Download Non-Euclidean Geometry and a Little on How We Got Here

Non-Euclidean Geometry
and a Little on
How We Got Here
David C. Royster
UNC Charlotte
November 20, 2005
2
MATH 6118-090
Spring 2006
Contents
1 The Origins of Geometry
1.1 Introduction . . . . . . . . .
1.2 A History of the Value of π
1.2.1 Indiana and π . . . .
1.3 Egyptian Geometry . . . . .
1.4 Early Greek Geometry . . .
1.5 Euclidean Geometry . . . .
1.6 Geometry after Euclid . . .
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3 Transformation Geometry: First View
3.1 The Axioms for a Euclidean Model . . . . . . . . . . . . . . . . . . . . . . .
3.2 Triangle Congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Euclid’s Real Fifth Postulate . . . . . . . . . . . . . . . . . . . . . . . . . .
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4 To Boldly Go Where No Man Has Gone Before
4.1 The Star Trek Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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5 Euclidean Geometry: Revisited
5.1 Similar Triangles . . . . . . . . . . . .
5.2 Graph of a Linear Equation . . . . . .
5.3 Power of a Point . . . . . . . . . . . .
5.4 Medians and Centroid . . . . . . . . .
5.5 Incircle, Excircles, and Law of Cosines
5.6 The Circumcenter and its Spawn . . .
5.7 The Euler Line . . . . . . . . . . . . .
5.8 Feuerbach’s Circle . . . . . . . . . . .
5.9 Pedal Triangles and the Simson Line .
5.10 Triangle Centers and Relative Lines .
5.11 Morley’s Theorem . . . . . . . . . . .
5.12 More Triangle Centers . . . . . . . . .
5.12.1 Fermat Point . . . . . . . . . .
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2 Euclidean Geometry
2.1 The Pythagoreans . . . . . . . . . . . . . . . .
2.2 Euclid’s Axioms for Geometry . . . . . . . . . .
2.2.1 Hilbert’s Axioms for Neutral Geometry
2.2.2 Birkhoff’s Axioms for Neutral Geometry
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4
CONTENTS
5.12.2 Gergonne Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.13 Isogons and Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 Euclidean Constructions
6.1 Different Compasses? . . . . . . . . . . .
6.2 Euclidean Constructions . . . . . . . . .
6.3 The Algebra of Constructible Numbers .
6.4 Famous Problems of Greek Mathematics
6.4.1 Squaring the Circle . . . . . . . .
6.4.2 Duplicating the Cube . . . . . .
6.4.3 Trisecting an Arbitrary Angle . .
6.5 The Regular Pentagon . . . . . . . . . .
6.6 Other Constructible Figures . . . . . . .
7 Introduction to Hyperbolic Geometry
7.1 Neutral Geometry . . . . . . . . . . . .
7.1.1 Alternate Interior Angles . . . .
7.2 Weak Exterior Angle Theorem . . . . .
7.2.1 Measure of Angles and Segments
7.3 Saccheri-Legendre Theorem . . . . . . .
7.3.1 The Defect of a Triangle . . . . .
7.4 Hyperbolic Axiom Results . . . . . . . .
7.5 Angle Sums (again) . . . . . . . . . . .
7.6 Saccheri Quadrilaterals . . . . . . . . . .
7.7 Similar Triangles . . . . . . . . . . . . .
7.8 Horoparallel and Hyperparallel Lines . .
7.9 Singly Asymptotic Triangles . . . . . . .
7.10 Doubly Asymptotic Triangles . . . . . .
8 Classification of Parallels
8.1 Fan Angles . . . . . . . .
8.2 Limiting Parallel Rays . .
8.3 Hyperparallel Lines . . . .
8.4 Classification of Parallels
8.5 Proof of Claim . . . . . .
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9 Other Geometries
9.1 The Idea of Parallelism . . . . . . . . . . . . . .
9.2 Saccheri’s Work . . . . . . . . . . . . . . . . . . .
9.3 Poincaré’s Disk Model for Hyperbolic Geometry .
9.3.1 Construction of Lines . . . . . . . . . . .
9.3.2 Distance . . . . . . . . . . . . . . . . . . .
9.3.3 Parallel Lines . . . . . . . . . . . . . . . .
9.3.4 Hyperbolic Circles . . . . . . . . . . . . .
9.3.5 Similarities with Euclidean Geometry . .
MATH 6118-090
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Spring 2006
CONTENTS
10 Area in Hyperbolic Geometry
10.1 Preliminaries . . . . . . . . . . . . . . . . .
10.2 Requirements for an Area Function . . . . .
10.3 The Uniqueness of Hyperbolic Area Theory
10.4 Angle of Parallelism . . . . . . . . . . . . .
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115
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11 More in the Poincaré Disk Model
121
11.0.1 Distance between Lines . . . . . . . . . . . . . . . . . . . . . . . . . 123
11.1 Congruence in the Hyperbolic Plane . . . . . . . . . . . . . . . . . . . . . . 124
12 Poincaré Upper Half Plane Model
12.1 Vertical Lines . . . . . . . . . . . . . . . . . . . .
12.2 Isometries . . . . . . . . . . . . . . . . . . . . . .
12.3 Inversion in the Circle: Euclidean Considerations
12.4 Lines in the Poincaré Half Plane . . . . . . . . .
12.5 Fractional Linear Transformations . . . . . . . .
12.6 Cross Ratio . . . . . . . . . . . . . . . . . . . . .
12.7 Translations . . . . . . . . . . . . . . . . . . . . .
12.8 Rotations . . . . . . . . . . . . . . . . . . . . . .
12.9 Reflections . . . . . . . . . . . . . . . . . . . . . .
12.10Distance and Lengths . . . . . . . . . . . . . . .
12.11The Hyperbolic Axioms . . . . . . . . . . . . . .
12.12The Area of Triangles . . . . . . . . . . . . . . .
12.13The Poincaré Disk Model . . . . . . . . . . . . .
12.14Angle of Parallelism . . . . . . . . . . . . . . . .
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125
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13 The Pseudosphere
147
14 Hypercycles and Horocycles
151
15 Hyperbolic Trigonometry
155
15.1 Circumference and Area of a Circle . . . . . . . . . . . . . . . . . . . . . . . 162
16 Hyperbolic Analytic Geometry
165
16.1 Saccheri Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
16.2 More on Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
16.3 Coordinate Geometry in the Hyperbolic Plane . . . . . . . . . . . . . . . . . 167
A Inversion in Euclidean Circles
173
B An Overview of Babylonian Mathematics
179
B.1 Introduction to Babylonian Mathematics . . . . . . . . . . . . . . . . . . . . 179
B.2 Pythagoras’s theorem in Babylonian mathematics . . . . . . . . . . . . . . . 183
C An Overview of Egyptian Mathematics
MATH 6118-090
189
Spring 2006
6
MATH 6118-090
CONTENTS
Spring 2006
Chapter 1
The Origins of Geometry
1.1
Introduction
In the beginning geometry was a collection of rules for computing lengths, areas, and volumes. Many were crude approximations derived by trial and error. This body of knowledge, developed and used in construction, navigation, and surveying by the Babylonians
and Egyptians, was passed to the Greeks. The Greek historian Herodotus (5th century BC)
credits the Egyptians with having originated the subject, but there is much evidence that
the Babylonians, the Hindu civilization, and the Chinese knew much of what was passed
along to the Egyptians.
The Babylonians of 2,000 to 1,600 BC knew much about navigation and astronomy,
which required a knowledge of geometry. Clay tablets from the Sumerian (2100 BC) and the
Babylonian cultures (1600 BC) include tables for computing products, reciprocals, squares,
square roots, and other mathematical functions useful in financial calculations. Babylonians were able to compute areas of rectangles, right and isosceles triangles, trapezoids and
circles. They computed the area of a circle as the square of the circumference divided by
twelve. The Babylonians were also responsible for dividing the circumference of a circle into
360 equal parts. They also used the Pythagorean Theorem (long before Pythagoras), performed calculations involving ratio and proportion, and studies the relationships between
the elements of various triangles. See Appendices A and B for more about the mathematics
of the Babylonians.
1.2
A History of the Value of π
The Babylonians also considered the circumference of the circle to be three times the diameter. Of course, this would make π = 3 — a small problem. This value for π carried along
to later times. The Roman architect Vitruvius took π = 3. Prior to this it seems that the
Chinese mathematicians had taken the same value for π. This value for π was sanctified by
the ancient Jewish civilization and sanctioned in the scriptures. In I Kings 7:23 we find:
He then made the sea of cast metal: it was round in shape, the
diameter rim to rim being ten cubits: it stood five cubits high,
and it took a line thirty cubits long to go round it.
— The New English Bible
7
8
CHAPTER 1. THE ORIGINS OF GEOMETRY
The same verse can be found in II Chronicles 4:2. It occurs in a list of specifications
for the great temple of Solomon, built around 950 BC and its interest here is that it gives
π = 3. Not a very accurate value of course and not even
√ very accurate in its day, for the
25
Egyptian and Mesopotamian values of 8 = 3.125 and 10 = 3.162 have been traced to
much earlier dates. Now in defense of Solomon’s craftsmen it should be noted that the
item being described seems to have been a very large brass casting, where a high degree
of geometrical precision is neither possible nor necessary. Rabbi Nehemiah attempted to
change the value of π to 22/7 but was rejected.
The fact that the ratio of the circumference to the diameter of a circle is constant has
been known for so long that it is quite untraceable. The earliest values of π including the
’Biblical’ value of 3, were almost certainly found by measurement. In the Egyptian Rhind
Papyrus, which is dated about 1650 BC, there is good evidence for 4( 98 )2 = 3.16 as a value
for π.
The first theoretical calculation of π seems to have been carried out by Archimedes of
Syracuse (287-212 BC). He obtained the approximation
223
22
<π< .
71
7
Before giving an indication of his proof, notice that very considerable sophistication involved
in the use of inequalities here. Archimedes knew, what so many people to this day do not,
that π does not equal 22/7, and made no claim to have discovered the exact value. If we
take his best estimate as the average of his two bounds we obtain 3.1418, an error of about
0.0002π.
The following is Archimedes’ argument.
Consider a circle of radius 1, in which we inscribe a regular polygon of 3×2n−1 sides, with
semiperimeter bn , and superscribe a regular polygon of 3 × 2n−1 sides, with semiperimeter
an .
The diagram for the case n = 2 is on the
right.
T
The effect of this procedure is to define
an increasing sequence {b1 , b2 , b3 , . . .} and a
B
decreasing sequence {a1 , a2 , a3 , . . .} so that
A
both sequences have limit π.
We are going to use some trigonometπ/k
ric notation which was not available to
Archimedes. We can see that the two semiO
perimeters are given by
an = K tan(π/K), bn = K sin(π/K),
where K = 3 × 2n−1 . Likewise, we have
an+1 = 2K tan(π/2K), bn+1 = 2K sin(π/2K).
Now, you can use a couple of trigonometric
identities to show that
Figure 1.1: Archimedes Argument
1
1
2
+
=
(1.1)
an bn
an+1
an+1 bn = (bn+1 )2 .
(1.2)
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Spring 2006
1.2. A HISTORY OF THE VALUE OF π
9
√
√
Archimedes started from a1 = 3 tan(π/3) = 3 3 and b1 = 3 sin(π/3) = 3 3/2 and
calculated a2 using Equation (1.1), then b2 using (1.2), then a3 using (1.1), then b3 using
(1.2), and so forth. He continued until he had calculated a6 and b6 . His conclusion was that
b6 < π < a6 .
Archimedes did not have the advantage of an algebraic and trigonometric notation
and had to derive (1.1) and (1.2) purely by geometry. Moreover he did not even have
the advantage of our decimal notation for numbers, so that the calculation of a6 and b6
from (1.1) and (1.2) was not an easy task. So this was a pretty stupendous feat both of
imagination and of calculation. Our real wonder is not that he stopped with polygons of
96 sides, but that he went so far.
Now, if we can compute π to this accuracy, we should be able to compute it to greater
accuracy. Various people did, including:
Ptolemy
Zu Chongzhi
al-Khwarizmi
al-Kashi
Vite
Roomen
Van Ceulen
(c. 150 AD)
(430-501 AD)
(c. 800 )
(c. 1430)
(1540-1603)
(1561-1615)
(c. 1600)
3.1416
355/113
3.1416
14 places
9 places
17 places
35 places
Table 1.1: Early Calculations of π
Except for Zu Chongzhi, about whom little to nothing is known and who is very unlikely
to have known about Archimedes’ work, there was no theoretical progress involved in these
improvements, only greater stamina in calculation. Notice how the all of this work, as in
all scientific matters, passed from Europe to the East for the millennium 400 to 1400 AD.
Al-Khwarizmi lived in Baghdad, and incidentally gave his name to algorithm, while the
words al jabr in the title of one of his books gave us the word algebra. Al-Kashi lived still
further east, in Samarkand, while Zu Chongzhi, of course, lived in China.
The European Renaissance brought about in due course a whole new mathematical
world. It was also a move from the geometry calculation of π seen earlier to an analytic
computation. Among the first effects of this reawakening was the emergence of mathematical
formulæ for π. One of the earliest was the work of Wallis (1616-1703)
1 · 3 · 3 · 5 · 5 · 7 · ...
2
=
.
π
2 · 2 · 4 · 4 · 6 · 6 · ...
One of the best-known formulæ is
π
1 1 1
= 1 − + − + ....
4
3 5 7
This formula is sometimes attributed to Leibniz (1646-1716) but it seems to have been
first discovered by James Gregory (1638-1675).
These are both dramatic and astonishing formulæ, because the expressions on the righthand side are completely arithmetical in nature, while π arises from geometry.
From the point of view of the calculation of π, however, neither of these is of any use
at all. In Gregory’s series, for example, to get 4 decimal places correct we require the error
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10
CHAPTER 1. THE ORIGINS OF GEOMETRY
to be less than 0.00005 = 1/20, 000, and so we need about 10,000 terms of the series to get
this type of accuracy. However, Gregory also showed the more general result than
tan−1 (x) = x −
x3 x5
+
− . . . (−1 ≤ x ≤ 1)
3
5
(1.3)
We get the first series from this one by substituting x = 1. Now, use the fact that
µ
¶
1
π
−1
√
tan
=
6
3
we get
π
1
=√
6
3
¶
µ
1
1
1
+
−
+ ...
1−
3·3 5·3·3 7·3·3·3
which converges much more quickly. The 10th term is
1
√ ,
19 × 39 3
which is less than 0.00005, and so we have at least 4 places correct after just 9 terms.
An even better idea is to take the formula
µ ¶
µ ¶
π
−1 1
−1 1
= tan
+ tan
(1.4)
4
2
3
and then calculate the two series obtained by putting first 12 and the 13 into (1.3).
We could get even faster convergence if we could find some large integers a and b so that
µ ¶
µ ¶
π
−1 1
−1 1
= tan
+ tan
4
a
b
In 1706 Machin found such a formula:
π
= 4 tan−1
4
µ ¶
µ
¶
1
1
−1
+ tan
5
239
(1.5)
This is not too hard to prove. If you can prove (1.4) then there is no real extra difficulty
about (1.5), except that the arithmetic is worse. The amazing fact is that someone came
up with it in the first place.
When you have a formula like this, the only difficulty in computing π is the monotony
of completing the calculation. A few people did devote vast amounts of time and effort to
this tedious effort. One of them, an Englishman named Shanks, used Machin’s formula to
calculate π to 707 places, publishing the results of his many years of labor in 1873. Here is
a summary of how the improvement went:
Shanks knew that π was irrational since this had been proved in 1761 by Lambert.
Shortly after Shanks’ calculation it was shown by Lindemann in 1882 that π is transcendental, that is, π is not the solution of any polynomial equation with integer coefficients.
In fact this result of Lindemann showed that squaring the circle is impossible. The transcendentality of π implies that there is no ruler and compass construction to construct a
MATH 6118-090
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1.2. A HISTORY OF THE VALUE OF π
Sharp
Machin
1699
1701
de Lagny
Vega
Rutherford
Shanks
1719
1789
1841
1873
11
71 digits using Gregory’s result
used an improvement to get 100 digits
and the following used his methods:
112 correct digits
126 places and in 1794 got 136
calculated 152 digits and in 1853 got 440
calculated 707 places of which 527 were correct
Table 1.2: Calculation of decimal value of π
square equal in area to a given circle. Note that this takes us back to the geometry from
the analytic and answers a geometry question via algebraic and analytic means.
Very soon after Shanks’ calculation a curious statistical anomaly was noticed by De
Morgan, who found that in the last of 707 digits there was a suspicious shortage of 7’s.
He mentions this in his Budget of Paradoxes of 1872 and it remained a curiosity until 1945
when Ferguson discovered that Shanks had made an error in the 528th place, after which
all his digits were wrong. In 1949 a computer was used to calculate π to 2000 places. In
this and all subsequent computer expansions the number of 7’s does not differ significantly
from its expectation, and indeed the sequence of digits has so far passed all statistical tests
for randomness.
We do know a lot more about π, and all of it is analytic information. There are very
interesting algorithms for computing π with great accuracy and speed. Since the beginning
of the computer age we have π computed to greater and greater numbers of digits. The
current record is over 1.2 trillion digits. See Table 1.3.
Just a note on how the notation π arose. Oughtred in 1647 used the symbol d/π for the
ratio of the diameter of a circle to its circumference. David Gregory (1697) used π/r for
the ratio of the circumference of a circle to its radius. The first to use π with its present
meaning was an Welsh mathematician William Jones in 1706 when he states 3.14159 and
c = π. Euler adopted the symbol in 1737 and it quickly became a standard notation.
1.2.1
Indiana and π
There are many false claims about the value of π, most of them coming from attempts to
square the circle — one of the great unsolved problems of Greek mathematics. One of the
best stories is that of the Indiana legislature in 1897. In the State of Indiana in 1897 the
House of Representatives unanimously passed a Bill introducing a new mathematical truth.
B e it enacted by the General Assembly of the State of Indiana: It has been found that
a circular area is to the square on a line equal to the quadrant of the circumference, as the
area of an equilateral rectangle is to the square of one side.
(Section I, House Bill No. 246, 1897)
The author of the bill was Edwin J. Goodwin, an M.D., of Solitude, Indiana. It seems
that he was a crank (or amateur) mathematician who had been working on a procedure to
square the circle. He contacted his Representative, one Taylor I. Record of Posey County
on January 18, 1897, with his epoch-making suggestion: if the State would pass an Act
recognizing his discovery, he would allow all Indiana textbooks to use it without paying him
a royalty.
What a novel idea! How could the legislature pass up such an offer? Nobody in the
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CHAPTER 1. THE ORIGINS OF GEOMETRY
Mathematician
Ferguson
Ferguson & Wrench
Smith & Wrench
Reitwiesner et al.
Nicholson & Jeenel
Felton
Genuys
Felton
Guilloud
Shanks & Wrench
Guilloud & Filliatre
Guilloud & Dichampt
Guilloud & Bouyer
Miyoshi & Kanada
Guilloud
Tamura
Tamura & Kanada
Tamura & Kanada
Kanada, Yoshino & Tamura
Ushiro & Kanada
Gosper
Bailey
Kanada & Tamura
Kanada & Tamura
Kanada, Tamura & Kubo
Kanada & Tamura
Chudnovskys
Chudnovskys
Kanada & Tamura
Chudnovskys
Kanada & Tamura
Chudnovskys
Chudnovskys
Kanada & Takahashi
Kanada & Takahashi
Kanada & Takahashi
Chudnovskys
Kanada & Takahashi
Kanada & Takahashi
Kanada & Takahashi
Kanada & Takahashi
Kanada et al.
Date
01/1947
09/1947
1949
1949
1954
1957
01/1958
05/1958
1959
1961
1966
1967
1973
1981
1982
1982
1982
1982
1982
10/1983
10/1985
01/1986
09/1986
10/1986
01/1987
01/1988
05/1989
06/1989
07/1989
08/1989
11/1989
08/1991
05/1994
06/1995
08/1995
10/1995
03/1996
04/1997
06/1997
04/1999
09/1999
12/2002
Places
710
808
1,120
2,037
3,092
7,480
10,000
10,021
16,167
10,0265
250,000
500,000
1,001,250
2,000,036
2,000,050
2,097,144
4,194,288
8,388,576
16,777,206
10,013,395
17,526,200
29,360,111
33,554,414
67,108,839
134,217,700
201,326,551
480,000,000
525,229,270
53,6870,898
1,011,196,691
1,073,741,799
2,260,000,000
4,044,000,000
3,221,220,000
4,294,967,286
6,442,450,000
8,000,000,000
17,179,869,142
51,539,600,000
68,719,470,000
206,158,430,000
1,241,100,000,000
Type of computer
Desk calculator
Desk calculator
Desk calculator
ENIAC
NORAC
PEGASUS
IBM 704
PEGASUS
IBM 704
IBM 7090
IBM 7030
CDC 6600
CDC 7600
FACOM M-200
unknown
MELCOM 900II
HITACHI M-280H
HITACHI M-280H
HITACHI M-280H
HITACHI S-810/20
SYMBOLICS 3670
CRAY-2
HITACHI S-810/20
HITACHI S-810/20
NEC SX-2
HITACHI S-820/80
CRAY-2
IBM 3090
Hitachi S-820/80
m-zero
m-zero
Hitachi S-3800/480
Hitachi S-3800/480
Hitachi S-3800/480
Hitachi S-3800/480
Hitachi S-3800/480
m-zero
Hitachi SR2201
Hitachi SR2201
Hitachi SR8000
Hitachi SR8000
Hitachi SR8000/MP
Table 1.3: Computer calculations of π
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1.2. A HISTORY OF THE VALUE OF π
13
Indiana Legislature knew enough mathematics to know that the discovery was nonsense, as
it had been proved in the 1700’s that it is impossible to square the circle. In due course
the bill had its third House reading, and passed 67-0. At this point the text of the bill was
published and, of course, became the target for ridicule, in this and other states.
The next day, the following article appeared in the Indianapolis Sentinel:
To SQUARE THE CIRCLE
Claims Made That This Old Problem Has Been Solved.
“The bill telling how to square a circle, introduced in the House by Mr. Record, is not
intended to be a hoax. Mr. Record knows nothing of the bill with the exception that he
introduced it by request of Dr.Edwin Goodwin of Posey County, who is the author of the
demonstration. The latter and State Superintendent of Public Instruction Geeting believe
that it is the long-sought solution of the problem, and they are seeking to have it adopted by
the legislature. Dr. Goodwin, the author, is a mathematician of note. He has it copyrighted
and his proposition is that if the legislature will endorse the solution, he will allow the state
to use the demonstration in its textbooks free of charge. The author is lobbying for the
bill.”
On “February 2, 1897, ...Representative S.E. Nicholson, of Howard County, chairman of
the Committee on Education, reported to the House.
“Your Committee on Education, to which was referred House Bill No. 246, entitled a a
bill for an act entitled an act introducing a new mathematical truth, has had same under
consideration, and begs leave to report the same back to the House with the recommendation
that said bill do pass.
“The report was concurred in, and on February 8,1897, it was brought up for the second
reading, following which it was considered engrossed. Then ’Mr. Nicholson moved that the
constitutional rule requiring bills to be read on three days be suspended, that the bill may
be read a third time now.’ The constitutional rule was suspended by a vote of 72 to 0 and
the bill was then read a third time. It was passed by a vote of 67 to 0, and the Clerk of the
House was directed to inform the Senate of the passage of the bill.”
The newspapers reported the suspension of the constitutional rules and the unanimous
passage of the bill matter-of-factly, except for one line in the Indianapolis Journal to the
effect that “this is the strangest bill that has ever passed an Indiana Assembly.”
The bill was referred to the Senate on Feb. 10,1897 and was read for the first time on
Feb. 11 and referred to the Committee on Temperance. “On Feb.12 Senator Harry S. New,
of Marion County, Chairman of the Committee on Temperance, made the following report
to the Senate:
“Your committee on Temperance, to which was referred House Bill No. 246, introduced
by Mr.Record, has had the same under consideration and begs leave to report the same
back to the Senate with the recommendation that said bill do pass.”
The Senate Journal mentions only that the bill was read a second time on Feb. 12,
1897, that there was an unsuccessful attempt to amend the bill by strike out the enacting
clause, and finally it was postponed indefinitely. That the bill was killed appears to be a
matter of dumb luck rather than the superior education or wisdom of the Senate. It is true
that the bill was widely ridiculed in Indiana and other states, but what actually brought
about the defeat of the bill is recorded by Professor C. A. Waldo in an article he wrote
for the Proceedings of the Indiana Academy of Science in 1916. The reason he knows is
that he happened to be at the State Capitol lobbying for the appropriation of the Indiana
Academy of Science, on the day the Housed passed House Bill 246. When he walked in the
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CHAPTER 1. THE ORIGINS OF GEOMETRY
found the debate on House Bill 246 already in progress. In his article, he writes (according
to Edington):
“An ex-teacher from the eastern part of the state was saying: ‘The case is perfectly
simple. If we pass this bill which establishes a new and correct value for π, the author offers
to our state without cost the use of his discovery and its free publication in our school text
books, while everyone else must pay him a royalty.’ The roll was then called and the bill
passed its third and final reading in the lower house. A member then showed the writer [i.e.
Waldo -AA] a copy of the bill just passed and asked him if he would like an introduction
to the learned doctor, its author. He declined the courtesy with thanks remarking that he
was acquainted with as many crazy people as he cared to know.
“That evening the senators were properly coached and shortly thereafter as it came to
its final reading in the upper house they threw out with much merriment the epoch making
discovery of the Wise Man from the Pocket.”
Note that this value of π is mentioned in the Guiness Book of Records — as the most
inaccurate value for π.
It appears that Dr. Goodwin was not satisfied with just one value of π, as the following
values appear in the bill:
“Since the rule in present use [presumably π equals 3.14159...] fails to work ..., it should
be discarded as wholly wanting and misleading in the practical applications,” the bill declared. Instead, mathematically inclined Hoosiers could take their pick among the following
formulae:
1. The ratio of the diameter of a circle to its circumference is 5/4 to 4. In other words,
π = 16/5 = 3.2.
2. The area of a circle equals the area of a square whose side is 1/4 the circumference of
the circle. Working this out algebraically, we see that π must be equal to 4.
3. The ratio of the length of a 90◦ arc to the√length of a segment connecting the arc’s
two endpoints is 8 to 7. This gives us π = 2 × 16/7, or about 3.23.
There may have been other values for π as well; the bill was so confusingly written that
it’s impossible to tell exactly what Dr. Goodwin was getting at. Mathematician David
Singmaster says he found six different values in the bill, plus three more in Goodwin’s other
writings and comments, for a total of nine.
1.3
Egyptian Geometry
Now, the Egyptians were not nearly as inventive as the Babylonians but they were extensive
users of mathematics, especially geometry. They were extremely accurate in their construction, making the right angles in the Great Pyramid of Giza accurate to one part in 27,000.
From the above approximation they computed the area of a circle to be the square of 8/9
of the diameter.
µ ¶2 µ ¶2
8
16
A=
d =
r2
9
9
They also knew the Pythagorean Theorem and were able to compute volumes and dihedral
angles of pyramids and cylinders. See Appendices B and C
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1.4. EARLY GREEK GEOMETRY
1.4
15
Early Greek Geometry
The ancient knowledge of geometry was passed on to the Greeks. Maybe we should say
that they gathered all that they could find about geometry. They seemed to be blessed
with an inclination toward speculative thinking and the leisure to pursue this inclination.
They insisted that geometric statements be established by deductive reasoning rather than
trial and error. This began with Thales of Miletus (624–547 BC). He was familiar with the
computations, right or wrong, handed down from Egyptian and Babylonian mathematics. In
determining which of the computations were correct, he developed the first logical geometry.
This orderly development of theorems by proof was the distinctive characteristic of Greek
mathematics and was new. He is credited with proving the following results:
1. A circle is bisected by any diameter.
2. The base angles of an isosceles triangle are equal.
3. The angles between two intersecting straight lines are equal.
4. Two triangles are congruent if they have two angles and one side equal.
5. An angle inscribed in a semicircle is a right angle.
This new mathematics of Thales was continued over the next two centuries by Pythagoras of Samos (569–475 BC) and his disciples. Pythagoras is regarded as the first pure
mathematician to logically deduce geometric facts from basic principles. He is credited
with proving many theorems such as the angles of a triangle summing to 180◦ , and the
infamous Pythagorean Theorem for a right-angled triangle (which had been known experimentally in Egypt and Babylon for over 1000 years). The Pythagorean school is considered
as the (first documented) source of logic and deductive thought, and may be regarded as
the birthplace of reason itself. As philosophers, they speculated about the structure and
nature of the universe: matter, music, numbers, and geometry. The Pythagoreans, as a
religious sect, believed that the elevation of the soul and union with God were achieved by
the study of music and mathematics. They developed a large body of mathematics by using
the deductive method.
Their foundation of plane geometry was brought to a conclusion around 440 BC in the
Elements by the mathematician Hippocrates of Chios (470-410 BC). This treatise has been
lost, but many historians agree that it probably covered most of Books I-IV of Euclid’s
Elements, which appeared about a century later, circa 300 BC. In this first Elements,
Hippocrates included geometric solutions to quadratic equations and early methods of integration. He studied the classic problem of squaring the circle showing how to square a lune.
He worked on duplicating the cube which he showed equivalent to constructing two mean
proportionals between a number and its double. Hippocrates was also the first to show that
the ratio of the areas of two circles was equal to the ratio of the squares of their radii.
Plato (427-347 BC) founded “The Academy” in 387 BC which flourished until 529 AD.
He developed a theory of Forms, in his book Phaedo, which considers mathematical objects
as perfect forms (such as a line having length but no breadth). He emphasized the idea of
proof and insisted on accurate definitions and clear hypotheses, paving the way for Euclid,
but we cannot attribute any major mathematical discoveries to him.
Theætetus of Athens (417-369 BC) was a student of Plato’s, and the creator of solid
geometry. He was the first to study the octahedron and the icosahedron, and thus construct
MATH 6118-090
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CHAPTER 1. THE ORIGINS OF GEOMETRY
all five regular (or Platonic) solids. This work of his formed Book XIII of Euclid’s Elements.
His work about rational and irrational quantities also formed Book X of Euclid.
Eudoxus of Cnidus (408-355 BC) developed a precursor to algebra by developing a theory
of proportion which is presented in Book V of Euclid’s Elements in which Definitions 4 and
5 establish Eudoxus’ landmark concept of proportion. In 1872, Dedekind stated that his
work on cuts for the real number system was inspired by the ideas of Eudoxus. Eudoxus
also did early work on integration using his method of exhaustion by which he determined
the area of circles and the volumes of pyramids and cones. This was the first seed from
which the calculus grew two thousand years later.
Menaechmus (380-320 BC) was a pupil of Eudoxus, and discovered the conic sections.
He was the first to show that ellipses, parabolas, and hyperbolas are obtained by cutting a
cone in a plane not parallel to the base.
Euclid of Alexandria (325-265 BC) is best known for his 13 Book treatise The Elements
(∼300 BC), collecting the theorems of Pythagoras, Hippocrates, Thætetus, Eudoxus and
other predecessors into a logically connected whole.
Euclid was a disciple of the Platonic school. Around 300 BC he produced the
definitive treatment of Greek geometry and number theory in his thirteenvolume Elements. In compiling this masterpiece Euclid built on the experience and achievements of his predecessors in preceding centuries: on the
Pythagoreans for Books I-IV, VII, and IX, on Archytas for Book VIII, on
Eudoxus for Books V, VI, and XII, and on Theætetus for Books X and
XIII. So completely did Euclid’s work supersede earlier attempts at presenting geometry that few traces remain of these efforts. It’s a pity that Euclid’s
heirs have not been able to collect royalties on his work, for he is the most
widely read author in the history of mankind. His approach to geometry has
dominated the teaching of the subject for over two thousand years. Moreover,
the axiomatic method used by Euclid is the prototype for all of what we now
call pure mathematics. It is pure in the sense of pure thought: no physical
experiments need be performed to verify that the statements are correct-only
the reasoning in the demonstrations need be checked.
In this treatise, he organized this large body of known mathematics, including discoveries
of his own, into the first formal system of mathematics. This formalness was exhibited by
the fact that the Elements began with an explicit statement of assumptions called axioms or
postulates, together with definitions. The other statements — theorems , lemmæ, corollaries
— were then shown to follow logically from these axioms and definitions. Books I-IV, VII,
and IX of the work dealt primarily with mathematics which we now classify as geometry,
and the entire structure is what we now call Euclidean geometry.
1.5
Euclidean Geometry
Euclidean geometry was certainly conceived by its creators as an idealization of physical
geometry. The entities of the mathematical system are concepts, suggested by, or abstracted
from, physical experience but differing from physical entities as an idea of an object differs
from the object itself. However, a remarkable correlation existed between the two systems.
The angle sum of a mathematical triangle was stated to be 180◦ , if one measured the angles
MATH 6118-090
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1.6. GEOMETRY AFTER EUCLID
17
of a physical triangle the angle sum did indeed seem to be 180◦ , and so it went for a
multitude of other relations. Because of this agreement between theory and practice, it is
not surprising that many writers came to think of Euclid’s axioms as self evident truths.
Centuries later, the philosopher Immanuel Kant even took the position that the human
mind is essentially Euclidean and can only conceive of space in Euclidean terms. Thus,
almost from its inception, Euclidean geometry had something of the character of dogma.
Euclid based his geometry on five fundamental assumptions:
Postulate I: For every point P and for every point Q not equal to P there exists a unique
line ` that passes through P and Q.
Postulate II: For every segment AB and for every segment CD there exists a unique point
E such that B is between A and E and segment CD is congruent to segment BE.
Postulate III: For every point O and every point A not equal to O there exists a circle
with center O and radius OA.
Postulate IV: All right angles are congruent to each other.
Before we study the Fifth Postulate, let me say a few words about his definitions.
Euclid’s methods are imperfect by modern standards. He attempted to define everything
in terms of a more familiar notion, sometimes creating more confusion than he removed.
As an example:
A point is that which has no part.
A line is breadthless length. A straight line is a line which lies evenly with the points on
itself.
A plane angle is the inclination to one another of two lines which meet. When a straight
line set upon a straight line makes adjacent angles equal to one another, each of the equal
angles is a right angle.
Euclid did not define length, distance, inclination, or set upon. Once having made the
above definitions, Euclid never used them. He used instead the rules of interaction between
the defined objects as set forth in his five postulates and other postulates that he implicitly
assumed but did not state.
Postulate V: If a straight line falling on two straight lines makes the interior angles on
the same side less than two right angles, the two straight lines, if produced indefinitely, meet
on that side on which the angles are less than two right angles.
1.6
Geometry after Euclid
Archimedes of Syracuse (287-212 BC) is regarded as the greatest of Greek mathematicians,
and was also an inventor of many mechanical devices (including the screw, the pulley, and
the lever). He perfected integration using Eudoxus’ method of exhaustion, and found the
areas and volumes of many objects. A famous result of his is that the volume of a sphere
is two-thirds the volume of its circumscribed cylinder, a picture of which was inscribed on
his tomb. He gave accurate approximations to π and square roots. In his treatise On Plane
Equilibriums, he set out the fundamental principles of mechanics, using the methods of
geometry, and proved many fundamental theorems concerning the center of gravity of plane
figures. In On Spirals, he defined and gave the fundamental properties of a spiral connecting
radius lengths with angles as well as results about tangents and the area of portions of the
curve. He also investigated surfaces of revolution, and discovered the 13 semi-regular (or
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CHAPTER 1. THE ORIGINS OF GEOMETRY
Archimedean) polyhedra whose faces are all regular polygons. He was a great inventor of
war machines for the soldiers of Syracuse and is credited with many other inventions. He
was killed by a Roman soldier 212 BC when the Romans invaded Syracuse.
Apollonius of Perga (262-190 BC) was called ‘The Great Geometer’. His famous work
was Conics consisting of 8 Books In Books 5 to 7, he studied normals to conics, and
determined the center of curvature and the evolute of the ellipse, parabola, and hyperbola.
In another work Tangencies, he showed how to construct the circle which is tangent to three
objects (points, lines or circles). He also computed an approximation for π better than the
one of Archimedes.
Hipparchus of Rhodes (190-120 BC) is the first to systematically use and document
the foundations of trigonometry, and may have invented it. He published several books of
trigonometric tables and the methods for calculating them. He based his tables on dividing
a circle into 360 degrees with each degree divided into 60 minutes. This is the first recorded
use of this subdivision. In other work, he applied trigonometry to astronomy making it a
practical predictive science.
Heron of Alexandria (10-75 AD) wrote Metrica which gives methods for computing
areas and volumes. Book I considers areas of plane figures and
p surfaces of 3D objects, and
contains his now-famous formula for the area of a triangle = s(s − a)(s − b)(s − c) where
s = (a + b + c)/2 is the semiperimeter. Book II considers volumes of three-dimensional
solids. Book III deals with dividing areas and volumes according to a given ratio, and gives
a method to find the cube root of a number.
Menelaus of Alexandria (70-130 AD) developed spherical geometry in his only surviving
work Sphaerica. In Book I, he defines spherical triangles using arcs of great circles. This
marked a turning point in the development of spherical trigonometry. Book II applies
spherical geometry to astronomy; and Book III deals with spherical trigonometry including
Menelaus’s theorem about how a straight line cuts the three sides of a triangle in proportions
whose product is −1.
Claudius Ptolemy (85-165 AD) wrote the Almagest 1 giving the mathematics for the
geocentric theory of planetary motion. The books are believed to have been written in 150
AD. The work is considered to be one of the great masterpieces of early mathematical and
scientific works. The Almagest remained the major work in astronomy for 1400 years until it
was superceded by the heliocentric theory of Copernicus. In Books I and II Ptolemy refined
the foundations of trigonometry based on the chords of a circle established by Hipparchus.
In this treatise is one of the results for which he is known , known as Ptolemy’s Theorem,
that states that for a quadrilateral inscribed in a circle, the product of its diagonals is equal
to the sum of the products of its opposite sides. From this, he derived the chord formulæ
for sin(a + b), sin(a − b), and sin(a/2), and used these to compute detailed trigonometric
tables.
Pappus of Alexandria (290-350 AD) was the last of the great Greek geometers. His major
work in geometry is Synagoge or The Collection, a handbook on a wide variety of topics:
arithmetic, mean proportionals, geometrical paradoxes, regular polyhedra, the spiral and
quadratrix, trisection, honeycombs, semiregular solids, minimal surfaces, astronomy, and
mechanics. In Book VII, he proved Pappus’ Theorem which forms the basis of modern
projective geometry; and also proved Guldin’s Theorem (rediscovered in 1640 by Guldin)
to compute a volume of revolution.
1
The name Almagest is the Latin form of the shortened title “al mijisti” of the Arabic title ”al-kitabu-lmijisti”, meaning “The Great Book”.
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Hypatia of Alexandria (370-415 AD) was the first woman to make a substantial contribution to the development of mathematics. She learned mathematics and philosophy from
her father Theon of Alexandria, and assisted him in writing an eleven part commentary
on Ptolemy’s Almagest, and a new version of Euclid’s Elements. Hypatia also wrote commentaries on Diophantus’s Arithmetica, Apollonius’s Conics and Ptolemy’s astronomical
works. About 400 AD, Hypatia became head of the Platonist school at Alexandria, and
lectured there on mathematics and philosophy. Although she had many prominent Christians as students, she ended up being brutally murdered by a fanatical Christian sect that
regarded science and mathematics to be pagan. Nevertheless, she is the first woman in
history recognized as a professional geometer and mathematician.
One of the themes that ran through Geometry following Euclid was the search for a
replacement or a proof of dependence of his Fifth Postulate. No one seemed to like this
Fifth Postulate, possibly not even Euclid himself—he did not use it until Proposition 29.
The reason that this statement seems out of place is that the first four postulates seem to
follow from experience—try to draw more than one line through 2 different points. The
Fifth Postulate is nonintuitive. It does come from the study of parallel lines, though. An
equivalent statement to this postulate is :
Playfair’s Postulate: Given a line and a point not on that line, there exists one and only
one line through that point parallel to the given line.
Euclid’s Fifth Postulate seemed to be too burdensome. If it is so complicated, then it
should follow from the other axioms. Since it is not intuitive, we should be able to prove it
as a theorem. We should be able to prove that it is dependent in this Axiom system. If we
have a set of axioms A1 , A2 , . . . ,An for our mathematical system and we can prove that
Axiom An is derivable, or provable, from the other axioms, then An is indeed redundant.
In a sense analogous to linear algebra, we are looking for a basis for this mathematical
system. Unlike vector spaces and linear algebra, however, there is not a unique number of
elements in this basis, for it includes the axioms, definitions, and the rules of logic that you
use.
Many people have tried to prove the Fifth Postulate. The first known attempt to prove
Euclid V, as it became known, was by Posidonius(1st century BC). He proposed to replaced
the definition of parallel lines (those that do not intersect) by defining them as coplanar
lines that are everywhere equidistant from one another. It turns out that without Euclid V
you cannot prove that such lines exist. It is true that such a statement that parallel lines
are equidistant from one another is equivalent to Euclid V.
Ptolemy followed with a proof that used the following assumption:
For every line ` and every point P not on `, there exists at most one line m through P such
that m is parallel to l.
We will show in Section 3.3 that this statement is equivalent to Euclid V, and therefore this
did not constitute a proof of Euclid V.
Proclus (410-485 AD) also attempted to prove Euclid V. His argument used a limiting
process. He retained all of Euclid’s definitions, all of his assumptions except Euclid V, and
hence all of his propositions which did not depend on Euclid V. His plan was (1) to prove
on this basis that a line which meets one of two parallels also meets the other, and (2)
to deduce Euclid V from this proposition. His handling of step (2) was correctly handled.
The argument in step (1) runs substantially as follows. Let g and h be parallel lines and
let another line k meet h at P . From Q, a point of k situated between g and h, drop a
perpendicular to h. As Q recedes indefinitely far from P , its distance QR from h increases
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CHAPTER 1. THE ORIGINS OF GEOMETRY
and exceeds any value, however great. In particular, QR will exceed the distance between
g and h. For some position of Q, then, QR will equal the distance between g and h. When
this occurs, k will meet g.
There are a number of assumptions here which go beyond those found in Euclid. I will
mention only the following two:
• the distance from one of two intersecting lines to the other increases beyond all bounds
as we recede from their common point,
• the distance between two parallels never exceeds some finite value.
The first of the two assumptions is not a grave error on the part of Proclus, for it can
be proved as a theorem on the basis of what he assumed from Euclid. Unfortunately for
Proclus, his second assumption is equivalent to Euclid V.
Nasiraddin (1201-1274), John Wallis (1616-1703), Legendre (1752-1833), Wolfgang Bolyai,
Girolamo Saccheri (1667-1733), Johann Heinrich Lambert (1728-1777), and many others
tried to prove Euclid V, and failed. In these failures there developed a goodly number of
substitutes for Euclid V; i.e., statements that were equivalent to the statement of Euclid
V. The following is a list of some of these that are more common:
1. Through a point not on a given line there passes not more than one parallel to the
line.
2. Two lines that are parallel to the same line are parallel to each other.
3. A line that meets one of two parallels also meets the other.
4. If two parallels are cut by a transversal, the alternate interior angles are equal.
5. There exists a triangle whose angle-sum is two right angles.
6. Parallel lines are equidistant from one another.
7. There exist two parallel lines whose distance apart never exceeds some finite value.
8. Similar triangles exist which are not congruent.
9. Through any three non-collinear points there passes a circle.
10. Through any point within any angle a line can be drawn which meets both sides of the
angle.
11. There exists a quadrilateral whose angle-sum is four right angles.
12. Any two parallel lines have a common perpendicular.
It fell to three different mathematicians to independently show that Euclid V is not
provable from the other axioms and what is derivable from them. These were Karl Friedrich
Gauss, János Bolyai, and Nicolai Ivanovich Lobachevskii. Once these men broke the ice,
the pieces of geometry began to fall into place. More was learned about non-Euclidean
geometries—hyperbolic and elliptic or doubly elliptic. The elliptic geometry was studied by
Riemann, gave rise to riemannian geometry and manifolds, which gave rise to differential
geometry which gave rise to relativity theory, et al.
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21
This gives us something to anticipate as we learn more about geometry. We will spend
our time studying hyperbolic geometry, for it lends itself to better study—not requiring
major changes in the axiom system that we have chosen. We may have an opportunity to
see that hyperbolic geometry is now lending itself to considerations in the latest research
areas of mathematics.
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CHAPTER 1. THE ORIGINS OF GEOMETRY
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Chapter 2
Euclidean Geometry
2.1
The Pythagoreans
Consider possibly the best known theorem in geometry.
