. See Figure 6.8. But what is the
inside o f a triangle on a sphere?
The restriction that the area on the inside has to be finite does not
work for the spherical triangles because all areas on a sphere are finite.
So what is it about the large triangle that challenges our view o f
triangle? You might try to resolve the triangle definition problem by
specifying that each side m ust be the shortest geodesic betw een the
endpoints. However, be aware that antipodal points (that is, a pair o f
points that are at diam etrically opposite poles) on a sphere do not have a
unique shortest geodesic joining them. On a cylinder we can have a
triangle for which all the sides are the shortest possible segments, yet the
triangle does not have finite area. Try to find such an example. In
addition, a triangle on a cone will always bound one region that has
finite area, but a triangle that encircles the cone point m ay cause
problems.

Problem 6.5 Angle-Side-Angle (ASA) 85
P roblem 6.5 A ngle-S ide-A ngle (ASA)
Are two triangles congruent i f one side and the adjacent angles
o f one are congruent to one side and the adjacent angles o f
another? See Figure 6.9.
Suggestions
This problem is sim ilar in many ways to the previous one. As before,
look for counterexam ples on all surfaces; and i f ASA does not hold for
all triangles, see if it works for small triangles. If you find that you must
restrict yourself to small triangles, see if your previous definition o f
“small” still works; if it does not work here, then modify it.
It is also important to keep in mind when considering ASA that both
o f the angles m ust be on the same side — the interior o f the triangle. For
example, see Figure 6.10.
Figure 6.10 Angles of a triangle must be on same side
Let us look at a p ro o f o f ASA on the plane as depicted in Figure
6. 11.
The planar p ro o f for ASA does not w ork on spheres, cylinders, and
cones because, in general, geodesics on these surfaces intersect in more
than one point. But can you make the planar p ro o f work on a hyperbolic
plane?

86 Chapter 6 Triangles and Congruencies
Step 1: Two triangles with ASA Step 2: Reflect about the
Figure 6.11 ASA on the plane
As was the case for SAS, we m ust ask ourselves if we can find a
class o f small triangles on each o f the different surfaces for which the
above argument is valid. Y ou should check if your previous definitions
o f small triangle are too weak, too strong, or ju st right to make ASA true
on spheres, cylinders, cones, and hyperbolic planes. It is also important
to look at cases for which ASA does not hold. Just as with SAS, some
interesting counterexam ples arise.
Figure 6.12 Possible counterexample to ASA

Problem 6.5 Angle-Side-Angle (ASA) 87
In particular, try out the configuration in Figure 6.12 on a sphere.
To see what happens you will need to try this on an actual sphere. I f you
extend the two sides to great circles, what happens? You may instinc­
tively say that it is not possible for this to be a triangle, and on the plane
m ost people w ould agree, but try it on an physical sphere and see what
happens. Does it define a unique triangle? Rem ember that on a sphere
two geodesics always intersect twice.
Finally, notice that in our p ro o f o f ASA on the plane, we did not use
the fact that the sum o f the angles in a triangle is 180°. We avoided this
for two reasons. For one thing, to use this “fact” we would have to prove
it first. This is both time consuming and unnecessary. We will prove it
later (in different ways) in Chapters 7 and 10. M ore importantly, such a
p ro o f will not w ork on spheres and hyperbolic planes because the sum o f
the angles o f triangles on spheres and hyperbolic planes is not always
180° — see the triangles depicted in Figures 6.13 and 6.14. W e will
explore further the sum o f the angles o f a triangle in Chapter 7.
Rem ember that it is best to come up with a p ro o f that will w ork for
all surfaces because this will be more powerful, and, in general, will tell
us more about the relationship between the plane and the other surfaces.

88____ Chapter 6 Triangles and Congruencies
Figure 6.14 Hyperbolic triangle

_________ Chapter 7
A rea and Holonomy
We [my student and I] are both greatly amazed; and my share
in the satisfaction is a double one, for he sees twice over who
makes others see.
— Jean Henri Fabre, The Life o f the Fly,
New York: Dodd, Mead and Co., 1915, p. 300.
There are m any things in this chapter that have amazed us and our
students. W e hope you, the reader, will also be amazed by them. We will
find a form ula for the area o f triangles on spheres and hyperbolic planes.
We will then investigate the connections betw een area and parallel
transport, a notion o f local parallelism that is definable on all surfaces.
We will also introduce the notion o f holonomy, which has m any applica­
tions in m odem differential geometry and engineering.
Figure 7.1 Lune or biangle
89

9 0 Chapter 7 Area and Holonomy
Definition: A lune or biangle is any o f the four regions deter­
mined by two (not coinciding) great circles (see Figure 7.1).
The two angles o f the lune are congruent. ( Why?)
P roblem 7.1 T he A rea of a T riangle on a S phere
a. The two sides o f each interior angle o f a triangle A on a sphere
determine two congruent lunes with lune angle the same as the
interior angle. ‘Show how the three pairs o f lunes determined by
the three interior angles, a, f , y, cover the sphere with some
overlap. ( What is the overlap?)
Draw this on a physical sphere, as in Figure 7.2.
b. Find a form ula fo r the area o f a lune with lune angle 6 in terms
o f 6 and the (surface) area o f the sphere (o f radius p), which you
can call Sp. Use radian measure fo r angles.
Figure 7.2 Finding the area of a spherical triangle
Hint: W hat i f <9is n? ji/2? c. Find a form ula fo r the area o f a triangle on a sphere o f radius p. Problem 7.1 The Area of a Triangle on a Sphere 91 S uggestions This is one o f the problem s that you almost certainly m ust do on an actual sphere. There are simply too m any things to see, and the drawings we make on paper distort lines and angles too much. The best way to start is to make a small triangle on a sphere and extend the sides o f the triangle to complete great circles. Then look at w hat you’ve got. You will find an identical triangle on the other side o f the sphere, and you can see several lunes that extend out from the triangles. The key to this problem is to put everything in term s o f areas that you know. We will see later (Problem 14.3) that the area o f the whole sphere with radius p is Sp = 4tif?, or you may find a derivation o f this form ula in a m ultivari­ able calculus text, or you can ju st leave your answer in terms o f Sp. Historical Note Formulas expressing the area o f a spherical triangle and polygon in terms o f their respective angular excesses appeared in print for the first time in the paper “On a new ly discovered m easure o f area o f spherical triangles and polygons” published as an appendix to A new invention in algebra (Invention nouvelle en I ’algebre, Amsterdam, 1629) by the Flemish m athem atician A lbert Girard (1595-1632). For G irard’s proof, see [HI: Rosenfeld], pp. 27-31. A p ro o f sim ilar to the one indicated here was first published in 1781 by the m athem atician Leonhard Euler (1707-1783). Problem 7 .2 A rea of Hyperbolic T riangles Before we start to explore the area o f a general triangle on the hyper­ bolic plane, we first look for triangles with large area. a. On your hyperbolic plane draw as large a triangle as you can find. Compare your triangle with the large triangles that others have found. What do you notice? This part o f the problem is best to do communicating with other people. We can try to mimic the derivation o f the area o f spherical tri­ angles, but o f course there are no lunes and the area o f the hyperbolic plane is evidently infinite. Nevertheless, i f we focus on the exterior 92 Chapter 7 Area and Holonomy angles o f a hyperbolic triangle and look at the regions formed, we obtain a picture o f the situation in the annular hyperbolic plane. See Figure 7.3. Draw this picture on your hyperbolic plane. In Figure 7.3, a triangle is drawn with its interior angles, a, (3, y, and exterior angles, n - a , n-(3, n - y The three extra lines are geodesics that are asymptotic at both ends to an extended side o f the triangle. We call the region enclosed by these three extra geodesics an ideal triangle. In the annular hyperbolic plane these are not actually triangles because their vertices are at infinity. In Figure 7.3 we see that the ideal triangle is divided into the original triangle and three “triangles” that have two o f their vertices at infinity. We call a “triangle” with two vertices at infinity (and all sides geodesics) a 2/3-ideal triangle. You can use this decom po­ sition to determine the area o f a hyperbolic triangle in much the same way you determined the area o f a spherical triangle. So first we must investigate the areas o f ideal and 2/3-ideal triangles. Figure 7.3 Triangle with an ideal triangle and three 2/3-ideal triangles Problem 7.2 Area of Hyperbolic Triangles 93 Now let us look at 2/3-ideal triangles. b. Show that on the same hyperbolic plane, all H'i-ideal triangles with the same angle 6 are congruent. Think o f the p ro o f o f SAS (Problem 6.4). I f you have two 2/3-ideal trian­ gles with angle 6, then by reflections you can place one o f the Wangles on top o f the other. The triangles will then coincide except possibly for the third side, which is asymptotic to the two sides o f the angle 9. Now you m ust argue that these third sides must coincide. Or, in other words, w hy is the situation in Figure 7.4 impossible for 2/3-ideal triangles on a hyperbolic plane? Note, from Problem 5.4, that we can rotate so that any geodesic we pick is (after rotation) a radial geodesic. Problem 5.2 may be helpful. Figure 7.4 Are 2/3-ideal triangles with angle Q congruent? Because all the 2/3-ideal triangles are congruent, we can define an area function as A ,/a ) = area o f a 2/3 ideal triangle with exterior angle a on a hyperbolic plane with radius p. 94 Chapter 7 Area and Holonomy c. Show that the area function Ap is an additive function. That is, A/f a + p) = A f a ) + AfJ3). Look at the picture in Figure 7.5 and show that the area o f AADE is the sum o f the areas o f triangles AABC and AACE by showing that APDE is congruent to APBC. So we now have shown that the area function Ap is additive and it is also clearly continuous. T heorem 7 .2 . A continuous additive function (from the real numbers to the real numbers) is linear. Because the area function is additive, it also is true that it is linear over the rational numbers. For example, f 2 A f a ) = A f a ) + A f a ) = A f a + a ) = A f l a ) , and, if you set f t = 2 a, then the same equations show that ' A A f f f = Aff/ip). Thus, because the area function is continuous, the function must be linear (over the real numbers). Problem 7.2 Area of Hyperbolic Triangles____ 95 Therefore, A f a ) = constant x a, for 0 < a < n. We can conclude that A^O) = 0. If we let the finite vertex o f 2/3-ideal triangle go to infin­ ity, then the interior angle will go to zero and the exterior angle will go to n. Thus A fin:) must be the area o f an ideal triangle. So we have proved the following: All ideal triangles on the same hyperbolic plane have the same area, which we can call Ip. So we can write the area function as A f a ) = a x (Ip/ri). In fact, we will show in Problem 17.4 that all ideal triangles (on the same hyperbolic plane) are congruent. This is a result you may have guessed from your work in part a, you can also prove it using part b. We will also show in Problem 17.4 that the form ula for the area o f an ideal triangle is Ip = u p 1. Then it follows that A f a ) = a ft . Notice that it is only after 17.4 that we know for certain that 2/3-ideal triangles (and ideal triangles) have finite area, though you may have surmised that from your w ork on part a. d. Find a form ula fo r the area o f a hyperbolic triangle. Look at Figure 7.3 and put it together with what we have ju st proved. Historical Note The p ro o f is based on a p ro o f that C. F. Gauss included in a 1832 letter to J. B olyai’s father that is published in his collected works. P roblem 7.3 S um of the A ngles of a T riangle a. What can you say about the sum o f the interior angles o f trian­ gles on spheres and hyperbolic planes'? Are there maximum and/or minimum values fo r the sum? Look at triangles with non-zero area and use your formulas from Prob­ lems 7.1 and 7.2. b. What is the sum o f the (interior) angles o f a planar triangle? 96 Chapter 7 Area and Holonomy L et AABC be a triangle on the plane and imagine a sphere o f radius p passing through the points A, B, C. These three points also determine, a small spherical triangle on the sphere. Now imagine the radius p grow­ ing to infinity and the spherical triangle converging to the planar friangle. This result for the plane is normally proved after invoking a parallel postulate. Here, we are making the assum ption that the plane is a sphere o f infinite radius. We will turn to a discussion o f the various parallel postulates in Chapter 10. I ntroducing Parallel T ransport Imagine that you are walking along a straight line or geodesic carrying a horizontal stick that makes a fixed angle with the line you are walking on. If you w alk along the line m aintaining the direction o f the stick relative to the line constant, then you are performing a parallel transport o f that “direction” along the path. (See Figure 7.6.) “Parallel transport” is sometimes called “parallel displacem ent” or “parallel transfer” . Introducing Parallel Transport 97 To express the parallel transport idea, it is common term inology to say that ♦ r' is a parallel transport o f r along /; ♦ r is a parallel transport o f r' along /; ♦ r and r' are parallel transports along /; ♦ r can be parallel transported along / to /■'; or ♦ r ' can be parallel transported along / to r. On the plane there is a global notion o f parallelism — two lines in the same plane are parallel i f they do not intersect when extended. As we will see in Problem 8.2 (or, for the plane, from standard results in high school geometry), i f two lines are parallel transports along another line in the plane or the hyperbolic plane, then they are also parallel in the sense that they will not intersect i f extended. On a sphere this is not true — any two great circles on the same sphere intersect and intersect twice: In Problem 10.1 you will show that i f two lines in the plane are parallel transports along a third line, then they are parallel transports along every line that transverses them. This is also not true on a sphere and not true on a hyperbolic plane. For example, any two great circles (longitudes) through the north pole are parallel transports o f each other along the equator, but they are not parallel transports along great circles near the north pole. We will explore this aspect o f parallel transport more in Chapters 8 and 10. Parallel transport has become an important notion in differential geometry, physics, and m echanics. One important aspect o f differential geometry is the study o f properties o f spaces (surfaces) from an intrinsic point o f view. As we have seen, it is not in general possible to have a global notion o f direction that will determine when a direction (vector) at one point is the same as a direction (vector) at another point. However, we can say that they have the same direction with respect to a geodesic g i f they are parallel transports o f each other along g. Parallel transport can be extended to arbitrary curves, as we shall discuss at the end o f this chapter. There is even a m echanical device (first developed in third-century China!), called the “South-Seeking 98 Chapter 7 Area and Holonomy Chariot,” which will perform parallel transport along a curve on a surface. See [DG: Santander], W ith this notion it is possible to talk about the rate at which a particular vector quantity changes intrinsically along a curve (covariant differentiation). In general, covariant differentiation is useful in the areas o f physics and mechanics. In physics, the notion o f parallel transport is * central to some o f the theories that have been put forward as possible candidates for a “unified field theory,” a hoped-for but as yet unrealized theory that would unify "all known physical laws about forces o f nature. Historical Note According to [HI: Kline], p. 1132, parallel transport was first introduced in 1906 by L. E. J. Brouwer (1881-1966) in the context o f surfaces that are locally Euclidean, spherical, or hyperbolic. The general notion o f parallel transport was introduced in 1917 independently by Tullio Levi- Civita (1873-1941) and Gerhard Hessenberg (1874-1925). I ntroducing Holonomy Figure 7.7 The holonomy of a double-right triangle on a sphere Let us explore what happens when we parallel transport a line segment around a triangle. For example, consider on a sphere an isosceles trian­ gle with base on the equator and opposite vertex on the north pole (see Figure 7.7). N ote that the base angles are right angles. N ow start at the north pole with a vector (a directed geodesic segment — the gray arrows in Figure 7.7) and parallel transport it along one o f the sides o f the Introducing Holonomy 99 triangle until it reaches the base. Then parallel transport it along the base to the third side. Then parallel transport back to the north pole along the third side. Notice that the vector now points in a different direction than it did originally. You can follow a sim ilar story for the right hyperbolic triangle represented in Figure 7.8 and see that here also there is a differ­ ence between the starting vector and the ending parallel transported vector. This difference is called the holonomy o f the triangle. N ote that the difference angle is counterclockwise on the sphere and clockwise in the hyperbolic plane. Figure 7.8 Holonomy of a hyperbolic triangle This works for any small triangle, (that is, a triangle that is contained in an open hem isphere) on a sphere and for all triangles in a hyperbolic plane. We can define the holonom y o f a {small, i f on a sphere) triangle, X( A) , as follows: I f you parallel transport a vector (a directed geodesi$ segment) counterclockwise around the three sides o f a small triangle, then the holonomy o f the triangle is the smallest angle from the original position o f the vector to its fin a l position with counterclockwise being positive and clockwise being negative. 100 Chapter 7 Area and Holonomy For the spherical triangle in Figure 7.7 we see that the holonom y is positive and equal to the upper angle o f the triangle. For the hyperbolic triangle in Figure 7.8 we see that the holonomy is negative (clockwise). Holonom y can also be defined for large triangles on a sphere, but it is more complicated because o f the confusion as to what angle to measure. For example, w hat should be the holonomy when you parallel transport around the equator — 0 radians or 2tc radians? Compare with the form ula for the area o f a spherical triangle from Problem 7.2. ~ S IT , bos/- .favbs, D _ _ , " I f " ' ) ‘ f d i s - t i ^ _ P roblem 7 .4 T he Holonomy of a S mall T riangle Vind a form ula that expresses the holonomy o f a small triangle 'on a sphere and a form ula that expresses the holonomy o f any triangle on a hyperbolic plane. What is the holonomy o f a trian- 'gle on the plane? Suggestions W hat happens to the holonomy when you change the angle at the north pole o f the triangle in Figure 7.7? W hat happens i f you parallel transport around the triangle a vector pointing in a different direction? Parallel transport vectors around different triangles on your model o f a sphere. Try it on triangles that are very nearly the whole hemisphere and try it .on very small triangles. W hat do you notice? Try this also on your models o f the hyperbolic plane, again for different size triangles. A good way to approach t h l formula for general triangles is to start -Ovith any geodesic segment at one o f the angles o f the triangle and follow ij as it is parallel transported around the triangle. Keep track o f the relationships between the angles this segment makes with the sides and 'the exterior angles. See Figure 7.9, which is drawn for spherical triangles; the reader should be able to draw an analogous picture for a general hyperbolic triangle. Problem 7.4 The Holonomy of a Small Triangle 101 Pause, explore, and write out your ideas for this problem before reading further. 102 Chapter 7 Area and Holonomy T h e G a u s s - B o n n e t F o r m u l a f o r T r i a n g l e s In working on Problem 7.4 you should find (among other things) that 'The holonomy o f a {small, i f on a sphere) triangle is equal to 2n minus the sum o f the exterior angles or equal to the sum o f 'the interior angles minus n. Let fix, f f , fh be the interior angles o f the triangle and a\, a2, a} the exterior angles. Then algebraically the statement above can be w ritten as TC{A) = 27r - (a i + a 2 + a2) = {fi\ + fh + fh ) - n. The quantity [ Z/7, - n ] = [2 n - Xa, ] is also called the excess o f A, and when the excess is negative, the positive quantity [ jt - £/?, ] = [X «, - 27t ] v is called the defect o f A. If you have not already seen it, pote now the close connection between the holonomy, the excess, and the area o f a triangle. Note that the holonomy is positive for triangles on a sphere and negative for trian­ gles in a hyperbolic plane (and zero for triangles on a plane). One conse­ quence o f this formula is that the holonom y does not depend on either the vertex or the vector we start with. This is to be expected because parallel transport does not change the relative angles o f any figure. Following Problems 7.1, 7.2, and 7.4, we can write the result for triangles on a sphere with radius p in this form: Sphere: H {S) = (J3\ + fh + fh ) - n = Area(A) 4n/Sp = Area(A) p~2, For a hyperbolic plane made with annuli with radius p, we get: Hyperbolic: K{A) = (/? i+ fh + f h ) - n - -Area(A) n/Ip = -A rea(A ) p 1. The quantity p 1 is traditionally called the Gaussian curvature or ju st plain curvature o f the sphere and - p 2 is called the (Gaussian) curvature o f the hyperbolic plane. I f K denotes the {Gaussian) curvature as ju st defined, then the formula {J3x + ft.i + fh ) - 7i = Area(A) K The Gauss-Bonnet Formula for Triangles 103 is called the Gauss-Bonnet Formula (for trian g les)., The formula is originally due to C. F. Gauss (1777-1855, German) and was extended by P. O. Bonnet (1819-1892, French), as we will describe at the end o f this chapter. Gan you see how this result gives a bug on the surface an intrinsic way o f determining the quantity K and thus also determining the extrin-' sic radius p i ' The Gauss-Bonnet Form ula not only holds for triangles in an open hemisphere or in a hyperbolic plane but can also be extended to anyi simple (that is, non-intersecting) polygon (that is, a closed curve made* up o f a finite num ber o f geodesic segments) contained in an open hemisphere or in a hyperbolic plane. ♦ P r o b l e m 7.5 G a u s s - B o n n e t F o r m u l a f o r P o l y g o n s Definition. The holonomy o f a simple polygon, K{T), in an open hemisphere or in a hyperbolic plane is defined as follows! I f you parallel transport a vector (a directed geodesic seg­ ment) counterclockwise around the sides o f the simple poly­ gon, then the holonomy o f the polygon is the smallest angle measured counterclockwise from the original position o f the vector and its fin a l position. F ig u r e 7 . 1 0 E x te r io r a n g le s 104 Chapter 7 Area and Holonomy If you w alk around a polygon with the interior o f the polygon on the left, the exterior angle at a vertex is the change in the direction at that vertex. This change is positive i f you turn counterclockwise and negative i f yoy turn clockwise. (See Figure 7.10.) We will first look at convex polygons because this is the only case we will need later and it is easier to understand. A region is called convex if every pair o f points in the region can be joined by a geodesiij Segment lying w holly in the region. a. (S'how that i f T is a convex polygon in an open hemisphere or in a hyperbolic plane, then , , ‘ = 2 ti- l a , = X fi - (n - 2)n= Area(T) K, ^ where Xa, is the sum o f the exterior angles, £/?, is the sum o f the \interior angles, n is the number o f sides, and K is the Gaussian curvature. Divide the convex polygon into triangles as in Figure 7.11. Now apply 7.4 to each triangle and carefully add up the results. You can check \ directly that X ( V) = 2 n - l a , . b. Prove that every simple polygon on the plane or on a hemisphere or on a hyperbolic plane can be dissected into triangles without adding extra vertices. F ig u r e 7 . 1 1 D iv id in g a c o n v e x p o ly g o n in to tr ia n g le s *Problem 7.5 Gauss-Bonnet Formula for Polygons 105 S uggestions Look at this on the plane, hemispheres,’and hyperbolic planes. The diffi­ culty in this problem is coming up with a method that works for all * polygons, including very general or complex ones, such as the polygori in Figure 7.12. Figure 7.12 General polygon You may be tem pted to try to connect nearby vertices to create triangles, but how do we know that this is always possible? How do you know that in any polygon there is even one pair o f vertices that can be joined in the interior? The polygon may be so complex that parts o f it get in the way o f what y ou’re trying to connect. So you might start by giving a convincing argument that there is at least one pair o f vertices that can be joined by a segment in the interior o f the polygon. To see that there is som ething to prove here, there are examples o f' polyhedra in 3-space with no pair o f vertices that can be joined in the interior? This interesting fact was first published in 1911 by N. J. Lennes; thus such polyhedra are often called Lennes Polyhedra. One example o f a Lennes Polyhedron in depicted in Figure 7.13. The polyhe­ dron consists o f eight triangular faces and six vertices. Each vertex is joined by an edge to four o f the other vertices, and the straight line segment jo ining it to the fifth vertex lies in the exterior o f the polygon. Therefore, it is impossible to dissect this polyhedron into tetrahedra without adding extra vertices. This example and some history o f the problem are discussed in [DI: Eves, p. 211] and [DI: Ho]. 106 Chapter 7 Area and Holonomy Note that there is at least one convex vertex (a vertex with interior angle less than n) on every polygon (in fact, it is not too hard to see that there must be at least three such vertices). To see this, pick any geodesic ih the exterior o f the polygon and parallel transport it toward the polygon until it first touches the'polygon. It is easy to see that the line must now be intersecting the polygon at a convex vertex. Figure 7.13 A polyhedron with vertices not joinable in the interior c. -Show that i f Y is a simple polygon in an open hemisphere or in a hyperbolic plane, then K(T) = 2 n - Iff, = I/?, - ( n - 2)n = Area(T) K \ where 'La, is the sum o f the exterior angles, Lfi, is the sum o f thd interior angles, and K is the Gaussian curvature. Start by applying part b. Then proceed as in part a, but for this part you,’ may find it easier to show that the holonomy o f the polygon is the sum o f the holonomies o f the triangles by removing one triangle at a time! Again, you can check directly that 3{{T) = 2re - * * G a u s s - B o n n e t F o r m u l a f o r P o l y g o n s o n S u r f a c e s The above discussion o f holonomy is in the context o f an open hem i­ sphere and a hyperbolic plane, but the results have a much more general applicability and constitute an important aspect o f differential geometry. In particular, 1 we can extend this result even further to general surfaces, *Gauss-Bonnet Formula for Polygons on Surfaces 107 even those o f non-constant curvature. In fact, Gauss defined the (Gaussian) curvature K(p) at a point p on any surface to be i K(p) = lim ^ p K{A) / A (A) , « where the limit is taken over a sequence o f small (geodesic) triangles that converge to p. The reader can check that the Gaussian curvature o f a sphere (with radius p) is \! ( f and that the Gaussian curvature o f a hyper­ bolic plane (with radius p, the radius o f the annular strips) is - M ( f . This definition leads us to another formula, namely, Theorem 7.5a. The Gauss-Bonnet Formula fo r Polygons on Surfaces On any smooth surface (2-manifold), i f T is a (geodesic) polygon that bounds a contractible region, then K{Y) = 2 : 1 - 1 a, = JI(d K(p)dA,>
where the integral is the {surface) integral over 7(T), th e ,
interior o f the polygon:
A region is said to be contractible i f it can be continuously deformed to „
a point in its interior. See Figure 7.14 for examplesr
Not contractible
F ig u r e 7 . 1 4

108 Chapter 7 Area and Holonomy
The p ro o f o f this form ula involves dividing the interior o f T into
rrtany triangles, each so small that the curvature K is essentially constant
‘over its interior, and then applying the Gauss-Bonnet Formula for
spheres and hyperbolic planes to each o f the triangles.
In addition, all o f the versions o f the Gauss-Bonnet Formula given
thus far can be extended to arbitrary, simple, piecewise smooth, closed
curves. (It is this extension that was B onnet’s contribution to the Gauss-
Bonnet Formula.) I f y is such a curve, then we can define the holonomy
,K{y) = lim X(y,), where the lim it is over a sequence (which converges
point-wise to y) o f geodesic polygons {y;} whose vertices lie on y. Using
this definition, the Gauss-Bonnet Formula can be extended even further.
Theorem 7 .5 b . The Gauss-Bonnet Form ula fo r Curves That
B ound a Contractible Region
On a sphere or hyperbolic plane, with (constant) curvature K,
K { f ) = A ( / ) K ,
\ where A(y) is the area o f the region bounded by y
On general surfaces,
X { f ) = \\HnK(p) dA,
where I(y) is the interior o f the region bounded by y.
Another version o f the Gauss-Bonnet Formula is discussed in Prob­
lem 17.6, where the integral is over the whole surface.
For further discussions, see Differential Geometry. A Geometric
Introduction [DG: Henderson], Chapters 5 and 6, especially Problems
5.4 and 6.4.

_________ Chapter 8
Parallel T ransport
Parallel straight lines are straight lines lying in a plane which
do not meet if continued indefinitely in both directions.
— Euclid, Elements, Definition 23 [Appendix A]
In this chapter we will further develop the notion o f parallel transport
that was introduced in Chapter 7. This chapter may be studied independ­
ently o f Chapter 7 if you read the section in Chapter 7 entitled Introduc­
ing Parallel Transport. The basic idea o f Chapter 8 is to collect all the
results related to parallelism that can be examined without assum ing any
special properties on the plane about parallel lines or about the sum o f
the angles on the plane. These properties (postulates) will be discussed
in detail in Chapter 10.
P roblem 8.1 Euclid’s Exterior A ngle T heorem
(EEAT)
109

110 Chapter 8 Parallel Transport
a. Any exterior angle o f a triangle is greater than each o f the oppo­
site interior angles. W a rn in g : E uclid’s EAT is not the same as
the Exterior Angle Theorem usually studied in high school.
b. When is EE A T true on the plane, on a sphere, and on a hyper­
bolic plane?
S uggestions
You may find the following hint (which is found in E uclid’s proof)
useful: Draw a line from the vertex o f a to the midpoint, M, o f the
opposite side, BC. Extend that line beyond M to a point A ‘ in such a way
that A M = M A ‘. Join A 1 to C. This hint will be referred to as E uclid ’s hint
and is pictured in Figure 8.2.
Be cautious transferring this hint to a sphere. It will probably help
to draw E uclid’s hint directly on a physical sphere.
It is not necessary to use E uclid’s hint to prove EEAT, and in fact
many people don’t “see” the hint. Another perfectly good way to prove
EEAT is to use Problem 8.2. Problems 8.1 and 8.2 are very closely
related, and they can be done in either order. It is also fine to use 8.1 to
prove 8.2 or use 8.2 to prove 8.1, but o f course d on’t do both. As a final
note, rem ember you do not have to look at figures using only one orien­
tation — rotations and reflections o f a figure do not change its
properties, so if you have trouble “seeing” something, check to see if it’s
something you’re fam iliar with by orienting it differently on the page.
EEAT is not always true on a sphere, even for small triangles. Look
at a counterexam ple as depicted in Figure 8.3. Then look at your p ro o f o f

P r o b l e m 8 .1 Euclid’s Exterior Angle Theorem (EEAT) 111
EEAT on the plane. It is very likely that your p ro o f uses properties o f
angles and triangles that are true for small triangles on the sphere. Thus
it may appear to you that your planar proof is also a valid proof o f EEAT
for small triangles on the sphere. But there is a counterexample.
This could be, potentially, a very creative situation for you —
whenever you have a proof and counterexample o f the same result,
you have an opportunity to learn something deep and meaningful.
So try out your planar proof o f EEAT on the counterexam ple in Figure
8.3 and see what happens. Then try it on both large and small spherical
triangles. If you can determine exactly which triangles satisfy EEAT and
which triangles d on’t satisfy EEAT, then this information will be useful
(but not crucial) to you in later problems.
P roblem 8 .2 S ymmetries of Parallel T ransported
Lines
Consider two lines, r and r’, that are parallel transports o f each
other along a third line, l. Consider now the geometric figure
that is form ed by the three lines and look fo r the symmetries o f
that geometric figure. See Figure 8.4.

112 Chapter 8 Parallel Transport
What can you say about the lines r and r l Do they intersect? I f
so, where? Look at the plane, spheres, and hyperbolic planes.
Figure 8.4 What can you say about r and r l
Suggestions
Parallel transport was already informally introduced in Chapter 7. In
Problem 8.2 you have an opportunity to explore the concept further and
prove its implications on the plane and a sphere. You will study the
relationship between parallel transport and parallelism , as well.
It is common in high school to use E uclid’s definition o f parallel
lines as “straight lines lying in a plane which do not m eet i f continued
indefinitely in both directions.” But this is an inhuman definition —
there is no way to check all points on both lines to see i f they ever meet.
This definition is also irrelevant on a sphere because we know that all
geodesics on a sphere will cross each other. But we can m easure the

Problem 8.2 Symmetries of Parallel Transported Lines 113
angles o f a transversal. This is w hy it is more useful to talk about lines as
parallel transports o f one another rather than as parallel. So the question
becomes
I f a transversal cuts two lines at congruent angles, are the lines,
in fa c t parallel in the sense o f not intersecting?
There are m any ways to approach this problem. First, be sure to
look at the symmetries o f the local portion o f the figure formed by the
three lines’. See w hat you can say about global symmetries from what
you find locally. For the question o f parallelism , you can use EEAT, but
not i f you used this problem to prove EEAT previously. Also, don’t
underestim ate the power o f symmetry when considering this problem.
M any ideas that w ork on the plane will also be useful on a sphere and a
hyperbolic plane, so try your planar p ro o f on a sphere and a hyperbolic
plane before attempting something completely different.
In Chapter 1, we said that an isometry is a symmetry o f a geometric
figure i f it transform s that figure into itself. That is, the figure looks the
same before and after the isometry. Here, we are looking for the symme­
tries o f the figure on the plane and sphere and hyperbolic plane.
From Figure 8.5, we can see that on a sphere we are looking for the
symmetries o f a lune cut at congruent angles by a geodesic. A lune is a
spherical region bounded by two h a lf great circles (see Problem 7.1).
You may be inclined to use one or both o f the following results that
are true on the plane: Any transversal o f a pair o f parallel lines cuts
these lines at congruent angles (Problem 10.1). And, the angles o f any
triangle add up to a straight angle (Problems 7.3b and 10.2). The use o f
these results should be avoided for now, as they are both false on both a
Sphere Plane
Figure 8.5 What are the symmetries of these figures?

114 Chapter 8 Parallel Transport
sphere and a hyperbolic plane. We have been investigating what is
common between the plane, spheres, hyperbolic planes — trying to use
common proofs whenever possible. You may be tempted to use other
properties o f parallel lines that seem familiar to you, but in each case ask
yourself whether or not the property is true on a sphere and on a hyper­
bolic plane. If it is not true on these surfaces, then don’t use it here
because it is not needed.
P roblem 8.3 T ransversals through a M idpoint
a. Prove: I f two geodesics r and r ‘ are parallel transports along
another geodesic l, then they are also parallel transports along
any transversal passing through the midpoint o f the segment o f l
between r and r’. Does this hold fo r the plane, spheres, and
hyperbolic planes? See Figure 8.6.
b. On a sphere or hyperbolic plane, are these the only lines that
will cut r and r ‘ at congruent angles? Why?
c. Prove: Two geodesics (on the plane, spheres, or hyperbolic
planes) are parallel transports o f each other i f and only i f they
have a common perpendicular.
Is there only one common perpendicular?
All parts o f this problem continue the ideas presented in Problem 8.2. In
fact, you may have proven this problem while working on 8.2 without
even knowing it. There are many ways to approach this problem; Using

Problem 8.3 Transversals through a Midpoint 1 15
symmetry, is always a good way to start. You can also use some o f the
triangle congruence theorems that you have been working with in
Chapter 6\ Look at the things you have discovered about transversals
from Problems 8.1 and 8.2; they are very applicable here. For the hyper­
bolic plane you may want to use results from Chapters 5 and/or 7.
P roblem 8 .4 W hat Is “P arallel”?
Since Chapter 7, you have been dealing with issues o f parallelism. Paral­
lel transport gives you a way to check parallelism. Even though parallel
transported lines intersect on the sphere, there is a feeling o f local paral­
lelness about them. In most applications o f parallel lines the issue is not
whether the lines ever intersect, but whether a transversal intersects them
at congruent angles at certain points; that is, whether the lines are parall­
el transports o f each other along the transversal. You may choose to
avoid definitions o f “parallel” that do not give you a direct method o f
verification, such as these common definitions for parallel lines in the
plane:
1. Parallel lines are lines that never intersect;
2. Parallel lines are lines such that every transversal cuts them at
congruent angles; or
3. Parallel lines are lines that are everywhere equidistant.
a. Check out fo r each o f these three definitions whether they apply
to parallel transported lines on a sphere or on a hyperbolic
plane.
This is closely related to Problems 8.2 and 8.3.
b. Show that there are pairs o f geodesics on a hyperbolic plane that
do not intersect and yet there are no transversals that cut at
congruent angle’s. That is, the geodesics are parallel {in the
sense o f not intersecting) but not parallel transports o f each
other.
Use the results o f 8.2 and 8.3 and look on your hyperbolic plane for the
boundary between geodesics that intersect and geodesics that are parallel
transports. We call a pair o f geodesics that satisfy part b by the name

116 Chapter 8 Parallel Transport
asymptotic geodesics. Be warned that in many texts these geodesics are
called simply parallel.
*c. Show that there are pairs o f geodesics on any cone with cone
angle greater than 360° that do not intersect and yet there are
no transversals that cut at congruent angles. That is, the geodes­
ics are parallel {in the sense o f not intersecting) but not parallel
transports o f each other along any straight (in the sense o f
symmetry) line.
Experiment with a paper cone with cone angle greater than 360°.
These examples should help you realize that parallelism is not ju st
about non-intersecting lines and that the m eaning o f parallel is different
pn different surfaces. You will explore and discuss these various notions
o f parallelism and parallel transport further in Chapters 9 -1 3 . Because
we have so many (often unconscious) connotations and assumptions
attached to the word “parallel,” we find it best to avoid using the term
“parallel” as much as possible in this discussion. Instead we will use
terms such as “parallel transport,” “non-intersecting,” and “equidistant,”
which make explicit the m eaning that is intended.
In Chapter 9, we will continue our explorations o f triangle congru­
ence theorems, some o f which involve parallel transport.
In Chapter 10, we will consider various parallel postulates and
explore how they apply on the plane, spheres, and hyperbolic planes. We
will assume the parallel postulates on the plane and use them to prove
the properties o f non-intersecting and parallel transported lines on the
plane. In the process, you may learn som ething about the history and
philosophy o f parallel lines and the postulates that have been used in
attempts to understand parallelism.
In Chapter 11, our understanding o f different notions o f “parallel”
will help us to explore isometries and patterns.
In Chapter 12, we will study parallelogram s and rectangles (and
their analogues on spheres and hyperbolic planes) and in the process
show that, on the plane, non-intersecting lines are equidistant.
In Chapter 13, we will use the results from Chapter 12 to explore
results that are only true on the plane, such as the Pythagorean Theorem
and results about sim ilar triangles.

_______________ Chapter 9
SSS, ASS, SAA, and AAA
Things which coincide with one another are equal to one
another.
— Euclid, Elements, Common Notion 4 [Appendix A]
This chapter is a continuation o f the triangle congruence properties stud­
ied in Chapter 6.
P r o b l e m 9.1 S i d e – S i d e – S i d e ( S S S )
Are two triangles congruent i f the two triangles have congruent
corresponding sides? Look at plane, spheres, and hyperbolic
planes. See Figure 9.1.
Figure 9.1 SSS
S uggestions
Start investigating SSS by making two triangles coincide as m uch as
possible, and see w hat happens. For example, in Figure 9.2, i f we line up
one pair o f corresponding sides o f the triangles, we have two different
orientations for the other pairs o f sides as depicted in Figure 9.2. O f
course, it is up to you to determine if each o f these orientations is
actually possible, and to prove or disprove SSS. Again, symmetry can be
very useful here.
117

118 Chapter 9 SSS, ASS, SAA, and AAA
Figure 9.2 Are these possible?
6 n a sphere, SSS doesn’t w ork for all triangles. The counterexam ­
ple in Figure 9.3 shows that no m atter how small the sides o f the triangle
are, SSS does not hold because the three sides always determine two
different triangles on a sphere. Thus, it is necessary to restrict the size o f
more than ju st the sides in order for SSS to hold on a sphere. W hatever
argument you used for the plane should w ork for suitably defined small
triangles on the sphere and all triangles on a hyperbolic plane. M ake sure
you see w hat it is in your argum ent that doesn’t w ork for large triangles
on a sphere.
Figure 9.3 A large triangle with small sides
There are also other types o f counterexam ples to SSS on a sphere.
Can you find them?

Problem 9.2 Angle-Side-Side (ASS) 119
P r o b l e m 9.2 A n g l e – S i d e – S i d e (ASS)
a. Are two triangles congruent i f an angle, an adjacent side, and
the opposite side o f one triangle are congruent to an angle, an ,
adjacent side, and the opposite side o f the other! Look at plane,
spheres, and hyperbolic planes. See Figure 9.4.
Figure 9.4 ASS
S uggestions
Suppose you have two triangles with the above congruencies. We will
call them ASS triangles. We w ould like to see if, in fact, the triangles are
congruent. We can line up the angle and the first side, and we know the
length o f the second side (BC or B ‘C j, but we don’t know where the
second and third sides will meet. See Figure 9.5.
Figure 9.5 ASS is not true, in general
Here, the circle that has as its radius the second side o f the triangle
intersects the ray that goes from A along the angle a to B twice. So ASS
doesn’t work for all triangles on the plane or spheres or hyperbolic
plane6. Try this for yourself on these surfaces to see what happens. Can
you make ASS w ork for an appropriately restricted class o f triangles?
On a sphere, also look at triangles with m ultiple right angles, and, again,
define “small triangles” as necessary. Your definition o f “small triangle”

120 Chapter 9 SSS, ASS, SAA, and AAA
here may be very different from your definitions in Problems 6.4 and
6.5.
There are numerous collections o f triangles for which ASS is true.
Explore. See what you find on all three surfaces.
b. Show that ASS holds fo r right triangles on the plane {where the
Angle in Angle-Side-Side is right).
This result is often called the Right-Leg-Hypotenuse Theorem
(RLH), which can be expressed in the following way:
RLH: On the plane, i f the leg and hypotenuse o f one right
triangle are congruent to the leg and hypotenuse o f another
right triangle, then the triangles are congruent.
What happens on a sphere and a hyperbolic plane?
At this point, you might conclude that RLH is true for small trian­
gles on a sphere. But there are small triangle counterexam ples to RLH
‘on spheres! The counterexam ple in Figure 9.6 will help you to see some
ways in which spheres are intrinsically very different from the plane. We
can see that the second leg o f the triangle intersects the geodesic that
contains the third side an infinite num ber o f times. So on a sphere there
are small triangles that satisfy the conditions o f RLH although they are
not congruent. W hat about RLH on a hyperbolic plane?
Figure 9.6 Counterexample to RLH on a sphere
However, i f you look at your argument for RLH on the plane, you
should be able to show the following:

P roblem 9.2 Angle-Side-Side (ASS) 121
On a sphere, RLH is valid fo r a triangle with all sides less than
1/4 o f a great circle: .
RLH is also true for a m uch larger collection o f triangles on a
sphere. Can you find such a collection? W hat about on a hyperbolic
plane?
P r o b l e m 9.3 S i d e – A n g l e – A n g l e (SAA)
Are two triangles congruent i f one side, an adjacent angle, and
the opposite angle o f one triangle are congruent, respectively, to
one side, an adjacent angle, and the opposite angle o f the other
triangle? Look at plane, spheres, and hyperbolic planes.
S uggestions
As a general strategy when investigating these problem s, start by making
the two triangles coincide as m uch as possible. You did this when inves­
tigating SSS and ASS. Let us try it as an initial step in our p ro o f o f SAA.
Line up the first sides and the first angles. Because we don’t know the
length o f the second side, we m ight end up with a picture sim ilar to
Figure 9.7.
The situation shown in Figure 9.7 m ay seem to you to be
impossible. You may be asking yourself, “Can this happen?” I f your
tem ptation is to argue that a and f cannot be congruent angles and that it
is not possible to construct such a figure, behold Figure 9.8.

122 Chapter 9 SSS, ASS, SAA, and AAA
A A ‘
You may be suspicious o f this example because it is not a counter­
example on the plane. You m ay feel certain that it is the only counterex­
ample to SAA on a sphere. In fact, we can find other counterexam ples
for SAA on a sphere.
W ith the help o f parallel transport, you can construct m any counter­
examples for SAA on a sphere. If you look back to the first counterex­
ample given for SAA, you can see how this problem involves parallel
transport, or similarly how it involves E uclid’s Exterior Angle Theorem,
which we looked at in Problem 8.1.
N Can we make restrictions such that SAA is true on a sphere? You
should be able to answer this question by using the fuller understanding
o f parallel transport you gained in Problems 8.1 and 8.2. You m ay be
tem pted to use the result, the sum o f the interior angles o f a triangle is
180°, in order to prove SAA on the plane. This result will be proven
later (Problem 10.2) for the plane, but we saw in Problem 7.3 that it does
not hold on spheres and hyperbolic planes. Thus, .we encourage you to
avoid using it and to use the concept o f parallel transport instead. This
suggestion stems from our desire to see what is common betw een the
plane and the other two surfaces, as m uch as possible. In addition, before
we can prove that the sum o f the angles o f a triangle is 180°, we will
have to make some additional assumptions on the plane that are not
needed for SAA.

Problem 9.4 Angle-Angle-Angle (AAA) 123
P r o b l e m 9 . 4 A n g l e – A n g l e – A n g l e ( A A A )
Are two triangles congruent i f their corresponding angles are
congruent? Look at plane, spheres, and hyperbolic planes.
Two triangles that have corresponding angles congruent are called
similar triangles. We will discuss sim ilar triangles in Problem 13.4.
As with the three previous problems, make the two AAA triangles
coincide as much as possible. W e know that we can line up one o f the
angles, but we don’t know the lengths o f either o f the sides coming from
this angle. So there are two possibilities: (1) Both sides o f one are longer
than both sides o f the other, as the example in Figure 9.10 shows on the
plane, or (2) one side o f the first triangle is longer than the correspond­
ing side o f the second triangle and vice versa, as the example in Figure
9.11 shows on a sphere.
As with Problem 9.3, you may think that the example in Figure 9.11
cannot happen on a plane, a sphere, or a hyperbolic plane. The possible
existence o f a counterexam ple relies heavily on parallel transport — you
can identify the parallel transports in each o f the examples given. Try
each counterexam ple on the plane, on a sphere, and on a hyperbolic
plane and see what happens. If these examples are not possible, explain
why, and i f they are possible, see if you can restrict the triangles suffi­
ciently so that AAA does hold.

124 Chapter 9 SSS, ASS, SAA, and AAA
Parallel transport shows up in AAA, sim ilar to how it did in SAA,
b u t here it happens sim ultaneously in two places. In this case, you will
recognize that parallel transport produces sim ilar triangles that are not
necessarily congruent. Are these constructions (in Figures 9.10 and 9.11)
possible? How? Are the triangles not congruent? W hy? These construc­
tions and non-congruencies may seem intuitively possible to you, but
you should ju stify why in each case. Again you may need properties o f
angle sums from Problem 7.3 and properties o f parallel transport from
Problems 8.1 through 8.4. Y ou m ay also use the property o f parallel
transport on the plane stated in Problem 10.1 — you can assume this
property now as long as you are sure not to use AAA when proving it
•later.
On a sphere or on a hyperbolic plane, is it possible to make the two
parallel transport constructions shown in Figure 9.11 and thus get two
non-congruent triangles? Try it and see. It is important that you make
such constructions and that you study them on a model o f a sphere and
on a model o f a hyperbolic plane.