Theorem 2.1 (The Pythagorean Theorem) Suppose a right angle triangle 4ABC has
a right angle at C, hypotenuse c, and sides a and b. Then
c2 = a2 + b2 .
b
a
b
a
c
c
A
c
c
b
C
a
a
B
b
C'
Figure 2.1: Pythagorean Theorem
Proof: On the side AB of 4ABC, construct a square of side c. Draw congruent triangles
on each of the other three sides of this square, as in Figure 2.1.
Since the angles at A and B sum to 90◦ , the angle CBC 0 is 180◦ . That means that we
have a line. Thus, the resulting figure is a square. The area of the larger square can be
23
24
CHAPTER 2. EUCLIDEAN GEOMETRY
calculated in two different ways. First, it is a square of side a + b. Second, we add together
the area of the square and the four triangles.
1
(a + b)2 = 4( ab) + c2
2
a2 + 2ab + b2 = 2ab + c2
a2 + b2 = c2
as we wanted.
What assumptions did we accept in this proof? There were several.
• The area of a square of side s is s2 .
• The interior angles of a triangle sum to 180◦ .
Aren’t these reasonable assumptions,
though? Clearly, any time I draw a square
North Pole
the area will be equal to the square of the
side. Is that true? Is it always true that the
interior angles of a triangle sum to 180◦ ?
Consider drawing the square on a piece of
paper and then lay the paper on a globe. If
90 W Longitude
the globe is big, with respect to the paper,
then the square looks pretty much like it
Prime Meridian
does on the flat paper. On the other hand,
consider the triangle on the globe that is
Equator
made from the Prime Meridian, the Equator and the line of longitude at 90◦ W. Each
of these lines meets the other at a 90◦ angle.
Thus the sum of the interior angles is 270◦
— much more than 180◦ . In fact, any triangle drawn on the surface of the globe will
Figure 2.2: A triangle with 3 right angles
have an angle sum more than 180◦ . Also,
the area of the square drawn on the globe
will be slightly more than the square of the side. In our flat frame of reference the errors
are too small to detect.
Theorem 2.2 (The Converse of the Pythagorean Theorem) Suppose we are in a geometry where the Pythagorean theorem is valid. Suppose that in triangle 4ABC we have
a2 + b2 = c2 .
Then the angle at C is a right angle.
Proof: As in Figure 2.3 let the perpendicular at A intersect the line BC at the point
D. Let r = |AD| and s = |DC|. Then by the Pythagorean theorem, r2 + s2 = b2 and
r2 + (a ± s)2 = c2 . The choice of sign depends on whether C is acute or obtuse. Thus,
expanding the second equation and substituting the first gives
a2 ± 2sa + b2 = c2 .
Since c2 = a2 + b2 , we have that 2sa = 0. Thus s = 0 and D = C making C a right angle.
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2.2. EUCLID’S AXIOMS FOR GEOMETRY
25
A
c
b
r
C
s
D
B
Figure 2.3: Converse of Pythagorean Theorem
2.2
Euclid’s Axioms for Geometry
I mentioned Euclid’s Axioms earlier. Now, we want to be more careful in the way that we
frame the axioms and make our definitions. This is the basis with which we must work for
the rest of the semester. If we do a bad job here, we are stuck with it for a long time.
Since we do not want to have to second guess everything that we prove, we will want
to agree on some facts that are absolute and unquestionable. These should be accepted by
all and should be easily stated. These will be our axioms. Euclid chose to work with five
axioms. (Hilbert in his later work chose to work with 16 axioms.)
Postulate 1: We can draw a unique line segment between any two points.
Postulate 2: Any line segment can be continued indefinitely.
Postulate 3: A circle of any radius and any center can be drawn.
Postulate 4: Any two right angles are congruent.
Postulate 5: Given a line ` and a point P not on `, there exists a unique line `2 through
P which does not intersect `.1
What assumptions have we made here? First of all, we have assumed that a set of points,
called the Euclidean plane exists. With this assumption comes the concept of length, of
lines, of circles, of angular measure, and of congruence. It also assumes that the plane is
two-dimensional. All this in five little sentences.
Let’s consider what Hilbert does in his choices, and then what Birkhoff chose.
2.2.1
Hilbert’s Axioms for Neutral Geometry
GROUP I : Incidence Axioms
I–1: For every point P and for every point Q not equal to P there exists a unique line `
that passes through P and Q.
I–2: For every line ` there exist at least two distinct points incident with `.
I–3: There exist three distinct points with the property that no line is incident with all
three of them.
1
This is actually Playfair’s postulate. We will give the statement of Euclid later.
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CHAPTER 2. EUCLIDEAN GEOMETRY
GROUP II : Betweeness Axioms
B–1: If A ∗ B ∗ C 2 , then A, B, and C are three distinct points all lying on the same line,
and C ∗ B ∗ A.
←→
B–2: Given any two distinct points B and D, there exist points A, C, and E lying on BD
such that A ∗ B ∗ D, B ∗ C ∗ D, and B ∗ D ∗ E.
B–3: If A, B, and C are three distinct points lying on the same line, then one and only one
of the points is between the other two.
B–4: (plane separation axiom) For every line ` and for any three points A, B, and C
not lying on `:
(a) if A and B are on the same side of ` and B and C are on the same side of `, then
A and C are on the same side of `.
(b) if A and B are on opposite sides of ` and B and C are on opposite sides of `,
then A and C are on the same side of `.
GROUP III : Congruence Axioms
C–1: If A and B are distinct points and if A0 is any point, then for each ray r emanating
from A0 there is a unique point B 0 on r such that B 0 6= A0 and AB ∼
= A0 B 0 .
C–2: If AB ∼
= CD and AB ∼
= EF , then CD ∼
= EF . Moreover, every segment is congruent
to itself.
C–3: If A ∗ B ∗ C, A0 * B 0 * C 0 , AB ∼
= A0 B 0 , and BC ∼
= B 0 C 0 , then AC ∼
= A0 C 0 .
−−→
C–4: Given any ∠BAC and given any ray A0 B 0 emanating from a point A0 , then there is a
−−→
←−→
unique ray A0 C 0 on a given side of line A0 B 0 such that ∠B 0 A0 C 0 ∼
= ∠BAC.
C–5: If ∠A ∼
= ∠B and ∠A ∼
= ∠C, then ∠B ∼
= ∠C. Moreover, every angle is congruent to
itself.
C–6: (SAS ) If two sides and the included angle of one triangle are congruent respectively
to two sides and the included angle of another triangle, then the two triangles are
congruent.
GROUP IV: Continuity Axioms
Archimedes’ Axiom: If AB and CD are any segments, then there is a number n
−−→
such that if segment CD is laid off n times on the ray AB emanating from A, then a
point E is reached where n · CD ∼
= AE and B is between A and E.
Dedekind’s Axiom: Suppose that the set of all points on a line ` is the union Σ1 ∪Σ2
of two nonempty subsets such that no point of Σ1 is between two points of Σ2 and
vice versa. Then there is a unique point, O, lying on ` such that P1 ∗ O ∗ P2 if and
only if P1 ∈ Σ1 and P2 ∈ Σ2 and O 6= P1 , P2 .
2
A ∗ B ∗ C denotes that B lies between A and C
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2.2. EUCLID’S AXIOMS FOR GEOMETRY
27
(The following two Principles follow from Dedekind’s Axiom, yet are at times more
useful.)
Circular Continuity Principle: If a circle γ has one point inside and one point
outside another circle γ 0 , then the two circles intersect in two points.
Elementary Continuity Principle: If one endpoint of a segment is inside a circle
and the other outside the circle, then the segment intersects the circle.
Hilbert also used five undefined terms: point, line, incidence, betweenness, and
congruence.
2.2.2
Birkhoff ’s Axioms for Neutral Geometry
The setting for these axioms is the “Absolute (or Neutral) Plane”. It is universal in the
sense that all points belong to this plane. It is denoted by A2 .
Axiom 1: There exist nonempty subsets of A2 called “lines,” with the property that each two
points belong to exactly one line.
Axiom 2: Corresponding to any two points A, B ∈ A2 there exists a unique number d(AB) =
d(BA) ∈ R, the distance from A to B, which is 0 if and only if A = B.
Axiom 3: (Birkhoff Ruler Axiom) If k is a line and R denotes the set of real numbers, there exists
a one-to-one correspondence (X ↔ x) between the points X ∈ k and the numbers
x ∈ R such that
d(A, B) = |a − b|
where A ↔ a and B ↔ b.
Axiom 4: For each line k there are exactly two nonempty convex sets R0 and R00 satisfying
(a) A2 = R0 ∪ k ∪ R00
(b) R0 ∩ R00 = φ, R0 ∩ k = φ, and R00 ∩ k = φ. That is, they are pairwise disjoint.
(c) If X ∈ R0 and Y ∈ R00 then XY ∩ k 6= φ.
Axiom 5: To each angle ∠ABC there exists a unique real number x with 0 ≤ x ≤ 180 which is
the (degree) measure of the angle
x = ∠ABC ◦ .
−−→
Axiom 6: If BD ⊂ Int (∠ABC), then
∠ABD◦ + ∠DBC ◦ = ∠ABC ◦ .
−−→
Axiom 7: If AB is a ray in the edge, k, of an open half plane H(k; P ) then there exist a one-toone correspondence between the open rays in H(k; P ) emanating from A and the set
−−→
of real numbers between 0 and 180 so that if AX ↔ x then
∠BAX ◦ = x.
Axiom 8: (SAS ) If a correspondence of two triangles, or a triangle with itself, is such that two
sides and the angle between them are respectively congruent to the corresponding two
sides and the angle between them, the correspondence is a congruence of triangles.
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Chapter 3
Transformation Geometry: First
View
3.1
The Axioms for a Euclidean Model
Since we do not want to have to second guess everything that we prove, we will want to
agree on some facts that are absolute and unquestionable. These should be accepted by all
and should be easily stated. These will be our axioms.
Postulate 1: We can draw a unique line segment between any two points.
Postulate 2: Any line segment can be continued indefinitely.
Postulate 3: A circle of any radius and any center can be drawn.
Postulate 4: Any two right angles are congruent.
Postulate 5: Given a line ` and a point P not on `, there exists a unique line `2 through
P which does not intersect `.
We need some definitions to work with our choice of Euclid’s Axioms.
Definition 3.1 Distance is a real-valued function which assigns to any pair of points in
the plane a non-negative real number satisfying the following properties:
1) d(P, Q) = d(Q, P )
2) d(P, Q) ≥ 0, and d(P, Q) = 0 if and only if P = Q
3) d(P, R) ≤ d(P, Q) + d(Q, R) (the Triangle Inequality).
We call such a function a metric. We say that the distance from P to Q is d(P, Q).
In our usual high school geometry classes you might call this P Q, P Q, or |P Q|.
A line segment is the shortest path between two points. A line is an indefinite continuation of a line segment.
The circle CP (r) centered at P of radius r is the set of points
CP (r) = {Q | d(P, Q) = r}.
Definition 3.2 A transformation T of the plane P is a mapping (or function) taking
points x ∈ P in the plane to points y ∈ P in the plane.
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CHAPTER 3. TRANSFORMATION GEOMETRY: FIRST VIEW
We will introduce the concept of congruence using the idea of isometries.
Definition 3.3 An isometry of the plane is a map from the plane to itself which preserves
distances. That is, f is an isometry if for any two points P and Q in the plane we have
d(f (P ), f (Q)) = d(P, Q).
In your dealings with the Euclidean plane you have run across several isometries: translations, rotations, and reflections. We will formalize these definitions a little later.
Definition 3.4 Two sets of points (defining a triangle, angle, or some other figure) are
congruent if there exists an isometry which maps one set to the other.
This idea of congruence is completed by the following axioms, which guarantee the
existence of the isometries we will need.
Postulate 6. Given any points P and Q, there exists an isometry f so that f (P ) = Q.
(Translations are examples of such.)
Postulate 7. Given a point P and two points Q and R which are equidistant from P ,
there exists an isometry which fixes P and maps Q to R. (Rotations and reflections are
examples of these.)
Postulate 8. Given any line `, there exists an isometry which fixes every point in ` but
fixes no other points in the plane. (A reflection through ` is such an example.)
Please note that, for example, Postulate 6 does not guarantee the existence of translations. In fact, depending on how translations are defined, translations may not exist in
spherical geometry, but Axiom 6 will hold.
Definition 3.5 Two lines `1 and `2 intersect at right angles if any two adjacent angles
at the point of intersection are congruent. That is, they intersect at right angles if there
exists an isometry which sends an angle to one of its adjacent angles.
3.2
Triangle Congruence
You will notice that in our list of eight postulates we do not mention how to determine
if two triangles are congruent, other than the definition. We had several methods at our
disposal when we studied geometry earlier. What happened to them?
In all honesty, we are looking at geometry in a different way. We are using the group of
isometries to determine the geometry. This is the approach that Felix Klein advocated in
his Erlangen Programme. We will see that by determining just exactly what the isometries
are in a particular situation, we will be able to describe the geometry of the situation.
We are used to having at least three congruence criteria for triangles: side-angle-side
(SAS), angle-side-angle (ASA), and side-side-side (SSS). Shouldn’t those hold here? Of
course, but we will have to prove them from our postulates. We will want to do so without
using the parallel postulate so that we know that they will be valid in our neutral geometry,
or a geometry without a parallel axiom chosen.
If you will remember, the other choices for corresponding parts of triangles did not
form congruence criteria when you studied them before: side-side-angle (SSA) and angleangle-angle (AAA). However, angle-angle-side (AAS) did set up a congruence. If you can
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3.2. TRIANGLE CONGRUENCE
31
remember you should remember that AAS worked because we appealed to the fact that
all triangles summed to 180◦ , so we could then state that the third angles were congruent
and we had reduced this situation to ASA. This proof depends on the Euclidean parallel
postulate, so we would want to try to prove this differently, if it is going to be true in neutral
geometry.
The proof that we will give depends on a consequence of one of the Continuity Axioms,
which we will take up later. Suffice it to say that we will accept the following lemma without
proof at this time.1
Lemma 3.1 Two distinct circles intersect in zero, one, or two points. If there is exactly
one point of intersection, then that point lies on the line joining the two centers.
Theorem 3.1 (SSS) If the corresponding sides of two triangles 4ABC and 4DEF have
equal lengths, then the two triangles are congruent.
Proof: Recall that the definition of congruence requires us to produce an isometry φ so
that φ(A) = D, φ(B) = E, and φ(C) = F .
First, assume that the triangles are not degenerate (i.e., that each of {A, B, C} and
{D, E, F } form sets of non-collinear points). If they are degenerate, then you should be
able to prove this relatively easily from the Triangle Inequality.
E
D
B
A
C
F
Figure 3.1: Initial positions
Now, by Axiom 6 there must be an isometry f1 that sends A to D. This f1 is an isometry
and |AB| = |DE|, so
|Df1 (B)| = |f1 (A)f1 (B)| = |AB| = |DE|.
Since f1 (B) and E are equidistant from B, by Postulate 7 there is an isometry f2 such that
f2 (D) = D and f2 (f1 (B)) = E. If we should be so lucky that f2 (f1 (C)) = F , then we are
done because the isometry f2 ◦ f1 is the necessary isometry. However, we cannot assume
that we are always that lucky.
To complete the proof, we need to assume that f2 (f1 (C)) 6= F and consider the circle
centered at D with radius AC and the circle centered at E with radius BC. By Lemma 3.1
1
Its proof does not depend on the result of the theorem we are going to prove.
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CHAPTER 3. TRANSFORMATION GEOMETRY: FIRST VIEW
C
B
A=D
C
E
A=D
B=E
F
F
Figure 3.2: Final isometries for SSS
these two circles intersect in at most two points. One of these points is F and the other
must be f2 (f1 (C)). By Postulate 8 there is an isometry f3 which fixes every point on DE
but fixes no other point. Since F is not on DE it must be mapped to another point and
that point must be f2 (f1 (C)), and vice versa. Let φ = f3 ◦ f2 ◦ f1 . Then
φ(A) = D
φ(B) = E
φ(C) = F
and the two triangles are congruent, by definition.
When one triangle is congruent to another, we will write 4ABC ≡ 4DEF .
Since we indicated earlier that the isometries are going to help us determine the geometry, we need to categorize the isometries. To do this, we need to understand that there
is an orientation, that is a right hand and a left hand. We must choose which direction is
positive, and it is a choice — it is not something determined a priori. This choice is called
the orientation. More formally, a nondegenerate triangle 4ABC is said to be oriented
clockwise if the path from A to B to C is oriented clockwise. If a nondegenerate triangle is
not oriented clockwise, then we say it is oriented counterclockwise.
Definition 3.6 An isometry is a direct isometry or a proper isometry if the image
of every clockwise triangle is oriented clockwise. An isometry which is not direct is called
a improper isometry. Due to the baggage associated with this terminology (and due to
tradition) we will call these orientation preserving and orientation reversing isometries.
Definition 3.7 An isometry ϕ is a translation if ϕ is an orientation preserving isometry
and either ϕ is the identity or ϕ has no fixed points.
Definition 3.8 An isometry ϕ is a rotation if ϕ is an orientation preserving isometry
and either ϕ is the identity or there is exactly one point P such that ϕ(P ) = P . We call P
the center of rotation for ϕ.
Definition 3.9 An isometry ϕ is a reflection through the line ` if ϕ(P ) = P for every
P ∈ ` and if ϕ(P ) 6= P if P 6∈ `.
MATH 6118-090
Spring 2006
3.3. EUCLID’S REAL FIFTH POSTULATE
3.3
33
Euclid’s Real Fifth Postulate
Euclid stated his postulate in a less favorable form. We believe this for several reasons.
First, Euclid’s statement is:
Postulate 5: Suppose a line meets two other lines so that the sum of the angles on one
side is less than two right angles. Then the two other lines meet at a point on that side.
This is not a nice, simple statement. Apparently, Euclid did not like the statement
because he did not use it until Proposition 29 in his elements. We stated it earlier in the
form of Playfair’s Postulate. We need to show that these are equivalent.
Theorem 3.2 Let P be a point not on `, and let Q lie on ` so that P Q is perpendicular to
`2 Let `2 be the line through P which is parallel to ` (as guaranteed by Playfair’s Postulate).
Then `2 intersects P Q at a right angle.
Proof: Assume that `2 is not perpendicular to P Q. Let `3 denote the reflection of `2
through the line P Q. Now, `3 6= `2 . Since ` is perpendicular to P Q, the reflection of
` through P Q is itself. Thus, `3 cannot intersect `, because if it did then the reflection
of the point of intersection would be a point of intersection between `2 and `, which do
not intersect. This gives us more than one line through P which is parallel to `, which
contradicts Postulate 5, so `2 must be perpendicular to P Q.
You might ask how we know that this point Q exists. If is does not exist, we would not
have a contradiction. Never fear, we have its existence due to the next lemma.
Lemma 3.2 Let ` be a line and P a point not on `. Then there exists a point Q on ` so
that ` is perpendicular to P Q.
Proof: Left to the reader.
Note that the converse of Theorem 3.2 is also true.
Theorem 3.3 Suppose that ` is perpendicular to `1 and `2 , then `1 is parallel to `2 .
Proof: Left to the reader.
Corollary 1 Suppose that ` intersects two other lines `1 and `2 so that the alternate interior
angles are congruent. Then `1 and `2 are parallel.
Proof: We will prove this, but I want you to note that we do not need to appeal to a
parallel postulate in order to prove this theorem.
Let k intersect `1 and `2 at P and Q, respectively, as in Figure 3.3. Let M be the
midpoint of P Q. Drop a perpendicular from O to `1 and let the foot of the perpendicular
be R, thus M R is perpendicular to `1 . Now, consider the rotation centered at M which
sends P to Q. Let the image of R under this rotation be R0 . Now R0 does not a priori lie
on the line `2 . However, we do have that 4M RP ≡ 4M R0 Q since one is the image of the
other under an isometry. Now, ∠M P R ≡ ∠M QR0 , since the alternate interior angles are
congruent. Thus, R0 must lie on `2 . Therefore, ∠P RM ≡ ∠QR0 M are right angles. Thus,
by Theorem 3.2 `1 and `2 are parallel.
2
Q is called the foot of P in `.
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34
CHAPTER 3. TRANSFORMATION GEOMETRY: FIRST VIEW
R'
Q
l2
k
P
O
R
l1
Figure 3.3: Corollary 1
Note that this is not the usual manner in which you encounter this theorem. You usually
encounter its converse:
Theorem 3.4 Suppose that `1 and `2 are parallel and that k is a transversal intersecting
`1 and `2 . Then, alternate interior angles are congruent.
This statement is equivalent to Euclid’s Fifth Postulate, whereas Corollary 1 is true in
Neutral Geometry. You must be careful in stating these results. One is true in much more
generality than the other, yet both seem so much the same.
Corollary 1 (Euclid’s Axiom V) Suppose a line ` meets two other lines `1 and `2 so
that the sum of the angles on one side is less than two right angles. Then the two other
lines meet at a point on that side.
Theorem 3.5 The three angles of a triangle sum to two right angles.
The best proof of this is the one that you give to students of this result.
MATH 6118-090
Spring 2006
Chapter 4
To Boldly Go Where No Man Has
Gone Before
4.1
The Star Trek Lemma
This is a very common result in Euclidean geometry. It is only true in Euclidean geometry.
It is used often and its proof gives us more practice in theorem-proving.
Let A, B, and C be points on a circle centered at O. We will call angle ∠BAC an
inscribed angle since it is inscribed in a circle. The angular measure of the arc BC is the
measure of the central angle ∠BOC, where the angle is measured on the same side of O as
the arc. We say that ∠BAC subtends the arc BC.
Lemma 4.1 (Star Trek Lemma) The measure of the inscribed angle is half of the angular measure of the arc it subtends.
Proof: There are several cases to the proof of the lemma. We will look only at the case
where ∠BAC is an acute angle and the center, O, lies in the interior of the angle, as in
Figure 4.1.
A
O
B
C
Figure 4.1: Star Trek Lemma
Note that OA, OB and OC are all radii, so we have several isosceles triangles. Extend
the segment OA until it meets the circle at a point D. Since 4AOB is isosceles, ∠BAO ≡
35
36
CHAPTER 4. TO BOLDLY GO WHERE NO MAN HAS GONE BEFORE
∠OBA. Also, since the sum of the angles is 180◦ ,
∠BOD = ∠OBA + ∠BAO = 2∠BAO.
Similarly, ∠COD = 2∠CAO. Thus, adding these together, we have ∠BOC = 2∠BAC.
Figure 4.2: The Reason for the Name
MATH 6118-090
Spring 2006
Chapter 5
Euclidean Geometry: Revisited
5.1
Similar Triangles
The concept of similar triangles seems so innocuous and so basic, it cannot be related to the
Parallel Axiom, can it? It is. It is extremely important in Euclidean geometry. There are
numbers of theorems and concepts that rely on similar triangles: slope and trigonometry
are just two of these concepts.
For this chapter and most of the rest of the text, we will use the notation |AB| = d(A, B)
to denote the distance from A to B and the length of the segment AB.
Theorem 5.1 Let B 0 and C 0 be on AB and AC, respectively, on the triangle 4ABC. Then
B 0 C 0 is parallel to BC if and only if
|AC 0 |
|AB 0 |
=
.
|AB|
|AC|
Let’s defer the proof of this until later — we only want to use it now to see what some of
the definitions need to be. Note that since B 0 C 0 is parallel to BC, then the corresponding
angles of 4ABC and 4AB 0 C 0 are congruent. We call such triangles similar.
Definition 5.1 We say that two triangles 4ABC and 4A0 B 0 C 0 are similar if their corresponding angles are congruent, in which case we write 4ABC ∼ 4A0 B 0 C 0 .
Lemma 5.1 If 4ABC ∼ 4A0 B 0 C 0 , then
|A0 B 0 |
|A0 C 0 |
|B 0 C 0 |
=
=
.
|AB|
|AC|
|BC|
Proof: Since ∠BAC ∼
= ∠B 0 A0 C 0 , there is an isometry which sends A0 to A and sends B 0
0
and C to points on AB and AC, respectively. Since ∠ABC ∼
= ∠A0 B 0 C 0 , the line B 0 C 0 is
parallel to BC, so by Theorem 5.1,
|A0 B 0 |
|A0 C 0 |
=
.
|AB|
|AC|
Similarly, by sending B 0 to B, we can show that
|B 0 C 0 |
|A0 B 0 |
=
.
|AB|
|BC|
Combining, we get the desired result.
37
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CHAPTER 5. EUCLIDEAN GEOMETRY: REVISITED
5.2
Graph of a Linear Equation
How powerful is this idea of similarity? Here we will consider the proof of a seemingly
obvious fact — but it only seems obvious until we think about the idea in different ways.
First, a linear equation in two variables x and y is an equation of the form
ax + by + c = 0.
We often rearrange this equation to take the form y = mx+b thus defining y as a function of
x. What is the graph of this linear function in the standard coordinate plane? Immediately
you answered, “A line.” You are of course right, but why? If you will recall the equation
r = θ in polar coordinates for the cartesian plane, its graph is the Archimedean spiral —
not a line! So just because you have a linear equation in some set of variables, you may
or may not get the graph of a line from it. However, so that not all is lost, it is true that
the graph of r = θ in the (r, θ)-plane is a line. Why is this always true, then? Also, we
might want to ask if for every coordinate system we might end up using, will the equation
of a line always be a linear equation? Again, you want to say yes, but now you have
some doubts? If you draw a line in the plane and are forced to use polar coordinates, will
it’s equation be a linear equation? In this case, not always! The lines through the origin
have equation θ = a constant, but the equation of the line vertical line through (1, 0) is
r = sec(θ). Okay, we know that polar coordinates are not going to work all the time, but
what about “reasonable” choices for coordinate systems? Can we say something about the
relationship between linear equations and lines?
Theorem 5.2 For every pair of real numbers m and b, there is a unique non-vertical line
such that a point is on that line if and only if its coordinates satisfy the linear equation
y = mx + b. Conversely, for every non-vertical line there is a unique pair of real numbers
m and b such that a point is on the line if and only if its coordinates satisfy y = mx + b.
This theorem follows directly from the following two lemmæ.
Lemma 5.2 For a given choice of m and b, all points (x, y) whose coordinates satisfy
y = mx + b belong to a unique vertical line.
Lemma 5.3 For a given non-vertical line, there is a unique pair of real numbers m, b such
that all points on that line satisfy the equation y = mx + b.
Proof: [Proof of Lemma 5.2] We will have to break this into two parts: m 6= 0 and m = 0.
First, if m 6= 0, choose two values of x — x1 and x2 so that x1 < x2 . Let y1 = mx1 + b
and y2 = mx2 + b. The points P = (x1 , y1 ) and Q = (x2 , y2 ) are two points on the graph
of y = mx + b. What we need to
do now is to show that an arbitrary point A = (x, y)
←→
with y = mx + b lies on the line P Q. We assume here that x 6= x1 and x 6= x2 , since it is
obviously true in those two cases.
We have three real numbers so one of the following must be true: x < x1 , x1 < x < x2 ,
or x2 < x.
Assume that x < x1 and plot the points A, P , and Q. As in the figure to the right,
construct the right triangles 4ABP and 4P CQ. The legs of the right triangles are as
displayed: x1 − x, y1 − y, x2 − x1 , y2 − y1 .
MATH 6118-090
Spring 2006
5.2. GRAPH OF A LINEAR EQUATION
39
Q
y2 - y1
P
x2 - x1
C
y 1-y
A
B
x 1-x
←→
To show that the point A is on the line P Q we need to show that ∠AP Q = 180◦ . To
do that we need to show that the triangles 4ABP and 4P CQ are similar.
Note that
y1 − y = mx1 + b − (mx + b)
= m(x1 − x)
y2 − y1 = mx2 + b − (mx1 + b)
= m(x2 − x1 )
Thus,
m(x1 − x)
m(x2 − x1 )
QC
PB
=
=m=
=
.
AB
x1 − x
x2 − x1
PC
Therefore, the triangles are similar, since the legs of the right triangles are proportional.
Therefore, ∠AP B ∼
= ∠P QC. We know that ∠BP C is a right angle because the segments
P B and P C are drawn parallel to the coordinate axes. Therefore,
∠AP B + ∠BP C + ∠CP Q = ∠P QC + ∠BP C + ∠CP Q
= ∠P QC + ∠CP Q + ∠BP C
= 90◦ + 90◦ = 180◦ .
←→
Thus, the point A lies on the line P Q. This proof works similarly in the other two cases:
x1 < x < x2 and x2 < x with the same calculations.
If m = 0, then y = b. Look at the quadrilateral made up of the points (0, 0), (0, b),
(x, 0) and (x, b). The distance from the x-axis to (x, b) is b, so this quadrilateral has two
equal and parallel sides. Hence the other two sides are parallel, making (x, b) lie on the line
through (0, b) that is parallel to the x-axis.
Proof: [Proof of Lemma
←→ 5.3] Let P = (x1 , y1 ) and Q = (x2 , y2 ) be two points in the plane
with x1 6= x2 . Let ` = P Q be the line passing through P and Q. Note that ` is not vertical,
since the x-coordinates are different. Because x1 6= x2 , let
m=
y2 − y1
.
x2 − x1
Let C = (x, y) and D = (x0 , y 0 ) be any two points on `. Construct triangles 4CDF and
4ABE so that BE and CF are parallel to the y-axis and AE and DF are parallel to the
MATH 6118-090
Spring 2006
40
CHAPTER 5. EUCLIDEAN GEOMETRY: REVISITED
x-axis. From the usual Alternate Interior Angles Theorem, you can see that corresponding
angles are congruent so that 4CDF ∼ 4ABE. Thus,
BE
CF
=
,
AE
DF
which means that
m=
y2 − y1
y0 − y
= 0
,
x2 − x1
x −x
and this number m depends only on the line ` and not which points we have chosen from `.
Let b denote the y-coordinate of the point where ` crosses the y-axis. Since the line is not
vertical, such a point exists.
Let (x, y) be an arbitrary point on the line `. Since we can compute the same slope
regardless of which two points we choose, using (x, y) and (0, b) we have that
y−b
= m
x−0
y − b = m(x − 0)
y = mx = b
Thus, the coordinates of the point satisfy the equation y = mx + b.
5.3
Power of a Point
Theorem 5.3 Let P be a point inside a circle Γ. Let QQ0 and RR0 be two chords which
intersect at P . Then
|P Q| · |P Q0 | = |P R| · |P R0 |.
Proof: By the Star Trek Lemma, ∠RR0 Q ∼
= ∠RQ0 Q and ∠Q0 RR0 ∼
= ∠Q0 QR0 . Since the
0
0
0
angles at P are vertical angles, the triangles 4P Q R and 4P R Q are similar. Thus,
|P R|
|P Q0 |
=
,
|P Q|
|P R0 |
and our result follows by cross-multiplication.
Theorem 5.4 Let P be a point outside a circle Γ. Let QQ0 and RR0 be two chords which
intersect at P . Then
|P Q| · |P Q0 | = |P R| · |P R0 |.
Proof: This time our setup is slightly different.
However, it is easy to see how to proceed. From Theorem 5.1 we have that ∠P Q0 R ∼
=
∠P R0 Q. The angle ∠P is shared by the two triangles. Using the fact that the angle sum
of a triangle is 180◦ , then the third angles are equal. Thus, 4P Q0 R ∼
= 4P R0 Q. Setting up
the appropriate ratios gives us the result.
MATH 6118-090
Spring 2006
5.4. MEDIANS AND CENTROID
41
R'
Q'
P
Q
R
Figure 5.1: Theorem 5.3
Thus, for any point P and any chord, QQ0 , of the circle Γ the product |P Q||P Q0 | is a
constant. We define the value Π(P ) = ±|P Q| · |P Q0 | to be the power of a point with respect
to a circle. We choose the sign to be positive if P is outside the circle and negative if P is
inside the circle.
Assume that the circle has center O and radius r. Then, choose QQ0 to be a diameter
of Γ that goes through P . If P is outside Γ, It then follows that
|P Q| = |OP | − |OR| = |OP | − r and |P Q0 | = |OP | + |OR| = |OP | + r.
Thus, Π(P ) = |P Q||P Q0 | = |OP | − r2 . I leave it to you to check that the same is true if P
lies inside Γ.
5.4
Medians and Centroid
In 4ABC let A0 , B 0 , and C 0 be the midpoints of the sides BC, AC, and AB respectively.
The line segments AA0 , BB 0 , and CC 0 are called the medians of 4ABC.
Q'
Q
P
R
R'
Figure 5.2: Chords for the Power of P
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42
CHAPTER 5. EUCLIDEAN GEOMETRY: REVISITED
Lemma 5.4 Let 4ABC be a triangle, and let D, E be the midpoints of AB and AC,
respectively. Then the line DE is parallel to the base BC. If F is the midpoint of BC, then
DE ∼
= BF .
Proof: Let D be the midpoint of AB, and draw lines through D parallel to AC and BC.
Let them meet the opposite sides in points E and F . Since DE is parallel to BC, the
angles ∠DBF 0 ∼
= ∠ADE 0 are congruent. Similarly, since AC is parallel to DF , the angles
∠DAE 0 ∼
= ∠BDF 0 are congruent.
Now, AD ∼
= DB, and the angles of the triangles 4ADE 0 and 4DBF 0 are congruent,
so by ASA the two triangles are congruent. Therefore, AE 0 ∼
= DF 0 and DE 0 ∼
= BF 0 .
Look at the parallelogram DEF C. Since it is a parallelogram, opposite sides are congruent. Thus, DF 0 ∼
= CE 0 and DE 0 ∼
= CF 0 . Thus, E and F are the midpoints of the sides
0
AC and BC, respectively. Thus, E = E and the line DE is the line DE. Thus, DE is
parallel to BC as claimed. Furthermore,
1
DE = DE 0 ∼
= BF 0 = BF = BC.
2
Corollary 1 Let 4ABC be a triangle and let D, E, and F be the midpoints of the sides.
Then the sides of the triangle 4DEF are parallel to the sides of 4ABC, and the four
triangles so formed are all congruent to one another.
We say a triangle 4ABC is congruent to the double of 4DEF , symbolically 4ABC ∼
=
24DEF , if the two triangles are similar and the ratio of the sides is 2:1.
Lemma 5.5 (2 ASA) Let 4ABC and 4A0 B 0 C 0 be two triangles, and assume that ∠ABC ∼
=
0
0
0
0
0
0
0
0
0
0
0
∼
∼
∼
∠A B C and ∠ACB = ∠A C B , and that BC = 2B C . Then 4ABC = 24A B C .
Proof: Let D, E, and F be the midpoints of the sides of 4ABC, and draw the triangle
4DEF . From above we know that DE ∼
= 12 BC ∼
= B 0 C 0 . Furthermore, since DE is parallel
to BC, the angles of 4ADE at D and E are congruent to the angles of 4A0 B 0 C 0 at B and
C. Thus by ASA 4ADE ∼
= 4A0 B 0 C 0 . Since 4ABC is the double of 4ADE, the result
follows.
Similar theorems for SSS and SAS can be proven
Theorem 5.5 The three medians of a triangle 4ABC intersect at a common point G.
Furthermore,
|BG|
|CG|
|AG|
= 0 = 0 = 2.
0
|A G|
|B G|
|C G|
The common point of intersection is called the centroid of the triangle 4ABC.
Proof: Let 4ABC be the triangle, let D and E be the midpoints of AB and AC, and
construct DE. Let the two medians BE and CD meet at a point G. Since DE is parallel to
BC, we see that ∠DEG ∼
= ∠CBG and ∠EDG ∼
= ∠BCG . Also we know that BC ∼
= 2DE .
∼
Thus, by our previous proposition, we see that 4BGC = 24EGD. In particular, it follows
that BG ∼
= 2GE . Therefore G is the point on the median BE that is 2/3 of the way from
B to E. If we draw the medians BE and AF, we find that they meet in a point that is also
2/3 of the way from B to E. Thus, this must be the same point and all three medians meet
in a single point, G.
Corollary 1 The centroid G lies on each median 2/3 of the way from the vertex to the
midpoint of the opposite side.
MATH 6118-090
Spring 2006
5.5. INCIRCLE, EXCIRCLES, AND LAW OF COSINES
43
A
A
C'
G
B
B'
G
C
B
C
Figure 5.3: Medians
5.5
Incircle, Excircles, and Law of Cosines
Theorem 5.6 The angle bisectors of a triangle intersect at a common point I called the
incenter, which is the center of the unique circle inscribed in the triangle (called the incircle).
Proof: Consider the angle ∠ABC and let D be a point on the angle bisector. Let E and
E 0 be the points on BA and BC, respectively, so that ∠BED and ∠BE 0 D are right angles.
Thus, 4BED ∼
= 4BE 0 D by AAS, since they share BD. Thus, |DE| = |DE 0 | and the circle
centered at D with radius |DE| is tangent to both BA and BC.
Let I be the intersection of the angle bisectors of ∠ABC and ∠ACB. The perpendiculars
from I to AB and BC are congruent from what we saw above. Likewise, the perpendiculars
from I to BC and AC are congruent. Thus, the perpendiculars from I to AB and AC are
congruent, so I lies on the angle bisector of ∠BAC. (Why?) Thus, the three angle bisectors
intersect at a common point.
For a triangle 4ABC we can define two angle bisectors at each vertex. We have the
interior angle bisector at A, about which we just studied. We also have an exterior angle
bisector at A. Note, that this is NOT the extension of the interior angle bisector to the
exterior of the triangle. It is the angle bisector of the angle supplementary to the angle,
∠BAC.
We define three excenters, Ia , Ib , and Ic , as follows. The excenter Ia is the point of
intersection of the interior angle bisector of A and the exterior angle bisectors at B and C.
It is the center of a circle which is tangent to BC and the extended sides AB and AC, and
lies outside 4ABC. This circle is called an excircle.
Let the inradius r be the radius of the incircle. Let s = 12 (a+b+c) be the semiperimeter
of 4ABC.
Theorem 5.7 If r is the inradius of 4ABC, and s is the semiperimeter of 4ABC. Then
area(4ABC) = |4ABC| = rs.
Proof: Left for the reader.
Theorem 5.8 (Law of Cosines) For any triangle 4ABC, we have
c2 = a2 + b2 − 2ab cos(C).
Proof: Let D be the altitude dropped from A to BC. Then by the Pythagorean Theorem
c2 = |AD|2 + |DB|2 .
MATH 6118-090
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44
CHAPTER 5. EUCLIDEAN GEOMETRY: REVISITED
Ic
A
Ic
I
B
C
Ia
Figure 5.4: The triangle, its incircle, and its excircles
MATH 6118-090
Spring 2006
5.6. THE CIRCUMCENTER AND ITS SPAWN
45
Now,
|AD| = b sin(C)
|DB| = |a − b cos(C)|
Thus,
c2 = b2 sin2 (C) + a2 − 2ab cos(C) + b2 cos2 (C)
c2 = a2 + b2 − 2ab cos(C)
as we needed.
Theorem 5.9 (Heron’s Formula) For any triangle 4ABC
p
|4ABC| = s(s − a)(s − b)(s − c).
Proof: Note that
1
|4ABC| = ab sin(C).
2
By the Law of Cosines,
cos(C) =
a2 + b2 − c2
2ab
Thus, applying some algebra
|4ABC| =
=
=
=
=
=
=
1 p
ab 1 − cos2 (C)
2 p
4a2 b2 − (a2 + b2 − c2 )2
1
ab
2
2ab
1p
(2ab + a2 + b2 − c2 )(2ab − a2 − b2 + c2 )
4
1p
((a + b)2 − c2 )(c2 − (a − b)2 )
4
1p
(a + b + c)(a + b − c)(c − a + b)(c + a − b)
4
r
a + b + c a + b − c −a + b + c a − b + c
2
2
2
2
p
s(s − a)(s − b)(s − c)
Heron’s formula is named for Heron of Alexandria, who lived sometime between 100 BC
and 300 AD. Scholars state that the formula dates back to at least Archimedes (ca. 250
BC).
5.6
The Circumcenter and its Spawn
We have seen the centroid —center of mass — and the incenter. There is yet another center
of a triangle. We remember that given any three points there is a unique circle passing
through them. How do you find that circle?
Take the perpendicular bisectors of the sides of a triangle formed by the three points.
These bisectors meet in a common point, called the circumcenter. The radius of the circumcircle is called the circumradius.
MATH 6118-090
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46
CHAPTER 5. EUCLIDEAN GEOMETRY: REVISITED
A
C'
B'
O
R
B
a/2
A'
C
Figure 5.5: Circumcenter
Theorem 5.10 Given a triangle 4ABC, the perpendicular bisectors of the sides are concurrent. The point is the center of a circle which passes through the vertices of the triangle.
The point is called the circumcenter of the triangle.
Proof: We must have that two of the perpendicular bisectors intersect. Let p1 and p2
denote the perpendicular bisectors of AB and AC respectively. If p1 is parallel to p2 , then
since AC is perpendicular to p2 , AC is perpendicular to p1 . Since AB is perpendicular to
p1 , then AB must be parallel to AC or they coincide. Thus, we would not have a triangle.1
Thus, two perpendicular bisectors intersect in a point O. Let M denote the midpoint of
AB. Then 4AOM ∼
= 4BOM , since the angle at M is a right angle, AM ∼
= BM , and
∼
OM = OM . Hence, AO ∼
= BO. Using AC we can also show that AO ∼
= CO. Thus, the
triangles 4BON and 4CON are congruent, where N is the midpoint of BC. Hence, ON
is perpendicular to BC and we are done.
Let R denote the radius of the circumcircle.
Theorem 5.11 (Extended Law of Sines) In triangle 4ABC
b
c
a
=
=
= 2R.
sin A
sin B
sin C
Proof: In 4ABC, let ON be the perpendicular bisector of BC. Then 4BOC is isosceles,
∠BON ∼
= ∠CON and BN = CN = a/2. By the Star Trek Lemma ∠BOC = 2A. Thus,
∠BON = ∠A. Thus,
a
R sin A =
2
and
a
2R =
.
sin A
Similarly,
b
c
2R =
=
,
sin B
sin C
as we needed.
1
This actually uses a result that is equivalent to Euclid’s fifth postulate.
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5.7. THE EULER LINE
47
Most of us remember the Law of Sines, but few of us ever ask “What is the common
ratio given in the Law of Sines?” Now you know, that common ratio is twice the radius of
the circumcircle.
5.7
The Euler Line
What happens if the circumcenter, O, coincides with the centroid, G? That would
mean that the medians of the triangle are
the perpendicular bisectors as well. This
will force the triangle to be an equilateral
triangle. What happens if the triangle is
not equilateral? Is there any relationship
between the circumcenter and the centroid?
They will be distinct.
Theorem 5.12 In an arbitrary triangle,
the three altitudes intersect in a common
point, called the orthocenter.
A
H
G
O
B
A'
C
There are several ways to prove this.
You can do it in a very straightforward
manner. However, you will miss a neat result that follows from the following type of
proof.
Figure 5.6: Euler line
Proof: If 4ABC is equilateral, then the altitudes are the perpendicular bisectors and
medians, so the altitudes all meet at G = O. Assume that 4ABC is not equilateral, so
that O 6= G.
Let H be the point on the line OG so that |GH| = 2|OG| and the points O, G, and
H appear in that order. Let A0 be the midpoint of BC, so that G ∈ AA0 and OA0 is the
perpendicular bisector of BC.
Consider the triangles 4GOA0 and 4GHA. Now, ∠OGA0 = ∠HGA since they are
vertical angles. We have proven that the centroid divides the median in a 2:1 ratio, so
|AG| = 2|GA0 |. We constructed the point H so that |GH| = 2|OG|, so the two triangles
are similar. Hence, AH is parallel to OA0 . If we extend AH to where it intersects BC in
a point D, then AD is perpendicular to BC and AD is an altitude of 4ABC. A similar
argument works for the other sides.
Theorem 5.13 (The Euler Line) The circumcenter O, the centroid G, and the orthocenter H are collinear. Furthermore, G lies between O and H and
|OG|
1
= .
|GH|
2
This line is called the Euler line. It was not discovered in any ancient writings and
apparently, Leonhard Euler (1707–1783) was the first to discover this result.
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CHAPTER 5. EUCLIDEAN GEOMETRY: REVISITED
5.8
Feuerbach’s Circle
The following theorem is not extremely important, but it is “fun”. Let A0 , B 0 , and C 0 be the
midpoints of the sides of a triangle 4ABC. Let D, E, and F be the bases of the altitudes.
Let H be the orthocenter, and let A00 , B 00 , and C 00 be the midpoints of AH, BH, and CH,
respectively.
Theorem 5.14 (The Nine Point Circle Theorem) The nine points A0 , B 0 , C 0 , A00 ,
B 00 , C 00 , D, E, and F all lie on a circle.
Proof: B 0 and C 0 are midpoints. Therefore, B 0 C 0 is parallel to BC.
Consider the triangle 4AHB. B 00 is the
midpoint of HB and C 0 is the midpoint of
C
AB. Thus, B 00 C 0 is parallel to AH. Now,
remember that AH is perpendicular to BC,
D
so it is perpendicular to B 0 C 0 . Therefore
C''
B 00 C 0 is perpendicular B 0 C 0 .