________ Chapter 10
Parallel Postulates
[Euclid’s Fifth Postulate] ought to be struck from the postu­
lates altogether. For it is a theorem — one that invites many
questions … — and requires for its demonstration a number of
definitions as well as theorems…. it lacks the special character
of a postulate.
— Proclus (Greek, 410-485) [AT: Proclus], p. 151
P a r a l l e l L i n e s o n t h e P l a n e A r e S p e c i a l
Up to this point we have not had to assume anything about parallel
(non-intersecting) lines. No version o f a parallel postulate has been nec­
essary, on the plane, on a sphere, or on a hyperbolic plane. We defined
the concrete notion o f parallel transport and proved in Problem 8.2 that,
on the plane (and hyperbolic planes), parallel transported lines do not
intersect. Now in this chapter we will look at three important properties
on the plane that require further assumptions and that will be needed in
later chapters. If you are w illing to assume these three statements, you
may skip this chapter, but we urge you to finish reading through at least
the next page.
I f two lines on the plane are parallel transports o f each other
along some transversal, then they are parallel transports
along any transversal. (Contained in Problem 10.1)
On the plane, the sum o f the interior angles o f a triangle is
always 180°. (Contained in Problem 7.3b or Problem 10.2)
125

126 Chapter 10 Parallel Postulates
On the plane, non-intersecting lines are parallel transports
hlong all transversals. (Contained in Problem 10.3d)
W e have already seen that none o f these properties is true on a
sphere or a hyperbolic plane. Thus all three need for their proofs some
property o f the plane that does not hold on spheres and hyperbolic
planes. The various properties that permit proofs o f these three state­
ments are collectively term ed the Parallel Postulates.
\ Only the three statements above are needed from this chapter for
The rest o f the book. Therefore, it is possible to omit this chapter and
assume one o f the above three statements and then prove the others.
However, parallel postulates have a historical importance and a central
position in m any geometry textbooks and in many expositions about
non-Euclidean geometries. The problems in this chapter are an attempt
to help people unravel and enhance their understanding o f parallel postu­
lates. Com paring situations on the plane with situations on a sphere and
on a hyperbolic plane is a powerful tool for unearthing our hidden
assumptions and m isconceptions about the notion o f “parallel” on the
plane.
As we discussed already in Chapter 8, because we have so many
(often unconscious) connotations and assumptions attached to the word
“parallel,” we find it best to avoid using the term “parallel” as much as
possible in this discussion. Instead we will use terms such as “parallel
transport,” “non-intersecting,” and “equidistant,” which make explicit
the m eaning that is intended.
P r o b l e m 10.1 P a r a l l e l T r a n s p o r t o n t h e P l a n e
Show that i f l\ and h are lines on the plane such that they are
parallel transports along a transversal l, then they are parallel
transports along any transversal. Prove this using any assump­
tions you fin d necessary. Make as fe w assumptions as you can,
and make them as simple as possible. Be sure to state your
assumptions clearly.
What part o f your p r o o f does not work on a sphere or on a
hyperbolic plane?

Problem 10.1 Parallel Transport on the Plane 127
S uggestions
This problem is by no means as trivial as it at first may appear. In order
to prove this theorem, you will have to assume som ething — there are
many possible assumptions* so use your imagination. But at the same
time, try not to assume any more than is necessary. If you are having
trouble deciding w hat to assume, try to solve the problem in a way that
seems natural to you and then see what develops while making explicit
any assumptions you are using.
On spheres and hyperbolic planes, try the same construction and
p ro o f you used for the plane. W hat happens? Y ou should find that your
p roof does not work on these surfaces. So what is it about your p ro o f (on
a sphere and hyperbolic plane) that creates difficulties?
Problem 10.1 emphasizes the differences between parallelism on
the plane and parallelism on spheres and hyperbolic planes. On the
plane, non-intersecting lines exist, and one can “parallel transport”
everywhere. Yet, as was seen in Problems 8.2 and 8.3, on spheres and
hyperbolic planes two lines are cut at congruent angles if and only if the
transversal line goes through the center o f symmetry formed by the two
lines. That is, on spheres and hyperbolic planes two lines are parallel
transports only when they can be parallel transported through the center
o f symmetry formed by them. Be sure to draw a picture locating the
center o f symmetry and the transversal. On spheres and hyperbolic
planes it is impossible to slide the transversal along two parallel trans­
ported lines while keeping both angles constant (som ething you can do
on the plane). In Figures 10.1, the line r’ is a parallel transport o f line r
along line /, but it is not a parallel transport o f r along /’.
Figure 10.1a Parallel transport on a hyperbolic plane along / but not along /’

128 Chapter 10 Parallel Postulates
Figure 10.1b Parallel transport on a sphere along /, but not along /’
Pause, explore, and write out your ideas before reading
further.
We will now divide parallel postulates into three groups: those
involving m ostly parallel transport, those involving m ostly equidistance,
and those involving m ostly intersecting or non-intersecting lines. This
division is useful even though it is rough and unlikely to fit every
conceivable parallel postulate. Which o f these three groups is the most
appropriate fo r your assumption from Problem 10.1? We will call the
assumption you made in Problem 10.1 “your parallel postulate.’’’’
P r o b l e m 10.2 P a r a l l e l P o s t u l a t e s N o t I n v o l v i n g
( N o n – ) I n t e r s e c t i n g L i n e s
Commonly used postulates o f this sort are
H = 0: The holonomy o f triangles is zero.
A = 180: The sum o f the angles o f a triangle is equal to 180°.

P roblem 1 0 .2 Parallel Postulates Not Involving (Non-)Intersecting Lines 129
PT !: I f two lines are parallel transports (PT) along one line
then they are P T along ALL transversals.
Note that these are false on spheres and on hyperbolic planes. The last
two properties (A = 180 and P T !) will be needed crucially in almost all
the rem aining chapters (and you may have already used one o f them in
Problem 9.4). These properties are needed to study (on the plane) paral­
lelograms, rectangles, and sim ilar triangles (Chapters 12 and 13), circles
and inversion (Chapters 15 and 16), projections o f spheres and hyper­
bolic planes onto the Euclidean plane (Chapters 14 and 17), Euclidean
m anifolds (Chapters 18 and 24), solutions to quadratic and cubic
equations (Chapter 19), trigonom etry (Chapter 20), and polyhedra in
3-space (Chapter 23). It is these properties and their uses that we most
often associate with consequences o f the parallel postulates.
In Problem 7.3b we proved A = 180 on the plane, so we can
a. Ask what assumption about the plane was used in proving 7.3b.
b. Use A = 180 to prove PT!.
Look first at transversals that intersect the line along which the two lines
are parallel transports.
c. Using PT ! (Problem 10.1), prove A = 180 without using results
from Chapter 7.
Start by parallel transporting one side o f the triangle.
d. Show that, on the plane, H = 0 <=> A = 180 <=> P T !. I f your p o s­
tulate from 10.1 is in this group, then show that it is also equiva­
lent to the others.
This is usually accom plished m ost efficiently by proving H = 0 =>
A = 180 => PT ! => Your Postulate => H = 0, or in any other order.
e. Prove that, on the plane, two parallel transported lines are
equidistant.
Look for rectangles or parallelograms.

130 Chapter 10 Parallel Postulates
E q u i d i s t a n t C u r v e s o n S p h e r e s
a n d H y p e r b o l i c P l a n e s
The latitude circles on the earth are sometimes called “parallels o f
latitude.” They are parallel in the sense that they are everywhere equidis­
tant as are concentric circles on the plane. In general, transversals do not
cut equidistant circles at congruent angles. However, there is one impor­
tant case where transversals do cut the circles at congruent angles. Let l
dnd /’ be latitude circles the same distance from the equator on opposite
sides o f it. See Figure 10.2. Then every point on the equator is a center
o f half-turn symmetry for these pair o f latitudes. Thus, as in Problems
8.3 and 10.1, every transversal cuts these latitude circles in congruent
angles, even though these latitude circles are not geodesics. In the first
section o f Chapter 2 we noted that Euclid, in his Phenomena, discussed
such equidistant circles.
The same ideas work on a hyperbolic plane: I f g is a geodesic and /
and /’ are the (two) curves (not geodesics) that are a distance d from g,
then / and V are equidistant from each other and every transversal cuts
them at congruent angles. This follows from the fact that g has half-turn
symmetry at every point.

P roblem 10.3 Parallel Postulates Involving (Non-)Intersecting Lines 131
P r o b l e m 10.3 P a r a l l e l P o s t u l a t e s I n v o l v i n g
( N o n – ) I n t e r s e c t i n g L i n e s
One o f E uclid’s assumptions constitutes E u clid ’s Fifth (or Parallel)
Postulate (EFP), which says
EFP: I f a straight line intersecting two straight lines makes
the interior angles on the same side less than two right angles,
then the two lines ( if extended indefinitely) will meet on that
side on which are the angles less than two right angles.
For a picture o f EFP, see Figure 10.3.
You probably did not assume EFP in your proof o f Problem 10.1.
You are in good company — m any mathematicians, including Euclid,
have tried to avoid using it as m uch as possible. However, we will
explore EFP because, historically, it is important, and because it has
some very interesting properties, as you will see in Problem 10.3. On a
sphere, all straight lines intersect twice, which means that EFP is trivi­
ally true on a sphere. But in Problem 10.4, you will show that EFP is
also true in a stronger sense on spheres. You will also be able to prove
that EFP is false on a hyperbolic plane.
Thus, EFP does not have to be assumed on a sphere — it can be
proved! However, in m ost high school geometry textbooks, EFP is
replaced by another postulate, claimed to be equivalent to EFP. This

132 Chapter 10 Parallel Postulates
postulate we will call the H igh School Parallel Postulate (H SP ), and it
can be expressed in the following way:
HSP: For every line l and every point P not on l, there is a
unique line l ‘ that passes through P and does not intersect (is
parallel to) l.
Figure 10.4 High School Parallel Postulate
In m any books the H S P is called “Playfair’s Parallel Postulate”, but
this is an inappropriate name, as we will show in the historical notes in
the last two sections o f this chapter.
Note that, because any two great circles on a sphere intersect, there
are no lines / ‘th a t are parallel to / in the “not intersecting” sense. There­
fore, H S P is not true on spheres. On the other hand, if we change “paral­
lel” to “parallel transport” then every great circle through P is a parallel
transport o f / along some transversal. W hat happens on a hyperbolic
plane? In Problem 10.3, you will explore the relationships among E F P ,
H S P, and your postulate from Problem 10.1. In Problem 10.4 we will
explore these postulates on spheres and hyperbolic planes.
a. Show that, on the plane, E F P and H S P are equivalent. I f your
postulate from 10.1 is in this group, is it equivalent to the
others? Why or why not?
To show that E F P and H S P are equivalent on the plane, you need
to show that you can prove E F P if you assum e H S P and vice versa. If
the three postulates are equivalent, then you can prove the equivalence
by showing that
E F P => H S P => Your Postulate => E F P

P roblem 10.3 Parallel Postulates Involving (Non-)Intersecting Lines 133
or in any other order. It w ill probably help you to draw lots o f pictures o f
what is going on. Also, rem em ber that we proved in Problem 8.2
(without using any parallel postulate) that parallel (non-intersecting)
lines exist. Note that H S P is not true on a sphere but E F P is true, so
your p ro o f that E F P implies H S P on the plane m ust use some property
o f the plane that does not hold on a sphere. Look for it.
b. Prove that either E F P or H S P can be used to prove (without
using 10.1 or 7.3b) one o / H = 0, A = 180, or PT!. (It does not
m atter which. Why?)
So, are all these postulates equivalent to each other on the plane?
The answer is almost, but not quite! In order for A = 180 (or H = 0 or
P T !) to imply E F P (or H S P ), we have to make an additional as­
sumption, the Archimedean Postulate (AP) (in some books this is called
the Axiom o f Continuity), nam ed after the Greek m athematician, Archi­
medes (who lived in Sicily, 2877-212 b .c .):
*AP: On a line, i f the segment AB is less than (contained in)
the segment AC, then there is a finite (positive) integer, n, such
that, i f we p u t n copies o f AB end-to-end (see Figure 10.5),
then the n,h copy will contain the point C.
9 d * d * d * d * d * d * d * d * d \ d ^ >
d = \AB\
Figure 10.5 The Archimedean Postulate
The Archimedean Postulate can also be interpreted to rule out the
existence o f infinitesimal lengths. The reason these postulates are
“almost but not quite” equivalent to each other on the plane is that,
though A P is needed, it is assumed by m ost people to be true on the
plane, spheres, and hyperbolic planes. But this is the first time we have
needed this assumption in this book.
*c. Show that, on the plane, A P and A = 180 imply EFP.
Look at the situation o f E F P , which we redraw in Figure 10.6. Pick a
sequence o f equally spaced points A, A’, A”,… on n (on the side o f the

134 Chapter 10 Parallel Postulates
angle a). Next, parallel transport / along m to lines /’, /”,… that intersect
•n at the points A’, A”,… . And parallel transport m along 1 to m ’, m”,…,
which also intersect n at the points A’, A”,… . Look for congruent angles
and congruent triangles. Use AP to argue that n and m will eventually
intersect.
Figure 10.6 AP and A = 180 imply EFP
Clearly, this proof feels different from the others proofs in this
chapter. And, as already pointed out, most o f the applications in this
book o f parallel postulates only need A = 180 and PT!. So is it possible
for us not to bother with EFP and HSP? Or are they needed? Yes, we
need them, but only in a few places, such as in Problem 11.3, where what
we need to know is
d. Prove using EFP: On the plane, non-intersecting lines are paral­
le l transports along every transversal.
Cbmpare with EFP.
P r o b l e m 10.4 EFP a n d HSP o n S p h e r e
a n d H y p e r b o l i c P l a n e
a. Show that E F P is true on a sphere in a strong sense; that is, i f
lines l and l are cut by a transversal t such that the sum o f the
interior angles a + f on one side is less than two right angles,
then not only do l and /’ intersect, but they also intersect closest

P roblem 10.4 EFP and HSP on Sphere and Hyperbolic Plane 135
to t on the side o f a and P You will have to determine an appro­
priate meaning fo r “closest.”
To help visualize the postulates, draw these “parallels” on an actual
sphere. There are really two parts in this p ro o f — first, you m ust come
up with a definition o f “closest,” and then prove that E F P is true for this
definition. The two parts may come about sim ultaneously as you come
up with a proof. This problem is closely related to Euclid’s Exterior
Angle Theorem but can also be proved without using EE A T . One case
that you should look at specifically is pictured in Figure 10.7. It is not
necessarily obvious how to define the “closest” intersection.
b. On a hyperbolic plane let l be a geodesic and let P be a point
not on /; then show that there is an angle 6 with the property
that any line i passing through P is parallel to (not intersecting)
/ i f the line V does not form an angle less than 6 with the line
from P that is perpendicular to l (Figure 10.8).
Figure 10.8 Multiple parallels on a hyperbolic plane

136 Chapter 10 Parallel Postulates
Look at a variable line through P that does intersect / and then look at
what happens as you move the intersection point out to infinity.
c. Using the notion o f parallel transport, change the High School
Postulate so that the changed postulate is true on both spheres
and hyperbolic planes. Make as fe w alterations as possible and
keep some form o f uniqueness.
Try to limit your alterations so that the new postulate preserves the spirit
o f the old one. You can draw ideas from any o f the previous problem s to
obtain suitable m odifications. Prove that your m odified versions o f the
postulate are true on spheres and hyperbolic planes.
d. ..Either prove your postulate from Problem 10.1 on a sphere and
on a hyperbolic plane or change it, with as fe w alterations as
possible, so that it is true on these surfaces. You may need to
make different changes fo r the two surfaces.
In Problem 10.1 you should have decided w hether or not your postulate
is true on spheres or on hyperbolic spaces.
C o m p a r i s o n s o f P l a n e , S p h e r e s ,
a n d H y p e r b o l i c P l a n e s
Figure 10.9 is an attempt to represent the relationships among parallel
transport, non-intersecting lines, E F P , and H SP. Can you fit your postu­
late into the diagram?
The High School Postulate (H SP) assumes both the existence and
uniqueness o f parallel lines. In Problem 8.2, you proved that i f one line
is a parallel transport o f another, then the lines do not intersect on the
plane or on a hyperbolic plane. Thus, it is not necessary to assume the
existence o f parallel lines (non-intersecting lines) on the plane or hyper­
bolic plane.
On a sphere any two lines intersect. However, in Problem 8.4 we
saw that there are non-intersecting lines that are not parallel transports
o f each other on a hyperbolic plane and on any cone with cone angle
larger than 360°.

E u c lid e a n H y p e r b o lic S p h e r ic a l
Parallel
transported
lines
do not intersect
and are
equidistant
8.2 & 10.2
diverge in both
directions
8.2 & 8.3
always intersect
2.1
Parallel
transported
lines
are parallel
transports
along all
transversals
10.1
are parallel transports along any
transversal that passes through the
center o f symmetry o f the lines
8.3
N on­
intersecting
lines
are parallel
transports
along all
transversals
10.3d
sometimes are
not parallel
transports
8.4b
do not exist
2.1
Euclid’s 5th
Postulate
m ust be
assumed
Chapter 10
is false
8.4b
is provable in a
strong sense
10.4a
High School
Parallel
Postulate
a unique line
through a point
not intersecting
a given line
10.3
m any lines
through a point
not intersecting
a given line
10.4b
no
non-intersecting
lines
2.1
Y our postulate
from 10.1
Two points
determine
a unique line and line segment
6.1d
at least two line
segments
6.1d
Sum o f the
angles o f a
triangle
= 180°
7.3b & 10.2
< 180° 7.1 > 180°
7.2
Holonomy = 0° 10.2
< 0° 7.4 > 0°
7.4
VAT & ITT are alwavs true, 3.2 & 6.2
SAS, ASA,
and SSS
hold for all triangles
6.4, 6.5, & 9.1
hold for small
triangles
6.4, 6.5, & 9.1
AAA
is false, 9.4,
similar
triangles
is true, 9.4,
no sim ilar triangles
Figure 10.9 Comparisons of the three geometries
137

138 Chapter 10 Parallel Postulates
P a r a l l e l P o s t u l a t e s w i t h i n t h e
B u i l d i n g S t r u c t u r e s S t r a n d
The problem o f parallels puzzled Greek geometers (see the quote at the
beginning o f this chapter). The prevailing b e lie f was that it surely
follows from the straightness o f lines that lines like n and m in Figure
10.10 had to intersect, and so E uclid’s Fifth Postulate w ould turn out to
be unnecessary.
Figure 10.10 Surely these intersect!
During and since the Greek era, there have been attempts to derive
the postulate from the rest o f elementary geometry; attempts to reform u­
late the postulate or the definition o f parallels into something less objec­
tionable; and descriptions o f what geometry w ould be like if the postu­
late was in some way denied. Part o f these attempts was to study what
geometric properties could be proved without using a parallel postulate,
but with the other postulates (properties) o f the plane. The results o f
these studies became known as absolute geometry. Both the plane and a
hyperbolic plane are examples o f absolute geometry. Included in
absolute geometry are all the results about the plane and hyperbolic
planes that we have discussed in Chapters 1, 3, 5, 6, 8, and 9 (except
AAA, Problem 9.4). In general, absolute geometry includes everything
that is true o f both the plane and hyperbolic planes.
Claudius Ptolem y (Greek, 100-178) implicitly used what we now
call Playfair’s Postulate (P P ) to prove E F P . This is reported to us by
Proclus (Greek, 410^485) [AT: Proclus], who explicitly states PP.
P P . Two straight lines that intersect one another cannot be both
‘-parallel to the same straight line.
Or, in another common wording,

Parallel Postulates within the Building Structures Strand 139
PP. Given a straight line and a point not on the line, there is at
most one straight line through the point that does not intersect
the given line.
Playfair’s Parallel Postulate got its current name from the Scottish
mathematician John Playfair (1748-1819), who brought out successful
editions o f E uclid’s Elements in the years following 1795. A fter 1800
many comm entators referred to Playfair’s postulate (P P ) as the best
statement o f Euclid’s postulate, so it became a tradition in m any geome­
try books to use P P instead o f E F P . The reader can see easily that both
P P and E F P assert the uniqueness o f parallel lines and neither asserts
the existence o f parallel lines. H S P seems to have been first stated by
Hilbert in the first edition o f his Grundiagen der Geometrie (but not in
the later editions, where he used P P ) and in [EG: Hilbert]. But later,
apparently starting in the 1960s, m any (but not all) textbooks (including
earlier editions o f this book!) started calling H S P by the name
“Playfair’s Postulate” . In absolute geometry H S P is equivalent to P P
and E F P (as you showed in 10.3a), and because absolute geometry was
the focus o f m any investigations, the statement “H S P is equivalent to
E F P ” was made and is still being repeated in many textbooks and
expository writings about geometry even when the context is not
absolute geometry. As you showed in 10.4a, E F P is true on spheres and
so is P P , and so they cannot be equivalent to H S P, which is clearly false
on the sphere.
During the 9th through 12th centuries, parallel postulates were
explored by m athematicians in the Islamic world. al-Hasan ibn
al-Haytham (Persian, 965-1040) proved E F P by assuming, A quadrilat­
eral with three right angles must have all right angles. Quadrilaterals
with three right angles are known later in W estern literature as Lambert
quadrilaterals after Johann Lam bert (G erm an-Sw iss, 1728-1777), who
studied them and also was the first to extensively investigate hyperbolic
trigonometric functions. The Persian poet and geom eter Omar Khayyam
(1048-1131) wrote a book that in translation is entitled Discussion o f
Difficulties in Euclid [AT: Khayyam 1958], Khayyam introduces a new
postulate (which he attributes to Aristotle, though it has not been found
among the surviving works o f Aristotle), which says Two straight lines
that start to converge continue to converge. In his work on parallel lines
he studied the Khayyam quadrilaterals (later in the W est to be called
Saccheri quadrilaterals), which are discussed here in Chapter 12. N assir

140 Chapter 10 Parallel Postulates
al-Din Al-Tusi (Khorasan, 1201-1274) furthered the study o f parallel
postulates and is credited (though some say it may have been his son)
with first proving that A = 180 is equivalent to E F P , our 10.3bc.
A l-Tusi’s works were the first Islamic m athem atical works to be discov­
ered in the W estern Renaissance and were published in Rome in 1594.
The assumption parallel lines are equidistant (see our 10.2d) was
discussed in various forms by Aristotle (Greek, 383-322 b .c . ) , Posi­
donius (Greek, 135-51 b .c . ) , Proclus, ibn Sina (Uzbek, 980-1037), Omar
Khayyam, and Saccheri (Italian, 1667-1733).
John W allis (English, 1616-1703) proved that E F P followed from
the assum ption To every triangle, there exists a similar triangle o f
arbitrary magnitude. (See our 9.4.) The works, already mentioned, o f
W allis, Saccheri, Lambert in the 17th and 18th centuries were continued
by the French school into the early 19th century by Joseph Fourier
(French, 1768-1830), who concluded that geometry was a physical
science and could not be established a priori, and Adrien-M arie Legen­
dre (1752-1833), who proved that in absolute geometry the sum o f the
interior angles o f a triangle is always less than or equal to 180 °.
The breakthrough in the study o f parallel postulates came in the 19th
century when, apparently independently, Janos Bolyai (Hungarian,
1802-1860), and N. I. Lobachevsky (Russian, 1792-1856) finally devel­
oped hyperbolic geometry as an absolute geometry that did not satisfy
‘ E F P . See Chapters 5 and 17 for more discussions o f hyperbolic geome­
try. These discoveries o f hyperbolic geometry showed that the quest for
a proof o f E F P from within absolute geometry is impossible.
For details o f the relevant histories, see [HI: Gray], [DG:
M cCleary], [HI: Rosenfeld], and the H eath’s editorial notes in [AT:
Euclid, Elements].
N o n – E u c l i d e a n G e o m e t r i e s w i t h i n t h e
H i s t o r i c a l S t r a n d s
Building Structures Strand. Because o f this long history o f investigation
into parallel postulates w ithin the Building Structures Strand, many
books misleadingly call hyperbolic geometry “the non-Euclidean geome­
try.” As pointed out in the introduction to Chapter 2, spherical geometry
has been studied since ancient tim es within the Navigation/Stargazing
Strand, but it did not fit into absolute geometry and thus was (and still is)
left out o f m any discussions o f non-Euclidean geometry, especially those

Non-Euclidean Geometries within the Historical Strands 141
that take place in the context o f the Building Structures Strand. In those
discussions that do include spherical geometry, hyperbolic geometry is
usually called Lobachevskian geometry or Lobachevskian/Bolyai
geometry, and spherical geometry is usually called Riemannian geome­
try. Bernhard Riemann (German, 1826-1866) pioneered an intrinsic and
analytic view for surfaces and space and suggested that the intrinsic
geometry o f the sphere could be thought o f as a non-Euclidean
geometry. He also introduced an intrinsic analytic view o f the sphere
that became known as the Riemann sphere. The Riemann sphere is
usually studied in a course on complex analysis. The use o f the term
“Riemannian geometry” to denote the non-Euclidean geometry o f the
sphere leads to confusions since “Riemannian geometry” more com ­
monly refers to the part o f differential geometry that gives intrinsic ana­
lytic descriptions o f the local geometry o f general surfaces and 3-space.
Art/Pattern Strand. In this strand, artists’ experimentations with
perspective in art led to the m athematical theory o f projective geometry
that developed initially independently o f both spherical and hyperbolic
geometries. Then in the last part o f the 19th century, projective geometry
was used to provide a unified treatm ent o f the three geometries. It was
within this development that Felix Klein introduced the names double
elliptic geometry (for the geometry o f the sphere), elliptic geometry (for
the geometry o f the sphere with antipodal points identified — this the
same as the real projective plane, RP2, discussed in Problem 18.3),
parabolic geometry (for Euclidean geometry), and hyperbolic geometry
(for the Bolyai/Lobachevsky geometry). In Chapter 11, we will discuss
K lein’s developm ent o f various geometries in term s o f their isometries.
A geometric connection between spherical and projective geometry can
be seen in the image onto the plane o f the sphere (with antipodal points
identified) under gnomic projection. (See Problems 14.2 and 20.6.)
For a fuller discussion o f spherical and hyperbolic geometry in this
strand, see [HI: Kline], Chapter 38, and [HI: Katz], Section 17.3.
In Kline we also find a statement (page 913) that represented the
view o f some mathematicians:
By the early 1870s several basic non-Euclidean geometries, the
hyperbolic, and the two elliptic geometries, had been introduced
and intensively studied. The fundamental question which had yet
to be answered in order to make these geometries legitimate
branches of mathematics was whether they were consistent.

142 Chapter 10 Parallel Postulates
So for those who believe that the only “legitim ate” branches o f m athe­
matics are those that have been included in some rigorous formal system,
then non-Euclidean geometries did not becam e a part o f mathematics
until the late 1800s.
Navigation/Stargazing Strand. In the introduction to Chapter 2 we
saw that spherical geometry has been studied within this strand for more
than 2000 years, and in this study were developed (spherical) coordinate
systems and trigonometric functions. W e hope that the reader agrees that
this was (and still is) legitimate mathematics.
For more readings (and references) on this history o f spherical
geometry within the Navigation/Stargazing Strand, see [HI: Katz],
Chapter 4, and [HI: Rosenfeld], Chapter 1. Into the 19th century, spheri­
cal geometry occurred almost entirely in the Navigation/Stargazing
Strand and was used by Brahe and Kepler in studying the m otion o f stars
and planets and by navigators and surveyors. The popular book (which is
also available online) Spherical Trigonometry: For the use o f colleges
and schools (1886) [SP: Todhunter] contains several discussions o f the
use o f spherical geometry in surveying and was used in British schools
before hyperbolic geometry was widely known in the British Isles. On
the continent, C. F. Gauss (1777-1855) was using spherical geometry in
various large-scale surveying projects before the advent o f hyperbolic
geometry. W ithin this context, Gauss initiated (and Riemann expanded)
the analytic study o f surfaces that led to w hat we now call differential
geometry. Gauss introduced Gaussian curvature, which was discussed in
Chapter 7. W ithin this theory o f surfaces, spherical geometry is des­
cribed locally as the geometry o f surfaces with constant positive
(Gaussian) curvature and hyperbolic geometry is described locally as the
geometry o f surfaces with constant negative curvature. As we discussed
in Chapter 5, there are no analytic surfaces in 3-space with the complete
hyperbolic geometry, but the pseudosphere (Problem 5.3) locally has
hyperbolic geometry. Because o f this, in the Navigation/Stargazing
strand, hyperbolic geometry is called by some authors pseudosphere
geometry.

Chapter 11
Isometries and Patterns
What geometrician or arithmetician could fail to take pleasure
in the symmetries, correspondences and principles of order
observed in visible things?
— Plotinus, The Enneads, II.9.16 [AT: Plotinus]
[Geometry is] the study of the properties of a space which are
invariant under a given group of transformations.
F. Klein, Erlangen Program
Life forms illogical patterns. It is haphazard and full of
beauties which I try to catch as they fly by, for who knows
whether any of them will ever return? — Margot Fonteyn
Recall that in Chapter 1 we gave the following:
Definition. Xn isometry is a transform ation that preserves
distances and angle measures.
In this chapter we will show (for the plane, spheres, and hyperbolic
planes) that every isometry is the composition (product) o f (not more
than three) reflections, and we will determine all the different types o f
isometries. This finishes the study o f reflections and rotations we started
in Problem 5.4. We will note the differences betw een the kinds o f isom e­
tries that appear in the three geometries.
Then we will study patterns in these three spaces. Along the way
we will look at some group theory through its origins, that is, geometri­
cally.
This m aterial originated prim arily within the A rt/Pattem Strand.
143

144 Chapter 11 Isometries and Patterns
It would be good for the reader to start by reviewing Problem 5.4.
We will start the chapter with a further investigation o f isometries and
then with a discussion o f definitions and terminology. We advise the
reader to investigate this introductory m aterial as concretely as possible,
making drawings and/or moving about paper triangles.
P r o b l e m 11.1 I s o m e t r i e s
Definitions:
A reflection through the line (geodesic) l is an isometry R/
such that it fixes only those points that lie on l and, fo r each
point P not on l, l is the perpendicular bisector o f the geodesic
segment joining P to Ri(P).
Reflections are closely related to our notion o f geodesic (instrinsically
straight). We say in Chapters 1, 2, 4, and 5 that the existence o f (local)
reflections through any geodesic is a m ajor determining property o f
geodesic in the plane, spheres, cylinders, cones, and hyperbolic planes.
A directed angle is an angle with one o f its sides designated as
the initial side (the other side is then the terminal side). It is
usual to indicate the direction with an arrow, as in Figure
11.1. The usual convention is to say that angles with a counter­
clockwise direction are positive and those with a clockwise
direction are negative. When we write ZAPB fo r a directed
angle, then we consider AP to be on the initial side
F ig u r e 1 1 . 1 D ire c te d a n g le a n d ro ta tio n a b o u t P th r o u g h a n g le Z A P B

P roblem 1 1 .1 Isometries 145
A rotation about P through the directed angle 6 is an isome­
try Stf that leaves the point P fixed and is such that, fo r every
Q * P, SdQ) is on the same circle with center P that Q is on,
and the angle Z Q P S fQ ) is congruent to 6 and in the same
direction. See Figure 11.1.
a. Prove: I f P is a point on the plane, sphere, or hyperbolic plane
and ZAPB is any directed angle at P, then there is a rotation
about P through the angle 6 = ZAPB.
Refer to Problem 5.4 and Figure 5.13. Rem ember to show that the com ­
position o f these two reflections has the desired angle property. Rem em ­
ber to check that every point is rotated.
But, it is still not yet clear that anything we would w ant to call a
translation exists on spheres or hyperbolic planes.
b. Let m and n be two geodesics on the plane, a sphere, or a hyper­
bolic plane with a common perpendicular l. Look at the compo­
sition o f the reflection Rm through m with the reflection R„
through n. Show why this composition R„Rm could be called a
translation o f the surface along l. How fa r are points on l
moved”? What happens to points not on /? See Figure 11.2.
Remember that R„Rm denotes: First reflect about m and then reflect
about n. Let Q be an arbitrary point on / (but not on m or n). Investigate
where Q is sent by R m and then by R„Rm. Then investigate what happens
to points not on / — note that they stay the same distance from /. Be sure
to draw pictures.
We can now formulate a definition that works on all surfaces:

146 Chapter 11 Isometries and Patterns
A translation o f distance d along the line {geodesic) / is an isome­
try Trf that takes each point on l to a point on l at the distance {along l)
o f d and takes each point not on l to another point on the same side o f l
and at the same distance from l. See Figure 11.2.
Note that on a sphere a translation along a great circle / is the same
as a rotation about the poles o f that great circle.
c. Prove: I f l is a geodesic on the plane, a sphere, or a hyperbolic
plane and d is a distance, then there is a translation o f distance
d along l.
Use part b.
In Figure 11.3 (which can be considered to be on either the plane, a
sphere, or a hyperbolic plane), two congruent geometric figures, S?
and are given, but there is not a single reflection, or rotation, or trans­
lation that will take one onto the other.
Figure 11.3 Glide reflection
However, it is clear that there is some com position o f translations,
rotations, and reflections that will take & onto <&. In fact, the composi­ tion o f a reflection through the line / and a translation along / will take @ onto <8. This isometry is called a glide reflection along l. A glide reflection (or ju st plain glide) o f distance d along the line {geodesic) l is an isometry that takes each point on l to a point on l at the distance {along 1) o f d and takes each point not on l to another point on the other side o f l and at the same distance from l. P roblem 1 1 .1 Isometries 147 d. I f l is a geodesic on the plane, sphere, or hyperbolic plane and if d is a distance, then there is a glide o f distance d along l. L a t e r i n t h i s c h a p t e r w e w i l l s h o w t h a t t h e s e a r e t h e o n l y i s o m e t r i e s o f t h e p l a n e a n d s p h e r e s . H o w e v e r , o n t h e h y p e r b o l i c p l a n e t h e r e is a n o t h e r i s o m e t r y t h a t i s n o t a r e f l e c t i o n , r o t a t i o n , t r a n s l a t i o n , o r g l i d e . O n t h e h y p e r b o l i c p l a n e t h e r e a r e s o m e p a i r s o f g e o d e s i c s , c a l l e d asymptotic geodesics, t h a t d o n o t i n t e r s e c t a n d a l s o d o n o t h a v e a c o m m o n p e r p e n d i c u l a r ( a n d t h u s a r e n o t p a r a l l e l t r a n s p o r t s ) . S e e P r o b ­ l e m 8.4b. F o r e x a m p l e , t w o r a d i a l g e o d e s i c s i n t h e a n n u l a r h y p e r b o l i c p l a n e a r e a s y m p t o t i c . T h e c o m p o s i t i o n o f r e f l e c t i o n s a b o u t t w o a s y m p ­ t o t i c g e o d e s i c s i s d e f i n e d t o b e a horolation. S e e F i g u r e 1 1 . 4 , w h e r e A' = R„,(A), A " = R „ ( R , „ ( 4 ) ) , B" = R „ ( R C" = R „ ( R „ , ( Q ) . Figure 11.4 Horolation N o t e t h a t e v e r y p o i n t m o v e s a l o n g a n a n n u l a r l i n e ( n o t a g e o d e s i c ) s u c h a s a i n F i g u r e 1 1 . 4 . T h e a n n u l a r l i n e s c o u l d b e c a l l e d “ c i r c l e s o f i n f i n i t e r a d i u s ” . [Do you see why? I f you are given a circular arc in the plane how to you fin d its center?] I n m u c h o f t h e l i t e r a t u r e t h e s e a n n u l a r l i n e s , o r c i r c l e s o f i n f i n i t e r a d i u s , a r e c a l l e d horocycles. [ I n t h e p a s t , s o m e s t u d e n t s h a v e a f f e c t i o n a t e l y c a l l e d t h e m “ h o r r o r c y c l e s . ” ] N o t e t h a t on the plane circles o f infinite radius are straight lines. 148 Chapter 11 Isometries and Patterns For the horolation depicted in Figure 11.4 the two reflection lines are radial geodesics. This is not really the special case it looks to be: If / and m are any two asymptotic (but not radial) geodesics, then / must intersect a radial geodesic r, in fact, infinitely many radial geodesics. Reflect the whole hyperbolic plane through the bisector b o f the angle betw een the end o f / at which it is asymptotic to m and the end o f r at which it is asymptotic to other radial geodesics. See Figure 11.5. The images o f / and m under the reflection are now radial geodesics, r = R*(/) and R*(m). A horolation is an isometry that is neither a rotation nor a transla­ tion but can be thought o f as a rotation about the point at infinity where the two asymptotic lines converge. In the case o f radial geodesics, a horolation will take each annulus to itse lf because the radial geodesics are perpendicular to the annuli. See Figure 11.4. In Chapter 17 we will also see that a horolation corresponds to a translation parallel to the x-axis in the hyperbolic coordinate system introduced in Problem 5.2 and the upper half-plane model introduced in Problems 17.1 and 17.2. In the context o f the models o f the hyperbolic plane introduced in Chapter 17 many authors use the terms “elliptic isom etry”, “parabolic isom etry”, and “hyperbolic isom etry” to refer to what we are calling, “rotation”, “horolation”, and “translation.” Problem 11.2 T hree Points D etermine an Isometry In order to analyze w hat all isometries are, we need the following very important property o f isometries: P roblem 1 1 . 2 Three Points Determine an Isometry 149 Prove the following: On the plane, spheres, or hyperbolic planes, i f f and g are isometries and A, B, C are three non-collinear points, such that f{A) = g{A), f(B) = g(B), and f{C) — g(C), then f and g are the same isometry, that is, f[X) = g(X ) fo r every point X. Let us see an example o f how this works before you see why this property holds. In Figure 11.6, H/> represents the half-turn around P, Rm
represents the reflection through line m, G/ represents a glide reflection
along /, and
& = {A, B, C}, & = {A’, B’, C’} and {A”, B ” , C ” ).
Figure 11.6 A glide equals a half-turn followed by a reflection
We can see that Hp(&i) = ^ , Rm(&i) = &s, and G /(^i) = # j. But
then G; and RmHp perform the same action on the three points. I f we
apply the above result, we can say that G/ = RmH/»; that is, Gi{X)
= RmH/>(Z), for all points X on the plane. You can see now the usefulness
o f proving Problem 11.2.
If you have trouble getting started with this problem, then take a
specific example o f two congruent triangles tsABC and A A ‘B ‘C such as
in Figure 11.7. Pick another point X and convince yourself as concretely
as possible that there is only one location that X will be taken to by any
isometry that takes A, B, C onto A’, B \ C’.
P roblem 11.3. C lassification of Isometries
a. Prove that on the plane, spheres, or hyperbolic planes, every
isometry is the composition o f one, two, or three reflections.

150 Chapter 11 Isometries and Patterns
Look back at what we did in the proofs o f SAS and ASA. Also use
Problem 11.2, which allows you to be more concrete when thinking
about isometries because you only need to look at the effect o f the
isom etries on any three non-collinear points that you pick. To get started
on 11.3, cut a triangle out o f an index card and use it to draw two
congruent triangles in different orientations on a sheet o f paper. For
example, see Figure 11.7. Can you move one triangle to the other by
three (or fewer) reflections? You can use your cutout triangle for the
intermediate steps.
We have already showed in Problems 5.4 and 11.1 that the product
o f two reflections is a rotation ( if the two reflection lines intersect) and a
translation (if the two reflection lines have a common perpendicular, that
is, if the two lines are parallel transports). Thus we can
b. Prove that on the plane and spheres, every composition o f two
reflections is either the identity, a translation, or a rotation.
What are the other possibilities on a hyperbolic plane? Why are
you sure that these are the only isometries that are the composi­
tion o f two reflections? What happens i f you switch the order o f
the two reflections?
In this part start with the two reflections as given. W hat are the different
ways in which the two reflection lines can intersect? To fully answer this
part on a hyperbolic plane, you will have to use the result (found infor­
mally in Chapters 5 and 8 and proved in Problem 17.3) that Two geodes­
ics in a hyperbolic plane either intersect in a point, or have a common
perpendicular, or are asymptotic.
Y our pro o f o f part b or Problem 11.1 probably already shows that
T heorem 1 1 .3 . On the plane, spheres, and hyperbolic planes,
a rotation is determined by two intersecting reflection lines
(,geodesics). The lines are determined only by their point o f

P roblem 11.3 Classification of Isometries 151
intersection and the angle between them; that is, any two lines
with the same intersection point and the same angle between
them will produce the same rotation. See Figure 11.8.
Figure 11.8 Two pairs of reflection lines, which determine the same rotation
c. On the plane, spheres, and hyperbolic plane, the product o f two
rotations (in general, about different points) is a single rotation,
translation, or horolation. Show how to determine geometrically
which specific isometry you obtain, including the center and
angle o f any rotation.
Use Theorem 11.3 and write the composition o f the two rotations with
each rotation the composition o f two reflections in a way that the middle
two reflections cancel out. Be careful to keep track o f the directions o f
the rotations. See Figure 11.9.
Figure 11.9 Composition of two rotations

152 Chapter 11 Isometries and Patterns
d. On the plane, spheres, and hyperbolic planes, every composition
o f three reflections is either a reflection or a glide reflection.
How can you tell which one?
Approach using Theorem 11.3: W rite the composition o f three reflec­
tions, R;, Rm, R„, as either (R;Rm)R„ or R/(RmR„). If the reflection lines, /,
m, n, intersect, then use part c to replace the reflection lines within the
parentheses with reflection lines that produce the same result. Try to
produce the situation where the first (or last) reflection line is perpen­
dicular to the other two. W hat is the situation i f none o f /, m, n intersect
each other? This approach works well on the plane and spheres but is
more difficult on a hyperbolic plane because the situation when all three
lines do not intersect is m ore complicated.
Approach using Problem 11.2 and triangles: Let M B C be a trian­
gle and let E A ‘B ‘C be its image under the composition o f the three
reflections. Note that the two triangles cannot be directly congruent (see
the discussion around Figure 6.5). Extend two corresponding sides (say
AB and A ‘B ’) o f the triangles to lines (geodesics) / and m. Then there are
three cases: The lines / and m intersect, or have a common perpen­
dicular, or are asymptotic.
If parts a, b, and d are true, then
e. Every isometry o f the plane or a sphere or a hyperbolic plane is
either a reflection, a translation, a rotation, a horolation, a
glide reflection, or the identity.
Notice that in our proofs o f SAS and ASA and (probably) in your
p ro o f o f part b we only need to use reflections about the perpendicular
bisectors or segments joining two points. Because, on a sphere, we only
need great circles to jo in two points and for perpendicular bisectors, we
can restate part b on a sphere to read
Every isometry o f a sphere is the composition o f one, two, or
three reflections through great circles.
Thus, in particular, i f there were a reflection o f the sphere that was
not through a great circle, then that reflection (being an isometry) would
also be composition o f one, two, or three reflections through great
circles.