A'
B'
Similarly, B 0 C 00 is parallel to AH, out of
E
4AHC. Also, B 00 C 00 is parallel to BC and
H
0
0
0
0
00
00
hence to B C . Therefore, B C B C is a
B''
rectangle.
A''
Construct the circle with diameter
A
F
C'
B
C 0 C 00 . Since, ∠C 0 B 00 C 00 and ∠C 0 B 0 C 00 are
0
00
right angles, B and B lie on this circle,
and since |B 0 B 00 | = |C 0 C 00 | we have B 0 B 00 is
Figure 5.7: 9 Point Circle
a diameter.
Since CF is an altitude, C 0 F C 00 is a right angle, placing F on the circle.
Since B 0 B 00 is a diameter and ∠B 0 EB 00 is a right angle, E lies on the circle.
Now, make a similar argument to show that C 0 A00 C 00 A0 is a rectangle, so A0 and A00 lie
on the circle, and A0 A00 is a diameter, so D lies on the circle.
We mention the following results without proof.
Theorem 5.15 (Feuerbach’s Theorem) The nine point circle of 4ABC is tangent to
the incircle and the excircles of 4ABC.
Theorem 5.16
1. The nine-point circle is the circumcircle of the medial triangle.
2. The nine-point circle has radius one-half that of the circumcircle.
3. The nine-point circle is the circumcircle of the triangle whose vertices are the midpoints of the segments joining 4ABC’s vertices to the orthocenter.
4. The nine-point circle passes through the points where 4ABC’s sides are cut by the
lines that join 4ABC’s vertices with its orthocenter.
5.9
Pedal Triangles and the Simson Line
The Euler line is not unique in the study of triangles. There are other interesting points
and lines associated to any triangle.
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5.9. PEDAL TRIANGLES AND THE SIMSON LINE
49
Z
P
A
A
Y
Y
Z
C
X
P
C
B
X
B
Figure 5.8: Pedal Triangle
Figure 5.9: Simson Line
A cyclic quadrilateral is a quadrilateral that can be inscribed in a circle. The following
is left for homework.
Theorem 5.17 A convex quadrilateral ABCD is a cyclic quadrilateral if and only if ∠ABC+
∠CDA = 180◦ .
Let 4ABC be an arbitrary triangle and let P be a point either inside or outside the
triangle. Let X be the foot of the perpendicular to the extended side BC and through
P . Define points Y and Z on the extended sides AC and AB respectively, similarly. The
triangle 4XY Z is called the pedal triangle with respect to the point P and the triangle
4ABC.(cf. Figure 5.8
Lemma 5.6 Let P be a point inside 4ABC, and let 4XY Z be the pedal triangle with
respect to P . Then ∠AP B = ∠ACB + ∠XZY .
Proof: Let CP intersect AB at C 0 . Then write
∠AP B = ∠AP C 0 + ∠C 0 P B.
Since ∠ABC 0 is an exterior angle of 4AP C, we have that ∠AP C 0 = ∠P AC + ∠ACP .
Now, ∠P ZA = ∠AY P = 90◦ , so they sum to 180◦ and AY P Z is a cyclic quadrilateral.
Thus,
∠P AC = ∠P AY = ∠P ZY,
which implies ∠AP C 0 = ∠P ZY + ∠ACP . Similarly, ∠C 0 P B = ∠XZP + ∠P CB. Thus,
∠AP B = ∠AP C 0 + ∠C 0 P B
= (∠P ZY + ∠XZP ) + (∠ACP + ∠P CB)
= ∠XZY + ∠ACB,
as desired.
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CHAPTER 5. EUCLIDEAN GEOMETRY: REVISITED
Theorem 5.18 (The Simson Line) Let Γ be the circumcircle for 4ABC. Let P be a
point on Γ, and let 4XY Z be the pedal triangle with respect to P . Then 4XY Z is a
degenerate triangle, i.e. the points X, Y , Z are collinear. This line is called the Simson
line.
Proof: Without loss of generality, we may assume P lies on the arc AC. Then ∠AP B =
∠ACB, since they subtend the same arc. Hence, by Lemma 5.6 ∠XZY = 0. That is
4XY Z is degenerate. Thus, X,Y , and Z are collinear.
5.10
Triangle Centers and Relative Lines
Recall that an excircle of a triangle 4ABC is a circle outside the triangle that is tangent
to all three of the lines that extend the sides of the triangle. We have three such circles,
each tangent to a side and the extensions of the other two sides.
Lemma 5.7 The lines connecting the point of tangency of each excircle of 4ABC to the
opposite vertex will intersect in a point, called the Nagel point, N . (cf. Figure 5.10)
One more point of interest is the center of the incircle for 4ABC’s medial triangle. This circle is called the
Spieker circle and its center is called the Spieker point, S.
Lemma 5.8 The Nagel segment is a line segment from
the incenter, I, to the Nagel point, N , which contains the
Spieker point, S, and the centroid, G.(cf. Figure 5.11)
Ic
A
Nb
Nc
N
Na
B
C
We display the following results about the Nagel segment and the Spieker circle without proof.
Ia
Lemma 5.9 For 4ABC,
1. The Spieker circle is the incircle of 4ABC’s medial
triangle.
2. The Spieker circle has radius one-half of 4ABC’s
incircle.
Figure 5.10: Nagel Point
3. The Spieker circle is the incircle of the triangle whose
vertices are the midpoints of the segments that join 4ABC’s vertices with its Nagel
point.
4. The Spieker circle is tangent to the sides of 4ABC’s medial triangle where that triangle’s sides are cut by the lines that join 4ABC’s vertices with its Nagel point.
Note the similarity to the nine-point circle. In addition,
we have the following.
A
S
G
Lemma 5.10 The Spieker point is the midpoint of the
Nagel segment. The centroid is one-third of the way from
the incenter to the Nagel point.
I
N
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Figure 5.11: Nagel Segment
C
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5.11. MORLEY’S THEOREM
51
These theorems of concurrence we have considered to
now are related to the concurrence of three lines. Lines
are not the only items of interest in geometry. Miquel’s
Theorem considers the concurrence of sets of three circles
associated with a triangle.
Theorem 5.19 (Miquel’s Theorem) If three points are chosen, one on each side of a
triangle, then the three circles determined by a vertex and the two points on the adjacent
sides meet at a point called the Miquel point.
Proof: Let 4ABC be our triangle and let D,E, and F be
arbitrary points on the sides of the triangle. Construct the
circles determined by pairs of these points and a vertex.
Consider two of the circles, C1 and C2 , with centers I and
J. They must intersect at D, so they must intersect at
a second point, call it G. In circle C2 , we have that the
angles ∠EGD and ∠ECD are supplementary. In circle C1
∠F GD and ∠ABD are supplementary.
Then,
A
F
G
E
B
D
C
∠EGD◦ + ∠DGF ◦ + ∠EGF ◦ = 360◦
(180◦ − ∠C ◦ ) + (180◦ − ∠B ◦ ) + ∠EGF ◦ = 360◦
∠EGF ◦ = ∠C ◦ + ∠B ◦
= 180◦ − ∠A◦
Figure 5.12: Miquel Point
so that ∠EGF and ∠EAF are supplementary, and hence E, A, F , and G form a cyclic
quadrilateral. Thus, all three circles are concurrent. Note that you must modify this proof,
slightly, if the Miquel point is outside of the triangle.
5.11
Morley’s Theorem
Theorem 5.20 (Morley’s Theorem) The adjacent trisectors of the angles of a triangle
are concurrent by pairs at the vertices of an equilateral triangle.
The following proof is due to John Conway.
Proof: Let the angles A,B,and C measure 3α, 3β, and 3γ respectively. Let x+ mean
x + 60◦ . Now, we have that α + β + γ = 60◦ , since 3α + 3β + 3γ = 180◦ . Then there
certainly exist seven abstract triangles having the angles:
1
α++ , β, γ
2
α, β ++ , γ
3
α, β, γ ++
4
α, β + , γ +
5
α+ , β, γ +
6
α+ , β + , γ
7
0+ , 0+ , 0+
since in every case the triple of angles adds to 180 degrees. Now these triangles are only
determined up to scale, i.e., up to similarity. Determine the scale by saying that certain
lines are all to have the same length.
Triangle number 7, with angles 0+ , 0+ , 0+ , is clearly equilateral, so we can take all its
edges to have some fixed length L. Then arrange the edges joining B + to C + in triangle 4,
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CHAPTER 5. EUCLIDEAN GEOMETRY: REVISITED
B
C
A
Figure 5.13: Morley’s Theorem
C + to A+ in triangle 5, and A+ to B + in triangle 6 also to have length L. We will scale the
other triangles appropriately.
Then it’s easy to see that these all fit together to make up a triangle whose angles are
3A, 3B, 3C, and which is therefore similar to the original one, so proving Morley’s theorem.
To see this, you just have to check that any two sides that come together have the same
length, and that the angles around any internal vertex add to 360 degrees. The latter is
easy, and the former is proved using congruences such as that that takes the vertices A,
C + , B + of triangle number 4 to the points A, B ++ , Y of triangle number 2.
5.12
More Triangle Centers
The few centers we have seen only begin to scratch the surface of what is known about the
different triangle centers and central lines of triangles. I will mention only a few more here.
The best location to find information about triangle centers is
http://cedar.evansville.edu/~ck6/tcenters/: the Triangle Centers website.
5.12.1
Fermat Point
Let 4ABC be an arbitrary triangle. We want to consider the equilateral triangle constructed on each side of the triangle 4ABC. That is 4A0 BC is the equilateral triangle
on side BC, 4AB 0 C is the equilateral triangle on side AC, and 4ABC 0 is the equilateral
triangle on side AB.
The lines AA0 , BB 0 , and CC 0 meet in the Fermat point. This is said to be the first
triangle center discovered after ancient Greek times. It arose from a problem posed by the
great French mathematician, Pierre Fermat. The problem requests the solver to find the
point P in the triangle for which the sum P A + P B + P C is minimal. Torricelli proved that
the Fermat point is the solution if each angle of the triangle 4ABC is less than 120◦ . The
Fermat point is also known as the first isogonic center. This is because the angles ∠BF C,
∠CF A and ∠AF B are all equal.
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5.13. ISOGONS AND SYMMETRY
53
A'
C
B'
F
B
A
C'
Figure 5.14: Fermat point
5.12.2
Gergonne Point
For any triangle 4ABC let A0 be the point where the incircle meets side BC, B 0 be the
point where the incircle meets side AC, and C 0 be the point where the incircle meets side
AB. The segments AA0 , BB 0 , and CC 0 meet in a point, called the Gergonne point, Ge, of
the triangle.
C
A'
B'
X
C'
B
A
Figure 5.15: Gergonne point
5.13
Isogons and Symmetry
How do all of these relate to one another, or do they? Of course they do — somehow. We
take a short time here to point out some relations.
The word isogon has almost completely been replaced by the word equiangular. It
describes a polygon with all angles congruent. Obvious examples are rectangles and the
regular polygons. The word comes from the Greek isos for same, and gon for knee or corner.
Isogonic is a related word that describes a type of symmetry between lines, passing through
the vertex of an angle, and the angle bisector.
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CHAPTER 5. EUCLIDEAN GEOMETRY: REVISITED
−−→
In the figure BB 0 is the bisector of ∠ABC. The rays
A
−−→
−−→
BX and BY are isogonal because they make the same angle
X
with the angle bisector — one is a reflection of the other
B'
in the angle bisector. We often say that one is the isogonal
reflection of the other. It is clear that if L2 is the isogonic
Y
reflection of L1 , then L1 is the isogonic reflection of L2 .
Two points on these rays, such as X and Y , are called
isogonal points.
B
If three lines in a triangle are concurrent,
then their isogonic lines are also concurrent.
A
B'
In the figure the segments AA0 , BB 0 , and
C'
CC 0 intersect at the point X. The three
rays are the isogonic lines for the three segX
ments, which are reflected about the angle
bisectors (dashed rays). The rays intersect
in a single point also, labeled X 0 . Points X
I
and X 0 are called isogonal conjugates.
X'
One pair of isogonal conjugates is the
C
B
A'
orthocenter (intersection of the altitudes)
and the circumcenter (center of the circle
which circumscribes a triangle). These are
the points X and X 0 in the above figure. In
any triangle find these two points, C and
−−→
O, and draw the angle bisector, AX, from
a vertex, A, of the triangle, then the angles ∠P AX and ∠QAX are congruent.
The pedal triangle of the orthocenter
has all three vertices laying on the nineA
point circle. The “feet” of the circumcenter
of 4ABC are the midpoints of the sides a,
b, and c, and so the pedal triangle of the
circumcenter also has vertices laying on the
nine-point circle. Even more the center of
the nine-point circle is the center of the line
segment joining the orthocenter and the circumcenter. What is perhaps more amazing
G
K
is that a similar situation exists for any two
isogonal conjugates X and Y . The circumcircle of the pedal triangle of X is also the
C
circumcircle of the pedal triangle of Y , and
B
the center of the circle is the midpoint of
the segment XY .
Some isogonal reflections have special
names. If we reflect the medians of a triangle in the angle bisectors, the reflections are called symmedians, a contraction of the
term symmetric medians. The symmedians intersect in a point called the Lemoine point,
for French engineer and mathematician Emile Lemoine. It is also known as the symmedian
point of the triangle.
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5.13. ISOGONS AND SYMMETRY
55
In the figure, the medians are thick, the angle bisectors are dashed, and the symmedians
are thin. The Lemoine point is point K. If we call X the point where the symmedian from
A intersects side BC, then the ratio of
BX
AB 2
=
.
XC
AC 2
Thus in a triangle 4ABC, the three points on side BC cut by the median, the angle bisector,
and the symmedian respectively, allow us to divide side c of the triangle into ratios of 1 : 1,
a : b, or a2 : b2 . The Lemoine circle is defined by three lines through the Lemoine point
parallel to the sides of the triangle. These lines intersect the triangle in six points, which
all lie on the Lemoine Circle. The center of the Lemoine circle is the midpoint of a line
segment from the Lemoine point to the circumcenter.
Theorem 5.21 The symmedian (Lemoine) point is the isogonal reflection of the triangle’s
centroid.
The symmedians are related to another triangle center called the Gergonne point, named
for Joseph Gergonne (1771-1859). The Gergonne point (recall from above) is the point of
intersection of the lines from the vertices of a triangle to the points where the inscribed
circle is tangent to the triangle. This triangle connecting the three points of tangency is
called the Gergonne triangle or the contact triangle. If you construct the symmedians of
the Gergonne triangle, you will find that the isogonic reflections of the medians are the lines
from the vertices to the points of tangency.
Theorem 5.22 The Lemoine point of the Gergonne triangle is the Gergonne point of the
original triangle, 4ABC.
The center of the incircle together with the three excircle centers form an orthocentric
system2 . The radii of the in- and excircles are closely related to the area of the triangle. If K
is the triangle’s area and its sides are a, b and c, then the radius of the incircle (also known
as the inradius) is 2K/(a+b+c), the excircle at side a has radius 2K/(−a+b+c), the excircle
at side b has radius 2S/(a − b + c) and the excircle at side c has radius 2S/(a + b − c). From
these formulas we see in particular that the excircles are always larger than the incircle,
and that the largest excircle is the one attached to the longest side.
The triangle’s nine point circle is tangent to the three excircles as well as to the incircle.
The triangle’s Feuerbach point lies on the incircle. Denoting the three vertices of the triangle
by A, B and C and the three points where the incircle touches the triangle by TA , TB and TC
(where TA is opposite A, etc.), the triangle 4TA TB TC is the contact triangle or Gergonne
triangle of 4ABC. The incircle of 4ABC is the circumcircle of 4TA TB TC . The three lines
ATA , BTB and CTC intersect in a single point, the triangle’s Gergonne point G.
The point of all of this is to show that there is still research going on in Euclidean
geometry. There are numerous articles each year about triangle centers, different ratios,
and certain associated triangles and circles. The era of Euclidean geometry did not stop
with the Greeks, nor with Newton, nor with Gauss.
2
In geometry, an orthocentric system is a set of four points in the plane where one point is the orthocenter
of the triangle formed by the other three. If four points form an orthocentric system, then each of the four
points is the orthocenter of the other three. These four possible triangles will all have the same nine-point
circle.
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Spring 2006
Chapter 6
Euclidean Constructions
The idea of constructions comes from a need to create certain objects in our proofs. A
construction is, in some sense, a physical manifestation of the abstract.
In Greek times, geometric constructions of figures and lengths were restricted to the
use of only a straightedge and compass (or in Plato’s case, a compass only). No markings
could be placed on the straightedge to be used to make measurements. While there is only
one type of straightedge - a ruler without markings - there are two types of compasses.
The Greeks, following Plato, use dividers, or what we now call a collapsing compass. This
compass loses the radius once the instrument is moved. This would mean that you could
not construct a circle and then move the compass and construct a circle of the same radius.
Furthermore, this compass could not be used even to mark off distances by setting it and
then walking it along.
Because of the prominent place Greek geometric constructions held in Euclid’s Elements,
these constructions are sometimes also known as Euclidean constructions. Such constructions lay at the heart of the three classical geometric problems (1) squaring the circle (also
known as quadrature of the circle), (2) duplicating the cube, and (3) trisecting an arbitrary
angle. The Greeks were unable to solve these problems. It was not until the eighteenth
and nineteenth century that the problems were proved to be actually impossible under the
limitations imposed.
The constructions are all related to Euclid’s first three axioms:
1. To draw a straight line from any point to any point.
2. To produce a finite straight line continuously in a straight line.
3. To draw a circle with any center and radius.
6.1
Different Compasses?
The straightedge is given as having no length markings because none of these postulates
provides us with the ability to measure lengths. Thus you may use the straightedge to
connect a pair of given points or extend a given line segment arbitrarily in a straight line.
However, you cannot use the straightedge to extend a line segment 3 inches. Since the
straightedge is the same, this does not bother us. However, since the Euclidean compass is
different from the modern, or fixed, compass, we do have some concerns. Can we make the
same constructions? Are the rules the same for both compasses? Well, we can.
57
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CHAPTER 6. EUCLIDEAN CONSTRUCTIONS
Theorem 6.1 (The Compass Equivalence Theorem) A circle C(B, r) can be congruently copied (using only a straightedge and collapsing compass) so that a given point A serves
as the center of the copy.
Proof: Consider the given circle and a point A as shown to the right.
We need to construct a circle of radius r centered at
A, using only a straightedge and collapsing compass. First
construct the circle centered at A with radius AB and the
circle centered at B with radius BA. These circles intersect in two points. Call the points C and D. C(B, r)
and C(A, AB) intersect in a point E. Consider the circle A
C(C, CE). It will intersect C(B, BA) in a point P . We
claim that |AP | = r.
4P CB ∼
= 4ECA by SSS. Thus, ∠P CB ∼
= ∠ECA, and
B
∠P CB − ∠ACB = ∠ECA − ∠ACB.
Therefore, ∠P CA ∼
= ∠ECB.
Now, CP ∼
= CE and AC ∼
= BC. Thus, by
∼
SAS, 4AP C = 4BEC, which implies that |AP | =
|BE| = r.
C
P
E
6.2
Euclidean Constructions
Now, we can discuss the constructions using the
Euclidean axioms, but our compass. Constructions
1–9 and the proof that the constructions work are
left to the reader.
Construction 1: We can construct the perpendicD
ular bisector of a line segment.
Construction 2: We can construct a perpendicular line from a given point to a line not containing
that point.
Construction 3: We can construct a line perpendicular to a given line at a given point P
on `.
Construction 4: We can bisect an arbitrary angle.
Construction 5: We can copy a given angle to a given ray with a given vertex.
Construction 6: Given a line ` and a point P not on `, we can construct a line through
P parallel to the given line.
Construction 7: We can construct an equilateral triangle.
Construction 8: We can construct a square.
Construction 9: We can construct a regular hexagon.
r
A
B
These are the usual constructions. We have many more. In fact, the process of constructions can (and does) form an algebraic system. We need to know how to add, subtract,
multiply and divide segments. First, we will recall a construction that you may or may not
recall from your high school geometry course.
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59
Let us start with two points. We will normalize our measurement of length by defining
the distance between those two points to be equal to one unit. We say that a length a
is a constructible length if there are two constructible points P and Q so that |P Q| = a.
Another, equivalent, definition is that a is constructible if one can construct a line segment of
length a in a finite number os steps from a segment of unit length using only a straightedge
and compass.
Construction 10: We can partition a line segment into three (or more) congruent pieces.
In other words, if a is constructible then so is a/n for any positive integer, n.
C
D3
D2
D1
P
Q
A
B
It is quite simple, so let us illustrate it with n = 3. Let AB denote the segment that we
wish to divide into three congruent sub-segments. Choose a point, C, not on the line AB.
We would prefer that you choose C so that the angle ∠CAB is an acute or right angle,
but this is not necessary. Choose a point, D1 , on the ray AC. This could be the point
C, but need not be. Normally, we would choose the segment, AD1 , to have unit length,
but again this is not necessary. Copy AD1 twice more on the ray AC, to get two more
segments, D1 D2 and D2 D3 , each of which is congruent to AD1 . Thus the final segment has
the property that |AD3 | = 3|AD1 |. Now, construct the segment D3 B. Construct the lines
through D2 and D1 parallel to the line BD3 . Each will intersect AB in a point. Call these
points P and Q. Then, AP ∼
= PQ ∼
= QB and you have trisected the segment.
Construction 11: If a and b are constructible, then so are a + b and a − b.
This one should be obvious, but let’s go over the proof in order to see how we need to
prove these results.
Proof: Let A and A0 be points so that |AA0 | = a and B and B 0 be points so that |BB 0 | = b.
Draw line AA0 and construct the circle centered at A0 of radius b. This circle will intersect
the line AA0 in two points, C and C 0 . Let C be on the opposite side of A from A0 , then
|AC| = a + b and |AC 0 | = a − b.
Construction 12: If a and b are constructible, then so is ab.
Proof: We will use similar triangles. Suppose that AB has length a and CD has length
b. How do we construct a segment that has length ab?
On a constructible line P Q containing point O, find points I and E so that |OI| = 1
and |OE| = a. Now choose a point K on the perpendicular to P Q at I so that |KI| = b.
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Construct the line OK and the perpendicular to P Q at E. These lines intersect in a point
F . Then the triangles 4OIK and 4OEF are similar and
b
|EF |
=
,
1
c
and we are done.
By replacing the roles of c and 1 in the above
proof, you should easily find a proof for the followF
ing.
K
Construction 13: If a and b are constructible,
then so is a/b.
ab
b
We have the set of constructible numbers. How
big is it? Clearly, we can perform addition, subtraction, multiplication and division. This means that
O
1
I a E
we can construct all integers and rational numbers.
Can we construct
We suspect so. We can
√ more?
√
surely construct 2, 3 and others. Given a num√
ber a, that is constructible, can we construct a?
√
Construction 14: If a is constructible, then a is constructible.
First, assume that AB has length a and BC has length 1. We construct a circle with
AC as a diameter and let the line perpendicular to AC passing through B intersect the
circle at the point D. Then, 4ADB ∼ 4DCB and
|DB|
|AB|
=
|BD|
|BC|
a
|BD|
=
|BD|
1
a = |BD|2
√
|BD| = a
Note that the tools of our constructions
D
are a straightedge and compass. These draw
lines and circles. Lines can be described by
linear equations of the form ax + by + c = 0,
while circles can be described by quadratic
equations of the form x2 +y 2 +cx+dy +e =
0. Constructions are entirely focused on the
intersections of lines with lines, lines with
circles, and circles with circles. The interB
C
sections of these circles and lines are then, A
algebraically, just the solutions to systems
of linear and/or quadratic equations.
Solutions to such systems are always expressed in terms of linear or quadratic equations,
never cubic or higher order equations. For example, any solution to the linear equation looks
like y = −(ax − c)/b. Plugging this into the equation for a circle yields a quadratic equation
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in x whose roots can be found by the quadratic formula. These roots are then expressed in
terms of square roots — but not higher order roots.
What this tells us then is that if we start with a unit length, every counting number is
constructible by addition. From these it follows that all rational numbers are constructible
by division. Add to these their square roots, fourth roots (square roots of square roots),
eighth roots, and so on. There is a large family of numbers that are constructible. However,
we have left out a large number of numbers as well.
6.3
The Algebra of Constructible Numbers
The theory of constructible numbers is pretty well covered by the next two theorems.
Theorem 6.2 The use of a straightedge alone can never yield segments of numbers outside
the original number field.
√
If a is an element of the number field so that a is not an element of our number field,
√
then our construction technique can extend this number field by making a constructible.
√
Then we can construct all numbers of the form b + c a, where b and c are in the number
field. Now, the set of all numbers of this form also forms a field, called an extension field
of the original field of numbers.
√
Theorem 6.3 A real number is constructible if and only if it is of the form b + c a, where
each of a, b, and c are either rational numbers or are constructed from repeated extraction
of square roots of rational numbers.
The best way to state this is that this number must lie in an extension field of the
rational numbers of degree 2n .
Construction 15: Construct a regular pentagon with a given side. (We defer this until
the next section.)
Construction 16: Construct a regular n-gon with a given side. Is this possible for all n?
For which n is it possible?
Gauss proved the following result:
Theorem 6.4 (Gauss) A regular polygon can be inscribed in a circle by means of a straightedge and compass alone if and only if the number of sides, n, can be expressed as n =
r
2k · p1 p2 . . . pm , for a nonnegative integer k and each pi a distinct prime of the form 22 + 1,
for r > 0.
Some of the regular polygons that are constructible, according to this theorem, are those
with 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, or 24 sides. Note that the theorem does not tell us
how to do the construction, only that it can or cannot be done.
6.4
6.4.1
Famous Problems of Greek Mathematics
Squaring the Circle
Given a circle with a given radius, can you construct a square with the same area?
This appears to be an old problem, dating back to the time before the Rhind papyrus.
In the Rhind papyrus, the Egyptian scribe Ahmes gives a rule for constructing a square
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with area equal to that of a circle. The method is to take 1/9 off the circle’s diameter and
construct a square from what is remaining. What does this give for the area?
µ
A2 =
¶2
8
256 2
(2r) =
r
9
81
A° = πr2 , so
256
π=
= 3.160493827
81
which is close, but not correct.
If we could do this construction, then we would have to be able to construct a square
√
with side π. We would have to know that π is a rational number in order to do so. It
was not until the late 1880’s that it was shown that π is transcendental. We need to know
that it is transcendental to know that it cannot be written as the 2n th root of some rational
number.
Hippocrates was the first to actually use a plane construction to find a square with area
equal to a figure with circular sides. He squared certain lunes1 , and also the sum of a lune
and a circle. Now although he squared certain lunes, he had not shown that every lune can
be squared. In particular the lune that he squared in his plane construction of a square
of area equal to that of a certain lune and a circle was one he could not square by plane
methods.
Hippias and Dinostratus are associated with the method of squaring the circle using a
quadratrix. The curve it thought to be the invention of Hippias while its application to
squaring the circle appears to be due to Dinostratus.
Step by step description of a quadratrix:
Start with a square 2ABCD in the first quadrant with A at the origin. Take the circle
centered at A with radius AB. As the radius rotates from the y-axis counterclockwise to
the x-axis, the segment DC moves at the same rate parallel to itself down to the x-axis.
The intersections of the line and the angle is quadratrix of Hippias.
This curve in Figure 6.1 can be written as a formula in
one of the following ways.
Parametric Form: r(t) = (t cot(t), t), −2π ≤ t ≤ 2π;
Polar: r = θ/ sin(θ);
Cartesian: x = y cot(y).
Pappus relates how the quadratrix was used to square
the circle. The quadratrix meets the line _
AB, the bottom
of the square, in the point G and the arc BD is the quarter
arc of the circle, we know that
_
BD
AB
=
;
AB
AG
so the length of the circumference of the circle is expressed
Figure 6.1: Quadratrix of Hipin terms of the lengths of straight lines. This leads to the
pias
construction of a square equal in area to the circle.
This curve solves the problem of squaring the circle but,
as given by Hippias, the curve is constructed by mechanical means given by a uniform motion
1
A lune is a planar figure bounded by two circular arcs.
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of a line in a time equal to the rotating radius of a circle. The construction was rightly
criticized as requiring the knowledge of the ratio of a line and an arc of a circle, so one
must assume that you know the property that is required to square the circle in order to
show that you can square the circle. Dinostratus never claimed that the quadratrix gave a
plane construction method to square the circle. Nicomedes many years later also used the
quadratrix to square the circle.
Next consider the contributions of Archimedes to the problem of squaring the circle.
Archimedes is famed for his introduction of the spiral curve. The Archimedean spiral is the
spiral given in polar form by r = θ. Archimedes gives the following definition of the spiral
in his work On spirals:- If a straight line drawn in a plane revolves uniformly any number
of times about a fixed extremity until it returns to its original position, and if, at the same
time as the line revolves, a point moves uniformly along the straight line beginning at the
fixed extremity, the point will describe a spiral in the plane.
To square the circle Archimedes gives the following construction. Let P be the point on the spiral when it has completed one turn. Let the tangent at P cut the line perpendicular to OP at T . Then Archimedes proves in Proposition 19
of On spirals that OT is the length of the circumference of the
circle with radius OP . Now it may not be clear that this is
solved the problem of squaring the circle but Archimedes had
already proved as the first proposition of Measurement of the
circle that the area of a circle is equal to a right-angled triangle
having the two shorter sides equal to the radius of the circle
and the circumference of the circle. So the area of the circle
with radius OP is equal to the area of the triangle 4OP T .
Figure 6.2: Archimedean
Both Apollonius and Carpus used curves to square the cir- spiral
cle but it is not clear exactly what these curves were. The one
used by Apollonius is called by Iamblichus ‘sister of the cochloid’ and this has led to various
guesses as to what the curve might have been. Again the curve used by Carpus of Antioch
is called the ‘curve of double motion’ which some have argued was the cycloid.
Now leave the ancient Greek period and look at later developments. First we should
comment that the Greeks were certainly not the only ones to be interested in squaring the
circle at this time. Mathematicians in India were interested in the problem while in China
mathematicians such as Liu Hsiao of the Han Dynasty showed himself to be one of the most
prominent of those attempting to square the circle around 25 AD.
Some time later the Arab mathematicians were, like the Greeks, fascinated by the
problem. In his work the mathematician al-Haytham tried to convince people that squaring
the circle was possible by a plane construction. His promised treatise on the topic never
appeared, so he finally must have realized that he could not solve the problem.
Not long after the work of al-Haytham, Franco of Lige in 1050 wrote a treatise De
quadratura circuli(On squaring the circle). In it Franco examines three earlier methods
based on the assumption that π is 25/8 , 49/16 or 4. Franco states (reasonably enough)
that these are false, then gives his own construction which is based on the assumption that
π is 22/7 . Although this treatise is of great historical interest, it does show European
mathematics at the time was behind the ancient Greeks in depth of understanding.
Moving forward to about 1450, Cusa attempted to prove that the circle could be squared
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scribed polygons is quite fallacious, it is one of the first serious attempts in ‘modern’ Europe
to solve the problem. Again it is worth commenting that the ancient Greeks basically knew
that the circle could not be squared by plane methods, although they stood no chance of
proving it. Regiomontanus, who brought a new impetus to European mathematics, was
quick to point out the error in Cusa’s arguments.
The mechanical methods of the Greeks certainly appealed to Leonardo da Vinci who
thought about mathematics in a very mechanical way. He devised several new mechanical
methods to square the circle. Many mathematicians in the sixteenth century studied the
problem, including Oronce Fine and Giambattista della Porta. The proof by Fine was
shown to be incorrect by Pedro Nunes soon after he produced it. The beginnings of the
differential and integral calculus led to an increased interest in squaring the circle, but
the new era of mathematics still produced fallacious ‘proofs’ of plane methods to square
the circle. One such false proof, given by Saint-Vincent in a book published in 1647, was
based on an early type of integration. The problem was still providing much impetus for
mathematical development.
James Gregory developed a deep understanding of infinite sequences and convergence.
He applied these ideas to the sequences of areas of the inscribed and circumscribed polygons
of a circle and tried to use the method to prove that there was no plane construction for
squaring the circle. His proof essentially attempted to prove that π was transcendental
— that is not the root of a rational polynomial equation. Although he was correct in
what he tried to prove, his proof was certainly not correct. At the same time others, such
as Huygens, believed that π was algebraic — that it is the root of a rational polynomial
equation.
There was still an interest in obtaining methods to square the circle which were not plane
methods. For example Johann Bernoulli gave a method of squaring the circle through the
formation of evolvents.
The historian of mathematics, Montucla, made squaring the circle the topic of his first
historical work published in 1754. This was written at a time long before the problem was
finally resolved, so was very outdated. However, the work is a classic.
A major step forward in proving that the circle could not be squared using ruler and
compasses occurred in 1761 when Lambert proved that π was irrational. This essentially
was a Calculus proof using iterated integration by parts and proving that if π were rational
then there is an integer between 0 and 1, exclusive. This was not enough to prove the
impossibility of squaring the circle with ruler and compass since
√ certain algebraic numbers
can be constructed with ruler and compass — for example 2. It only led to a greater
flood of amateur solutions to the problem of “squaring the circle” and in 1775 the Paris
Academy passed a resolution which stated that no further attempted solutions submitted
to them would be examined. A few years later the Royal Society in London also banned
consideration of any further ‘proofs’ of squaring the circle as large numbers of amateur
mathematicians tried to achieve fame by presenting the Society with a solution. This
decision of the Royal Society was described by De Morgan about 100 years later as the
“official blow to circle-squarers.”
The popularity of the problem continued and there are many amusing stories told by
De Morgan on this topic in his book Budget of Paradoxes which was edited and published
by his wife in 1872, the year after his death. De Morgan suggests that St. Vitus be made
the patron saint of circle-squarers. This is a reference to St. Vitus’ dance, a wild leaping
dance in which people screamed and shouted and which led to a kind of mass hysteria. De
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Morgan also suggested the term morbus cyclometricus as being the circle squaring disease.
Clearly De Morgan found himself having to try to persuade these circle-squarers that their
methods were incorrect, yet many stubbornly held to their views despite the best efforts
of the professional mathematicians. For example a certain Mr James Smith wrote several
books attempting to prove that π = 25/8 . Of course Mr. Smith was able to deduce from
this that the circle could be squared but neither Hamilton, De Morgan nor others could
convince him of his errors.
The final solution to the problem of whether the circle could be squared using ruler and
compass methods came in 1880 when Lindemann proved that π was transcendental, that is it
is not the root of any polynomial equation with rational coefficients. The transcendentality
of π finally proves that there is no ruler and compass construction to square the circle.
6.4.2
Duplicating the Cube
Given a cube with a given volume, can you construct a cube with
Z
twice that volume?
This is not as esoteric as it may first appear. First, let’s ask a
D
C
similar question about a square. If you are given a square of side
1 and area 1, can you construct a square of area 2? The
Y
√ answer W
is yes and you need to construct a square with side 2, which
we can do. Compare Figure 6.3 where 2ABCD is the original
A
B
square and 2W XY Z has area twice that of 2ABCD. However,
√
X
to construct a cube of volume 2, we must be able to construct 3 2,
which according to the previous work is impossible.
Eratosthenes, in his work entitled Platonicus tells the story Figure 6.3: Doubling
that when Apollo proclaimed to the inhabitants of Delos through the Square
his oracle that, in order to get rid of a plague, they would have
to construct an altar double that of the existing one. Their craftsmen had great difficulty
in their efforts to discover how a solid could be made the double of a similar solid. They
therefore went to ask Plato about it, and he in true academician fashion replied that the
oracle meant, not that the god wanted an altar of double the size, but that by giving
them this task he wanted to shame the Greeks for their neglect of mathematics and their
contempt of geometry. In a separate story, Eratosthenes, this time in a letter to Ptolemy
Euergetes, tells the history of the problem in a slightly different manner. The inhabitants
of Delos, suffering a terrible pestilence, consulted their oracles, and were ordered to double
the volume of the altar to their god, Apollo. An altar was built having an edge double the
length of the original; but the plague was unabated — the oracles had not been obeyed.
The error was discovered, and the Delians appealed to Plato for his advice. Plato referred
them to Eudoxus. These stories are mere fable, for the problem is far older than Plato. We
do know that there was a terrible plague in Athens and Delos about 430 BC, so it is likely
that the inhabitants went to the oracles for advice. What the oracles truly said cannot be
known for sure. It is doubtful that the oracles gave them a geometry problem to solve!
Hippocrates of Chios, the discoverer of the square of a lune, showed that the problem
reduced to the determination of two mean proportionals between two given lengths, one
of them being twice the length of the other. That is, given two lengths, find two mean
proportionals between them, or i.e. given lengths a, b find x, y such that
a
x
y
= = ;
x
y
b
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or as the Greeks would have written it a : x = x : y = y : b.
Now we can see that if we can find such an x and y, we will have
³ a ´3 ³ a ´ µ x ¶ ³ y ´ a
a3
=
=
·
·
= .
x3
x
x
y
b
b
Thus if we are given a cube with side a and want to construct a cube b : a times the volume,
we need to construct the cube of side x.
Although Hippocrates could not determine the proportionals, his statement of the problem in this form was a great advance, for it was thought that the problem of trisecting an
angle was reducible to a similar form which, in the language of algebraic geometry, is to
solve geometrically a cubic equation. According to Proclus, a man named Hippias, probably
Hippias of Elis, trisected an angle with a mechanical curve, named the quadratrix. Archytas of Tarentum (c. 430 BC) solved the problems by means of sections of a half cylinder.
According to Eutocius, Menaechmus solved them by means of the intersections of conic
sections. Eudoxus also gave a solution.
All these solutions were condemned by Plato on the ground that they were mechanical
and not geometrical, i.e. they were not effected by means of circles and lines.
The pursuit of mechanical methods furnished a stimulus to the study of mechanical loci:
for example, the locus of a point carried on a rod which is caused to move according to a
definite rule.
In these endeavors Nicomedes invented the conchoid and Diocles, the cissoid. Dinostratus studied the quadratrix invented by Hippias. All these curves furnished solutions to
this problem, as did the trisectrix — a special form, of Pascal’s limaçon. The problem of
duplicating a cube — or geometrically solving a cubic equation — was also attacked by the
Arabian mathematicians. Tobit ben Korra (836-901) is credited with a solution, while Abul
Gud solved it by means of a parabola and an “equilateral” hyperbola.
6.4.3
Trisecting an Arbitrary Angle
Given any angle α using a straightedge and compass construct an angle β so that 3β = α.
This problem doesn’t seem to have a nice story behind it as do the others. The Greeks
did know how to trisect an arbitrary segment, so it seems reasonable that one should be able
to trisect an arbitrary angle. The most confusing part about this problem is that while we
will answer it in the negative – that is that there is NO method that will trisect EVERY angle
– there are methods that will work for some angles. For example, one can trisect an angle
of 27◦ . This causes lots of ‘false’ proofs to be submitted to many mathematics departments
worldwide. For an interesting history of trisectors see The Trisectors by Underwood Dudley
published by the Mathematical Association of America in 1994.
Now, if we can construct an angle α, then we can construct a segment of length | cos(α)|.
In fact an angle α is constructible if and only if a segment of length | cos(α)| is constructible.
Thus, the angle 3α is constructible if and only if a segment of length | cos 3α| is constructible.
Therefore, we can trisect angle 3α if and only if we can construct a segment of length | cos α|.
Now, applying a few trigonometric identities we get
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cos 3α = cos(2α + α)
= cos 2α cos α − sin 2α sin α
= (2 cos2 α − 1)(cos α) − (2 sin α cos α)(sin α)
= (2 cos2 α − 1)(cos α) − (2 cos α(1 − cos2 α)
= 4 cos3 α − 3 cos α
Now, let’s look at a specific angle. Let’s start with an angle of 3α = 60◦ . Can we trisect
an angle of 60◦ ? Then let α = 20◦ and x = cos α. We know that cos 3α = cos 60◦ = 1/2.
Then our ”trisection equation” becomes:
1
4x3 − 3x = .
2
8x3 − 6x − 1 = 0
So, if this equation does not have any rational roots, then we cannot construct cos α.
By the Rational Root Theorem, the possible rational roots are
1 1 1
{±1, ± , ± , ± }.
2 4 8
Checking each one shows that none of these rational numbers is a root of the equation, so
there are NO rational roots to this equation. Therefore, any irrational roots would have
to be cubic roots rather than roots a power of 2, so | cos 3α| is not constructible and you
cannot trisect every angle using only a straightedge and compass.
However, there are ways if you allow other tools.
There is a fairly straightforward way to trisect any angle which was known to Hippocrates of Chios using a straightedge with two marks on it. Because of this construction
the problem of trisecting an angle may have attracted less attention by the best ancient
Greek mathematicians. Although not possible with an unmarked straight edge and compass, it is easy to carry out in practice. So as a practical problem there was little left to do
although the Greeks still were not satisfied in general with mechanical, i.e. non compass
and straightedge, solutions from a purely mathematical point of view. As Plato said, “In
proceeding in [a mechanical] way, did not one lose irredeemably the best of geometry...”
There is another mechanical solution attributed to Archimedes in the Arabic work called
the Book of Lemmas. Most historians of mathematics believe that many of the results given
in the Book of Lemmas are indeed due to Archimedes and the result given on trisecting an
angle is so much in the spirit of the work On spirals that it is widely accepted that this
method is indeed due to Archimedes. The construction proceeds as follows.
Theorem 6.5 (Archimedes) If we are in possession of a compass and a straightedge that
is notched in two places, then it is possible to trisect an arbitrary angle.
All I need to do is to make two little marks on my straightedge and I can solve the
problem! How?
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B
Q
r
r
r
P
O
A
Proof: Let the arbitrary angle be given by the intersection of lines `1 and `2 , which
intersect at O. This proof shows how to trisect the acute angle in the intersection.
Let r denote the distance between the two notches on the straightedge. Draw a circle
centered at O of radius r. It will intersect `1 and `2 at A and B so that the acute angle
∠AOB is the angle to be trisected.
Now, put one notch of the straightedge on the line `2 and the other notch on the circle,
and move the straightedge until it goes through B. So we have a point P on `2 and Q on
the circle so that |P Q| = r.
Let ∠QP O = α. Since 4QP O is isosceles, ∠QOP = α. Thus, ∠OQB = 2α, since
it is an exterior angle to 4QOP . Since 4QOB is isosceles, ∠QBO = 2α and ∠BOQ =
180◦ − 4α. Now, we also have that
∠P OQ + ∠QOB + ∠AOB = 180,
so substituting we get
∠AOB = 180◦ − (∠P OQ + ∠QOB)
= 180◦ − α − (180◦ − 4α)
= 3α.
So, ∠QOP = 13 ∠AOB, as desired.
Nicomedes lived at about the same time as Archimedes (in the second century BC). He
is the mathematician for whom the Conchoid of Nicomedes is named. In fact this curve
was invented by Nicomedes precisely to formalize the process which we have described of
rotating a ruler keeping a point on a line. The ruler has a fixed distance marked on it
and one mark is kept on a given line while the other traces the conchoid curve. Now
this is exactly the curve needed to solve the trisection of an angle and Nicomedes solved
the problem with his curve. However, in practice the method of moving a ruler until the
required configuration was found was on the whole much easier than drawing the conchoid
so Nicomedes method was more of theoretical rather than practical interest.
Pappus tell us about the conchoid of Nicomedes in his Mathematical collection. In
the same work Pappus writes about how the problem of trisecting an angle was solved by
Apollonius using conics. Pappus gives two solutions which both involve the drawing of a
hyperbola.
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The constructions of Pappus show how the Greeks ‘improved’ their solutions to the
problem of trisecting an angle. From a mechanical solution they had progressed to a solution
involving conic sections. The proof of the impossibility had to await the mathematics of
the 19th century. The final pieces of the argument were put together by Pierre Wantzel in
1837. Gauss had stated that the problems of doubling a cube and trisecting an angle could
not be solved with ruler and compasses but he gave no proofs. In this 1837 paper Wantzel
was the first to prove these results.
6.5
The Regular Pentagon
Rather than simple directions for constructing a regular pentagon, this is an opportunity
to tie together two areas of mathematics and make some connections.
Consider the regular pentagon ABCDE inscribed in a circle of radius one centered at
O, as above. Let the chord BE intersect the radius OA at the point F . Then we have
|OF | = cos ∠BOA = cos 72◦ .
Therefore, if we can construct the number, cos 72◦ , then we can construct F , and hence the
regular pentagon. Once we have constructed F , we would then construct the perpendicular
to OA at F . This will intersect the circle in two points, giving us three of the vertices of the
regular pentagon.
Using circles we get the other two vertices.
How will we construct cos 72◦ ? We will appeal to the geometry of complex numbers. Recall
that from DeMoivre’s theorem we can represent any
complex number as:
B
C
a + bi = r · eiθ = r(cos θ + i sin θ),
where θ is the principal angle between the positive real axis and the given complex number. Since
we are on the circle of radius 1, we have that any
number on the unit circle can be represented by
D
◦
eiθ = cos θ + i sin θ. Set θ = 2π
5 = 72 . For notational simplicity, we will use ω to denote ei(2π/5) .
Note that:
³
´5
ei(2π/5) = e2πi = 1
O
F
A
E
so that ω 5 − 1 = 0. Factoring, we have
(ω − 1)(ω 4 + ω 3 + ω 2 + ω + 1) = 0.
Since ω 6= 1, we have to have that the second factor must be zero. Divide both sides of this
equation by ω 2 .
ω 2 + ω + 1 + ω −1 + ω −2 = 0.
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Set x = ω + ω −1 . Now,
x = ω + ω −1
= e2πi/5 + e−2πi/5
= cos(2π/5) + i sin(2π/5) + cos(−2π/5) + i sin(−2π/5)
= cos(2π/5) + i sin(2π/5) + cos(2π/5) − i sin(2π/5)
= 2 cos(2π/5)
Note that
x2 = (ω + ω −1 )2 = ω 2 + 2 + ω −2
Thus,
x2 + x = ω 2 + 2 + ω −2 + ω + ω −1 = (ω 2 + ω + 1 + ω −1 + ω −2 ) + 1
It then follows from above that
x2 + x − 1 = 0.
From the quadratic formula we get
−1 ±
x=
2
√
5
.
Since x must be positive, we get
−1 +
2 cos(2π/5) = x =
√2
−1 + 5
cos(2π/5) =
4
√
5
This is clearly constructible, so the regular pentagon is constructible.
The following construction is due to H.W. Richmond.
Theorem 6.6 Let |OA| have length 1, and construct the circle with center O and radius
1. Let E be the intersection of this circle with the perpendicular to OA at O. Let B be the
midpoint of OE, and let the angle bisector of ∠ABO intersect OA at C. Then
√
−1 + 5
.
|OC| =
4
Proof: Let ∠CBO = α and let τ = tan α. Then
τ = tan α =
|OC|
|OC|
=
= 2|OC|.
|OB|
1/2
Also,
tan(2α) =
|OA|
1
=
= 2.
|OB|
1/2
Recall the double angle formula for tangents:
tan 2θ =
MATH 6118-090
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.
1 − tan2 θ
Spring 2006
6.6. OTHER CONSTRUCTIBLE FIGURES
71
E
E
Q
B
B
O
C
P
O
A
C
A
Q'
P'
Figure 6.4: Richmond’s construction of the regular pentagon
So,
2τ
1 − τ2
1 − τ2 = τ
2=
τ2 + τ − 1 = 0
−1 ±
τ=
2
√
5
.
√
Since |OC| > 0, we then get that |OC| = −1+4 5 , as we needed.
To finish the construction of the regular pentagon, just find the perpendicular to OA at
C which will intersect the circle at points P and P 0 . Use the length AP to find the points
Q and Q0 to get the regular pentagon.
6.6
Other Constructible Figures
We now know constructions for the regular n-gons with n = 3, 4, 5, 6, and 8. What other
regular n-gons are constructible?
Theorem 6.7 It is possible to construct the regular 2r -gon for any integer r ≥ 2.
To prove this, you simply keep bisecting angles. From the square you get the octogon,
from it you get the 16-gon, etc.
Theorem 6.8 We can construct the regular 15-gon.
Proof: On a unit circle, place the vertices of the regular 15-gon and label them P0 , ..., P14 .
These points exist, but at this time we do not know that they are constructible. We may
assume that P0 is constructible, otherwise rotate the circle to where P0 is constructible.
Now, P0 , P5 and P10 for a regular triangle so that P5 and P10 are constructible. Also,
P0 , P3 , P6 , P9 , and P12 for a regular pentagon and hence are constructible. Thus, we
can construct the segment P5 P6 which is a side of the regular 15-gon. Use this length to
construct the other sides.
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CHAPTER 6. EUCLIDEAN CONSTRUCTIONS
We can generalize this result, using the important fact here that 3 and 5 are relatively
prime.
Theorem 6.9 If we can construct a regular m-gon and a regular n-gon where m and n are
relatively prime, then we can construct a regular mn-gon.
Theorem 6.10 Suppose a length x > 0 is the root of an irreducible polynomial of degree n
and suppose that n has an odd prime factor. Then x is not constructible.
Theorem 6.11 Let p be an odd prime. We cannot construct a regular pr -gon for any r ≥ 2,
and we can construct a p-gon if and only ifp is of the form
k
p = 22 + 1.
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Chapter 7
Introduction to Hyperbolic
Geometry
The major difference that we have stressed is that there is one small difference in the parallel
postulate between Euclidean and hyperbolic geometry. We have been working with eight
axioms. Let’s recall the first seven and then add our new parallel postulate.
Postulate 1: We can draw a unique line segment between any two points.
Postulate 2: Any line segment may be continued indefinitely.
Postulate 3: A circle of any radius and any center can be drawn.
Postulate 4: Any two right angles are congruent.
Postulate 6: Given any two points P and Q, there exists an isometry f such that f (P ) =
Q.
Postulate 7: Given a point P and any two points Q and R which are equidistant from P ,
there exists an isometry which fixes P and sends Q to R.
Postulate 8: Given any line `, there exists a map which fixes every point in ` and fixes no
other points.
Our new postulate is one of the negations of Playfair’s Postulate.
Postulate Not 5: There exists a line ` and there exists a point P not on `, so that there
are two distinct lines `1 and `2 through P which do not intersect `.
Note that in negating Playfair’s Postulate, we have to choose whether we want to have
no parallel lines (leading us to elliptic geometry) or more than one parallel line through
the given point. We shall show that the existence of two distinct parallel lines leads to the
existence of an infinite number of distinct parallel lines.
What could such an animal look like? How could we have multiple parallels? Before we
take that up, we need to see just how dependent we are on the Euclidean postulate. We
will see what we can prove without resorting to any parallel postulate, then we will move
into hyperbolic geometry.
7.1
Neutral Geometry
We have not spent too much time considering the ramifications of the axioms unrelated to
the Parallel Axiom. What can we derive from these alone? Remember, the purpose of a
lot of mathematics in the time between Euclid and Bolyai-Lobachevskii-Gauss was to prove
that the Parallel Postulate did depend on the others.
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CHAPTER 7. INTRODUCTION TO HYPERBOLIC GEOMETRY
One of the common notions that we accept is that a line divides the plane into two
disjoint sets, or sides of the line. This is part of the axiom set that Hilbert chose but is not
explicitly stated in the other axiom sets. It is provable using these other systems of axioms.
As it is, we will accept it as a fact that the line divides the plane into two disjoint sets and
accept the following:
Lemma 7.1 Let ` be a line in the plane and let A and B be two points in the plane that
do not lie on `. Then A and B are on opposite sides of the line ` if and only if the segment
AB intersects `.
7.1.1
Alternate Interior Angles
Definition 7.1 Let L be a set of lines in the plane. A line ` is transversal of L if
1. ` 6∈ L , and
2. ` ∩ m 6= ∅ for all m ∈ L .
Let ` be transversal to m and n at points A and B, respectively. We say that each
of the angles of intersection of ` and m and of ` and n has a transversal side in ` and a
non-transversal side not contained in `.
A
B
m
n
l
Definition 7.2 An angle of intersection of m and ` and one of n and ` are alternate
interior angles if their transversal sides are opposite directed and intersecting, and if their
non-transversal sides lie on opposite sides of `. Two of these angles are corresponding
angles if their transversal sides have like directions and their non-transversal sides lie on
the same side of `.
Definition 7.3 Two line k and ` are parallel if k ∩ ` = ∅.
Theorem 7.1 (Alternate Interior Angle Theorem) If two lines cut by a transversal
have a pair of congruent alternate interior angles, then the two lines are parallel.
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7.1. NEUTRAL GEOMETRY
75
E
A'
B'
C'
A
B
C
n
m
D
l
Figure 7.1: Alternate Interior Angles
Proof: Let m and n be two lines cut by the transversal `. Let the points of intersection
be B and B 0 , respectively. Choose a point A on m on one side of `, and choose A0 ∈ n on
the same side of ` as A. Likewise, choose C ∈ m on the opposite side of ` from A. Choose
C 0 ∈ n on the same side of ` as C. Then it is on the opposite side of ` from A0 .
We are given that ∠A0 B 0 B ∼
= ∠CBB 0 . Assume that the lines m and n are not parallel;
i.e., they have a nonempty intersection. Let us denote this point of intersection by D. D is
on one side of `, so by changing the labeling, if necessary, we may assume that D lies on the
same side of ` as C and C 0 . There is a unique point E on the ray B 0 A0 so that B 0 E ∼
= BD.
0
0
∼
Since, BB = BB , we may apply the SAS Axiom to prove that
4EBB 0 ∼
= 4DBB 0 .
From the definition of congruent triangles, it follows that ∠DB 0 B ∼
= ∠EBB 0 . Now, the sup0
0
plement of ∠DBB is congruent to the supplement of ∠EB B. The supplement of ∠EB 0 B
is ∠DB 0 B and ∠DB 0 B ∼
= ∠EBB 0 . Therefore, ∠EBB 0 is congruent to the supplement of
∠DBB 0 . Since the angles share a side, they are themselves supplementary. Thus, E ∈ n
and we have shown that {D, E} ⊂ n or that m∩n is more that one point. This contradiction
gives us that m and n must be parallel.
Corollary 1 If m and n are distinct lines both perpendicular to the line `, then m and n
are parallel.
Proof: ` is the transversal to m and n. The alternate interior angles are right angles. All
right angles are congruent, so the Alternate Interior Angle Theorem applies. m and n are
parallel.
Corollary 2 If P is a point not on `, then the perpendicular dropped from P to ` is unique.
Proof: Assume that m is a perpendicular to ` through P , intersecting ` at Q. If n is
another perpendicular to ` through P intersecting ` at R, then m and n are two distinct
lines perpendicular to `. By the above corollary, they are parallel, but each contains P .
Thus, the second line cannot be distinct, and the perpendicular is unique.
The point at which this perpendicular intersects the line `, is called the foot of the
perpendicular.
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A
E
B
G
C
D
Figure 7.2: Exterior Angle Theorem
Corollary 3 If ` is any line and P is any point not on `, there exists at least one line m
through P which does not intersect `.
Proof: By Corollary 2 there is a unique line, m, through P perpendicular to `. Now there
is a unique line, n, through P perpendicular to m. By Corollary 1 ` and n are parallel.
Note that while we have proved that there is a line through P which does not intersect
`, we have not (and cannot) proved that it is unique.
7.2
Weak Exterior Angle Theorem
Let 4ABC be any triangle in the plane. This triangle gives us not just three segments, but
in fact three lines.
Definition 7.4 An angle supplementary to an angle of a triangle is called an exterior
angle of the triangle. The two angles of the triangle not adjacent to this exterior angle are
called the remote interior angles.
Theorem 7.2 (Exterior Angle Theorem) An exterior angle of a triangle is greater than
either remote interior angle. (See Figure 7.2)
Proof: We shall show that ∠ACD > ∠A. In a like manner, you can show that ∠ACD >
∠B. Then by using the same techniques, you can prove the same for the other two exterior
angles.
Now, either:
∠A < ∠ACD ∠A ∼
= ∠ACD or ∠A > ∠ACD.
If ∠A = ∠BAC ∼
= ∠ACD, then by the Alternate Interior Angle Theorem, lines AB and
CD are parallel. This is impossible, since they both contain B.
−→
−−→
Assume, then, that ∠A > ∠ACD. Then there exists a ray AE between rays AB and
−→
AC so that
∠CAE ∼
= ∠ACD.
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7.2. WEAK EXTERIOR ANGLE THEOREM
77
−→
By the Crossbar Theorem 1 , ray AE intersects BC in a point G. Again by the Alternate
Interior Angle Theorem lines AE and CD are parallel. This is a contradiction.
Thus, ∠A < ∠ACD.
Lemma 7.2 (SAA Congruence) In triangles 4ABC and 4DEF given that AC ∼
= DF ,
∠A ∼
= ∠D, and ∠B ∼
= ∠E, then 4ABC ∼
= 4DEF .
C
A
F
G
B D
H
E
Figure 7.3: SAA Congruence
Proof: If AB ∼
= DE, we are done by Angle-Side-Angle. Thus, let us assume that AB 6∼
=
DE. Then, by we must have that either AB < DE or AB > DE.
If AB < DE, then there is a point H ∈ DE so that AB ∼
= DH. Then by the SAS
∼
∼
Theorem 4ABC = 4DHF . Thus, ∠B = ∠DHF . But ∠DHF is exterior to 4F HE, so
by the Exterior Angle Theorem ∠DHF > ∠E ∼
= ∠B. Thus, ∠DHF > ∠B, and we have
a contradiction. Therefore, AB is not less than DE. By a similar argument, we can show
that assuming that AB > DE leads to a similar contradiction.
Thus, our hypothesis that AB ∼
6 DE cannot be valid. Thus, AB ∼
=
= DE and 4ABC ∼
=
4DEF by ASA.
Lemma 7.3 Two right triangles are congruent if the hypotenuse and a leg of one are congruent respectively to the hypotenuse and a leg of the other.
Lemma 7.4 Every segment has a unique midpoint.
Proof: Let AB be any segment in the plane, and let C be any point not on line AB. There
−−→
exists a unique ray BX on the opposite side of line AB from P such that ∠P AB ∼
= ∠XBA.
−−→
There is a unique point Q on the ray BX so that AP ∼
= BQ. Q is on the opposite side of
line AB from P . Since P and Q are on opposite sides of line AB, P Q ∩ AB 6= ∅. Let M
denote this point of intersection. Either M lies between A and B, A lies between M and
B, B lies between A and M , M = A, or M = B.
We want to show that M lies between A and B, so assume not. Since ∠P AB ∼
= ∠QBA,
by construction, we have from the Alternate Interior Angle Theorem that lines AP and BQ
1
−−→
−→
−→
−−→
Crossbar Theorem: If the ray AD is between rays AB and AC, then AD intersects the segment BC.
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CHAPTER 7. INTRODUCTION TO HYPERBOLIC GEOMETRY
P
A
M
B
Q
Figure 7.4: Uniqueness of the midpoint
are parallel. If M = A then A, P , and M are collinear on the line AP and lines AP = AB
which intersects line BQ. We can dispose of the case M = B similarly.
Thus, assume that A lies between M and B. This will mean that the line P A will
intersect side M B of 4M BQ at a point between M and B. Thus, by Pasch’s Theorem 2
it must intersect either M Q or BQ. It cannot intersect side BQ as lines AP and BQ are
parallel. If line AP intersects M Q then it must contain M Q for P , Q, and M are collinear.
Thus, M = A which we have already shown is impossible. Thus, we have shown that A
cannot lie between M and B.
In the same manner, we can show that B cannot lie between A and M . Thus, we have
that M must lie between A and B. This means that ∠AM P ∼
= ∠BM Q since they are
∼
vertical angles. By Angle-Angle-Side we have that 4AM P = 4BM Q. Thus, AM ∼
= MB
and M is the midpoint of AB.
We state the following results without proof. The proof is left to the reader.
Lemma 7.5
i) Every angle has a unique bisector.
ii) Every segment has a unique perpendicular bisector.
Lemma 7.6 In a triangle 4ABC the greater angle lies opposite the greater side and the
greater side lies opposite the greater angle; i.e., AB > BC if and only if ∠C > ∠A.
Lemma 7.7 Given 4ABC and 4A0 B 0 C 0 , if AB ∼
= A0 B 0 and BC ∼
= B 0 C 0 , then ∠B < ∠B 0
0
0
if and only if AC < A C .
2
Pasch’s Theorem: If 4ABC is any triangle and ` is any line interescting side AB in a point between
A and B, then ` also intersects either side AC or side BC. If C 3 `, then ` does not intersect both AC and
BC.
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7.2. WEAK EXTERIOR ANGLE THEOREM
7.2.1
79
Measure of Angles and Segments
At some point we have to introduce a measure for angles and for segments.
The proofs of these results require the axioms of continuity. We need the measurement
of angles and segments by real numbers, and for such measurement Archimedes’s axiom is
required.
Theorem 7.3
(A) There is a unique way of assigning a degree measure to each
angle such that the following properties hold:
i) ∠A is a real number such that 0 < ∠A < 180◦ .
ii) ∠A = 90◦ if and only if ∠A is a right angle.
iii) ∠A = ∠B if and only if ∠A ∼
= ∠B.
iv) If the ray AC is interior to ∠DAB, then ∠DAB = ∠DAC + ∠CAB.
v) For every real number x between 0 and 180, there exists an angle ∠A
such that ∠A = x◦ .
vi) If ∠B is supplementary to ∠A, then ∠A + ∠B = 180◦.
(B) Given a segment OI, called the unit segment. Then there is a unique way of
assigning a length |AB| to each segment AB such that the following properties
hold:
i) |AB| is a positive real number and |OI| = 1.
∼ CD.
ii) |AB| = |CD| if and only if AB =
iii) B lies between A and C if and only if |AC| = |AB| + |BC|.
iv) |AB| < |CD| if and only if AB < CD.
v) For every positive real number x, there exists a segment AB such that
|AB| = x.
Definition 7.5 An angle ∠A is acute if ∠A < 90◦ , and is obtuse if ∠A > 90◦ .
Corollary 1 The sum of the degree measures of any two angles of a triangle is less than
180◦ .
This follows from the Exterior Angle Theorem.
Proof: We want to show that ∠A + ∠B < 180◦ . From the Exterior Angle Theorem,
∠A < ∠CBD
∠A + ∠B < ∠CBD + ∠B = 180◦ ,
since they are supplementary angles.
Corollary 2 (Triangle Inequality) If A, B, and C are three noncollinear points, then
|AC| < |AB| + |BC|.
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CHAPTER 7. INTRODUCTION TO HYPERBOLIC GEOMETRY
C
¡
¡
¡
¡
¡
¡QQ
Q
Q
Q
Q
Q
Q
Q
A
-
B
B
E
D
A
C
Figure 7.5: First step in Saccheri-Legendre Theorem
7.3
Saccheri-Legendre Theorem
Theorem 7.4 (Saccheri-Legendre Theorem) The sum of the degree measures of the
three angles in any triangle is less than or equal to 180◦ ;
∠A + ∠B + ∠C ≤ 180◦ .
Proof: Let us assume not; i.e., assume that we have a triangle 4ABC in which ∠A +
∠B + ∠C > 180◦ . So there is an x ∈ R+ so that
∠A + ∠B + ∠C = 180◦ + x.
Compare Figure 7.5. Let D be the midpoint of BC and let E be the unique point on
the ray AD so that DE ∼
= AD. Then by SAS 4BAD ∼
= 4CED. This makes
∠B = ∠DCE
∠E = ∠BAD.
Thus,
∠A + ∠B + ∠C = (∠BAD + ∠EAC) + ∠B + ∠ACB
= ∠E + ∠EAC + (∠DCE + ∠ACD)
= ∠E + ∠A + ∠C
So, 4ABC and 4ACE have the same angle sum, even though they need not be congruent.
Note that ∠BAE + ∠CAE = ∠BAC, hence
∠CEA + ∠CAE = ∠BAC.
It is impossible for both of the angles ∠CEA and ∠CAE to have angle measure greater than
1/2∠BAC, so at least one of the angles has angle measure less than or equal to 1/2∠BAC.
Therefore, there is a triangle 4ACE so that the angle sum is 180◦ + x but in which one
angle has measure less than or equal to 1/2∠A◦ . Repeat this construction to get another
MATH 6118-090
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7.3. SACCHERI-LEGENDRE THEOREM
81
triangle with angle sum 180◦ + x but in which one angle has measure less than or equal to
1/4∠A◦ . Now there is an n ∈ Z+ so that
1
∠A ≤ x,
2n
by the Archimedean property of the real numbers. Thus, after a finite number of iterations
of the above construction we obtain a triangle with angle sum 180◦ + x in which one angle
has measure less than or equal to
1
∠A ≤ x.
2n
Then the other two angles must sum to a number greater than 180◦ contradicting Corollary
1.
Corollary 1 In 4ABC the sum of the degree measures of two angles is less than or equal
to the degree measure of their remote exterior angle.
7.3.1
The Defect of a Triangle
Since the angle sum of any triangle in neutral geometry is not more than 180◦ , we can
compute the difference between the number 180 and the angle sum of a given triangle.
Definition 7.6 The defect of a triangle 4ABC is the number
defect (4ABC) = 180◦ − (∠A + ∠B + ∠C).
In euclidean geometry we are accustomed to having triangles whose defect is zero. Is
this always the case? The Saccheri-Legrendre Theorem indicates that it may not be so. Is
the defect of triangles preserved? That is, if we have one defective triangle, then are all
of the sub and super-triangles defective? By defective, we mean that the triangles have
positive defect.
Theorem 7.5 (Additivity of Defect) Let 4ABC be any triangle and let D be a point
between A and B. Then defect (4ABC) = defect (4ACD) + defect (4BCD) .
C
¡AHHH
¡ A
HH
¡
HH
A
¡
HH
A
¡
HH
A
¡
H
A
A
D
B
Figure 7.6: Additivity of Defect
Proof: Since the ray CD lies in ∠ACB, we know that
∠ACB = ∠ACD + ∠BCD,
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CHAPTER 7. INTRODUCTION TO HYPERBOLIC GEOMETRY
C
D
A
B
Figure 7.7: Right triangle defect
and since ∠ADC and ∠BDC are supplementary angles ∠ADC +∠BDC = 180◦ . Therefore,
defect (4ABC)
= 180◦ − (∠A + ∠B + ∠C)
= 180◦ − (∠A + ∠B + ∠ACD + ∠BCD)
= 180◦ + 180◦ − (∠A + ∠B
+ ∠ACD + ∠BCD + ∠ADC + ∠BDC)
= defect (4ACD) + defect (4BCD) .
Corollary 1 defect (4ABC) = 0 if and only if defect (4ACD) = defect (4BCD) = 0.
A rectangle is a quadrilateral all of whose angles are right angles. We cannot prove the
existence or non-existence of rectangles in Neutral Geometry. Nonetheless, the following
result is extremely useful.
Theorem 7.6 If there exists a triangle of defect 0, then a rectangle exists. If a rectangle
exists, then every triangle has defect 0.
Let us first outline the proof in five steps.
1. Construct a right triangle having defect 0.
2. From a right triangle of defect 0, construct a rectangle.
3. From one rectangle, construct arbitrarily large rectangles.
4. Prove that all right triangles have defect 0.
5. If every right triangle has defect 0, then every triangle has defect 0.
Having outlined the proof, each of the steps is relatively straightforward.
1. Construct a right triangle having defect 0.
Let us assume that we have a triangle 4ABC so that defect (4ABC) = 0. We may
assume that 4ABC is not a right triangle, or we are done. Now, at least two angles
are acute since the angle sum of any two angles is always less than 180◦ . Let us assume
that ∠A and ∠B are acute. Also, let D be the foot of C on line AB. We need to
know that D lies between A and B.
Assume not; i.e., assume that A lies between D and B. (See Figure 7.7.) This means
that ∠CAB is exterior to 4CAD and, therefore, ∠A > ∠CDA = 90◦ . This makes
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7.3. SACCHERI-LEGENDRE THEOREM
83
C
E
D
B
X
Figure 7.8: Existence of Rectangles
∠A obtuse, a contradiction. Similarly, if B lies between A and D we can show that
∠B is obtuse. Thus, we must have that D lies between A and B.
This makes 4ADC and 4BDC right triangles. By Corollary 1 above, since 4ABC
has defect 0, each of them has defect 0, and we have two right triangles with defect 0.
2. From a right triangle of defect 0, construct a rectangle.
We now have a right triangle of defect 0. Take 4CBD from Step 1, which has a right
angle at D. There is a unique ray CX on the opposite side of BC from D so that
∠DBC ∼
= ∠BCX.
Then there is a unique point E on ray CX such that CE ∼
= BD.
Thus, 4CDB ∼
= 4BEC by SAS. Then ∠BEC = 90◦ and 4BEC must also have
defect 0. Now, clearly, since defect (4CDB) = 0
∠DBC + ∠BCD = 90◦
and, hence,
∠ECB + ∠BCD = ∠ECD = 90◦ .
Likewise, ∠EBD = 90◦ and 2CDBE is a rectangle.
3. From one rectangle, construct arbitrarily large rectangles.
Given any right triangle 4XY Z, we can construct a rectangle 2P QRS so that P S >
XZ and RS > Y Z. By applying Archimedes Axiom, we can find a number n so that
we copy segment BD in the above rectangle on the ray ZX to reach the point P so
that n · BD ∼
= P Z and X lies between P and Z. We make n copies of our rectangle
sitting on P Z = P S. This gives us a rectangle with vertices P , Z = S, Y , and some
other point. Now, using the same technique, we can find a number m and a point R
on the ray ZY so that m·BE ∼
= RZ and Y lying between R and Z. Now, constructing
m copies of the long rectangle, gives us the requisite rectangle containing 4XY Z.
4. Prove that all right triangles have defect 0.
Let 4XY Z be an arbitrary right triangle. By Step 3 we can embed it in a rectangle
2P QRS.
∼ 4P SR, we have that ∠RP S + ∠P RS = 90◦ and then, 4P RS has
Since 4P QR =
defect 0. Using Corollary 1 to Theorem 7.5 we find defect (4RXY ) = 0 and thus,
defect (4XY Z) = 0. Therefore, each triangle has defect 0.
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CHAPTER 7. INTRODUCTION TO HYPERBOLIC GEOMETRY
R
½
­
½­
½
½ ­
½ ­
½
­
½
Z
­
¡
½
½
­
¡
½
­
¡
½
­ ¡
½
½
­ ¡
½
­ ¡
½
­¡
½
½
­¡
½
­¡
½
­
½
¡
P
X
S=Y
Figure 7.9: Step 1
5. If every right triangle has defect 0, then every triangle has defect 0.
As in the first step, use the foot of a vertex to decompose the triangle into two right
triangles, each of which has defect 0, from Step 4. Thus, the original triangle has
defect 0.
Corollary 1 If there is a triangle with positive defect, then all triangles have positive defect.
7.4
Hyperbolic Axiom Results
Hyperbolic geometry is often called Bolyai-Lobachevskiian geometry after two of its discovers
János Bolyai and Nikolai Ivanovich Lobachevskii. Bolyai first announced his discoveries in
a 26 page appendix to a book by his father, the Tentamen, in 1831. Another of the great
mathematicians who seems to have preceded Bolyai in his work is Karl Friedreich Gauss.
He seems to have done some work in the area dating from 1792, but never published it. The
first to publish a complete account of non-Euclidean geometry was Lobachevskii in 1829.
It was first published in Russian and was not widely read. In 1840 he published a treatise
in German.
We shall call our added axiom the Hyperbolic Axiom.
PPPr ³³
1
i
³
)
³³PPP
q
¾
-`
We shall denote the set of all points in the plane by H2 , and call this the hyperbolic
plane.
Lemma 7.8 There exists a triangle whose angle sum is less than 180◦ .
Proof: Let ` be a line and P a point not on ` such that two parallels to ` pass through P .
We can construct one of these parallels as previously done using perpendiculars. Let Q be
MATH 6118-090
Spring 2006
7.4. HYPERBOLIC AXIOM RESULTS
85
P
Yr - m
XXX
H
A H
X
HHXXX
A
XXX
H
XXX
HH
Ab
XXXX
X? A
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rXX
zn
HH
AU
HH
H
H
-`
Q
R
the foot of the perpendicular to ` through P . Let m be the perpendicular to the line P Q
through P . Then m and ` are parallel. Let n be another line through P which does not
intersect `. This line exists by the Hyperbolic Axiom. Let P X be a ray of n lying between
P Q and a ray P Y of m.
claim: There is a point R ∈ ` on the same side of the line P Q as X and Y so that
∠QRP < ∠XP Y .
proof of claim. The idea is to construct a sequence of angles
∠QR1 P, ∠QR2 P, . . . , ∠QRn P, . . .
so that ∠QRj+1 P < 12 ∠QRj P . We will then apply Archimedes Axiom for real numbers to
complete the proof.
There is a point R1 ∈ ` so that QR1 ∼
= P Q. Then 4QR1 P is isosceles and ∠QR1 P ◦ ≤
◦
45 . Also, there is a point R2 ∈ ` so that R1 lies between F and R2 and R1 R2 ∼
= P R1 .
Then 4P R1 R2 is isosceles and ∠R1 P R2 ∼
= ∠QR2 P . Since ∠QR1 P is exterior to 4P R1 R2
it follows that
∠R1 P R2 + ∠QR2 P ≤ ∠QR1 P ,
◦
so then ∠QR2 P ≤ 22 12 . Continuing with this construction, we find a point Rn ∈ ` so that
Rn−1 lies between A and Rn and
µ ¶◦
45
∠QRn P ≤
.
2n
Applying the Archimedean axiom we see that for any positive real number, for example
∠XP Y , there is a point R ∈ ` so that R is on the same side of the line P Q as X and Y
and ∠QRP < ∠XP Y . Thus, we have proved our claim.
Now, the ray P R lies in the interior of ∠QP X, for if not then the ray P X is in the
interior of ∠QRP . By the Crossbar Theorem it follows that the ray P X ∩ ` 6= ∅ which
implies that n and ` are not parallel—a contradiction. Thus, ∠RP Q < ∠XP Q. Then,
∠RP Q + ∠QRP < ∠XP Q + ∠QRP < ∠XP Q + ∠XP Y = 90◦ .
Therefore, ∠P + ∠Q + ∠R < 180◦ and defect (4P QR) > 0.
The Hyperbolic Axiom only hypothesizes the existence of one line and one point not on
that line for which there are two parallel lines. With the above theorem we can now prove
a much stronger theorem.
Theorem 7.7 (Universal Hyperbolic Theorem) In H2 for every line ` and for every
point P not on ` there pass through P at least two distinct lines, neither of which intersect
`.
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86
CHAPTER 7. INTRODUCTION TO HYPERBOLIC GEOMETRY
6
¾ P``
```
``
- m
`S
¾
Q
R
- `
?
t
Proof: Drop a perpendicular P Q to ` and construct a line m through P perpendicular to
P Q. Let R be any other point on `, and construct a perpendicular t to ` through R. Now,
let S be the foot of the perpendicular to t through P . Now, the line P S does not intersect
` since both are perpendicular to t. At the same time P S 6= m. Assume that S ∈ m, then
2P QRS is a rectangle. By Theorem 7.6, if one rectangle exists all triangles have defect 0.
We have a contradiction to Lemma 7.8. Thus, P S 6= m, and we are done.
7.5
Angle Sums (again)
We have just proven the following theorem.
Theorem 7.8 In H2 rectangles do not exist and all triangles have angle sum less than 180◦ .
This tells us that in hyperbolic geometry the defect of any triangle is a positive real
number. We shall see that it is a very important quantity in hyperbolic geometry.
Corollary 1 In H2 all convex quadrilaterals have angle sum less than 360◦ .
7.6
Saccheri Quadrilaterals
Girolamo Saccheri was a Jesuit priest who lived from 1667 to 1733. Before he died he
published a book entitled Euclides ab omni nævo vindicatus ( Euclid Freed of Every Flaw ).
It sat unnoticed for over a century and a half until rediscovered by the Italian mathematician
Beltrami.
Saccheri wished to prove Euclid’s Fifth Postulate from the other axioms. To do so he
decided to use a reductio ad absurdum argument. He assumed the negation of the Parallel
Postulate and tried to arrive at a contradiction. He studied a family of quadrilaterals that
have come to be called Saccheri quadrilaterals. Let S be a convex quadrilateral in which two
adjacent angles are right angles. The segment joining these two vertices is called the base.
The side opposite the base is the summit and the other two sides are called the sides. If
the sides are congruent to one another then this is called a Saccheri quadrilateral. The
angles containing the summit are called the summit angles.
Theorem 7.9 In a Saccheri quadrilateral
i) the summit angles are congruent, and
MATH 6118-090
Spring 2006
7.6. SACCHERI QUADRILATERALS
87
ii) the line joining the midpoints of the base and the summit—called the altitude—
is perpendicular to both.
D
HH
N
C
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©
©©
H¡
©
H ©@
¡©H
©HH@
¡
H@
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HH
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@
A
M
B
Proof: Let M be the midpoint of AB and let N be the midpoint of CD.
1. We are given that
∠DAB = ∠ABC = 90◦ .
Now, AD ∼
= BC and AB ∼
= AB, so that by SAS 4DAB ∼
= 4CBA, which implies
∼
∼
that BD = AC. Also, since CD = CD then we may apply the SSS criterion to see
that 4CDB ∼
= 4DCA. Then, it is clear that ∠D ∼
= ∠C.
2. We need to show that the line M N is perpendicular to both lines AB and CD.
Now DN ∼
= CN , AD ∼
= BC, and ∠D ∼
= ∠C. Thus by SAS 4ADN ∼
= 4BCN .
This means then that AN ∼
= BN . Also, AM ∼
= BM and M N ∼
= M N . By SSS
4AN M ∼
= 4BN M and it follows that ∠AM N ∼
= ∠BM N . They are supplementary
angles, hence they must be right angles. Therefore M N is perpendicular to AB.
Using the analogous proof and triangles 4DM N and 4CM N , we can show that M N
is perpendicular to CD.
Thus, we are done.
Lemma 7.9 In a Saccheri quadrilateral the summit angles are acute.
Proof: Recall from Corollary 1 to Theorem 7.8 that the angle sum for any convex quadrilateral is less that 360◦ . Thus, since the Saccheri quadrilateral is convex,
∠A + ∠B + ∠C + ∠D < 360◦
2∠C < 180◦
∠C < 90◦
Thus, ∠C and ∠D are acute.
A convex quadrilateral three of whose angles are right angles is called a Lambert
quadrilateral.
Lemma 7.10 The fourth angle of a Lambert quadrilateral is acute.
Proof: If the fourth angle were obtuse, our quadrilateral would have an angle sum greater
than 360◦ , which cannot happen. If the angle were a right angle, then a rectangle would
exist and all triangles would have to have defect 0. Since there is a triangle with angle sum
less than 180◦ , we have a triangle with positive defect. Thus, the fourth angle cannot be a
right angle either.
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88
CHAPTER 7. INTRODUCTION TO HYPERBOLIC GEOMETRY
Lemma 7.11 The side adjacent to the acute angle of a Lambert quadrilateral is greater
than its opposite side.
Proof: Left to the reader.
Lemma 7.12 In a Saccheri quadrilateral the summit is greater than the base and the sides
are greater than the altitude.
Proof: Using Theorem 7.9 if M is the midpoint of AB and N is the midpoint of CD, then
2AM N D is a Lambert quadrilateral. Thus, AB > M N and, since BC ∼
= AB, both sides
are greater than the altitude.
Also, applying Theorem 7.9 DN > AM . Since CD ∼
= 2DN and AB ∼
= 2AM it follows
that CD > AB, so that the summit is greater than the base.
7.7
Similar Triangles
In Euclidean geometry we are used to having two triangles similar if their angles are congruent. It is obvious that we can construct two non-congruent, yet similar, triangles. In fact
John Wallis attempted to prove the Parallel Postulate of Euclid by adding another postulate.
D
A
wallis’ postulate: Given any triangle 4ABC and given
££ B
£B
any segment DE. There exists a triangle 4DEF having
B
£
B
£ B
£
B
DE as one of its sides that is similar to 4ABC.
£
BB E 0 £
BF0
However Wallis’ Postulate is equivalent to Euclid’s Par- £
£
B
F
allel Postulate. Thus, we know that the negation of Wallis’ E
£
B
£
B
Postulate must hold in hyperbolic geometry. That is, un£
B
der certain circumstances similar triangles do not exist. We
£
B
can prove a much stronger statement.
B
C
Theorem 7.10 (AAA Criterion) In H2 if ∠A ∼
= ∠D,
∠B ∼
= ∠E, and ∠C ∼
= ∠F , then 4ABC ∼
= 4DEF . That is, if two triangles are similar,
then they are congruent.
proof: Since ∠BAC ∼
= ∠EDF , there exists an isometry which sends D to A, the ray
DE to the ray AB, and the ray DF to the ray AC. Let the image of E and F under
this isometry be E 0 and F 0 , respectively. If the two triangles are not congruent, then we
may assume that E 0 6= B and that E 0 lies between A and B. Then BC and E 0 F 0 cannot
intersect by the Alternate Interior Angles Theorem. Then BCE 0 F 0 forms a quadrilateral.
The quadrilateral has the following angles:
∠E 0 BC = ∠ABC
∠F 0 CB = ∠ACB
∠BE 0 F 0 = 180◦ − ∠ABC
∠CF 0 E 0 = 180◦ − ∠ACB
which sum to 360◦ . This contradiction leads us to the fact that E 0 = B and F 0 = C and
the two triangles are congruent.
MATH 6118-090
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7.8. HOROPARALLEL AND HYPERPARALLEL LINES
89
As a consequence of Theorem 7.10 we shall see that in hyperbolic geometry a segment
can be determined with the aid of an angle. For example, an angle of an equilateral triangle
determines the length of a side uniquely. Thus in hyperbolic geometry there is an absolute
unit of length.
7.8
Horoparallel and Hyperparallel Lines
Let ` be a line and let P 6∈ `. By the Hyperbolic Axiom, there are two lines through P
which do not intersect `. Note that every
line passing through P which is between
these two lines cannot intersect `. Thus, the
P
Hyperbolic Axiom implies the existence of
an infinite number of lines through P which
β β
do not intersect ` — not just two.
In fact, there is more. There must be
two lines through P which do not intersect
` and all lines through P which do not inQ
tersect ` lie between these two lines. These Θ
boundary lines through P and the line ` are
called a number of things: horoparallel,
Ω
limiting parallel, or boundary parallel. The lines that lie between the boundary lines are called hyperparallel or ultraparallel to `. The lines P Ω and ΘΩ
are horoparallel, in Figure 7.10. Let P be a
Figure 7.10: Poincaré Disk Model
point not on `, which in Figure 7.10 is represented by the line ΘΩ. Note that the rays P Ω and P Θ are limiting parallel to `. Let Q
be the foot of P in `. Let β = ∠ΩP Q. Note that ∠ΘP Q = β as well by reflecting through
P Q.
Theorem 7.11 The angle β depends only on the length |P Q|.
Proof: Let `0 be another line, Q0 a point on `0 , and P 0 a point not on `0 such that P 0 Q0 is
perpendicular to `0 and |P 0 Q0 | = |P Q|. Let `0 have endpoints of Ω0 and Θ0 at infinity and
let β 0 = ∠Ω0 P 0 Q0 . We need to show that β = β 0 .
Assume not. Then one of these numbers is larger than the other. Without loss of
generality we may take β 0 < β. Construct the ray r making an angle of β 0 with P Q. Since
β 0 < β, the ray must intersect ` at some point R. Applying the axioms on isometries, since
we have |P Q| = |P 0 Q0 |, there exists an isometry, ϕ, which sends P to P 0 and Q to Q0 . Let
R0 = ϕ(R). Now, since isometries preserve angles, we know that ∠R0 P 0 Q0 = β 0 . This means
that the ray P 0 R0 and the ray P 0 Ω0 coincide, forcing P 0 Ω0 to intersect `0 , a contradiction.
Thus, β = β 0 .
Since the angle will depend only on the length, there must be a function which assigns
to each length an angle:
Π(|P Q|) = β.
MATH 6118-090
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90
CHAPTER 7. INTRODUCTION TO HYPERBOLIC GEOMETRY
The angle β is called the angle of parallelism. Note that we have to have that
0 < Π(|P Q|) < π/2
because we have multiple parallels.
Theorem 7.12 Two hyperparallel lines have a common perpendicular.
Proof: Let `1 and `2 be two hyperparallel lines. Let P ∈ `2 and Q ∈ `1 so that P Q is
perpendicular to `1 ; i.e., Q is the foot of P in `1 . Let Ω and Θ be the points at infinity at
the ends of `1 and `2 on the side where the angle `2 makes with P Q is smaller. Since `2 is
hyperparallel to `1 , the angle ΘP Q > β = Π(P Q). Let R be a point on `1 on the same side
of Q as Ω. As R moves toward Q, ∠P RQ approaches a right angle. By moving R towards
Ω, we can make ∠P RQ as small as we like. This means that
lim ∠P RΩ = 180◦
R→Ω
and
lim ∠RP Ω = 0.
R→Ω
Thus, if R = Q, then ∠ΘP R = ∠ΘP Q, and as R goes to Ω, ∠ΘP R approaches ∠ΘP Ω.
Thus, at some point we must have ∠ΘP R = ∠P RQ. Fix the point R so that this is the
case. Let T be the midpoint of P R, and let S be the point on `1 so that T S is perpendicular
to `1 . Let S 0 be the point on `2 between P and Θ so that |P S 0 | = |RS|. Then by SAS,
4P S 0 T ∼
= 4RST . Therefore, ∠S 0 T P = ∠ST R, and because P T R is a straight line, the
angle ∠S 0 T S must also be 180◦ . What is more, ∠P S 0 T = ∠RST , which is a right angle.
Thus, SS 0 is a common perpendicular to both `1 and `2 .
Lemma 7.13 If alternate interior angles of a transversal ` to `1 and `2 are congruent, then
`1 and `2 are hyperparallel.
7.9
Singly Asymptotic Triangles
Let Ω be an ideal point (point at infinity),
Θ'
and let A and B be two points in the hyperbolic plane. The region bounded by AB,
BΩ, and AΩ is called a singly asymptotic
Ω
triangle or trilateral . The angles of a trilateral are the angles ∠ΩAB and ∠ΩBA,
B
and these are called its interior angles. The
angle at Ω has measure 0. This follows from the following result.
A
Θ
Lemma 7.14 Let Ω be an ideal point and let ∠P QΩ = 90◦ . Given any ² > 0 there exists
a point R ∈ QΩ so that ∠P RQ < ².
Theorem 7.13 The angles in a singly asymptotic triangle sum to less than 180◦ .
MATH 6118-090
Spring 2006
7.9. SINGLY ASYMPTOTIC TRIANGLES
91
Proof Let 4ABΩ be an asymptotic triangle. Construct the line ΘΘ0 through A such that
Θ is on the same side of AB as Ω and ∠Θ0 AB ∼
= ∠ABΩ. Then by Lemma 7.13, ΘΘ0 and
BΩ are hyperparallel. Thus, the ray AΩ lies between the ray AΘ and AB. Therefore,
∠BAΩ < ∠BAΘ and
∠BAΩ + ∠ABΩ = ∠BAΩ + ∠Θ0 AB
< ∠BAΘ + ∠Θ0 AB = 180◦
This is what we needed.
Many of the properties of singly asymptotic triangles are analogous to those of triangles.
Lemma 7.15 (Crossbar Theorem for Singly Asymptotic Triangles) A line which subdivides an angle of a singly asymptotic triangle intersects the opposite side.
Lemma 7.16 (Pasch’s Theorem for Singly Asymptotic Triangles) A line which intersects a side of the singly asymptotic triangle 4ABΩ but does not pass through a vertex
will intersect another side, provided the line is not limiting parallel to either AΩ or BΩ at
Ω.
Lemma 7.17 (Exterior Angle Theorem) If 4P QΩ is a singly asymptotic triangle, the
exterior angles at P and Q are greater than their respective opposite interior angles.
A P
¦
»
:Ω
»»»
¦
»
»»
M¦
»»»
»
¦
»»
»»»
¦
»
Q »»
-D
U ¾ ¦»
¦B
R
−−→
Proof: Choose R ∈ P Q so that G P QR. We need to show
that ∠RQΩ > ∠QP Ω.
←→
−−→
∼
There
←→is a unique ray QD lying on the same side of P Q as Ω so that ∠RQD = ∠QP Ω.
∼
∼
If U ∈ QD is so that G U QD, then ∠U QP = ∠RQD = ∠QP Ω. Thus, we have that the
alternate interior←→
angles are
←→ congruent. Let M be the midpoint of P Q and let A and B be
∼
the feet of M in P Ω and QD, respectively. By AAS 4AP M ∼
so that AM
= 4BQM←→
←→ = BM
∼
and angleAM P = BM Q. Thus,
M , and B are collinear. Thus, P Ω and QD have a
←→ A, ←→
−−→
−→
−−→
common perpendicular. Thus, P Ω )( QD. Thus QD lies between QΩ and QR, from which
it follows that ∠RQΩ > ∠QP Ω.
Two singly asymptotic triangles are congruent if the angles and segment of one are
congruent, respectively, to the angles and segment of the other.
Lemma 7.18 If in the trilateral 4ABΩ and 4CDΘ we have ∠BAΩ ∼
= ∠DCΘ, then
∠ABΩ ∼
= ∠CDΘ if and only if AB ∼
= CD.