P roblem 11.3 Classification of Isometries 153
You now have powerful tools to make a classification o f discrete
strip patterns on the plane, spheres, and hyperbolic planes and finite
patterns on the plane and hyperbolic planes.
K lein’s Erlangen P rogram
In 1872, on the occasion o f his appointment as professor in the Univer­
sity o f Erlangen at the age o f 23, Felix Klein (1849-1925) proposed that
the view o f symmetries should be radically extended. He proposed
viewing geometry as “the study o f the properties o f a space which are
invariant under a given group o f transform ations.” The proposal is now
universally known as the Erlangen Program.
Klein provided several examples o f geometries and their associated
transformations. In Euclidean plane geometry the transform ations are all
the isometries (reflections, rotations, translations, and glides) and includ­
ing the sim ilarity transform ations (dilations), that are not isometries
because they do not preserve distances but do preserve angles and thus
take triangles to similar triangles (see Problem 13.4).
W ithin the Erlangen Program, spherical geometry is the study o f the
isometries o f the sphere (reflections, rotation/translations, and glides)
and hyperbolic geometry is the study o f its isometries (reflections,
rotations, translations, glides, and horolations). In general, symmetry
was used as an underlying principle and it was shown that different
geometries could coexist because they dealt with different types o f
propositions and invariances related to different types o f (symmetry)
transformations. Topology, projective geometry, and other theories have
all been placed within the classification o f geometries in the Erlangen
Program. The long-term effects o f the Erlangen Program can be seen all
over pure mathematics, and the idea o f transform ations and o f synthesis
using groups o f symmetry is now standard also in physics. K lein’s
demonstration o f the relationship o f geometry to groups o f transform a­
tions helped to provide impetus for the development o f the abstract
notion o f a group by the end o f 19th century. For more discussion o f
K lein’s program, see [HM : Yaglom].
S ymmetries and Patterns
In Chapter 1 we talked about symmetries o f the line. All o f those sym­
m etries can be seen as isometries o f the plane except for similarity

154 Chapter 11 Isometries and Patterns
symmetry and 3-D rotation symmetry (through any angle not an integer
m ultiple o f 180°). Similarity symmetry changes lengths between points
o f the geometric figure and thus is not an isometry. Three-dimensional
rotation symmetry is an isom etry o f 3-space, but it moves any plane o ff
itself and thus cannot be an isometry o f a plane (unless the angle o f
rotation is a multiple o f 180°). The notion o f symmetry grew out o f the
A rt/Pattem Strand o f history. W ell before written history, symmetry and
patterns were part o f the hum an experience o f weaving and decorating.
W hat is a symmetry o f a geometric figure? A symmetry o f a geo­
metric figure is an isometry that takes the figure onto itself. For example,
reflection through any m edian is a symmetry o f an equilateral triangle.
See Figure 11.10.
B ‘
Figure 11.10 Reflection symmetries
It is easy to see that rotations through 1/3 and 2/3 o f a revolution,
S1/3 and S 2 /3 , are also symmetries o f the equilateral triangle. In addition,
the identity, Id, is (trivially, by the definition) an isometry and, thus, is
also a symmetry o f the equilateral triangle. Therefore, the equilateral
triangle has six symmetries: R<, R& Rc, S 1 /3 , S 2 /3 , Id, where R*. Ra Rc, denote the reflections through the medians from A, B, C. These are the only symmetries o f the equilateral triangle. N ow look at the geometric figure in Figure 11.11. It has exactly the same symmetries as an equilateral triangle. Though the two figures look very different, we say they are isomorphic patterns. A pattern is a figure together with all its symmetries; and we call the collection o f all symmetries o f a geometric figure its symmetry group. W e want some way to denote that two different patterns have the same symmetries, as is the case with the pattern in Figure 11.11 and the equilateral triangle in Figure 11.10. We do this by saying that two patterns are isomorphic i f they have the same symmetries. This should become clearer through more examples. Symmetries and Patterns 155 Figure 11.11 Same symmetries as the equilateral triangle The letters S and N each have only half-turn and the identity as symmetries; thus we say they are isomorphic patterns with symmetry group {Id, S1/2}. Similarly, the letters A and M each have only vertical reflection and the identity as symmetries and thus are isomorphic patterns with symmetry group {Id,R}. Note that the letters S and A each have the same number o f symmetries, but we do n o t call them isom or­ phic patterns because the symmetries are different symmetries. The reader should check their understanding by finding isomorphic patterns among the other letters o f the alphabet in normal printing by hand. To construct further examples, we can start with a geometric figure, often called a motif, that has no symmetry (except the identity). For example, the geometric figures in Figure 11.12 are possible motifs. To make examples o f patterns with a specific isometry, we can start with any m otif and then use the isometry and its inverse to make additional copies o f the m o tif over and over again. In the process we obtain another geometric figure for which the initial isometry is a symmetry. Let us look at an example: If we start with the first m o tif in Figure 11.12 and the isometry is translation to the right through a distance d, then, using this isometry and its inverse (translation to the 156 Chapter 11 Isometries and Patterns left through a distance d) and repeating them over and over, we obtain the pattern in Figure 11.13. Figure 11.13 Pattern with translation symmetries The symmetry group o f the pattern in Figure 11.13 is {Id (the identity), T nd (where n = ±1, ±2, ± 3 ,...)} . If the isometry is clockwise rotation through 1/3 o f a revolution about the lower endpoint o f the motif, then we obtain the pattern in Figure 11.14. This figure is a pattern with symmetries {Id, S 1/3, S2/3}. Note that this pattern is not isomorphic with the equilateral triangle pattern. If, in the constructions depicted in Figures 11.13 and 11.14, we replace the m otif with any other m otif (with no non-trivial symmetries), then we will get other patterns that are isomorphic to the original ones because the symmetries are the same. We call the collection o f all symmetries o f a geometric figure its symmetry group. If g, h are symmetries o f a figure then you can easily see that Symmetries and Patterns 157 The composition gh {first transform by h and then follow it by g ) is also a symmetry. For example, for an equilateral trian­ gle the composition o f the reflection RA with the reflection R B is a rotation, S2/3- In symbols, RbRa= S2/3 ■ See Figure 11.15. Composition o f symmetries is associative: That is, i f h, g, k are symmetries o f the same figure, then (hg)k = h(gk) = hgk. The identity transformation, Id (the transformation that takes every point to itself), is a symmetry. For every symmetry g o f a figure there is another symmetry f such that g f a n d fg are the identity — in this case we call f the inverse o f g. For example, the inverse o f the rotation S2/3 is the rotation S 1 /3 and vice versa. In symbols, S 1 /3 S 2 /3 = S 2 /3 S 1 /3 = Id. Those readers who are fam iliar with abstract group theory will recognize the above as the axioms for an abstract group. We will discuss further connections with group theory at the end o f this chapter. If two patterns are isomorphic, then their symmetry groups are isomorphic as abstract groups. The converse is often, but not always, true. For example, the symmetry groups o f the letter A and the letter S are isom or­ phic (as abstract groups) to Z 2 , but they are not isomorphic as patterns because they have different symmetries. A strip (or linear, or frieze) pattern is a pattern that has a transla­ tion symmetry, with all o f its symmetries also symmetries o f a given 158 Chapter 11 Isometries and Patterns line. For example, the pattern in Figure 11.13 is a strip pattern as is the pattern in Figure 11.11. The strip pattern in Figure 11.16 has symmetry group: { Id , R / , T„* G„d = R/T„d (where n = 0, ±1, ±2, ±3, •■•)}• :ccccccecc Figure 11.16 A strip pattern You are now able to start to study properties about patterns and isometries. P roblem 11.4 Examples of Patterns a. Go through all the letters o f the alphabet (in normal printing by hand) and decide which are isomorphic as patterns. b. Find as many (non-isomorphic) patterns as you can that have only finitely many symmetries. List all the symmetries o f each pattern you find. c. Find as many (non-isomorphic) strip patterns as you can. List all the symmetries o f each strip pattern you find. The purpose o f this problem is to get you looking at and thinking about patterns. Examples o f different strip patterns and m any finite patterns can be found on buildings everywhere: houses o f worship, courthouses, and m ost older buildings. Also look for other decorations on plates or on wallpaper edging. In the next two problem s we will determine i f your list o f strip patterns and finite patterns contains all possible strip and finite patterns. * P roblem 11.5 C lassification of D iscrete Strip Patterns A strip pattern is discrete if every translation symmetry o f the strip pattern is a m ultiple o f some shortest translation. ^ P roblem 11.4 Examples o f Patterns 159 a. Prove there are only seven non-isomorphic strip patterns on the plane that are discrete. b. What are some non-discrete strip patterns'? c. What happens with strip patterns on spheres and hyperbolic planes? Hint: Use Problem 11.3. * P roblem 11.6 C lassification of Finite P lane Patterns Look at all the finite patterns on the plane that you found in Problem 11.5b. Do you notice that for each there is a point (not necessary on the figure) such that every symmetry o f the pattern leaves the point fixed? See Figure 11.17. Figure 11.17 Finite patterns have centers This leads to the following: a. Show that any pattern on the plane with only finitely many symmetries has a center. That is, there is a point in the plane (not necessarily on the figure) such that every symmetry o f the pattern leaves the point fixed. Is this true on spheres and hyper­ bolic planes? This was first proved by Leonardo da Vinci (1452-1519, Italian), and you can prove it too! H int: Start by looking at what happens if there is a translation or glide symmetry. 160 Chapter 11 Isometries and Patterns b. Describe all the patterns on the plane and hyperbolic planes with only finitely many symmetries. Hint: Use part a. W hat rotations are possible i f there are only finitely many symmetries? c. Describe all the patterns on the sphere that are finite and that have centers. Hint: If there is one center, then its antipodal point is necessarily also a center. Whyl I f you take a cube with its vertices on a sphere and project from the center o f the sphere the edges o f the cube onto the sphere, then the result is a pattern on the sphere with only finitely many symmetries. This pattern does not fit with the patterns you found in part c because this pattern has no center (on the sphere). See Problems 11.7b and 23.5 for more examples. * Problem 11.7 R egular T ilings with Polygons We will now consider some special infinite patterns o f the plane, spheres, and hyperbolic planes. These special patterns are called regular tilings (or regular tessellations, or mosaics). Definition. A regular tiling {n, k} o f a geometric space (the plane, a sphere, or a hyperbolic plane) is made by taking identical copies o f a regular n-gon (a polygon with n edges) and using these n-gons to cover every point in the space so that there are no overlaps except each edge o f one n-gon is also the edge o f another n-gon and each vertex is a vertex o f k n-gons. You probably know the three familiar regular tilings o f the plane: {3, 6}, the regular tiling o f the plane by triangles with six triangles coming together at a vertex; {4, 4}, the regular tiling o f the plane by squares with four squares coming together at a vertex; and {6, 3}, the regular tiling o f the plane by hexagons with three hexagons coming together at each vertex. There is only one way to tile a plane with regular *P roblem 11.7 Regular Tilings with Polygons 161 hexagons; however, there are other ways to tile a plane with regular triangles and squares but only one for each that is a regular tiling. N ote that each regular tiling can also be thought o f as a (infinite) pattern. The notation {n, k) is called the Schlafli symbol o f the tiling, named after Ludwig Schlafli (1814—1895, Swiss). W e now study the possible regular tilings. a. Show that, {3, 6}, {4, 4}, {6, 3} are the only Schlafli symbols that represent regular tilings o f the plane. Focus on what happens at the vertices. b. Find all the regular tilings o f a sphere. Again, focus on what happens at the vertices. Rem ember that angles o f a regular w-gon on the sphere are larger than the angles o f the correspond­ ing regular «-gon on the plane ( Why?); and use Problems 7.1 and 7.4. There are more finite patterns o f the sphere besides those given in Problem 11.6c and 11.7b. See a soccer ball for an example and [SG: M ontesinos] for the complete classification. c. Show that each o f the Schlafli symbols {n, k } with both n and k greater than 1 represents a regular tiling o f the plane (part a), or o f a sphere (part b), or o f a hyperbolic plane. Again, focus on w hat happens at the vertices. ♦ Ot h e r P e r i o d i c ( a n d N o n - P e r i o d i c ) P a t t e r n s There are numerous non-regular tilings and other patterns other than regular tilings. Besides the three that come from regular tilings, there are 14 more (non-isomorphic) infinite periodic patterns on the plane. (“Periodic” means that the pattern has a minimal translation symmetry.) See Figure 11.18. These 17 periodic patterns in the plane are often called wall-paper patterns. See E scher’s (M aurits Cornelius Escher, 1898-1972, Dutch) drawings for examples and [SG: Budden] for proofs and more examples. For a complete exposition on periodic patterns and tilings on the plane, see [SG: Griinbaum], 162 Chapter 11 Isometries and Patterns & & & & & & % & a 9 c n c n ^ 9 C / Figure 11.18 A repeating pattern that is not a regular tiling — the thin straight lines (forming squares) are lines of reflection symmetry, there is half-turn symmetry about corners of the squares, and there is 4-fold rotation symmetry about the center of each square. There also exist non-periodic tilings o f the plane; see Figure 11.19. Notice the 5-fold local rotation symmetry at the center — this will never repeat anywhere in the pattern and thus the pattern cannot be periodic. Figure 11.19 Non-periodic tiling of the plane For m ore discussion o f non-periodic tilings, see [SG: Senechal] and [SG: Griinbaum], See [SG: M ontesinos] for a discussion about other patterns and tilings on hyperbolic spaces. *Other Periodic (and Non-Periodic) Patterns 163 There are 230 non-isomorphic repeating patterns in Euclidean 3-space. These are the patterns that appear in crystals. However, in protein crystals there are no reflection (through planes) or central (reflection through a point) symmetries — as a consequence there are only 65 repeating patterns without plane reflections or central symme­ tries in 3-space. The corresponding groups are call chiral groups. See [SG: Giacovazzo] for more information on crystallographic groups. ♦ G e o m e t r i c M e a n i n g o f A b s t r a c t G r o u p T e r m i n o l o g y The collection o f symmetries o f a geometric figure with the operation o f composition is an abstract group. We showed above how the usual axioms o f a group are satisfied. For example, the pattern in Figure 11.20 has symmetry group {R^, Rs, Rc, Si/3, S2/3, Id} that is isomorphic as an abstract group to what is usually called D3, the third dihedral group. Figure 11.20 Isomorphic to D3 Figure 11.21 Subgroups If a figure has a subfigure and i f the symmetries o f the subfigure are all also symmetries o f the original figure, then the symmetry group o f the 164 Chapter 11 Isometries and Patterns subfigure is a subgroup o f the symmetry group o f the original figure. See Figure 11.21. The symmetry group o f the subfigure is {S1/3, S2/3, Id}, which is isomorphic as an abstract group to Z3. This group is a subgroup o f the symmetry group o f the original figure, {S1/3, S2/3, R i, Rs, Rc, Id}, which is isomorphic to D3. In D3 the two cosets o f Z 3, Id Z 3 and R 1Z 3 = RbZ 3 = R rZ 3, correspond to the two copies o f the subfigure in Figure 11.22. In general, if G is the symmetry group o f a figure and H is a subgroup that is the symmetry group o f a subfigure in the figure, then the cosets o f H correspond to the several congruent copies o f the subfigure that exist within the larger figure. The term group was coined (as groupe in French) by E. Galois (1811-1832) in 1830. The modem definition o f a group is somewhat different from that o f Galois, for whom the term denoted a subgroup of the group o f permutations o f the roots o f a given polynomial. F. Klein and S. Lie (1842-1899) used the term closed system in their earliest writing on the subject o f groups. Group appears in English in an article by Arthur Cayley, who wrote "A set o f symbols, 1, a , (S, ..., all o f them different, and such that the product o f any two o f them (no matter what order), or the product o f any one o f them into itself, belongs to the set, is said to be a group." See [SG: Lyndon] for more discussion on groups. Chapter 12 Dissection T heory Oh, come with old Khayyam, and leave the Wise To talk, one thing is certain, that Life flies; One thing is certain, and the Rest is Lies; The flower that once has blown for ever dies. — Omar Khayyam, Rubaiyat From the translation by Edward Fitzgerald W h a t I s D i s s e c t i o n T h e o r y ? Figure 12.1 Parallelogram In showing that the parallelogram in Figure 12.1 has the same area as a rectangle with the same base and height (altitude), we can easily cut the parallelogram into two pieces and rearrange them to form the rectangle in Figure 12.2. 165 166 Chapter 12 Dissection Theory We say that two figures (F and G) are equivalent by dissection ( F =d G) i f one can be cut up into a finite num ber o f pieces and the pieces rearranged to form the other. Some authors use the term “equide- com posable” instead o f “equivalent by dissection” . Question: If two planar polygons have the same area, are they equivalent by dissection? Answer: Yes! For all (finite) polygons on either the plane, or on a sphere, or on a hyperbolic plane. You will prove these results about dissections in this chapter and the next and use them to look at the m eaning o f area. In this chapter you will show how to dissect any triangle or parallelogram into a rectangle with the same base. Then you will do analogous dissections on spheres and hyperbolic planes after first defining an appropriate analog o f parallelo­ grams and rectangles. A fter that you will show that two polygons on a sphere or on a hyperbolic plane that have the same area are equivalent by dissection to each other. The analogous result on the plane m ust wait until the next chapter. The proofs and solutions to all the problem s can be done using “ =d” , but if you wish you can use the weaker notion o f “=s” : We say that two figures (F and G) are equivalent by subtraction (F = s G) i f there are two other figures, S and S', such that S = d S ' and F u S =d G u S ', where F ans S and G and S' intersect at m ost in their boundaries. Some authors use the term “o f equal content” instead o f “equivalent by subtraction” . Saying two figures are equivalent by subtraction means that they can be arrived at by rem oving equivalent parts from two initially equivalent figures, as in Figure 12.3. Figure 12.3 Equivalent by subtraction What Is Dissection Theory? 167 If we cut out the two small squares as shown in Figure 12.3, we can see that the shaded portions o f the rectangle and the parallelogram are equivalent by subtraction, but it is not at all obvious that one can be cut up and rearranged to form the other. Equivalence by dissection is generally preferable to equivalence by subtraction because it is gives a direct way o f seeing that two figures have the same area. However, sometimes it is easier to find a p ro o f o f equivalence by subtraction. In addition, equivalence by subtraction has the advantage (as we will see) that in some situations equivalence by dissection is only true if one assumes the Archim edean Postulate (which we first met in Problem 10.3), while equivalence by subtraction does not need the Archim edean Postulate. However, we w ould urge you to prove equivalence by dissection w herever you can. The Archimedean Postulate (in some books this is called the Axiom o f Continuity), named after the Greek mathematician Archim edes (who lived in Sicily, 2877-212 b .c . ) , is as follows: *AP: On a line, i f the segment AB is less than (contained in) the segment AC, then there is a finite (positive) integer, n, such that i f we p u t n copies o f AB end to end (see Figure 10.5), then the n‘h copy will contain the point C. The Archim edean Postulate can also be interpreted to rule out the ex­ istence o f infinitesimal lengths. It is true that A P is needed to prove some results about equivalence by dissection; however, m ost people assume A P to be true on the plane, spheres, and hyperbolic planes. A D i s s e c t i o n P u z z l e f r o m 250 b . c . S o l v e d i n 2003 About 250 b .c ., Archim edes wrote a treatise entitled Stomachion, which was lost, though from commentaries it was clear that the work discussed the puzzle pictured in Figure 12.4. This is a puzzle, sim ilar to Tangrams, that consists o f 14 pieces that fit into a square (all o f the vertices lie on a 12 x 12 grid). However, A rchim edes’ interest in the puzzle was not known. Then in 2003, Reviel Netz, a Stanford historian o f science, deciphered parts o f the Palimpsest, which consists o f pieces o f parch­ m ent that originally (about 1000 a .d . ) contained several works o f A rchi­ medes, but in the 13th century the words o f Archim edes were scraped o ff and the parchm ent used to write a prayer book. Netz uncovered the 168 Chapter 12 Dissection Theory introduction to the Stomachion treatise and discovered that Archimedes asked the question How many different ways can these 14 pieces be rearranged to f i t exactly in the square? Then, on Novem ber 17, 2003, it was announced in the MAA OnLine column Math Games http://ww w.m aa.org/editorial/m athgam es/m athgam es_l 1 17 03 .html that Bill Cutler, a puzzle designer with a Ph.D. in m athematics from Cornell University, found using a computer that there were 17,152 distinct solutions (or only 536 i f you counted as the same solutions that varied by rotation or reflection o f the square or differed only in the inter­ change o f the congruent pairs o f pieces 7 and 14 or 6 and 13). Then, the story was continued on the front page o f the New York Times, December 14, 2003, where it was announced that the University o f C alifom ia-S an Diego m athematicians Ronald Graham and Fan Chung had independ­ ently solved the problem using combinatorics. See MAA OnLine and the New York Times article for more details on the fascinating history o f A rchim edes’ Palimpsest and the Stomachion. Figure 12.4 Archimedes' Stomachion puzzle H i s t o r y o f D i s s e c t i o n s i n t h e T h e o r y o f A r e a Dissections have been the basis, through history, o f many proofs for the Pythagorean Theorem, including in Ancient India and China; see also Chapter 13 and Problem 13.2. In the Plato’s Meno [AT: Plato] there is a http://www.maa.org/editorial/mathgames/mathgames_l History of Dissections in the Theory of Area 169 Socratic dialogue in which is described the dissection p ro o f that the diagonal o f a square is the side o f a square o f tw ice the area. One o f the wall mosaics in the famous A lham bra palace is based on dissection o f the square to prove the Pythagorean Theorem. Euclid in his Elements implicitly used equivalence by dissection and equivalence by subtraction when he proved propositions about the area (he used the term “equal”) o f polygons. The use o f dissection to find areas continued after Euclid, but it was not until 1902 that David H ilbert [FO: Hilbert] developed E uclid’s Postulates and dissection theory into a rigorous theory o f area. Hilbert discusses (in his Chapter IV) both equivalence by dissection (zerle- gungsgleich) and equivalence by subtraction (erganzungsgleich) and when the Archim edean Postulate was necessary. In a footnote, Hilbert gives credit for similar discussions o f the theory o f areas to M. Gerard (in 1895-1898), F. Schur (in 1898), and O. Stolz (in 1894). A recent detailed discussion o f the dissection theory o f area can be found in [TX: H artshom e]. A recent history o f dissections is contained in Chapter 13. P r o b l e m 12.1 D i s s e c t P l a n e T r i a n g l e a n d P a r a l l e l o g r a m Some o f the dissection problem s ahead are very simple, while some are rather difficult. I f you think that a particular problem was so easy to solve that you may have m issed something, chances are you hit the nail right on the head. M ost o f the dissection proofs will consist o f two parts: First show where to make the necessary cuts, and then prove that your construction works, that is, that all the pieces do in fact fit together as you say they do. a. Show that on the plane every triangle is equivalent by dissection to a parallelogram with the same base no matter which base o f the triangle you pick. Part a is fairly straightforward, so don’t try anything complicated. You only have to prove it for the plane — a p ro o f for spheres and hyperbolic planes will come in a later problem after we find out w hat to use in place o f parallelograms. Make paper models, and make sure your method works for all possible triangles with any side taken as the base. In particular, make sure that your p ro o f works for triangles whose heights 170 Chapter 12 Dissection Theory are much longer than their bases. Also, you need to show that the result­ ing figure actually is a parallelogram. b. Show that, i f you assume A P, then on a plane every paral­ lelogram is equivalent by dissection to a rectangle with the same base and height. Show equivalence by subtraction without assuming AP. A partial p ro o f o f this was given in the introduction at the beginning o f this chapter. But for this problem, your p ro o f m ust also work for tall, skinny parallelogram s, as shown in Figure 12.5, for which the given construction does not work. You may say that you can simply change the orientation o f the parallelogram and use a long side as the base; but, as for part a, we want a p ro o f that will w ork no m atter which side you choose as the base. Be sure to note where you use A P. Again, do not try anything too com plicated, and you only have to w ork on the plane. D i s s e c t i o n T h e o r y o n S p h e r e s a n d H y p e r b o l i c P l a n e s The above statements take on a different flavor when working on spheres and hyperbolic planes because we cannot construct parallelo­ grams and rectangles, as such, on these spaces. We can define two types o f polygons on spheres and hyperbolic spaces and then restate the above two problem s for these spaces. The two types o f polygons are the Khayyam quadrilateral and the Khayyam parallelogram. These defini­ tions were first put forth by the Persian geometer-poet Om ar Khayyam Dissection Theory on Spheres and Hyperbolic Planes 171 (1048-1131) in the 11th century AD [AT: Khayyam 1958], Through a bit o f W estern chauvinism, geometry books generally refer to these quadri­ laterals as Saccheri quadrilaterals after the Italian priest and professor Gerolamo Saccheri (1667-1733), who translated into Latin and extended the works o f Khayyam and others. A Khayyam quadrilateral (KQ ) is a quadrilateral such that AB = CD and Z B A D = Z A D C = /z/2. A Khayyam parallelogram (KP) is a quadrilateral such that AB = CD and AB is a parallel transport o f DC along AD. In both cases, BC is called the base and the angles at its ends are called the base angles. See Figure 12.6. Figure 12.6 Khayyam quadrilaterals and parallelograms P r o b l e m 12.2 K h a y y a m Q u a d r i l a t e r a l s a. Prove that the base angles o f a KQ are congruent. b. Prove that the perpendicular bisector o f the top o f a KQ is also the perpendicular bisector o f the base. 172 Chapter 12 Dissection Theory c. Show that the base angles are greater than a right angle on a sphere and less than a right angle on a hyperbolic plane. d. A KQ on the plane is a rectangle and a KP on the plane is a parallelogram. To begin this problem, note that the definitions o f KP and KQ make sense on the plane as well as on spheres and hyperbolic planes. The pictures in Figure 12.6 are deliberately drawn with a curved line for the base to emphasize the fact that, on spheres and hyperbolic planes, KPs and KQs do have the same properties as rectangles and parallelograms. You should think o f these quadrilaterals and parallelogram s in terms o f parallel transport instead o f parallel lines. Everything you have learned about parallel transport and triangles on spheres and hyperbolic planes can be helpful for this problem. Symmetry can also be useful. Now we are prepared to modify Problem 12.1 so that it will apply to spheres and hyperbolic planes. P r o b l e m 12.3 D i s s e c t S p h e r i c a l a n d H y p e r b o l i c T r i a n g l e s a n d K h a y y a m P a r a l l e l o g r a m s a. Show that every hyperbolic triangle, and every small spherical triangle, is equivalent by dissection to a Khayyam parallelo­ gram with the same base as the triangle. Try your p ro o f from Problem 12.1 as a first stab at this problem. You only need to look at a sphere. You should also look at the different proofs given for Problem 12.1. The only difference betw een the plane and spheres and hyperbolic planes as far as this problem is concerned is that you must be more careful on spheres and hyperbolic planes because there are no parallel lines; there is only parallel transport. Some o f the proofs for Problem 12.1 work well on a sphere or on a hyperbolic plane, and others do not. Rem ember that the base o f a KP is the side opposite the given congruent angles. b. Prove that every Khayyam parallelogram is equivalent by dissection ( i f you assume AP), or equivalent by subtraction P roblem 12.3 Dissect Triangles and Khayyam Parallelograms 173 {without assuming AP), to a Khayyam quadrilateral with the same base. As with part a, start with your planar p ro o f and work from there. As before, your method must work for tall, skinny KPs. Once you have come up with a construction, you must then show that the pieces actually fit together as you say they do and prove that the angles at the top are right angles. See Figure 12.7. * Figure 12.7 Dissecting KP into KQ * P r o b l e m 12.4 S p h e r i c a l P o l y g o n s D i s s e c t t o L u n e s In the next chapter you will show (under the assumption o f A P ) that every polygon on the plane is equivalent by dissection to a square, and then we will use this and the Pythagorean Theorem to show that any two polygons with the same area are equivalent by dissection. This does not apply to spheres and hyperbolic planes because there are no squares on these surfaces. However, we have already shown in Problems 7.1 and 7.4 that two polygons (or triangles) on the same sphere have the same area if they have the same holonomy. Thus, every polygon on a sphere m ust have the same area as some lune with the same holonomy. N ow we can show that not only do they have the same area but they are also equivalent by dissection. 174 Chapter 12 Dissection Theory Assuming AP, show that every simple (sides intersect only at the vertices) small polygon on a sphere is equivalent by dissec­ tion to a lune with the same holonomy. That is, the angle o f the lune is equal to (V2)(2n - sum o f the exterior angles o f the polygon). Consequently, two simple small polygons with the same area on the same sphere are equivalent by dissection. O utline of Proof The p ro o f o f this result can be com pleted by proving the following steps (or lemmas). (This p ro o f was first suggested to David by his daughter, Becky, now Rebecca Wynne.) 1. Every simple small polygon can be dissected into a finite number o f small triangles, such that the holonomy o f the polygon is the sum o f the holonomies o f the triangles. See Problem 7.5, but what is needed here is easier than 7.5. 2. Each small triangle is equivalent by dissection to a KQ with the same base and same holonomy. Check your solutions for Problems 12.2 and 12.3. 3. Two KQs with the same base and the same holonomy {or base angles) are congruent. M atch up the bases and see what you get. 4. I f two As have the same base and the same holonomy, then they are equivalent by dissection. Put together the previous steps. 5. Any A is equivalent by dissection to a lune with K {A ) = 77(lune) = {twice the angle o f the lune). Hint: A lune can also be considered as a triangle. ♦ P roblem 12.4 Spherical Polygons Dissect to Lunes 175 6. Two simple small polygons on a sphere with the same area are equivalent by dissection to the same lune and therefore are equivalent by dissection to each other. W hat is the union o f two lunes? The first four steps above will also work (with essentially the same proofs) on a hyperbolic plane. But there is no clear replacem ent for the biangles (which do not exist on a hyperbolic plane). There is a p ro o f o f the following: T heorem 12.4. On a hyperbolic space, two simple (the sides intersect only at the vertices) polygons with the same area are equivalent by dissection. Tw o published proofs in English are in [TX: M illman & Parker], page 267, and [DI: Boltyanski 1978], page 62. These proofs are similar, and both use the first four steps above and use the completeness o f the real num bers (in the form o f a version o f the Intermediate Value Theorem). You can check that B ecky’s p ro o f above does not use completeness. In addition, the p ro o f o f the same result on the plane (see the discussion between Problems 13.2 and 13.3) also does not need the use o f completeness axiom. Figure 12.8 Triangles with same base and same area 176 Chapter 12 Dissection Theory In the plane all the triangles with the same base and the same height have the same area and the vertices opposite the base o f these triangles form a (straight) line not intersecting the line determined by the base. On a sphere, the situation is different. Your p ro o f above should show that midpoints o f the (non-base) edges lie on a great circle and the vertices opposite the base m ust lie on a curve equidistant from this great circle. See Figure 12.8. Chapter 13 Square Roots, Pythagoras, and Similar T riangles The diagonal of an oblong produces by itself both the areas which the two sides of the oblong produce separately. — Baudhayana, Sulbasutram, Sutra 48 [AT: Baudhayana] This chapter is devoted to a subject that is not only o f great theoretical interest but also o f recreational interest, for dissection theory plays an important role in a rigorous developm ent o f area and volume and also furnishes a seemingly endless variety o f attractive and challenging puzzles. In the last chapter, we showed that two polygons on a sphere with the same area are equivalent by dissection because both are equivalent to the same biangle. In this chapter, we will prove an analogous result on the plane by showing that every planar polygon is equivalent by dissec­ tion to a square. This result is the basis for m any popular dissection puzzles. Before reading this chapter you should go through the Introduc­ tion to Dissection Theory and Problem 12.1 at the start o f Chapter 12. In the process o f exploring this dissection theory, we will follow a path through a com er o f the forest o f m athematics — a path that has delighted and surprised the authors many times. We will bring with us the question, W hat are square roots? Along the way we will confront relationships betw een geometry and algebra o f real numbers, in addition to sim ilar triangles, the Pythagorean Theorem (the quote above is a state­ ment o f this theorem written before Pythagoras), and possibly the oldest written p ro o f in geometry (at least 2600 years old). This path will lead to 177 178 Chapter 13 Square Roots, Pythagoras, and Similar Triangles the solutions o f quadratic and cubic equations in Chapter 19. We will let D avid’s personal experience start us on this path. S q u a r e R o o t s W hen David was in eighth grade, he asked his teacher, “W hat is a square root?” He knew that the square root o f N was a num ber whose square was equal to N, but where w ould one find it? (Hidden in that question is “How does one know it always exists?”) He knew w hat the square roots o f 4 and 9 were — no problem there. He even knew that the square root o f 2 was the length o f the diagonal o f a unit square, but what o f the square root o f 2.5 or o f 7i? At first, the teacher showed David a square root table (a table o f numerical square roots), but he soon discovered that i f one took the num ber listed in the table as the square root o f 2 and squared it, you got 1.999396, not 2. (M odern-day pocket calculators give rise to the same problem but hide it reasonably effectively by calculating more digits than they display and then rounding off.) So David persisted in asking his question, W hat is the square root? Then the teacher answered by giving him THE A N SW ER — the Square Root Algorithm. Do you re­ member the Square Root Algorithm — that procedure, sim ilar to long division, by which it is possible to calculate the square root? Or perhaps more recently you were taught the “divide and average” method, a more efficient method that was known to Archytas o f Tarentum (428-350 b .c ., Greek) but today is often called N ew ton’s Method after Isaac Newton (English, 1643-1727). I f A i is an approxim ation o f the square root o f N, then the average o f A\ and N/A\ is an even better approxim ation, which we could call A2 . And then the next approxim ation A3 is the average o f A2 and N/A2. In equation form this becomes A r t = (1/2 )(A„+(N/An)). For example, i f A, = 1.5 is an approxim ation o f the square root o f 2, then A 2 = 1.417— , ^ 3 = 1 .4 1 4 2 1 6 - and so forth are better and better approximations. Square Roots 179 But wait! M ost o f the time these algorithms do not calculate the square root — they only calculate approxim ations to the square root. The algorithms have an advantage over the tables because we could, at least in theory, calculate approxim ations as close as we wished. However, they are still only approxim ations and D avid’s question still rem ained — What is the square root these algorithms approximate? D avid’s eighth-grade teacher then gave up, but later in college David found out that some m odem mathematicians answer his question in the following way: “We make an assum ption (the Completeness Axiom) that implies that the sequence o f approxim ations from the Square Root Algorithm m ust converge to some real num ber.” And, when he continued to ask his question, he found that in m odem m athematics the square root is a certain equivalence class o f Cauchy (Augustin-Louis Cauchy, 1789-1857, French) sequences o f rational numbers, or a certain Dedekind (Julius Dedekind, 1831-1916, German) cut. Finally, David let go o f the question and forgot it in the turmoil o f graduate school, writing his thesis, and beginning his m athematical career. Later, David started teaching the geometry course that is the basis for this book. One o f the problem s in the course was the following problem. [For this chapter you will need, at tim es, to assume the A rchi­ m edean Postulate, A P — when you do, note it.] P r o b l e m 13.1 A R e c t a n g l e D i s s e c t s i n t o a S q u a r e Show that, on the plane, every rectangle is equivalent by dissec­ tion to a square. b Figure 13.1 Dissecting a rectangle into a square S uggestions Problems 13.1 and 13.2 can be done in any order. So i f you get stuck on one problem, you can still go on to the other. In Problems 13.1 and 13.2, it is especially important to make accurate models and constructions — 180 Chapter 13 Square Roots, Pythagoras, and Similar Triangles rough drawings will not show the necessary length and angle relation­ ships. Problem 13.1 is one o f the oldest problem s in geometry, so you may have guessed (correctly!) that it is one o f the more complex ones. This problem is interesting for more than ju st historical reasons. You are asked to prove that you can cut up any rectangle (into a finite num ber o f pieces) and rearrange the pieces to form a square, as in Figure 13.1. Because you are neither adding anything to the rectangle nor re­ moving anything, the area m ust rem ain the same, so ab = x2, or x = J a b . W hat you are really finding is a geometric interpretation o f a square root. This was done already by Pythagoreans (6th century b.c.), who called x the geometric mean o f a and b. Let us look at a p ro o f that is similar to the proofs in m any standard geometry textbooks: Let s = Ja b be the side o f the square equivalent by dissection to the rectangle with sides a and b. Place the square, AEFH, on the rectangle, ABCD, as shown in Figure 13.2. Draw ED to intersect BC in R and HF in K. Let BC intersect HF in G. [In the case that ABCD is so long and skinny that K ends up betw een G and F, we can, by cutting ABCD in h a lf and stack­ ing the halves, reduce the p ro o f to the case above. N ote that A P is im plicitly being used here.] From the sim ilar trian­ gles A KDH and A EDA we have HK/AE = HD/AD, or H K = ( AE)(HD)/AD = s(a - s)/a = 5 - s 2/a = s - b . Therefore, we have A EFK = A RCD, A EBR = A KHD. F ig u r e 1 3 . 2 T e x t b o o k p r o o f Problem 13.1 A Rectangle Dissects into a Square 181 David was satisfied with this proof until, in the second year o f teaching the geometry course, he started sensing student uneasiness with it. As he listened to their comments, he noticed questions being asked by the students: “What is -fab ?” “How do you find it?” Those used to be his questions! The students and David also noticed that the facts used about similar triangles in the p roof above are usually proved using the theory o f areas o f triangles. Thus, the proof could not be used as part o f a concrete the­ ory o f areas o f polygons, which was our purpose in studying dissection theory in the first place. Notice that the above proof also assumes the existence o f the square, which in analysis is based on what is called the Completeness Axiom. The conclusion here seems to be that it would be desirable instead to construct the square root x. That started David on an exploration that continued on and o ff for many years. Now let us solve a few problems in dissection theory. Here are three methods for constructing x. For all three construc­ tions we will use a rectangle like the one shown in Figure 13.1, with the longer side b as the base and the shorter side a as the height. For both the first and second constructions, you need to know that a triangle with all vertices on the circumference o f a circle and one side a diameter o f the circle is a right triangle. See Figure 13.3. That the angle not on the diameter is a right angle follows from a more general result, which we will prove in Problem 15.1a, but the special case we need is easy to see now. In particular, draw the line segment from the vertex to the center o f the circle and note that two isosceles triangles are formed. By ITT, Problem 6.2, there are congruent angles as marked; and, because 182 Chapter 13 Square Roots, Pythagoras, and Similar Triangles the sum o f the angles in a planar triangle is 180° (Problems 7.3b or 10.2) we can compute that (2 a + y) + (2/3 + ju) = 360° and, because y+ ju= 180°, it follows immediately that a + J3= 90°. See Problem 15.1 for a more general result. For the first construction, Figure 13.4, take the rectangle and lay a out to the left o f b. Use this base line as the diam eter o f a circle. The length x that you are looking for is the perpendicular line from the left side o f the rectangle to where it intersects the circle. The second construction, Figure 13.5, is similar, but this time put a on the inside o f the base o f the rectangle. N ow the side x you are looking for is the segment from the lower left com er o f the rectangle to the point o f intersection on the circle with the perpendicular rising from the point a in from the com er as indicated in the figure. F ig u r e 1 3 . 5 S e c o n d c o n s tr u c tio n o f x Problem 13.1 A Rectangle Dissects into a Square 183 The third construction, Figure 13.6, is a bit m ore algebraic than the others, and doesn’t directly involve a circle. b - a Figure 13.6 Third construction of x This construction can be used together with the result o f Problem 13.2 in order to obtain a p ro o f by subtraction. For all o f these constructions, it is im perative that you use accurate models. W hatever method you choose, make the rectangle and the square overlap as much as possible, and see how to fit the other pieces in. Then you have to prove that all o f the sides and angles line up properly. N ote that it is m uch better to solve this problem geometrically rather than by only trying to w ork out the algebra — actually do the construction and proceed from there. Finally, you don’t have to use one o f the constructions shown here. If these don’t make sense to you, then find one o f your own that does. And rem em ber not to use results about sim ilar triangles because we normally need results about areas in order to prove these results as we will do in Problem 13.4. Pause, explore, and write out your ideas before reading further. 184 Chapter 13 Square Roots, Pythagoras, and Similar Triangles B a u d h a y a n a 's S u l b a s u t r a m W hile reading an article, David ran across an item that said that the problem o f changing a rectangle into a square appeared in the Sulba­ sutram. “ Sulbasutram ” means “rules o f the cord,” and the several San­ skrit texts collectively called the Sulbasutra were written by the Vedic Hindus starting before 600 b .c . and are thought to be com pilations o f oral wisdom that may go back to 2000 b .c . (See, for example, A. Seiden- berg The Ritual Origin o f Geometry [H I: Seidenberg].) These texts have prescriptions for building fire altars, or Agni. However, contained in the Sulbasutra are sections that constitute a geometry textbook detailing the geometry necessary for designing and constructing the altars. As far as we have been able to determine, these are the oldest geometry (or even m athematics) textbooks in existence. There are at least four versions o f the Sulbasutram by Baudhayana, Apastamba, Katyayana, and Manava. The geometric descriptions are very sim ilar in these four books, and we will only use Baudhayana’s version here (see [AT: Baudhayana]). The first chapter o f Baudhayana’s Sulbasutram contains geometric statements called “Sutra.” Sutra 54 is what we asked you to prove in Problem 13.1. It states the following: If you wish to turn an oblong [here “oblong” means “rectangle”] into a square, take the shorter side of the oblong for the side of square. Divide the remainder into two parts and inverting join those two parts to two sides of the square. Fill the empty place by adding a piece. It has been taught how to deduct it. [See Figure 13.7.] I b I I b a I I I a i * I I i n b a a F ig u r e 1 3 . 7 S u tr a 5 4 Baudhayana’s Sulbasutram 185 So our rectangle has been changed into a figure with an “empty place,” which can be filled “by adding a piece” (a small square). The result is a large square from which a small square has to be “deducted.” Now Sutra 51 (Refer to Figure 13.8.): If you wish to deduct one square from another square, cut off a piece from the larger square by making a mark on the ground with the side of the smaller square which you wish to deduct; draw one of the sides across the oblong so that it touches the other side; by this line which has been cut off the small square is deducted from the large one. D C Figure 13.8 Sutra 51: Construction of side of square We wish to deduct the small square (a2) from the large square. Sutra 51 tells us to “scratch up” with the side o f the smaller square — this produces the line AB and the oblong ABCD. Now, i f we “draw ” the side CD o f the large square to produce an arc, then this arc intersects the other side at the point E. The sutra then claims that BE is side o f the desired square whose area equals the area o f the large square minus the area o f the small square. This last assertion follows from Sutra 50, which we ask you to prove in Problem 13.2. See Figure 13.9 for the drawing that goes with Sutra 50: If you wish to combine two squares of different size into one, scratch up with the side of the smaller square a piece cut off from the larger one. The diagonal of this cutoff piece is the side of the combined squares. A a 186 Chapter 13 Square Roots, Pythagoras, and Similar Triangles Be sure you see why Sutra 50 is a statement o f what we call the Pythago­ rean Theorem. A. Seidenberg, in an article entitled The Ritual Origin o f Geometry [HI: Seidenberg], gives a detailed discussion o f the significance o f Baudhayana’s Sulbasutram. He argues that it was written before 600 b .c . (Pythagoras lived about 500 b .c . and Euclid about 300 b .c .) . He gives evidence to support his claim that it contains codification o f knowledge going “far back o f 1700 b .c .” and that knowledge o f this kind was the common source o f Indian, Egyptian, Babylonian, and Greek m ath­ ematics. Together Sutras 50, 51, and 54 describe a construction o f a square with the same area as a given rectangle (oblong) and a proof (based on the Pythagorean Theorem) that this construction is correct. You can find stated in m any books and articles that the ancient Hindus, in general, and the Sulbasutram, in particular, did not have proofs or dem onstrations, or they are dismissed as being “rare.” However, there are several sutras in [AT: Baudhayana] sim ilar to the ones discussed above. W e suggest you decide for yourself to what extent they constitute proofs or demonstrations. Baudhayana avoids the Completeness Axiom by giving an explicit construction o f the side o f the square. The construction can be summa­ rized in Figure 13.10. This is the same as E uclid’s construction in Proposition 11-14 (see [AT: Euclid, Elements], page 409). But Euclid’s p ro o f is m uch more complicated. Note that neither Baudhayana nor Euclid gives a p ro o f o f Problem 13.1 because the use o f the Pythagorean Theorem obscures the Baudhayana’s Sulbasutram 187 dissection. However, they do give a concrete construction and a p ro o f that the construction works. Both Baudhayana and Euclid prove the following theorem, which uses equivalence by subtraction. Theorem. For every rectangle R there are squares S\ and Si such that R + S2 is equivalent by dissection to S\ + S2 and thus R and S\ have the same area. Notice that both Baudhayana’s and E uclid’s proofs o f this theorem and your p ro o f o f Problem 13.1 avoid assuming that the square root exists (and, thus, avoid the Completeness Axiom). They also avoid using any facts about similar triangles. These proofs explicitly construct the square and show in an elem entary way that its area is the same as the area o f the rectangle. There is no need for the area or the sides o f the rectangle to be expressed in numbers. Also given a real number, b, the square root o f b can be constructed by using a rectangle with sides b and 1. In Problem 13.3 we will use these techniques to prove basic properties about sim ilar triangles. So, finally, we have an answer to our question, W hat is a square root? Also, see D avid’s chapter, “ Square Roots in the Sulba Sutra,” in the book [EG: Gorini], which describes a geometric method based on Baudhayana that finds arbitrarily accurate num erical approxim ations to many square roots and does so in a way that is computationally more efficient than the “divide and average” method (Archytas/Newton’s Method) described at the beginning o f this chapter. 188 Chapter 13 Square Roots, Pythagoras, and Similar Triangles P r o b l e m 13.2 E q u i v a l e n c e o f S q u a r e s Prove the following: On the plane, the union o f two squares is equivalent by dissection to another square. S uggestions This is closely related to the Pythagorean Theorem. There are two general ways to approach this problem: You can use Problem 13.1 or you can prove it on its own, which will result in a p ro o f o f the Pythago­ rean Theorem — hence, you can’t use the Pythagorean Theorem to solve this problem because you will be proving it! To see how this problem relates to the Pythagorean Theorem, think about the following statement o f the Pythagorean Theorem: The square on the hypotenuse is equal to the sum o f the squares on the other two sides. This is not ju st an algebraic equation — the squares referred to are actual geometric squares. Figure 13.11 Pythagorean Theorem As in the other dissection problems, make the three squares coincide as m uch as possible, and see how you might get the remaining pieces to overlap. You might start by reflecting the square with side c over the side c on the triangle. Then prove that the construction works as you say it does. Problem 13.2 Equivalence of Squares 189 A n y P o l y g o n C a n B e D i s s e c t e d i n t o a S q u a r e I f we put together Problems 12.1,13.1, and 13.2, we get the surprising Theorem 1 3 . 2 a On the plane, every simple polygon is equivalent by dissection to a square. By 7.5b, the polygon can be dissected into triangles. dissected > By 12.1b, all
parallelograms
can be dissected
into rectangles.
By 13.1, every
rectangle
can be dissected
into a square.
Using 13.2, transform every pair of squares into
a single square and continue until only one square is left.
F ig u r e 1 3 . 1 2 E v e r y p o ly g o n d is s e c ts in to a s q u a re

190 Chapter 13 Square Roots, Pythagoras, and Similar Triangles
H i s t o r y o f D i s s e c t i o n s
We have now seen dissection proofs in Chapter 12, Problems 13.1 and
13.2, and we will see more in later chapters (for example, the dissection
proof o f the Law o f Cosines in 20.2a). In the section History o f D issec­
tions in the Theory o f A rea o f Chapter 12 we outlined how dissection
have been used to prove results about area and to develop a rigorous
theory o f area. In this chapter we have seen more examples o f this.
N ote that the result in Theorem 13.2a implies, on the plane, that
Theorem 1 3 . 2 b Any two polygons with the same area are equivalent
by dissection to each other.
This theorem is true not only on the plane but also on spheres (Problem
12.4) and hyperbolic planes (Theorem 12.4). On the plane the theorem is
often called the Bolyai-Gerwien-Wallace Theorem (with sometimes
W allace’s name left off) because apparently it was first proved by them:
W allace in 1831, Bolyai in 1832, and Gerwien in 1833.
Note that Theorem 13.2 does not tell us anything about how many
pieces w ould be needed in any particular dissection. In fact, the number
will, in general, be very large. So this leads to the question o f making
dissections with the least num ber o f pieces, or with the m ost symmetry,
or, in some other way, with the most elegance. A large num ber o f
polygonal dissection puzzles and recreations arises from these questions.
There is a surprisingly rich history associated with recreational dissec­
tions. The m ost recent accounts o f dissection puzzles and their history is
in [DI: Frederickson 1997] and [DI: Frederickson 2002]. Refer to these
books for further history and about 1000 dissection pictures. There are
also further references in the Dissection section o f the Annotated
Bibliography.
As an example, the dissection in Figure 13.13 is a famous dissec­
tion o f a equilateral triangle into a square that requires only four pieces.
The discovery o f this dissection is usually credited to Henry Dudeney
(English, 1857-1930) in 1902, but [DI: Frederickson 1997], p. 136,
raises some doubts about this attribution.
In this dissection, the segments AD, DB, BE, EC, FG are all equal to
h a lf the side o f the triangle; EF is equal to the side o f the equivalent
square; DJ and GK are each perpendicular to EF. I f the four pieces are
successively hinged to one another at the points, D, E, G, then holding

History of Dissections 191
piece I fixed and swinging the connected set o f pieces IV-III-II clock­
wise, the equilateral triangle is neatly carried into the square.
Figure 13.13 Four-piece dissection of triangle to square
We could build a set o f four connected tables based on this fact; swing­
ing the tables in one direction causes the tops to fit together into a single
equilateral triangular table and swinging them in the other direction
causes the tops to fit together into a single square table. This is an
example o f a hinged dissection. If you go back to your dissections in
Chapter 12 and 13.1 and 13.2, you will probably find that some o f them
can also be hinged. The [DI: Frederickson 2002] book is entirely about
hinged dissections and what he calls “twisting dissections.”
The 18th and 19th centuries produced a variety o f puzzle books that
contained dissection puzzles. Articles about dissections also appeared in
scientific journals. In the early 19th century the Bolyai-Gerwien-W allace
Theorem was proved. Dissection puzzles had great popularity during
1880s and 1890s, and at the end o f 19th century puzzle columns by Sam
Loyd and H enry E. D udeney began to appear in newspapers and
magazines. The 20th century brought an increase in sophistication to
dissections. A rticles and problem s appeared in periodicals such as
Mathematical Gazette, American Mathematical Monthly, and Scientific
American. Geometric dissection has become a frequent topic in books on
m athematical recreations, but there are books that focus exclusively on
dissections, [DI: Lindgren 1964 & 1972] and [DI: Boltyanski 1980].
See the references in the Dissections section o f the Bibliography for
more about dissections.

192 Chapter 13 Square Roots, Pythagoras, and Similar Triangles
P r o b l e m 13.3 M o r e D i s s e c t i o n – R e l a t e d P r o b l e m s
Some o f these dissection problem s are easy and some are hard. We do
not tell you which; but, in any case, what is hard for us may be easy for
you.
a. Show that the four-piece dissection o f an equilateral triangle to
a square depicted in Figure 13.13 actually works as claimed.
b. Show that any obtuse triangle (one angle is greater than a right
angle) can be dissected into acute triangles (all angles are less
than a right angle). What is the dissection with the smallest
number o f pieces you can find?
c. Prove that the figure form ed when the midpoints o f the sides o f a
quadrilateral are jo in ed in order is a parallelogram, and its
area is h a lf that o f the quadrangle. Rem ember that quadrilater­
als need not be convex; see Figure 13.14.
This parallelogram is often referred to as the Varignon parallelogram o f
the quadrilateral, because the result first appeared in a 1731 posthumous
publication by Pierre Varignon (1654-1722).