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92
CHAPTER 7. INTRODUCTION TO HYPERBOLIC GEOMETRY
A
PP X
P
¡ PP
q
¢ ¡
1Ω
¢ ¡
³
³ ³
¢¡³³³
³
¡
¢ ³
C
¢
B
D
P
¢ PPP
P
q
¢
Θ
³
¢
³
³ 1
¢ ³³³
¢³³
6 ∠CDΘ. We may suppose
Proof: First, let us assume that AB ∼
= CD but ∠ABΩ ∼
=
−−→
−→
−−→
that ∠ABΩ > ∠CDΘ. Then, there is a unique ray BW between BΩ and BA so that
−→
−→
−−→
−→
∠ABW ∼
= ∠CDΘ. Since AΩ and BΩ are limiting parallel, BW must intersect AΩ in a
−−→
point X. There is a point Y ∈ CΘ so that AX ∼
= CY . Then by SAS 4ABX ∼
= 4CDY .
−−→
−−→
∼
Therefore, ∠CDY ∼
∠ABX
∠CDΘ.
Then,
CΘ
and
DΘ
are
not
limiting
parallel, a
=
=
contradiction.
Now assume that ∠ABΩ ∼
6 CD. We may then assume that AB >
= ∠CDΘ and AB ∼
=
−→
CD. There is a unique point P ∈ AB so that BP ∼
= CD. Let P Ω be the limiting parallel
−→
−→
−→
ray to AΩ from P . Thus, P Ω is also parallel to BΩ. The first part of this proof now implies
∠BP Ω ∼
= ∠DCΘ ∼
= ∠BAΩ. But ∠jBP Ω is exterior to 4P AΩ, by the Exterior Angle
Theorem ∠BP Ω > ∠BAΩ, a contradiction.
7.10
Doubly Asymptotic Triangles
Given a line ` with ideal points Ω and Λ, and a point A not on `, we will have a doubly
asymptotic triangle, 4AΩΛ. The rays AΩ and AΛ will be the boundary rays for the angle
of parallelism at A to `.
Let’s change the question. If I am given to rays AΩ and AΛ is it possible to construct
the triangle 4AΩΛ? Note that we do not have the guarantee that the line ΩΛ exists because
the “points” Ω and Λ are not points of the hyperbolic plane, so Postulate 1 does not have
to apply.
Theorem 7.14 Two intersecting rays have a unique common parallel.
A
A'
E
F
Ω
Ω'
a
a'
Λ
Proof: Let OΩ and OΩ0
be the two intersecting rays.
Choose A on OΩ and A0 ∈
OΩ0 so that |OA| = |OA0 |.
Construct the rays A0 Ω and
AΩ0 . Let a and a0 be the angle bisectors of ∠ΩAΩ0 and
∠ΩA0 Ω0 , respectively. Let E
be the point of intersection
of AΩ0 and A0 Ω, and let Λ
be the ideal point in the direction OE. Finally, let A0 Ω
intersect a at F .
We need to show a and a0 are hyperparallel. We will do this by showing that they do
not intersect and are not boundary parallel.
Suppose that a and a0 do intersect at a point D. By the symmetry of the situation this
point D must lie on OΛ. Consider the two triangles 4ADΩ and 4A0 DΩ. Since AD ∼
= A0 D,
MATH 6118-090
Spring 2006
7.10. DOUBLY ASYMPTOTIC TRIANGLES
93
there exists an isometry fixing D and sending A to A0 . Since ∠DAΩ ∼
= ∠DA0 Ω, we know
0
that AΩ is sent to A Ω under this isometry. Thus the two singly asymptotic triangles
are congruent, 4ADΩ ∼
= 4A0 DΩ. This implies that ∠ADΩ ∼
= ∠A0 DΩ, which cannot be
possible. Thus, the point D cannot exist and the two rays do not intersect.
Now, suppose that a and a0 are boundary parallel. This will make them intersect at
infinity at Λ. Now consider the two singly asymptotic triangles 4AF Ω and 4A0 F Λ. By
construction ∠F AΩ ∼
= ∠F A0 Λ — recall that they are both bisectors of the angles ∠ΩAΩ0
0
0
and ∠ΩA Ω . Now ∠A0 F Λ ∼
= ∠AF Ω since they are vertical angles. Thus 4AF Ω and
0
4A F Λ, and hence are congruent. Thus, AF ∼
= A0 F . Thus, F = E which is impossible.
0
Thus, a and a are not boundary parallel.
Since a and a0 do not intersect and are not boundA
A'
ary parallel, they must be
E
hyperparallel. Thus, there is
F
Ω'
a line perpendicular to both Ω
a and a0 . Call these points
a
a'
of intersection C and C 0 , respectively. We want to show
that the line CC 0 is boundΘ'
Θ
C
C'
ary parallel to OΩ and OΩ0 ,
or that CC 0 intersects infinΛ
ity at Ω and Ω0 .
0
Assume that CC intersects infinity at Θ and Θ0 . The triangles 4CAΩ and 4C 0 A0 Ω are similar, since ∠CAΩ ∼
=
∠C 0 A0 Ω and AC ∼
= A0 C 0 . Thus, ∠ΘCΩ ∼
= ∠ΘC 0 Ω. If Θ 6= Ω then 4C 0 CΩ is a singly
asymptotic triangle whose angle sum is more than 180◦ . This contradiction tells us that
Ω = Θ. We show Ω0 = Θ0 similarly.
Theorem 7.15 Two similar doubly asymptotic triangles are congruent. That is, if 4AΩΛ
and 4A0 Ω0 Λ0 are two doubly asymptotic triangles with ∠ΩAΛ ∼
= ∠Ω0 A0 Λ0 , then there is an
isometry which sends 4AΩΛ to 4A0 Ω0 Λ0 .
MATH 6118-090
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94
MATH 6118-090
CHAPTER 7. INTRODUCTION TO HYPERBOLIC GEOMETRY
Spring 2006
Chapter 8
Classification of Parallels
8.1
Fan Angles
A line ` subdivides an angle if it passes through the vertex of the angle and intersects the
interior of the angle.
Let P be an arbitrary point in the hyperbolic plane. The pencil of lines through P ,
denoted P(P ) is the set of all lines coincident with P .
Theorem 8.1 If P 6∈ ` then there exists an angle with vertex at P whose interior contains
` and whose bisector is perpendicular to `.
```
```
Pr
``
`
m
```
```
n
`
Proof: In the pencil P(P ) let p be the perpendicular to ` and let m ∈ P(P ) be perpendicular to p at P . Thus, we know that m ∩ ` = ∅. By the Universal Hyperbolic Theorem
there is another line n ∈ P(P ) which does not intersect `. Since n is distinct from m, n is
not perpendicular to p. Thus, one pair of the vertical angles formed by p and n is a pair of
−→
acute angles. Thus, there exists a ray P A ⊆ n so that
∠QP A◦ = α◦ < 90◦ ,
where Q is the foot of P in `.
−−→
∼
There
←→ is a ray P B on the opposite side of p from A so that ∠QP B = ∠QP A. The
line P B does not intersect `. If it did, then let R be the point of intersection. Choose
R0 on the same side of p as A so that QR ∼
= QR0 . Then by SAS 4P RQ ∼
= 4P R0 Q.
0
0
∼
∼
Thus, ∠QP R = ∠QP R =
←→∠QP A and R ∈ n. This implies that n ∩ ` 6= ∅, which is a
contradiction. Therefore, P B ∩ ` = ∅.
−−→
Clearly, P Q bisects ∠BP A. We only need to show that ` is contained in the interior of
∠BP A, or that every point on ` lies in the interior of ∠BP A.
95
96
CHAPTER 8. CLASSIFICATION OF PARALLELS
Let X ∈ ` be on the same side of p as A, and let Y be the unique point on ` on the
−−→
opposite side of p as X so that QX ∼
= QY . Now, we know that the ray P Q lies in the
interior of ∠AP B and, in fact, using an abuse of notation, ∠BP A = ∠QP B ∪ ∠QP A.
−−→
−−→
Claim: QX lies in the interior of ∠QP A. Likewise, QY lies in the interior←→
of ∠QP B.
To prove this claim, first note that since X is on the same side of p = P Q as A, every
−−→
−−→
point of QX lies on the same side of ←→
p as A. Thus, we are left only to show that if R ∈ QX
then R is on the same side of n = P A as Q. This must be the case, for if not we would
have that n ∩ ` 6= ∅, a contradiction. Thus, the above claim is true.
We then have that
−−→
−−→
` = QY ∪ {Q} ∪ QX,
which from above must lie in the interior of ∠AP B.
Corollary 1 Every line in P(P ) that intersects ` subdivides the angle ∠AP B.
We have almost proven the following result.
Theorem 8.2 If P 6∈ ` then there is a unique angle with the following properties:
i) the angle contains ` in its interior;
ii) the bisector of this angle is perpendicular to `;
iii) every line passing through P that subdivides the angle intersects `.
S
P
@
¾
¡
­
¡­
@
¡­
@
¡­
@
Y¡ ­
@
¡@­
¡ ­
X@
¡
­
@
¡
­
@
@
¡
­
V
U
Q
-`
Proof: To prove that this angle exists, we need to construct the sides of the angle. We
already know that there are angles with vertex at P that contain the line `. We want to
prove much more in this theorem. In some sense we are proving that there is a smallest
such angle.
Let S be a point on the line m←→
which is the perpendicular at P to the perpendicular
through P to `. Consider the line SQ. Let Σ1 be the set of all points T on SQ, such
←→ that
−→
−→
P T intersects `, together with all points on the ray opposite to QS. Let Σ2 = SQ \ Σ1 .
By the Crossbar Theorem if T ∈ SQ and T ∈ Σ1 , then the entire subsegment T Q ⊂ ←→
Σ1 .
Hence, (Σ1 , Σ2 ) is a Dedekind
cut.
By
Dedekind’s
Axiom
there
is
a
unique
point
X
on
SQ
←→
such that for P1 , P2 ∈ SQ, X lies between P1 and P2 if and only if X 6= P1 and X 6= P2 ,
P1 ∈ Σ1 and P2 ∈ Σ2 .
MATH 6118-090
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8.2. LIMITING PARALLEL RAYS
97
−−→
−−→
By the definition of Σ1 and Σ2 rays below P X all intersect ` and rays above P X do
not. Now, we wish to prove that
−−→
P X ∩ ` = ∅.
−−→
Assume on the contrary that P X ∩ ` = {U }. Choose
←→ a point V on ` so that U lies between
V and Q. Since V and U are on the same side of SQ, V and P are on opposite sides, so that
V P meets SQ in a point Y . We can show that X lies between Y and Q. Thus, Y ∈ Σ2 .
−−→
−−→
This is impossible, for if Y ∈ Σ2 , then P Y ∩ ` = ∅. Thus,
P X does not meet `.
←→
−−→
Doing the same construction on the opposite side of P Q at P gives us a similar ray P X 0
with similar properties. We need to show that ∠QP X ∼
not,
= ∠QP X 0 . Assume −
−→0 then we
0
may
assume
that
∠QP
X
<
∠QP
X
.
By
Congruence
Axiom
4
there
is
a
ray
P
R between
−−→0 −−→0
−−→0
−−→
P X and P Q such that ∠QP R0 ∼
the
properties
of
P
X
,
P
R
must
intersect `.
= ∠QP X. By
←→
0
0
∼
Let R be the point on ` on the opposite of P Q from R so that QR = QR . Then by SAS
4QP R0 ∼
= 4QP R. This implies that ∠QP R ∼
= ∠QP R0 ∼
= ∠QP X. By Congruence Axiom
−−→
−−→
4 we have to have that R ∈ P X or P X ∩ ` 6= ∅, a contradiction. Thus, ∠QP X ∼
= ∠QP X 0 .
−−→
Checking the properties, clearly P Q bisects the angle, the angle contains ` and by the
−−→
definition of the ray P X any line that subdivides the angle intersects `.
For P 6∈ ` this angle at P is called the fan angle of P and `, and will be denoted by
∠(P, `).
8.2
Limiting Parallel Rays
←→
−−→
−−→
Consider a ray AB and a point P 6∈ AB. WE need to characterize all rays P X satisfying:
←→
←→
1. P X ∩ AB = ∅,
−−→
−→
−−→
2. every ray between P X and P A intersects AB.
In euclidean geometry, there is only
one such ray. Is this the case in H2 ?
←→
There is a←→
fan angle of P and AB, call it ∠XP Y , and ←→
one side
on the
←→of this angle lies−→
−−→
same side of AP as B. Let us
say
that
it
is
P
X.
Now,
P
X
∩
AB
=
∅.
Also,
if
P
R lies
←→
−−→
−→
−→ ←→
between P X and P A, then P R subdivides ∠XP Y . Thus, P R ∩ AB 6= ∅, but due to its
−→
−−→
−−→
relative position with respect to P A and P X it must, in fact, intersect AB. Thus, the side
of the fan angle satisfies the two aforementioned conditions. Do any others?
−→
−−→
Let P T be different from P X.
←→
←→
←→
−→
−→
If P T is on the line P A, then P T intersects AB and the ray P T does not have property
(1).
←→
−→
−−→
−→
−→
If
P T lies on the same side of P A as Y , let P U be between P A and P T . Then the line
←→
−→
−−→
−−→
−−→ −−→
−→
P A separates P T and P U from AB, which means that P U ∩ AB = ∅, so that P T does not
satisfy property (2).
←→
−→
−→
−−→
If P T lies on the same side of P A as X, then one of P T and P X is between the other and
−→
P A.
←→
−→
−−→
−→
−→
(i ) If P T is between P X and P A then P T must intersect AB by the properties of the fan
−→
angle. Thus, P T would not satisfy property (1).
−−→
−→
−→
−→
−−→ −−→
(ii ) If P X is between P A and P T then P T cannot satisfy property (2), for P X ∩ AB = ∅.
MATH 6118-090
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98
CHAPTER 8. CLASSIFICATION OF PARALLELS
−−→
Thus, P X is the only ray at P with these two properties. This gives rise to the following
definition.
−−→
−−→
Definition 8.1 A ray P Q is parallel to a ray AB if
←→
←→
i) P Q ∩ AB = ∅,
−−→
−→
−−→
ii) every ray between P Q and P A intersects AB.
−−→
−−→
We have just shown that this ray is unique. In this case the rays P Q and AB are called
−−→
−−→
limiting parallel rays and we say that P Q is limiting parallel to AB.
−−→
−−→
−−→
−−→
Question: If P Q is limiting parallel to AB, is AB is limiting parallel to P Q?
We need to following lemma for the proof of the main theorem. The proof is left as an
exercise.
←→
−−→
Lemma 8.1 If ∠P AB is acute, then the foot of P in the line AB lies in the ray AB and
is different from A.
We state the following theorems without proof.
−−→
−−→
−−→
Theorem 8.3 If P Q is limiting parallel to AB, then AB lies in the interior of the angle
∠AP Q.
−−→
Theorem 8.4 Let P 6∈ ` and let B lie between A and C in `. Then P Q is limiting parallel
−−→
−−→
−−→
to AB if and only if P Q is limiting parallel to BC.
−−→
−−→
Theorem 8.5 Let B lie between A and C on `. AB is limiting parallel to DE if and only
−−→
−−→
if BC is limiting parallel to DE.
−−→
−−→
−−→
−−→
Theorem 8.6 If P Q is limiting parallel to AB, then AB is limiting to P Q.
PQ
Q
A Q
M
A Q
©
©
A QQ
©©
A
Q
©
Q
©
A
Q
©
A
Q©©
Q
A
X ©© Q
Q
A ©©
Q
A©
Q
©G
©
Q
H ©
Q
©
Q
A B F
C
DQ
Q
←→
−−→
−−→
Proof: Let F be the foot of P in AB = `. Let C ∈ ` so that F C is like directed to AB.
−−→ −−→
−−→ −−→
−−→
This means that either AB ⊂ F C or F C ⊂ AB. It follows
P Q is
←→from Theorem 8.5
←→that
←→
−−→
limiting parallel to F C, so that C lies on the same side of P F as does Q and F C ∩ P Q = ∅.
−−→
Let F X lie in the interior of ∠P F C.
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8.2. LIMITING PARALLEL RAYS
99
−−→ −−→
Claim: F X ∩ P Q 6= ∅.
−−→
−−→
From this it follows that F C is limiting parallel to P Q. Applying Theorem 8.5 again,
−−→
−−→
we have that AB is limiting parallel to P Q. Thus, we need to establish this claim to finish
the proof.
◦
F C ◦ = 90◦ , ∠P F X is acute. Thus, by Lemma 8.1 the foot of P in
←→Since ∠P F X < ∠P−
−→
F
X must lie in the ray F X. Label this point G. It follows that G lies on the same side of
←→
P F as C. ←→
−−→
−−→ −−→
If G ∈ P Q, then it must lie in P Q.
It follows that F X ∩ P Q 6= ∅, and we are done.
←→
−−→
−−→
If G lies on the opposite side of P Q from F , then again F X intersects P Q and we are
done.
←→
Assume then, as the final case, that G lies on the same side of P Q as F . Then, this
puts G in the interior of ∠QP F . Let α◦ = ∠Q◦ P G. We have that α◦ < ∠QP F ◦ . By
−→
Congruence Axiom 4 there is a unique ray P Z in the interior of ∠QP F so that
∠F P Z ◦ = α◦ = ∠QP G◦ .
−−→
−−→ −→
−−→
Since P Q is limiting parallel to F C, P Z must intersect F C in some point D. From
Lemma 8.1 G 6= F , so that P G < P F by an Exercise (the hypotenuse of a right trian∼
gle is longer
than either leg). Thus,
there is a point
H ∈ P F so that P H←→
= P G.
←→
←→
←→
Let
HY
be
perpendicular
to
P
F
at
H.
Since
AB
is
perpendicular
to
P
F it follows that
←→ ←→
←→
AB∩HY = ∅. Now, HY does intersect one side of the triangle 4P F D. By Pasch’s Theorem
it must intersect a second side. That side must be P D. Let the point of intersection be E.
Now, E 6= P and E 6= D.
−−→
On P Q there is a unique point M so that P M ∼
= P E. Recalling that P H ∼
= PG
∼
and ∠HP E = ∠F P Z ∼
that
4P
HE
4P
GM
by
SAS.
Thus,
= ∠QP G, we then have
=
←→
←→
∼
∠P
is a right angle. Thus, GM = F X since both are perpendicular
to
←→GM = ∠P HE and ←→
←→
−−→
P G at G. Therefore, F X ∩ P Q = {M }. Since M lies on the same side of F C as Q, it
−−→
−−→ −−→
follows, finally, that M ∈ F X and F X ∩ P Q 6= ∅.
−
→
→
−
A ray X is parallel to a line ` if R is a limiting parallel ray to some ray in `. This
makes the line k horoparallel to a line ` if some ray in k is a limiting parallel ray to some
ray in `. We have just proven that these parallelisms are symmetric and we may denote
them by
−
→
−
→
X k`
`kX
`kk
k k `.
←→
←→
−−→ −−→
−−→
If k = AB, ` = CD, and AB k CD, then AB is said to be parallel to ` in the direction
−−→
−−→
of CD on `. Furthermore, k and ` are said to be parallel in the direction of AB on k and
−−→
in the direction of CD on `.
Theorem 8.7 IF P 6∈ ` then there are exactly two lines through P that are limiting parallel
to `. Each contains an arm of the fan angle ∠(P, `) and they are limiting parallel to ` in
opposite directions.
As we have mentioned several times already, there is no simple transitivity of parallelism
in hyperbolic geometry. There is a weak form of transitivity.
Theorem 8.8 (Weak Transitivity of Parallels) Two lines parallel to a third in the
same direction on the third are parallel to each other.
MATH 6118-090
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100
8.3
CHAPTER 8. CLASSIFICATION OF PARALLELS
Hyperparallel Lines
Let k and ` be lines in H2 and let An , n = 1, 2, . . . be points on k. Let A0n be the foot of An
in `. We say that the points {An } are equidistant from ` if Ai A0i ∼
= Aj A0j for all choices of
i and j.
Theorem 8.9 If ` and `0 are non-intersecting lines in H2 , then any set of points on `
equidistant from `0 has at most two elements.
Proof: Assume not; i.e., assume that A, B, C ∈ ` are equidistant from `0 . Thus, the
quadrilaterals 2AA0 B 0 B, 2AA0 C 0 C, and 2BB 0 C 0 C are all Saccheri quadrilaterals. Thus,
∠A0 AB ∼
= ∠B 0 BA, ∠A0 AC ∼
= ∠C 0 CA, and ∠B 0 BC ∼
= C 0 CB. Stringing these together, we
have
∠B 0 BA ∼
= ∠A0 AB ∼
= ∠C 0 CA ∼
= ∠B 0 BC.
Since ∠B 0 BA and ∠B 0 BC are supplementary, they are right angles. Thus all three quadrilaterals are rectangles, a contradiction.
This theorem states that at most two points at a time on ` can be equidistant from
It does not put a limit on the number of pairs of such points, though. We may have
pairs (A, B) and (C, D) so that AA0 ∼
= BB 0 and CC 0 ∼
= DD0 , but it is not possible that
AA0 ∼
= CC 0 or any of the other possibilities. It may also occur that the number of pairs of
such points will be zero. There is no guarantee that such a pair will exist. This will help
us distinguish types of parallel lines.
`0 .
Theorem 8.10 Let ` and `0 be non-intersecting lines for which there is a pair of points
A, B ∈ ` which are equidistant from `0 . Then, ` and `0 have a common perpendicular
segment which is the shortest segment between ` and `0 .
Proof: 2AA0 B 0 B is a Saccheri quadrilateral. Its altitude is perpendicular to both ` and
`0 by Theorem 7.9. Also, the quadrilateral formed by this altitude and any other segment
from a point X ∈ ` and its foot X 0 ∈ `0 is a Lambert quadrilateral. By Theorem 7.11 the
altitude is less than XX 0 and hence is the shortest segment between ` and `0 .
Theorem 8.11 In H2 if lines ` and `0 have a common perpendicular segment M M 0 , then
they are non-intersecting and M M 0 is unique. Moreover, if A, B ∈ ` so that M is the
midpoint of AB then A and B are equidistant from `0 .
Proof: We already know that the lines do not intersect from Corollary 1 to the Alternate
Interior Angles Theorem. If P P 0 were another common perpendicular then 2M M 0 P 0 P
would be a rectangle, which cannot exist. Thus, M M 0 is unique.
A'
A
MATH 6118-090
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M
B'
B
Spring 2006
8.4. CLASSIFICATION OF PARALLELS
101
Let A, B ∈ ` be so that M is the midpoint of AB. We have that AM ∼
= BM
and ∠AM M 0 ∼
= ∠BM M 0 , so by SAS 4AM M 0 ∼
= 4BM M 0 . Thus, AM 0 ∼
= BM 0 and
∠AM 0 M ∼
= ∠BM 0 M . By Angle Subtraction ∠A0 M 0 A ∼
= ∠B 0 M 0 B. Then by AAS 4AA0 M 0 ∼
=
4BB 0 M 0 and AA0 ∼
= BB 0 .
Lemma 8.2 Let M M 0 be the common perpendicular to ` and `0 . Let A, B ∈ ` so that A
lies between M and B, then AA0 < BB 0 .
If k and ` are non-intersecting lines which admit a common perpendicular then they are
said to be hyperparallel. This is denoted by k )( `. They are sometimes called divergently
parallel lines.
Recall that for P 6∈ ` there is a unique fan angle ∠(P, `) for P and `. Let F be the foot of
P in ` and let X and Y be on opposite arms of the fan angle. The angle ∠F P X ∼
= ∠F P Y
is called the angle of parallelism at P with respect to ` and its angle measure is denoted
by Π(P F )◦ . We know that its measure must be less than 90◦ .
Theorem 8.12 If ` is any line in H2 and 0 < α < 90 then there is a point P 6∈ ` so that
Π(P F )◦ = α◦ .
8.4
Classification of Parallels
We have defined two types of parallel lines—hyperparallel and horoparallel. Are these
the only types of parallel lines in hyperbolic geometry? Are there lines that are both
hyperparallel and horoparallel? These are the questions that we will answer in this section.
−−→
Theorem 8.13 Let P X be the limiting parallel ray to ` through P and let Q and X 0 be the
feet of P and X, respectively, in `. Then P Q > XX 0 .
PX
XX
Q
XXXX
X0
XXX
X
z
`
−−→
Proof: If X lies between P and Y then XY k ` by Theorem 8.5. Thus, ∠X 0 XY is the
angle of parallelism for X and `. Thus, it is acute. Therefore, ∠X 0 XP is obtuse. This
implies that ∠QP X < ∠X 0 XP and therefore P Q > XX 0 .
This shows that the distance from X to ` decreases as the distance from P to X increases along the limiting parallel ray. In fact, one can show that the distance from X to
` approaches 0. Note that we have seen that for hyperparallel lines the further one gets
from the common perpendicular, the greater the distance between the lines. This gives one
the impression that these types of parallel lines should be distinct.
Corollary 1 If ` and k are horoparallel in a given direction, then ` and k have no common
perpendicular.
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CHAPTER 8. CLASSIFICATION OF PARALLELS
−−→
Proof: Let P ∈ ` so that P X k k and assume that there is a point M ∈ ` so that
(i) M lies between P and X, and
(ii) M M 0 is a common perpendicular to ` and k.
From Theorem 8.13 P Q > M M 0 > XX 0 . By Theorem 8.10 M M 0 < XX 0 , a contradiction.
Thus, ` and k have no common perpendicular.
This tells us that if two lines are limiting parallel, then they are not hyperparallel. We
now need to show that if two lines are non-intersecting and are not limiting parallel, then
they must be hyperparallel. Having done this we will then know that there are only two
types of parallel lines.
Theorem 8.14 Given lines m and n so that
i) m ∩ n = ∅, and
ii) m does not contain a limiting parallel ray to n in either direction.
Then m is hyperparallel to n, m )( n.
A
B
H
E
A'
K
B'
H'
K'
Proof: We need to find two points H, K ∈ n that are equidistant from m; i.e., HH 0 ∼
=
KK 0 . If this is true, then by Theorem 7.9 the perpendicular bisector of HK is the common
perpendicular to m and n. By Theorem 8.11 this common perpendicular is unique. It
follows then that m )( n.
(i) The search for H and K.
Let A, B ∈ n with feet A0 , B 0 ∈ m. If AA0 ∼
= BB 0 we are done. Assume that these
segments are not congruent. Then, we may assume that AA0 > BB 0 , since one must be
greater than the other. There is a point E ∈ AA0 ←→
so that A0 E ∼
= BB 0 .
−−→
0
There is a unique ray EF on the same side of AA as B so that ∠A0 EF ∼
= ∠B 0 BG with
B between A and G.
−−→ −→
Claim: EF ∩ AG 6= ∅. (Proof will follow.)
−−→
Call this point of intersection H. The there is a unique point K ∈ BG so that BK ∼
=
0
0
0
0
0
∼
∼
EH. Then, by SAS 4EHA = 4BKB . Thus, A H = B K. We also have that ∠EA H ∼
=
∠BB 0 K, so that by Angle Subtraction it follows that ∠HA0 H 0 ∼
= ∠KB 0 K 0 . Since HH 0 and
MATH 6118-090
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8.5. PROOF OF CLAIM
103
KK 0 are perpendicular to m, we have by Hypotenuse-Angle that 4HA0 H 0 ∼
= 4KB 0 K 0 .
0
0
∼
Therefore, HH = KK , and H and K are equidistant from m.
Thus, we are finished once we have proved the claim.
It turns out that the claim is the hardest item to prove in the whole theorem. It takes
us back to Section 7.9.
8.5
Proof of Claim
−−→ −→
We are now
in a position
to show that −
EF
6= ∅.
−−
→
−
−
→
−→ ∩ AG
−
−
→
−
→
−−→
Let A0 M k EF , A0 N k AG, and B 0 P k BG.
Recall that EA0 ∼
= BB 0 and ∠A0 EF ∼
= ∠B 0 BG.
F M
0
0
∼
A
Now ∠EA M = ∠BB P . Choose a point L ∈ m so
G,N,P
B
that B 0 lies between A0 and L.
H
E
K
Since m and
n were
not limiting
parallel
−−→
−−0→
−−→
−−0→ lines,
0
0
we have that B L 6= B P and A L 6= A N . Now,
0
0
∼
∠M
by Angle−−
Subtraction.
Since
= ∠P
−−0→A L−→
−−0→B L −−
→
→
−→
0P k −
A N k AG,−−A
N
k
BG,
and
B
BG,
we
must
→ −−→
have that A0 N k B 0 P . Thus, ∠P B 0 L is exterior
to the triangle 4A0 B 0 Ω. −Therefore,
∠N A0 L
<
−0→
−−
→
A'
B'
H'
K'
0
0M
∠P B−−L.
This
means
that
A
N
lies
between
A
→
−
−
→
−
−
→
0 L, which implies that A0 M lies between A0 N
and A
−−0→
−−→ −→
and A A. But since A0 N k AG, we have that
−−
→ −→
A0 M ∩ AG = {J}.
←→
−−
→
Since J ∈
A0 M , it lies on the
A0 . Thus, A and J are on opposite
←→
←→ same side of EF as does−
−→
sides
←→ of EF . Thus, AJ ∩ EF = {H} 6= ∅. But, H ∈ EF since H lies on the same side of
AA0 as does J.
It is possible to construct the limiting parallel ray to a line through a given point with a
compass
←→ and straightedge. Let P 6∈ ` and let Q be the foot of P in `. Let m be perpendicular
to P Q at P so that m )( `. Let R ∈ `, R 6= Q and let S be the foot of R in m. Then
2P QRS is a Lambert quadrilateral and P S < QR. QR < P R since P R is a hypotenuse.
Take a circle centered at P of radius QR. S is inside this circle and R is outside this circle.
There is a unique point X ∈ RS where RS intersects the circle. One can show that the
−−→
ray P X is limiting parallel to `!
A
B
H
E
A'
K
B'
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CHAPTER 8. CLASSIFICATION OF PARALLELS
Spring 2006
Chapter 9
Other Geometries
9.1
The Idea of Parallelism
We have agreed that we would work with a reasonable set of axioms for our geometry. We
require that our sets of axioms be consistent, independent and complete.
Definition 9.1 A set of axioms is said to be consistent if neither the axioms nor the propositions of the system contradict one another.
Definition 9.2 A set of axioms is said to be independent if none of the axioms can be
derived from any of the other axioms.
Definition 9.3 A set of axioms is said to be complete if it is not possible to add a new
independent axiom to the system.
We have looked at Euclid’s axioms and have commented on how the first four differ
from the Fifth Axiom in that they are direct, concise, and easy to read. The efforts of
mathematicians since Euclid’s time to show that the Fifth Axiom is dependent on the first
four have all met with failure. Geometer’s have offered ”proofs” of the Fifth Axiom, but
each has been found to be flawed. This is one of the times that failure has been helpful, in
that they all turn out to be equivalent statements to Euclid’s Fifth Axiom.
We are able to show that each of the following statements is logically equivalent to
Euclid’s Fifth Axiom.
1. (Playfair’s Postulate) Through a point not on a given line, exactly one parallel may
be drawn to the given line.
2. The sum of the angles in a triangle is equal to two right angles.
3. There exists a pair of similar triangles that are not congruent.
4. There exists a pair of lines everywhere equidistant from one another.
5. If three angles of a quadrilateral are right angles, then the fourth angle is also a right
angle.
6. If a line intersects one of two parallel lines, it will intersect the other.
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CHAPTER 9. OTHER GEOMETRIES
7. Lines parallel to the same line are parallel to one another.
8. Two lines which intersect one another cannot both be parallel to the same line.
Now, we would need to show that each of these is equivalent to Euclid’s Fifth Axiom by
showing that each implies that Euclid’s Fifth Axiom holds and that Euclid’s Fifth Axiom
implies that each of these is true. You are allowed to use any of the Propositions that Euclid
proved without using his Fifth Axiom.
9.2
Saccheri’s Work
Recall that Saccheri introduced a certain family of quadrilaterals. Look again at Section 7.6
to remind yourself of the properties of these quadrilaterals. Saccheri studied the three
different possibilities for the summit angles of these quadrilaterals.
Hypothesis of the Acute Angle (HAA)
The summit angles are acute
Hypothesis of the Right Angle (HRA)
The summit angles are right angles
Hypothesis of the Obtuse Angle (HOA) The summit angles are obtuse
Saccheri intended to show that the first and last could not happen, hence he would have
found a proof for Euclid’s Fifth Axiom. He was able to show that the Hypothesis of the
Obtuse Angle led to a contradiction. This result is now know as the Saccheri-Legendre
Theorem (Theorem 7.6). He was unable to arrive at a contradiction when he looked at
the Hypothesis of the Acute Angle. He gave up rather than accept that there was another
geometry available to study. It has been said that he wrote that the Hypothesis of the
Acute Angle must be false “because God wants it that way.”
9.3
Poincaré’s Disk Model for Hyperbolic Geometry
When we adopt the Hyperbolic Axiomthen there are certain ramifications:
1. The sum of the angles in a triangle is less than two right angles.
2. All similar triangles that are congruent, i.e. AAA is a congruence criterion.
3. There are no lines everywhere equidistant from one another.
4. If three angles of a quadrilateral are right angles, then the fourth angle is less than a
right angle.
5. If a line intersects one of two parallel lines, it may not intersect the other.
6. Lines parallel to the same line need not be parallel to one another.
7. Two lines which intersect one another may both be parallel to the same line.
How can we visualize this? Surely it cannot be by just looking at the Euclidean plane in
a slightly different way. We need a model with which we could study the hyperbolic plane.
If it is to be a Euclidean object that we use to study the hyperbolic plane, H2 , then we
must have to make some major changes in our concept of point, line, and/or distance.
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107
We need a model to see what H2 looks like. We know that it will not be too easy, but
we do not want some extremely difficult model to construct. We will work with a small
subset of the plane, but give it a different way of measuring distance.
There are three traditional models for H2 . They are known as the Klein model, the
Poincaré Disk model, and the Poincaré Half-Plane model. We will start with the Disk model
and move to the Half-Plane model later. There are geometric “isomorphisms” between these
models, it is just that some properties are easier to see in one model than the other. The
two Poincaré models tend to give us the opportunity to do computations more easily.
In order to give a model for H2 , we need to determine the set of points, then determine
what lines are and how to measure distance. For Poincaré’s Disk Model we take the set of
points that lie inside the unit circle, i.e., the set
H2 = {(x, y) | x2 + y 2 < 1}.
Note that points on the circle itself are not in the hyperbolic plane. However they do play
an important part in determining our model. Euclidean points on the circle itself are called
ideal points, omega points, vanishing points, or points at infinity.
A unit circle is any circle in the Euclidean plane is a circle with radius one.
Definition 9.4 Given a unit circle Γ in the Euclidean plane, points of the hyperbolic plane
are the points in the interior of Γ. Points on this unit circle are called omega points (Ω) of
the hyperbolic plane.
If we take Γ to be the unit circle centered at the origin, then we would think of the
hyperbolic plane as H2 = {(x, y) | x2 + y 2 < 1} and the omega points are the points
Ω = {(x, y) | x2 +y 2 = 1}. The points in the Euclidean plane satisfying {(x, y) | x2 +y 2 > 1}
are called ultraideal points.
We now have what our points will be. We see that we are going to have to modify our
concept of line in order to have the Hyperbolic Axiom to hold.
Definition 9.5 Given a unit circle Γ in the Euclidean plane, lines of the hyperbolic plane
are arcs of circles drawn orthogonal1 to Γ and located in the interior of Γ.
9.3.1
Construction of Lines
This sounds nice, but how do you draw them?
−→
1. Start with a circle Γ centered at O and consider the ray OA, where A lies on the
circle, Γ.
−→
2. Construct the line perpendicular to OA at A.
3. Choose a point P on this perpendicular line for the center of the second circle and
make P A the radius of a circle centered at P .
4. Let B denote the second point of intersection with circle Γ. Then the arc AB represents a line in this model.
1
Circles are orthogonal to one another when their radii at the points of intersection are perpendicular.
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CHAPTER 9. OTHER GEOMETRIES
Now, how do you construct these lines in more general
circumstances? There are three cases we need to consider.
Γ
Case I :A, B ∈ Γ
Case II : A ∈ Γ and B lies inside Γ
Case III : A and B both lie inside Γ.
P
−→
−−→
Case I : Construct rays P A and P B where P is the center
Α
−→
of the circle Γ. Construct the lines perpendicular to P A
−−→
and P B at A and B respectively. Let Q be the point of
Β
intersection of those two lines. The circle Ω centered at Q
with radius QA intersects Γ at A and B. The line between
Q
A and B is the arc of Ω that lies inside Γ.
Note that this arc is clearly orthogonal to Γ by its conFigure 9.1: Poincaré line
struction.
−→
−−→
Case II : Construct rays P A and P B where P is the center
−→
of the circle Γ. Construct the line perpendicular to P A at A. Draw segment AB and
construct its perpendicular bisector. Let Q be the point of intersection of this line and the
tangent line to Γ at A. The circle Ω centered at Q with radius QA contains A and B. The
line containing A and B is the arc of Ω that lies inside Γ.
This arc, as constructed is orthogonal to Γ at A. We want to see that it is orthogonal at
the other point of intersection with the circle. Let that point of intersection be X. Then,
X ∈ Γ means that P A ∼
= P X. Since X lies on our second circle it follows that QX ∼
= QA.
∼
∼
Since P Q = P Q, we have that 4P AQ = 4P XQ, which means that ∠P XQ is a right
A
P
Figure 9.2: Poincaré lines through A
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109
angle, as we wanted to show.
−→
−→
Case III : Construct the ray P A and then construct the line perpendicular to P A at A. This
intersects Γ in points X and Y. Construct the tangents to Γ at X and at Y . These tangent
lines intersect at a point C. The circle Ω centered at Q is the circle passing through A, B,
and C. The line containing A and B is the arc of Ω that lies inside Γ.
From our construction, we have that
X
4P XC ∼ 4P AX and it follows that
C
2
2
|P A||P C| = |P X| = r . Now, Q lies on
T
the perpendicular bisectors of AC and AB
as Ω is the circumcircle for 4ABC. There
A
is a point T on the circle Ω so that the tanP
G1
gent line to Ω at T passes through P .
Q
G2
Construct the line through P and Q
B
which intersects the circle in two points G1
and G2 so that G1 lies between P and Q.
Now,
|P T |2 = |P Q|2 − |QT |2
Figure 9.3: Poincaré line in Case III
= (|P Q| − |QT |) (|P Q| + |QT |)
= (|P Q| − |QG1 |) (|P Q| + |QG2 |)
= |P G1 ||P G2 | which by Theorem 5.3,
= |P A||P C| = r2
Therefore, T lies on the circle Γ and Γ and Ω are orthogonal at that point. A similar
argument shows that they are orthogonal at the other point of intersection.
9.3.2
Distance
Now, this area inside the unit circle must represent the infinite hyperbolic plane. This
means that our standard distance formula will not work. We introduce a distance metric
by
2dr
dρ =
1 − r2
where ρ represents the hyperbolic distance and r is the Euclidean distance from the center
of the circle. Note that dρ → ∞ as r → 1. This means that lines are going to have infinite
extent.
The relationship between the Euclidean distance of a point from the center of the circle
and the hyperbolic distance is:
Z
ρ=
0
r
2du
= ln
1 − u2
µ
1+r
1−r
¶
= 2 tanh−1 r,
ρ
or r = tanh .
2
Now, for those of you who don’t remember ever having seen this function tanh(x), we
give a little review. The hyperbolic trigonometric functions cosh(x) and sinh(x) are defined
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CHAPTER 9. OTHER GEOMETRIES
by:
ex − e−x
2
ex + e−x
cosh(x) =
2
sinh(x) =
and
tanh(x) =
sinh(x)
e2x − 1
ex − e−x
=
= x
.
cosh(x)
e + e−x
e2x + 1
We will study these in more depth later.
Now, we can use this to define the distance between two points on a Poincaré line. Given
two hyperbolic points A and B, let the Poincaré line intersect the circle in the omega points
P and Q. Define
AP · BQ
AP/AQ
=
,
(AB, P Q) =
BP/BQ
AQ · BP
to be the cross ratio of A and B with respect to P and Q, where AP denotes the the
Euclidean arclength. Define the hyperbolic distance from A to B to be
d(A, B) = log |AB, P Q|.
We will prove the following later.
Theorem 9.1 If a point A in the interior of Γ is located at a Euclidean distance r < 1
from the center O, its hyperbolic distance from the center is given by
d(A, O) = log
1+r
.
1−r
Lemma 9.1 The hyperbolic distance from any point in the interior of Γ to the circle itself
is infinite.
9.3.3
Parallel Lines
←→
It is easy to
see that the Hyperbolic Axiom works in this model. Given a line AB ←→
and a
←→
point D ∈
/ AB, then we can draw at least two lines through D that do not intersect AB.
Call these two lines through D lines `1 and `2 . Notice←→
now how two of←→
our previous
results do not hold, as we remarked earlier. We have that AB and `1 and AB and `2 are
parallel, but `1 and `2 are not parallel. Note also that `2←→
intersects one of a pair of parallel
lines (`1 ), but does not intersect the other parallel line (AB).
As we now know, the hyperbolic plane has two types of parallel lines. The definition
that we will give here
←→will depend explicitly on the model that we have chosen. Consider
the hyperbolic
line
AB
which intersects the circle Σ in the ideal points Λ and Ω. Take a
←→
point←→
D∈
/ AB. Construct the line through Λ and D. Since this line does not intersect the
line AB inside the circle, these two hyperbolic lines are parallel. However, they seem to be
approaching
one another as we go ”to infinity”. Since
there
_
_
_ are two ”ends” of the Poincaré
line AB, there are two of these lines. The line AB and DΛ are horoparallel. The defining
property is as follows.
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111
Γ
D
Α
Β
Figure 9.4: Multiple parallels through A
_
_
Definition 9.6 Let P ∈ AB. Consider the collection of lines DP
_ as P goes to Ω or Λ. The
first line_through D in this collection that does not intersect AB in H2 is the horoparallel
line to AB in that direction.
_
Drop a perpendicular from D to AB and label this point of intersection M . Angles
∠ΛDM and ∠ΩDM are called angles of parallelism.
Theorem 9.2 The angles of parallelism associated with a given line and point are congruent.
Proof: Assume not, i.e., assume ∠ΛDM 6= ∠ΩDM .
Then one angle is greater than the other. Without loss
of generality, we may assume that ∠ΛDM < ∠ΩDM .
Then there is a point E in the interior
_ of ∠ΩDM such
that
_ ∠ΛDM
_ = ∠EDM . The line ED must_intersect
AB since DΩ is the limiting parallel line to AB in that
Α
direction.
Let
the
point
of
intersection
be
F
.
Choose
Β
_
_
G on AB on the opposite side of DM from F so that
F M = GM . Then 4GM D ∼
= 4F M D. This implies
that
_ ∠GDM =_∠F DM = ∠ΛDM . This means that
DΩ intersects
AB at G. This contradicts
the condition
_
_
Limiting Parallel
that DΩ is limiting parallel to AB. Thus, the angles of Figure 9.5:
Poincaré Lines
parallelism are congruent.
Γ
D
M
Ω
Λ
Theorem 9.3 The angles of parallelism associated
with a given line and point are acute.
Proof: Assume not, i.e., assume that ∠M DΩ_> 90◦ . Then
there is a point E interior to
_
◦
∠M DΩ so that ∠M DE = 90 ._ Then, since DE and _
AB are perpendicular to the same
line, they are parallel. Thus, DE does not intersect AB which contradicts the condition
that DΩ is the limiting parallel line.
If the angle of parallelism is 90◦ then we can show that we have Euclidean geometry.
Thus, in H2 the angle of parallelism is acute.
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CHAPTER 9. OTHER GEOMETRIES
Theorem 9.4 (Lobachevskii’s
Theorem) Given a point P at a hyperbolic distance ρ
_
from a hyperbolic line AB (i.e., d(P, M ) = ρ), the angle of parallelism, θ, associated with
the line and the point satisfies
µ ¶
θ
e−ρ = tan
.
2
Note then that
lim θ =
ρ→0
π
and lim θ = 0.
ρ→∞
2
Proof: The proof of this is interesting in that we play one geometry against the other in
order to arrive at our conclusion.
B
B
P
Γ
R
P
Γ
R
Q
A
A
Figure 9.6:
Theorem
Lobachevskii’s
Figure 9.7: After the first translation
_
We are given a line AB_and a point P not on the line. Construct the line through P
which is perpendicular to AB. Call the point of intersection R as in Figure 9.6. Then we
have that ρ = d(P, R). We can translate P to the center of the unit circle and translate
our line to a line so that our line perpendicular to AB is a radius of Γ as we have done in
Figure 9.7. Construct the radii from P to the ideal points A and B and construct the lines
−→
tangent to Γ at these points. These tangent lines intersect at a point Q which lies on P R.
Now, since we have moved our problem to the center of the circle, we can use our previous
result to see that if r is the Euclidean distance from P to R, then we have
ρ = log
1+r
,
1−r
or rewriting this we have
eρ =
1−r
1+r
or e−ρ =
.
1−r
1+r
Now, we are talking about Euclidean distances (with r) and using our Euclidean right
triangles with radius 1 we have that:
r = QP − QR = QP − QA = sec ∠QP A − tan ∠QP A = sec θ − tan θ =
MATH 6118-090
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.
cos θ
Spring 2006
9.3. POINCARÉ’S DISK MODEL FOR HYPERBOLIC GEOMETRY
113
Now, algebra leads us to:
1−r
1+r
cos θ + sin θ − 1
=
cos θ − sin θ + 1
cos θ + sin θ − 1 cos θ + sin θ + 1
=
cos θ − sin θ + 1 cos θ + sin θ + 1
cos2 θ + 2 cos θ sin θ + sin2 θ − 1
=
cos2 θ + 2 cos θ − sin2 θ + 1
2 sin θ cos θ
sin θ
=
=
2
2 cos θ + 2 cos θ
1 + cos θ
¡ ¢
¡ ¢
2 sin 2θ cos 2θ
¡ ¢
¢
=¡
2 cos2 2θ − 1 + 1
µ ¶
θ
= tan
2
e−ρ =
9.3.4
Hyperbolic Circles
Now, if we have a center of a circle that is not at the center P of the unit circle Σ, we know
that the hyperbolic distance in one direction looks skewed with respect to the Euclidean
distance. That would lead us to expect that a circle in this model might take on an elliptic
or oval shape. We will prove later that this is not the case. In fact, hyperbolic circles
embedded in Euclidean space retain their circular appearance — their centers are offset!
Theorem 9.5 Given a hyperbolic circle with radius R, the circumference C of the circle is
given by C = 2π sinh(R).
9.3.5
Similarities with Euclidean Geometry
Because we only changed the Fifth Axiom and not the others, everything that holds in
Neutral Geometry (or geometry without a parallel postulate) also holds in the Hyperbolic
Plane. Other things may also hold, but may require a different proof!! Examples of theorems
that are still true are:
Theorem 9.6 Supplements of congruent angles are congruent.
Theorem 9.7 Vertical angles are congruent to each other.
Theorem 9.8 An angle congruent to a right angle is a right angle.
Theorem 9.9 For every line ` and every point P there exists a line through P perpendicular
to `.
Theorem 9.10 (ASA) Given 4ABC and 4DEF with ∠A ∼
= ∠D, ∠C ∼
= ∠F , and AC ∼
=
∼
DF . Then 4ABC = 4DEF .
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CHAPTER 9. OTHER GEOMETRIES
Theorem 9.11 (SSS) Given triangles 4ABC and 4DEF . If AB ∼
= DE, AC ∼
= DF ,
∼
∼
and BC = EF , then 4ABC = 4DEF .
Theorem 9.12 (Alternate Interior Angles Theorem) If two lines cut by a transversal have a pair of congruent alternate interior angles, then the two lines are non-intersecting.
Theorem 9.13 If m and n are distinct lines both perpendicular to the line `, then m and
n are non-intersecting.
Theorem 9.14 If P is a point not on `, then the perpendicular dropped from P to ` is
unique.
Theorem 9.15 If ` is any line and P is any point not on `, there exists at least one line
m through P which does not intersect `.
Theorem 9.16 (Exterior Angle Theorem) An exterior angle of a triangle is greater
than either remote interior angle.
Theorem 9.17 (SAA Congruence) In triangles 4ABC and 4DEF given that AC ∼
=
DF , ∠A ∼
= ∠D, and ∠B ∼
= ∠E , then 4ABC ∼
= 4DEF .
Theorem 9.18 Two right triangles are congruent if the hypotenuse and a leg of one are
congruent respectively to the hypotenuse and a leg of the other.
Theorem 9.19 Every segment has a unique midpoint.
Theorem 9.20 Every angle has a unique bisector.
Theorem 9.21 Every segment has a unique perpendicular bisector.
Theorem 9.22 In a triangle the greater angle lies opposite the greater side and the greater
side lies opposite the greater angle; i.e., AB > BC if and only if ∠C > ∠A.
Theorem 9.23 Given 4ABC and 4A0 B 0 C 0 , if AB ∼
= A0 B 0 and BC ∼
= B 0 C 0 , then ∠B <
0
0
0
∠B if and only if AC < A C .
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Chapter 10
Area in Hyperbolic Geometry
10.1
Preliminaries
A polygonal region is a plane figure which can be expressed as the union of a finite number
of triangular regions, in such a way that if two of the triangular regions intersect, their
intersection is an edge or a vertex of each of them.
Let R be a polygonal region. A triangulation of R is a finite collection,
K = {T1 , T2 , . . . , Tn }
of triangular regions Ti , such that
1. the Ti ’s intersect only at edges and vertices, and
2. their union is R.
Let R1 and R2 be polygonal regions. Suppose that they have triangulations
K1 = {T1 , T2 , . . . , Tn },
K2 = {T10 , T20 , . . . , Tn0 },
such that for each i we have Ti ∼
= Ti0 . Then we say that R1 and R2 are equivalent by finite
decomposition, and we write R1 ≡ R2 .
Let δT = defect (4(T )) for any triangular region T .
Theorem 10.1 If K1 and K2 are triangulations of the same polygonal region R, then
δK1 = δK2 .
Theorem 10.2 If two Saccheri quadrilaterals have the congruent summits and equal defects, then their summit angles are congruent, in which case, the two Saccheri quadrilaterals
are congruent.
Given 4ABC, with BC considered as the base. Let D and E be the midpoints of AB
and AC;
let F , G, and H be the feet of the perpendiculars from B, A, and C, respectively,
←→
to ` = DE. As you have proven in the homework, 2HCBF is a Saccheri quadrilateral. It
is known as the quadrilateral associated with 4ABC. It depends on the choice of the base,
but it should be clear which base we mean.
115
116
CHAPTER 10. AREA IN HYPERBOLIC GEOMETRY
A
¢QQ
¢
Q
Q
¢
¢
¾
Q
Q
Q
¢
F
D¢
¢
¢
G
Q
QE
Q
Q
HQ
Q
¢
¢
Q
Q
Q
¢
Q
Q
¢
B
C
Figure 10.1: The Saccheri quadrilateral associated with 4ABC
Theorem 10.3 Every triangular region is equivalent by finite decomposition to its associated quadrilateral region.
Theorem 10.4 Every triangular region has the same defect as its associated quadrilateral
region.
Theorem 10.5 If 4ABC and 4DEF have the same defect and a pair of congruent sides,
then the two triangular regions are equivalent by finite decomposition.
Theorem 10.6 (Bolyai’s Theorem in the Hyperbolic Plane) If T1 and T2 are triangular regions, and δT1 = δT2 , then T1 ≡ T2 .
Theorem 10.7 If δ4ABC > δ4DEF , then there is a point P between A and C such that
δ4ABP = δ4DEF .
10.2
Requirements for an Area Function
Now, how do we define the area of a polygonal region in the hyperbolic plane? Should we
define the area in the same way that we do in the Euclidean plane? If so, what are the
minimum requirements for an area function of the Euclidean plane? Minimally it should
satisfy the following: Let R be the set of all polygonal regions in H2 . If an area function is
to be exactly like our Euclidean area function, then we need a function
α :R→R
such that
1. αR > 0 for every R;
2. if R1 and R2 intersect only in edges and vertices, then
α(R1 ∪ R2 ) = αR1 + αR2 ;
3. if T1 and T2 are triangular regions with the same base and altitude, then αT1 = αT2 .
If there is such a function α, then by (2) and (3) it will satisfy
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10.2. REQUIREMENTS FOR AN AREA FUNCTION
117
4. if R1 ≡ R2 , then αR1 = αR2 , because congruent triangles have the same bases and
altitudes.
In Euclidean geometry we can show that this area function is unique and it must satisfy
the formula αT = 12 bh for each triangular region, T , where b is the length of the base and h
is the length of the altitude. Look at Elementary Geometry from an Advanced Standpoint
by E.E. Moise for a proof of this result.
Theorem 10.8 In the hyperbolic plane there is no such function α : R → R satisfying
(1), (2), (3), and, hence, (4).
6
XP
P
AH
XX
@ HH
XX
PP
XXX
PP
@ HH
P XXX
HHPPP
@
XX
1
@
HH PPPXXXX
XX
P
@
T1
T2 HH T3 PPPT4 XXXX
@
PP
HH
XXX
PP
XXX
@
H
@
P
¾
H
1
1
1
1
P0
P1
P2
P3
P4
?
-
Proof: Consider the right angle ∠AP0 P1 , with AP0 = P0 P1 = 1. For each n, let Pn be
−−−→
the point of P0 P1 such that P0 Pn = n. This gives a sequence of triangles
4AP0 P1 , 4AP1 P2 , . . . ,
and a corresponding sequence of triangular regions
T1 , T2 , . . . .
By condition (3) all the regions Ti have the same “area” αTi = A.
Now consider the defects of these triangles and let di = δTi . For each n
d1 + d2 + · · · + dn = δ4AP0 Pn < 180.
Since the partial sums d1 + d2 + · · · + dn are bounded, we have that the infinite series,
∞
X
dn
n=0
is convergent. Therefore,
lim dn = 0.
n→∞
Hence dn < d1 for some n.
By Theorem 10.7 there is a point, B, between A and P0 such that
δ4BP0 P1 = δ4APn−1 Pn = dn .
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CHAPTER 10. AREA IN HYPERBOLIC GEOMETRY
6
PP
AH
@HPP
H PP
B Z@ HH
P
HHPPP
Z@
PP
H
Z@
PP
HH
Z@
PP
HH
Z@
PP
Z
PP
H
PP
@
H
Z
Z
¾
@
H
...
P0
P1
Pn−1
Pn
?
-
Therefore by Bolyai’s Theorem the regions T and Tn determined by these triangles are
equivalent by finite decomposition. By condition (4) this means that αT = αTn . But
αTn = αT1 = A.
Therefore,
αT = αT1 .
Because α4ABP1 > 0 and
αT1 = αT + α4ABP1 .
this must be impossible.
10.3
The Uniqueness of Hyperbolic Area Theory
Okay, we can’t have that the area of a triangle in the hyperbolic plane depends on the
“height” and “base” of the triangle. What would be reasonable, then? The first two
properties must hold. Thus, we need to change the third property above for an area function.
An area function α should have the following properties:
1. αR > 0 for every R;
2. if R1 and R2 intersect only in edges and vertices, then
α(R1 ∪ R2 ) = αR1 + αR2 ;
3. if R1 ≡ R2 , then αR1 = αR2 .
There is such a function—the defect. Recall that we used the notation defect (4ABC) =
δ4ABC. We know that δ satisfies (1), (2), and (3). In fact, we have the following theorem:
Theorem 10.9 Let
α :R→R
be an area function satisfying (1), (2), and (3). Then there is a positive constant k > 0
such that
π 2
αR =
k δR
180
for every R ∈ R.
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119
We are not able to prove this at this time. The proof is within your grasp, but we do
not have time to do it.
Corollary 1 In H2 the area of any triangle is at most πk 2 .
There is no finite triangle whose area equals the maximal value πk 2 , although you can
approach this area as closely as you wish (and achieve it with a trebly asymptotic triangle).
J. Bolyai proved that you can construct a circle of area πk 2 and a regular 4-sided polygon
with a 45◦ angle that also has this area.
10.4
Angle of Parallelism
In the Poincaré model of hyperbolic geometry, we found a formula relating the angle measurement of the angle of parallelism and the distance to which it is associated. This formula
is a special case of the general formula discovered independently by Bolyai and Lobachevsky.
Theorem 10.10 [Formula of Bolyai-Lobachevsky]
tan
α
= e−d/k ,
2
where k is the proportionality constant from Theorem 10.9.
We have proven this formula in the Poincaré model where k = 1.
We will return to the area of triangles in the Poincaré half plane model later.
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Chapter 11
More in the Poincaré Disk Model
We have seen how we can define a distance function that will lead us to a geometry that
models the Hyperbolic plane. Now, we would like to look further at what must happen in
this model and what other results may be true in H2 .
Let’s look again at the Saccheri quadrilateral. We
have shown that the summit angles are congruent to
one another. Saccheri wanted to show that they must
be right angles. What happens in H2 ?
Theorem 11.1 The summit angles of a Saccheri
quadrilateral are acute.
E
F
C
B
Proof: We have already proven this, but look at how
A
Ω
the proof can be constructed in this specific model. The
proof we give here is tied to this model.
The
point
Ω
←→
is an ideal point associated←→
to the line
AB.
Recall that
←→
this means that the lines EΩ and F Ω are horoparallel
at Ω.
Figure 11.1: Poincaré Disk SacNow, recall that BE ∼
= AF . So ∠BEΩ is the angle cheri quadrilateral
of parallelism associated to the length BE. Likewise, ∠AF Ω is the angle of parallelism
associated to the length AF . Since the distances are congruent the angles of parallelism must
be the same by Lobachevskii’s Theorem. By Theorem 7.9 it follows that ∠BEΩ + ∠ΩEC =
∠AF E.
Since they form a straight angle we have that:
∠AF E + ∠AF Ω + ∠ΩF C = 180◦ .
Substituting we get
(∠BEω + ∠ΩEC) + ∠BEω + ∠ΩF C = 180◦ .
Claim: ∠ΩEC < ∠ΩF C.
If so, then
2∠BED + 2∠DEC < 180◦ .
This implies that ∠BEF = ∠BED + ∠DEC < 90◦ and the summit angles of a Saccheri
quadrilateral are acute.
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CHAPTER 11. MORE IN THE POINCARÉ DISK MODEL
We now have to prove this claim. First, we know that ∠ΩEC + ∠BEΩ = ∠BEF
and ∠AF Ω + ∠ΩF C = ∠AF C. Also, ∠BEΩ ∼
= ∠AF Ω. Thus, if we can show that
∠BEF < ∠AF C we will be done.
First, ∠BEF ∼
6 ∠AF C for if they were congruent, then this would force ∠EF A ∼
=
=
∠BEF ∼
∠AF
C
and
this
would
force
∠EF
A
and
∠AF
C
to
be
right
angles
and
this
would
=
mean that a rectangle would exist, which is impossible.
So, assume that ∠BEF > ∠AF C.
Note: There is an Exterior Angle Theorem for singly asymptotic triangles. If we had a
proof of that here, we can immediately see that ∠ΩEC < ∠ΩF C.
We can now use this to find the angle sum of any triangle in H2 .
Theorem 11.2 (Hyperbolic Triangle Angle Sum) The angle sum of any hyperbolic
triangle is less than 180◦ .
Proof: The three cases to consider are 4ABC being (i) an acute, (ii) an obtuse, or (iii)
a right triangle.
A
A
H
G
D
F E
C
B
C
B
Figure 11.2: Acute Triangle
Figure 11.3: Obtuse Triangle
Case 1 : 4ABC is acute.
In this case let D and E be←→
the midpoints of AB and AC, respectively. Construct BG, AF ,
and CH perpendicular to DE.
By vertical angles ∠BDG ∼
= ∠ADF . Then by SAS we have that 4BGD ∼
= 4ADF ,
∼
which implies that BG = AF . Likewise, we can show that 4CHE ∼
= 4AEF . It thus
follows that BG ∼
= CH and 2BGHC is a Saccheri quadrilateral.
Note also that
∠ABC + ∠ACB + ∠BAC = ∠ABC + ∠ACB + (∠DAF + ∠EAF )
= (∠ABC + ∠DBG) + (∠ACB + ∠ECH)
= ∠GBC + ∠HCB
Therefore the angle sum of 4ABC is the same as the sum of the summit angles of a
Saccheri quadrilateral. Since both are acute and congruent, then the angle sum of 4ABC
must be less than 180◦ .
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Case 2 : 4ABC is obtuse.
In this case let ∠ACB be the obtuse angle. Drop a perpendicular from C to AB. This
will intersect AB in a point D between A and B. This divides the obtuse triangle, 4ABC,
into two acute triangles: 4ACD and 4BCD.
Now, we know that
∠ABC + ∠ACB + ∠BAC = ∠DBC + ∠ACD + ∠DCA + ∠DCB
= (∠DBC + ∠DCB) + (∠ACD + ∠DCA) + (∠ADC + ∠BDC − 180◦ )
= (∠DBC + ∠DCB + ∠BDC) + (∠ACD + ∠DCA + ∠ADC) − 180◦ )
< 180◦ + 180◦ − 180◦ = 180◦
The third case can be proven similarly.
Theorem 11.3 The sum of the acute angles in a right triangle is less than 90◦ .
Theorem 11.4 The sum of the interior angles of a convex hyperbolic polygon is less than
(n − 2)180◦ .
Theorem 11.5 The angle sum of a Saccheri quadrilateral is less than 360◦ .
Theorem 11.6 Rectangles do not exist in the hyperbolic plane.
Theorem 11.7 (Hyperbolic Pythagorean Theorem) In a hyperbolic right triangle 4ABC
with sides a, b, and hypotenuse c, then
cosh(c) = cosh(a) cosh(b).
We will prove this in the next chapter.
11.0.1
Distance between Lines
How do we measure the distance from a point to a line? How do we measure the distance
between two lines?
To answer the first, we measure the distance of the line segment from the point that
is perpendicular to the line. To answer the second, we need to know that the lines are
not intersecting. If they intersect, then the distance between them is 0 at one point and
goes to infinity as we move away. In Euclidean geometry, we know that parallel lines are
everywhere equidistant. That is, that the distance between the two lines is a constant.
Things are not as simple in the hyperbolic plane. We can still measure the distance
from a point to a line. However, how do we handle the distance between two lines? We
shall look at the shortest distance between two parallel lines. It is going to have to be the
case that these two lines admit a common perpendicular. Else, the distance measured from
m to n would be different from the distance as measured from n to m.
Definition 11.1 The shortest distance between two parallel lines is measured along a common perpendicular between the two lines.
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11.1
Congruence in the Hyperbolic Plane
The concept of congruence stays the same in H2 , but the conditions that imply congruence
do change. We still have the conditions of SAS, SSS, and ASA. Remember that we proved
the following result earlier.
Theorem 11.8 (AAA) If three angles of one triangle are congruent respectively to three
angles of another triangle, then the two triangles are congruent.
From this result it follows that:
Theorem 11.9 Saccheri quadrilaterals with congruent summit angles and congruent summits are congruent.
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Chapter 12
Poincaré Upper Half Plane Model
The next model of the hyperbolic plane that we will consider is also due to Henri Poincaré.
We will be using the upper half plane, or {(x, y) | y > 0}. We will want to think of this
with a different distance metric on it.
Let H = {x + iy | y > 0} together with the arclength element
p
dx2 + dy 2
.
ds =
y
Note that we have changed the arclength element for this model!!!
12.1
Vertical Lines
Let x(t) = (x(t), y(t)) be a piecewise smooth parametrization of a curve between the points
x(t0 ) and x(t1 ).
Recall that in order to find the length of a curve we break the curve into small pieces
and approximate the curve by multiple line
p segments. In the limiting process we find that
the Euclidean arclength element is ds = dx2 + dy 2 . We then find the length of a curve
by integrating the arclength over the parametrization of the curve.
s
Z t1 µ ¶2 µ ¶2
dx
dy
s=
+
dt.
dt
dt
t0
Now, we want to work in the Poincaré Half Plane model. In this case the length of this
same curve would be
r
¡ dx ¢2 ³ dy ´2
Z t1
+ dt
dt
sP =
dt.
y
t0
Let’s look at this for a vertical line segment from (x0 , y0 ) to (x0 , y1 ). We need to parameterize the curve, and then use the arclength element to find its length. Its parametrization
is:
x(t) = (x0 , y), y ∈ [y0 , y1 ].
The Poincaré arclength is then
r
¡ dx ¢2 ³ dy ´2
Z t1
Z t1
+ dt
dt
1
sP =
dt =
dy = ln(y)|yy10 = ln(y1 ) − ln(y0 ) = ln(y1 /y0 )
y
t0
t0 y
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CHAPTER 12. POINCARÉ UPPER HALF PLANE MODEL
Now, consider any piecewise smooth curve x(t) = (x(t), y(t)) starting at (x0 , y0 ) and
ending at (x0 , y1 ). So this curves starts and ends at the same points as this vertical line
segment. Suppose that y(t) is an increasing function. This is reasonable. Now, we have
r
¡ dx ¢2 ³ dy ´2
Z t1
+ dt
dt
s=
dt
y
t0
r³ ´
2
dy
Z t1
dt
≥
dt
y
t0
Z y(t1 )
dy
≥
y(t0 ) y
≥ ln(y(t1 )) − ln(y(t0 )).
This means that this curve is longer than the vertical line segment which joins the two
points. Therefore, the shortest path that joins these two points is a vertical (Euclidean)
line segment. Thus, vertical (Euclidean) lines in the upper half plane are lines in the
Poincaré model.
Let’s find the distance from (1, 1) to (1, 0) which would be the distance to the real axis.
Now, since (1, 0) is NOT a point of H , we need to find lim d((1, 1), (1, δ)). According to
δ→0
what we have above,
dP ((1, 1), (1, δ)) = ln(1) − ln(δ) = − ln(δ).
Now, in the limit we find that
dP ((1, 1), (1, 0)) = lim dP ((1, 1), (1, δ)) = lim − ln(δ) = +∞
δ→0
δ→0
This tells us that a vertical line has infinite extent in either direction.
12.2
Isometries
Recall that an isometry is a map that preserves distance. What are the isometries of H ?
The arclength element must be preserved under the action of any isometry. That is, a
map
(u(x, y), v(x, y))
is an isometry if
dx2 + dy 2
du2 + dv 2
=
.
2
v
y2
Some maps will be obvious candidates for isometries and some will not.
Let’s start with the following candidate:
Ta (x, y) = (u, v) = (x + a, y).
Now, clearly du = dx and dv = dy, so
du2 + dv 2
dx2 + dy 2
=
.
v2
y2
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12.3. INVERSION IN THE CIRCLE: EUCLIDEAN CONSIDERATIONS
127
Thus, Ta is an isometry. What does it do? It translates the point a units in the horizontal
direction. This is called the horizontal translation by a.
Let’s try:
Rb (x, y) = (u, v) = (2b − x, y).
Again, du = −dx, dv = dy and our arclength element is preserved. This isometry is a
reflection through the vertical line x = b.
We need to consider the following map:
µ
¶
x
y
Φ(x, y) = (u, v) =
,
.
x2 + y 2 x2 + y 2
First, let’s check that it is a Poincaré isometry. Let r2 = x2 + y 2 . Then
õ
¶2 µ 2
¶2 !
du2 + dv 2
r4
r2 dx − 2x2 dx − 2xydy
r dy − 2xydx − 2y 2 dy
= 2
+
v2
y
r4
r4
µ 2
¶
1 ((y − x2 )dx − 2xydy)2 − ((x2 − y 2 )dy − 2xydx)2
= 2
y
r4
¢
1 ¡
= 4 2 (x4 − 2x2 y 2 + y 4 + 4x2 y 2 )dx2 − (2xy(y 2 − x2 ) + 2xy(x2 − y 2 )dxdy + r4 dy 2
r y
dx2 + dy 2
=
y2
We will study this function further. It is called inversion in the unit circle.
12.3
Inversion in the Circle: Euclidean Considerations
We are building a tool that we will use in studying H . This is a Euclidean tool, so we will
be working in Euclidean geometry to prove results about this tool. There is more about
inversions in the circle in Appendix A.
Let’s look at this last isometry. We would like to understand what this function does.
For each point (x, y), let r2 = x2 + y 2 . This makes r the distance from the origin to (x, y).
This function sends (x, y) to (x/r2 , y/r2 ). The distance from Φ(x, y) = (x/r2 , y/r2 ) to the
origin is 1/r2 . Thus, if r > 1 then the image of the point is on the same ray, but its distance
to the origin is now less than one. Likewise, if r < 1, then the image lies on the same
ray but the image point lies at a distance greater than 1 from the origin. If r = 1, then
Φ(x, y) = (x, y). Thus, Φ leaves the unit circle fixed and sends every point inside the unit
circle outside the circle and every point outside the unit circle gets sent inside the unit
circle. In other words, Φ turns the circle inside out.
What does Φ do to a line? What does it do to a circle? Let’s see.
The image of a point P under inversion in a circle centered at O and with radius r is
the point P 0 on the ray OP and such that
|OP 0 | =
r2
.
|OP |
Lemma 12.1 Let ` be a line which does not go through the origin O. The image of ` under
inversion in the unit circle is a circle which goes through the origin O.
Proof: We will prove this for a line ` not intersecting the unit circle.
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CHAPTER 12. POINCARÉ UPPER HALF PLANE MODEL
Let A be the foot of O on ` and
let |OA| = a. Find A0 on OA so
that |OA0 | = 1/a. Construct the circle with diameter OA0 . We want to
show that this circle is the image of
` under inversion.
Let P ∈ ` and let |OP | = p.
Let P 0 be the intersection of the segment OP with the circle with diameter OA0 . Let |OP 0 | = x. Now,
look at the two triangles 4OAP and
4OP 0 A0 . These two Euclidean triangles are similar, so
O
P'
P
A'
A
|OP 0 |
|OA|
=
0
|OA |
|OP |
a
x
=
1/a
p
1
x=
p
Therefore, P 0 is the image of P under inversion in the unit circle.
Lemma 12.2 Suppose Γ is a circle which does not go through the origin O. Then the image
of Γ under inversion in the unit circle is a circle.
Proof: Again, we prove this for just one case: the case where Γ does not intersect the unit
circle.
Let the line through O and the center of Γ intersect Γ at points A and B. Let |OA| = a
and |OB| = b. Let Γ0 be the image of Γ under dilation by the factor 1/ab. This dilation is
∆ : (x, y) 7→ (x/ab, y/ab).
Let B 0 and A0 be the images of A and B, respectively, under this dilation, i.e. ∆(A) = B 0
and ∆(B) = A0 . Then |OA0 | = (1/ab)b = 1/a and |OB 0 | = (1/ab)a = 1/b. Thus, A0 is the
image of A under inversion in the unit circle. Likewise, B 0 is the image of B. Let `0 be an
arbitrary ra through O which intersects Γ at P and Q. Let Q0 and P 0 be the images of P
and Q, respectively, under the dilation, ∆.
Now, 4OA0 P 0 ∼ 4OBQ, since
one is the dilation of the other. Note
that ∠QBA ∼
= ∠QP A by the Star
P
Trek
lemma,
and hence 4OBQ ∼
Q
4OP
A.
Thus,
4OA0 P 0 ∼ 4OP A.
Q'
P'
From this it follows that
O
B'
A'
A
B
Γ′
Γ
MATH 6118-090
|OA0 |
|OP 0 |
=
|OP |
|OA|
1/a
|OP 0 |
=
|OP |
a
1
|OP 0 | =
|OP |
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12.4. LINES IN THE POINCARÉ HALF PLANE
129
Thus, P 0 is the image of P under inversion, and Γ0 is the image of Γ under inversion.
Lemma 12.3 Inversions preserve angles.
Proof: We will just consider the
case of an angle α created by the in{ α
tersection of a line ` not intersecting
0
A {'
the unit circle, and a line ` through
O.
{''
Let A be the vertex of the angle
A'
β
α. Let P be the foot of O in `. Let
O
P 0 be the image of P under inversion. Then the image of ` is a circle
P'
Γ whose diameter isTOP 0 . The image of A is A0 = Γ `0 . Let `00 be
P
the tangent to Γ at A0 . Then β, the
angle formed by `0 and `00 at A0 is
the image of α under inversion. We
need to show that α ∼
= β.
First, 4OAP ∼ 4OP 0 A0 , since they are both right triangles and share the angle O.
Thus, ∠A0 P 0 O ∼
= ∠OAP ∼
= α. By the tangential case of the Star Trek lemma, β ∼
= ∠A0 P 0 O.
∼
Thus, α = β.
12.4
Lines in the Poincaré Half Plane
From what we have just seen we can now prove the following.
Lemma 12.4 Lines in the Poincaré upper half plane model are (Euclidean) lines and
(Euclidean) half circles that are perpendicular to the x-axis.
Proof: Let P and Q be points in H not on the same vertical line. Let Γ be the circle
through P and Q whose center lies on the x-axis. Let Γ intersect the x-axis at M and
N . Now consider the mapping ϕ which is the composition of a horizontal translation by
−M followed by inversion in the unit circle. This map ϕ is an isometry because it is the
composition of two isometries. Note that M is first sent to O and then to ∞ by inversion.
Thus, the image of Γ is a (Euclidean) line. Since the center of the circle is on the real axis,
the circle intersects the axis at right angles. Since inversion preserves angles, the image of
Γ is a vertical (Euclidean) line. Since vertical lines are lines in the model, and isometries
preserve arclength, it follows that Γ is a line through P and Q.
Problem: Let P = 4 + 4i and Q = 5 + 3i. We want to find M , N , and the distance from
P to Q.
First we need to find Γ. We need to find the perpendicular bisector of the segment P Q
and then find where this intersects the real axis. The midpoint of P Q is the point (9+7i)/2,
or (9/2, 7/2). The equation of the line through P Q is y = 8 − x. Thus, the equation of the
perpendicular bisector is y = x − 1. This intersects the x-axis at x = 1, so the center of the
circle is 1 + 0i. The circle has to go through the points 4 + 4i and 5 + 3i. Thus the radius is
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CHAPTER 12. POINCARÉ UPPER HALF PLANE MODEL
5, using the Pythagorean theorem. Hence, the circle meets the x-axis at M = −4 + 0i and
N = 6 + 0i.
We need to translate the line Γ so that M goes to the origin. Thus, we need to translate
by 4 and we need to apply the isometry T4 : (x, y) → (x + 4, y). Then, P 0 = T4 (P ) = (8, 4)
and Q0 = T4 (Q) = (9, 3). Now, we need to invert in the unit circle and need to find the
images of P 0 and Q0 . We know what Φ does:
¶ µ
¶
1 1
8 4
=
Φ(P ) = Φ((8, 4)) =
,
,
80 80
10 20
µ
¶ µ
¶
9 3
1 1
0
Φ(Q ) = Φ((9, 3)) =
,
=
,
90 90
10 30
µ
0
Note that we now have these two images on a vertical (Euclidean) line. So the distance
between the points dP (Φ(P 0 ), Φ(Q0 )) = ln(1/20) − ln(1/30) = ln(3/2). Thus, the points P
and Q are the same distance apart.
Φ(T4(Γ))
Γ
T4(Γ)
P=4+4i
T4(P)
Q=5+3i
M
T4(Q)
N
Figure 12.1: Isometries in H
12.5
Fractional Linear Transformations
We want to be able to classify all of the isometries of the Poincaré half plane. It turns out
that the group of direct isometries is easy to describe. We will describe them and then see
why they are isometries.
A fractional linear transformation is a function of the form
T (z) =
az + b
cz + d
where a, b, c, and d are complex numbers and ad − bc 6= 0. The domain of this function is
the set of all complex numbers C together with the symbol, ∞, which will represent a point
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12.5. FRACTIONAL LINEAR TRANSFORMATIONS
131
at infinity. Extend the definition of T to include the following
T (−d/c) = lim
z→− dc
az + b
= ∞,
cz + d
if c 6= 0,
az + b
a
=
if c 6= 0,
cz + d
c
az + b
T (∞) = lim
= ∞ if c = 0.
z→∞ cz + d
T (∞) = lim
z→∞
The fractional linear transformation, T , is usually represented by a 2 × 2 matrix
·
¸
a b
γ=
c d
and write T = Tγ . The matrix representation for T is not unique, since T is also represented
by
·
¸
ka kb
kγ =
kc kd
for any scalar k 6= 0. We define two matrices to be equivalent if they represent the same
fractional linear transformation. We will write γ ≡ γ 0 .
Lemma 12.5
Tγ1 γ2 = Tγ1 (Tγ2 (z)).
From this the following theorem follows.
Theorem 12.1 The set of fractional linear transformations forms a group under composition (matrix-multiplication).
Proof: Theorem 12.5 shows us that this set is closed under our operation. The identity
element is given by the identity matrix,
·
¸
1 0
I=
.
0 1
The fractional linear transformation associated with this is
TI (z) =
z+0
= z.
0z + 1
The inverse of an element is
Tγ−1 = Tγ −1 ,
since
Tγ (Tγ −1 (z)) = TI (z) = z.
We can also see that to find Tγ −1 we set w = Tγ (z) and solve for z.
az + b
cz + d
(cz + d)w = az + b
dw − b
z=
.
−cw + a
w=
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CHAPTER 12. POINCARÉ UPPER HALF PLANE MODEL
That is Tγ −1 is represented by
·
¸
·
¸
1
d −b
d −b
≡
= γ −1 .
−c a
ad − bc −c a
Here we must use the condition that ad − bc 6= 0.
In mathematical circles when we have such an interplay between two objects — matrices
and fractional linear transformations — we write γz when Tγ (z) is meant. Under this
convention we may write
·
¸
az + b
a b
.
γz =
z=
c d
cz + d
Note that the second “=” is not equals in the usual sense, but is instead an assignment or
a definition.
This follows the result of Theorem 12.5 in that
(γ1 γ2 )z = γ1 (γ2 z),
however in general k(γz) 6= (kγ)z. Note that
k(γz) =
k(az + b)
,
cz + d
while
az + b
.
cz + d
Recall the following definitions for any ring R:
½·
¸
¾
a b
M2×2 (R) =
| a, b, c, d ∈ R
c d
(kγz) = γz =
GL2 (R) = {γ ∈ M2×2 (R) | det(γ) 6= 0}
SL2 (R) = {γ ∈ GL2 (R) | det(γ) = 1} .
For the purposes of what we will be doing, we prefer the ring to be the field of complex
numbers, C, the field of real numbers, R, the field of rational numbers, Q, or the ring of
integers Z. GL2 (R) is called the general linear group over R, and SL2 (R) is called the special
linear group over R.
There is another group, which is not as well known. This is the projective special
linear group denoted by PSL2 (R). PSL2 (R) is obtained from GL2 (R) by identifying γ with
kγ for any k 6= 0. The group PSL2 (C) is isomorphic to the group of fractional linear
transformations.
Remember that we wanted to classify the group of direct isometries on the upper half
plane. We want to show that any 2 × 2 matrix with real coefficients and determinant 1
represents a fractional linear transformation which is an isometry of the Poincaré upper
half plane.
Lemma 12.6 The horizontal translation by a
Ta (x, y) = (x + a, y),
can be thought of as a fractional linear transformation, represented by an element of SL2 (R).
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12.5. FRACTIONAL LINEAR TRANSFORMATIONS
133
Proof: If a ∈ R, then
Ta (x, y) = Ta (z) = z + a,
and this is represented by
z ∈ C,
·
¸
1 a
τa =
.
0 1
This is what we needed.
Lemma 12.7 The map
µ
ϕ(x, y) =
y
−x
, 2
2
2
x + y x + y2
¶
,
which is inversion in the unit circle followed by reflection through x = 0, can be thought of
as a fractional linear transformation which is represented by an element of SL2 (R).
Proof: As a function of complex numbers, the map ϕ is
ϕ(z) = ϕ(x + iy) =
This map is generated by
−x + iy
−(x − iy)
1
=
=− .
2
2
x +y
(x + iy)(x − iy)
z
·
¸
0 −1
σ=
.
1 0
Theorem 12.2 The group SL2 (R) is generated by σ and the maps τa for a ∈ R.
Proof: Note that
so
and
·
¸·
¸
0 −1 1 r
στr =
1 0
0 1
·
¸
0 −1
=
1 r
·
¸·
¸
0 −1 0 −1
στs στr =
1 s
1 r
·
¸
−1
−r
=
s rs − 1
·
¸·
¸
0 −1 −1
−r
στt στs στr =
1 t
s rs − 1
·
¸
−s
1 − rs
=
st − 1 rst − r − t
Now, we know what the composition of these transformations will look like. To say that
these generate SL)2 (R) means is that for an arbitrary
·
¸
a b
γ=
∈ SL)2 (R)
c d
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CHAPTER 12. POINCARÉ UPPER HALF PLANE MODEL
we need to find r, s, t ∈ R so that
·
¸ ·
¸
−s
1 − rs
a b
.
γ=
=
c d
st − 1 rst − r − t
This cannot be too difficult for a 6= 0. Set s = −a, solve b = 1 − rs = 1 + ra for r and solve
c = st − 1 = −at − 1 for t. This gives
r=
b−1
a
and t =
−1 − c
.
a
Since det(γ) = 1, this forces d = rst − r − t. Thus, for a 6= 0, then γ can be written as a
product involving only σ and translations. If a = 0, then c 6= 0, since ad − bc = 1, and
·
¸
−c −d
σγ =
,
a
b
which can be written as a suitable product. Thus SL2 (R) is generated by the translations
and σ.
Lemma 12.8 The group SL2 (R), when thought of as a group of fractional linear transformations, is a subgroup of the isometries of the Poincaré upper half plane.
Lemma 12.9 If γ ∈ GL2 (R) and detγ > 0, then γ is an isometry of the Poincaré upper
half plane.
Theorem 12.3 The image of a circle or line in C under the action of a fractional linear
transformation γ ∈ SL2 (C) is again a circle or a line.
12.6
Cross Ratio
Henri Poincaré was studying this cross ratio when he discovered this particular representation of the hyperbolic plane.
S
Let a, b, c, d be elements of the extended complex numbers, C {∞}, at least three of
which are distinct. The cross ratio of a, b, c, and d is defined to be
a−c
(a − c)(b − d)
[a, b; c, d] = a − d =
.
b−c
(a − d)(b − c)
b−d
The algebra for the element ∞ and division by zero is the same as it is for fractional linear
transformations.
If a = d, [a, b; c, d] = ∞
b−d
[∞, b; c, d] =
= [a, b; c, ∞]
b−c
a−c
= [a, b; ∞, d]
[a, ∞; c, d] =
a−d
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12.6. CROSS RATIO
135
S
If we fix three distinct elements a, b, and c ∈ C {∞}, and consider the fourth element
as a variable z, then we get a fractional linear transformation:
z−b
z
T (z) = (z, a; b, c) = − c .
a−b
a−c
This is the unique fractional linear transformation T with the property that
T (a) = 1,
T (b) = 0,
and T (c) = ∞,
that is, it sends a to 1, b to 0 and c to ∞.
We need to look at several examples to see why we want to use the cross ratio.
Example 12.1 Find the fractional linear transformation which sends 1 to 1, −i to 0 and
−1 to ∞.
From above we need to take: a = 1, b = −i, and c = −1. Thus, set
w = (z, 1; −i, −1)
z+i 1+i
=
/
z+1 1+1
2z + 2i
=
(1 + i)(z + 1)
In matrix notation,
·
¸
2
2i
w=
z.
1+i 1+i
Example 12.2 Find the fractional linear transformation which fixes i, sends ∞ to 3, and
0 to −1/3.
This doesn’t seem to fit our model. However, we can combine two of our transformations
to get this one. First think of sending
i→1
∞ → 0 and
0→∞
i→1
3→0
1
− →∞
3
If we go by the first transformation followed by the inverse of the second we will send:
i→1 →
i
∞→0 →
3
1
0 → ∞→ −
3
Let
γ1 z = (z, i; ∞, 0)
and
γ2 w = (w, i; 3, −1/3).
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So, γ1 (i) = 1, γ1 (∞) = 0, γ1 (0) = ∞, γ2 (i) = 1, γ2 (3) = 0, and γ2 (−1/3) = ∞. Therefore,
γ2−1 (1) = i, γ2−1 (0) = 3, and γ2−1 (∞) = −1/3. Now, compose these functions:
γ = γ2−1 γ1 .
Let’s check what γ does: γ(i) = i, γ(∞) = 3 and γ(0) = −1/3, as desired.
Now, set w = γ(z) and
w = γ2−1 γ1 (z)
γ2 (w) = γ1 (z)
(w, i; 3, −1/3) = (z, i; ∞, 0).
Now, we need to solve for z:
w−3
i−3
z−∞ i−∞
/
=
/
w + 1/3 i + 1/3
z−0 i−0
(3i + 1)w − 3(3i + 1)
i
=
3(i − 3)w + i − 3
z
(−3(3i + 1)w − (3i + 1)
z=
(3i + 1)w − 3(3i + 1)
3w + 1
=
−w + 3
·
¸
3 1
=
−1 3
Then, using our identification, we will get that
·
¸ ·
¸
1 3 −1
3 −1
≡
z
w=
1 3
10 1 3
12.7
Translations
Now, we have claimed that the Poincaré upper half plane is a model for the hyperbolic
plane. We have not checked this. Let’s start with the sixth postulate:
Postulate 6: Given any two points P and Q, there exists an isometry f such that f (P ) =
Q.
Let P = a+bi and Q = c+di. We have many choices, but we will start with an isometry
that also fixes the point at ∞. In some sense, this is a nice isometry, since it does not map
any regular point to infinity nor infinity to any regular point. Now, since f (∞) = ∞ and
f (P ) = Q, f must send the line through P and ∞ to the line through Q and ∞. This
means that the vertical line at x = a is sent to the vertical line at x = c. Thus, f (a) = c.
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12.8. ROTATIONS
137
This now means that we have to have
(w, c + di; c, ∞) = (z, a + bi; a, ∞)
w−c
z−a
=
di
bi
d(z − a)
w=
+c
· b
¸
d bc − ad
=
z.
0
b
Since b > 0 and d > 0, then the determinant of this matrix is positive. That and the fact
that all of the entries are real means that it is an element of PSL2 (R) and is an isometry of
the Poincaré upper half plane.
We claim that this map that we have chosen is a translation. Now, recall that translations are direct isometries with no fixed points. How do we show that it has no fixed
points? A fixed point would be a point z0 so that f (z0 ) = z0 . If this is the case, then solve
for z below:
d(z0 − a)
+ c = z0
b
ad − bc
z0 =
d−b
But, note that a, b, c, and d are all real numbers. Thus, if b 6= d then z0 is a real number
and is not in the upper half plane. Thus, this map has no fixed points in H and is a
translation. If b = d, then z0 = ∞, and again there are no solutions in the upper half plane,
so the map is a translation.
In the Poincaré upper half plane, we classify
S our translations by how many fixed points
there are on the line at infinity (that is, in R ∞.) Let
·
¸
a b
γ=
.
c d
Then γ(z) = z if
cz 2 + (d − a)z − b = 0.
Now, if c 6= 0, then this is a quadratic equation with discriminant
∆ = (d − a)2 − 4bc.
Thus, there is a fixed point in H if ∆ < 0, and no fixed points if ∆ ≥ 0. If ∆ = 0 then
there is exactly one fixed point on the line at infinity. In this case the translation is called
a parabolic translation. If ∆ > 0 the translation is called a hyperbolic translation.
12.8
Rotations
What are the rotations in the Poincaré upper half plane? What fractional linear transformations represent rotations?
A rotation will fix only one point. Let P = a + bi. We want to find the rotation that
fixes P and rotates counterclockwise through an angle of θ.
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CHAPTER 12. POINCARÉ UPPER HALF PLANE MODEL
First, find the (Euclidean) line through P which makes an angle θ with the vertical line
through P . Find the perpendicular to this line, and find where it intersects the x-axis. The
circle centered at this intersection and through P is the image of the vertical line under the
rotation. Let this circle intersect the x-axis at points M and N . Then the rotation is given
by
(w, P ; N, M ) = (z, P ; a, ∞).
We want to find an easy point to rotate, then we can do this in general. It turns out
that the simplest case is to rotate about P = i.
Here let the center of the half
circle be at −x, and let the (Euclidean) radius of the circle be r. Then
x = r cos θ, r sin θ = 1, M = −r − x,
θ
P=i
and N = r − x. So we have to solve
(w, i; r − x, −r − x) = (z, i; 0, ∞).
r
1
After quite a bit of algebraic manipulation, we get
·
¸
cos 2θ sin 2θ
w = ρθ z =
z
sin 2θ cos 2θ
M
x
N
For an arbitrary point P = a + bi we need to apply a translation that sends P to i and
then apply the rotation, and then translate back. The translation from P = a + bi to 0 + i
is
·
¸
1 −a
γ=
.
0 b
The inverse translation is
·
γ
−1
¸
b a
=
.
0 1
Thus, the rotation about P is
·
γ
12.9
−1
¸·
¸·
¸
b a cos 2θ sin 2θ 1 −a
ρθ γ =
0 1 sin 2θ cos 2θ 0 b
·
¸
b cos 2θ − a sin 2θ (a2 + b2 ) sin 2θ
=
− sin 2θ
a sin 2θ + b cos 2θ
Reflections
Not all isometries are direct isometries. We have not yet described all of the orientationreversing isometries of the Poincaré upper half plane. We did see that the reflection through
the imaginary axis is given by
R0 (x, y) = (−x, y),
which is expressed in complex coordinates as
R0 (z) = −z.