Problem 13.3 More Dissection-Related Problems 193
d. Show that i f one diagonal divides a quadrangle into two trian­
gles o f equal area, it bisects the other diagonal. Conversely, i f
one diagonal bisects the other, it bisects the area o f the
quadrangle.
e. Divide a triangle by two straight lines into three parts, which
when properly arranged shall form a parallelogram whose
angles are o f given magnitudes.
f. From the arbitrary point within a triangle, draw three lines to
the sides o f the triangle that shall trisect its area.
g. Dissect a regular hexagon by straight cuts into six pieces that
can be reassembled to form an equilateral triangle. Does there
exist a five-piece solution?
h. Given 6 > n, show that it is possible to f i t together n isosceles
right triangles, all o f different sizes, so as to make a single
isosceles right triangle. Can the problem be solved fo r n = 5?
Can a right isosceles triangle be dissected into a fin ite number
o f right isosceles triangles, no two having a common side?
i. Show that any given triangle can be dissected by straight cuts
into fo u r pieces that can be arranged to form two triangles
similar to the given triangle.
j. Dissect a regular pentagon by straight cuts into six pieces that
can be p u t together to form an equilateral triangle.
k. Draw a straight line that will bisect both the area and the
perimeter o f a given quadrilateral.
^ T h r e e – D i m e n s i o n a l D i s s e c t i o n s
a n d H i l b e r t ‘ s T h i r d P r o b l e m
In 1900, David Hilbert delivered a lecture before the International
Congress o f M athem aticians in which he listed ten problem s “from the
discussion o f which an advancem ent o f science may be expected.” In a

194 Chapter 13 Square Roots, Pythagoras, and Similar Triangles
later paper he expanded these to 23 problem s, which are now called
Hilbert’s Problems.
H i l b e r t ‘s T h i r d P r o b l e m . Is it possible to specify two tetra-
hedra o f equal bases and equal altitudes that can in no way be
split up into congruent tetrahedra and that cannot be com­
bined with congruent tetrahedra to form two polyhedra that
themselves could not be split up into congruent tetrahedral
See [DI: Sah], [DI: Boltyanski 1978], and [TX: Hartshom e]
for a discussion o f this problem and its history.
Shortly after H ilbert’s lecture, M ax Dehn (1878-1952, Germany,
United States) found such tetrahedra and also proved that a regular tetra­
hedron is not equivalent by dissection to a cube. Thus there is no possi­
bility o f dissecting polyhedra into cubes. To show these results, Dehn
proved the following:
Theorem 1 3 . 4 I f P and Q are two polyhedra in 3-space that
are equivalent by dissection (or by subtraction), then the
dihedral angles (see Chapter 21) o f P are, mod n, a rational
linear combination o f the dihedral angles o f Q. That is, if a,
are the dihedral angles o f P and f are the dihedral angles
o f Q, then there are integers n„ ms, a, b such that
X, n,a, + an = Zymjf3j + bn.
This interesting fact is o f far-reaching importance in the theory o f
volumes. It implies that although a theory o f areas o f polygons can be
constructed without continuity considerations, such is not possible for a
complete theory o f volumes o f polyhedra. Thus the ordinary treatm ent o f
volumes, such as we find in Euclid’s Elements, involves infinitesimals,
which do not occur in the theory o f polygonal areas. In addition, we
showed in Problem 7.5b that though it is always possible to dissect a
polygon into a finite num ber o f triangles having their vertices only at the
vertices o f the polygon, there exist polyhedra, called Lennes polyhedra,
which cannot be dissected into finite num bers o f tetrahedra having their
vertices only at the vertices o f the polyhedra.

P roblem 13.4 Similar Triangles 195
P r o b l e m 13.4 S i m i l a r T r i a n g l e s
N ear the beginning o f this chapter we gave a textbook p ro o f o f Problem
13.1, which used properties o f sim ilar triangles. Later you found a p ro o f
that did not need to use sim ilar triangles. Rem ember that in our discus­
sion o f AAA we said that two triangles were similar if their correspond­
ing angles are congruent. N ow you are ready to give a dissection p roof
o f the following:
a. AAA s i m i l a r i t y c r i t e r i o n : I f two triangles are similar, then the
corresponding sides o f the triangles are in the same proportion
to one another.
S uggestions
Look at your p ro o f o f Problem 13.1. It probably shows im plicitly that
Problem 13.4 holds for a pair o f sim ilar triangles in your construction.
For more generality, let # b e one o f the angles o f the triangles and place
the two 6 ’s in VAT position in such a way that they form two parallelo­
grams as in Figure 13.15.
D

196 Chapter 13 Square Roots, Pythagoras, and Similar Triangles
Show that the two parallelograms are equivalent by dissection and
use that result to show that ac = bd or, in equivalent form, a/d = b/c. You
may find it clearer if you start by looking at the special case o f 6= n il.
b. SAS similarity criterion: I f two triangles on a plane have an
angle in common and i f the corresponding sides o f the angle are
in the same proportion to each other, then the triangles are
similar.
Draw the two triangles with their common angle coinciding. Then show
that the opposite sides are parallel. You will probably have to use part a.
These results about similar triangles will be used in most o f the
remaining chapters o f the book.

Chapter 14
Projections of a S phere
onto a Plane
Geography is a representation in picture of the whole
known world together with the phenomena which are con­
tained therein.
… The task of Geography is to survey the whole in its
proportions, as one would the entire head. For as in an entire
painting we must first put in the larger features, and afterward
those detailed features which portraits and pictures may
require, giving them proportion in relation to one another so
that their correct measure apart can be seen by examining
them, to note whether they form the whole or a part of the
picture. … Geography looks at the position rather than the
quality, noting the relation of distances everywhere,…
It is the great and the exquisite accomplishment of math­
ematics to show all these things to the human intelligence …
— Claudius Ptolemy, Geographia, Book One, Chapter I
A m ajor problem for map makers (cartographers) since Ptolemy (approx.
85-165 a.d ., Alexandria, Egypt) and before is how to represent accu­
rately a portion o f the surface o f a sphere on the plane. It is the same
problem we have encountered when making drawings to accom pany our
discussions o f the geometry o f the sphere. We shall use the term inology
used by cartographers and differential geometers to call any one-to-one
function from a portion o f a sphere onto a portion o f a plane a chart. As
Ptolem y states in the quote above, we would like to represent the sphere
on the plane so that proportions (and thus angles) are preserved and the
relative distances are accurate. For a history and m athematical descrip­
tions o f charts o f the sphere, see [CE: Snyder], For a history and
197

198 Chapter 14 Projections o f a Sphere onto a Plane
discussion o f the political, social, and ethnic controversies that have been
and continue to be connected with maps and map making, see [CE:
Monmonier]. For a discussion o f how maps and the stars were used to
determine the position o f the earth, see the delightful book [CE: Sobel],
History o f specific map projections is given in the last section o f this
chapter.
In this chapter we will study various charts for spheres. We will
need properties o f similar triangles that are investigated in Problem 13.4.
P roblem 14.1 C harts M ust Distort
It is impossible to make a chart without some distortions.
Which results that you have studied so fa r show that there must
be distortions when attempting to represent a portion o f a sphere
on the plane?
Nevertheless, there are projections (charts) from a part o f a sphere to the
plane that do take geodesic segments to straight lines, that is, that
preserve the shape o f straight lines. There are other projections that pre­
serve all areas. There are still other projections that preserve the measure
o f all angles. In this chapter, we will study these three types o f projec­
tions on a sphere, and in Chapter 17 we will look at projections o f hyper­
bolic planes.
P roblem 14.2 G nomic P rojection
Imagine a sphere resting on a horizontal plane. See Figure 14.1. A
gnomic projection is obtained by projecting from the center o f a sphere
onto the plane. Note that only the lower open hemisphere is projected
onto the plane; that is, if A is a point in the lower open hemisphere, then
its gnomic projection is the point, g(X), where the ray from the center
through X intersects the plane.
a. Show that a gnomic projection takes the portions o f great circles
in the lower hemisphere onto straight lines in the plane. (A
mapping that takes geodesic segments to geodesic segments is
called a geodesic mapping.)

P roblem 14.2 Gnomic Projection 199
b. Gnomic projection is often used to make navigational charts fo r
airplanes and ships. Why would this be appropriate?
Hint: Start with our extrinsic definition o f “great circle.”
P r o b l e m 14.3 C y l i n d r i c a l P r o j e c t i o n
Figure 14.2 Cylindrical projection
Imagine a sphere o f radius r, but this time center it in a vertical cylinder
o f radius r and height 2r. The cylindrical projection is obtained by pro­
jecting from the axis o f the cylinder, which is also a diam eter o f the

2 0 0 Chapter 14 Projections o f a Sphere onto a Plane
sphere; that is, i f X is a point (not the north or south poles) on the sphere
and 0(X ) is the point on the axis at the same height as X, then X is
projected onto the intersection o f the cylinder with the ray from 0(X ) to
X. See Figure 14.2.
a. Show that cylindrical projection preserves areas. (Mappings
that preserve area are variously called area-preserving or
equiareal.)
Geometric Approach; Look at an infinitesimal piece o f area on
the sphere bounded by longitudes and latitudes. Check that when it is
projected onto the cylinder the horizontal dim ension becomes longer but
the vertical dimension becomes shorter. Do these compensate for each
other?
♦ Analytic Approach: Find a function / from a rectangle in the
{z,9)-plane onto the sphere and a function h from the same rectangle
onto the cylinder such that c(f{z, 0)) = h(z, 9). Then use the techniques o f
finding surface area from vector analysis. (For two vectors A, B, the
m agnitude o f the cross product \A * B\ is the area o f the parallelogram
spanned by A and B. An element o f surface area on the sphere can be
represented by \fz x fe \d z d 9 , the cross product o f the partial derivatives.)
We can easily flatten the cylinder onto a plane and find its area to
be A nr1. We thus conclude the following:
b. The {surface) area o f a sphere o f radius r is A nt2.
P r o b l e m 14.4 S t e r e o g r a p h i c P r o j e c t i o n
Imagine the same sphere and plane, only this time project from the
upperm ost point (north pole) o f the sphere onto the plane. This is called
stereographic projection.
a. Show that stereographic projection preserves the sizes o f angles.
(M appings that preserve angles are variously called angle­
preserving, isogonal, or conform al.)
S uggestions
There are several approaches for exploring this problem. Using a purely
geometric approach requires visualization but only very basic geometry.

P roblem 14.4 Stereographic Projection 201
An analytic approach requires knowledge o f the differential o f a function
from R 2 into R 3. See Figure 14.3.
Figure 14.3 Stereographic projection is angle-preserving
Geometric Approach: An angle at a point A o n the sphere is deter­
m ined by two great circles intersecting at X. Look at the two planes that
are determined by the north pole N and vectors tangent to the great
circles at X. Notice that the intersection o f these two planes with the
horizontal image plane determines the image o f the angle. Because the
3-dimensional figure is difficult for many o f us to imagine in full detail,
you may find it helpful to consider what is contained in various
2-dimensional planes. In particular, consider the plane determined by X
and the north and south poles, the plane tangent to the sphere at X, and
the planes tangent to the sphere at the north and south poles. Determine
the relationships among these planes.
^Analytic Approach: Introduce a coordinate system and find a
formula for the function s~’ from the plane to the sphere, which is the
inverse o f the stereographic projection s. Use the differential o f s~l to
examine the effect o f s ‘ 1 on angles. You will need to use the dot (inner)
product and the fact that the differential o f s_1 is a linear transform ation
from the (tangent) vectors at s(X) to the tangent vectors at X. *
*b. Show that stereographic projection takes circles through N to
straight lines and circles not through N to circles.
(Such mappings are called circle-preserving.)

2 02 Chapter 14 Projections o f a Sphere onto a Plane
N
Figure 14.4 Stereographic projection is circle-preserving
S uggestions
Let y b e a circle on the sphere with points A and B and let y ‘, A’, B’ be
their images under stereographic projection. Form the cone that is
tangent to the sphere along the circle y and let P be its cone point (note
that P is not on the sphere). See Figure 14.4. Thus the segments BP and
AP are tangent to the sphere and have the same length r. Look in the
plane determined by N, A, and P and show that ZPAA’ is congruent to
ZAA’P’. You probably have already proved this in part a; i f not, look at
the plane determined by N, A, and P and its intersections with the plane
tangent to the north pole N and the image plane /Z In this plane draw
line PA” parallel to P’A’. Then use similar triangles and 6.2c to show
and thus y’ is a circle with center at P ‘.
H i s t o r y o f S t e r e o g r a p h i c P r o j e c t i o n a n d A s t r o l a b e
The earliest known uses o f projections o f a sphere onto the plane were in
Greece and were for purposes o f map making (as described in the quote
at the beginning o f this chapter from Ptolem y’s Geographia). In addi­
tion, as we discussed in Chapter 2, the Greeks (following the Babyloni­
ans) in 4th century b.c. considered the visible cosmos to be a sphere with
three different coordinate systems: celestial, ecliptic, and horizon. At

History o f Stereographic Projection and Astrolabe 203
least as early as 2nd century b.c ., the Greek and later the Arab m athem ati­
cians used the sphere projections that we have studied to represent these
three spheres on the a plane.
The earliest references to stereographic projection in literature are
given by Vitruvius (Roman, -1 0 0 b.c .), Ten Books on Architecture, and
in Ptolem y’s Representation o f the Sphere in the Plane. The ancients
knew how to prove (using propositions from A pollonius’ Conics [AT:
Appollonius]) some properties o f stereographic projection, such as the
following: Circles through the pole are m apped onto straight lines and
all other circles are m apped onto circles. (See Problem 14.4b.)
In Ptolem y’s work the description o f stereographic projection was
used for a horoscopic instrument for determining time. Later the word
horoscope denoted the point o f intersection o f the ecliptic and the
eastern part o f the horizon determined by means o f this instrument.
Probably Theon o f Alexandria (Greek, ?335-?405) was the first to
combine stereographic projections o f the three-sphere m odel onto a
single compact planar instrument called an astrolabe, as pictured in
Figure 13.15. T heon’s w ork has not survived, but we know o f it because
o f two surviving works on the astrolabe: On the Construction and Use o f
the Astrolabe by Philoponus (6th century Alexandria) and the Treatise on
the Astrolabe by Severus Sebokt (7th century Syria).
The astrolabe was widely known in the m edieval East and Europe
from the 9th century until the 19th century and used to solve problems
concerning the apparent positions o f the stars, sun, moon, and planets for
use in navigation, tim e-telling, and astrology. In the M iddle Ages stereo­
graphic projection was often called “astrolabe projection.” The astrolabe
allowed for determining relative positions to about 1° accuracy.
The term “ stereographic projection” was apparently first introduced
by Francois D ’Aguillon (1566-1617) in his Six Books o f Optics. The
earliest exposition o f the theory o f stereographic projection with proofs
was the Book on the Construction o f the Astrolabe by the 9th-century
Baghdad scholar al-Farghani. M athem aticians in the medieval East also
tried to use other geometric transform ations for constructing astrolabes.
The 10lh-century scholar al-Saghani suggested a projection from an
arbitrary point on the axis — i f that point is the center o f the sphere, then
this projection is the gnomic projection o f Problem 14.2. In such a
projection circles on the sphere are m apped onto conics. Al-Biruni
(973-1048, now Uzbekistan) described several ways o f constructing an
astrolabe; including the use a cylindrical projection (Problem 14.3).

2 0 4 Chapter 14 Projections o f a Sphere onto a Plane
Figure 14.5 Astrolabe
In the Figure 14.5 the concentric circles on the background plate
about the center represent stereographic projection o f the celestial sphere
from its south pole — the center o f the astrolabe represents the north
star, the outer rim represents the southern tropic (the furthest the sun
goes south, about -24°), and the concentric circle about two-thirds o f the
way out is the celestial equator. The metal annulus that is off-center is
the stereographic projection o f the ecliptic (the path o f the sun and
planets) — this projection is a circle but the equal divisions o f the eclip­
tic (the 360 degrees o f the ecliptic) are not projected to equal arcs. The
tight system o f circles on the background plate o f the astrolabe (and
m ostly above the center) is the projection o f the coordinates o f the
visible hemisphere with the horizon (not complete) on the outside and
the image o f the zenith (the point in the sky directly over head) where
the coordinate circles converge. Since the relationship betw een the
North Star and the zenith changes with latitude, a given astrolabe is only
accurate at one latitude (about 46° north for the one in the photo).
For more about astrolabes and map making, see [HI: Rosenfeld],
pp. 121-130, [HI: Evans], pp. 141-162, and [HI: Berggren], pp.
165-186.

Chapter 15
C ircles
… the Power of the World always works in circles, and every­
thing tries to be round.
— Black Elk in Black Elk Speaks [GC: Neihardt]
We now study some important properties (Problem 15.1) o f circles in the
plane that are stated and proved in E uclid’s Elements. These planar
results will be used in later chapters, and they are also often studied for
their own interest. In Problem 15.2, we will explore a recent (June 2003)
extension o f these results to circles on spheres (and later to hyperbolic
planes). We will end the chapter with applications o f these properties o f
circles (Problem 15.3) to the ancient problem o f trisecting angles
(Problem 15.4).
For Chapter 15, the only results needed from Chapters 9 -1 4 are
Problem 1 3 . 4 a : The AAA similarity criterion for triangles
on the (Euclidean) plane: I f two triangles are similar (have
congruent angles), then the corresponding sides o f the trian­
gles are in the same proportion to one another. [Needed
throughout this and later chapters.]
Problem 9.1: Side-Side-Side: I f two triangles (small triangles
i f on a sphere) have congruent corresponding sides, then the
triangles are congruent. [Needed throughout this and later
chapters.]
P roblem 14.4: Stereographic projection o f a sphere onto a
plane preserves angles, takes circles to circles (or to straight
lines). [Needed only for Problem 15.2.]
205

2 06 Chapter 15 Circles
I f you are willing to assume these results, then you can work through
Chapter 15 without Chapters 9-14.
P r o b l e m 1 5 . 1 A n g l e s a n d P o w e r o f P o i n t s
f o r C i r c l e s i n t h e P l a n e
These results are all stated and proved in E uclid’s Elements. They are
contained in Propositions 27, 32, and 35-37 o f E uclid’s Book III, which
is entirely devoted to properties o f planar circles.
a. I f an arc o f a circle subtends an angle 2 a from the center o f the
circle, then the same arc subtends an angle a from any point on
the circumference. In particular, two angles that subtend (from
different points on the circumference) the same arc are
congruent.
Use Figures 15.1 and 15.2. Draw a segment from the center o f the circle
to the point A and use ITT. N ote the four different locations for A.
F ig u r e 1 5 . 2 A n g le s s u b te n d e d f r o m o n t h e a rc

P roblem 15.1 Angles and Power o f Points for Plane Circles 207
b. On a plane, i f two lines through a point P intersect a circle at
points A, A’ {possibly coincident) and B, B’ {possibly coin­
cident)i, then
\PA\ x \PA’\ = \PB\ x \PB’\.
This product is called the power o f the point P with respect to
the circle.
Use Figures 15.3 and 15.4 and draw the segment joining A to B’ and the
segment joining A’ to B. Then apply part a and look for sim ilar triangles.
Figure 15.3 Power of a point outside with respect to a circle
For more discussion on power o f a point on the plane, see the
delightful little book [EG: Coxeter & Greitzer],
In Problem 15.2 we will state and prove results about the power o f a
point on spheres and hyperbolic planes that were discovered by Robin
Hartshom e and published in 2003.

2 08 Chapter 15 Circles
s s ‘ = tt’ = d d
Figure 15.4 Power of a point inside with respect to a circle
♦ P r o b l e m 15.2 P o w e r o f P o i n t s
f o r C i r c l e s o n S p h e r e s
There are no similar triangles on spheres and hyperbolic planes, so it
seems surprising that there could be a notion o f power o f a point for
circles on spheres or hyperbolic planes. However, Robin Hartshom e,
[SP: H artshom e], recently found a way to provide a unified definition o f
power o f a point for circles in non-Euclidean geometries — this problem
is based on his paper. H artshom e starts by pointing out that the equality
o f the products \PA\ x \PA’\ = \PB\ x \PB’\ in 15.1b that we used to define
the power o f a point for plane circles can be w ritten as an equality o f
areas o f the rectangles with sides, PA, PA’, and PB, PB’, respectively —
this is the way Euclid considered the pow er o f a point. O f course, there
are no rectangles on spheres and hyperbolic planes. W hat can we use in
their place? In Chapter 12, for our work in dissection theory, we intro­
duced Khayyam quadrilaterals as an appropriate analogue o f rectangles
on spheres and hyperbolic planes. Hartshom e instead introduces another
analogue o f a rectangle: A semi-rectangle is a quadrilateral with oppo­
site sides equal and at least one right angle.

P roblem 15.2. Power of Points for Circles on Spheres 2 09
a. On the plane, semi-rectangles are rectangles. On spheres and
hyperbolic planes, semi-rectangles have the angle opposite the
right angle also right and the two remaining angles are congru­
ent. See Figure 15.5.
Figure 15.5 Semi-rectangles
There is a close relationship betw een semi-rectangles and Khayyam
quadrilaterals. In fact, the interested reader can show that a semi­
rectangle can be dissected, with only one straight cut, into a Khayyam
quadrilateral whose base angles are the non-right angles in the semi­
rectangle.
Using semi-rectangles, we can now restate 15.1b as follows:
Theorem 15.2 On a plane, sphere, or hyperbolic plane, i f two lines
through a point P intersect a circle at points A, A’ {possibly
coincident) and B, B’ {possibly coincident), then the area o f the
semi-rectangle with side lengths PA and PA’ is equal to the area
o f the semi-rectangle with side lengths PB and PB ‘. This area is
called the power o f the p o in t P with respect to the circle.
On the plane this theorem reduces directly to 15.1b. Do you see why?
We will now prove Theorem 15.2 on spheres (part c), but first we need
to prove the following result, which is interesting in its own right:
b. On a sphere with radius p, a spherical right triangle with excess
S a n d legs (sides adjacent to the right angle) a and b satisfies the
follow ing:
tan(<5/2) = tan(a/2/?)tan(Z>/2/?).

2 1 0 Chapter 15 Circles
Thus, i f R is a semi-rectangle with sides a and b and with
non-right angles equal to y, as in Figure 15.5, and i f we set let S
= y – k/2, then
Area(R) = 2 d p 1 and tan(£/2) = tan(a/2/?)tan(6/2p).
The proofs o f parts b and c (below) rely on properties o f stereographic
projection which is discussed in Problem 14.4. The properties we need
here are that stereographic projection preserves all angles and takes
circles on the sphere to circles (or lines) on the plane.
O utline of Proof of 15.2b:
1. Rotate the sphere until the right angled vertex on the triangle is
at the south pole, S. Now, using stereographic projection from
the north pole, project A onto the plane. The image o f the trian­
gle on the plane is the figure SHI depicted in Figure 15.6. The
sides SH and SI are straight. {Why?) Let A be the image o f the
great circle that contains the hypotenuse o f A and let C be the
center o f A. Let T be the image o f the equator; then T is a circle
with center S and radius 2p. (Why?) The circle A (that contains
the arc HI) intersects the circle T at diam etrically opposite points
DE. (Why?)
F ig u r e 1 5 . 6 S te r e o g r a p h ic im a g e o f a s p h e ric a l rig h t t r ia n g le

P roblem 1 5 . 2 . Power of Points for Circles on Spheres 211
2. Referring to Figure 15.6, show that Z IC H = 8 and that ZIH ‘H =
812. (Hint: Show that ZFC1 = n/2 – (3.)
3. Show that \SH\ x \SH’\ = |SD| x \SE\ and thus \SH’\ = A fr/a.
Conclude that 4/9 tan (£ /2 ) = a b \ Rem ember that a and b* are
the projections o f the sides o f our original semi-rectangle R and
that 8 p 2 is the area o f the triangle.
4. Let R be a semi-rectangle (as in Figure 15.5) whose non-right
angles are both y, then use a diagonal to divide R into two
congruent right triangles with legs a and b and non-right angles
a and /?. Note that a+ p = y. This is enough to conclude (Why?)
the last equations in part b.
Now we are ready to
c. Prove that Theorem 15.2 is true on a sphere.
O utline of proof of 15.2c
F ig u r e 1 5 . 7 P o w e r o f a p o in t o n a s p h e r e

212 Chapter 15 Circles
1. N ow let P, A, A’, B, B’ be as in Theorem 15.2 and Figure 15.7.
By rotating the sphere, i f necessary, we may consider P to be the
south pole S. Let T denote the given circle on sphere. Project
stereographically everything onto the plane and let A \ A ‘ \ B \
B ‘\ and F ‘ denote the corresponding images on the plane after
stereographic projection. Note that the image o f P = S is the
same S and that A’-S-A1’, and B’-S-B’* are straight lines. (Why?)
Using 15.1b, conclude that
|&4*| x |&4’*| = |SB*| x
2. Putting this together with Steps 3 and 4 o f the p ro o f o f 15.2b, we
conclude (Why?) that
tan(<5/2) = tan(S72), where the area o f the semi-rectangle on PA and PA' is 2 8 p 2 and the area o f the semi-rectangle on PB and PB' is 28'p 1. See Step 1. Thus the areas are equal. (Why?) There is a p ro o f o f Theorem 15.3 on a hyperbolic plane that is very similar to the p ro o f above on a sphere. Instead o f stereographic projec­ tion, the p ro o f on a hyperbolic \ ne uses the Poincare disk model thought o f as a projection (see Pro 'm 17.5), which also preserves angles and takes circles on the hyperbc plane to circles (or lines) on the plane. P r o b l e m 15.3 A p p l i c a t i o n s o f P o . ~r o f a P o i n t Here are a few applications o f the notion o f pow er o. ooint that hold on the plane, spheres, and hyperbolic planes. We will n, ‘ other applica­ tions later, especially in Chapter 16 and 19, but these M be applied only in the case o f the Euclidean plane. The applications , relate to the notion o f radical axis o f two circles, which we define to b he locus o f points P such that the pow er o f the point P is the same with rc ^ct to both circles. You may assume that Theorem 15.2 is true on hype, lie planes. a. I f two circles intersect in two points, then their radical axis is the fu ll line determined by the two points o f intersection. I f two Problem 15.3 Applications of Power of a Point 213 circles are tangent, then the radical axis is their common tangent line. b. I f P is a point on the radical axis o f two circles, C and D, then any circle with center P that intersects C at right angles also intersects D at right angles. c. I f three circles intersect each o f the other two circles in two points, then the three chords so defined intersect in a common point. See Figure 15.8. d. What happens if, in part c, some o f the pairs o f circles are tangent instead o f intersecting in two points? P r o b l e m 1 5 . 4 T r i s e c t i n g A n g l e s n d O t h e r C o n s t r u c t i o n ^ In Problem 6.3 you showed how to bisect angles and find the perpen­ dicular bisector o f a line segment by only using a compass (for drawing circles) and an unm arked straightedge (for drawing line segments joining two points but not used for measuring). We will now extend these constructions and discuss in what sense angles can and cannot be trisected. 2 1 4 Chapter 15 Circles a. Show, using a compass and unmarked straightedge, how to i. Construct a line from a given point perpendicular to a given line. ii. Construct the line tangent to a given circle at a given point. iii. Construct the two lines tangent to a given circle from a given point outside the circle. [Hint: Use 15.1b.] iv. Construct a line that is a parallel transport o f a given line along a given transversal. b. On the plane, show how to n-sect a given line segment using only a compass and (unmarked) straightedge. Will your construction work on spheres or hyperbolic planes? Hint: Look at Figure 15.9, where the given segment is AB and AC, is any segment forming an acute angle with AB. Duplicate AC, n tim es. (How?) Draw a line through Ci parallel to BC„. In the figure, n equals 5. Figure 15.9 Dividing a given line segment into 5 congruent pieces It is often stated in popular literature that it is impossible to trisect angles with straightedge and compass. See [TX: M artin], page 49. However, it has been known since ancient Greek tim es that any angle can be tri­ sected using only a (marked) straightedge and compass. For example, Archim edes (287-212 b .c .) showed how to trisect any angle (less than 135°) using a m arked straightedge (a straightedge with two points marked on it). c. A rch im ed es C o n s tru c tio n (for angles less than 90°): Referring to Figure 15.10, let ^ A B C be an angle less than 90°. Assume you have a straightedge with two marks on it that are a distance r apart. Draw the circle o f radius r with center B and, keeping Problem 15.4 Trisecting Angles and Other Constructions 215 the angle the same, move A to the intersection o f BA with the circle. Lay the straightedge on the figure so that it passes through A; and one mark, D, is on the circle and the other mark, E, is on the extension o f BC. Show that Z A E C is 1/3 o f ZABC. Figure 15.10 Archimedes' construction: ZAEC is 1/3 of ZABC d. Show that the mechanism in Figure 15.11 will trisect an angle. Will it work on spheres? F ig u r e 1 5 . 1 1 M e c h a n is m t o tr is e c t a n a n g le 2 1 6 Chapter 15 Circles OK, so then maybe it is impossible to trisect angles with a compass and an unmarked straightedge? e. Construct on a transparent sheet the figure on the left in Figure 15.12 with compass and unmarked straightedge starting with any two points AB. Lay the transparent sheet on the angle a (see the figure on the right in Figure 15.12) so the vertex, V, o f the angle lies on DB, one side contains C, and the other is tangent to the semicircle with center A. Prove that Z B VC trisects angle a. In order to give a correct statement o f w hat is impossible, it is necessary first to define a compass and unm arked straightedge sequence to be a finite sequence o f points, lines, and circles that starts with two distinct given points and such that (1) Each o f the other points in the sequence is the intersection o f lines or circles that occur before it in the sequence. (2) Each circle has its center and one point on the boundary occurring before it in the sequence. (3) Each line contains two distinct points that occur before it in the sequence. T h e o rem 15.4, It is impossible to trisect a 60° angle with a compass and unmarked straightedge sequence. For a p ro o f o f this theorem and related discussions about various compass and straightedge constructions, see [TX: Martin] or [TX: Hart- shom e], Section 28. __________ Chapter 16 Inversions in C ircles Q: How does a geometer capture a lion in the desert? A: Build a circular cage in the desert, enter it, and lock it. Now perform an inversion with respect to the cage. Then you are outside and the lion is locked in the cage. — A mathematical joke from before 1938 We now study inversions in a circle (which is the analogue o f reflection in a line) and its applications. Though inversion in a circle can be defined on spheres and hyperbolic planes, it seems to have no significant applications on these surfaces. Therefore, in this chapter we will only consider the case o f the Euclidean plane. To study Chapter 16, the only results needed in Chapters 10-15 are Problem 1 3 .4 : The AAA s im ila rity and SAS s im ila rity criteria for triangles on the plane. Problem 1 5 .1 b : On a plane, i f two lines through a point P intersect a circle at points A, A' {possibly coincident) and B, B' {possibly coincident), then \PA\ x \PA'\ = \PB\ x \PB'\. If you are willing to assume these criteria for sim ilar triangles, then you can work through Chapter 16 without Chapters 10-15. E a r l y H i s t o r y o f I n v e r s i o n s Apollonius o f Perga (c. 250-175 b.c.) was famous in his time for work on astronomy (Navigation/Stargazing Strand), before his now 2 17 2 1 8 Chapter 16 Inversions in Circles well-known w ork on conic sections. Unfortunately, A pollonius’ original w ork on astronomy and m ost o f his mathematical w ork (except for Conics, [AT: Apollonius]) has been lost and we only know about it from a comm entary by Pappus o f Alexandria (290-350 a.d.). According to Pappus, Apollonius investigated one particular family o f circles and straight lines. Apollonius defined the curve: Ck(A, B ) is the locus o f points P such that PA = k x PB, where A and B are two points in the Euclidean plane, and k is a positive constant. Now this curve is a straight line if k = 1 and a circle otherwise and is usually called an Apollonian circle. Apollonius proved (see Problem 16.1b) that a circle c (with center C and radius r) belongs to the family {ck{A,B)} if and only if BC x AC = r2 and A and B are on the same ray from C. In m odem terms, we use “BC x AC = r2 and A and B are on the same ray from C ” as the definition o f A and B being inversions o f each other with respect to the circle c. There is indirect evidence (see [HI: Calinger], page 181) that Apollonius used inversion in a circle to solve an astronomical problem s concerning celestial orbits. The theory o f inversions was apparently not carried on in a systematic w ay until the 19th century, when the theory was developed purely geometrically from Euclid's Book III, but this (as far as we know) was not done in ancient times. W e suggest that this was because E uclid’s Elements and Apollo­ n iu s’ circles were parts o f different historical strands. In Problem 16.4 we w ill explore a problem o f Apollonius that uses inversions for its solutions. P r o b l e m 16.1 I n v e r s i o n s i n C i r c l e s Definitions. An inversion with respect to a circle T is a trans­ form ation from the extended plane (the plane with <», the “point at infinity,” added) to itse lf that takes C, the center o f the circle, to oo and vice versa, and that takes a point at a distance s from the center to the point on the same ray (from the center) that is at a distance o f r^/s from the center, where r is the radius o f the circle. See Figure 16.1. We call ( P ,P r) an inversive pair because (as the reader can check) P and P ' are Problem 16.1 Inversions in Circles 2 1 9 taken to each other by the inversion. The circle T is called the circle o f inversion. Note that an inversion takes the inside o f the circle to the outside and vice versa and that the inversion takes any line through the center to itself. Because o f this, inversion can be thought o f as a reflection in the circle. See also part c for the close connection betw een inversions in the plane and reflections on a sphere. We strongly suggest that the reader play with inversions by using dynamic geometry software such as Geometers Sketchpad®, C a b rf\ or Cinderella9. Sample constructions can be found at the Experiencing Geometry W eb site: ww w.m ath.com ell.edu/~henderson/ExpG eom /. You may construct the image, P', o f P under the inversion through the circle T as follows (see Figure 16.2): I f P is inside T, then draw through P the line perpendicular to the ray CP. Let S and R be the intersections o f this line with T. Then P' is the intersection o f the lines tangent t o Y at S and R. (To construct the tangents, note that lines tangent to a circle are perpendicular to the radius o f the circle.) I f P is outside T, then draw the two tangent lines from P to T. Let S and R be the points o f tangency on T. Then P' is the intersection o f the line SR with CP. (The points S and R are the intersections o f T with the circle with diam eter CP. Why?) http://www.math.comell.edu/~henderson/ExpGeom/ 2 2 0 Chapter 16 Inversions in Circles a. Prove that these constructions do construct inversive pairs. The purpose o f this part is to explore and better understand inversion, but it will not be directly used in the other parts o f this problem. W hen P is inside T, you need to prove that C-P-P' are collinear. W hen P' is outside T, you need to prove that SR is perpendicular to CP. Next we prove *b. A pollo n iu s’ T h e o rem . Define the curve ct(A, B) to be the locus o f points P such that PA = kx. PB, where A and B are two points in the plane and k is a positive constant. A circle c (with center C and radius r) belongs to the fam ily { cfA , B)} i f and only i f A and B are an inversive pair with respect to c. The following result demonstrates the close connection between inversion through a circle in the plane and reflections through great circles on a sphere. If you have studied Problem 14.4 (or assume it), then you can use this part c in your analysis o f inversions in 16.2. * *c. Let I be a sphere tangent at its south pole to the plane 17and let f: Z —> ILbe a stereographic projection from the north pole. I f T
is the circle that is the image under f o f the equator and i f g is
the intrinsic (or extrinsic) reflection o f the sphere through its
equator (or equatorial plane), then show that the transforma­
tion f o g °/ ” ‘ is the inversion o f the plane with respect to the
circle F See Figure 16.3.

Problem 16.1 Inversions in Circles 221
Imagine a sphere tangent to a plane at its south pole, S. Now use stereo­
graphic projection to project the sphere from the north pole, N, onto the
plane; see Problem 14.4. Stereographic projection was known already to
Hipparchus (Greek, second century b.c.). Show that the triangle ASNP is
similar to ARNS, which is congruent to AQSN, which is sim ilar to
ASP’N.
N
Figure 16.3 Stereographic projection and inversion
P r o b l e m 16.2 I n v e r s i o n s P r e s e r v e A n g l e s a n d
P r e s e r v e C i r c l e s ( a n d L i n e s )
Part c provides another route to prove that inversions are conformal
(preserves angles, see Problem 14.4). In Problem 14.4 you showed that / ,
stereographic projection, is conformal. In addition, g (being an isometry)
is conformal. Thus, the i n v e r s i o n i s conformal.
a. Show that an inversion takes each circle orthogonal to the circle
o f inversion to itself. See Figure 16.4.
Two circles are orthogonal if, at each point o f intersection, the angle
between the tangent lines is 90°. (Note that, at these points, the radius o f
one circle is tangent to the other circle.)

2 2 2 Chapter 16 Inversions in Circles
b. Show that an inversion takes a circle through the center o f
inversion to a line not through the center, and vice versa. What
happens in the special cases when either the circle or the
straight line intersects the circle o f inversion?
F ig u r e 1 6 . 5 C irc le s th r o u g h t h e c e n t e r in v e r t t o lin e s

Problem 16.2 Inversions Preserve Angles and Circles (and Lines) 223
Look at Figure 16.5 (where CP is a diam eter o f the circle) and prove that
ACPQ and ACQ’P’ are sim ilar triangles. Note that the line is parallel to
the line tangent to the circle at C.
c. An inversion takes circles not through the center o f inversion to
circles not through the center. Note: The (circum ference o f a)
circle inverts to another circle but the centers o f these circles are
on the same ray from C though n o t an inversive pair.
Look at Figure 16.6, where PQ is a diam eter o f the circle. If P, Q, X
invert to P’, Q’, X , then show that
Z P ‘ X Q’ = Z P X Q = right angle
by looking for sim ilar triangles. Thus, argue that as X varies around the
circle with diam eter PQ, then X varies around the circle with diam eter
Q’P’ ■
d. Inversions are conformal.
Look at two lines that intersect and form an angle at P. Look at the
images o f these lines.
Inversions were used in the 19th century to solve a long-standing
engineering problem (from the M otion/M achines Strand) that is the

2 2 4 Chapter 16 Inversions in Circles
subject o f Problem 16.3. Other applications (that started with
Apollonius) are discussed in Problem 16.4. M ore history and further
expansions o f the notion o f inversion are contained in the last section.
P r o b l e m 16.3 U s i n g I n v e r s i o n s t o D r a w
S t r a i g h t L i n e s
At the beginning o f Chapter 1, there is a b rie f history o f attempts to find
linkages that would draw a straight line. In this problem we explore the
mathematics behind this linkage.
a. Show that fo r the linkage in Figure 16.7 the points P and Q are
the inversions o f each other through the circle o f inversion with
center at C and radius r = J s 2 — d 2 .
Draw the circle with center R and radius d and note that C, P, Q are
collinear.
■— = F ix e d p o in t w ith h in g e
• — = V a ria b le p o in t w ith h in g e
F ig u r e 1 6 . 7 A lin k a g e f o r c o n s tr u c tin g a n in v e rs io n

Problem 16.3 Using Inversions to Draw Straight Lines 225
b. Show that the point Q in the linkage in Figure 16.8 always traces
a straight line.
Figure 16.8 Linkage for drawing a straight line
If we modify the Peaucellier-Lipkin linkage by changing the
distance betw een the anchor points, then
c. The point Q in the linkage in Figure 16.9 always traces the arc o f
a circle. Why? Show that the radius o f the circle is expressed by
r 2/ / (g2 – f 2), where r is as in part a.
F ig u r e 1 6 . 9 P e a u c e llie r-L ip k in lin k a g e m o d ifie d t o d r a w t h e a r c o f a c irc le

2 2 6 Chapter 16 Inversions in Circles
* P r o b l e m 16.4 A p o l l o n i u s ‘ P r o b l e m
In Book IV o f the Elements, Euclid shows how to construct the circle
that passes through three given (non-collinear) points and also how to
construct a circle tangent to three given straight lines (not passing the
same point). Apollonius o f Perga (c. 250-175 b .c .) generalized this to
A p o llo n iu s’ P ro b le m : Given three objects, each o f which may be a
point, a line, or a circle, construct a circle that passes through
each o f the given points and is tangent to the given lines and
circles.
Solutions to this problem are discussed in Apollonius’ On Tangencies
{De Tactionibus). Unfortunately, A pollonius’ work has not survived, but
it has been “reconstructed” from both Arabic and Greek commentaries
especially through the description o f its contents from Pappus o f Alex­
andria (ca. a .d . 300). In Book 7 o f the Mathematical Collection, Pappus
described the contents o f various works by Apollonius. Pappus presents
a list o f problem s in A pollonius’ lost work, and on the basis o f this infor­
m ation the w ork has been reconstructed at least four times.
Francois Viete (1540-1603) restored Apollonius De tactionibus and
published it under the title Apollonius Gallus in 1600. Frans van Schoo-
ten, in a 1657 reconstruction, showed that Apollonius’ problem can be
solved by the algebraic methods o f Descartes’ Geometrie (1637).
Joachim Jungius and W oldeck W eland (1622 -1641), in a reconstruction
titled Apollonius Saxonicus, used a purely geometrical m ethod that they
called “metagoge,” that is, the reduction o f the general case o f a problem
to a special case o f the same problem or a sim pler problem (such as in
16.4c below). Another reconstruction was com pleted by Robert Simson
(1749). The authors o f these reconstructions were doing mathematics,
not the history o f mathematics, as can be inferred from the fact that the
“reconstructions” differ from each other and sometimes deal with gener­
alizations o f the problems that had actually been treated by Apollonius.
In addition to the attempted reconstructions, there were m any and
varied solutions o f the Apollonius problem produced by later m athem ati­
cians, including Isaac Newton (1643-1727), A. van Roomen
(1561-1615), J. Casey (1820-1881), R. Descartes, P. Fermat, Princess
Elizabeth (1596-1662), L. Euler (1707-1783), N. Fuss (1755-1826), L.
N. M. Carnot (1753-1823), J. D. Gergonne (1771-1859), C. F. Gauss
(1777-1855), J. V. Poncelet (1788-1867), A. L. Cauchy (1789-1857),

*Problem 16.4 Apollonius’ Problem 2 2 7
and Eduard Study (1862-1930). Y ou should recognize some o f these
names. Poncelet and Cauchy solved A pollonius’ problem while first year
students at the Ecole Polytechnique.
In 1679 P. Ferm at form ulated and solved an extension o f A pollo­
nius problem to 3-space: Construct a sphere tangent to four given
spheres. This is called Ferm at’s problem by some later authors. A large
number o f mathematicians were discussing this problem in 19th century.
Some further generalizations o f A pollonius’ problem are discussed in H.
S. M. Coxeter, “The Problem o f Apollonius,” American Mathematical
Monthly 75 (1968), pp. 5-15.
In 2003, R. H. Lewis and S. Bridged, in a paper entitled “Conic
tangency equations and Apollonius problem s in biochem istry and
pharm acology” (Mathematics and Computers in Simulation, vol. 61; Jan.
2003, pp. 101-114), discuss current applications o f A pollonius’ prob­
lems. The applications involve bonding interactions in hum an bodies
between protein m olecules and hormone, drug, and other molecules.
A pollonius’ problem can be discussed in 10 possible cases (letting
P = point, L = line, C = circle):
PPP, PPL, PPC, PLL, PLC, PCC, LLL, LLC, LCC, CCC.
a. Solve the case P P P . Show that the solution implies that the
perpendicular bisectors o f the sides o f a triangle all pass
through the same point. We call this point the circumcenter o f
the triangle. See Figure 16.10.
F ig u r e 1 6 . 1 0 P P P

2 28 Chapter 16 Inversions in Circles
Hint: I f the three points are on a line, then that line (as a circle with
infinite radius) is the solution. Otherwise the three points determine a
triangle.
b. Solve the case L L L . Show that the solution implies that the
angle bisectors o f a triangle pass all through the same point.
This point is called the incenter o f the triangle.
Hint: If the three lines intersect in the same point, then there is no
solution. If the three lines form a triangle, then there are four solutions.
See Figure 16.11. W hat happens in other cases?
c. Solve the cases P P L and PPC .
Outline o f solution: If both points are on the given line or circle, the only
solution is the given line or circle. If point A is not on the given line (or
circle), then we can take A as the center o f an inversion. Denote the other
point B and the given line or circle T. U nder the inversion, T goes to
another circle T’ (never a line, Why?) and B goes to another point B’. Let
/ be a line through B’ that is tangent to F . Inverting this tangent / back to
the original picture, we will have a solution. {Why?) I f B is on T, then
there exists one solution; i f neither A or B are on T, then there are two
solutions.
Figure 16.11 LLL
d. Solve the cases PLL , PLC , and PC C .

♦Problem 16.4 Apollonius’ Problem 2 2 9
Hint: Choose the given point as the center o f an inversion. A fter the
inversion the solution will be a line (in order to contain the inversive
image o f the point). See Figure 16.12. D epending on the original
location o f the given point, we get the following subcases:
1. The point is on neither o f the lines (line and circle, circles). Then
those circles (line and circle, lines) in inversion will go to two circles
and the problem reduces to constructing a tangent to two given circles.
In this subcase there are either 0, 1 ,2 , 3, or 4 solutions possible. {Why?)
2. The point is the point o f tangency o f circles (or circle and line).
3. The point is the point o f intersection o f circles (or circle and line).
4. The point is on one circle (or line) but not on the other.
Figure 16.12 Case PCC: invert through r and then find tangent to C’and D ‘
e. A p p lica tio n o f P C C . There is a story that in World War 1 troops
used P C C to pinpoint the location o f large enemy guns. The
three separate observation points synchronize their clocks to the
second and then note the exact second that they hear the gun’s
sound. How could these timings be used to pinpoint the location
o f the gun?
Hint: The speed o f sound is approxim ately 340 m eters/second (it varies
±5% in somewhat predictable ways with tem perature and atm ospheric
conditions). Draw a picture o f the instant in time when the sound reaches
the first observation point.
f. Solve cases L L C , L C C , and C C C .

2 3 0 Chapter 16 Inversions in Circles
Outline o f solution: Each o f these three cases can be reduced to either
CCC or PCC by using an appropriate inversion. {Do you see how?)
There are m any subcases depending on how the circles and lines relate
to each other: inside, outside, intersecting. However, the overall strategy
is to reduce these cases to P L C , P L L , or P C C in part d. We illustrate
this with the subcase o f L C C , where all circles and lines are disjoint and
the two circles are on the same side o f the line and exterior to each other.
See Figure 16.13.
Let r be the radius o f the smaller circle, C. Shrink the other circle, D, to
the circle, D ’, with the same center but with radius reduced by r.
Construct another line parallel to L, on the side opposite to C and D, and
at a distance o f r. N ow apply P L C to the point P, the circle D’, and the
line L’, to get a circle E’ that is tangent to P, D’, and L’. The required
solution to the original L C C is the circle E with the same center as E’
but with radius decreased by r. See Figure 16.13.
E x p a n s i o n s o f t h e N o t i o n o f I n v e r s i o n s
There was a rebirth o f interest in inversions in the 19th century. Jakob
Steiner (1796-1863) was among the first to start extensively using the
technique o f inversions in circles to solve geometric problems. Steiner
had no early schooling and did not learn to read or w rite until he was age
14. Against the wishes o f his parents, at age 18 he w ent to the Pestalozzi
School at Yverdon, Switzerland, where his extraordinary geometric

Expansions o f the Notion o f Inversions 2 3 1
intuition was discovered. By age 28 he was making m any geometric
discoveries using inversions. At age 38 he occupied the chair o f geome­
try established for him at the University o f Berlin, a post he held until
his death.
Steiner defined an inversive transformation to be any transform a­
tion that is the composition o f inversions and initiated inversive geome­
try, which is the study o f properties o f the extended plane that are
preserved by inversive transform ations. It follows from Problem 16.2
that inversive transform ations preserve angles and take circles and lines
to circles and lines.
Jean V ictor Poncelet (1788-1867) showed that an inversion is a
birational transformation, that is, a one-to-one transform ation o f the
extended plane such that both the transform ation and its inverse are o f
the form
x ‘ =f[x,y), y ‘ = g(x,y), w here/ and g are rational functions.
There were m any other European m athematicians in the 19th century
who studied inversions and inversive geometry. A pplications o f inver­
sions in physics were used by Lord Kelvin (Sir W illiam Thomson)
(1824-1907) in 1845, and also by Joseph Liouville (1809-1882) in
1847, who called inversions the transformations by reciprocal radii.
In 1854 Luigi Cremona (1830-1903) made a systematic study o f
birational transform ations that carry the entire extended plane onto itself.
These transform ations are now often called Cremona transformations.
Since inversions take the entire extended plane to itself, they are
Cremona transform ations. These were subsequently studied by M ax
Noether (1844-1921), who proved that a plane Cremona transform ation
(and thus inversions) could be constructed by a sequence o f quadratic
and linear transform ations.
In 1855, August Ferdinand M obius (1790-1868) undertook a
systematic study o f circular transform ations (conformal transform ations
that map points on a circle to points on a circle) by purely geometrical
means. He defined what are now called Mobius transformations, which
are often studied today in courses on complex analysis:
A Mobius transformation is any transformation o f the extended
complex plane onto itself o f the form
M(z) = f h T f j ‘ where a, b, c, d are complex num bers and ad – b e * 0.

2 3 2 Chapter 16 Inversions in Circles
The following properties o f M obius transform ations are proved in
Sections 5.3 and 5.4 o f [TX: Brannan]:
♦ Every Mobius transformation is an inversive transformation
(but no inversion can be a M obius transform ation because
M obius transform ations preserve orientation and inversions
do not).
♦ M obius transformations form a subgroup o f the group o f
inversive transformation.
♦ Every inversion F can be written in the form F(z) = M(z),
where M is a Mobius transformation.
♦ Given any two sets o f three points, z\, z% z% and w i, Wi, w_i,
there is a unique Mobius transformation that maps z\ to w\,
Z2 to W2, and zi to w .̂
M obius geometry is also connected to Laguerre geometry initiated
by Edm ond Laguerre (1834-1868) and M inkowskian geometry initiated
by Hermann M inkowski (1864—1909), which is the geometry that
Einstein used in the Theory o f Relativity and Space/Time. In 1900,
Edward Kasner (1878-1955) was apparently the first to study inversive
geometry in accordance with Klein’s Erlanger Program (see Chapter 11,
page 153). Research on transform ations that preserve circles continued
into at least the m iddle o f the 20th century. Continuing to today, inver­
sions are used in the two Poincare models o f hyperbolic geometry (see
Problems 17.2 through 17.5).
For more discussion o f inversive and related geometries, see [TX:
Brannan], Chapter 5, and [HI: Kline], Section 39.3. Kline also makes
connections to algebraic geometry. See also [HM : M archisotto] for
related history in the 20th century.

Chapter 17
Projections (M odels)
of Hyperbolic Planes
In recent times the mathematical public has begun to occupy
itself with some new concepts which seem to be destined, in
the case they prevail, to profoundly change the entire order of
classical geometry.
— E. Beltrami (1868), when he developed
the projective disk model (Problem 17.5)
In this chapter we will study projections o f a hyperbolic plane onto the
plane and use these “m odels” to prove some results about the geometry
o f hyperbolic planes. In the case o f hyperbolic planes it is custom ary to
call these “models” instead o f “projections” because it was thought that
there were no surfaces that were hyperbolic planes. As in the case o f
spherical projections, any projections (models) o f the hyperbolic plane
m ust distort some geometric properties; and with models it is more diffi­
cult to gain the intrinsic and intuitive experiences that are possible with
the hyperbolic surfaces discussed in Chapter 5. N evertheless, these
models do give the most analytically accurate picture o f hyperbolic
planes and allow for more accurate and precise constructions and proofs.
We take as our starting point the geodesic rectangular coordinates
presented in Problem 5.2. In order to connect these coordinates to the
study o f the m odels, we will need the results on circles from Chapter 14
and an analytic sophistication that is not necessary in other chapters in
this book. However, no technical results from analysis are needed. The
reader may bypass most o f the analytic technicalities (which occur in
Problems 17.1 and 17.2) i f the reader is willing to assume the results o f
Problem 17.2, which make the connections betw een an annular hyper­
bolic plane and the upper-half-plane model and prove which curves in
233

2 3 4 Chapter 17 Projections (Models) o f Hyperbolic Planes
the upper half-plane correspond to geodesics in the annular hyperbolic
plane. The basic properties o f geodesics and constructions in the upper-
half-plane m odel (and therefore in annular hyperbolic planes) are inves­
tigated in Problem 17.3. We continue our w ork on the area o f triangles
by investigating in Problem 17.4 ideal and 2/3-ideal triangles. Other
popular m odels o f hyperbolic planes are contained in 17.5 (Poincare disk
model) and 17.6 (projective disk model).
D i s t o r t i o n o f C o o r d i n a t e S y s t e m s
The reader should review the description o f the annular hyperbolic plane
in Chapter 5 and the discussion in 5.2. In Problem 5.2, we defined
geodesic rectangular coordinates on the annular hyperbolic plane as the
map x: R 2 – » H2 defined as indicated in Figure 17.1.
Figure 17.1 Geodesic rectangular coordinates on annular hyperbolic plane
W e showed in Problem 5.2 that the coordinate map x is one-to-one
and onto from the whole o f R 2 onto the whole o f the annular hyperbolic
plane. Horizontal lines map onto the annular strips and vertical lines map
onto radial geodesics. Then we showed the following:
5.2b. Let A and p be two radial geodesics on a hyperbolic plane
with radius p. I f the distance between A and p along the base
curve is w, then the distance between them at a distance s from
the base curve is
w exp(-s/p).

Distortion o f Coordinate Systems 2 3 5
Thus, the coordinate chart jc preserves (does not distort) distances
along the (vertical) second coordinate curves but at x(a, b) the distances
along the first coordinate curve are distorted by the factor o f exp(~b/p)
when compared to the distances in R2. To be m ore precise,
Definition. Let y: A —> B be a map, and let t A(t) be a curve
in A. Then the distortion o f y along A at the point p = MO) is
defined as
the arc length along y(A) from y(A(t)) to j(/.(0 ))
™ the arc length along A from A{t) to A(0)
In the case o f the above coordinate curves, A is the path in R2,
t (a + 1, b) or t h>. (a, b + 1), and the distortions o f x along the coordinate
curves are
We seek a change o f coordinates that will distort distances equally
in both coordinate directions. The reason for seeking this change (as we
will see below) is that i f distances are distorted the same in both coordi­
nate directions, then the chart will preserve angles. (Remember, we call
such a chart conformal.)
We cannot hope to have no distortion in both coordinate directions
(if there were no distortion, then the chart w ould be an isometry), so we
try to make the distortion in the second coordinate direction the same as
the distortion in the first coordinate direction. A fter a little experim enta­
tion we find that the desired change is
lim
(- 0
the arc length from x(a + 1, b) to x(a, b)
\(a + t,b )~ (a, 6)|
and
lim
the arc length from x(a, b + t) to x(a, b) /
\{a,b + t)-{ a ,b )\ t L
z(x,y) = x(x, p In(y/p)),
with the domain o f z being the upper half-plane

2 36 Chapter 17 Projections (Models) o f Hyperbolic Planes
where x is the geodesic rectangular coordinates defined above. This is
usually called the upper-half-plane model o f the hyperbolic plane. The
upper-half-plane model is a convenient way to study the hyperbolic
plane — think o f it as a map o f the hyperbolic plane in the same way
that we use planar maps o f the spherical surface o f the earth.
* P r o b l e m 17.1 A C o n f o r m a l C o o r d i n a t e S y s t e m
Show that the distortion o f z along both coordinate curves
x —> z(x, b) and y —» z{a,y)
at the point z(a, b) is plb.
It may be best to first try this for p = 1. For the first coordinate direction,
use the result o f Problem 5.2b. For the second coordinate direction, use
the fact that the second coordinate curves in geodesic rectangular coordi­
nates are param etrized by arc length. Use first-sem ester calculus where
necessary.
L em m a 1 7 .1 . I f the distortion o f z at the point p = (a, b) is the
same [say A(p)] along each coordinate curve, then at (a, b) the dis­
tortion o f z has the same value along any other curve Mf) =
z(x(t), y(t)) that passes through p; and z preserves angles at p (that
is, z is conformal).
P ro o f. Suppose that A,(0) = (x(0), >(0)) = (a, b) = p . Assum ing that the
annular hyperbolic plane can be locally isom etrically (that is, preserving
distances and angles) embedded in 3-space (see Problem 5.3), the distor­
tion o f z along X at p is
the arc length along z(X) from z(X(t)) to z(X(0)) _
™ the arc length along X from X(t) to 2(0)
])j-[the arc length along z(2) from z(X(t)) to z(2(0))]
‘”° ITT [the arc length along X from X(t) to 2(0)]
speed o f z(X(t)) at t = 0 | jjz(X(t) ) \,=0 \(z0 A)'(0)|
speed o f 2(t) at t = 0 – \-jX(t)\t=0 ~ U ‘(0)|
Therefore, along the first coordinate curve t>~* (a + t, b), the distortion is

* P roblem 17.1 A Conformal Coordinate System 2 37
——– ’—p – 2- = the norm o f the partial derivative, \z\ip)\.
d t ( a ^ ) l t = 0
Similarly, the distortion along the second coordinate curve is |z2(p)|. The
velocity vector o f the curve = z(x(t),y(t)) at p is
Thus, the velocity vector, (z ° A)’ at t = 0 is a linear com bination o f the
partial derivative vectors, zi(p) and ziip) — note that these vectors are
orthogonal. Therefore, the velocity vectors o f curves through p = 1(0) all
lie in the same plane called the tangent plane at zip). Also note that the
velocity vector, (z° /1)’, depends only on the velocity vector, l'( 0 ) , and
not on the curve A. Thus, z induces a linear map (called the differential
dz) that takes vectors at p = 1(0) to vectors in the tangent plane at zip).
This differential is a sim ilarity that m ultiples all length by A ip) and thus
preserves angles. The distortion o f z along 1 is also A ip).
Definition. In the above situation we call A ip) the distortion
o f the map z at the p o in t p and denote it dist(z)(p).
♦ P r o b l e m 17.2 U p p e r H a l f – P l a n e I s M o d e l
o f A n n u l a r H y p e r b o l i c P l a n e
We were able to prove in Problem 5.1 that there are reflections about the
radial geodesics but only assum ed (based on our physical experience
with physical m odels) the existence o f other geodesics and reflections
through them. To assist us in looking at transform ations o f the annular
hyperbolic space (with radius p), we use the upper-half-plane model. If /
is a transform ation taking the upper h a lf plane R 2+ to itself, then we have
iz ( x if) ,y it) ) t=0 = zi(p)-£x(t)t,o + Ziip)j;yit),=o.
H 2 —– 8 …. » H2
z, – i z
/
We see that g = z ° f ° z ~ x is a transform ation from the annular hyperbolic
plane to itself. We call g the transform ation o f M2 that corresponds t o /

238 Chapter 17 Projections (Models) of Hyperbolic Planes
W e will call / an isometry o f the upper-half-plane model i f the corre­
sponding g is an isometry o f the annular hyperbolic plane. To show that
g is an isometry, you m ust show that the transform ation g = z ° f ° z ~1
preserves distances. Rem ember that distance along a curve is equal to
the integral o f the speed along the curve. Thus, it is enough to check that
the distortion o f g at each point is equal to 1. Before we do this we must
first show that
a. The distortion o f an inversion ic with respect to a circle T at a
point P, which is a distance s from the center C o f T, is equal to
r / s 2, where r is the radius o f the circle. See Figure 17.2.
Hint: Because the inversion is conformal, the distortion is the same in all
directions. Thus check the distortion along the ray from C through P.
The distance along this ray o f an arbitrary point can be param etrized by
t ts. Use the definition o f distortion given in Problem 17.1.
b. Let f be the inversion in a circle whose center is on the x-axis.
Show that f takes R 2+ to itself and that g = z ° f ° Z ~’ has distor­
tion 1 at every point and is thus an isometry.
O utline of a proof:
1. Note that each o f the maps z, z ~ \ f are conformal and have at each
point a distortion that is the same for all curves at that point. If

♦ P roblem 17.2 Upper Half-Plane Is Model of Annular Hyperbolic Plane 239
dist(£)(p) denotes the distortion o f the function k at the point p,
then argue that
dist(g)(p) = dist(z_1)(p) x d ist(/)(z H(p)) x dist(z)(/(z_1(^)).
1. If z(a, b) = p , then show (using 5.1c) that dist(z l)(p) = blp, where
/?is the radius o f the annuli.
2. Show (using part a) that d ist(/’)(z ‘(/?)) = t^/s2, where r is the
radius o f the circle C that defines / and s is the distance from the
center o f C to (a, b).
3. Then show that dist(z)(/(z (p)) =
64
We call these inversions (or the corresponding transform ations in
the annular hyperbolic plane) hyperbolic reflections. We also call reflec­
tions through vertical half-lines (corresponding to radial geodesics)
hyperbolic reflections.
Now you can prove that
c. I f y is a semicircle in the upper half-plane with center on the
x-axis or a straight half-line in the upper half-plane perpendicu­
lar to x-axis, then z(y) is a geodesic in the annular hyperbolic
plane.
Because o f this, we say that such yare geodesics in the upper-half­
plane model. Because the compositions o f two isometries is an isometry,
we see im m ediately that any composition o f inversions in semicircles
(whose centers are on the x-axis) is an isometry in the upper-half-plane
model (that is, the corresponding transform ation in the annular hyper­
bolic plane is an isometry).
P r o b l e m 17.3 P r o p e r t i e s o f H y p e r b o l i c G e o d e s i c s
a. Any similarity (dilations) o f the upper half-plane corresponds to
an isometry o f an annular hyperbolic. Such similarities must
have their centers on the x-axis. ( Why!)
Look at the composition o f inversions in two concentric semicircles.

2 4 0 Chapter 17 Projections (Models) of Hyperbolic Planes
b. I f y is a semicircle in the upper half-plane with center on the
x-axis, then there is an inversion (in another semicircle) that
takes yto a vertical line that is tangent to y.
Hint: An inversion takes any circle through the center o f the inversion to
a straight line (see Problem 16.2).
Each o f the following three parts is concerned with finding a
geodesic. Each problem should be looked at in both the annular hyper­
bolic plane and in the upper-half-plane model. In a crocheted annular
hyperbolic plane we can construct geodesics by folding m uch the same
way we can on a piece o f (planar) paper. Geodesics in the upper-half-
plane m odel can be constructed using properties o f circles and inversions
(see Problem 16.2). You will also find part b very useful.
c. Given two points A and B in a hyperbolic plane, there is a
unique geodesic joining A to B; and there is an isometry that
takes this geodesic to a radial geodesic (or vertical line in the
upper-half-plane model).
In the upper-half-plane model, construct a circle with center on the
x-axis that passes through A and B. Then use part b.
We use AB to denote the unique geodesic segment joining A to B.
d. Given a geodesic segment AB with endpoints points A and B in a
hyperbolic plane, there is a unique geodesic that is a perpen­
dicular bisector o f AB.
Use appropriate folding in annular hyperbolic plane. In the upper-half-
plane model, make use o f the properties o f a reflection through a perpen­
dicular bisector.
e. Given an angle Z A B C in a hyperbolic plane, there is a unique
geodesic that bisects the angle.
In the upper-half-plane model, again use the properties o f a reflection
through the bisector o f an angle.
f. Any two geodesics on a hyperbolic plane either intersect, are
asymptotic, or have a common perpendicular.

P roblem 17.3 Properties of Hyperbolic Geodesics 241
Look at two geodesics in the upper-half-plane model that do not intersect
in the upper half-plane nor on the bounding x-axis.
P r o b l e m 1 7 . 4 H y p e r b o l i c I d e a l T r i a n g l e s
In Problem 7.2 we investigated the area o f triangles in a hyperbolic
plane. In the process we looked at ideal triangles and 2/3-ideal triangles.
We can look more analytically at the ideal triangles. It is impossible to
picture the whole o f an ideal triangle in an annular hyperbolic plane, but
it is easy to picture ideal triangles in the upper-half-plane model. In the
upper-half-plane model an ideal triangle is a triangle with all three verti­
ces either on the x-axis or at infinity. See Figure 17.3.
At first glance it appears that there must be m any different ideal
triangles. However,
a. Prove that all ideal triangles on the same hyperbolic plane are
congruent.
Review your work on Problem 16.2. Perform an inversion that takes one
o f the vertices to infinity and the two sides from that vertex to vertical
lines. Then apply a sim ilarity to the upper half-plane, taking it to the
standard ideal triangle with vertices (-1 ,0 ), (0,1), and .
b. Show that the area o f an ideal triangle is nff. (Remem ber this p
is the radius o f the annuli.)

242 Chapter 17 Projections (Models) of Hyperbolic Planes
Hint: Because the distortion dist(z)(o, b) is p/b, the desired area is
We now picture in Figure 17.4 2/3-ideal triangles in the upper-half­
plane model.
c. Prove that all 2/3-ideal triangles with angle 9 are congruent
and have area ( n – 9)(f.
Show, using Problem 16.2, that all 2/3-ideal triangles with angle 6 are
congruent to the standard one at the right o f Figure 17.4 and show that
the area is the double integral
Figure 17.4 2/3-ideal triangles in the upper-half-plane model
P r o b l e m 17.5 P o i n c a r e D i s k M o d e l
You showed in Problem 17.1c that the coordinate map x from a hyper­
bolic plane to the upper half-plane preserves angles (is conform al); this
we called the upper-half-plane model. Now we will study other models
o f the hyperbolic plane.
Let z : R 2+ – » H 1 be the coordinate map defined in Problem 17.2 that
defines the upper-half-plane model. We will now transform the upper-
half-plane model to a disk model that was first discussed by Poincare in
A
(- 1,0) ( 1,0)
1882.

P roblem 1 7 .5 P o in c a r e D is k M o d e l 243
a. Show that any inversion through a circle whose center is in the
lower half-plane (that is, y < 0) will transform the upper half­ plane onto an open (without its boundary) disk. Show that the hyperbolic geodesics in the upper half-plane are transformed by this inversion into circular arcs (or line segments) perpendicu­ lar to the boundary o f the disk. Review the m aterial on inversions discussed in Problem 16.2. b. I f w. D 2 R 2+ is the inverse o f a map from the upper h a lf plane to a (open) disk from p a rt a, then show that the composition z ° w : D 2 -+ H 2 is conformal. We call this the (Poincare) disk model, after Henri Poincare (1854—1912, French). Review the material on inversions in 16.2 and on the upper-half-plane model in 17.2. c. Show that any inversion through a circular arc (or line seg­ ments) perpendicular to the boundary o f D2 takes D2 to itself. Show that these inversions correspond to isometries in the (annular) hyperbolic plane. Thus, we call these circular arcs (or line segments) hyperbolic geodesics and call the inversions hyperbolic reflections in the Poincare disk model. Review Problem 17.2. See Figure 17.5 for a drawing o f geodesics and a triangle in the Poincare disk and the projective disk model (Problem 17.6). In Poincare disk model In projective disk model Figure 17.5 Geodesics and a triangle 2 4 4 Chapter 17 Projections (Models) of Hyperbolic Planes P r o b l e m 17.6 P r o j e c t i v e D i s k M o d e l Let D 2 be the disk model o f a hyperbolic plane and assume its radius is 2. Then place a sphere o f radius 1 tangent to the disk at its center. Call this point o f tangency the south pole S. See Figure 17.6. N ow let s be the stereographic projection from the sphere to the plane containing D2, and note that s(equator) is the boundary o f D2, and thus s takes the Southern Hem isphere onto D2. Now let h be the orthogo­ nal projection o f the southern hemisphere onto the disk, B2, o f radius 1. N Figure 17.6 Obtaining the projective disk model from the Poincare disk model Show that the mapping h ° s~' takes D2 to B2 and takes each circle {or diameter) o f D 2 to a {straight) cord o f B2. Thus h o s _i ° {z ° w)_1 is a map from the hyperbolic plane to B2, which takes geodesics to straight line segments {cords) in B2. We call this the projective disk model, but in the literature it is also called the Beltrami/Klein m odel or the Klein model, named after Eugenio Beltrami (1835-1900, Italian), who described the m odel in 1868, and Felix Klein (1849-1925, German), who fully developed it in 1871. See Figure 17.5 for a drawing o f geodesics and a triangle in the projective disk model. Chapter 18 G eometric 2-M anifolds The concept “two-dimensional manifold” or “surface” will not be associated with points in three-dimensional space; rather it will be a much more general abstract idea. — Hermann Weyl (1913) There is clearly a large variety o f different surfaces around in our experi­ ential world. The study o f the geometry o f general surfaces is the subject o f differential geometry. In this chapter we will study geometric 2-manifolds, that is, a connected space that locally is isometric to either the (Euclidean) plane, a sphere, or a hyperbolic plane. The surface o f a cylinder (no top or bottom and indefinitely long) and a cone (with the cone point removed) are examples o f geometric 2-manifolds. We study these because their geometry is simpler and closely related to the geome­ try we have been studying o f the plane, spheres, and hyperbolic planes. In addition, the study o f these surfaces will lead us to the study o f the possible global shapes o f our physical universe. There are no prerequisites for this chapter from after Chapter 7, but some o f the ideas may be difficult the first time around. Problems 18.1, 18.3, and 18.6 are the m inimum that is needed from this chapter before you study Chapter 24 (3-M anifolds — Shape o f Space); the other prob­ lems can be skipped. We use the term “m anifold” here instead o f “surface” because we usually think o f surfaces as sitting extrinsically in 3-space. Here we want to study only the intrinsic geometry; and thus any particular extrinsic embedding does not matter. M oreover, we will study some geometric 2-manifolds (for example, the flat torus) that cannot be (isometrically) embedded in 3-space. We ask what is the intrinsic geometric experience 245 2 4 6 Chapter 18 Geometric 2-Manifolds and Coverings on geometric 2-manifolds o f a 2-dimensional bug. How will the bug view geodesics (intrinsically straight lines) and triangles? How can a bug on a geometric 2-manifold discover the global shape o f its universe? These questions will help us as we think about how we as hum an beings can think about our 3-dimensional physical universe, where we are the (3-dimensional) bugs. This chapter will only be an introduction to these ideas. For a geometric introduction to differential geometry, see [DG: Henderson]. For more details about geometric 2-manifolds, see [DG: Weeks] and Chapter 1 o f [DG: Thurston], For the classification o f (triangulated) 2-manifolds, see the recent [TP: Francis & W eeks], which contains an accessible p ro o f due to John H. Conway. In Problem 4.1, we have already studied two examples o f geometric 2-manifolds — cylinders and cones (without the cone point). Because these surfaces are locally isometric to the Euclidean plane, these types of geometric m anifolds are called fl a t (or Euclidean) 2-manifolds. It would be good at this point for you to review what you know from Chapter 4 about cylinder and cones. P r o b l e m 18.1 F l a t T o r u s a n d F l a t K l e i n B o t t l e Flat T orus C E Figure 18.1 1-sheeted covering of cylinder and flat torus A nother example o f a flat (Euclidean) 2-manifold is provided by a video game that was popular a while ago. A blip on the video screen represent­ ing a ball travels in a straight line until it hits an edge o f the screen. Then the blip reappears traveling parallel to its original direction from a point at the same position on the opposite edge. Is this a representation o f some surface? If so, what surface? First, imagine rolling the screen into a C E ♦P roblem 18.1 Flat Torus and Flat Klein Bottle 247 tube where the top and bottom edges are glued (Figure 18.1). This is a representation o f the screen as a 1-sheeted covering o f the cylinder. A blip on the screen that goes o ff the top edge and reappears on the bottom is the lift o f a point on the cylinder that travels around the cylinder, crossing the line that corresponds to the jo ining o f the top and bottom o f the screen. Now let us further imagine that the cylinder can be stretched and bent so that we can glue the two ends to make a torus. N ow the screen represents a 1-sheeted covering o f the torus. If the blip goes o ff on one side and comes back on the other at the same height, this represents the lift o f a point m oving around the torus and crossing the circle that corre­ sponds to the place where the two ends o f the cylinder are joined. The possible m otions o f a point on the torus are represented by the motions on the video screen! You can ’t make a m odel in 3-space o f a flat torus from a flat piece o f paper without distorting it. Such a torus is called a fl a t torus. It is best not to call this a “surface,” because there is no way to realize it isometri- cally in 3-space and it is not the surface o f anything. But the question o f whether or not you can make an isometric model in 3-space is not impor­ tant — the point is that the gluings in Figure 18.1 intrinsically define a flat 2-manifold. I f you distort the cylinder in Figure 18.1 in 3-space, you can get the torus pictured in Figure 18.2. This is not a geometric 2-manifold because the original flat (Euclidean) geometry has been distorted and it is also not exactly either spherical or hyperbolic. a. Show that the fla t torus is locally isometric to the plane and thus, is a geometric 2-manifold, in particular, a fla t (,Euclidean) 2-manifold. 248 Chapter 18 Geometric 2-Manifolds and Coverings Note that each point on the interior o f an edge o f the screen is the lift o f a point that has another lift on the opposite edge. Thus, a lift o f a neigh­ borhood o f that is in two pieces (one near each o f the two opposite edges). W hat happens at the four com ers o f the computer screen (which are lifts o f the same point)? The torus in Figure 18.2 and the flat tom s are related in that there is a continuous one-to-one mapping from either to the other. We say that they are homeomorphic, or topologically equivalent. We can further express this situation by saying that the tom s in Figure 18.2 and the flat tom s are both topological tori. Figure 18.3 Flat torus from a hexagon There is another representation o f the flat tom s based on a hexagon. Start with a regular hexagon in the plane and glue opposite sides as indicated in Figure 18.3. *b. Show that gluing the edges o f the hexagon as in Figure 18.3 form s a fla t 2-manifold (called the hexagonal torus). In order for a flat 2-manifold to be formed from a regular polygon, the interior angle o f the polygon m ust be an integral factor o f 360°. {Do you see why?) Thus it is not possible to use other regular polygons to create flat 2-manifolds; see Problem 11.7. However, we can see from 11.7 that it may be possible for spherical and/or hyperbolic 2-manifolds to be created from regular polygons; see Problems 18.3 and 18.4. ♦P roblem 18.1 Flat Torus and Flat Klein Bottle 2 4 9 ♦ Flat K lein Bottle Now we describe a related geometric 2-manifold, traditionally called a fla t Klein bottle, named after Felix Klein (1849-1925, German). Imagine the same video screen, again with a traveling blip representing a ball that travels in a straight line until it hits an edge o f the screen. W hen it hits the top edge then the blip proceeds exactly the same as for the flat torus (traveling parallel to its original direction from a point at the same position on the opposite edge). However, when the blip hits a vertical edge o f the screen it reappears on the opposite edge but in the diametri­ cally opposite position and travels in a direction with slope that is the negative o f the original slope. (See Figure 18.4.) As before, imagine rolling the screen into a tube where the top and bottom edges are joined. This is again a representation o f the screen as a 1-sheeted covering o f the cylinder. Figure 18.4 1-sheeted covering of a flat Klein bottle A blip on the screen that goes o ff the top edge and reappears on the bottom is the lift o f a point on the cylinder that travels around the cylin­ der crossing the line that corresponds to the joining o f the top and bottom o f the screen. N ow the screen represents a 1-sheeted covering o f the Klein bottle. The possible m otions o f a point on the K lein bottle are represented by the motions on the video screen! Now, let us further imagine that the cylinder can be stretched and bent so that we can jo in the two ends to make a topological Klein bottle; see Figure 18.5. The figure in Figure 18.5 is not a geometric 2-manifold, because m ost points do not have neighborhoods that are isomorphic to either the plane, a sphere, or a hyperbolic plane. Y ou also can’t make an isometric model in 3-space o f a flat Klein bottle w ithout distorting it and having self-intersections. But the gluings in Figure 18.4 define intrinsically a flat 2-manifold. 2 5 0 Chapter 18 Geometric 2-Manifolds and Coverings *c. Show that the fla t Klein bottle is locally isometric to the plane and thus is a geometric 2-manifold, in particular, a fla t (Euclid­ ean)i 2-manifold. Note that the four com ers o f the video screen are lifts o f the same point and that a neighborhood o f this point has 360° — that is, 90° from each o f the four comers. It can be shown that Theorem 1 8 .1 . Flat tori and fla t Klein bottles are the only fla t (Euclidean) 2-manifolds that are fin ite and geodesically complete (every geodesic can be extended indefinitely). See the last section in Chapter 4, Relations to Differential Geometry. For a detailed discussion, see [DG: Thurston], pages 25-28. For a more elem entary discussion, see [DG: W eeks], Chapters 4 and 11. N ote that a finite cylinder is not geodesically complete; and if it is extended indefinitely, then it is geodesically complete but not finite. A cone with the cone point is not a flat m anifold at the cone point; with the cone point removed the cone is not geodesically complete. N ote that we get a flat torus for each size rectangle in the plane. These flat tori are different geom etrically because there are different distances around the tori. However, topologically they are all the same as (homeomorphic to) the surface o f a doughnut. Note that i f you move a right-hand glove (which we stylize by O ) around the flat tom s, it will always stay right handed; however, i f you move it around the flat Klein bottle horizontally, it will become left ♦P roblem 18.1 Flat Torus and Flat Klein Bottle 251 handed. See Figure 18.6. We describe these phenomena by saying that the flat torus is orientable and a Klein bottle is non-orientable. * P roblem 18.2 U niversal Covering of F lat 2-M anifolds Problem 4.2 contains an introduction to some o f the ideas in this problem and thus we urge the reader to look at 4.2 before proceeding with this chapter. a. On a fla t torus or fla t Klein bottle, how do we determine the different geodesics connecting two points'? How many are there? How can we justify our conjectures? Look at straight lines in the universal coverings introduced below. b. Show that some geodesics on the fla t torus or fla t Klein bottle are closed curves (in the sense that they come back and continue along themselves like great circles), though possibly self- intersecting. How can you fin d them ? 252 Chapter 18 Geometric 2-Manifolds and Coverings Look in the universal coverings introduced below. c. Show that there are geodesics on the fla t torus and fla t Klein bottle that never come back and continue along themselves. Look at the slopes o f the geodesics found in part b. The geodesics found in part c can be shown to come arbitrarily close to every point on the manifold. Such curves are said to be dense in the manifold. Suggestions We suggest that you use coverings ju st as you did for cones and cylin­ ders. The difference is that in this case the sheets o f the coverings extend in two directions. See Figure 18.7 for a covering o f the flat torus. I f this covering is continued indefinitely in all directions, then the whole plane covers the flat torus with each point in the torus having infinitely many lifts. W hen a covering is the whole o f either the (Euclidean) plane, a sphere, or a hyperbolic plane, it is called the universal covering. See * P roblem 18.2 Universal Covering of Flat 2-Manifolds 253 Figure 18.8 for a universal covering o f a flat K lein bottle. These cover­ ings are called “universal” because there are no coverings o f the plane, spheres, or hyperbolic planes that have more than one sheet; see the next section for a discussion o f this for a sphere. \ y d * a P x A X / t \ Z) P / ' f y 5 (b/2)2, this geometric
solution is impossible. When x > (b/2), Khayyam uses the drawings
shown in Figure 19.4.
( x – b /2 ) 2
b/2
Figure 19.4 Type 5, x>(b/2)

272 Chapter 19 Geometric Solutions of Quadratic and Cubic Equations
For solutions o f Type 6, Khayyam uses the drawing in Figure 19.5.
< ---------- x ---------- > x – b l 2
x – bl 2
bl 2
Figure 19.5 Type 6
Do the above solutions f i n d the negative roots? The answer is
clearly “no” if you mean, “Did a l’Khowarizmi and Khayyam mention
negative roots?” But let us not be too hasty. Suppose – r (r positive) is
the negative root o f x 2 + bx = c. Then { – r f + b (-r) = c or r 2 = br + c.
Thus r is a positive root o f x 1 = bx + c\ The absolute value o f the nega­
tive root o f x2 + bx = c is the positive root o f x2 = bx + c and vice versa.
Also, the absolute values o f the negative roots o f x2 + bx + c = 0 are the
positive roots o f x 2 + c = bx. So, in this sense, Yes, the above geometric
solutions do fin d all the real roots o f all quadratic equations. Thus it is
m isleading to state, as most historical accounts do, that the geometric
methods failed to find the negative roots. The users o f these methods did
not find negative roots because they did not conceive o f them. However,
the methods can be used directly to find all the positive and negative
roots o f all quadratics.
c. Use Khayyam’s methods to fin d all roots o f the follow ing equa­
tions: x 2 + 2x = 2, x 2 = 2x + 2, x2 + 2>x + 1 = 0.
bx c
b
P r o b l e m 19.2 C o n i c S e c t i o n s a n d C u b e R o o t s
The Greeks (for example, Archytas o f Tarentum, 428-347 b.c., who was
a Pythagorean in southern Italy, and Hippocrates o f Chios in Asia M inor,
5th century b.c.) noticed that, i f a/c = c/d = d/b, then ( a /c f = (c/d)(d/b) =
(ic/b) and, thus, c3 = cfb. (For more historical discussion, see [HI: van
der W aerden], page 150, and [HI: Katz], Chapter 2.) Now setting a = 1,
we see that we can find the cube root o f b if we can find c and d such
that c2 = d and d2 = be. If we think o f c and d as being variables and b as

P roblem 19.2 Conic Sections and Cube Roots 273
a constant, then we see these equations as the equations o f two parabolas
with perpendicular axes and the same vertex. The Greeks also saw it this
way, but first they had to develop the concept o f a parabola! The first
general construction o f conic sections was done by M enaechmus (a
member o f P lato’s Academy) in the 4th century b.c.
To the Greeks, and later Khayyam, if AB is a line segment, then the
parabola with vertex B and parameter AB is the curve P such that, if C
is on P, then the rectangle BDCE (see Figure 19.6) has the property that
(BE)2 = BD • AB. Because in Cartesian coordinates the coordinates o f C
are (BE, BD), this last equation becomes a fam iliar equation for a para­
bola.
Points o f the parabola may be constructed by using the construction
for the square root given in Chapter 13. In particular, E is the intersec­
tion o f the semicircle on AD with the line perpendicular to AB at B. (The
construction can also be done by finding D ‘ such that AB = DD’. The
semicircle on BD’ then intersects P at C.) We encourage you to try this
construction yourself; it is very easy to do i f you use a compass and
graph paper.
Now we can fi n d the cube root. Let b be a positive num ber or
length and let AB = b and construct C so that CB is perpendicular to AB
Figure 19.6 Construction of parabola

2 7 4 Chapter 19 Geometric Solutions of Quadratic and Cubic Equations
and such that CB = 1. See Figure 19.7. Construct a parabola with vertex
B and param eter AB and construct another parabola with vertex B and
param eter CB. Let E be the intersection o f the two parabolas. Draw the
rectangle BGEF. Then
( E F f = B F A B and (G E f = GBCB.
But, setting c = GE = BF and d – GB = EF, we have
d2 = cb and c2 = d. Thus c3 = b.
I f you use a fine graph paper, it is easy to get three-digit accuracy in this
construction.
The Greeks did a thorough study o f conic sections and their proper­
ties, culm inating in A pollonius’ (c. 260-170 b .c ., Greek) book Conics,
which appeared around 200 b .c . Y o u can read this book in English trans­
lation (see [AT: Apollonius]).
a. Use the above geometric methods with a fin e graph paper to fin d
the cube root o f 10.
To find roots o f cubic equations, we shall also need to know the
{rectangular) hyperbola with vertex B and parameter AB. This is the
curve such that, if E is on the curve and ACED is the determined rectan­
gle (see Figure 19.8), then (EC)2 = BC-AC.

P roblem 19.2 Conic Sections and Cube Roots 275
The point E can be determined using the construction from Chapter
13. Let F be the bisector o f AB. Then the circle with center F and radius
FC will intersect at D the line perpendicular to AB at A. From the draw ­
ing it is clear how these circles also construct the other branch o f the
hyperbola (with vertex A).
Figure 19.8 Construction of hyperbola
b. Use the above method with graph paper to construct the graph
o f the hyperbola with parameter 5. What is an algebraic equa­
tion that represents this hyperbola?
Notice how these descriptions and constructions o f the parabola and
hyperbola look very much as if they were done in Cartesian coordinates.
The ancestral form s o f Cartesian coordinates and analytic geometry are
evident here. They are also evident in the solutions o f cubic equations in
the next section. The ideas o f Cartesian coordinates did not appear to
Rene Descartes (1596-1650, French) out o f nowhere. The underlying
concepts were developing in Greek and Muslim mathematics. One o f the
apparent reasons that full developm ent did not occur until Descartes is
that, as we have seen, negative numbers were not accepted. The full use
o f negative numbers is essential for the realization o f Cartesian coordi­
nates. However, even Descartes seems to have avoided negatives as
m uch as possible when he was studying curves — he w ould start with a

2 76 Chapter 19 Geometric Solutions of Quadratic and Cubic Equations
curve (constructed by some geometric or m echanical procedure) and
then choose axes so that the important parts o f the curve had both coor­
dinates positive. However, it is not true (as asserted in some history o f
m athematics book) that D escartes always used x and y to stand for posi­
tive values. For example, in Book II o f Geometrie [AT: Descartes] he
describes the construction o f a locus generated by a point C and defines
on page 60 y = CB, where B is a given point, and derives an equation
satisfied by y and other variables; and, in the same paragraph on page 63,
he continues, “I f y is zero or less than nothing in this equation …”
P r o b l e m 19.3 S o l v i n g C u b i c E q u a t i o n s
G e o m e t r i c a l l y
In his Al-jabr w a ’l muqabalah, Omar Khayyam also gave geometric
solutions to cubic equations. You will see that his methods are sufficient
to find geometrically all real (positive or negative) roots o f cubic equa­
tions; however; in the first chapter Khayyam says (see [AT: Khayyam
1931], p. 49),
When, however, the object of the problem is an absolute number, nei­
ther we, nor any of those who are concerned with algebra, have been
able to prove this equation — perhaps others who follow us will be
able to fill the gap — except when it contains only the three first de­
grees, namely, the number, the thing and the square.
By “absolute num ber,” Khayyam is referring to what we call alge­
braic solutions, as opposed to geometric ones. This quotation suggests,
contrary to w hat many historical accounts say, that Khayyam expected
that algebraic solutions w ould be found.
Khayyam found 19 types o f cubic equations (when expressed with
only positive coefficients). (See [AT: Khayyam 1931], p. 51.) O f these
19, 5 reduce to quadratic equations (for example, x 3 + ax = bx reduces to
x 2 + ax = b). The rem aining 14 types Khayyam solved by using conic
sections. His methods find all the positive roots o f each type, although
he failed to m ention some o f the roots in a few cases, and, o f course, he
ignored the negative roots. Instead o f going through his 14 types, let us
look how a simple reduction will reduce them to only four types in addi­
tion to types already solved, such as x3 = b. Then we will look at
K hayyam ’s solutions to these four types.

P roblem 19.3 Roots of Cubic Equations 2 77
In the cubic y 3 + py2 + gy + r = 0 (where p, g, r, are positive, nega­
tive, or zero), set y = x – (p/.3). Try it! The resulting equation in x will
have the form x3 + sx + t = 0, (where s and t are positive, negative, or
zero). I f we rearrange this equation so that all the coefficients are posi­
tive, we get the following four types that have not been previously
solved:
(1) x 3 + ax = b, (2) x 3 + b = ax,
(3) jc3 = ax + b, and (4) x3 + ax + b = 0,
where a and b are positive, in addition to the types previously solved.
a. Show that in order to fin d all the roots o f all cubic equations we
need only have a method that finds the roots o f Types 1, 2, and 3.
K hayyam’s Solution for T ype 1: x 3 + ax = b
A cube and sides are equal to a number. Let the line AB (see Figure
19.9) be the side of a square equal to the given number of roots [that
is, (AB)2 = a, the coefficient]. Construct a solid whose base is equal to
the square on AB, equal in volume to the given number [6], The
construction has been shown previously. Let BC be the height of the

278 Chapter 19 Geometric Solutions of Quadratic and Cubic Equations
solid. [That is, BC – (AB? = b.] Let BC be perpendicular to AB … .
Construct a parabola whose vertex is the point B … and parameter AB.
Then the position of the conic HBD will be tangent to BC. Describe
on BC a semicircle. It necessarily intersects the conic. Let the point of
intersection be D\ drop from D, whose position is known, two perpen­
diculars DZ and DE on BZ and BC. Both the position and magnitude
of these lines are known.
The root is EB. K hayyam ’s p ro o f (using a more m odem , compact
notation) is as follows: From the properties o f the parabola (Problem
19.2) and circle (Problem 15.1), we have
(DZ)2 = (EB)2 = BZ ■ AB and (ED)2 = (BZ)2 = EC ■ EB,
thus
EB ■ (BZ)2 = (EB)2-EC = B Z -A B -E C
and, therefore,
and
A B -E C = EB- BZ,
(EB? = EB ■ (BZ AB) = (A B -E C )-A B = (AB? ■ EC.
So
(EB? + a(EB) = (AB? ■ EC + (AB)2 • (EB) = (AB? CB = b.
Thus EB is a root o f x3 + ax = b. Because x 2 + ax increases as x increases,
there can be only this one root.
Khayyam’s Solutions for T ypes 2 and 3:
x 3 + b = a x and Xs = a x + b
Khayyam treated these equations separately; but, by allowing negative
horizontal lengths, we can combine his two solutions into one solution o f
x 2± b = ax. Let AB be perpendicular to BC and as before let (AB? = a
and (AB? ■ BC = b. Place BC to the left if the sign in front o f b is nega­
tive (Type 3) and place BC to the right if the sign in front o f b is positive
(Type 2). Construct a parabola with vertex B and param eter AB.

P roblem 19.3 Roots of Cubic Equations 279
Construct both branches o f the hyperbola with vertices B and C and
param eter BC. See Figure 19.10.
Each intersection o f the hyperbola and the parabola (except for B)
gives a root o f the cubic. Suppose they meet at D. Then drop perpendicu­
lars DE and DZ. The root is BE (negative i f to the left and positive if to
the right). Again, if you use fine graph paper, it is possible to get three-
digit accuracy here. We leave it for you, the reader, to provide the proof,
which is very sim ilar to Type 1.
b. Verify that Khayyam ’s method described above works fo r Types
2 and 3. Can you see from your verification why the extraneous
root given by B appears?
c. Use K hayyam ’s method to fin d all roots o f the cubic
x3= 15* + 4.
Use fin e graph paper and try fo r three-place accuracy.

2 8 0 Chapter 19 Geometric Solutions of Quadratic and Cubic Equations
P r o b l e m 19.4 A l g e b r a i c S o l u t i o n o f C u b i c s
A little more history: M ost historical accounts assert correctly that
Khayyam did not find the negative roots o f cubics. However, they are
m isleading in that they all fail to m ention that his methods are fully
sufficient to find the negative roots, as we have seen above. This is in
contrast to the common assertion (see, for example, [EM : Davis &
Hersh]) that Girolamo Cardano (1501-1578, Italian) was the first to
publish the general solution o f cubic equations. In fact, as we shall see,
Cardano him self admitted that his methods are insufficient to find the
real roots o f m any cubics.
Cardano published his algebraic solutions in the book Artis Magnae
(The Great Art) in 1545. For a readable English translation and historical
summary, see [AT: Cardano]. Cardano used only positive coefficients
and thus divided the cubic equations into the same 13 types (excluding x3
= c and equations reducible to quadratics) used earlier by Khayyam.
Cardano also used geometry to prove his solutions for each type. As we
did above, we can make a substitution to reduce these to the same types
as above:
(1) x3 + ax = b, (2) x 3 + b = ax,
(3) x 3 = ax + b, and (4) x 3 + ax + b = 0.
If we allow ourselves the convenience o f using negative numbers and
lengths, then we can reduce these to one type: x 3 + ax + b = 0, where
now we allow a and b to be either negative or positive.
The m ain “trick” that Cardano used was to assume that there is a
solution o f x 3 + ax + b = 0 o f the form x = t m + wl/3. Plugging this into
the cubic, we get
(tm + M1/3)3 + a{tV3 + w1/3) + b = 0.
If you expand and simplify this, you get to
t + u + b + (3 t’l3u’13 + a)(tU3 + u’13) = 0.
(Cardano did this expansion and sim plification geometrically by imagin­
ing a cube with sides t113 + um.) Thus x = tm + uv3 is a root if
t + u = – b and t’13 uV3 = -{a t3).
Solving, we find that t and u are the roots o f the quadratic equation

P roblem 19.4 Algebraic Solution of Cubics 281
z2 + b z – (a/3)3 = 0,
which Cardano solved geom etrically (and so can you, Problem 19.1) to
get
t = -b /2 + J(b /2 )2 + (a/3)3 and u = -b /2 – J(b /2 )2 + (a/3)3 .
Thus the cubic has roots
X = tm + um
= {-b/2 + J(b /2 )2 + (a/3)3 }1/3 + { – b / 2 – J (b/2)2 + (a/3)3 } m.
This is Cardano’s cubic formula. But a strange thing happened.
Cardano noticed that the cubic x3 = 15x + 4 has a positive real root 4 but,
for this equation, a = -1 5 and b = – 4 , and i f we put these values into his
cubic formula, we get that the roots o f x3 = 15x + 4 are
x = {2 + J ^ \ 2 l } m + { 2 – ^ Y 2 \ y ‘ K
But these are the sum o f two complex num bers even though you have
shown in Problem 19.3 that all three roots are real. How can this expres­
sion yield 4?
In Cardano’s time there was no theory o f complex numbers, and so
he reasonably concluded that his method w ould not w ork for this equa­
tion, even though he did investigate expressions such as V -1 2 1 . Car­
dano writes ([A T: Cardano, p. 103]),
When the cube of one-third the coefficient of x is greater than the
square of one-half the constant of the equation … then the solution of
this can be found by the aliza problem which is discussed in the book
of geometrical problems.
It is not clear what book he is referring to, but the “aliza problem ”
presum ably refers to the mathematician known as a l’Hazen, Abu Ali
a l’Hasan ibu a l’Haitam (965-1039), who was bom in Persia and worked
in Egypt and whose works were known in Europe in Cardano’s time.
A l’Hazen had used intersecting conics to solve specific cubic equations
and the problem o f describing the image seen in a spherical m irror —
this latter problem in some books is called “A lhazen’s problem .”
In addition, we know today that each complex num ber has three
cube roots and so the formula

282 Chapter 19 Geometric Solutions of Quadratic and Cubic Equations
X = { 2 + }w + {2 – 1/3
is ambiguous. In fact, some choices for the two cube roots give roots o f
the cubic and some do not. (Experiment with x3 = 15x + 4.) Faced with
Cardano’s formula and equations such as x3 = 15x + 4, Cardano and
other m athematicians o f the time started exploring the possible meanings
o f these complex num bers and thus started the theory o f complex
numbers.
a. Solve the cubic x3 = 15x + 4 using Cardano’s form ula and your
knowledge o f complex numbers.
Rem ember that on the previous two pages we showed that x = tm + um is
a root o f the equation i f t + u = – b and t 113 um = – (a /3).
b. Solve x3 = 15x + 4 by dividing through by x – 4 and then solving
the resulting quadratic.
c. Compare your answers and methods o f solution from Problems
19.3c, 19.4a, and 19.4b.
W h a t D o e s T h i s A l l P o i n t T o ?
W hat does the experience o f this chapter point to? It points to different
things for each o f us. W e conclude that it is worthwhile to pay attention
to the m eaning in mathematics. Often in our haste to get to the m odem ,
powerful analytic tools, we ignore and trod upon the m eanings and
images that are there. Sometimes it is hard even to get a glimpse that
some m eaning is missing. One way to get this glimpse and find meaning
is to listen to and follow questions o f “W hat does it m ean?” that come up
in ourselves, in our friends, and in our students. We m ust listen crea­
tively because we and others often do not know how to express precisely
what is bothering us.
Another way to find m eaning is to read the m athematics o f old and
keep asking, “W hy did they do that?” or “W hy didn’t they do this?”
W hy did the early algebraists (up until at least 1600 and m uch later, we
think) insist on geometric proofs? We have suggested some reasons
above. Today, we norm ally pass over geometric proofs in favor o f
analytic ones based on the 150-year-old notion o f Cauchy sequences and
the Axiom o f Completeness. However, for m ost students and, we think,

So What Does This All Point To? 283
most mathematicians, our intuitive understanding o f the real num bers is
based on the geometric real line. As an example, think about m ultiplica­
tion: W hat does a x b m ean? Compare the geometric images o f a x b
with the m ultiplication o f two infinite, non-repeating, decimal fractions.
W hat is 72″ X7r?
There is another reason why a geometric solution may be more
meaningful: Sometimes we actually desire a geometric result instead o f a
numerical one. For example, David and a friend were building a small
house using wood. The ro o f o f the house consisted o f 12 isosceles trian­
gles that together formed a 12-sided cone (or pyramid). It was necessary
for them to determine the angle between two adjacent triangles in the
ro o f so they could appropriately cut the log rafters. David immediately
started to calculate the angle using (numerical) trigonometry and
algebra. But then he ran into a problem. He had only a slide rule with
three-place accuracy for finding square roots and values o f trigonometric
functions. At one point in the calculation he had to subtract two numbers
that differed only in the third place (for example, 5.68 – 5.65); thus his
result had little accuracy. As he started to figure out a different com puta­
tional procedure that would avoid the subtraction, he suddenly realized
he d id n ’t want a number, he wanted a physical angle. In fact, a num eri­
cal angle would be essentially useless — imagine taking two rough
boards and putting them at a given numerical angle apart using only an
ordinary protractor! W hat he needed was the physical angle, full size. So
David and his friend constructed the angle on the floor o f the house
using a rope as a compass. This geometric solution had the following
advantages over a numerical solution:
♦ The geometric solution resulted in the desired physical angle,
while the numerical solution resulted in a number.
♦ The geometric solution was quicker than the numerical solu­
tion.
♦ The geometric solution was immediately understood and
trusted by D avid’s friend (and fellow builder), who had
alm ost no m athematical training, while the numerical solu­
tion was beyond the friend’s understanding because it
involved trigonometry (such as the Law o f Cosines).