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12.10. DISTANCE AND LENGTHS
139
Note that in terms of a matrix representation, we can represent R0 (z) by
·
¸
−1 0
R0 (z) = µz =
z
0 1
Now, to reflect through the line ` in H , first use the appropriate isometry, γ1 to move
the line ` to the imaginary axis, then reflect and move the imaginary axis back to `:
γ1−1 µγ1 z = γ1−1 µγ1 z.
Note that µ2 = 1 and that µγµ ∈ SL2 (R) for all γ ∈ SL2 (R), since detµ = −1. Therefore,
γ1−1 µγ1 z = γ1−1 (µγ1 µ)µz = γ2 µz = γ2 (−z),
where γ2 ∈ SL2 (R). Thus, every reflection can be written in the form γ(−z) for some
γ ∈ SL2 (R).
Theorem 12.4 Every isometry f of H which is not direct can be written in the form
f (z) = γ(−z)
for some γ ∈ SL2 (R). Furthermore, if
·
¸
a b
γ=
c d
then f (z) is a reflection if and only if a = d.
12.10
Distance and Lengths
We want a formula for the distance between two points or the length of any line segment.
We know this for two points on the same vertical line: if P = a + bi and Q = a + ci, then
¯Z c ¯
¯
dy ¯¯
|P Q| = ¯¯
¯
b y
= | ln(c/b)|
Now, what will happen if P and Q don’t lie on a vertical line segment. Then there is a
half circle with center on the x-axis which goes through both P and Q. Let this half circle
have endpoints M and N . Since isometries preserve distance, we will look at the image of
σ which sends P to i and P Q to a vertical line. This is the transformation that sends P to
i, M to 0 and N to ∞. Since the image of Q will lie on this line, Q is sent to some point
0 + ci for some c. Then
|P Q| = | ln(c/1)| = | ln(c)|.
Note that
(σz, i; 0, ∞) = (z, P ; M, N )
σz
and in particular, since σ(Q) = ci and (σz, i; 0, ∞) =
, we get
i
c = (Q, P ; M, N ),
so
|P Q| = | ln(Q, P ; M, N )|.
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CHAPTER 12. POINCARÉ UPPER HALF PLANE MODEL
The Hyperbolic Axioms
We have not checked yet that the Poincaré upper half plane really meets all of the axioms
for a hyperbolic geometry. We need to check that all of the axioms are valid.
Postulate 1: We can draw a unique line segment between any two points.
Postulate 2: A line segment can be continued indefinitely.
We checked earlier that Postulate 2 is satisfied. Since there exists a half circle or vertical
line through any two points in the plane.
Postulate 3: A circle of any radius and any center can be drawn.
This follows from the definition. Once we know how to measure distance, we can create
circles.
Postulate 4: Any two right angles are congruent.
Our isometries preserve Euclidean angle measurement, so define the angle measure in H
to be the same as the Euclidean angle measure. Then any two right angles are congruent.
Postulate 6: Given any two points P and Q, there exists an isometry f such that f (P ) =
Q.
Postulate 7: Given a point P and any two points Q and R such that |P Q| = |P R|, there
is an isometry which fixes P and sends Q to R.
Postulate 8: Given any line `, there exists a map which fixes every point in ` and leaves
no other point fixed.
These we established in our last 4 sections.
Postulate 5: Given any line ` and any point P 6∈ `, there exist two distinct lines `1 and
`2 through P which do not intersect `.
This follows easily using non-vertical Poincaré lines.
12.12
The Area of Triangles
We have stated previously that the area of an asymptotic triangle is finite. It can be shown
that all trebly asymptotic triangles are congruent. This means that the area of all trebly
asymptotic triangles is the same. What is this common value in the Poincaré upper half
plane?
First, let’s compute the area of a doubly asymptotic triangle. We want to compute the
area of the doubly asymptotic triangle with vertices at P = ei(π−θ) in H , and vertices at
infinity of 1 and ∞. The angle at P for this doubly asymptotic triangle has measure θ.
Consider Figure 12.2.
The area element for the Poincaré upper half plane model is derived by taking a small
(Euclidean) rectangle with sides oriented horizontally and vertically. The sides approximate hyperbolic segments, since the rectangle is very small. The area would then be a
product of the height and width (measured with the hyperbolic arclength element). The
vertical sides of the rectangle have Euclidean length ∆y, and since y is essentially unMATH 6118-090
Spring 2006
12.12. THE AREA OF TRIANGLES
141
∆y
. The horizontal sides have Euclidean length ∆x
y
∆x
dxdy
and hence hyperbolic length
. This means that the area element is given by
.
y
y2
changed, the hyperbolic length is
Lemma 12.10 The area of a
doubly asymptotic triangle P ΩΘ
with points Ω and Θ at infinity
and with angle ΩP Θ = P has
area
θ
P
|4P ΩΘ| = π − P,
θ
- cos θ
1
0
where P is measured in radians.
Figure 12.2: Doubly Asymptotic Triangle
Proof: Let the angle at P have
measure θ. Then 4P ΩΘ is similar to the triangle in Figure 12.2
and is hence congruent to it. Thus, they have the same area. The area of the triangle in
Figure 12.2 is given by
Z
1
A(θ) =
− cos θ
1
Z
Z
∞
√
1−x2
1
dxdy
y2
dx
√
1 − x2
− cos θ
= arccos(−x)|1− cos θ = π − θ
=
Corollary 1 The area of a trebly asymptotic triangle is π.
P
Ω
Θ
Σ
Figure 12.3: Trebly Asymptotic Triangle
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CHAPTER 12. POINCARÉ UPPER HALF PLANE MODEL
Proof: : Let 4ΩΘΣ be a trebly asymptotic triangle, and let P be a point in the interior.
Then
|4ΩΘΣ| = |4P ΩΣ| + |4P ΘΣ| + |4P ΩΘ|
= (π − ∠ΩP Σ) + (π − ∠ΘP Σ) + (π − ∠ΩP Θ)
= 3π − 2π = π
Corollary 2 Let 4ABC be a triangle in H with angle measures A, B, and C. Then the
area of 4ABC is
|4ABC| = π − (A + B + C),
where the angles are measured in radians.
In the figure below, the figure on the left is just an abstract picture from the hyperbolic
plane. The figure on the right comes from the Poincaré model, H .
Σ
A
A
B
C
C
Θ
B
Ω
Ω
Θ
Σ
Proof: Construct the triangle 4ABC and continue the sides as rays AB, BC, and CA.
Let these approach the ideal points Ω, Θ, and Σ, respectively. Now, construct the common
parallels ΩΘ, ΘΣ, and ΣΩ. These form a trebly asymptotic triangle whose area is π. Thus,
|4ABC| = π − |4AΣΩ| − |4BΩΘ| − |4CΘΣ|
= π − (π − (π − A)) − (π − (π − B)) − (π − (π − C))
= π − (A + B + C).
12.13
The Poincaré Disk Model
Consider the fractional linear transformation in matrix form
·
¸
1 −i
φ=
−i 1
or
w=
MATH 6118-090
z−i
.
1 − iz
Spring 2006
12.14. ANGLE OF PARALLELISM
143
This map sends 0 to −i, 1 to 1, and ∞ to i. This map sends the upper half plane to the
interior of the unit disk. The image of H under this map is the Poincaré disk model, D.
Under this map lines and circles perpendicular to the real line are sent to circles which
are perpendicular to the boundary of D. Thus, hyperbolic lines in the Poincaré disk model
are the portions of Euclidean circles in D which are perpendicular to the boundary of D.
There are several ways to deal with points in this model. We can express points in terms
of polar coordinates:
D = {reiθ | 0 ≤ r < 1}.
We can show that the arclength segment is
√
2 dr2 + r2 dθ2
ds =
.
1 − r2
The group of proper isometries in D has a description similar to the description on H .
It is the group
½
·
¸¾
a b
Γ = γ ∈ SL2 (C) | γ =
b a
All improper isometries of D can be written in the form γ(−z) where γ ∈ Γ.
Lemma 12.11 If dp (O, B) = x, then
d(O, B) =
ex − 1
.
ex + 1
Proof: If Ω and Λ are the ends of the diameter through OB then
x = log(O, B; Ω, Λ)
OΩ · BΛ
ex =
OΛ · BΩ
1 + OB
BΛ
=
=
BΩ
1 − OB
ex + 1
OB = x
e −1
which is what was to be proven.
12.14
Angle of Parallelism
Let Π(d) denote the radian measure of the angle of parallelism corresponding to the hyperbolic distance d. We can define the standard trigonometric functions, not as before—using
right triangles—but in a standard way. Define
sin x =
cos x =
∞
X
(−1)n
n=0
∞
X
(−1)n
n=0
tan x =
MATH 6118-090
sin x
.
cos x
x2n+1
(2n + 1)!
(12.1)
x2n
(2n)!
(12.2)
(12.3)
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CHAPTER 12. POINCARÉ UPPER HALF PLANE MODEL
In this way we have avoided the problem of the lack of similarity in triangles, the premise
upon which all of real Euclidean trigonometry is based. What we have done is to define these
functions analytically, in terms of a power series expansion. These functions are defined for
all real numbers x and satisfy the usual properties of the trigonometric functions.
Theorem 12.5 (Bolyai-Lobachevskii Theorem) In the Poincaré model of hyperbolic
geometry the angle of parallelism satisfies the equation
µ
¶
Π(ρ)
−ρ
e = tan
.
2
Proof: By the definition of the angle of parallelism, d = dp (P, Q) for some point P to its
foot Q in some p-line `. Now, Π(d) is half of the radian measure of the fan angle at P , or
is the radian measure of ∠QP Ω, where P Ω is the limiting parallel ray to ` through P .
We may choose ` to be a diameter of the unit disk and Q = O, the center of the disk,
so that P lies on a diameter of the disk perpendicular to `.
The limiting parallel ray through P is the arc of a circle δ so that
(i) δ is orthogonal to Γ,
1. ` is tangent to δ at Ω.
P
α
Ω
α
Λ
Q
α
P
α
Q
Ω
Λ
Figure 12.4: Angle of Parallelism: left in D, right in H
The tangent line to δ at P must meet ` at a point R inside the disk. Now ∠QP Ω =
∠QΩP = β radians. Let us denote Π(d) = α. Then in 4P QΩ, α + 2β = π2 or
β=
Now, d(P, Q) = r tan β = r tan
¡π
ed =
MATH 6118-090
4
−
α
2
π α
− .
4
2
¢
. Applying Lemma 12.11 we have
1 + tan β
r + d(P, Q)
=
.
r − d(P, Q)
1 − tan β
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12.14. ANGLE OF PARALLELISM
Using the identity tan( π4 − α2 ) =
145
1 − tan α/2
it follows that
1 + tan α/2
ed =
1
.
tan α/2
Simplifying this it becomes
µ
e
−d
= tan
Π(d)
2
¶
.
Also, we can write this as Π(d) = 2 arctan(e−d ).
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Chapter 13
The Pseudosphere
Some of the difficulties of the different plane models for the hyperbolic plane are readily
apparent. The Poincaré models, while conformal, distort distances—we cannot use a Euclidean ruler. People have searched for another model which is conformal and that represents
hyperbolic lengths faithfully by Euclidean lengths. Such a model would be called isometric.
We are able to construct a model of elliptic geometry on the surface of the sphere in 3-space.
Is there a surface in 3-space on which we can isometrically model hyperbolic geometry? If
so, the lines of the hyperbolic plane would then be represented by geodesics 1 on the surface.
Since there seems to be some feeling that hyperbolic lines cannot be straight (i.e., modeled
on Euclidean lines), we might expect this surface to be curved in Euclidean 3-space.
A theorem first proved by David Hilbert states that it is impossible to embed the
entire hyperbolic plane isometrically as a surface in Euclidean space. On the other hand it
is possible to embed the Euclidean plane isometrically in hyperbolic space, as the surface
of the horosphere.
It is possible to embed a portion of the hyperbolic plane isometrically in Euclidean
space, the portion called a horocyclic sector, bounded by an arc of a horocycle and the
two diameters of the horocycle cutting off this arc. Figure 13.1 shows such a sector in the
Poincaré Disk model.
We must embed this sector isometrically onto a surface in Euclidean 3-space. The
surface that represents this region is called a pseudosphere. It is obtained by rotating a
curved known as the tractrix around its asymptote. The tractrix is the curve characterized
1
A geodesic segment between two points is the shortest path on the surface between those two points.
Figure 13.1: Horocyclic Sector in the Poincaré Disk model
147
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CHAPTER 13. THE PSEUDOSPHERE
Figure 13.2: Tractrix curve
by the condition that the length of the segment of the tangent line to the curve from the
curve to the y-axis is constant. It has the following equation in the Euclidean cartesian
plane for a given constant a:
p
p
a + a2 − y 2
) − a2 − y 2 .
x = a ln(
y
and has graph shown in Figure 13.2.
The representation on the pseudosphere gives us the opportunity to give some geometric
meaning to the fundamental constant k in the theorem on area. Gauss discovered a way
in which we can measure the curvature of a surface. Intuitively, it is a number K which
measures how much a surface bends. The sphere of radius R, for example, has constant
curvature 1/R2 . This makes sense as you think about, because the larger a sphere is the
less it bends. In general the curvature will change from point to point, or region to region,
on a surface — think about the surface of a doughnut. It should be zero at points where
the surface is flat and it should be large where the surface bends sharply. Surfaces where
the curvature is the same at all points are called surfaces of constant curvature. The sphere
is such an example.
For a surface of constant curvature K, Gauss found a fundamental formula which relates
the curvature, area, and angular measure. Take a geodesic triangle, 4ABC—a triangle
where the sides are geodesic segments. Let (∠A)r denote the radian measure of angle A.
Gauss showed that
K × area(4ABC) = (∠A)r + (∠B)r + (∠C)r − π.
What does this mean? There are three possible cases.
Case 1. Let K be positive. This means that both sides of the above equation are positive.
This means that the angle sum in radians of a geodesic triangle is greater than π( or in
degrees, greater than 180◦ ), and the area of the triangle is proportional to the excess. On
a sphere of radius R, then we know that the area of a geodesic triangle is
1
× ((∠A)r + (∠B)r + (∠C)r − π)
K
= R2 × ((∠A)r + (∠B)r + (∠C)r − π)
area(4ABC) =
If we have a geodesic triangle on the surface of the sphere with three right angles, then the
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excess of this triangle is
π
. Thus, the area of this geodesic triangle is
2
1
area(4ABC) = πR2 .
2
A theorem by Liebmann, Hopf, and Rinow proves that spheres are the only complete surfaces
of constant positive curvature in Euclidean three-space. It follows from this information
that elliptic geometry cannot be embedded isometrically in Euclidean three-space.
Case 2. K = 0. Gauss’ formula then says that the angle sum of a geodesic triangle, in
radians, is exactly π. An example is the Euclidean plane. Another example is an infinitely
long cylinder.
Case 3. K < 0. Now Gauss’ formula shows that the angle sum in radians is less than π, and
the area is proportional to the defect. The pseudosphere is a surface of constant negative
curvature. Since the pseudosphere represents a portion of the hyperbolic plane isometrically,
we can compare Gauss’ formula with the formula for the area in the hyperbolic plane. The
comparison gives K = −1/k 2 . Thus,√−1/k 2 is the curvature of the hyperbolic plane.
Of interest, let r = ik where i = −1. Then K = 1/r2 , so that the hyperbolic plane can
be described as a sphere of imaginary radius r = ik. This was noted by the mathematician
Lambert. This description is ver reasonable, for all the formulæ of hyperbolic trigonometry
can be obtained from the formulas known for spherical trigonometry by replacing r with
ik.
Finally, notice that as k gets very large, the curvature K approaches zero, and the
geometry of the surface resembles more and more the geometry of the Euclidean plane. It
is in this sense that Euclidean geometry is a limiting case of hyperbolic geometry.
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CHAPTER 13. THE PSEUDOSPHERE
Spring 2006
Chapter 14
Hypercycles and Horocycles
There is a curve peculiar to hyperbolic geometry, called the horocycle. Consider two
horoparallel lines, ` and m, with a common direction, say Ω. Let P be a point on one
of these lines P ∈ `. If there exists a point Q ∈ m such that the singly asymptotic triangle,
4P QΩ, has the property that
∠P QΩ ∼
= ∠QP Ω
then we say that Q corresponds to P . If the singly asymptotic triangle 4P QΩ has the above
property we shall say that it is equiangular . Note that it is obvious from the definition that
if Q corresponds to P , then P corresponds to Q. The points P and Q are called a pair of
corresponding points.
Theorem 14.1 If points P and Q lie on two limiting parallel lines in the direction of the
ideal point, Ω, they are corresponding points on these lines if and only if the perpendicular
bisector of P Q is limiting parallel to the lines in the direction of Ω.
Proof: Left as an exercise.
Theorem 14.2 Given any two horoparallel lines, there exists a line each of whose points
is equidistant from them. The line is limiting parallel to them in their common direction.
Proof: Let ` and m be horoparallel lines with common direction Ω. Let A ∈ ` and
B ∈ m. The bisector of ∠BAΩ in the singly asymptotic triangle 4ABΩ meets side BΩ
in a point X and the bisector of ∠ABΩ meets side AX of the triangle 4ABX in a point
C. Thus the bisectors of the angles of the singly asymptotic triangle 4ABΩ meet in a
point C. Drop perpendiculars from C to each of ` and m, say P and Q, respectively. By
Hypothesis-Angle 4CAP ∼
= 4CAM (M is the midpoint of AB) and 4CBQ ∼
= 4CBM .
∼
∼
Thus, CP = CM = CQ. Thus, by SAS for singly asymptotic triangles, we have that
4CP Ω ∼
= 4CQΩ
←→
and thus the angles at C are congruent. Now, consider the line CΩ and let F be any point
on it other than C. By SAS we have 4CP F ∼
= 4CQF . If S and T are the feet of F in `
∼
and m, then we get that 4P SF = 4QT F and F S ∼
= F T . Thus, every point on the line
CΩ is equidistant from ` and m.
This line is called the equidistant line.
The following results are left for the reader to prove.
151
152
CHAPTER 14. HYPERCYCLES AND HOROCYCLES
Lemma 14.1 Given any point on one of two horoparallel lines, there is a unique point on
the other which corresponds to it.
Lemma 14.2 If three points P , Q, and R lie on three parallels in the same direction so
that P and Q are corresponding points on their parallels and Q and R are corresponding
points on theirs, then P , Q, and R are noncollinear.
Lemma 14.3 If three points P , Q, and R lie on three parallels in the same direction so
that P and Q are corresponding points on their parallels and Q and R are corresponding
points on theirs, then P and R are corresponding points on their parallels.
Consider any line `, any point P ∈ `, and an ideal point in one direction of `, say Ω. On
each line parallel to ` in the direction Ω there is a unique point Q that corresponds to P .
The set consisting of P and all such points Q is called a horocycle, or, more precisely, the
horocycle determined by `, P , and Ω. The lines parallel to ` in the direction Ω, together
with `, are called the radii of the horocycle. Since ` may be denoted by P Ω, we may regard
the horocycle as determined simply by P and Ω, and hence call it the horocycle through P
with direction Ω, or in symbols, the horocycle (P, Ω).
All the points of this horocycle are mutually corresponding points by Lemma 14.3 , so
the horocycle is equally well determined by any one of them and Ω. In other words if Q is
any point of horocycle (P, Ω) other than P , then horocycle (Q, Ω) is the same as horocycle
(P, Ω). If, however, P 0 is any point of ` other than P , then horocycle (P 0 , Ω) is different
from horocycle (P, Ω), even though they have the same direction and the same radii. Such
horocycles, having the same direction and the same radii, are called codirectional horocycles.
There are analogies between horocycles and circles. We mention a few below. The
corresponding results for circles is mentioned in parentheses.
Lemma 14.4 There is a unique horocycle with a given direction which passes through a
given point.
(There is a unique circle with a given center which passes through a given point.)
Lemma 14.5 Two codirectional horocycles have no common point.
(Two concentric circles have no common point.)
Lemma 14.6 A unique radius is associated with each point of a horocycle.
(A unique radius is associated with each point of a circle.)
A tangent to a horocycle at a point on the horocycle is defined to be the line through
the point which is perpendicular to the radius associated with the point.
No line can meet a horocycle in more than two points. This is a consequence of the
fact that no three points of a horocycle are collinear inasmuch as it is a set of mutually
corresponding points, cf. Lemma 14.2.
Lemma 14.7 The tangent at any point A of a horocycle meets the horocycle only in A.
Every other line through A except the radius meets the horocycle in one further point B.
If α is the acute angle between this line and the radius, then d(A, B) is twice the segment
which corresponds to α as angle of parallelism.
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Proof: Let t be the tangent to the horocycle at A and let Ω be the direction of the
horocycle. If t met the horocycle in another point B, we would have a singly asymptotic
triangle with two right angles, since A and B are corresponding points. In fact the entire
horocycle, except for A, lies on the same side of t, namely, the side containing the ray AΩ.
Let k be any line through A other than the tangent or radius. We need to show that k
meets the horocycle in some other point. Let α be the acute angle between k and the ray
AΩ. Let C be the point of k, on the side of t containing the horocycle, such that AC is a
segment corresponding to α as angle of parallelism. (recall: e−d = tan(α/2)). The line
perpendicular to k at C is then parallel to AΩ in the direction Ω. Let B be the point of
k such that C is the midpoint of AB. The singly asymptotic triangles 4ACΩ and 4BCΩ
are congruent. Hence ∠CBΩ = α, B corresponds to A, and B ∈ (A, Ω).
A chord of a horocycle is a segment joining two points of the horocycle.
Lemma 14.8 The line which bisects a chord of a horocycle at right angles is a radius of
the horocycle.
We can visualize a horocycle in the Poincaré disk model as follows. Let ` be the
diameter of the unit disk whose interior represents the hyperbolic plane, and let O be the
origin. It is a fact that the hyperbolic circle with hyperbolic center P is represented by a
Euclidean circle whose Euclidean center R lies between P and A.
As P recedes from A towards the ideal point Ω, R is pulled up to the Euclidean midpoint
of ΩA, so that the horocycle (A, Ω) is a Euclidean circle tangent to the unit disk at Ω and
tangent to ` at A. It can be shown that all horocycles are represented in the Poincaré model
by Euclidean circles inside the unit disk and tangent to boundary circle. For the Poincaré
upper half plane model, our horocycles will be circles that are tangent to the x-axis.
H2
C1
H1
Figure 14.1: H1 and H2 are horocycles and C1 is a circle in the Poincaré Disk model
Another curve found specifically in the hyperbolic plane and nowhere else is the equidistant curve, or hypercycle. Given a line ` and a point P not on `, consider the set of all
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CHAPTER 14. HYPERCYCLES AND HOROCYCLES
points Q on one side of ` so that the perpendicular distance from Q to ` is the same as the
perpendicular distance from P to `.
The line ` is called the axis, or base line, and the common length of the perpendicular
segments is called the distance. The perpendicular segments defining the hypercycle are
called its radii . The following statements about hypercycles are analogous to statements
about regular Euclidean circles.
1. Hypercycles with equal distances are congruent, those with unequal distances are not.
(Circles with equal radii are congruent, those with unequal radii are not.)
2. A line cannot cut a hypercycle in more than two points.
3. If a line cuts a hypercycle in one point, it will cut it in a second unless it is tangent
to the curve or parallel to it base line.
4. A tangent line to a hypercycle is defined to be the line perpendicular to the radius at
that point. Since the tangent line and the base line have a common perpendicular, they
must be hyperparallel. This perpendicular segment is the shortest distance between
the two lines. Thus, each point on the tangent line must be at a greater perpendicular
distance from the base line than the corresponding point on the hypercycle. Thus,
the hypercycle can intersect the hypercycle in only one point.
5. A line perpendicular to a chord of a hypercycle at its midpoint is a radius and it
bisects the arc subtended by the chord.
6. Two hypercycles intersect in at most two points.
7. No three points of a hypercycle are collinear.
In the Poincaré disk model let Ω and Λ be the ideal end points of `. It can be shown
that the hypercycle to ` through Ω is represented by the arc of the Euclidean circle passing
through A, B, and Ω. This curve is orthogonal to all Poincaré lines perpendicular to the
line `. In the Poincaré upper half plane model, the hypercycle will be represented by an arc
of a Euclidean circle passing through A, B, and Ω.
In the Poincaré disk model, D, a Euclidean circle represents:
(i) a hyperbolic circle if it is entirely inside the unit disk;
(ii) a horocycle if it is inside the unit disk except for one point where it is tangent to the
unit disk;
(iii) a hypercycle if it cuts the unit disk non-orthogonally in two points;
(iv) a hyperbolic line if it cuts the unit disk orthogonally.
A similar situation is true for the Poincaré upper half plane model, H .
It follows that in the hyperbolic plane three non-collinear points lie either on a circle,
a horocycle, or a hypercycle accordingly, as the perpendicular bisectors of the triangle are
concurrent in an ordinary point, an ideal point, or an ultra-ideal point.
MATH 6118-090
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Chapter 15
Hyperbolic Trigonometry
Trigonometry is the study of the relationships among sides and angles of a triangle. In
Euclidean geometry we use similar triangles to define the trigonometric functions—but the
theory of similar triangles in not valid in hyperbolic geometry. Thus, we must them in
terms of their power series expansion for any real number, as in Equations 12.1.
We define the hyperbolic trigonometric functions
ex − e−x
2
ex + e−x
cosh x =
2
sinh x
ex − e−x
tanh x =
= x
cosh x
e + e−x
sinh x =
P
xn
Since ex = ∞
n=0 n! converges for all x ∈ R, the power series expansions of the hyperbolic
trigonometric functions are
∞
X
x2n+1
sinh x =
(2n + 1)!
cosh x =
n=0
∞
X
n=0
x2n
(2n)!
and converge for all real x. In fact, using complex analysis and letting i =
easily see that
³x´
sinh x = −i sin(ix) = i sin
³x´ i
cosh x = cos(ix) = cos
i
There are also the usual collections of hyperbolic trigonometry identities:
cosh2 x − sinh2 x = 1
cosh(x + y) = cosh x cosh y + sinh x sinh y
sinh(x + y) = sinh x cosh y + cosh x sinh y
155
√
−1, we can
156
CHAPTER 15. HYPERBOLIC TRIGONOMETRY
A straightforward calculation using double angle formulas for the circular functions gives
the following formulas:
sin Π(x) = sech x
(15.1)
cos Π(x) = tanh x
(15.2)
tan Π(x) = csch x
(15.3)
For example, to derive the first equation:
sin Π(x) = sin(2 arctan e−x )
= 2 sin(arctan e−x ) cos(arctan e−x )
µ
¶
µ
¶
e−x
1
=2
=2 x
1 + e−2x
e + e−x
1
=
= sech x.
cosh x
This function Π : (0, π2 ) → H2 provides a connection between the hyperbolic and circular
functions.
Given 4ABC, let a = dp (B, C), b = dp (A, C), and c = dp (A, B). First, we will derive
some formulæof hyperbolic geometry. Let B ∈ H2 . Let x = dp (O, B) be the Poincaré
distance from O to B and let t = d(O, B) be the Euclidean distance from O to B. From
Lemma 12.11 we have
1+t
.
ex =
1−t
We then get that
2t
1 − t2
1 + t2
cosh x =
1 − t2
2t
.
tanh x =
1 + t2
sinh x =
Theorem 15.1 Given any right triangle 4ABC with ∠C the right angle (having measure
π/2, then
sinh a
tanh b
cos A =
sinh c
tanh c
cosh c = cosh a cosh b = cot A cot B
cos A
cosh a =
sin B
sin A =
©
©©
c ©
©
©©
©©
A
©
b
(15.4)
(15.5)
(15.6)
B
a
π/2
C
Before we prove these equations, compare them with the formulæ for a right triangle in
Euclidean geometry.
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1. Equation 15.5 is the hyperbolic analogue of the Pythagorean theorem.
cosh c = cosh a cosh b
µ
¶ µ
¶
a2 a4
b2 b4
1+
+
+ ··· = 1 +
+
+ ··· · 1 +
+
+ ···
2!
4!
2!
4!
2!
4!
µ 2
¶
b2
a
+
+ ···
=1+
2!
2!
c2
c4
Thus, if 4ABC is sufficiently small so that higher powers of a, b, and c are negligible,
then
1+
c2
1
≈ 1 + (a2 + b2 )
2
2
2
2
c ≈ a + b2
2. Equation 15.4 says that for 4ABC sufficiently small sin A ≈ ac and cos A ≈ cb . How
close are these approximations? Consider right triangles with ∠A fixed and let c
approach 0. Since a < c, a → 0.
1
1 1
1
1
≈
= (1 − u + u2 − u3 + . . .)
3 =
c
sinh x
c1+u
c
c + 3!
where lim = 0. Thus,
c→0
sinh a
a
a2 a4
= (1 +
+
+ · · · )(1 − u + u2 − u3 + · · · )
sinh c
c
3!
5!
a
≈
c
Thus,
lim
c→0
a
sinh a
= lim
= sin A.
c c→0 sinh c
3. Equation 15.6 and the second equality in Equation 15.5 have no Euclidean parallels
for there the angles do not determine the lengths of the sides.
First recall that a point P = iep in H is a distance p from i. Also, remember that
·
¸
1 −i
φ=
−i 1
sends H to D. In this remember that φ(i) = 0, and note that
φ(iep ) =
iep − i
= i tanh(p/2).
ep + 1
Thus, a point that is a hyperbolic distance p away from zero in D is a Euclidean distance
tanh(p/2) away from zero.
Thus, we may choose our right triangle 4ABC with right angle C and sides of length
|AC| = b and |BC| = a to be the triangle in the Poincaré disk model D with vertices C = 0,
A = tanh(b/2) and B = i tanh(a/2).
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CHAPTER 15. HYPERBOLIC TRIGONOMETRY
B
B
A
C
C
A
Figure 15.1: Right triangles in D
However, for computation purposes, it turns out that we will be better off by making
A = 0 as in the right hand figure above. To do this we need to find a direct isometry γ
of D which sends A to 0 and the line AC to itself. So, we want γ(A) = 0, γ(1) = 1, and
γ(−1) = −1. Then
(z, A; 1, −1) = (γz, 0; 1, −1)
from which it follows that
·
¸
−1 A
γ=
A −1
Now, apply γ to B.
γ(B) =
−B + A
−i tanh(a/2) + tanh(b/2)
=
.
AB − 1
i tanh(a/2) tanh(b/2) − 1
Thus, for our proof we may use the triangle 4ABC in D with
A=0
−i tanh(a/2) + tanh(b/2)
B=
i tanh(a/2) tanh(b/2) − 1
C = − tanh(b/2).
Proof: To prove the Hyperbolic Pythagorean Theorem, we may assume that A = O is the
origin. Since B is a distance tanh(c/2) away from zero, we get
¯
¯
¯ −B + A ¯
¯
¯
tanh(c/2) = ¯
AB − 1 ¯
tanh2 (c/2) =
MATH 6118-090
tanh2 (a/2) + tanh2 (b/2)
tanh2 (a/2) tanh2 (b/2) + 1
Spring 2006
159
From our hyperbolic trigonometry identities, you can easily show that sech2 x = 1 −
tanh2 x, so
sech2 (c/2) = 1 − tanh2 (c/2)
tanh2 (a/2) tanh2 (b/2) − tanh2 (a/2) − tanh2 (b/2) + 1
tanh2 (a/2) tanh2 (b/2) + 1
(tanh2 (a/2) − 1)(tanh2 (b/2) − 1)
=
tanh2 (a/2) tanh2 (b/2) + 1
(tanh2 (a/2) tanh2 (b/2) + 1)
cosh2 (c/2) =
sech2 (a/2)sech2 (b/2)
= (tanh2 (a/2) tanh2 (b/2) + 1) cosh2 (a/2) cosh2 (b/2)
=
= sinh2 (a/2) sinh2 (b/2) + cosh2 (a/2) cosh2 (b/2)
= 2 cosh2 (a/2) cosh2 (b/2) − cosh2 (a/2) − cosh2 (b/2) + 1.
Again, using the hyperbolic trigonometry identity cosh 2x = 2 cosh2 x − 1, we have
cosh c = 2 cosh2 (c/2) − 1
= 4 cosh2 (a/2) cosh2 (b/2) − 2 cosh2 (a/2) − 2 cosh( b/2) + 1
= (2 cosh2 (a/2) − 1)(2 cosh2 (b/2) − 1)
= cosh a cosh b
Since A sits at the origin and angles are measured with a Euclidean protractor, we can
find the angle at A using regular trigonometry. We just need to know what the lengths of
the sides are
First, we need to find the coordinates for B. Rationalize the denominator for B.
B=
(tanh2 (a/2) + 1) tanh(b/2) + i tanh(a/2)(tanh2 (b/2) − 1)
tanh2 (a/2) tanh2 (b/2) + 1
Note that
tanh2 (b/2) − 1 = −sech2 (b/2)
tanh2 (a/2) + 1 =
sinh2 (a/2) + cosh2 (a/2)
cosh a
=
2
cosh (a/2)
cosh2 (a/2)
Thus,
(cosh a tanh(b/2) cosh2 (b/2) − i tanh(a/2) cosh2 (a/2)
sinh2 (a/2) sinh2 (b/2) + cosh2 (a/2) cosh2 (b/2)
cosh a sinh b − i sinh a
=
.
2
2(sinh (a/2) sinh2 (b/2) + cosh2 (a/2) cosh2 (b/2)
B=
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CHAPTER 15. HYPERBOLIC TRIGONOMETRY
Thus,
cos A =
Bx
|B|
cosh a sinh b
2(sinh (a/2) sinh (b/2) + cosh2 (a/2) cosh2 (b/2))
tanh2 (c/2)
s
cosh a sinh b
tanh2 (a/2) tanh2 (b/2) + 1
2
2
2
2
2(sinh (a/2) sinh (b/2) + cosh (a/2) cosh (b/2))
tanh2 (a/2) + tanh2 (b/2)
s
cosh a sinh b
sinh2 (a/2) cosh2 (b/2) + cosh2 (a/2) sinh2 (b/2)
2(sinh2 (a/2) sinh2 (b/2) + cosh2 (a/2) cosh2 (b/2)) sinh2 (a/2) sinh2 (b/2) + cosh2 (a/2) cosh2 (b/2)
cosh a sinh b
q
2
2
2
(sinh (a/2) sinh (b/2) + cosh (a/2) cosh2 (b/2))(sinh2 (a/2) cosh2 (b/2) + cosh2 (a/2) sinh2 (b/2))
2
=
=
=
=
2
2 cosh a sinh b
=p
[(cosh a − 1)(cosh b − 1) + (cosh a + 1)(cosh b + 1)][(cosh a − 1)(cosh b + 1) + (cosh b − 1)(cosh a + 1)]
cosh a sinh b
=p
(cosh a cosh b + 1)(cosh a cosh b − 1)
cosh a sinh b
=p
cosh2 a cosh2 b − 1
Likewise,
sinh a
sin A = p
2
cosh a cosh2 b − 1
Now,
cosh2 a cosh2 b − 1 = cosh2 c − 1
= sinh2 c
since cosh a cosh b = cosh c by the Hyperbolic Pythagorean Theorem. Thus,
cos A =
=
=
=
cos A =
sin A =
MATH 6118-090
cosh a sinh b
sinh c
cosh a sinh b
cosh b
sinh c
cosh b
tanh b cosh a cosh b
sinh c
tanh b cosh c
sinh c
tanh b
tanh c
sinh a
.
sinh c
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Theorem 15.2 For any triangle 4ABC in the hyperbolic plane
cosh c = cosh a cosh b − sinh a sinh b cos C
(15.7)
Hyperbolic Law of Cosines
sinh b
sinh c
sinh a
=
=
sin A
sin B
sin C
Hyperbolic Law of Sines
(15.8)
Theorem 15.3 (Hyperbolic Law of Cosines for Angles) Let 4ABC be a triangle in
H2 with sides a, b, and c opposite the angles A, B, and C. Then
cos C = − cos A cos B + sin A sin B cosh c.
Note that this theorem is the tool that allows us to solve for the sides of a triangle given
the three angles (AAA).
Since the area of a triangle is determined by its angles, and since the sides of a triangle
determine the angles, there must be a formula for the area of a triangle in terms of its sides:
a Heron’s Formula for Hyperbolic Geometry.
Theorem 15.4 (Heron’s Formula for Hyperbolic Geometry) Let 4ABC be a triangle in H2 with sides a, b, and c opposite the angles A, B, and C. Let
s=
a+b+c
2
be the semiperimeter. Let K = |4ABC| = area of 4ABC. Then
1 − cos(K) =
4 sinh s sinh(s − a) sinh(s − b) sinh(s − c)
.
(1 + cosh a)(1 + cosh b)(1 + cosh c)
Let us take a look at a specific example. Consider equilateral triangles. In Euclidean
geometry all are similar, since they all must have angles measuring 60◦ . If this were true in
hyperbolic geometry, they would have to be congruent by AAA. What then are the angles
in an equilateral triangle of differing sides? Look at the following table and see if you can
tell what is happening.
Sides
10
5
2.5
1.5
1.0
0.5
0.1
Radians
0.0135
0.1633
0.5359
0.7930
0.9188
1.0122
1.0458
Degrees
0.77
9.35
30.71
45.43
52.64
57.99
59.92
Table 15.1: Hyperbolic Equilateral Triangles
Other interesting examples would be how the angles in a right isosceles triangle vary
with the sides and what triangle is analogous to the standard 30-60-90 triangle in Euclidean
geometry.
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15.1
CHAPTER 15. HYPERBOLIC TRIGONOMETRY
Circumference and Area of a Circle
Theorem 15.5 The circumference C of a circle of radius r is
C = 2π sinh r.
π
n
r
a
p/2n
Figure 15.2: Hyperbolic Circle
In Euclidean geometry C = lim pn where pn is the perimeter of the regular n-gon
n→∞
inscribed in the circle.
·
¸
π
π
1 ³ π ´3
1 ³ π ´5
pn = 2nr sin = 2nr
−
+
− ···
n
n 3! n
5! n
· ³ ´
¸
1 π 2
1 ³ π ´4
= 2πr − 2πr
+
− ···
3! n
5! n
Thus, C = lim pn = 2πr.
n→∞
In hyperbolic geometry we can still compute the perimeter and compute the limit, but
we will use Theorem 15.1 to compute the perimeter.
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163
The proof is nothing but the following computation.
p
sinh 2n
π
sin
=
2n
sinh r
π
p
= sinh r sin
sinh
2n
 n
¡
¢
¡ π ¢2j+1 
2j+1
∞
∞
p
X
X
2n

= sinh r  (−1)j n
(2j + 1)!
(2j + 1)!
j=0
j=0



¡ π ¢2j 
∞ ¡ p ¢2j
∞
X
X
p 
2n
 = π sinh r  (−1)j n

2n
(2j + 1)!
n
(2j + 1)!
j=0
j=0
¶
µ
¶
µ
³
´
1 ³ π ´2
1 p 2
+ · · · = 2π sinh r 1 −
+ ···
p 1+
3! 2n
3! n
C = lim p = 2π sinh r.
n→∞
Let K be the area of 4ABC, so K = π − ∠A − ∠B − ∠C. Let 4ABC have a right
angle at C, then K = π2 − A − B.
Theorem 15.6 tan K/2 = tanh a/2 tanh b/2.
Once again the proof is a computation.
tanh2
a
b
cosh a − 1 cosh b − 1
tanh2 =
·
2
2
cosh a + 1 cosh b + 1
cos A − sin B cos B − sin A
=
·
cos A + sin B cos B + sin A
1 − sin(A + B) cos(A − B)
=
1 + sin(A + B) cos(A − B)
1 − cos K
K
=
= tan2 .
1 + cos K
2
Using this and our limiting approach we can now compute the area of a circle.
Theorem 15.7 The area, A, of a circle of radius r is
r
A = 4π sinh2 .
2
Proof: We do this just as before. If Kn is the area of the inscribed regular n-gon, then
A = lim Kn . In the right triangle in Figure 15.2 let K, a, and p denote Kn , an and pn .
n→∞
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CHAPTER 15. HYPERBOLIC TRIGONOMETRY
The area of the right triangle is Kn /2n.
K
a
p
= tanh tanh
4n
2
4n
K
a
p
4n tan
= (tanh ) 4n tanh
4n
2
4n
Now,
#
"
µ ¶
µ ¶
2 K 5
K
K
1 K 3
+
+ ···
4n tan
= 4n
+
4n
4n 3 4n
15 4n
µ ¶
µ ¶
1 K 2
2 K 4
= K[1 +
+
+ ···]
3 4n
15 4n
and
·
¸
p
p
1 ³ p ´3
2 ³ p ´5
4n tanh
= 4n
−
+
+ ···
4n
4n 3 4n
15 4n
2 ³ p ´4
1 ³ p ´2
−
+ ···]
= p[1 +
3 4n
15 4n
tan
Thus, we find that
K
= lim K = A
4n n→∞
p
lim 4n tanh
= lim p = C
n→∞
4n n→∞
lim a = r
lim 4n tan
n→∞
n→∞
Putting all of this together we have that
r
2
r
= 2π sinh r tanh
2
cosh r − 1
= 2π sinh r
sinh r
= 2π(cosh r − 1)
r
= 4π sinh2 .
2
A = C tanh
Thus, we have computed the area of a circle.
MATH 6118-090
Spring 2006
Chapter 16
Hyperbolic Analytic Geometry
16.1
Saccheri Quadrilaterals
Recall the results on Saccheri quadrilaterals from our previous chapter. Also recall that a
convex quadrilateral three of whose angles are right angles is called a Lambert quadrilateral.
D
C
N
A
M
B
• In a Saccheri quadrilateral the summit angles are congruent, and
• In a Saccheri quadrilateral the line joining the midpoints of the base and the summit—
called the altitude—is perpendicular to both.
• In a Saccheri quadrilateral the summit angles are acute.
• The fourth angle of a Lambert quadrilateral is acute.
• The side adjacent to the acute angle of a Lambert quadrilateral is greater than its
opposite side.
• In a Saccheri quadrilateral the summit is greater than the base and the sides are
greater than the altitude.
16.2
More on Quadrilaterals
Now we need to consider a Saccheri quadrilateral which has base b, sides each with length
a, and summit with length c. We know that c > a, but we would like to know
165
166
CHAPTER 16. HYPERBOLIC ANALYTIC GEOMETRY
• How much bigger?
• How are the relative sizes related to the lengths of the sides?
Theorem 16.1 For a Saccheri quadrilateral
sinh
c
b
= (cosh a) · (sinh ).
2
2
A'
B'
c
a
d
θ
b
A
a
B
Figure 16.1: Saccheri Quadrilateral
Proof: Compare Figure 16.1. Applying the Hyperbolic Law of Cosines from Theorem 15.2,
we have
cosh c = cosh a cosh d − sinh a sinh d cos θ.
(16.1)
From Theorem 15.1 we know that
sinh a
π
− θ) =
2
sinh d
cosh d = cosh a cosh b
cos(θ) = sin(
Using these in Equation 16.1 we eliminate the variable d and have
cosh c = cosh2 a cosh b − sinh2 a
= cosh2 a(cosh b − 1) + 1
Now, we need to apply the identity
x
2 sinh2 ( ) = cosh x − 1,
2
and we have the formula.
Corollary 3 Given a Lambert quadrilateral, if c is the length of a side adjacent to the acute
angle, a is the length of the other side adjacent to the acute angle, and b is the length of the
opposite side, then
sinh c = cosh a sinh b.
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167
Two segments are said to be complementary segments if their lengths x and x∗ are
related by the equation
π
Π(x) + Π(x∗ ) = .
2
The geometric meaning of this equation is shown in the following figure, Figure 16.2. These
lengths then are complementary if the angles of parallelism associated to the segments are
complementary angles. This is then an “ideal Lambert quadrilateral” with the fourth vertex
an ideal point Ω.
Ω
x
Π(x)
Π(x * )
x*
Figure 16.2: Complementary Segments
If we apply the earlier formulas for the angle of parallelism to these segments, we get
sinh x∗ = csch x
cosh x∗ = coth x
tanh x∗ = sech x
x∗
tanh
= e−x .
2
Theorem 16.2 (Engel’s Theorem) There is a right triangle with sides and angles as
shown in Figure 16.3 if and only if there is a Lambert quadrilateral with sides as shown
is Figure 16.3. Note that P Q is a complementary segment to the segment whose angle of
parallelism is ∠A.