2 8 4 Chapter 19 Geometric Solutions of Quadratic and Cubic Equations
♦ And, because the construction was done full size, the solution
autom atically had the degree o f accuracy appropriate for the
application.
Meaning is important in mathematics, and geometry is an important
source of that meaning.

______________ Chapter 20
T rigonometry and Duality
After we have found the equations [The Laws of Cosines and
Sines for a Hyperbolic Plane] which represent the dependence
of the angles and sides of a triangle; when, finally, we have
given general expressions for elements of lines, areas and
volumes of solids, all else in the [Hyperbolic] Geometry is a
matter of analytics, where calculations must necessarily agree
with each other, and we cannot discover anything new that is
not included in these first equations from which must be taken
all relations of geometric magnitudes, one to another. … We
note however, that these equations become equations of
spherical Trigonometry as soon as, instead of the sides a, b, c
we put a ,f- l ,b j^-l …
— N. Lobachevsky, quoted in [HY: Greenberg]
In this chapter, we will first derive, geometrically, expressions for the
circumference o f a circle on a sphere, the Law o f Cosines on the plane,
and its analog on a sphere. Then we will talk about duality on a sphere.
On a sphere, duality will enable us to derive other laws that will help our
two-dimensional bug to compute sides and/or angles o f a triangle given
ASA, RLH, SSS, or AAA. Finally, we will look at duality on the plane.
P r o b l e m 2 0 . 1 C i r c u m f e r e n c e o f a C i r c l e
a. Find a simple form ula fo r the circumference o f a circle on a
sphere in terms o f its intrinsic radius and make the form ula as
intrinsic as possible.
285

2 86 Chapter 20 Trigonometry and Duality
We suggest that you make an extrinsic drawing (sim ilar to Figure 20.1)
o f the circle, its intrinsic radius, its extrinsic radius, and the center o f the
sphere. You may well find it convenient to use trigonometric functions
to express your answer. N ote that the existence o f trigonometric func­
tions for right triangles follows from the properties o f sim ilar triangles
that were proved in Problem 13.4.
In Figure 20.1, rotating the segment o f length r (the extrinsic
radius) through a whole revolution produces the same circumference as
rotating r, which is an arc o f the great circle as well as the intrinsic
radius o f the circle on the sphere.
Even though the derivation o f the formula this way will be
extrinsic, it is possible, in the end, to express the circumference only in
terms o f intrinsic quantities. Thus, also think o f the following problem:
b. How could our 2-dimensional bug derive this form ula?
By looking at very small circles, the bug could certainly find uses for the
trigonometric functions they give rise to. Then the bug could discover

P roblem 2 0 . 1 Circumference of a Circle 2 87
that the geodesics are actually (intrinsic) circles, but circles that do not
have the same trigonometric properties as very small circles. Then what?
Using the expressions o f trigonometric and hyperbolic functions in
terms o f infinite series, it is proved (in [HY: Greenberg], p. 337) that
Theorem 20.1. In a hyperbolic plane o f radius 1, a circle with
intrinsic radius r has circumference c equal to
c = 2 ^ s in h (r).
c. Use the theorem to show that on a hyperbolic plane o f radius p,
a circle with intrinsic radius r has circumference c equal to
c = I n p smh{r/p).
W hen going from a hyperbolic plane o f radius 1 to a hyperbolic plane o f
radius p, all lengths scale by a factor o f p. Why?
The formula in part c should look very m uch like your formula for
part a (possibly with some algebraic m anipulations). This is precisely
w hat Lobachevsky was talking about in the quote at the beginning o f this
chapter. Check this out with part d.
d. Show that, i f you replace p by ip in the form ula o f p a rt c, then
you will get the form ula in part a.
Look up the definition o f sinh (hyperbolic sine) and express it as a
Taylor series and then compare with the Taylor series o f sine.
P r o b l e m 2 0 . 2 L a w o f C o s i n e s
I f we know two sides and the included angle o f a (small) triangle, then
according to SAS the third side is determined. I f we know the lengths o f
the two sides and the measure o f the included angle, how can we find the
length o f the third side? The various formulas that give this length are
called the Law o f Cosines.
a. Prove the Law o f Cosines fo r triangles in the plane (see Figure
20.2):
c2 = or2 + b2 – 2ab cos 6 .

288 Chapter 20 Trigonometry and Duality
For a geometric p ro o f o f this “law ,” look at the pictures in Figure 20.3.
We first saw the idea for these pictures in the marvelous book [EG:
Valens]. These pictures show the squares as rigid with hinges at all the
points m arked N ote that in the middle picture 9 is greater than
ti/ 2 . Y o u must draw a different picture for t?less than n/2. Prove the Law
o f Cosines on the plane using the pictures in Figure 20.3, or in any other
way you wish. Note the close relationship with the Pythagorean
Theorem.
b_____ a___
b2 ba
ab a 2
—– T>——– *— a—-
( a + b )2 = a2 + b2 + la b
F ig u r e 2 0 . 3 T h r e e r e la te d g e o m e t r ic p ro o fs

P roblem 2 0 . 2 Law of Cosines 2 8 9
b. Find the law o f cosines fo r small triangles on a sphere with
radius p:
cos c/p= cos a lp cos b/p+ sin a /p sin blp cos 6.
Hint: One approach that w ill w ork is to project the triangle by a gnomic
projection onto the plane tangent to the sphere at the vertex o f the given
angle. This projection will preserve the size o f the given angle ( Why?)
and, even though it will not preserve the lengths o f the sides o f the trian­
gle, you can determine what effect it has on these lengths. N ow apply the
planar Law o f Cosines to this projected triangle and turn the algebra
crank. It is very helpful to draw a 3-D picture o f this projection.
c. Derive a form ula fo r the distance between two points on the
earth in terms o f the latitude and longitude at the two points.
It is sometimes convenient to measure lengths o f great circle arcs on
the sphere in terms o f the radian measure. In particular,
radian measure o f the arc = (length o f the arc)//?,
where p is the radius o f the sphere. For example, the radian m easure o f
one quarter o f a great circle w ould be till and the radian m easure o f h a lf
a great circle w ould be ji. In Figure 20.4, the segment a is subtended by
the angle a at the pole and by the same angle a at the center o f the
sphere. The radian measure o f a is the radian m easure o f a. M ost other
texts and articles either use radian measure for lengths or assume that the
radius p is equal to 1. We will m ost o f the time keep the p explicit so the
connection with the radius will be clearly seen.
F ig u r e 2 0 . 4 R a d ia n m e a s u r e o f le n g th s

2 9 0 Chapter 20 Trigonometry and Duality
I f we m easure lengths in radians, then one possible form ula for the
spherical triangle o f radius p in part b is
cos c = cos a cos b + sin a sin b cos 6.
T heorem 2 0 .2 a . A Law o f Cosines fo r right triangles on a
sphere with radius p is
cos c/p = cos a /p cos b/p,
or, in radian measure o f lengths,
cos c = cos a cos b,
which can be considered as the spherical equivalent o f the
Pythagorean Theorem.
T heorem 2 0 .2 b . A Law o f Cosines fo r triangles on a hyper­
bolic plane with radius p is
cosh d p – cosh a /p cosh b/p + sinh a /p sinh b/p cos 6.
This theorem is proved in [HY: Stahl], p. 125, using analytic tech­
niques, and in [HY: Greenberg] using infinite series representations.
This is one o f the equations that Lobachevsky is talking about in the
quote at the beginning o f the chapter. It is important to realize that a
study o f hyperbolic trigonometric functions was started by V. Riccati
(1707-1775, Italy) and others, and continued by J. H. Lambert
(1728-1777, Germany) in 1761-68 — well before their use in hyper­
bolic geometry by Lobachevsky. For details on the history o f hyperbolic
functions, see the informative 2004 paper [HI: Barnett].
P roblem 20.3 Law of S ines
Closely related to the Law o f Cosines is the Law o f Sines. (Figure 20.5)
F ig u r e 2 0 . 5 L a w o f S in e s

P roblem 20.3 Law of Sines 291
a. I f AABC is a planar triangle with sides, a, b, c, and correspond­
ing opposite angles, a, (3, y, then o/sin a = 6/sin (3 = c/sin y.
The standard p ro o f for the Law o f Sines is to drop a perpendicular from
the vertex C to the side c and then to express the length o f this perpen­
dicular as both (b sin a) and {a sin (3). See Figure 20.6. From this the
result easily follows. Thus, on the plane the Law o f Sines follows from
an expression for the sine o f an angle in a right triangle.
Figure 20.6 Standard proof of Law of Sines on plane
For triangles on the sphere we can find a very sim ilar result.
b. Show that fo r a small triangle on a sphere with radius p,
(sin a//?)/sin a = (sin blp)lsm (3= (sin c//?)/sin y.
If AACD is a triangle on the sphere with the angle at D being a right
angle, then use gnomic projection to project AACD onto the plane that is
tangent to the sphere at A. Because the plane is tangent to the sphere
at A, the size o f the angle a is preserved under the projection. In general,
angles on the sphere not at A will not be projected to angles o f the same
size, but in this case the right angle at D will be projected to a right
angle. (Be sure you see why this is the case. A good drawing and the use
o f symmetry and similar triangles will help.) Now express the sine
o f a in terms o f the sides o f this projected triangle.
When we m easure the sides in radians, on the sphere the Law o f
Sines becomes
(sin a)/sin a = (sin b)/sin J3 = (sin c)/sin y.
For a right triangle this becomes (Figure 20.7)
sin a = (sin a)/(sin c).

292 Chapter 20 Trigonometry and Duality
D uality on a S phere
We can now ask, Is it possible to find expressions corresponding to the
other triangle congruence theorems that we proved in Chapters 6 and 9?
Let us see how we can be helped by a certain concept o f duality that we
will now develop.
W hen we were looking at SAS and ASA, we noticed a certain dual­
ity betw een points and lines (geodesics). SAS was true on the plane or
open hemisphere because two points determine a unique line segment
and ASA was true on the plane or open hemisphere because two (inter­
secting) lines determine a unique point. In this section we will make this
notion o f duality broader and deeper and look at it in such a way that it
applies to both the plane and the sphere.
On the whole sphere, two distinct points determine a unique straight
line (great circle) unless the points are antipodal. In addition, two
distinct great circles determine a unique pair o f antipodal points. Also, a
circle on the sphere has two centers that are antipodal. Rem ember also
that in m ost o f the triangle congruence theorems, we had trouble with
triangles that contained antipodal points. So our first step is to consider
not points on the sphere but rather point-pairs, pairs o f antipodal points.
W ith this definition in mind, check the following:
♦ Two distinct point-pairs determine a unique great circle
(geodesic).
♦ Two distinct great circles determine a unique point-pair.
♦ The intrinsic center o f a circle is a single point-pair.
♦ SAS, ASA, SSS, AAA are true fo r all triangles not containing
any point-pairs.

Duality on a Sphere 293
Now we can make the duality more definite.
♦ The dual o f a great circle is its poles (the point-pair that is
the intrinsic center o f the great circle). Some books use the
term “polar” in place o f “dual”.
♦ The dual o f a point-pair is its equator (the great circle whose
center is the point-pair).
Note: I f the point-pair P is on the great circle l, then the dual o f l is on
the dual o fP . See Figure 20.8.
If k is another great circle through P, then notice that the dual o f k is
also on the dual o f P. Because an angle can be viewed as a collection o f
lines (great circles) em anating from a point, the dual o f this angle is a
collection o f point-pairs lying on the dual o f the angle’s vertex. And,
vice versa, the dual o f the points on a segment o f a great circle are great
circles em anating from a point-pair that is the dual o f the original great
circle. Before going on, be sure to understand this relationship between
an angle and its dual. Draw pictures. Make models.
P roblem 2 0.4 T he D ual of a S mall T riangle
on a S phere
The dual o f the sm all triangle AABC is the small triangle
AA*B*C*, where A* is that pole o f the great circle o f BC that is on the
same side o f BC as the vertex A, similarly for B*, C*. See Figure 20.9.

2 9 4 Chapter 20 Trigonometry and Duality
Figure 20.9 The dual of a small triangle
a. Find the relationship between the sizes o f the angles and sides o f
a triangle and the corresponding sides and angles o f its dual.
b. Is there a triangle that is its own dual!
P r o b l e m 20.5 T r i g o n o m e t r y w i t h C o n g r u e n c e s
a. Find a dual o f the Law o f Cosines on the sphere. That is, fin d a
statement that results from replacing in the Law o f Cosines
every angle and every side by its dual.
There is an analogous dual Law o f Cosines for a hyperbolic plane,
proved in the same books cited with Theorem 20.2, the hyperbolic Law
o f Cosines.
b. For each o f ASA, RLH, SSS, AAA, i f you know the measures o f
the given sides and angles, how can you fin d the measures o f the
sides and angles that are not given? Do this fo r both spheres
and hyperbolic planes.
Use part a and the form ulas from Problems 20.1 and 20.2.
D u a l i t y o n t h e P r o j e c t i v e P l a n e
The gnomic projection, g (Problem 14.2), allows us to transfer the above
duality on the sphere to a duality on the plane. If P is a point on the

Duality on the Projective Plane 295
plane, then there is a point Q on the sphere such that g(Q) = P. The dual
o f Q is a great circle / on the sphere. If Q is not the south pole, then h a lf
o f / is in the southern hemisphere and its projection onto the plane, g(l),
is a line we can call the dual o f P. This defines a dual for every point on
the plane except for the point where the south pole o f the sphere rests.
See Figure 20.10. It is convenient to call this point the origin, O, o f the
plane.
Figure 20.10 Duality on the projective plane
N ote that O is the image o f S, the south pole, and that the dual o f S
is the equator, which is projected by G to infinity on the plane. Thus we
define the dual o f O to be the line at infinity. If l is any line in the plane,
then it is the image o f a great circle on the sphere that intersects the
equator in a point-pair. The image o f this point-pair is considered to be a
single point at infinity at the “end” o f the line /, the same point at both
ends. The plane with the line at infinity attached is called the projective
plane. This projective plane is the same as the real projective plane that
we investigated in Problem 18.3b, the only difference being that here we
are focusing on the gnomic projection onto the plane and in Chapter 18
we were considering it as a spherical 2-manifold.

2 96 Chapter 20 Trigonometry and Duality
* P r o b l e m 20.6 P r o p e r t i e s o n t h e P r o j e c t i v e P l a n e
a. Check that the follow ing hold on the projective p lane:
♦ Two points determine a unique line.
♦ Two parallel lines share the same point at infinity.
♦ Two lines determine a unique point.
♦ I f a point is on a line, then the dual o f the line is a
point that is on the dual o f the original point.
b. I f y is the circle with center at the origin and with radius the
same as the radius o f the sphere and i f (P,Q) is an inversive pair
with respect to y, then show that the dual o f P is the line perpen­
dicular to OP at the point -Q, the point in the plane that is
opposite (with respect to O) o f Q. See Figure 20.11.
c. Show that the dual o f a point P on y is a line tangent to y at the
diametrically opposite point -P.
Dual V iews of O ur Experience
Our usual point o f view for perceiving our w orld is from the origin
(because in this view we are the center o f our world). We look out in all
directions to observe what is outside and we strive to look out as close
toward infinity as we can. The dual o f this is the point o f view where we

♦ P roblem 20.6 Properties on the Projective Plane 297
are the line at infinity (the dual o f the origin), we view the whole world
(which is within us), and we strain to look in as close toward the center
as we can.
P e r s p e c t i v e D r a w i n g a n d H i s t o r y
Look at the perspective drawing in Figure 20.12.
Present-day theories o f projective geometry got their start from
Euclid’s Optics [AT: Euclid Optics] and later from the theories o f
perspective developed during the Renaissance by artists who studied the
geometry inherent in perspective drawings. The first m athematical rule
for getting the correct perspective was invented by Filippo Brunelleschi
(1377-1455) in or shortly before 1413. In the Renaissance there was no
such profession as artist. Some o f the m ost famous artists o f 14th and
15th centuries were trained as goldsmiths and were taught some m athe­
matics from books such as Leonardo o f Pisa’s (c .l 170-C.1240) book
Libro d ’abaco, m ostly known for its use o f Arabic numerals and origin
o f Fibonacci numbers, but there is m uch more mathematics in it. Brunel­
leschi’s work included painting and sculpture as well as the design o f
stage machinery and buildings. He invented his method o f “artificial per­
spective” for drawing different buildings in Rome in a way that appeared
realistic. But no written descriptions survive from Brunelleschi.
The earliest works to show parallel lines that converge to a point in
a more or less convincing m anner are not paintings, but relief panels by
a friend o f Brunelleschi’s, the sculptor Donatello (1 3 8 6 -1 4 6 6 ). Another
younger friend o f Brunelleschi’s who was also an early user o f perspec­
tive was the painter M asaccio (1401-C .1428). M asaccio’s fresco Trinity
( c .l 426) in Santa M aria Novella (Florence) provides the best opportunity
for studying M asaccio’s mathematics and hence making some educated
guesses about B runelleschi’s method.
The first surviving written account o f a method o f perspective
construction is by Leon Batista Alberti (1404-1472) in the beginning o f
a short treatise On Painting in 1435. The problem was how to construct
the perspective image o f a square checkerboard pavement. It is clear
from later literary evidence, m ainly dating from 16th century, that
A lberti’s construction was not the only one current in 15th century.
Among the first mathematical treatises on perspective we should
m ention also Pierro della Francesca (1412-1492), De prospectiva
pingendi. Pierro della Francesca is acknowledged as one o f the most

2 98 Chapter 20 Trigonometry and Duality
important painters o f the 15th century, but he had an independent reputa­
tion as a highly com petent m athematician. In A lbrecht Diirer’s
(1471-1528) Treatise on measurement with compasses and straightedge
(Nuremberg, 1525) is shown an instrument being used to draw a per­
spective image o f a lute. Another m echanical aid described by Diirer
goes back to the early 15th century and was described also by Alberti and
Leonardo da Vinci. It was a veil o f coarse netting, which was used to
provide something rather like a system o f coordinate lines. Such squar­
ing systems were fam iliar to artists also as a method o f scaling drawings
up for transfer to the painting surface. For more discussion on perspec­
tive and m athematics in the Renaissance, see [AD: Field].
The “vanishing point” where the lines that are parallel on the box
intersect on the horizon o f the drawing is an image on the drawing o f the
point at infinity on these parallel lines.
Horizon
One way to visualize this is to imagine yourself at the center o f a
transparent sphere looking out at the world. If you look at two parallel
straight lines and trace these lines on the sphere, you will be tracing
segments o f two great circles. If you followed this tracing indefinitely to
the ends o f the straight lines, you w ould have a tracing o f two h a lf great
circles on the sphere intersecting in their endpoints. These endpoints o f
the semicircles are the images o f the point at infinity that is the common
intersection o f the two parallel lines. I f you now use a gnomic projection
to project this onto a plane (for example, the artist’s canvas), then you
will obtain two straight line segments intersecting at one o f their
endpoints as in the drawing above.

Chapter 21
M echanisms
T h e m a t h e m a t i c a l i n v e s t i g a t i o n s r e f e r r e d t o b r i n g t h e w h o l e
a p p a r a t u s o f a g r e a t s c i e n c e t o t h e e x a m i n a t i o n o f t h e p r o p e r ­
t i e s o f a g i v e n m e c h a n i s m , a n d h a v e a c c u m u l a t e d i n t h i s d i r e c ­
t i o n r i c h m a t e r i a l , o f e n d u r i n g a n d i n c r e a s i n g v a l u e . W h a t is
l e f t u n e x a m i n e d is h o w e v e r t h e o t h e r , i m m e n s e l y d e e p e r p a r t
o f t h e p r o b l e m , t h e q u e s t i o n : H o w d i d t h e m e c h a n i s m , o r t h e
e l e m e n t s o f w h i c h it is c o m p o s e d , o r i g i n a t e ? W h a t la w s
g o v e r n i ts b u i l d i n g u p ?
— F . R e u l e a u x , Kinematics ( 1 8 7 6 ) , p . 3
In this chapter we will study mechanisms, which for our purposes we
define as collections o f rigid bodies with m oveable connections having
the purpose o f transform ing motion. W e have already studied two
mechanisms: an angle-trisecting mechanism in Problem 15.4d and the
Peaucellier-Lipkin straight-line m echanism in Problem 16.3. A machine
can be considered as a combination o f mechanism s connected together
in a way to do useful work.
In this chapter we will use the Law o f Cosines and the Law o f Sines
for the plane and sphere in Problems 21.1 and 21.2. Otherwise the
material demands only basic understanding o f plane and spheres.
I n t e r a c t i o n s o f M e c h a n i s m s w i t h M a t h e m a t i c s
We mentioned in Chapter 0 that one o f the strands in the history o f
geometry is the M otion/M achines Strand; and we showed in Chapter 1
how this strand led to m echanism for producing straight line motion. It is
known that the Greeks, in particular Aristotle, studied the so-called
simple machines: the wheel, lever, pulleys, and inclined plane. He also
2 9 9

3 0 0 Chapter 2 1 — Mechanisms
described gear wheel drive in windlasses and pointed out that the direc­
tion o f rotation is reversed when one gear wheel drives another gear
wheel. Archim edes made devices to m ultiply force or torque many times
and studied spirals and helices for m echanical purposes (see Figure 0.5).
We know H ero’s form ula for the area o f a triangle but in his time he was
better known as an engineer. M ost o f his inventions are known from the
writings o f the Roman engineer Vitruvius. However, there was little
application o f Greek natural science to engineering in antiquity. In fact,
engineering contributed far more to science than science did to engineer­
ing until the latter h a lf o f the 19th century. For more discussion o f this
history, see [M E: Kirby], p. 43.
One o f the sim plest mechanisms used in human activities are
linkages. Perhaps an idea o f using linkages came into somebody’s mind
because a linkage resem bles a human arm. We can find linkages in old
drawings o f various machines in 13th century; see Figure 1.3. Leonardo
da V inci’s Codex Madridi (1493) contained a collection o f machine
elements, Elementi Macchinali, and he invented a lathe for turning parts
with elliptical cross section, using a four-bar linkage (see Problem 21.1).
Georgius Agricola (1494—1555) is considered a founder o f geology as a
discipline but he gave descriptions o f m achines used in mining, and that
is where we can find pictures o f linkages used in these machines. Gigan­
tic linkages, principally for m ine pumping operations, connected water
wheels at the riverbank to pumps high up on the hillside. Such linkages
consisted m ostly o f what we call four-bar linkages; see Problem 21.1.
Figure 21.1 Mechanism for drawing a parabola (1657)

Interactions o f Mechanisms with Mathematics 301
M athem aticians got interested in linkages first for geometric
drawing purposes. We know about some such devices from ancient
Greek mathematics. (For example, see Problem 15.4 about devices for
trisecting an angle.) W hen Rene Descartes published his Geometry
(1637) he did not create a curve by plotting points from an equation.
There were always first given geometrical methods for drawing each
curve with some apparatus, and often these apparatus were linkages. See,
for example, Figure 21.1, which depicts the m echanism for drawing a
parabola that appear in the works o f Franz von Schooten (1615-1660)
that were a popularization o f D escartes’ work.
Isaac New ton developed m echanisms for the generation o f algebraic
curves o f the third degree. This tradition o f seeing curves as the result o f
geometric actions can be found also in works o f Roberval, Pascal, and
Leibniz. M echanical devices for drawing curves played a fundamental
role in creating new symbolic languages (for example, calculus) and
establishing their viability. The tangents, areas, and arc length associated
with many curves were known before any algebraic equations were
written. Critical experiments using curves allowed for the coordination
o f algebraic representations with independently established results from
geometry. For more detailed discussion o f the ideas in the last two
paragraphs, see [HI: Dennis], Chapter 2.
Linkages are closely related with kinematics or geometry o f motion.
First it was the random growth o f m achines and m echanisms under the
pressure o f necessity. M uch later, algebraic speculations on the genera­
tion o f curves were applied to physical problems.
Two great figures appeared in 18th century, Leonard Euler
(1707-1783) and James W att (1736-1819). Although their lives overlap
there was no known contact between them. But both o f them were
involved with the “geometry o f m otion.” W att, instrument m aker and
engineer, was concerned with designing m echanisms that produce
desired motions. W att’s search for a mechanism to relate circular m otion
with straight-line m otion is discussed in the historical introduction to
Chapter 1. E uler’s theoretical results were unnoticed for a century by the
engineers and m athematicians who were devising linkages to compete or
supersede W att’s mechanism. See Figure 1.4.
The fundamental idea o f the geometric analysis o f m otion (kinem at­
ics) stems from Euler, who wrote in 1775,
The investigation of the motion of a rigid body may be
conveniently separated into two parts, the one geometrical, the

302 Chapter 21 Mechanisms
other mechanical. In the first part, the transference of the body
from a given position to any other position must be investi­
gated without respect to the causes of motion, and must be
represented by analytical formulae, which will define the
position of each point of the body. This investigation will
therefore be referable solely to geometry … . [Euler, Novi
commentarii Academiae Petrop., vol. XX, 1775. Translation
in Willis, Principles o f Mechanism, 2nd ed., p. viii, 1870.]
These two parts are sometimes called kinem atics (geometry o f motion)
and kinetics (the mechanics o f motion). Here we can see beginnings o f
the separation o f the general problem o f dynamics into kinem atics and
kinetics.
Franz Reuleaux (1829-1905) divided the study o f machines into
several categories and one o f them was study o f the geometry o f m otion
([M E: Reuleaux], pp. 36-40). In the proliferation o f m achines at the
height o f the Industrial Revolution, Reuleaux was system atically analyz­
ing and classifying new m echanisms based on the way they constrained
motion. He hoped to achieve a logical order in engineering. The result
w ould be a library o f m echanism s that could be combined to create new
machines. He laid the foundation for a systematic study o f m achines by
determining the basic building blocks and developing a system for
classifying known m echanism types. Reuleaux created at Berlin a collec­
tion o f over 800 models o f m echanisms and authorized a German
company, Gustav Voigt, Mechanische Werkstatt, in Berlin, to m anufac­
ture these models so that technical schools could use them for teaching
engineers about machines.
Figure 21.2 Three Reuleaux models

Interactions o f Mechanisms with Mathematics 303
In 1882, Cornell University acquired 266 o f such models, and now
the rem aining 219 models is the largest collection o f Reuleuax kinematic
mechanisms in the world. See examples in Figure 21.2. W ith support
from the NSF, a team o f Cornell mathematicians, engineers, and librari­
ans has developed a digital Reuleaux kinematic model W eb site [ME:
KM ODDL] as the part o f N ational Science Digital Library. The publicly
available W eb site contains photos, mathematical descriptions, historical
descriptions, m oving virtual reality images, simulations, learning
modules (for m iddle school through undergraduate), and even download­
able files for 3D printing. In this chapter we will look at the m athematics
related to a few o f these m echanisms in order to show geometry in
M achine/M otion Strand.
There will be discussions o f more recent history later in the chapter.
P r o b l e m 21.1 F o u r – B a r L i n k a g e s
A four bar linkage is a mechanism that lies in a plane (or spherical
surface) and consists o f four bars connected by joints that allow rotation
only in the plane (or sphere) o f the mechanism. See Figure 21.3.
Planar Spherical
Figure 21.3 Reuleaux four-bar linkages
In normal practice one o f the links is fixed so that it does not move. In
Figure 21.4 we assume that the link OC is fixed and investigate the
possibilities o f m otion for the other three links . We call the link OA the
input crank and link CB the output crank. Similarly, we call the angle 6
the input angle and angle ^ th e output angle.

3 0 4 Chapter 21 Mechanisms
a. (Plane) The input crank will be able to swing opposite C (where
the input angle is jt = 180 degrees) only i f a + g < b + h. I f a + g > b + h, there will be a maximum input angle, dm ax,
satisfying
COS 0max
(g2 + a2) – (h + b)2
la g
What happens i f a + g = b + h l
You may find it helpful to experiment with four-bar linkages that you
make out o f strips o f cardboard; and/or to play with the online simula­
tions that are linked on the Experiencing Geometry W eb page. Apply the
Law o f Cosines (Problem 20.2).
We can now do the same analysis on the sphere, using the
spherical Law o f Cosines (20.2) and the special absolute value |/|s (see
the section Triangle Inequality ju st before Problem 6.3). On the plane,
|/|s = |/| = length o f l. On the sphere, |/|s = (shortest) distance betw een the
endpoints o f /.
b. (Sphere) The input crank will be able to swing opposite C
(where the input angle is n = 180 degrees) only i f \a + g|s
< b + h. I f \a + g\s > b + h, there will be a maximum input angle,
6Lx, satisfying
cos(/z + b) – cos a cos g
COS Ts max — • • fsin a sin g
where the edge lengths are measured by the radian measure o f
the angle they subtend at the center o f the sphere.
What happens if\a + g\% = b + h i

P roblem 2 1 .1 F o u r-B ar Linkages 3 0 5
W e can now finish this problem working with both the plane and the
sphere at the same time.
c. (Plane or Sphere) Similarly, whenever \b – h \ > \g – a \ there will
be a minimum input angle, 8mm, satisfying
The minimum angle is actualized when either B-A-C is straight
orA-C -B is straight, as in Figure 21.5.
Thus we have four types o f input cranks:
1. A crank i f the link OA can freely rotate com pletely around O. In
this case, b + h > |a + g|s and \a -g \ > \b~h\.
2. A 0-rocker i f there is a maximum input angle but the link OA can
rotate freely past 8= 0. Then b + h < \a + g\s and \a -g \ > \b – h\.
3. A Tv-rocker i f there is a m inimum input angle but the link OA can
rotate freely past 8— n. Then b + h > \a + gjs and \ a – g \ < \ b - h\. 4. A rocker i f there is a m aximum input angle and minimum input angle. In this case, b + h < \a + gjs and \a -g \ < \b -h \. The analysis o f the output crank is exactly symmetric to above with the lengths a and b interchanged. In particular, d. I f \b + gjs > a + h, then there is a maximum output angle (zW* that
cos \ b – h \ – cos a cos g
sin a s in g (sphere).
C
Figure 21.5 Minimum input angles
satisfies

3 0 6 Chapter 21 Mechanisms
(g2 + b2) – (h + a)2
cos Omax = ———- ^ ———-(plane)
. cos(h + a ) – c o s b c o s g , ,
c o s — ( s P h e r e ) ‘
I f |o – h\ > \g — b\, there is a minimum output angle that
satisfies
COS 0min
g 2 + b2 – \a – h \2
2 bg
(plane),
COS 0 njjn
cos \ a – h \ – cos b cos g
sin b sin g (sphere).
Thus we have four types o f output cranks:
1. A crank if the link CB can freely rotate completely around C. In
this case, a + h > \b + g|s and \b -g \ > \a -h \.
2. A 0-rocker if there is a maximum output angle but link CB can
rotate freely past =0. Then a + h < \b + g |s and |Z>-g| > \a -h \.
3. A jif-rocker if there is a minimum output angle but the link CB can
rotate freely past

\b + g | s and \b – g \< \a - h \. 4. A rocker if there is a maximum output angle and m inimum output angle. In this case, a + h <\b + g\% and \ b - g \ < \ a - h\. e. Putting these together we get eight types o f four-bar linkages. Make a model o f each type using cardboard strips. 1. A double crank in which the input and output links are cranks. b + h > |o + g|s, \a -g \ > \b – h \,a + h > |6 + g |s, and \b -g \ > \a -h \.
2. A crank-rocker if the input link is a crank and the output link is a
rocker, b + h > |o + g|s, \a -g \ > Ib -h \, a + h < \b + g |s, and \b -g \ < \ a - h \ . 3. A rocker-crank if the input link is a rocker and the output link is a crank, b + h < \a + g |s, \a -g \ < \ b - h \ , a + h > \b + g|s, and | b – g | >
\a -h \.
4. A rocker-rocker if both the input and the output link are rockers.
b + h < \a + g|s, | \b – h \,a + h < \b + g |s, and \b -g \ > \a -h \.

P roblem 21.1 Four-Bar Linkages 3 0 7
6. A On double rocker i f the input angle has a maximum and no
m inimum but the output angle has a m inimum but no maximum.
b + h < |a + g |s, \a -g \ > \b—h\, a + h > |6 + g |s, and \b -g \ < \a -h \. 1. A nO double rocker i f the input angle has a minimum and no maximum but the output angle has a m aximum but no minimum. b + h > \a + g |s, \ a – g \ < \ b - h \ ,a + h < \ b + g|s, and \ b - g \ > \ a – h\.
8. A tm double rocker i f the input and output angles both have
minimums but no maximums and move freely on the ends o f OC.
b + h > |a + g|s, \a -g \ < \b - h \ ,a + h > |6 + g|s, and |6 – g | < \a -h \. Check that the other eight combinations (a 0 or tc rocker combined with a crank or rocker) are not possible. This can be done either analytically (using the inequalities) or geom etrically (by noting symmetries). All the other four-bar linkages are the cases when one or m ore o f the inequalities become equalities, in each o f these cases the linkage can be folded. That is, the linkage has a configuration in which all the links line up with OC. Some four-bar linkages can be folded in m ore than one way; for example, the linkage with a = h = b = g can be folded in three different ways ( Try it!). P r o b l e m 2 1 .2 U n i v e r s a l J o i n t Almost all vehicles with an engine in front that drives the rear wheels have a drive shaft that transm its the pow er from the engine to the rear axle. It is important that the drive shaft be able to bend as the vehicle goes over bumps. The usual way to accom plish this “bending” is to put in the drive shaft a universal jo in t (also known as H ooke’s jo in t or C ardan’s joint). See Figure 21.6. To see this m odel in motion, go to the Experiencing Geometry W eb page, which will link to VR-movies on [M E: KM ODDL], In 1676, Robert Hooke (1635-1703) published a paper on an optical instrument that could be used to study the sun safely. In order to track the sun across the sky, the device featured a control handle fitted with a new type o f jo in t that allowed tw isting motion in one shaft to be passed on to another, no m atter how the two shafts were orientated. Hooke gave this the name “universal jo in t.” This jo in t was earlier suggested by Leonardo da Vinci and also is attributed to Girolamo Cardano. Therefore, on the Euorpean continent it got name “C ardan’s jo in t,” but in Britain the name o f “H ooke’s jo in t” was used. 308 Chapter 21 Mechanisms Figure 21.6 Universal joint a. The universal jo in t can be considered to be a spherical four-bar linkage with a = b = h = n il. The fix e d (grounded) link g is the angle between the input and output shafts that can be adjusted in the range jt/2 < g < 7t. The links a, h, and g are not actually links; however, the constraints o f the mechanism operate as if they were spherical links. See Figure 21.7. b. For what lengths o f the fix e d link (angles between the input and output shafts) is the universal jo in t a double crank (see 2 1 .l e ) ? This is o f utm ost importance in its automotive use because as an autom o­ bile goes over bumps the angle g between the shafts will change. We now look at the relationship betw een the rotation o f the input shaft to the rotation o f out shaft. One o f the problems with the universal jo in t is that, though one rotation o f the input shaft results in one rotation o f the output shaft, the rotations are not in sync during the revolution. c. Check that Figure 21.7 is correct. In particular, i. A is the pole fo r the great circle (dashed in the figure) passing through O and B; and the great circle arcs h and a intersect this great circle at right angles. ii. Likewise, B is the pole fo r the great circle passing through A and C; and the great circle arcs h and b also intersect this great circle at right angles. P roblem 21.2 Universal Joint 3 09 iii. The arc h must bisect the lune determined by these two dashed great circles. The angle o f this lune at P must be n il as marked. iv. I f we use a to label the angle at A and f t to label the angle at B, as in the figure, then radian measure o f the arc OB is a, and the radian measure o f the arc A C is p. The angle 6 measures the rotation o f the input shaft and the angle y/ measures the rotation o f the output shaft. Note that, when 6 = 0 then y/ = 0, also. As 0 changes in the positive counterclockwise direction, v|/ will be changing in the negative clockwise direction; thus, when 0 is positive, y/ will be negative and we will need to use - y/ when denoting the angle in the triangle OCA. d. The input and output angles satisfy ton ir V ) = ^ g . In practice, g is near n and - c o s g is positive. Note that when g = n, tan(-y/) = tan 6 and thus the two shafts turn in unison. Hint: A pply the Spherical Pythagorean Theorem (Theorem 20.2a) to the right triangle OPC, and the Law o f Sines (Problem 20.3b) to triangles OCA and OCB. 3 1 0 Chapter 21 Mechanisms P r o b l e m 21.3 R e u l e a u x T r i a n g l e a n d C o n s t a n t W i d t h C u r v e s What is this triangle? If an enorm ously heavy object has to be moved from one spot to another, it may not be practical to move it on wheels. Instead the object is placed on a flat platform that in turn rests on cylin­ drical rollers. As the platform is pushed forward, the rollers left behind are picked up and put down in front. An object moved this way over flat horizontal surface does not bob up and down as it rolls along. The reason is that cylindrical rollers have a circular cross section, and a circle is closed curve with constant width. W hat does this mean? I f a closed convex curve is placed betw een two parallel lines and the lines are moved together until they touch the curve, the distance betw een the parallel lines is the curve's width in one direction. Because a circle has the same width in all directions, it can be rotated between two parallel lines without altering the distance between the lines. Reuleaux model Wankel engine Figure 21.8 Mechanisms utilizing a Reuleaux triangle Is the circle the only curve with constant width? Actually there are infinitely m any such curves. The simplest noncircular such curve is named the Reuleaux triangle. M athem aticians knew it earlier (some authors refer to Leonard Euler in 18th century), but Reuleaux was the first to demonstrate and use o f its constant width properties. In Figure 21.8 there is a Reuleaux model using the Reuleaux triangle and an image P roblem 2 1 .3 Reuleaux Triangle and Constant Width Curves 311 o f inside o f a W ankel engine (sim ilar to that used in some M azda automobiles) showing the rotor in the shape o f a Reuleaux triangle. A Reuleaux triangle can be constructed starting with an equilateral triangle o f side s and then replacing each side by a circular arc with the other two original sides as radii. See Figure 21.9. a. Why will the Reuleaux triangle make a convenient roller but not a convenient wheel? Figure 21.10 Smoothed Reuleaux triangle The Reuleaux triangle has com ers, but if you w ant to smooth out the com ers you can extend a Reuleaux triangle a uniform distance d on every side as in Figure 21.10. Then you can b. Show that the resulting curve has constant width s + 2d. 312 Chapter 21 Mechanisms Figure 21.11 Constant width coins Other symmetrical curves o f constant width result if you start with a regular pentagon (or any regular polygon with an odd num ber o f sides) and follow sim ilar procedures. See Figure 21.11 for examples o f British coins that are the shape o f a constant width curve based on the heptagon. What advantages would these British coins have due to their shape? Figure 21.12 Creating irregular curves of constant width P roblem 2 1 .3 Reuleaux Triangle and Constant Width Curves 313 But here is one really surprising method o f constructing curves with constant width: Draw as m any straight lines as you please such that each line intersects all the others. See Figure 21.12. On one o f the lines start with a point sufficiently far away from the intersections. Now draw an arc from this point to an adjacent line, with the compass point at the intersection o f the two lines. Then, starting from the end o f this arc, draw another arc connecting to the next line with the compass point at the intersection o f these two lines. Proceed in this maimer from one line to the next, as indicated in Figure 21.12. If you do it carefully, the curve will close and will have a constant width. (You can try to prove it! It is not difficult at all.) The curves drawn in this way may have arcs o f as m any different circles as you wish. The example in Figure 21.12 shows steps in drawing such curves, but you will really enjoy making your own. A fter you have done that, you can make several more copies o f it and check that your wheels really roll! c. Prove that the procedure in the last paragraph produces a curve o f constant width as long as all the intersection points are inside the curve. Can you specify how fa r out you have to start in order fo r this to happen? The Reuleaux triangle has been used to make a drill bit that will drill a (alm ost) square hole. See Figure 21.13. Figure 21.13 Reuleaux triangle in a square d. A Reuleaux triangle o f width s can be turned completely around within a square o f side s in such a way that, at each time o f the motion, the Reuleaux triangle will be tangent simultaneously to 3 1 4 Chapter 21 Mechanisms all fo u r sides o f the square. Describe the small spaces in the corners o f the square that the Reuleaux triangle will not reach. The interested reader may find more information and references about Reuleaux triangles in [M E: KM ODDL]; go to the Experiencing Geometry W eb page for direct links. To sharpen your interest, we list some further properties o f Reuleaux triangles: 1. The inscribed and circumscribed circles o f an arbitrary curve o f constant width h are concentric and the sum o f their radii is equal to h. 2. Among curves with constant width h, the circle bounds the region o f greatest area and the Reuleaux triangle bounds the region o f least area. 3. Any curve o f constant width h has perim eter equal nh. 4. The com ers o f a Reuleaux triangle are the sharpest possible on a curve with constant width. 5. The circle is the only curve o f constant width with central symme­ try (see page 15). 6. For every point on a curve o f constant width there exists another point with the distance betw een the two equaling the width o f the curve, and the line joining these two points is perpendicular to the support lines at both points. {Support lines are lines that touch the curve and the curve lies totally on one side o f the line.) 7. There is at least one supporting line through every point o f a curve o f constant width. 8. If a circle has three (or more) points in common with a curve o f constant breadth h, then the length o f the radius o f the circle is at most h. I n v o l u t e s Reuleaux used the geometric idea o f involutes in the design o f several m echanisms including a pump (the middle image in Figure 21.2) and gear design (Figure 21.15). Before we discuss these we m ust describe the geometry o f the involute. Focus on one o f the four arms in the pump in Figure 21.14. Involutes 315 Figure 21.14 Involute arm Imagine that the outlined circle in picture is a spool that has a black thread with white edging wrapped around it in such a way that when fully w ound the end o f the thread is at the point A. Now imagine unwinding the thread, keeping the spool fixed, and keeping the thread pulled taut. The end o f the thread traces the outer edge o f the spiral arm. This curve is called the involute o f a circle (in this case, the outlined circle). Instead o f keeping the spool fixed and unwinding the thread, we could rotate the spool and pull the thread taut in the same direction. It is this latter view that we will use in analyzing the spiral pump. Examine the picture o f the spiral pum p in Figure 21.15. Now imagine that there is thread rolled around the left spool and then p u llt i taut and wrapped around the right spool as indicated in the picture. Place a small circle on the thread at the place that two arms touch each other. Now, instead o f unwrapping the thread, we will turn the two spools at the same rate, always keeping the thread taut. Since the spiral arms are in the shape o f an involute, the small circle will follow the outer edge o f both spiral arms. Thus, as the spools rotate, the spiral arms will stay in contact. 3 16 Chapter 21 Mechanisms Figure 21.15 Spiral pump with thread and small circle The two rotors in the spiral pump can be thought o f as gears with two teeth. The same discussion above illustrates why it is advantageous for gear teeth, in general, to be in the shape o f an involute curve so that the gear teeth stay in contact throughout uniform turning o f the gears. If the axes o f two engaging gears are parallel, then the involute is a planar involute, as described above. However, if the axes are not parallel, then the gears can be considered as on a sphere whose center is at the inter­ section o f the two axes. An involute o f a circle (on plane or sphere) can be described either as unrolling a taut string from the circle or as rolling o f a straight line along the circle. Figure 21.16 Spherical involute curves Involutes 317 Reuleaux designed several models to illustrate spherical involutes by rolling a straight line (great circle) on a small circle. See Figure 21.16 for photos o f these models, but more understanding will be possible if you experience the VR-movies on [ME: KM ODDL], see direct links on the Experiencing Geometry W eb page. L i n k a g e s I n t e r a c t w i t h M a t h e m a t i c s After Descartes and others used linkages to draw curves, it was natural for m athematicians to ask the question o f what curves could be drawn by linkages. In [ME: Kempe 1876], A. B. Kempe gave a p roof that any algebraic curve may be described by a linkage. The idea for K em pe’s proof, as discussed in [ME: Artobolevski], is as follows: Consider the algebraic curve fiyc, y) = 0, which can be expressed in the form YiAm„xmy"= 0, where the coefficients Amn are constant. Thus the generation o f the curve reduces to a series o f mathematical operations. K em pe’s idea was that each o f these mathematical operations can be fulfilled by individual linkages, which can then be linked together into a kinematic chain o f linkages. Linkages needed for this are as follows: a. Linkage for translating a point along a given straight line (for example, the Peaucellier-Lipkin linkage in Problem 16.3); b. Linkage for projecting a given point onto a given line; c. Linkage that cuts o ff on one axis a segment equal to a given segment on the other axis; d. Linkage that determines a straight line that passes through a given point and is parallel to a given line; e. Linkage that, given two segments r and 5 on one line and one segment t on another line, will obtain a second segment u on the second line such that r/s = t/u (M ultiplier); f. Linkage for the addition o f two given segments (Adder). For more details on this proof, see [ME: Artobolevski], For a different p ro o f with more explicit pictures o f the constructions o f linkages and their com binations, see [ME: Yates], Section 11. In a different direction, there was for many years an open question that appeared in robotics, topology, discrete geometry, and pattern recog­ nition: 318 Chapter 21 Mechanisms Given a linear chain o f links (each one connected to the next to form a polygonal path without self-intersections) or a cycle o f links (a linear chain with the first and last links joined), then is it possible to fin d a motion o f the chain or cycle during which there continue to be no self-intersections and, at the end o f the motion, the chain form s a straight line and the cycle form s a convex polygon. See Figure 21.17. Figure 21.17 Straightening and convexifying linkages In 2003, Robert Connelly, Erik D. Demaine, and Gunter Rote published a paper, “ Straightening Polygonal Arcs and Convexifying Polygonal Cycles” [M E: Connelly], in which they solved this problem positively and, in addition, proved that their motion is piecewise differentiable, does not decrease the distance between any pair o f vertices, and preserves any symmetry present in the initial configuration. Chapter 22 3 -S pheres and Hyperbolic 3 -S paces Let us, then, make a mental picture of our universe: ... as far as possible, a complete unity so that whatever comes into view, say the outer orb of the heavens, shall bring immedi­ ately with it the vision, on the one plane, of the sun and of all the stars with earth and sea and all living things as if exhibited upon a transparent globe. Bring this vision actually before your sight, so that there shall be in your mind the gleaming representation of a sphere, a picture holding all the things of the universe .... Keep this sphere before you, and from it imagine another, a sphere stripped of magnitude and of spatial differences; cast out your inborn sense of Matter, taking care not merely to attenuate it: call on God, maker of the sphere whose image you now hold, and pray Him to enter. And may He come bringing His own Universe .... — Plotinus, The Enneads, V.8.9, Burdette, NY: Larson, 1992 In this chapter you will explore hyperbolic 3-space and the 3-dimen- sional sphere that extrinsically sits in 4-space. But intrinsically, if we zoom in on a point in a 3-sphere or a hyperbolic 3-space, then locally the experience o f the space will become indistinguishable from an intrinsic and local experience o f Euclidean 3-space. This is also our hum an expe­ rience in our physical universe. We will study these 3-dimensional spaces both because they are possible geometries for our physical uni­ verse and in order to see that these geometries are closely related to their 2-dimensional versions. 3 19 3 2 0 Chapter 22 3-Spheres and Hyperbolic 3-Spaces The starred problem s will require some experience with abstract vector spaces but can be skipped as long as you experience Problem 22.1 and at least assume the results o f Problem 22.6. Try to imagine the possibility o f our physical universe being a 3-sphere in 4-space. It is the same kind o f imagination a 2-dimensional (2-D) being would need in order to imagine that it was on 2-sphere (ordi­ nary sphere) in 3-space. In Problem 18.6 we thought about how a 2-D bug could determine (intrinsically) that it was on a 2-sphere. Now, we want to explore how the bug could imagine the 2-sphere in 3-space, that is, how could the bug imagine an extrinsic view o f the 2-sphere in 3-space. In Problems 22.2 and 22.3 we will use linear algebra to help us talk about and analyze the 3-sphere in 4-space, but this will not solve the problem o f imagining the 3-sphere in 4-space. P roblem 22.1 Explain 2 -S phere in 3 -S pace t o a 2 -D imensional B u g How would you explain a 2-sphere in 2-space to a 2-D bug living in a {Euclidean) plane or on a 2-sphere so large that it appears fla t to the 2-D bug? Figure 22.1. Figure 22.1 2-D bug on a large 2-sphere P roblem 22.1 Explain 3-Space to 2-D Bug 321 S uggestions This bug’s 2-dimensional experience is very much like the experience o f an insect called a water strider that we talked about in Chapter 2. A water strider walks on the surface o f a pond and has a very 2-dim en­ sional perception o f the universe around it. To the w ater strider, there is no up or down; its whole universe consists o f the surface o f the water. Similarly, for the 2-D bug there is no front or back; the entire universe is the 2-dimensional plane. Living in a 2-D world, the bug can easily understand any notions in 2-space, including plane, angle, distance, perpendicular, circle, and so forth. You can assume the bug is smart and has been in geometry class. A bug living in a 2-D w orld cannot directly experience three dim en­ sions, ju st as we are unable to directly experience four dimensions. Yet, with some help from you, the 2-D bug can begin to imagine three dim en­ sions ju st as we can imagine four dimensions. One goal o f this problem is to try to gain a better understanding o f what our experience in our imagination o f 4-space might be. Think about what four dimensions might be like, and you may have ideas about the kinds o f questions the 2-D bug will have about three dimensions. You may know some answers, as well. The problem is finding a way to talk about them. Be creative! One important thing to keep in mind is that it is possible to have images o f things we cannot see. For example, when we look at a sphere, we can see only roughly h a lf o f it, but we can and do have an image o f the entire sphere in our minds. We even have an image o f the inside o f the sphere, but it is impossible to actually see the entire inside or outside o f the sphere all at once. Another example — sit in your room, close your eyes, and try to imagine the entire room. It is likely that you will have an image o f the entire room, even though you can never see it all at once. W ithout such images o f the whole room it would be difficult to m aneuver around the room. The same goes for your image o f the whole o f the chair you are sitting on or this book you are reading. Assum e that the 2-D bug also has images o f things that cannot be seen in their entirety. For example, the 2-D bug may have an image o f a circle. W ithin a 2-dimensional world, the entire circle cannot be seen all at once; the 2-D bug can only see approxim ately h a lf o f the outside o f the circle at a time and cannot see the inside at all unless the circle is broken. See Figure 22.2. 322 Chapter 22 3-Spheres and Hyperbolic 3-Spaces Figure 22.2 The 2-D bug sees a circle However, from our position in 3-space we can see the entire circle including its inside. Carrying the distinction between what we can see and w hat we can imagine one step further, the 2-D bug cannot see the entire circle but can imagine in the m ind the whole circle including in­ side and out. Thus, the 2-D bug can only imagine what we, from three dimensions, can directly see. So the 2-D bug’s image o f the entire circle is as if it were being viewed from the third dimension. It makes sense, then, that the image o f the entire sphere that we have in our minds is a 4-D view o f it, as if we were viewing it from the fourth dimension. When we talk about the fourth dim ension here, we are not talking about time, which is often considered the fourth dimension. Here, we are talking about a fourth spatial dimension. A fuller description o f our universe would require the addition o f a time dim ension onto whatever spatial dimensions one is considering. Try to come up with ways to help the 2-D bug imagine a 2-sphere in 3-space. It may help to think o f intersecting planes rotating with respect to each other: How will a 2-D bug in one o f the planes experience it? Draw on the b ug’s experience living in two dimensions as well as some o f your own experiences and attempts to imagine four dimensions. * * W h a t I s 4 - S p a c e ? V e c t o r S p a c e s a n d B a s e s We could think o f 4-space as R4: Let R 4 be the collection o f 4-tuples o f real numbers (x, y, z, w) with the distance function (metric) d{ (a, b, c, d), ( e , f g , h ) ) = J ( a - e ) 2 + (b - f ) 2 + ( c - g ) 2 + ( d - h ) 2 and dot product (a ,b ,c ,d ) • ( e , f g, h) = ae + b f+ c g + de. *What Is 4-Space? Vector Spaces and Bases 3 2 3 But this would be awkward sometimes because it fixes a given coordi­ nate system; and we will find it geometrically useful to be able to change coordinates (or basis) to fit a particular problem. We will find it more powerful to have a description o f space without given coordinates. So instead o f R4 we will think o f 4-space as an (abstract) vector space. A vector space has a point we call the origin O. Then designate every point P by a directed straight line segment v from O to P. These directed line segments are, o f course, what we call vectors. W e consider O itself also as a vector. If you have studied vector spaces only algebrai­ cally, then it may be difficult to see the geometric content. We assume that we can add vectors by geometrically defining u + v to be the diago­ nal o f the parallelogram determ ined by v and u, as in Figure 22.3. In order for this definition to make sense we need to assume that we can form the parallelogram s. It is also possible to not identify an origin and create what is called an affine space. You can read about geometric affine spaces in [DG: Henderson], Appendix A, or [DG: Dodson & Poston], Figure 22.3 Adding vectors We assume that addition o f vectors satisfies, for all u and v in V, 1. u + v is in V, and O + u = u; 2. u + v = v + u, and u + (v + w) = (u + v) + w; and 3. there is a - v in V such that v + (-v ) = O. Geometrically, we choose a segment whose length we designate as the unit length 1. W ith this unit length we can determine the length o f any vector v — we denote its length by |v|. For r a real num ber and v a vector, we define geom etrically the multiplication by scalars rv as the vector with length r|v| and lying on the straight line determined by v. 3 2 4 Chapter 22 3-Spheres and Hyperbolic 3-Spaces This m ultiplication by scalars satisfies, for all u and v in V and all reals r and s, 4. ru is in V, and Ou = O , and lu = u; 5. (r + s)u = ru + su, and r(u + v) = ru + rv; and 6. (rs)u = r(su). We also assume that we can find the angle 0, 0 < 0 < n, between any two vectors and we define the Euclidean inner product (sometimes called the standard inner product) o f two vectors to be: (v, w) = |v| |w| cos 0, w here 0 is the angle betw een v and w. Note that (v, w) is negative when 0 > till. We can check that this inner
product satisfies the following properties, for all u, v, w in V and all
reals r,
7. 0.
In a more abstract setting we could simply define a vector space
with inner product to be a set with {an origin (O), addition o f vectors
(+), m ultiplication by scalars, and real-valued function ( ( , ) ) } that satis­
fies the 10 properties above. Then we define
M = 7 and cos 0 = ^ j , 0 < 0 < n. Note that (v, w) = 0 implies that v and w are perpendicular. A subspace o f a vector space V is a subset Ucz V if ru + s \ is in U, for every pair o f vectors u, v in U and all real num bers r, s. If {ui, U2, ..., u„} is a finite collection o f vectors from V, then we call the span o f {ui, u2, ..., u„}, denoted by sp{ui, u2, ..., u„}, the smallest sub­ space o f V containing each o f Ui, u2, ..., u„. We say that {ui, u2, ..., u„} are linearly independent if, for each i, u, is not in sp{ui, U2, ...Um , U/+1, ..., u„}. I f {ui, u2, ..., u„} are linearly independent, then we say that *What Is 4-Space? Vector Spaces and Bases 325 U = sp{ui, U2, u „ } is an n-dimensional subspace and that {ui, u 2, u„} is a basis for U. In particular, we say V is m -d im en sio n al if V= sp { u b u2, u n} for some collection {ui, u2, um} o f m linearly independent vectors in V. If, in addition, (u, ,u,) = 0, tej, and (u, ,u,) = 1 for all 0< i , j 0} and call it the
upper half-space.
In Chapter 17 we started with the annular hyperbolic plane and then
defined a coordinate map z’ R2+ —> H 2. Now we do not have an isometric
model H3 but, instead, we have to start with the upper half-space and use
z to define H3. Recall that z: R2+ —> H2 has distortion p/b at the point
(a,b) in R2+, where p is the radius o f the annuli. As we saw in Chapter
17, we can study the geometry o f the hyperbolic plane H2 by considering
it to be the upper half-plane with angles as they are in R2+ and distances
distorted in R2+ by p/b at the point (a, b). So now we use this idea to
define hyperbolic 3-space H3.
D e f in it io n : Define the upper-half-space model o f hyperbolic
space H3 to be the upper half-space R3+ with angles as they are
in R3+ and with distances distorted by p/c at the point (a, b, c).
We call p the radius o f fP.
We define a great semicircle in H 3 to be the intersection o f
H3 with any circle that is in a plane perpendicular to the boun­
dary o f R3+ and whose center is in the boundary o f R3+ or the

* P roblem 2 2 .3 Hyperbolic 3-Space, Upper Half-Space 329
intersection o f R 3+ with any line perpendicular to the boundary
o f R 3+. The boundary o f R 3+ are those points in R 3 with z = 0.
We define a great hemisphere in R 3+ to be the intersection
o f R 3+ with a sphere whose center is on the boundary o f R 3+
in R 3 or the intersection o f R 3+ with any plane that is perpen­
dicular to the boundary o f R 3+ in R 3. See Figure 22.5.
Figure 22.5 Upper-half-space model of H3
a. Show that inversion through a great hemisphere in R 3+ has
distortion 1 in H 3 and, thus, is an isometry in H 3 and can be
called a (hyperbolic) reflection through the great hemisphere.
Look back at Problem 17.3. Note that any inversion in a sphere when
restricted to a plane containing the center o f the sphere is an inversion o f
the plane in the circle formed by the intersection o f the plane and the
sphere.
b. Show that, given a great semicircle [or great hemisphere], there
is a hyperbolic reflection (inversion through a great hemisphere)
that takes the great semicircle [hemisphere] to a vertical half­
line [half-plane] in the upper half-space.
Look back at Problem s 16.2b and 17.3b. Note that any inversion in a
sphere when restricted to a plane containing the center o f the sphere is
an inversion o f the plane in the circle formed by the intersection o f the
plane and the sphere.

330 Chapter 22 3-Spheres and Hyperbolic 3-Spaces
Any vertical half-plane is precisely an upper-half-plane model o f
H 2. Thus we conclude that each great hemisphere in H 3 has the geometry
o f H 2.
*c. Show that every great semicircle T has the symmetries in H 3 o f
reflection through any great hemisphere perpendicular to Y and
rotation about T through any angle. Because these are principal
symmetries o f a straight line in 3-space it makes sense to call
these great semicircles geodesics in H 3.
For the reflection, look in the half-plane containing T and in this plane
use the arguments o f Problem 16.2a. For the rotation, look at great hem i­
spheres containing T and restrict your attention to their intersections
with the vertical half-plane that passes through the center o f T and is
perpendicular to Y ; then refer back to Problem 5.4a.
d. I f two great semicircles in H 3 intersect, then they lie in the same
great hemisphere.
Use part b to assume that one o f the great semicircles is a vertical
half-line.
* P r o b l e m 22.4 D i s j o i n t E q u i d i s t a n t G r e a t C i r c l e s
a. Show that there are two great circles in S3 such that every point
on one is a distance o f one-fourth o f a great circle away from
every point on the other and vice versa.
Is there anything analogous to this in H 3 or in ordinary 3-space?
Why?
S u g g e s t i o n s
This problem is especially interesting because there is no equivalent
theorem on the 2-sphere; we know that on the 2-sphere, all great circles
intersect, so they can’t be everywhere equidistant. The closest analogy
on the 2-sphere is that a pole is everywhere equidistant from the equator.
W hen we go up to the next dimension, this pole “expands” to a great
circle such that every point on this great circle is everywhere equidistant
from the equator. W hile this may seem mind-boggling, there are ways o f
seeing what is happening.

*P r o b l e m 2 2 .4 Disjoint Equidistant Great Circles 331
An important difference created by adding the fourth dim ension lies
in the orthogonal complement to a plane. In 3-space, the orthogonal
complement o f a plane is a line that passes through a given point. This
means that for any given point on the plane (the origin is always a
convenient point), there is exactly one line that is perpendicular to the
plane at that point. Now w hat happens when you add the fourth dim en­
sion? In 4-space, the orthogonal complement to a plane is a plane. This
means that every line in one plane is perpendicular to every line in the
other plane. To understand how this is possible, think about how it
works in 3-space and refer to Figure 22.6, where we are depicting R 4 the
4-space with x, y, z, w coordinates. N ow look at the xy-plane and
the zw-plane. W hat do you notice? W hy is every line through the center
in one o f these planes perpendicular to every line through the center in
the other?
AZ A
Figure 22.6 Orthogonal planes
K now ing this, look at the two great circles in terms o f the planes in
which they lie, and look at the relationships between these two planes,
that is, where and how they intersect. Also, try to understand how great
circles can be everywhere equidistant.
I f we rotate along a great circle on a 2-sphere, all points o f the
sphere will move except for the two opposite poles o f the great circle. If
you rotate along a great circle on a 3-sphere, then the whole 3-sphere
will move except for those points that are a quarter great circle away
from the rotating great circle. Therefore, i f you rotate along one o f the
two great circles you found above, the other great circle will be left
fixed. But now rotate the 3-sphere sim ultaneously along both great
circles at the same speed. Now every point is m oved and is m oved along
a great circle!

3 3 2 Chapter 22 3-Spheres and Hyperbolic 3-Spaces
b. Write an equation fo r this rotation (in x, y, z, w coordinates) and
check that each point o f the 3-sphere is moved at the same speed
along some great circle. Show that all o f the great circles
obtained by this rotation are equidistant from each other (in the
sense that the perpendicular distance from every point on one
great circle to another o f the great circles is a constant).
These great circles are traditionally called Clifford parallels, named
after W illiam Clifford (1845-1879, English). See [DG: Thurston], pp
103-04, and [DG: Penrose] for readable discussions o f Clifford
parallels.
* P r o b l e m 22.5 H y p e r b o l i c a n d S p h e r i c a l
S y m m e t r i e s
We are now ready to see that the symmetries o f great circles and great
2-spheres in a 3-sphere [and great semicircles and great hem ispheres in a
hyperbolic 3-space] are the same as the symmetries o f straight lines and
(flat) planes in 3-space. If g is a great circle in the 3-sphere, then let
g1 denote the great circle (from Problem 22.4) every point o f which
is nl2 from every point o f g.
a. Check the entries in the table (Figure 22.7), which gives a sum­
mary o f various symmetries o f lines, great circles, and great
semicircles and o f {flat) planes, great 2-spheres, and great
hemispheres.
D e f in it io n : A surface in a 3-sphere or in a hyperbolic 3-space
is called totally geodesic if, for any every pair o f points on the
surface, there is a geodesic (with respect to S3 or H 3) that joins
the two points and also lies entirely in the surface.
b. Show that a great 2-sphere in S3 (with radius r) is a totally
geodesic surface and is itself a 2-sphere o f the same radius r.

♦ P roblem 2 2 .5 Hyperbolic and Spherical Symmetries 3 3 3
c. Show that a great hemisphere is a totally geodesic surface in H 3
(with radius r) and is isometric to a hyperbolic plane with the
same radius r.
In the upper-half-space m odel there is a hyperbolic reflection that takes
every great hemisphere to a plane perpendicular to the boundary. (See
Problem 22.3.)
Symmetries
of…
Reflection
through…
Reflection
through…
Half-turn
about…
Rotation
about…
Translation
along…
line / c E2 1 line _L / point in / NA /
great circle
gczS2 g
great circle
-L g
point/pair in g poles of g g
great
semicircle
g c H 2
g
great
semicircle
-L g
point/pair in g NA g
line / c E 3 plane c / plane L l line _L / intersecting / / /
great circle
f c S 3
great sphere
^ g
great sphere
-Lg
great circle
-Lg
intersecting g
g g
great
semicircle
g c H3
great
hemisphere
=5g
great
hemisphere
-Lg
great
semicircle
-Lg
intersecting g
g g
plane
P c E3
P plane -L P line in P line _L P line c P
great sphere
G c S 3
G
great sphere
L G
great circle
in G
great circle
1 G
great circle
c G
great
hemisphere
G c S 3
G
great
hemisphere
L G
great
semicircle
in G
great
semicircle
L G
great
semicircle
c G
Figure 22.7 Symmetries in Euclidean, spherical, and hyperbolic spaces

3 34 C h ap ter 22 3 -S p h eres an d H y p erb o lic 3-S p aces
Problem 22.6 Triangles i n 3-Dimensional Spaces
Show that i f A, B, C are three points in S3 [or in H3] that do not
all lie on the same geodesic, then there is a unique great
2-sphere [hemisphere], G 2, containing A, B, C.
Thus, we can define AABC as the (small) triangle in G 2 with
vertices A, B, C. With this definition, triangles in S3 [or in H3]
have all the properties that we have been studying o f small
triangles on a sphere [or triangles in a hyperbolic plane].
S u g g e s t i o n s
Think back to the suggestions in Problems 22.2 and 22.3 — they will
help you here, as well. Take two o f the points, A and B, and show that
they lie on a unique plane through the center, O, o f the 3-sphere [or a
unique plane perpendicular to the boundary o f R3+]. Then show that
there is a unique (shortest) geodesic in this plane. See Figure 22.8.
Think o f A, B, and C as defining three intersecting great circles [or
semicircles]. On a 3-sphere, look at the planes in which these great
circles lie and where the two planes lie in relation to one another. In
hyperbolic 3-space, use a hyperbolic reflection to send one o f the great
semicircles to a vertical line.
Be sure that you show that the great 2-sphere (hemisphere) con­
taining A, B, C is unique.

Chapter 23
POLYHEDRA
… if four equilateral triangles are put together, three of
their plane angles meet to form a single solid angle,… When
four such angles have been formed the result is the simplest
solid figure…
The second figure is composed of … eight equilateral
triangles, which yield a single solid angle from four planes.
The formation of six such solid angles completes the second
figure.
The third figure … has twelve solid angles, each
bounded by five equilateral triangles, and twenty faces, each of
which is an equilateral triangle.
… Six squares fitted together complete eight solid
angles, each composed by three plane right angles. The figure
of the resulting body is the cube …
There is still remained a fifth construction, which the
god used for arranging the constellations on the whole heaven.
— Plato, Timaeus, 54e-55c [AT: Plato]
Definitions and Terminology
[The text in the brackets applies to polyhedra on a 3-sphere or a hyper­
bolic 3-space.] A tetrahedron, AABCD, in 3-space [in a 3-sphere or a
hyperbolic 3-space] is determined by any four points, A, B, C, D, called
its vertices, such that all four points do not lie on the same plane [great
2-sphere, great hemisphere] and no three o f the points lie on the same
line [geodesic]. The faces o f the tetrahedron are the four [small]
triangles AABC, ABCD, ACDA, ADAB. The edges o f the tetrahedron are
the six line [geodesic] segments AB, AC, AD, BC, BD, CD. The interior
o f the tetrahedron is the [smallest] 3-dim ensional region that it bounds.
335

3 3 6 C h a p te r 23 P o ly h ed ra
Tetrahedra are to three dimensions as triangles are to two dimen­
sions. Every polyhedron can be dissected into tetrahedra, but the proofs
are considerably more difficult than the ones from Problem 7.5b, and in
the discussion to Problem 7.5b there is a polyhedron that is impossible
to dissect into tetrahedra without adding extra vertices. There are num er­
ous congruence theorems for tetrahedra, analogous to the congruence
theorems for triangles. We say two tetrahedra are congruent if one can
be transform ed by an isometry o f 3-space to coincide with the other. All
o f the problem s below apply to tetrahedra in Euclidean 3-space or a
3-sphere or a hyperbolic 3-space.
The dihedral angle, ZAB, at the edge AB is the angle form ed at AB
by A ABC and AABD. The dihedral angle is m easured by intersecting it
with a plane that is perpendicular to AB at a point betw een A and B. The
solid angle at A, Z A , is that portion o f the interior o f the tetrahedron “at”
the vertex A. See Figure 23.1.
You may find it helpful with these problem s to construct some
tetrahedra out o f cardboard.
Problem 23.1 Measure of a Solid Angle
The measure o f the solid angle is defined as the ratio

P roblem 2 3 .1 M easu re o f a S o lid A n g le 337
m (Z A ) = [limr -> o]area{(interior o f AABCD) n S }lr 2 ,
where S is any small 2-sphere with center at A whose radius, r, is smaller
than the distance from A to each o f the other vertices and to each o f the
edges and faces not containing A. Note that this definition is analogous
to the definition o f radian m easure o f an angle. Do you see why?
a. Show that the measures o f the solid and dihedral angles o f a
tetrahedron satisfy the following relationship:
m (Z A ) = m (AAB) + m(Z A C ) + m (ZA D ) – n.
b. Show that two solid angles with the same measure are not
necessarily congruent. We say the two solid angles are congru­
ent i f one can be transform ed by an isom etry to coincide with
the other.
S u g g e s t i o n s
Solid angles, whether in Euclidean 3-space or a 3-sphere or a hyperbolic
3-space, are closely related to spherical triangles on a small sphere
around the vertex. You can think o f starting with a sphere, S, and creat­
ing a solid angle by extending three sticks out from the center o f the
sphere. If you connect the ends o f these sticks, you will have a tetrahe­
dron. The important thing to notice is how the sticks intersect the sphere.
They will obviously intersect the sphere at three points, and you can
draw in the great circle arcs connecting these points. Look at the planes
in which the great circles lie. In this problem you need to figure out the
relationships betw een the angles o f the spherical triangle and the dihe­
dral angles.
The formula given above for the definition o f the solid angle uses
the intersection o f the interior o f the solid angle with any small sphere S.
This intersection is the small triangle that you ju st drew, and the area o f
the intersection is the area o f the triangle. Because the measure o f a solid
angle is defined in terms o f an area, it is possible for two solid angles to
have the same measure w ithout being congruent — they can have the
same area w ithout having the same shape.
W hat you are asked to prove here is the relationship betw een the
measure o f a solid angle and the m easures o f its dihedral angles. Because
they are closely related to spherical triangles on the small sphere, you
can use everything you know about small triangles on a sphere.

338 C h a p te r 23 P o ly h ed ra
Problem 23.2 Edges and Face Angles
We will study congruence theorem s for tetrahedra that can be thought o f
as the three-dim ensional analogue o f triangles. A tetrahedron has 4 verti­
ces, 4 faces, and 6 edges and we can denote it by AABCD, where A, B, C,
D are the vertices. Figure 23.2.
Show that i f AABCD and AA ‘B’C’D’ are two tetrahedra such that
Z B A C * Z B ‘A ’C’, ZC AD s ZC’A’D’, ZBAD = ZB’A’D’,
CA = C ‘ A B A = B ‘ A D A s D’A
then AABCD = AA’B’C’D’.
Part o f your p ro o f m ust be to show that the solid angles Z A and Z A ‘ are
congruent and not merely that they have the same measure.
Figure 23.2 Edges and faces
S u g g e s t i o n s
I f 5 is a small sphere with center at A and radius r, then
S n (interior o f AABCD)
is a spherical triangle whose sides have lengths
rZ B A C , r ZCAD, r ZBAD.
In the last problem , you saw how solid angles are related to spherical
triangles. This problem asks you to prove the congruence o f tetrahedra
based on certain angle and length measurements. (Note that the angles
shown above are not the dihedral angles o f the tetrahedron.) So, since
you can use spherical triangles to relate solid and dihedral angle

P roblem 2 3 .2 E d g es an d F ace A n g les 3 3 9
m easurem ents, why not use them to prove tetrahedra congruencies? Use
the hint given to see w hat m easurem ents o f the spherical triangle are
defined by m easurem ents o f the tetrahedron. Then see if the m easure­
m ents given do in fact show congruence, and show why.
Problem 23.3 Edges and Dihedral Angles
Show that i f
A B ~ A ‘B’, Z A B = Z A ‘B\ A C ^ A ‘C’,
Z A C = Z A C , AD = A’D’, Z A D = ZA’D’,
then AABCD s AA ‘B C D ‘. Figure 23.3
This is very sim ilar to the previous problem but uses different m easure­
ments — here we have the dihedral angles instead o f the angles on the
faces o f tetrahedron. Look at this problem the same way you looked at
the previous one — see how the measurem ents given relate to a spheri­
cal triangle, and then prove the congruence.
Problem 23.4 Other Tetrahedral Congruence
Theorems
Make up your own congruence theorems! Find and prove at
least two other sets o f conditions that will imply congruence fo r
tetrahedra; that is, make up and prove other theorems like those
in Problems 23.2 and 23.3.
It is important to make sure your conditions are sufficient to prove that
the solid angles are congruent, not ju st that they have the same measure.

340 Chapter 23 Polyhedra
P roblem 23.5 T he Five Regular Polyhedra
A regular polygon is a polygon lying in a plane or 2-sphere or hyper­
bolic plane such that all o f its edges are congruent and all o f its angles
are congruent. For example, on the plane a regular quadrilateral is a
square. On a 2-sphere and a hyperbolic plane a regular quadrilateral is
constructed as shown in Figure 23.4. See also Figure 18.16 for a regular
octagon on a hyperbolic plane.
N ote that h a lf o f a regular quadrilateral is a Khayyam quadrilateral
(see Chapter 12). On 2-spheres and hyperbolic planes there are no simi­
lar polygons; for example, a regular quadrilateral (congruent sides and
congruent angles) will have the same angles as another regular quadrilat­
eral i f and only i f they have the same area. (Do you see why?)
A polyhedron in 3-space [or in a 3-sphere or in a hyperbolic
3-space] is regular i f all o f its edges are congruent, all o f its face angles
are congruent, all o f its dihedral angles are congruent, and all o f its solid
angles are congruent The faces o f a polyhedron are assumed to be poly­
gons that lie on a plane [a great 2-sphere, a great hemisphere].
Figure 23.4 Regular quadrilaterals
Show that there are only five regular polyhedra. In Euclidean
3-space, to say “there are only fiv e regular polyhedra” is to
mean that any regular polyhedra is similar (same shape, but not
necessarily the same size) to one o f the five. It still makes sense
on a 3-sphere and a hyperbolic 3-space to say that “there are

P roblem 2 3 .5 The Five Regular Polyhedra 341
only fiv e regular polyhedra, ” but you need to make clear what
you mean by this phrase.
These polyhedra are often called the Platonic Solids, and are described
by Greek philosopher Plato (429-348 b .c.) as “forms o f bodies which
excel in beauty” (Timaeus, 53e [AT: Plato]), but there is considerable
evidence that they were known well before P lato’s time. See T. L.
H eath’s discussion in [AT: Euclid, Elements], Vol. 3, pp. 4 38-39, for
evidence that the five regular solids were known by Greeks before the
time o f Plato. In addition, there is a description o f the discovery in Scot­
land o f a complete set o f the five regular polyhedra carefully carved out
o f stone by Neolithic persons some 4000 to 6000 years ago in [HI:
Critchlow], pp. 148-49. The regular polyhedra are also the subject o f the
thirteenth (and last) book in [AT: Euclid, Elements].
S uggestions
Your argument should be essentially the same whether you are consider­
ing 3-space, or a 3-sphere, or a hyperbolic 3-space. There are many
widely different ways to do this problem. The following are some
approaches that we suggest:
First Approach: Note that the faces o f a regular polyhedron must
be regular polygons. Then focus on the vertices o f regular polyhedra.
Show that if the faces are regular quadrilaterals or regular pentagons,
then there m ust be precisely three faces intersecting at each vertex. Show
that it is impossible for regular hexagons to intersect at a vertex to form
the solid angle o f a regular polyhedron. I f the faces are regular (equilat­
eral) triangles, then show that there are three possibilities at the vertices.
Second Approach: Refer to Problem 18.5. Each regular polyhe­
dron can be considered to be projected out from its center onto a sphere
and thus determine a cell division o f the sphere. The Euler num ber o f
this spherical subdivision is v – e + / = 2, where v is the num ber o f verti­
ces, e is the num ber o f edges, a n d / i s the num ber o f faces. Then
2e = n f and 2e = k v . (W hyl)
Thus, deduce that
and rem ember that e must be a positive integer.

342 Chapter 23 Polyhedra
In both approaches you should then finish the problem by using
earlier problem s from this chapter to show that any two polyhedra
constructed from the same polygons, with the same num ber intersecting
at each vertex, must be congruent. This step is necessary because there
are polyhedra that are not rigid (that is, there are polyhedra that can be
continuously moved into a non-congruent polyhedra without changing
any o f the faces or changing the num ber o f faces coming together at each
vertex). See Robert Connelly’s “The Rigidity o f Polyhedral Surfaces”
(Mathematics Magazine, vol. 52, no. 5 (1979), pp. 275-83).
The five regular polyhedra are usually named the tetrahedron, the
cube, the octahedron, the dodecahedron, and the icosahedron. (See
Figure 23.5.) There is a duality (related to but not exactly the same as the
duality in Chapter 20, Trigonom etry and Duality) among regular polyhe­
dra: I f you pick the centers o f the faces o f a regular polyhedron, then
these points are the vertices o f a regular polyhedron, which is called the
dual o f the original polyhedron. You can see that the cube is dual to the
octahedron (and vice versa), that the icosahedron is dual to the dodeca­
hedron (and vice versa), and that the tetrahedron is dual to itself.
Figure 23.5 The five Platonic solids

Chapter 24
3 -Manifolds —
T he S hape of S pace
… if we look at the extreme points of the sky, all the visual
rays appear equal to us, and if diametrically opposed stars
describe a great circle, one is setting while the other is rising.
If the universe, instead of being spherical, were a cone or a
cylinder, or a pyramid or any other solid, it would not produce
this effect on earth: one of its parts would appear larger,
another smaller, and the distances from earth to heaven would
appear unequal.
— Theon of Smyrna (~70— 135, Greek), [AT: Theon]
It will be shown that a multiply extended quantity [three-
dimensional manifold] is susceptible of various metric
relations, so that Space constitutes only a special case of a
triply extended quantity. From this however it is a necessary
consequence that the theorems of geometry cannot be deduced
from general notions of quantity, but that those properties that
distinguish Space from other conceivable triply extended
quantities can only be inferred from experience.
— G. F. B. Riemann (1826-1866, German) On the
Hypotheses Which Lie at the Foundations o f Geometry, trans­
lated in [DG: Spivak], Vol. II, p. 135.
N ow we come to where we live. We live in a physical 3-dimensional
space, that is (at least) locally like Euclidean 3-space. The fundamental
question we will investigate in this chapter is, How can we tell w hat the
shape o f our Universe is? This is the same question both Theon o f
Smyrna and Riemann were attempting to answer in the above quotes.
343

3 44 Chapter 24 3-Manifolds — The Shape of Space
This is a very difficult question for which there is currently (as this is
written) no clear answer. However, there are several things we can say.
S pace a s an O riented G eometric 3-M anifold
Presum ably our physical Universe is globally a geometric 3-manifold. (A
geometric 3-m anifold is a space in which each point in the Universe has
a neighborhood that is isometric with a neighborhood o f either Euclidean
3-space, a 3-sphere, or a hyperbolic 3-space. The notion o f a 3-manifold
was introduced in the w ork by Riemann quoted above.) We say “is
globally a geometric 3-manifold” in the same sense in which we say that
globally Earth is a sphere (and spherical geometry is the appropriate
geometry for intercontinental airplane flights) even though it is clear
alm ost anywhere on the earth that locally there are m any hills and
valleys that make Earth not locally isometric to a sphere. However, the
highest point on Earth (M ount Everest) is 8.85 km above sea level and
the lowest point on the floor o f the ocean (the M ariana Trench) is 10.99
km below sea level — the difference is only about 0.3% o f the 6368 km
radius o f Earth (variations in the radius are o f the same magnitude).
It is known that locally our physical Universe is definitely not a
geometric 3-manifold. It was predicted by E instein’s general theory o f
relativity that the local curvature o f our physical Universe is affected by
any m ass (especially large m asses like our sun and other stars) [Albert
Einstein (1879-1955)]. This effect is fairly accurately illustrated by
imagining a 2-dimensional universe that is the surface o f a flat rubber
sheet. I f you place steel balls on this rubber sheet, the balls will locally
make dents or dimples in the sheet and thus will locally distort the flat
Euclidean geometry. E instein’s prediction has been confirmed in two
ways, as follows:
1. The orbits o f the planets M ercury, Jupiter, and Saturn are (quite
accurately) ellipses, and the m ajor axes o f these elliptical orbits
change directions (precess). Classical N ew tonian m echanics
(based on Euclidean geometry) predicts that in a century the
precession will be
M ercury: 1.48°, Jupiter: 1.20°, Saturn: 0.77°,
m easured in degrees o f an arc.
Astronom ers noticed that the observed amount o f preces­
sion agreed accurately with these values for Saturn and Jupiter;

Space as an Oriented Geometric 3-Manifold 3 4 5
however, for M ercury (the closest planet to the Sun) the ob­
served precession is 1.60°, which is 0.12° more than is predicted
by New tonian/Euclidean methods. But if one does the com puta­
tions based on the curvature o f space near the sun that is pre­
dicted by Einstein, the calculations agree accurately with the
observed precession. For a m athem atician’s description o f this
calculation (which uses formulas from differential geometry),
see [DG: M organ], Chapter 7.
S tarl Star 2 Star 1 Star 2
Figure 24.1 Observing local non-Euclidean geometry in the Universe
2. In 1919 British astronomers, led by A rthur Eddington (1882—
1944), measured the angle subtended by two stars from Earth:
once when the Sun was not near the path o f the light from the
stars to Earth and once when the path o f light from one o f the
stars went very close to the Sun. See Figure 24.1. In order to be
able to see the star when its light passes close to the Sun, they
had to make the second observation during a total eclipse o f the
Sun. They observed that the angle (a) m easured with the Sun
near was smaller than the angle (/?) m easured when the Sun was
not near. The difference between the two measured angles was
exactly what Einstein’s theory predicted. Some accounts o f this
experiment talk about the Sun “bending” the light rays, but it is
more accurate to say that light follows intrinsically straight paths

346 Chapter 24 3-Manifolds — The Shape of Space
(geodesics) and that the Sun distorts the geometry (curvature) o f
the space nearby. For more details o f this experiment and other
experiments that verified E instein’s theory, see
http://ww w.ncsa.uiuc.edu/Cyberia/Num Rel/EinsteinTest.htm l
But m ost o f our physical Universe is empty space with only scat­
tered planets, stars, galaxies, and the “dimples” that the stars make in
space is only a very small local effect near the star. The vast empty space
appears to locally (on a medium scale m uch larger than the scale o f the
distortions near stars) be a geometry o f the local symmetries o f Euclid­
ean 3-space. In particular, our physical space is observed to be locally
(on a medium scale) the same in all directions with isometric rotations,
reflections, and translations as in Euclidean 3-space.
Theorem 2 4 .0 . Euclidean 3-space, 3-spheres, and hyperbolic
3-spaces are the only simply connected 3-dimensional geome­
tries that locally have the same symmetries as Euclidean
3-space. (Sim ply connected means that every loop can be
continuously shrunk to a point in the space.)
The condition o f simply connected is to rule out general geometric
manifolds modeled on Euclidean 3-space, 3-spheres, or hyperbolic
3-spaces. There is a discussion o f the p ro o f (and more precise statement)
o f this theorem in [DG: Thurston], Section 3.8, where Thurston dis­
cusses eight possible 3-dimensional simply connected geometries, but
only three have the same symmetries as Euclidean 3-space. A more
elem entary discussion without proofs can be found in [DG: W eeks],
Chapter 18.
U nfortunately (or maybe fortunately!), geometric 3-manifolds are
not fully understood. At this point no one knows w hat all o f the geomet­
ric 3-manifolds are or how to distinguish one from the other. The theory
o f 3-manifolds is an area o f current active research. For a very accessible
discussion o f this research, see [DG: W eeks], Part III; for more detailed
discussions, see T hurston’s Three-Dimensional Geometry and Topology,
Volume 1, [DG: Thurston] and the second (and further?) volumes as
soon as they appear.

http://www.ncsa.uiuc.edu/Cyberia/NumRel/EinsteinTest.html

P roblem 24.1 Is Our Universe Non-Euclidean? 347
P roblem 24.1 Is O ur U niverse Non-E uclidean?
There is a widely reported story about the famous m athem atician Carl
Friedrich Gauss (1777-1855, German) that he tried to m easure the
angles o f a triangle whose vertices were three m ountain peaks in
Germany. I f the sum o f the angles had turned out to be other than 180°,
then he w ould have deduced that the Universe is not Euclidean (or that
light does not travel in straight lines). However, his m easurements were
inconclusive because he m easured the angles at 180° within the accuracy
o f his m easuring instruments. This story is apparently a myth [see
Breitenberger, “G auss’ Geodesy and the Axiom o f Parallels,” Archive
fo r the History o f the Exact Sciences, 31(1984), pp. 273-89], for though
Gauss did m easure the angles o f a large triangle o f m ountain peaks, the
purpose was to connect several grids o f triangles that were being used
for making a complete survey o f Europe. Nevertheless, the story still
leads to the question
a. Could we now show that the Universe is non-Euclidean by meas­
uring the angles o f a large triangle in our solar system? How
accurately would we have to measure the angles?
Note that, i f the Universe is a 3-sphere or a hyperbolic 3-space, the
radius R o f the Universe would have to be at least as large as the diam e­
ter o f our galaxy, which is about 1018 km. In the foreseeable future, the
largest triangle whose angles we could m easure has area less than the
area o f our solar system, which is about 8 x 1019 km2. Use the formulas
you found in Problem s 7.1 and 7.2.
You will find that in order to determine the geometry o f space, we
have to look further than our solar system.
b. I f the stars were distributed uniformly in space (it is not clear to
what extent this is actually true), how could you tell by looking
at stars at different distances whether space was locally Euclid­
ean, spherical, or hyperbolic?
I f you have trouble envisioning this, then start with the analogous prob­
lem for a 2-dim ensional bug on a plane, sphere, or hyperbolic plane.
W hat would this bug observe? Assume that you can tell how far away
each star is — this is something that astronomers know how to do.

348 Chapter 24 3-Manifolds— The Shape of Space
There are certain types o f stars (called by astronomers, s ta n d a r d
candles) that have a fixed known amount o f brightness. These include
the so-called Type la Supemovae.
c. Suppose we can see several o f these standard candles and can
determine accurately their distances from Earth. How could we
tell from this information whether the Universe is Euclidean,
spherical, or hyperbolic?
Hint: The apparent brightness o f a shining object in Euclidean space is
inversely proportional to the square o f the distance to the object.
d. It is impractical to measure the excess o f triangles in our solar
system by only taking measurements o f angles within our solar
system. What observations o f only distant stars and galaxies
would tell us that the Universe is not Euclidean 3-space?
I f you have trouble conceptualizing a 3-sphere or hyperbolic 3-space,
then you can do this problem, first, for a very small bug on a 2-sphere or
hyperbolic plane who can see distant points (stars) but who is restricted
to staying inside its “solar system,” which is so small that any triangle in
it has excess (or defect) too small to measure. Always think intrinsically!
You can assume, generally, that light will travel along geodesics, so
think about looking at various objects and the relationships you would
expect to find. For example, if the Universe were a 3-sphere and you
could see all the way around the Universe (the distance o f a great circle),
how w ould you know that the Universe is spherical? W hy? W hat if we
could see halfway around the Universe? Or a quarter o f the way around?
Think o f looking at stars at these distances.
All the ways discussed in parts a – d have been tried by astronomers,
but, up to now, none o f these observations has been accurate enough to
make a definite determination.
Recently, astronomers have attem pted to find at a far (but approxi­
m ately known) distance in the Universe a structure whose size is known
(based on physical assumptions). If such a structure is found, then it is
possible to m easure the angle subtended by this structure from Earth. If
the geometry o f the Universe is Euclidean, then the m easure o f the
observed angle is predicted by the Law o f Cosines. If the observed angle
is larger than the prediction, then the Universe w ould be spherical and if
it is smaller, then the Universe w ould be hyperbolic. {Do you see why?)

P roblem 2 4 .1 Is Our Universe Non-Euclidean? 3 49
In spring 2000, a group o f astronomers announced (see editorial and
article in Nature, volume 404, April 2000) that they have observed such
a structure in the cosmic background radiation and that their observa­
tions are “consistent with a flat, Euclidean U niverse” ; however, they also
wrote that the precise details o f their observations did not “agree with
any known physical theory.” But, even i f later observations and analysis
determine that the Universe is Euclidean, it will still leave open the
question o f which Euclidean 3-manifold the Universe is. W e will study
Euclidean 3-manifolds in the next section.
In June 2001 the U nited States’s National Aeronautic and Space
Adm inistration (NASA) launched the W ilkinson M icrowave Anisotropy
Probe (W M AP). See
http://map.gsfc.nasa.gov/
for more information. W M AP has been making m easurem ents o f the
cosmic background radiation since fall 2001. In 2007, the European
Space Agency is scheduled to launch the Planck satellite. See
http://ww w.esa.int/science/planck
for more information. Plank is being designed to make even more
accurate maps o f the cosmic background radiation. An analysis o f these
maps may provide the clues we need to definitely determine the global
geometry o f space. To describe how this will w ork we m ust first investi­
gate geometric 3-manifolds.
P roblem 2 4.2 Euclidean 3-M anifolds
We now consider the 3-dimensional analogue o f the flat torus. Consider
a cube in Euclidean 3-space with the opposite faces glued through a
reflection in the plane that is m idw ay betw een the opposite faces. See
Figure 24.2. In this figure, we have drawn a closed straight path that
starts from A on the bottom right edge and then hits the m iddle o f the
front face at B. It continues from the middle o f the back face and finishes
at the m iddle o f the top left edge at a point that is glued to A.
a. Show that the cube with opposite faces glued by a reflection
through the plane midway between is a Euclidean 3-manifold.
That is, check that a neighborhood o f each point is isometric to

http://map.gsfc.nasa.gov/

http://www.esa.int/science/planck

3 5 0 Chapter 24 3-Manifolds — The Shape of Space
a neighborhood in Euclidean 3-space. This Euclidean 3-mani­
fold is call the 3-torus.
Look separately at points (such as A) that are in the middle o f edges,
points (such as B ) that are the middle o f faces, and points that are verti­
ces (such as C).
C
C
Figure 24.2 A closed geodesic path on the 3-torus
N ext glue the vertical faces o f the cube the same way but glue the
top and bottom faces by a quarter turn. In this case we get the closed
geodesic depicted in Figure 24.3.
b. Show that you obtain a Euclidean 3-manifold from the cube with
vertically opposite faces glued by a reflection through the plane
midway between, and the top and bottom faces glued by a
quarter-turn rotation. This Euclidean 3-manifold is called the
quarter-turn manifold.

P roblem 24.2 Euclidean 3 -Manifolds 351
Figure 24.3 A closed geodesic in the quarter-turn manifold
c. Draw a picture similar to Figures 24.2 and 24.3, fo r the half­
turn manifold, which is the same as the quarter-turn manifold
except that it is obtained by gluing the top and bottom faces with
a half-turn.
In Problem 18.1 we represented the flat torus in two different ways
— one starting with a rectangle or square and the other starting from a
hexagon. The above discussion o f the 3-torus corresponds to the con­
struction o f the flat torus from a square. N ow we want to look at what
happens i f we use an analogue o f the hexagon construction.
Consider a hexagonal prism as in Figure 24.4. W e will make glu-
ings on the vertical sides by gluing each vertical face with its opposite in
such a way that each horizontal cross-section (which are all hexagons)
has the same gluings as the hexagonal flat torus (see Problem 18.2b and
Figure 18.3). The top and bottom face we glue in one o f three ways. If
we glue the top and bottom face through a reflection in the halfway
plane, then we obtain the hexagonal 3-torus. I f we glue the top and
bottom faces with a one-sixth rotation, then we obtain the one-sixth-turn

352 Chapter 24 3-Manifolds — The Shape of Space
manifold. If we glue the top and bottom with a one-third rotation then
we will get the one-third-turn manifold.
Figure 24.4 Hexagonal 3-manifolds
d. Show that the hexagonal 3-torus, the one-sixth-turn manifold,
and the one-third-turn manifold are Euclidean 3-manifolds and
that the hexagonal 3-torus is homeomorphic to the 3-torus. What
happens i f we consider the two-thirds-turn manifold and the
three-sixths-turn manifold and the five-sixths-turn manifold’?
It can be shown that
Theorem 2 4 .2 . There are exactly ten Euclidean 3-manifolds
up to homeomorphism. O f these, fo u r are non-orientable and
six are orientable. Five o f the six orientable Euclidean mani­
fo ld s are the 3-torus, the quarter-turn manifold, the half-turn
manifold, the one-sixth-turn manifold, and the one-third-turn
manifold.
See [DG: W eeks], page 252, for a discussion o f this theorem. For more
detail and a proof, see [DG: Thurston], Section 4.3.

P roblem 24.3 Dodecahedral 3-Manifolds 353
P roblem 24.3 Dodecahedral 3-M anifolds
Spherical and hyperbolic 3-manifolds are more com plicated than Euclid­
ean 3-manifolds. In fact, no one knows what all the hyperbolic 3-mani­
folds are. W e will only look at a few examples in order to get an idea o f
how to construct spherical and hyperbolic 3-manifolds in this problem
and the next.
There are two examples that can be obtained by making gluings o f
the faces o f a dodecahedron (see Problem 23.5). It will be best for this
problem for you to have a m odel o f the dodecahedron that you can look
at and touch. We want to glue the opposite faces o f the dodecahedron.
Looking at your dodecahedron (or Figure 24.5), you should see that the
opposite faces are not lined up but rather are rotated one-tenth o f a full
turn from each other. Thus there are three possibilities for gluings: We
can glue with a one-tenth rotation, or a three-tenths rotation, or a five-
tenths (= one-half) rotation. W hen making the rotations it is important
always to rotate in the same direction (say clockwise as you are facing
the face from the outside). You should check with your model that this is
the same as rotating clockwise while facing the opposite face.
Figure 24.5 Dodecahedron
a. When you glue the opposite faces o f a dodecahedron with a
one-tenth clockwise rotation, how many edges are glued to­
gether? What i f you use a three-tenths rotation? Or a one-half
rotation?
The best way to do this counting is to take a m odel and mark one edge
with tape. Then for each o f the two pentagon faces on which the edge
lies, the gluing glues the edge to another edge on the other side o f the

3 5 4 Chapter 24 3-Manifolds — The Shape of Space
dodecahedron — mark those edges also. Continue with those marked
edges until you have marked all the edges that are glued together.
Y our answers to part a should be 2, 3, 5 (not in that order!). Thus in
order for these m anifolds to be geometric m anifolds, we m ust use
dodecahedrons with dihedral angles o f 180°, 120°, and 72° in either
Euclidean space, a sphere, or a hyperbolic 3-space. But before we go
further we m ust figure out the size o f the dihedral angles o f the dodeca­
hedron in Euclidean space, which from our model seems to be close to
(if not equal to) 120°.
b. Calculate the size (j) o f the dihedral angle o f the (regular)
dodecahedron in Euclidean 3-space.
Imagine a small sphere with center at one o f the vertices o f the dodeca­
hedron. This sphere will intersect the dodecahedron in a spherical
equilateral triangle. The angles o f this triangle are the dihedral angles —
Do you see why? This triangle is called the lin k o f a vertex in the
dodecahedron. Determine the lengths o f the sides o f this triangle and
then use the Law o f Cosines (Problem 20.2).
Now imagine a very small dodecahedron in 3-sphere. Its dihedral
angles will be very close to the Euclidean angle (Why is this the
case?). If you now imagine the dodecahedron growing in the 3-sphere,
its dihedral angles will grow from . If you start with a very small
dodecahedron in a hyperbolic space, then its dihedral angles will start
very close to ^ a n d then decrease as the dodecahedron grows.
c. Show that the manifold from part a with three edges being glued
together is a spherical 3-manifold i f < 120°, or a Euclidean 3-manifold i f = 120° or a hyperbolic 3-manifold i f > 120°.
This geometric 3-manifold is called the Poincare dodecahedral
space in honor o f Henri Poincare (1854-1912, French), who
first described (not using the dodecahedron) a space homeomor-
phic to this geometric 3-manifold.
Show that each vertex o f the dodecahedron is glued to three other verti­
ces and that the four solid angles fit together to form a complete solid
angle in the m odel (either Euclidean 3-space, 3-sphere, or hyperbolic
3-space).

P roblem 24.3 Dodecahedral 3-Manifolds 355
d. Can the dihedral angles o f a dodecahedron in a 3-sphere grow
enough to be 180°? What does such a dodecahedron look like?
Is the manifold with two edges being glued together a spherical
3-manifold? This spherical 3-manifold is called the projective
3-space or R P i.
e. Can the dihedral angles o f a dodecahedron in a hyperbolic
3-space shrink enough to be equal to 72°? I f so, the dodecahe­
dral manifold with fiv e edges being glued together is a hyper­
bolic 3-manifold. This hyperbolic 3-manifold is called the
Seifert-W eber dodecahedral space, after H. Seifert (1907-1996)
and C. W eber, who first described both dodecahedral spaces in a
1933 article, H. Seifert and C. W eber, “Die beiden Dodekaeder-
raum e,” Mathematische Zeitschrift, vol. 37 (1933), no. 2, p. 237.
Imagine that the dodecahedron grows until its vertices are at infinity
(thus on the bounding plane in the upper h a lf space model). Use the fact
that angles are preserved in the upper-half-space model and look at the
three great hemispheres that are determined by the three faces coming
together at a vertex. Rem ember also to check that the solid angles at the
vertices o f the dodecahedron fit together to form a complete solid angle.
Two articles in the October 9, 2003, issue o f Nature (vol. 425, pp.
566-567 and 593-595) showed that the spherical dodecahedral space is
compatible with the latest observations o f the cosmic background radia­
tion and that it explains some m easurements better than other know
explanations. N ote what Plato says about the dodecahedron in the quote
at the beginning o f Chapter 23. W e will record on the Experiencing
Geometry W eb site the latest information as we are aware o f it.
Problem 2 4.4 Some Other G eometric 3-M anifolds
We now look at three more examples o f geometric 3-manifolds.
a. Start with a tetrahedron and glue the faces as indicated in
Figure 24.6. Does this gluing produce a manifold? Can the
tetrahedron be p u t in a 3-sphere or hyperbolic 3-space so that
the gluings produce a geometric 3-manifold? We call this the
tetrahedral space.

3 5 6 Chapter 24 3-Manifolds — The Shape of Space
Figure 24.6 Tetrahedral space
b. Start with a cube and glue each face to the opposite face with a
one-quarter-turn rotation. Does this gluing produce a manifolds
Can the cube be put in a 3-sphere or hyperbolic 3-space so that
the gluings produce a geometric 3-manifold! This is called the
quaternionic m anifold because its symmetries can be expressed
in the quaternions (a four-dim ensional version o f the complex
num bers with three imaginary axes and one real axis).
Again, investigate how m any edges are glued together and w hat happens
near the vertices.
c. Start with an octahedron and glue each face to the opposite face
with a one-sixth-turn rotation. Does this gluing produce a
manifold? Can the octahedron be put in a 3-sphere or hyper­
bolic 3-space so that the gluings produce a geometric 3-mani-
fo ld ! This is called the octahedral space.
Again, investigate how m any edges are glued together and w hat happens
near the vertices. You can also use your knowledge o f solid angles from
Chapter 23.
C o s m i c B a c k g r o u n d R a d i a t i o n
Astronom ers from earthbound observatories have noticed a radiation that
is rem arkably uniform coming to Earth from all directions o f space. In
1991, the United States’s Cosmic Background Explorer (COBE) m apped
large portions o f this radiation to a resolution o f about 10 degrees o f arc.
COBE determined that the radiation is uniform to nearly one part in
100,000, but there are slight variations (or texture) observed. It is this

C osm ic B ac k g ro u n d R ad ia tio n 3 5 7
texture that gives us the possibility to determine the global shape o f the
Universe. To understand how this determ ination may be possible, we
m ust first understand from where (and when!) this background radiation
came.
The generally accepted explanation o f the cosmic background
radiation is that in the early stages o f the Universe m atter was so dense
that no radiation could escape (the space was filled with m atter so dense
that all radiation was scattered). But at a certain point in time (about
300,000 years after the big bang and about 13 billion years ago), matter
in the Universe started coalescing and the density o f m atter decreased
enough that radiation could start traveling through the Universe in all
directions but was still dense enough that the radiating m atter (radiating
because it was hot) was fairly hom ogeneously distributed throughout the
Universe. It is this first escaped radiation that we see when we look at
the cosmic background radiation.
Remember that all radiation (light and others) travels at the speed o f
light (~1013 kilometers per year) and thus the cosmic radiation that
reaches us today has been traveling for about 13 billion years and has
traveled about 1.3 x 1022 km. Thus the cosmic background radiation
gives us a picture o f a sphere that was a slice o f the early Universe
roughly 13 billion years ago. The cosmologists call this the last scatter­
ing surface. It appears to us that Earth is at the center o f this sphere, but
this is talking in space-time. W hat we see is a picture o f a sphere in the
early universe whose center (at that time) was the point in the early
Universe where 13 billion years later the M ilky W ay galaxy, solar
system, and Earth would form. By the time the radiation from this sphere
reached that point, the Earth had formed and we humans are on the earth
recording the radiation.
The cosmologists say that the above discussion does not involve
any assumptions about the density o f m atter or the presence o f a “cosm o­
logical constant,” both o f which are hotly debated subjects among
cosmologists. In addition, the only assum ption made about the geometry
o f our physical Universe is the assumption that our physical Universe is
a geometric m anifold and that, even though it is expanding, the topology
is constant (or at least has been constant since the Universe was big
enough for the radiation to escape). In Theorem 24.5, we see that in fact
i f we know the topology we also know the geometry.
Before going on with the 3-dimensional discussion, let us look at an
analogous situation in two dimensions. Imagine that the 2-dimensional

3 5 8 C h ap ter 24 3 -M an ifo ld s — T he S hape o f Space
bug’s universe is a flat torus obtained from a square with opposite side
glued. We may assume that the bug is at the center o f the square looking
out in all directions at a textured circle from the center o f that circle, the
last scattering circle for the bug. I f this circle has diam eter larger than
the side o f the square, then the circle will intersect itself, as indicated in
Figure 24.7.
F
E
Figure 24.7 Seeing the 2-dimensional scattering circle
N ote, in Figure 24.7, that when the bug in the center looks toward
the point A, the bug w ill not see A but rather will see C. (The light from
A will have reached the center o f the square earlier!) Likewise, the bug
will see the point D on the circle but not the point B. The important
points to focus on are the points, E, F, G, H, where the circle intersects
itself. The bug will see these points in two different directions. See
Figure 24.7, where the point E is seen from both sides. If the texture is
unique enough, the bug should be able to tell that it is looking at the
same point in two different directions. The pattern o f these identical
point-pairs will indicate to the bug that its Universe is the flat torus. (See
Problem 24.5a.)
The three-dimensional situation is similar. Consider that our physi­
cal Universe is a 3-torus that is the result o f gluings on a cube as in
Figure 24.2. I f Earth is considered to be in the center o f this cube and if
the sphere o f last scatter has reached the faces o f the cube, then the inter­
section o f the sphere o f last scatter with the faces o f the cube will be
circles and (because o f the gluings) the circle on one face will be

C osm ic B ack g ro u n d R a d iatio n 3 5 9
identified (by a reflection) with the circle on the opposite face (see
Figure 24.8). The pattern o f circles shows the underlying cube and the
gluings. In Problem 24.5b you will explore what the pattern o f circles
will look like in the other geometric 3-manifolds we have discussed.
Figure 24.8 Self-intersections of the sphere of last scatter in the 3-torus
In our physical Universe, during 2002, NA SA launched the W ilkin­
son M icrowave Anisotropy Probe (W M AP). See
http://map.gsfc.nasa.gov/
for more information. In 2007, the European Space Agency is scheduled
to launch the Planck satellite. See
http://astro.estec.esa.nl/
for more information. These probes will produce detailed maps o f the
texture o f the microwave radiation. W M AP is producing about 0.3°
resolution, and the Planck satellite is hoped to provide about better than
0.1° resolution. Rem ember that the maps before 2002 had only 10°
resolution.

http://map.gsfc.nasa.gov/

http://astro.estec.esa.nl/

3 6 0 C h ap ter 24 3 -M an ifo ld s — T he S hape o f Space
For further discussion o f these m easurem ents, see Luminet, Stark-
man, and W eeks, “Is Space Finite?”, Scientific American, April 1999,
page 90; or Cornish and W eeks, “M easuring the Shape o f the U niverse,”
Notices o f the American Mathematical Society, 1998, or look for more
recent articles.
P r o b l e m 24.5 C i r c l e P a t t e r n s M a y S h o w t h e
S h a p e o f S p a c e
*a. For each o f the geometric 2-manifolds in Problems 18.1, 18.3,
18.4, what would the patterns o f matching point-pairs look like
i f the b u g ’s last scattering circle was large enough to intersect
itself!
Draw pictures analogous to Figure 24.7.
b. For each o f the geometric 3-manifolds in Problems 24.2, 24.3,
and 24.4, what would the pattern o f matching circles look like i f
our physical Universe were the shape o f that manifold and i f the
sphere o f first scatter reaches fa r enough around the Universe
fo r it to intersect itself!
Draw and examine, as best you can, pictures analogous to Figure 24.8.
We saw in Problem 18.5 that the area o f a spherical or hyperbolic
2 -manifold is determined by its topology and the radius (or curvature) of
the model. Similarly (but m uch more com plicatedly), we have for spheri­
cal and hyperbolic 3-manifolds,
T heorem 2 4 .5 . I f two orientable spherical or hyperbolic
3-manifolds are homeomorphic, then they are geometrically
similar. That is, two such homeomorphic manifolds are isomet­
ric i f the model spherical or hyperbolic spaces have the same
radius (curvature).
Thus, i f we are successful in finding a pattern o f circles that deter­
mines that the Universe is a spherical or hyperbolic 3-manifold, then we
will know the volume o f our physical Universe.

P roblem 2 4 .5 C ircle P attern s M ay S h o w the S hape o f Space 3 6 1
L a t e s t E v i d e n c e o n t h e S h a p e o f S p a c e
We will post updates o f these probes and the analyses o f the data that are
relevant to this text at the Experiencing Geometry W eb site:
http://ww w.m ath.com ell.edu/~henderson/ExpGeom
At the time this was w ritten there was still no conclusive evidence
as to what the shape o f the universe actually is. It is known that the
curvature o f the universe is very nearly zero and thus is very nearly a
Euclidean 3-manifold, but there is a distinct possibility that the universe
is a spherical 3-manifold with very large radius. However, even i f the
universe turns out to be a Euclidean 3-manifold, that does not mean that
it is Euclidean 3-space (see Problem 24.2). It fact, there is some evidence
that suggests that i f one could look far enough in the direction o f the
constellation Virgo, one w ould be able to see our galaxy! See the
Cosmology News Web site for more discussions and links:
http://ww w.geometrygam es.org/ESoS/Cosm ologyNews.htm l

http://www.math.comell.edu/~henderson/ExpGeom

http://www.geometrygames.org/ESoS/CosmologyNews.html

Appendix A
Euclid’s Definitions,
Postulates, and
Common Notions
At the age of eleven, I began Euclid, with my brother as my
tutor. This was one of the great events of my life, as dazzling
as first love. I had not imagined that there was anything so
delicious in the world.
— Bertrand Russell (1883),
Autobiography: 1872-1914, Allen & Unwin, 1967, p. 36
The following are the definitions, postulates, and common notions listed
by Euclid in the beginning o f his Elements, Book 1. These are H eath’s
translations from [AT: Euclid, Elements] except that we m odified them
to make the wording and usage m ore in line with word usage today. In
our m odifications we used H eath’s extensive notes on the translation in
order not to change the m eanings involved. But, rem ember that we can
not be sure o f the exact m eaning intended by Euclid — any translation
should be considered only as an approximation.
D e f i n i t i o n s
1. A p o in t is that which has no parts.
2. A curve is length without width.
[Heath translates this as “line,” but today we normally use the
term “curve” in place o f “ line” .]
363

36 4 A p p en d ix A E u c lid ’s D efin itio n s, P o stu lates, an d C o m m on N o tio n s
3. The ends o f a curve are points.
[It is not assum ed here that the curve has ends (for example, see
Definition 15 below); but, i f the line does have ends, then the ends
are points.]
4. A (straight) line is a curve that lies symmetrically with the points
on itself.
[The commonly quoted H eath’s translation says “lies evenly with
the points,” but in his notes he says “we can safely say that the
sort o f idea which Euclid wished to express was that o f a line …
without any irregular or unsymm etrical feature distinguishing one
part or side o f it from another.”]
5. A surface is that which has length and width only.
6 . The boundaries o f a surface are curves.
7. A p la n e is a surface that lies symmetrically with the straight lines
on itself.
[The comments for Definition 4 apply here as well.]
8 . An angle is the inclination to one another o f two curves in a plane
that meet on another and do not lie in a straight line.
[What Euclid meant by the term “inclination” is not clear to us or,
apparently, to Heath.]
9. The angle is called rectilinear when the two curves are straight
lines.
[O f course, we (and Euclid in most o f the Elements) call these
simply “angles.”]
10. When a straight line intersects another straight line such that the
adjacent angles are equal to one another, then the equal angles
are called right angles and the lines are called perpendicular
straight lines.
[As discussed in the last section o f Chapter 4, cones give us
examples o f spaces where right angles (as defined here) are not
always equal to 90 degrees.]
11 .A n obtuse angle is an angle greater than a right angle.
12. An acute angle is an angle less than a right angle.

D efin itio n s 3 6 5
13. A boundary o f anything is that which contains it.
14. A fig u r e is that which is contained by any boundary or boundaries.
15. A circle is a plane figure contained by one curve (called the cir­
cum ference) with a given point lying within the figure such that all
the straight lines joining the given point to the circumference are
equal to one another.
16. The given point o f a circle is called the center o f the circle.
17. A diam eter o f a circle is any straight line drawn through the center
and with its endpoints on the circumference; the straight line
bisects the circle.
18. A sem icircle is a figure contained by a diameter and the part o f the
circumference cut o ff by it. The center o f the semicircle is the same
as the center o f the circle.
19. Polygons are those figures whose boundaries are made o f straight
lines: triangles being those contained by three, quadrilaterals
those contained by four, and m ultilaterals those contained by more
than fo u r straight lines.
20. An equilateral triangle is a triangle that has three equal sides, an
isosceles triangle is a triangle that has only two o f its sides equal,
and a scalene triangle is a triangle that has all three sides
unequal.
21. A right triangle is a triangle that has a right angle, an obtuse
triangle is a triangle that has an obtuse angle, and an acute trian­
gle is a triangle that has all o f its angles acute.
22. A square is a quadrilateral that is equilateral (has all equal sides)
and right angled (has all right angles); a rectangle is a quadrilat­
eral that is right angled but not equilateral; a rhom bus is a
quadrilateral that is equilateral but not right angled; a rhom boid
is a quadrilateral that has opposite sides and angles equal to one
another but that is neither equilateral nor right angled. Let quadri­
laterals other than these be called trapezia.

366 A p p e n d ix A E u c lid ’s D efin itio n s, P o stu lates, an d C o m m o n N o tio n s
23. Parallel straight lines are straight lines lying in a plane that do
not meet i f continued indefinitely in both directions.
P o s t u l a t e s
1. A (unique) straight line may be drawn from any point to any other
point.
2. Every limited straight line can be extended indefinitely to a
(unique) straight line.
3. A circle may be drawn with any center and any distance.
4. All right angles are equal.
[Note that cones give us examples o f spaces in which all right
angles are not equal; see Chapter 4. Thus this postulate could be
rephrased: “There are no cone points.”]
5. I f a straight line intersecting two straight lines makes the interior
angles on the same side less than two right angles, then the two
lines (i f extended indefinitely) will meet on that side on which the
angles are less than two right angles.
[See Chapter 10 for m ore discussion o f this postulate.]
C o m m o n N o t i o n s
1. Things that are equal to the same thing are also equal to one
another.
2. I f equals are added to equals, then the results are equal.
3. I f equals are subtracted from equals, the remainders are equal.
4. Things that coincide with one another are equal to one another.
5. The whole is greater than any o f its parts.

Appendix B
C onstructions of
Hyperbolic Planes
We will describe four different isometric constructions o f hyperbolic
planes (or approxim ations to hyperbolic planes) as surfaces in 3-space. It
is very important that you actually perform at least one o f these
constructions. The act o f constructing the surface will give you a feel for
hyperbolic planes that is difficult to get any other way. Templates for all
the paper constructions (and information about possible availability o f
crocheted hyperbolic planes) can be found on the Experiencing Geome­
try Web site
w w w .m ath.com ell.edu/~henderson/ExpGeom /
T h e H y p e r b o l i c P l a n e f r o m P a p e r A n n u l i
A paper model o f the hyperbolic plane may be constructed as follows:
Cut out m any identical annular (“annulus” is the region betw een two
concentric circles) strips as in Figure B .l. Attach the strips to each other
by taping the inner circle o f one to the outer circle o f the other. It is
crucial that all the annular strips have the same inner radius and the same
outer radius, but the lengths o f the annular strips do not matter. You can
also cut an annular strip shorter or extend an annular strip by taping two
strips together along their straight ends. The resulting surface is o f
course only an approxim ation o f the desired surface. The actual hyper­
bolic plane is obtained by letting 8 -+ 0 while holding the radius p fixed.
Note that since the surface is constructed (as 8 – » 0) the same every­
where it is homogeneous (that is, intrinsically and geometrically, every
point has a neighborhood that is isometric to a neighborhood o f any
other point). We will call the results o f this construction the annular
367

http://www.math.comell.edu/~henderson/ExpGeom/

368 A p p e n d ix B — C o n stru ctio n s o f H y p erb o lic P lan es
hyperbolic plane. We strongly suggest that the reader take the time to
cut out carefully several such annuli and tape them together as
indicated.
How t o C r o c h e t t h e H y p e r b o l i c P l a n e
Once you have tried to make your annular hyperbolic plane from paper
annuli, you will certainly realize that it will take a lot o f time. Also, later
you will have to play with it carefully because it is fragile and tears and
creases easily — you may want ju st to have it sitting on your desk. But
there is another way to get a sturdier model o f the hyperbolic plane,
which you can work and play with as m uch as you wish. This is the
crocheted hyperbolic plane.
In order to make the crocheted hyperbolic plane, you need only
basic crocheting skills. All you need to know is how to make a chain (to

H o w to C ro ch et th e H y p erb o lic P lan e 3 6 9
start) and how to single crochet. T hat’s it! N ow you can start. See Figure
B.2 for a picture o f these stitches, and see their description in the list
below.
Figure B.2 Crochet stitches for the hyperbolic plane
First you should choose a yam that will not stretch a lot. Every yam
will stretch a little but you need one that will keep its shape. Now you
are ready to start the stitches:
1. Make your beginning chain stitches (Figure B.2a). About 20
chain stitches for the beginning will be enough.
2. For the first stitch in each row insert the hook into the 2nd chain
from the hook. Take yam over and pull through chain, leaving 2
loops on hook. Take yam over and pull through both loops. One
single crochet stitch has been completed (Figure B.2b).
3. For the next N stitches proceed exactly like the first stitch except
insert the hook into the next chain (instead o f the 2nd).
4. For the (N + l ) s‘ stitch proceed as before except insert the hook
into the same loop as the /Vth stitch. (For example, if your N = 12,
then crochet 12 stitches and then increase for the 13th stitch, and
then 12 again and the next an increase — keep this pattern
constant.)
5. Repeat Steps 3 and 4 until you reach the end o f the row.

3 7 0 A p p en d ix B — C o n stru ctio n s o f H y p erb o lic P lanes
6 . At the end of the row, before going to the next row do one extra
chain stitch.
7. When you have the model as big as you want, you can stop by
ju st pulling the yam through the last loop.
Be sure to crochet fairly tightly and evenly. That’s all you need
from crochet basics. Now you can go ahead and make your own hyper­
bolic plane. You have to increase (by the above procedure) the number
o f stitches from one row to the next in a constant ratio, N to N+ 1 — the
ratio and size o f the yam determine the radius (the p in the annular
hyperbolic plane) o f the hyperbolic plane. You can experiment with
different ratios but not in the same model. We suggest that you start with
a ratio o f 5 to 6. You will get a hyperbolic plane only i f you will be
increasing the number o f stitches in the same ratio all the time.
Crocheting will take some time, but later you can work with this
model without worrying about destroying it. The com pleted product is
pictured in Figure B.3. The N for this plane is 12.
Figure B.3 A crocheted annular hyperbolic plane

H o w to C ro ch et the H y p erb o lic P lan e 3 7 1
{ 3 ,7 } a n d {7,3} P o l y h e d r a l C o n s t r u c t i o n s
A polyhedral model can be constructed from equilateral triangles by
putting 7 triangles together at every vertex, or by putting 3 regular hepta­
gons (7-gons) together at every vertex. These are called the {3,7} poly­
hedral model and the {7, 3} polyhedral model because triangles (3-
gons) are put together 7 at a vertex, or heptagons (7-gons) are put
together 3 at a vertex. These models have the advantage o f being con­
structed more easily than the annular or crocheted models; however, one
cannot make better and better approxim ations by decreasing the size o f
the triangles. This is true because at each vertex the cone angle is (7 x
n/3) = 420° or (3 x 5n/7) = 385.71…°), no m atter what the size o f the
triangles and heptagons are; whereas the hyperbolic plane in the small
looks like the Euclidean plane with 360° cone angles. Another disadvan­
tage o f the polyhedral model is that it is not easy to describe the annuli
and related coordinates.
You can make these models less “pointy” by replacing the sides o f
the triangles with arcs o f circles in such a way that the new vertex angles
are 2nH or by replacing the sides o f the heptagons with arcs o f circles in
such a way that the new vertex angles are 2ji/3. But then the model is
less easy to construct because you are cutting and taping along curved
edges.
See Problem s 11.7 and 23.5 for more discussions o f regular polyhe­
dral tilings o f plane, spheres, and hyperbolic planes.
H y p e r b o l i c S o c c e r B a l l C o n s t r u c t i o n
We now explore a polyhedral construction that involves two different
regular polygons instead o f the single polygon used in the {3, 7} and
{7, 3} polyhedral constructions. A spherical soccer ball (outside the
North America, called a “football”) is constructed by using pentagons
surrounded by five hexagons or two hexagons and one pentagon together
around each vertex. The plane can be tiled by hexagons, each surrounded
by six other hexagons. The hyperbolic plane can be approxim ately
constructed by using heptagons (7-sided) surrounded by seven hexagons
and two hexagons and one heptagon together around each vertex. See
Figure B.4. This construction was discovered by Keith Henderson,
David’s son. Because a heptagon has interior angles with 5n!l radians (=
128.57.. .°), the vertices o f this construction have cone angles o f
3 6 8 .5 7 .. .° and thus are m uch smoother than the (3, 7} and (7, 3}

372 A p p en d ix B — C o n stru ctio n s o f H y p erb o lic P lan es
polyhedral constructions. It also has a nice appearance if you make the
heptagons a different color from the hexagons. It is also easy to construct
(as long as you have a tem plate — you can find a variety on the Experi­
encing Geometry Web site). As with any polyhedral construction, one
cannot get closer and closer approxim ations to the hyperbolic plane.
There is also no apparent way to see the annuli.
The hyperbolic soccer ball construction is related to the {3, 7}
construction in the sense that if a neighborhood o f each vertex in the
{3,7} construction is replaced by a heptagon, then the rem aining portion
o f each triangle is a hexagon.
Figure B.4 The hyperbolic soccer ball constructed by Keith Henderson
” { 3 , 6 V 2 } ” P o l y h e d r a l C o n s t r u c t i o n
We can avoid some o f the disadvantages o f the {3, 7} and soccer ball
constructions by constructing a polyhedral annulus. In this construction
we have seven triangles together only at every other vertex and six trian­
gles together at the others. This construction still has the disadvantage o f

“ {3,6 V2 }” Polyhedral Construction 373
not being able to produce closer and closer approxim ations and it also is
more “pointy” (larger cone angles) than the hyperbolic soccer ball.
The precise construction can be described in two different (but, in
the end, equivalent) ways:
1. Construct polyhedral annuli as in Figure B.5 and then tape them
together as with the annular hyperbolic plane.
2. The quickest way is to start with m any strips as pictured in Figure
B.6a — these strips can be as long as you wish. Then add four o f
the strips together as in Figure B.6b, using five additional trian­
gles. Next, add another strip every place there is a vertex with
five triangles and a gap (as at the marked vertices in Figure
B.6b). Every time a strip is added an additional vertex with
seven triangles is formed.
Figure B.6a Strips
The center o f each strip runs perpendicular to each annulus, and
you can show that these curves (the center lines o f the strip) are each
geodesics because they have local reflection symmetry.

3 7 4 Appendix B — Constructions of Hyperbolic Planes
Figure B.6b Forming the polyhedral annular hyperbolic plane

Bibliography
A (cut) Short Story:
Peter wanted to know the names of the birds.
He read a book and learned the names of the birds.
Peter wanted to learn how to swim.
He read a book and drowned.
— from E. C. Basar, R. A. Bonic, et al., Studying Freshman
Calculus, Lexington, MA: D. C. Heath and Co., 1976
This bibliography consists o f books (and other items) that are referenced
in this book. This is a subset o f a more complete, and frequently
updated, annotated bibliography with more than 500 listings on the W eb
at
http://w w w.m ath.com ell.edu/~henderson/biblio
Some o f the references below are indicated to have online versions: For
these there are clickable links in the bibliography on the Web. The
section names below correspond to a subset o f the section names in the
bibliography on the Web.
A D A rt and D esign
Keith Albam, Jenn Mial Smith, Stanford Steele, and Dinah Walker. The
Language of Pattern. New York: Harper & Row, 1974.
William Blackwell. Geometry in Architecture. New York: John Wiley & Sons,
1984.
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I ndex
The numbers in bold refer to pages on which there are definitions or
statements o f the indexed item.
A
AAA = Angle-Angle-Angle, 123-4,
137, 292, 294
AAA similarity criterion, 195-6, 205,
217
absolute value, special, 79
affine space, 323
al’Hazen, 281
al’Khowarizmi, 268, 272, 376
algebra, 177, 188, 267-9, 276, 282-3
abstract groups, 143, 163
angle, 19, 37-42
as geometric shape, 39
as measure, 38
bisector, 76, 78-9, 81, 86
cone, 46-50
congruent, 37-42, 145
definitions, 37-42
dihedral, 194, 283,336-9
directed, 39, 145
dynamic notion of, 38
right, 58, 134, 172, 291,366
solid, 336-341
sum of, 87, 95, 102, 113, 122, 125,
133, 137, 347
trisection of, 6, 79, 213-6
angle-preserving = conformal
annulus, 62-3, 367-70
antipodal, 77, 292
Apollonian Circle, 218
Apollonius, 203, 217, 218, 224, 226,
274, 376
circle, 218
Problems, 226-9
Theorem, 220
Archimedean Postulate = AP, 133-4,
167, 170, 180
Archimedes, 6, 7, 133, 167-8, 214-5,
300
Archytas, 178, 187, 272
area, 38, 89-95, 165-9, 177, 181, 183,
198, 200, 262, 268, 337, 340, 347
and dissection theory, 165-76, 181,
187
theory of, 168-9
area-preserving, 200
Art/Pattem Strand, 1, 10, 61, 141, 143,
154
385

386 Index
ASA = Angle-Side-Angle, 76, 85-88,
137, 152,265, 292, 294
ASS = Angle-Side-Side, 119-21
astrolabe, 203-204
B
Baudhayana, 4, 177, 184—8, 270, 376
biangle = lune, 89-90,173, 175
Bolyai, 59, 140
Building Structures Strand, 4, 6, 10,
138, 140
c
Cardano, 280-2, 307, 376
formula, 282
cartography, 4, 197
cell division, 262, 341
charts, 197-8, 235
circle-preserving, 201, 202
circles, 9, 17-8, 22, 29, 32-3, 36, 38,
76-8, 119, 130, 182, 205,217,
273-9, 285-92, 296, 321, 365
Appollonian, 218
great, 27, 35, 49, 78
intrinsic center, 285, 292-3
intrinsic radius, 77, 285-6
inversion with respect to a, 218
of inversion, 219, 222, 224
orthogonal, 221
subtended angle in, 206-9, 217
with infinite radius, 13, 147
circumcenter, 227
Clifford parallels, 332
compass, 11, 214, 216
completeness, 175, 179, 181, 186, 187,
282
complex numbers, 281-2
cone, 21, 36, 45-58, 72, 136, 283
conformal, 200, 221-3, 231, 235^13
congruent, 37-40, 73, 80, 85, 117-23,
146, 149-50, 164, 171, 194, 292,
336-9
directly, 80
conic sections, 5, 267, 270-81
constant width curves, 310, 314
constructions, 79, 213-216
convex, 104, 106
coordinate(s)
chart, 68, 236
conformal, 236
geodesics rectangular, 67
map, 67
system, 25, 63-7
cosets, 164
cosmic background radiation, 4, 349,
356-9
covariant (= intrinsic) differentiation,
98
covering space, 51-54, 264
branch points, 256, 257
of a sphere, 256
of cone, 53
of cylinder, 51
of flat 2-manifolds, 251
of other surfaces, 256, 258
universal, 252
Coxeter, 227, 379
crochet, 63, 69, 368-70
cube, 160, 194,280,342
cube roots, 267, 272, 274, 281
cubic equations, 178, 267, 274, 276-82
curvature, 30, 36, 102, 107-108
extrinsic, 30
Gaussian, 66, 102
intrinsic, 30,35-36, 74
curve, 22-3, 30
constant width, 310
locally straight, 23
smooth, 22-23
cylinder, 21, 44-55, 72, 200

Index 387
D
dense, 252
Descartes, 7, 26, 226, 275-6, 301, 317,
376
differentiable, 22, 24
differential, 200, 201
differential geometry, 35-6, 43, 57-8,
74, 89, 97, 106, 197
dihedral group, 163
dilations, 153
discrete, 159
dissection puzzles, 167, 190-3
dissections, 165-96
three-dimensional, 105, 193, 336
distance function, 322
distortion, 235-9, 242, 328-9
dodecahedral 3-manifold, 353
dodecahedron, 342, 353
dual (= polar), 293-6, 342
dual of the Law of Cosines, 294
duality, 285, 292-5, 342
E
Ecliptic, 26
EEAT = Euclid’s Exterior Angle
Theorem, 109-11, 113, 122, 135
EFP = Euclid’s Fifth Postulate, 60,
131-8, 140, 366
on sphere and hyperbolic plane, 134
Einstein, 232, 344, 345
equiareal, 200
equidistant, 126
curves, 130
great circles, 330
equivalent by dissection, 166-95
equivalent by subtraction, 166, 183,
187, 194
Euclid, 4, 5, 29, 37, 43, 58, 109, 110,
131, 139, 186, 187, 267,270,
297, 341,376
Elements, 5, 169, 186, 205, 363-6,
376
fifth (parallel) postulate, 5, 59, 366
fourth postulate, 58, 366
Phaenomena, 27
second postulate, 58, 366
Euclidean 3-Manifolds, 349
Euler, 91, 263, 301,310, 341
Euler number, 263, 341
extrinsic, 30, 34, 36, 44, 55, 71, 264,
286
extrinsic curvature, 30
F
Fermat, 226, 227
Oat
Klein bottle, 246, 249, 250
torus, 246, 250
2-manifold, 246-52
4-space, 319, 321-2,325-34
fractals, 20
G
Gauss, 103, 140, 142, 226, 347
Gauss-Bonnet Formula, 102-8, 262
Gaussian curvature, 66, 102
geodesic rectangular coordinates, 67,
234, 236
geodesically complete, 250
geodesics, 30, 32, 36, 43-4, 48-50,
73-4, 97, 198, 292
radial, 234
asymptotic, 66, 116, 147, 148
in H3, 330
in S3, 326
locally unique, 73
radial, 66-7, 93, 148, 237, 240
geometric 2-manifolds, 245-66,257
geometric 3-manifold, 319, 344,
349-60

388 Index
geometry
absolute, 138, 140
double elliptic, 141
elliptic, 141
hyperbolic, 60
inversive, 231-2
Laguerre, 232
Lobachevskian/Bolyai =
Lobachevskian, 141
Minkowskian, 232
parabolic, 141
projective, 3, 141, 153, 297
pseudosphere, 142
Riemannian, 141
spherical, 4, 27, 60
glide reflection = glide, 146, 149,
152-3, 160
global, 44, 51,71
great
circle, 35
circle, 29-33, 49, 97, 198, 201, 289,
292-5, 326-7, 330-4, 337
hemisphere, 329, 332-3, 340
semicircle, 328, 332
2-sphere, 326-8, 330-5, 340
group, 153
sub-, 163
symmetry, 157
theory, 1, 157
H
half-turn manifold, 351
half-turn, 18, 149
helix, 18, 49
heptagon, 371
hexagonal 3-torus, 351
High School Parallel Postulate = HSP,
132-9
on sphere, 134
Hilbert, David, 139,193-4, 379, 381
Hilbert’s Third Problem, 194
holonomy, 89, 98-103, 106-8, 128,
137, 173-4, 262
homeomorphic, 248, 250, 258, 264,
352, 354, 360
homogeneous, 63, 367
horocycles, 147
horolation, 147-148, 151-3
hyperbola, 274-5, 279
hyperbolic 2-manifold, 258, 261
hyperbolic 3-space, 319, 328
hyperbolic plane, 5, 43, 59-75, 85,
91-5,97,99-108,110,113-5,
118-27, 130-2, 134-8, 145-52,
160-1, 166, 172, 175, 205,
233-45, 252, 258-68, 285, 287,
290, 294, 328, 332, 334, 340,
347-8, 367-74
annular, 62-8, 234, 237, 367-8
crocheted, 62-63, 368-70
geodesics on, 66
Poincare disk model, 242
polyhedral, 371-4
projective disk model, 244
radius, 64, 65
upper-half-plane model, 67-8,
236-41
hyperbolic reflections, 239, 243
hyperbolic soccer ball, 64,371
I
icosahedron, 342
incenter of a triangle, 228
indistinguishable, 22-23
input angle, 303, 309
intrinsic, 30, 34, 43, 47, 55, 71, 97,
264-5,286,319
curvature, 30, 35-6
intrinsic (= covarient) derivative, 36,
98
inversion, 218-32, 238
circle of, 219
conformal, 221-223

Index 389
inversive pairs, 218-23
involutes, 314-6
spherical, 316, 317
isogonal, 200
isometric, locally, 55, 71
isometric embedding, 61
isometries (see also symmetry), 15,
141,143-56,238
classification of, 149-152
composition of, 145, 150
elliptic, 148
hyperbolic, 148
parabolic, 148
isomorphic
patterns, 154-5
groups, 163, 164
Isosceles Triangle Theorem = ITT,
75-76, 78, 137, 181,206
K
Kempe, 317, 382
Khayyam, 139-40, 170-1, 267-3,
276-80, 376
parallelograms = KP, 171-2
quadrilaterals = KQ, 171-4, 208-9,
340
Klein, 141, 143, 153, 164, 249
Erlanger Program, 153, 232
KMODDL, 303, 307, 314, 317, 382
L
Lambert quadrilaterals, 139
last scattering circle, 360
last scattering surface, 357, 359
Law of Cosines, 283, 285, 287-90, 299
hyperbolic, 290
spherical, 289-290, 299, 304
Law of Sines, 285, 290-1,
spherical, 291-292, 299, 304
Lennes polyhedra, 105, 194
Leonardo da Vinci, 160, 298, 300, 307
line at infinity, 295
link of a vertex, 354
linkage, 11
four-bar, 300-303
Lobachevsky, 59, 70, 140-1, 287, 290
local, 43, 71
logarithmic spiral, 20
lime = biangle, 90, 113, 127, 135,
173-4
M
manifold (see geometric manifold)
mechanisms, 11, 224-5, 299-318
million-dollar prize, 258
Mobius, 231,254
Mobius strip (=Mobius band), 254
Mobius transformations, 231-2
motif, 155-6
Motion/Machines Strand, 6, 11, 13,
223, 299
N
Navigation/Stargazing Strand, 3, 10,
61, 140, 142,217
Newton, 226, 301
non-orientable, 251
o
octahedral space, 356
octahedron, 342
one-sixth-tum manifold, 352
one-third-tum manifold, 352
orientable, 251, 360
orthogonal circles, 221
orthogonal complement, 325
output angle, 303, 309
P
parabola, 273, 275, 278-9

3 90 Index
parallel
as non-intersecting, 136-7
in perspective drawing, 297-8
postulates on H2, 134-5
postulates on plane, 124-33, 138-9,
366
postulates on sphere, 131-2, 134-5
transport, 36, 89, 96 -7, 102-3,
109-16, 122-8, 132, 134,
136-7, 171-2,214
parallel postulates
history, 138-42
parallelogram, 12, 165-6, 169-70, 172,
189, 195,200
Varignon, 192
parametrized by arc-length, 236
partial derivatives, 200
pattern, 1, 153-60
finite, 158
isomorphic, 154
strip = frieze, 158
periodic, 161
wall-paper, 162
Peaucellier, 13, 299, 317
Peaucellier-Lipkin linkage, 13, 225,
317
perpendicular bisector, 76-9, 86, 171
perspective drawings, 2-3, 297-8
Plato, 9, 168, 335, 341, 343, 355, 363,
377
Platonic solids, 341
Playfair’s Parallel Postulate = PPP, 60,
132,138-40
Plotinus, 143,319,377
Poincare, 258, 354
Conjecture, 257-8
disk model, 212, 234, 242-3
dodecahedral space, 354
point at infinity, 218, 295-8
point-pairs, 292-5
polygons, 73, 104, 107-8, 166, 173,
177, 181, 189, 340, 365
convex, 104
regular, 340-1, 371
simple, 104
polyhedra, 194, 335-342
Lennes, 105
regular, 340-2
power of a point, 207-12, 217
on spheres, 208
Proclus, 138, 140, 377
projection, 198
cylindrical, 199
gnomic, 198, 289, 291, 294, 298
stereographic, 200-4, 205, 210, 212,
220-1,244
projective 3-space, 355
projective disk model, 234, 244
projective plane, 255, 294-6
pseudosphere, 61, 68-70
Ptolemy, 25, 28, 138, 197, 202, 203
pumps, 11, 302, 314-315
Pythagorean Theorem, 5, 168-9, 177,
186, 188-189,288,290
on S2 and H2, 290
Q
quadratic equations, 178, 268-71, 280,
282
quarter-turn manifold, 350
quatemionic manifold, 356
R
radical axis of two circles, 212-3
rectangle, 165-6, 170, 172, 179-87,
189, 200, 274-5
reflection, 18, 20, 71-4, 82, 85, 110,
144, 146, 149-57
Reuleaux, 299, 302, 383
triangle, 8, 9, 310-14
Ribbon Test, 31
Riemann, 141-2, 3 4 3 ^
sphere, 141

Index 391
Right-Leg-Hypotenuse = RLH, 120,
294
rotation, 18, 20, 33, 38, 42, 71, 72, 81,
85, 110,145, 150, 153-7, 286
s
Saccheri quadrilaterals = Khayyam
quadrilaterals, 171
SAS similarity criterion, 196, 217
scale, 22
Schlafli symbol, 161
Seifert-Weber dodecahedral space, 355
self-similarity. See = similarity
symmetry
semi-rectangle, 208-212
Shape of Space
latest evidence, 361
shortest distance, 10, 14, 56
Side-Angle-Angle = SAA, 121-4
Side-Angle-Side = SAS, 76, 80-4, 137,
152, 265, 287, 292
Side-Side-Side = SSS, 117-8, 121,
137, 205, 292, 294
similarity, 2 0 ,123, 137, 153-4, 195 ,
217, 223
simply connected, 346
smooth, 22-24, 57
solid angle measure, 336
special absolute value, 79, 304
spherical
2-manifolds, 253
coordinates, 26, 202, 204
involutes, 316-7
projective plane, 255
spherical cone, 255
spiral pump, 315-6
square roots, 177-8, 181,187, 268,
270, 273, 283
standard candles, 348
Steiner, 230, 231
Stomachion, 167-8
straight, 13-24
definitions, 15-7
infinitesimally, 21-2
intrinsically (= geodesic), 30-1, 43,
73
locally, 16, 21-2, 44
on cones and cylinders, 44-50
on hyperbolic plane, 66
on sphere, 28-35
straightedge, 9, 13, 79,213-6
strip (=ffieze) pattern, 2, 153,157-9
subgroup, 163-4
subtend, 206
support lines, 314
surface of revolution, 68-9
symmetries, 154, 332
symmetry (see also isometry), 16-20,
32-4, 7 1 -2 ,154-9, 332-3
central, 19, 34
extrinsic, 34, 71
group, 154, 157-8, 163-4
half-turn, 14,18-9, 21, 33^1, 130,
333-4
identity, 150, 156-7
intrinsic, 30, 34, 45,48
local, 15,21,31,45,48, 71
point = central, 19-20, 332-3
reflection, 9, 15,17, 19, 22, 154,
332-3
reflection-in-the-line, 14,17, 21, 32,
71-4, 332-3
reflection-perpendicular-to-the-line,
17, 19,32,332-3
rigid-motion-along-itself, 18, 33,
332-3
rotation, 1&-9, 71-2, 156, 332-3
self-similarity, 20, 154
3-dimensional rotation, 35, 154,
332-3
translation, 18, 156, 158, 332-3
symmetry group, 154 -157

3 92 Index
T
tangent, 24, 35-6, 201, 296
tetrahedra, 105, 194, 335-9, 342
congruent, 336
tetrahedral space, 355
Theon of Alexandria, 203
Theon of Smyrna, 343, 377
3-manifold, see geometric 3-manifold
3-sphere, 35, 258, 319-28, 330-3,
335-41,347
3-torus, 350
tilings, 1, 160, 161
non-periodic, 162
regular, 160
Schlafli symbol, 161
torus
flat, 247-52, 266
hexagonal, 248
topological, 248
two-holed, 258, 261
totally geodesic, 332
tractrix, 69
transformation
birational, 231
Cremona, 231
inversive, 231-2
Mobius transformations, 231-2
translation, 18, 33, 81, 85, 145-6, 150,
153, 156, 159
transversal, 113-5, 125-7, 130, 132,
134, 137
triangle, 48, 73, 75, 80, 81, 84, 85
2/3-ideal, 92, 241-2
acute, 365
area, 90-1,95, 169, 241
defect, 102
double-right, 98, 119
equilateral, 154, 341, 365
excess, 102
hyperbolic, 88
ideal, 92, 95, 241
in three dimensions, 333
interior (inside), 84
isosceles, 75-8, 98, 283, 365
obtuse, 365
on geometric manifolds, 264
Reuleaux, 310
right, 120, 209, 286, 291,365
scalene, 365
similar, 123, 137, 180-1, 183, 187,
195, 286
small, 81-7, 99-102, 110, 118-20,
137, 174, 293,334,337
triple-right, 87, 119
very small, 100
Triangle Inequality, 78, 304
trigonometry, 3, 4, 27,283, 286-91
hyperbolic, 139
spherical, 142, 289-91, 294
u
universal joint, 307-9
upper-half-plane model, 236-41
upper half-space, 328-9
V
Vertical Angle Theorem = VAT, 39,
72, 137, 195
vector, 36, 201, 323
w
wall-paper patterns, 162
Wankel engine, 8, 310
Watt, 12, 301
Z
zoom, 22-24

T hird E d itio n
EXPERIENCING
GEOMETRY
Euclidean and N on-Euclidean w ith History
David W. Henderson
Daina Taimina
S o m e Q u o t e s f r o m R e v i e w e r s
“I like the authors’ writing style and I think the explanations are very
clear…. I know that my students read this book, which is certainly saying
something. Sometimes they think it’s hard to read, but it is at least
possible— and they could think it’s hard to read because it might be
one of the few times they have actually had to read the text…. I think
the historical illustrations will be useful for creating interest with the
students. This is a nice addition to this edition of the text.”
“I found this text to be a wonderful, fresh, and innovative treatment
of geometry. The Moore method of study by doing and involving the
student works very well indeed for this subject. The style of the text
is very friendly and encouraging and gets the student involved quickly
with a give-and-take approach. The student develops insights and skills
probably not obtainable in more traditional courses. This is a very fine
text that I would strongly recommend for a beginning course in
Euclidean and non-Euclidean geometry.”
“This book remains a treasure, an essential reference. I simply know of
no other book that attempts the broad vision of geometry that this book
does, that does it by-and-large so successfully, and that pays so much
attention to geometric intuition, student cognitive development, and
rigorous mathematics.”
— J u d y R o i t m a n , U n i v e r s i t y o f K a n s a s
— B a r b a r a E d w a r d s , O r e g o n S t a t e U n i v e r s i t y
— N o r m a n J o h n s o n , U n i v e r s i t y o f I o w a
S tu d en tA id . e d . g o v
F U N D I N G V O U R F U T U R E .
Upper Saddle River, NJ 07458
www.prenhall.com

http://www.prenhall.com

Contents
Preface
Chapter 0
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Chapter 13
Chapter 14
Chapter 15
Chapter 16
Chapter 17
Chapter 18
Chapter 19
Chapter 20
Chapter 21
Chapter 22
Chapter 23
Chapter 24
Appendix A
Appendix B
Bibliography
Index