16.3
Coordinate Geometry in the Hyperbolic Plane
In the hyperbolic plane choose a point O for the origin and choose two perpendicular lines
through O—OX and OY . In the Poincaré disk model we will use the Euclidean center of
our defining circle for this point O. We need to fix coordinate systems on each of these two
perpendicular lines. By this we need to choose a positive and a negative direction on each
line and a unit segment for each. There are other coordinate systems that can be used, but
this is standard. We will call these the u-axis and the v-axis. For any point P ∈ H 2 let U
and V be the feet of P on these axes, and let u and v be the respective coordinates of U
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CHAPTER 16. HYPERBOLIC ANALYTIC GEOMETRY
A
R
c
Q
Π(l)=α
b
c
m
l*
Π(m)=β
B
a
C
P
b
S
Figure 16.3: Engel’s Theorem
and V . Then the quadrilateral 2U OV P is a Lambert quadrilateral. If we label the length
of U P as w and that of V P as z, then by the Corollary to Theorem 16.1 we have
tanh w = tanh v · cosh u
tanh z = tanh u · cosh v
Let r = dh (OP ) be the hyperbolic distance from O to P and let θ be a real number so
that −π < θ < π. Then
tanh u = cos θ · tanh r
tanh v = sin θ · tanh r.
We also set
x = tanh u,
y = tanh v
T = cosh u cosh w, X = xT, Y = yT.
The ordered pair {OX, OY } is called a frame with axes OX and OY . With respect
to this frame, we say the point P has
• axial coordinates (u, v),
• polar coordinates (r, θ),
• Lobachevskii coordinates (u, w),
• Beltrami coordinates (x, y),
• Weierstrass coordinates (T, X, Y ).
If a point has Beltrami coordinates (x, y) and t = 1 +
p = x/t
p
1 − x2 − y 2 , put
q = y/t,
then (p, q) are the Poincaré coordinates of the point.
In Figure 16.5 we have:
u = 0.78
v = 0.51
w = 0.72
z = 0.94
r = 1.10
θ = 35.67◦ = 0.622radians
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16.3. COORDINATE GEOMETRY IN THE HYPERBOLIC PLANE
169
P
z
V
r
v
w
θ
O
U
u
Figure 16.4: Coordinates in Poincaré Plane
From which it follows that
x = tanh u = 0.653,
y = tanh v = 0.470
T = cosh u cosh w = 1.677
Y = yT = 0.788
p = x/t = 0.409
X = xT = 1.095
p
t = 1 + 1 − x2 − y 2 = 1.594
q = y/t = 0.295.
Thus the coordinates for P are:
• axial coordinates (u, v) = (0.78, 0.51),
• polar coordinates (r, θ) = (1.10, 0.622),
• Lobachevskii coordinates (u, w) = (0.78, 0.72),
• Beltrami coordinates (x, y) = (0.653, 0.470),
• Weierstrass coordinates (T, X, Y ) = (1.677, 1.095, 0.788).
• Poincaré coordinates (p, q) = (0.409, 0.295)
Every point has a unique ordered pair of Lobachevskii coordinates, and, conversely,
every ordered pair of real numbers is the pair of Lobachevskii coordinates for some unique
point. In Lobachevskii coordinates
1. for a 6= 0, u = a is the equation of a line;
2. for a 6= 0, w = a is the equation of a hypercycle;
3. e−u = tanh w is an equation of the line in the first quadrant that is horoparallel to
both axes.
−−→
4. eu = cosh w is an equation of the horocycle with radius OX.
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CHAPTER 16. HYPERBOLIC ANALYTIC GEOMETRY
V
v
O
z
r
θ
u
P
w
U
Figure 16.5: P in the Poincaré Disk
Thus, a line does not have a linear equation in Lobachevskii coordinates, and a linear
equation does not necessarily describe a line.
Every point has a unique ordered pair of axial coordinates. However, not every ordered
pair of real numbers is a pair of axial coordinates. Let U and V be points on the axes with
V 6= 0. Now the perpendiculars at U and V do not have to intersect. It is easy to see that
they might be horoparallel or hyperparallel, especially by looking in the Poincaré model.
If the two lines are limiting parallel (horoparallel) then that would make the segments OU
and OV complementary segments. It can be shown then that these perpendiculars to the
axes at U and V will intersect if and only if |u| < |v|∗ . It then can be shown that (u, v) are
the axial coordinates of a point if and only if tanh2 u + tanh2 v < 1.
Lemma 16.1 With respect to a given frame
i) Every point has a unique ordered pair of Beltrami coordinates, and (x, y) is an
ordered pair of Beltrami coordinates if and only if x2 + y 2 < 1.
ii) If the point P1 has Beltrami coordinates (x1 , y1 ) and point P2 has Beltrami
coordinates (x2 , y2 ), then the distance dh (P1 P2 ) = P1 P2 is given by the following
formulæ:
1 − x1 x2 − y1 y2
p
cosh P1 P2 = p
1 − x21 − y12 1 − x22 − y22
p
(x2 − x1 )2 + (y2 − y1 )2 − (x1 y2 − x2 y1 )2
tanh P1 P2 =
1 − x1 x2 − y1 y2
iii) Ax + By + C = 0 is an equation of a line in Beltrami coordinates if and only
if A2 + B 2 > C 2 , and every line has such an equation.
iv) Given an angle ∠P QR and given that the Beltrami coordinates of P are
(x1 , y1 ), of Q are (x2 , y2 ), and of R are (x3 , y3 ), then the cosine of this angle is
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171
given by
cos(∠P QR) =
(x2 − x1 )(x3 − x1 ) + (y2 − y1 )(y3 − y1 ) − (x2 y1 − x1 y2 )(x3 y1 − x1 y3 )
p
.
(x2 − x1 )2 + (y2 − y1 )2 − (x1 y2 − x2 y1 )2 (x3 − x1 )2 + (y3 − y1 )2 − (x1 y3 − x3 y1 )2
p
v) If Ax + By + C = 0 and Dx + Ey + F = 0 are equations of two intersecting
line in Beltrami coordinates and θ is the angle formed by their intersection, then
AD + BE − CF
√
.
cos θ = ± √
2
A + B 2 − C 2 D2 + E 2 − F 2
In particular the lines are perpendicular if and only if AD + BE = CF .
vi) If (x
p1 , y1 ) and (x2 , y2 ) are the
p Beltrami coordinates of two distinct points, let
t1 = 1 − x21 − y12 and t2 = 1 − x22 − y22 . Then the midpoint of the segment
joining the two points has Beltrami coordinates
¶
µ
x1 t2 + x2 t1 y1 t2 + y2 t1
,
t1 + t2
t1 + t2
and the perpendicular bisector of the two points has an equation
(x1 t2 − x2 t1 )x + (y1 t2 − y2 t1 )y + (t1 − t2 ) = 0.
vii) If A1 x + B1 y + C1 = 0 and A2 x + B2 y + C2 = 0 are equations of lines in
Beltrami coordinates and if A1 B2 = A2 B1 , then the two lines are hyperparallel.
viii) Every cycle has an equation in Beltrami coordinates that is of the form
p
1 − x2 − y 2 = ax + by + c.
1. The cycle is a circle if and only if −1 < a2 + b2 − c2 < 0 and c > 0.
2. The cycle is a horocycle if and only if a2 + b2 − c2 = 0 and c > 0.
3. The cycle is a hypercycle if and only if a2 + b2 − c2 > 0.
In Poincaré coordinates (p, q)
C((p2 + q 2 ) + 2Ap + 2Bq + C = 0
is an equation of a line if and only if A2 + B 2 > C 2 , and every line has such an equation.
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CHAPTER 16. HYPERBOLIC ANALYTIC GEOMETRY
The following is a representation of graph paper in the Poincaré disk model. Each line
is 1/4 unit apart. The distances are measured along the u and v axes.
Figure 16.6: Hyperbolic Graph Paper
MATH 6118-090
Spring 2006
Appendix A
Inversion in Euclidean Circles
In order to define congruence in the Poincaré model and verify the axioms of congruence,
we must study the operation of inversion in a Euclidean circle. This operation is the process
of reflecting across a line in the Poincaré disk model of the hyperbolic plane.
Let γ be a circle of radius r and center O. For any point P 6= O the inverse P 0 of P
−−→
with respect to the circle γ is the unique point P 0 ∈ OP so that OP · OP 0 = r2 . In this
case γ is called the circle of inversion.
These are Euclidean geometry theorems about circles, so we may use the facts that we
know about Euclidean geometry.
Proposition A.1 For P, P 0 and γ as above,
i) P = P 0 if and only if P ∈ γ.
ii) If P ∈ int γ then P 0 6∈ int γ, and vice versa.
iii) (P 0 )0 = P .
←→
Proposition A.2 Let P ∈ int γ and let T U be the chord through P perpendicular to OP .
Then P 0 = P (T U ), the pole of T U , i.e., the intersection of the tangents to γ at T and U .
T
O
P
P'
U
−−→
Proof: Suppose the tangent to γ at T cuts OP at the point P 0 . The right triangle 4OP T
is similar to right triangle OT P 0 (since they have ∠T OP in common and the angle sum is
180◦ ). Hence corresponding sides are proportional. Since `(OT ) = r, we get that
`(OP )
r
=
,
r
`(OP 0 )
173
174
APPENDIX A. INVERSION IN EUCLIDEAN CIRCLES
←→
which shows that P 0 is the inverse to P . Reflect across the line OP , and we see that the
tangent to γ at U also passes through P 0 , so that P 0 is the pole of T U .
Proposition A.3 If P is outside of γ, let Q be the midpoint of OP . Let σ be the circle of
radius QP centered at Q. Then
i) σ ∩ γ = {T, U }.
←→
←→
ii) P T and P U are tangent to γ.
iii) P 0 = T U ∩ OP .
T
O
P'
Q
P
U
Proof: By the circular continuity principle, σ and γ do intersect in two points T and U .
Since
∠OT
P and ∠OU P are inscribed in semicircles of σ, they are right angles. Therefore
←→
←→
P T and P U are tangent to γ. If T U intersects OP in a point P 0 , then P is the inverse of
P 0 , by the previous proposition. Thus, P 0 is the inverse of P in γ.
The next proposition shows how to construct the Poincaré line joining two ideal points—
the line of enclosure. Its proof shows that ∠OP 0 T in the previous figure is indeed a right
angle, as we needed.
Proposition A.4 Let T and U be points on γ that are not contained on a the same diameter, and let P be the pole of T U . Then
i) P T ∼
= PU,
ii) ∠P T U ∼
= ∠P U T ,
←→
←→
iii) OP ⊥ T U , and
iv) the circle δ with center P and radius P T cuts γ orthogonally at T and U .
Proof: By definition of pole, ∠OT P and ∠OU P are right angles, so by the HypotenuseLeg criterion, 4OT P ∼
= 4OU P . Therefore, P T ∼
= P U and ∠OP T ∼
= ∠OP U . The base
angles ∠P T U and ∠P U T of the isosceles triangle 4T P U are thus congruent, and the angle
−−→
bisector P O is perpendicular to the base T U . The circle δ is then
because
←→ well-defined
←→
PU ∼
= P T and δ intersects γ orthogonally by the hypothesis that P T and P U are tangent
to γ.
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175
T
O
P
U
Let P be a point in the plane and γ a circle with center O. The power of P with respect
to γ is defined to be
P w(P ) = (`(OP ))2 − r2 .
Proposition A.5 Assume P 6∈ γ and assume that two lines through P intersect γ in points
{R1 , R2 } and {S1 , S2 }. Then
i) (P R1 ) · (P R2 ) = (P S1 ) · (P S2 ) = P w(P ).
ii) If one of these lines through P is tangent to γ at T , then P w(P ) = P T 2 .
P2
Q1
P2
P1
O
O
Q1
Q2
Q2
P1
Proposition A.6 Let P 6∈ γ and P 6= O, the center of γ. Let δ be a circle through P . δ is
orthogonal to γ if and only if δ passes through P 0 , the inverse of P with respect to γ.
Proposition A.7 Let γ have radius r, δ have radius t and let P be the center of δ. δ is
orthogonal to γ if and only if P w(P ) = t2 , where the power is computed with respect to γ.
Let O be a point and k > 0. The dilation with center O and ratio k is a mapping of
−−→
the Euclidean plane that fixes O and maps every point P 6= O to a unique P ∗ ∈ OP such
that OP ∗ = kOP . Call this map DO,k .
Proposition A.8 Let δ be a circle with center C 6= O and radius s. DO,α maps δ to a
circle δ ∗ with center C ∗ = DO,α (C) and radius αs. If Q ∈ δ, the tangent to δ ∗ at Q∗ is
parallel to the tangent to δ at Q.
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APPENDIX A. INVERSION IN EUCLIDEAN CIRCLES
Proof: Choose rectangular coordinates so that O is the origin. Then the dilation is given
by the mapping (x, y) 7→ (αx, αy). The image of the line have equation ax + by = c is the
line
←→ax + by = αc, so the image is parallel to the original line. In particular,
←→ having equation
CQ is parallel to C ∗ Q∗ , and their perpendiculars at Q and Q∗ , respectively, are also parallel.
If δ has equation (x−h)2 +(y −k)2 = s2 , then δ ∗ has equation (x−αh)2 +(y −αk)2 = (αs)2 .
Proposition A.9 Let γ be the circle of radius r centered at O and let δ be the circle of
radius s centered at C. Assume O lies outside δ; let p = P w(O) with respect to δ and let
α = r2 /p. The image δ 0 of δ under inversion in γ is the circle of radius αs whose center is
C ∗ = DO,α (C). If P ∈ δ and P 0 is the inverse of P with respect to γ, then the tangent t0 to
δ 0 at P 0 is the reflection across the perpendicular bisector of P P 0 of the tangent to δ at P .
P
R
t
S
Q
C
t'
P'
C*
O
Corollary 1 δ is orthogonal to γ if and only if δ is mapped to itself by inversion in γ.
Proposition A.10 Let ` be a line so that O 6∈ `. The image of ` under inversion by γ is
a punctured circle with missing point O. The diameter through O of the completed circle δ
is perpendicular to `.
Proposition A.11 Let δ be a circle passing through the center O of γ. The image of δ
minus O under inversion in γ is a line ` so that O 6∈ ` and ` is parallel to the tangent to δ
at O.
Proposition A.12 A directed angle of intersection of two circles is preserved in magnitude
by an inversion. The same applies to the angle of intersection of a circle and a line or the
intersection of two lines.
O
P'
P
A'
A
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Proposition A.13 Let δ be orthogonal to Γ. Inversion in δ maps Γ onto Γ and the interior
of Γ onto itself. Inversion in δ preserves incidence, betweenness and congruence in the sense
of the Poincaré disk model.
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APPENDIX A. INVERSION IN EUCLIDEAN CIRCLES
Spring 2006
Appendix B
An Overview of Babylonian
Mathematics
Article by J. J. O’Connor and E. F. Robertson
December 2000
MacTutor History of Mathematics
B.1
Introduction to Babylonian Mathematics
The Babylonians lived in Mesopotamia, a fertile crescent between the Tigris and Euphrates
rivers. Here is a map of the region where the civilization flourished.
The region had been the center of the Sumerian civilization which flourished before 3500 BC. This was an advanced civilization building cities and supporting the people with irrigation systems, a legal system, administration,
and even a postal service. Writing developed and counting was based on a sexagesimal system, that is base 60.
Around 2300 BC the Akkadians invaded the area and for
some time the less civilized culture of the Akkadians mixed
with the more advanced culture of the Sumerians. The
Akkadians invented the abacus as a tool for counting and
they developed somewhat clumsy methods of arithmetic
with addition, subtraction, multiplication and division all playing a part. The Sumerians,
however, revolted against Akkadian rule and by 2100 BC they were back in control.
The Babylonian civilization, whose mathematics is the subject of this article, replaced
that of the Sumerians starting around 2000 BC. The Babylonians were a Semitic people
who invaded Mesopotamia, defeated the Sumerians and by about 1900 BC established their
capital at Babylon.
The Sumerians had developed an abstract form of writing based on cuneiform (i.e.
wedge-shaped) symbols. Their symbols were written on wet clay tablets which were baked
in the hot sun. Many thousands of these tablets have survived to this day. It was the use of
a stylus on a clay medium that led to the use of cuneiform symbols since curved lines could
not easily be drawn. The later Babylonians adopted the same style of cuneiform writing on
clay tablets. A picture of one of their tablets Babylonian mathematics is to the right.
179
180
APPENDIX B. AN OVERVIEW OF BABYLONIAN MATHEMATICS
Many of the tablets concern topics which are fascinating, although they do not contain deep mathematics. For
example we mentioned above the irrigation systems of the
early civilizations in Mesopotamia. Muroi1 writes:
It was an important task for the rulers of Mesopotamia
to dig canals and to maintain them, because canals were not
only necessary for irrigation but also useful for the transport of goods and armies. The rulers or high government
officials must have ordered Babylonian mathematicians to
calculate the number of workers and days necessary for the
building of a canal, and to calculate the total expenses of
wages of the workers.
There are several Old Babylonian mathematical texts in
which various quantities concerning the digging of a canal
are asked for. They are YBC 4666, 7164, and VAT 7528,
all of which are written in Sumerian ..., and YBC 9874 and BM 85196, No. 15, which
are written in Akkadian ... . From the mathematical point of view these problems are
comparatively simple ...
The Babylonians had an advanced number system, in some ways more advanced than
our present systems. It was a positional system with a base of 60 rather than the system
with base 10.
The Babylonians divided the day into 24 hours, each hour into 60 minutes, each minute
into 60 seconds. This form of counting has survived for 4000 years. To write 5h 25’ 30”,
i.e. 5 hours, 25 minutes, 30 seconds, is just to write the sexagesimal fraction, 5 25/60
30/3600. We adopt the notation 5;25,30 for this sexagesimal number. As a base 10 fraction
the sexagesimal number 5;25,30 is 5 4/10 2/100 5/1000 which is written as 5.425 in decimal
notation.
Perhaps the most amazing aspect of the Babylonian’s calculating skills was their construction of tables to aid calculation. Two tablets found at Senkerah on the Euphrates
in 1854 date from 2000 BC. They give squares of the numbers up to 59 and cubes of the
numbers up to 32. The table gives 82 = 1, 4 which stands for
82 = 1, 4 = 1 × 60 + 4 = 64
and so on up to 592 = 58, 1(= 58 × 60 + 1 = 3481). The Babylonians used the formula
ab =
(a + b)2 − a2 − b2
2
to make multiplication easier. Even better is their formula
ab =
(a + b)2 − (a − b)2
4
which shows that a table of squares is all that is necessary to multiply numbers, simply
taking the difference of the two squares that were looked up in the table then taking a
quarter of the answer.
1
K Muroi, Small canal problems of Babylonian mathematics, Historia Sci. (2) 1 (3) (1992), 173-180
MATH 6118-090
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B.1. INTRODUCTION TO BABYLONIAN MATHEMATICS
181
Division is a harder process. The Babylonians did not have an algorithm for long
division. Instead they based their method on the fact that
a
1
=a×
b
b
so all that was necessary was a table of reciprocals. We still have their reciprocal tables
going up to the reciprocals of numbers up to several billion. Of course these tables are
written in their numerals, but using the sexagesimal notation we introduced above, the
beginning of one of their tables would look like:
2
0; 30
1/2 = 0 + 30/60
3
0; 20
1/3 = 0 + 20/60
4
0; 15
1/4 = 0 + 15/60
5
0; 12
1/5 = 0 + 12/60
6
0; 10
1/6 = 0 + 10/60
8
0; 7, 30
1/7 = 0 + 7/60 + 30/3600
9
0; 6, 40
1/9 = 0 + 6/60 + 40/3600
10 0; 6
1/10 = 0 + 6/60
12 0; 5
1/12 = 0 + 5/60
15 0; 4
1/15 = 0 + 4/60
16 0; 3, 45
1/16 = 0 + 3/60 + 45/3600
18 0; 3, 20
1/18 = 0 + 3/60 + 20/3600
20 0; 3
1/20 = 0 + 3/60
24 0; 2, 30
1/24 = 0 + 2/60 + 30/3600
25 0; 2, 24
1/25 = 0 + 2/60 + 24/3600
27 0; 2, 13, 20 1/27 = 0 + 2/60 + 13/3600 + 20/216000
Now the table had gaps in it since 1/7, 1/11, 1/13, etc. are not finite base 60 fractions.
This did not mean that the Babylonians could not compute 1/13. They would write
7
1
1
1
=
=7×
∼7×
13
91
91
90
and these values, for example 1/90, were given in their tables. In fact there are fascinating
glimpses of the Babylonians coming to terms with the fact that division by 7 would lead to
an infinite sexagesimal fraction. A scribe would give a number close to 1/7 and then write
statements such as
... an approximation is given since 7 does not divide.2
Babylonian mathematics went far beyond arithmetical calculations. We now examine
some algebra which the Babylonians developed, particularly problems Babylonian mathematics which led to equations and their solution. The Babylonians were famed as constructors of tables. Now these could be used to solve equations. For example they constructed
tables for n3 + n2 then, with the aid of these tables, certain cubic equations could be solved.
For example, consider the equation
ax3 + bx2 = c.
Note that we are using modern notation, and nothing like this symbolic representation
existed in Babylonian times. The Babylonians could handle numerical examples of such
2
G. G. Joseph, The crest of the peacock (London, 1991)
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APPENDIX B. AN OVERVIEW OF BABYLONIAN MATHEMATICS
equations. They did this by using certain rules, which indicates that they did have the
concept of a typical problem of a given type and a typical method to solve it. For example
in the above case they would (in modern notation) multiply the equation by a2 and divide
it by b3 to get
³ ax ´3 ³ ax ´2 ca2
+
= 3 .
b
b
b
3
2
Setting y = ax/b gives the equation y + y = ca2 /b3 which could now be solved by
looking up the n3 + n2 table for the value of n satisfying n3 + n2 = ca2 /b3 . When a solution
was found for y then x was found by x = by/a. We cannot stress too much that all this was
done without algebraic notation.
Again a table would have been looked up to solve the linear equation ax = b. They
would consult the 1/n table to find 1/a and then multiply the sexagesimal number given in
the table by b. An example of a problem of this type is the following.
Suppose, writes a scribe, 2/3 of 2/3 of a certain quantity of
barley is taken, 100 units of barley are added and the original
quantity recovered.
The problem posed by the scribe is to find the quantity of barley. The solution given
by the scribe is to compute 0; 40 times 0; 40 to get 0; 26, 40. Subtract this from 1; 00 to get
0; 33, 20. Look up the reciprocal of 0; 33, 20 in a table to get 1; 48. Multiply 1; 48 by 1, 40
to get the answer 3, 0.
It is not that easy to understand these calculations by the scribe unless we translate
them into modern algebraic notation. We have to solve
2 2
× x + 100 = x
3 3
which is, as the scribe knew, equivalent to solving
¶
µ
4
x = 100.
1−
9
This is why the scribe computed 23 × 23 subtracted the answer from 1 to get 1 − 94 , then
looked up 1/(1 − 49 ) and so x was found from 1/(1 − 94 ) multiplied by 100 giving 180 (which
is 1; 48 times 1, 40 to get 3, 0 in sexagesimal).
To solve a quadratic equation the Babylonians essentially used the standard formula.
They considered two types of quadratic equation, namely x2 + bx = c and x2 − bx = c where
here b, c were positive but not necessarily integers. The form that their solutions took was,
respectively
!
õ ¶
!
õ ¶
b
b 2
b
b 2
+ c − and
+c + .
x=
2
2
2
2
Notice that in each case this is the positive root from the two roots of the quadratic
and the one which will make sense in solving “real” problems. For example problems which
led the Babylonians to equations of this type often concerned the area of a rectangle. For
example if the area is given and the amount by which the length exceeds the width is given,
then the width satisfies a quadratic equation and then they would apply the first version of
the formula above.
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B.2. PYTHAGORAS’S THEOREM IN BABYLONIAN MATHEMATICS
183
A problem on a tablet from Babylonian times states that the area of a rectangle is 1, 0
and its length exceeds its width by 7. The equation x2 + 7x = 1, 0 is, of course, not given
by the scribe who finds the answer as follows.
Compute half of 7, namely 3; 30, square it to get 12; 15. To
this the scribe adds 1, 0 to get 1; 12, 15. Take its square root
(from a table of squares) to get 8; 30. From this subtract 3; 30
to give the answer 5 for the breadth of the triangle.
Notice ¡that the scribe
has effectively solved an equation of the type x2 + bx = c by
¢
2
using x = (b/2) + c − (b/2). Berriman3 gives 13 typical examples of problems leading to
quadratic equations taken from Old Babylonian tablets.
If problems involving the area of rectangles lead to quadratic equations, then problems
involving the volume of rectangular excavation (a “cellar”) lead to cubic equations. The
clay tablet BM 85200+ containing 36 problems of this type, is the earliest known attempt
to set up and solve cubic equations4 . Of course the Babylonians did not reach a general
formula for solving cubics. This would not be found for well over three thousand years.
B.2
Pythagoras’s theorem in Babylonian mathematics
In this section we examine four Babylonian tablets which all have some connection with
Pythagoras’s theorem. Certainly the Babylonians were familiar with Pythagoras’ theorem.
A translation of a Babylonian tablet which is preserved in the British museum goes as
follows
4 is the length and 5 the diagonal. What is the breadth?
Its size is not known.
4 times 4 is 16.
5 times 5 is 25.
You take 16 from 25 and there remains 9.
What times what shall I take in order to get 9?
3 times 3 is 9.
3 is the breadth.
All the tablets come from roughly the same period, namely that of the Old Babylonian
Empire which flourished in Mesopotamia between 1900 BC and 1600 BC.
The four tablets which interest us here we will call the Yale tablet YBC 7289, Plimpton
322, the Susa tablet, and the Tell Dhibayi tablet. Let us say a little about these tablets
before describing the mathematics which they contain.
The Yale tablet YBC 7289 which we describe is one of a large collection of tablets held
in the Yale Babylonian collection of Yale University. It consists of a tablet on which a
diagram appears. The diagram is a square of side 30 with the diagonals drawn in. The
tablet and its significance was first discussed in Neugebauer5 and in Fowler and Robson6 .
3
A. E. Berriman, The Babylonian quadratic equation, Math. Gaz. 40 (1956), 185-192
J. Hoyrup, The Babylonian cellar text BM 85200+ VAT 6599. Retranslation and analysis, Amphora
(Basel, 1992), 315-358
5
O. Neugebauer and A. Sachs, Mathematical Cuneiform Texts (New Haven, CT, 1945).
6
D Fowler and E Robson, Square root approximations in old Babylonian mathematics: YBC 7289 in
context, Historia Math. 25 (4) (1998), 366-378
4
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APPENDIX B. AN OVERVIEW OF BABYLONIAN MATHEMATICS
Plimpton 322 is the tablet numbered 322 in the collection of G A Plimpton housed in
Columbia University. The top left hand corner of the tablet is damaged as and there is a
large chip out of the tablet around the middle of the right hand side. Its date is not known
accurately but it is put at between 1800 BC and 1650 BC. It is thought to be only part
of a larger tablet, the remainder of which has been destroyed, and at first it was thought
to be a record of commercial transactions. However Neugebauer and Sachs gave a new
interpretation and since then it has been the subject of a huge amount of interest.
The Susa tablet was discovered at the present town of Shush in the Khuzistan region of
Iran. The town is about 350 km from the ancient city of Babylon. W. K. Loftus identified
this as an important archaeological site as early as 1850 but excavations were not carried
out until much later. This particular tablet investigates how to calculate the radius of a
circle through the vertices of an isosceles triangle.
Finally the Tell Dhibayi tablet was one of about 500 tablets found near Baghdad by
archaeologists in 1962. Most relate to the administration of an ancient city which flourished
in the time of Ibalpiel II of Eshunna and date from around 1750 BC. The particular tablet
in which we are interested is not one relating to administration but one which presents a
geometrical problem which asks for the dimensions of a rectangle whose area and diagonal
are known.
Before looking at the mathematics contained in these four tablets we should say a little
about their significance in understanding the scope of Babylonian mathematics. First, you
must be careful not to read into early mathematics any ideas which we can see clearly
today yet which were never in the mind of that author. Conversely, we must be careful
not to underestimate the significance of the mathematics just because it has been produced
by mathematicians who thought very differently from today’s mathematicians. As a final
comment on what these four tablets tell us of Babylonian mathematics we must be careful
to realize that almost all of the mathematical achievements of the Babylonians, even if they
were all recorded on clay tablets, have been lost and, even if these four may be seen as
especially important among those surviving, they may not represent the best of Babylonian
mathematics.
There is no problem understanding what the Yale tablet YBC 7289 is about — a diagram
is on the right.
It has on it a diagram of a square with 30 on one
side, the diagonals are drawn in and near the center is
written 1, 24, 51, 10 and 42, 25, 35. These are numbers are
written in Babylonian numerals to base 60. The Babylonian numbers are always ambiguous and no indication
occurs as to where the integer part ends and the fractional
part begins. Assuming that the first number is 1; 24, 51, 10
then
√ converting this to a decimal gives 1.414212963 while
2 = 1.414213562. Calculating 30 × 1; 24, 51, 10 gives
42; 25, 35 which is the second number. The diagonal of
a square of side√30 is found by multiplying 30 by the approximation to 2.
30
1,24,51,10
42,25,35
This shows a nice understanding of Pythagoras’s theFigure B.1: Yale Tablet diaorem. However, even more significant is the question how
gram
the Babylonians found this remarkably good approximaMATH 6118-090
Spring 2006
B.2. PYTHAGORAS’S THEOREM IN BABYLONIAN MATHEMATICS
185
√
tion to 2. Several authors7 conjecture that the Babylonians used a method equivalent to
Heron’s method. The suggestion is that they started with a guess, say x. They then found
e = x2 − 2 which is the error. Then
³
³ e ´2
e ´2
x−
= x2 − e +
2x
2x
³ e ´2
= 2+
2x
and they had a better approximation since if e is small √
then (e/2x)2 will be very small.
Continuing the process with this better approximation to 2 yields a still better approximation and so on. In fact as Joseph points out, you need only two steps of the algorithm
if you start with x = 1 to obtain the approximation 1; 24, 51, 10.
This is certainly possible and the Babylonians’ understanding of quadratics adds some
weight to the claim. However there is no evidence of the algorithm being used in any other
cases and its use here must remain no more than a fairly remote possibility.
E. F. Robertson suggests an alternative. The Babylonians produced tables of squares,
in fact their whole understanding of multiplication was built round squares, so perhaps a
more obvious approach for them would have been to make two guesses, one high and one
low say a and b. Take their average (a + b)/2 and square it. If the square is greater than
2 then replace b by this better bound, while if the square is less than 2 then replace a by
(a + b)/2. Continue with the algorithm. Now this certainly takes many more steps to reach
the sexagesimal approximation 1; 24, 51, 10. In fact starting with a = 1 and b = 2 it takes
19 steps as the table below shows:
step
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
decimal
approx
1.500000000
1.250000000
1.375000000
1.437500000
1.406250000
1.421875000
1.414062500
1.417968750
1.416015625
1.415039063
1.414550781
1.414306641
1.414184570
1.414245605
1.414215088
1.414199829
1.414207458
1.414211273
1.414213181
sexagesimal
approx
1;29,59,59
1;14,59,59
1;22,29,59
1;26,14,59
1;24,22,29
1;25,18,44
1;24,50,37
1;25,04,41
1;24,57,39
1;24,54,08
1;24,52,22
1;24,51;30
1;24,51;03
1;24,51;17
1;24,51;10
1;24,51;07
1;24,51;08
1;24,51;09
1;24,51;10
7
R. Calinger, A conceptual history of mathematics (Upper Straddle River, N. J., 1999) and G. G. Joseph,
The crest of the peacock (London, 1991)
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APPENDIX B. AN OVERVIEW OF BABYLONIAN MATHEMATICS
The Babylonians were not afraid to undertake a computation and they may have been
prepared to continue this straightforward calculation until the answer was correct to the
third sexagesimal place.
Next we look again at Plimpton 322 The tablet has four columns with 15 rows. The
last column is the simplest to understand for it gives the row number and so contains 1, 2,
3, ... , 15. The remarkable fact which Neugebauer and Sachs pointed out is that in every
row the square of the number c in column 3 minus the square of the number b in column 2
is a perfect square, say h.
c2 − b2 = h2
So the table is a list of Pythagorean integer triples. Now this is not quite true since
Neugebauer and Sachs believe that the scribe made four transcription errors, two in each
column and this interpretation is required to make the rule work. The errors are readily
seen to be genuine errors, however, for example 8, 1 has been copied by the scribe as 9, 1.
The first column is harder to understand, particularly since damage to the tablet means
that part of it is missing. However, using the above notation, it is seen that the first column
is just (c/h)2 .
So far so good, but if one were writing down Pythagorean triples one would find much
easier ones than those which appear in the table. For example the Pythagorean triple
(3, 4, 5) does not appear, nor does (5, 12, 13) and in fact the smallest Pythagorean triple
which does appear is (45, 60, 75) (15 times (3, 4, 5)). Also the rows do not appear in any
logical order except that the numbers in column 1 decrease regularly. The puzzle then is
how the numbers were found and why are these particular Pythagorean triples are given
in the table. Several historians (see for example Calinger) have suggested that column 1 is
connected with the secant function. However, as Joseph comments
This interpretation is a trifle fanciful.
Zeeman has made a fascinating observation. He has pointed out that if the Babylonians
used the formulas h = 2mn, b = m2 − n2 , c = m2 + n2 to generate Pythagorean triples
then there are exactly 16 triples satisfying n ≤ 60, 30◦ ≤ t ≤ 45◦ , and tan2 t = h2 /b2 having
a finite sexagesimal expansion (which is equivalent to m, n, b having 2, 3, and 5 as their
only prime divisors). Now 15 of the 16 Pythagorean triples satisfying Zeeman’s conditions
appear in Plimpton 322. Is it the earliest known mathematical classification theorem? Are
we now reading too much into the mathematics of the Babylonians, though.
To give a fair discussion of Plimpton 322 we should add that not all historians agree
that this tablet concerns Pythagorean triples. For example Exarchakos8 claims that the
tablet is connected with the solution of quadratic equations and has nothing to do with
Pythagorean triples; “we prove that in this tablet there is no evidence whatsoever that the
Babylonians knew the Pythagorean theorem and the Pythagorean triads.” According to E.F.
Robertson there are numerous tablets which show that the Babylonians of this period had
a good understanding of Pythagoras’s theorem. Other authors, although accepting that
Plimpton 322 is a collection of Pythagorean triples, have argued that they had, as Viola9
writes a practical use in giving a; “general method for the approximate computation of areas
8
T. G. Exarchakos, Babylonian mathematics and Pythagorean triads, Bull. Greek Math. Soc. 37 (1995),
29-47
9
T Viola, On the list of Pythagorean triples (“Plimpton 322”) and on a possible use of it in old Babylonian
mathematics (Italian), Boll. Storia Sci. Mat. 1 (2) (1981), 103-132
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187
of triangles.”
The Susa tablet sets out a problem about an isosceles triangle with sides 50, 50 and 60.
The problem is to find the radius of the circle through the three vertices.
A
Here we have labeled the triangle A, B, C and the
center of the circle is O. The perpendicular AD is drawn
from A to meet the side BC. Now the triangle 4ABD is a
right angled triangle so, using Pythagoras’ theorem AD2 =
AB 2 − BD2 , so AD = 40. Let the radius of the circle be x.
Then AO = OB = x and OD = 40−x. Using Pythagoras’s
O
2
2
theorem again on the triangle 4OBD we have x = OD +
DB 2 . So x2 = (40−x)2 +302 giving x2 = 402 −80x+x2 +302
and so 80x = 2500 or, in sexagesimal, x = 31; 15.
C
D
B
Finally consider the problem from the Tell Dhibayi
tablet. It asks for the sides of a rectangle whose area is
0; 45 and whose diagonal is 1; 15. Now this to us is quite
an easy exercise in solving equations. If the sides are x and Figure B.2: Susa tablet diay we have xy = 0.75 and x2 + y 2 = (1.25)2 . We would gram
substitute y = 0.75/x into the second equation to obtain a
quadratic in x2 which is easily solved. This however is not the method of solution given
by the Babylonians and really that is not surprising since it rests heavily on our algebraic
understanding of equations. The way the Tell Dhibayi tablet solves the problem is actually
much more interesting than the modern method.
Here is the method from the Tell Dhibayi tablet. We preserve the modern notation
x and y as each step for clarity but we do the calculations in sexagesimal notation (as of
course does the tablet). Compute 2xy = 1; 30. Subtract from x2 + y 2 = 1; 33, 45 to get
x2 + y 2 − 2xy = 0; 3, 45. Take the square root to obtain x − y = 0; 15. Divide by 2 to get
(x−y)/2 = 0; 7, 30. Divide x2 +y 2 −2xy = 0; 3, 45 by 4 to get x2 /4+y 2 /4−xy/2 = 0; 0, 56, 15.
Add xy = 0; 45 to get x2 /4 + y 2 /4 + xy/2 = 0; 45, 56, 15. Take the square root to obtain
(x + y)/2 = 0; 52, 30. Add (x + y)/2 = 0; 52, 30 to (x − y)/2 = 0; 7, 30 to get x = 1. Subtract
(x − y)/2 = 0; 7, 30 from (x + y)/2 = 0; 52, 30 to get y = 0; 45. Hence the rectangle has
sides x = 1 and y = 0; 45. Remember that this piece of mathematics is 3750 years old.
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Spring 2006
Appendix C
An Overview of Egyptian
Mathematics
Article by J. J. O’Connor and E. F. Robertson
December 2000
MacTutor History of Mathematics
Civilization reached a high level in Egypt at an early period. The country was well
suited for the people, with a fertile land thanks to the river Nile. It was also a country
which was easily defended having few natural neighbors to attack it because the surrounding
deserts provided a natural barrier to invading forces. As a consequence Egypt enjoyed long
periods of peace when society advanced rapidly.
By 3000 BC two earlier nations had joined to form a single Egyptian nation under a
single ruler. Agriculture had been developed making heavy use of the regular wet and dry
periods of the year. The Nile flooded during the rainy season providing land that irrigation
systems made fertile. Knowing when the rainy season was about to arrive was vital and the
study of astronomy developed to provide calendar information. The large area covered by
the Egyptian nation required complex administration, a system of taxes, and armies had
to be supported. As the society became more complex, records required to be kept, and
computations done as the people bartered their goods. A need for counting arose, then
writing and numerals were needed to record transactions.
By 3000 BC the Egyptians had already developed their hieroglyphic writing. This
marks the beginning of the Old Kingdom period during which the pyramids were built. For
example the Great Pyramid at Giza was built around 2650 BC and it is a remarkable feat
of engineering. This provides the clearest of indications that the society of that period had
reached a high level of achievement.
Hieroglyphs for writing and counting gave way to a hieratic script for both writing and
numerals. We want to look at the arithmetical methods which they devised to work with
these numerals
The Egyptian number systems were not well suited for arithmetical calculations. We are
still familiar with Roman numerals and so it is easy to understand that although addition of
Roman numerals is quite satisfactory, multiplication and division are essentially impossible.
The Egyptian system had similar drawbacks to that of Roman numerals. However, the
Egyptians were very practical in their approach to mathematics and their trade required
that they could deal in fractions. Trade also required multiplication and division to be
189
190
APPENDIX C. AN OVERVIEW OF EGYPTIAN MATHEMATICS
possible so they devised remarkable methods to overcome the deficiencies in their number
systems. Basically they had to devise methods of multiplication and division which only
involved addition.
Early hieroglyphic numerals can be found on temples, stone monuments and vases. They
give little knowledge about any mathematical calculations which might have been done with
the number systems. While these hieroglyphs were being carved in stone there was no need
to develop symbols which could be written more quickly. However, once the Egyptians
began to use flattened sheets of the dried papyrus reed as “paper” and the tip of a reed as
a “pen” there was then a reason to develop more rapid means of writing. This prompted
the development of hieratic writing and numerals.
There must have been a large number of papyri, many dealing with mathematics in one
form or another, but sadly since the material is rather fragile almost all of it has perished.
It is remarkable that any have survived at all. The fact that they have is a consequence of
the dry climatic conditions in Egypt. Two major mathematical documents survive.
Two great example of Egyptian mathematics are the Rhind papyrus and another papyrus, the Moscow papyrus, with a translation into hieratic script. It is from these two
documents that most of our knowledge of Egyptian mathematics comes.
The Rhind papyrus is named after the Scottish Egyptologist, A. Henry Rhind, who
purchased it in Luxor in 1858. The papyrus, a scroll about 6 meters long and 1/3 of a
meter wide, was written around 1650 BC by the scribe Ahmes who states that he is copying
a document which is 200 years older. The original papyrus on which the Rhind papyrus is
based therefore dates from about 1850 BC.
The Moscow papyrus also dates from this time. It is now becoming more common to
call the Rhind papyrus after Ahmes rather than Rhind since it seems much fairer to name
it after the scribe than after the man who purchased it. The same is not possible for the
Moscow papyrus however, since the scribe who wrote this document did not record his
name. The Moscow papyrus is now in the Museum of Fine Arts in Moscow, while the
Rhind papyrus is in the British Museum in London.
The Rhind papyrus contains eighty-seven problems while the Moscow papyrus contains
twenty-five. The problems are mostly practical but a few are posed to teach manipulation
of the number system itself without a practical application in mind. For example the first
six problems of the Rhind papyrus ask how to divide n loaves between 10 men where n = 1
for Problem 1, n = 2 for Problem 2, n = 6 for Problem 3, n = 7 for Problem 4, n = 8 for
Problem 5, and n = 9 for Problem 6. Clearly fractions are involved here and, in fact, 81
of the 87 problems given involve operating with fractions. Rising1 discusses these problems
of fair division of loaves which were particularly important in the development of Egyptian
mathematics.
Some problems ask for the solution of an equation. For example Problem 26: a quantity
added to a quarter of that quantity become 15. What is the quantity? Other problems
involve geometric series such as Problem 64: divide 10 hekats of barley among 10 men so
that each gets 1/8 of a hekat more than the one before. Some problems involve geometry.
For example Problem 50: a round field has diameter 9 khet. What is its area? The Moscow
papyrus also contains geometrical problems.
Unlike the Greeks who thought abstractly about mathematical ideas, the Egyptians
were only concerned with practical arithmetic. Most historians believe that the Egyptians
1
G. R. Rising, The Egyptian use of unit fractions for equitable distribution, Historia Math. 1 (1) (1974),
93-94
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did not think of numbers as abstract quantities but always thought of a specific collection of
8 objects when 8 was mentioned. To overcome the deficiencies of their system of numerals
the Egyptians devised cunning ways round the fact that their numbers were poorly suited
for multiplication as is shown in the Rhind papyrus.
Joseph and many other authors gives some of the measurements of the Great Pyramid
which make some people believe that it was built with certain mathematical constants in
mind. The angle between the base and one of the faces is 51◦ 500 35”. The secant of this
angle is 1.61806 which is remarkably close to the golden ratio ϕ = 1.618034. Not that
anyone believes that the Egyptians knew of the secant function, but it is of course just the
ratio of the height of the sloping face to half the length of the side of the square base. On
the other hand the cotangent of the slope angle of 51◦ 500 35” is very close to π/4. Again
of course nobody believes that the Egyptians had invented the cotangent, but again it is
the ratio of the sides which it is believed was made to fit this number. Now the observant
reader will have realized that there must be some sort of relationship between the golden
ratio and π for these two claims to both be at least numerically accurate. In fact there is a
numerical coincidence: the square root of the golden ratio times π is close to 4, in fact this
product is 3.996168.
Robins2 argues against both the golden ratio or π being deliberately involved in the
construction of the pyramid. He claims that the ratio of the vertical rise to the horizontal
distance was chosen to be 5 21 to 7 and the fact that (11/14) × 4 = 3.1428 and is close to
π is nothing more than a coincidence. Similarly Robins claims the way that the golden
ratio comes in is also simply a coincidence. Robins claims that certain constructions were
made so that the triangle which was formed by the base, height and slope height of the
pyramid was a 3, 4, 5 triangle. Certainly it would seem more likely that the engineers would
use mathematical knowledge to construct right angles than that they would build in ratios
connected with the golden ratio and π.
Finally we mention some details of the ancient Egyptian calendar. As we mentioned
above, it was important for the Egyptians to know when the Nile would flood and so this
required calendar calculations. The beginning of the year was chosen as the heliacal rising
of Sirius, the brightest star in the sky. The heliacal rising is the first appearance of the star
after the period when it is too close to the sun to be seen. For Sirius this occurs in July
and this was taken to be the start of the year. The Nile flooded shortly after this so it was
a natural beginning for the year. The heliacal rising of Sirius would tell people to prepare
for the floods. The year was computed to be 365 days long and this was certainly known
by 2776 BC and this value was used for a civil calendar for recording dates. Later a more
accurate value of 365 14 days was worked out for the length of the year but the civil calendar
was never changed to take this into account. In fact two calendars ran in parallel, the one
which was used for practical purposes of sowing of crops, harvesting crops etc. being based
on the lunar month. Eventually the civil year was divided into 12 months, with a 5 day
extra period at the end of the year. The Egyptian calendar, although changed much over
time, was the basis for the Julian and Gregorian calendars.
2
G Robins and C C D Shute, Mathematical bases of ancient Egyptian architecture and graphic art,
Historia Math. 12 (2) (1985), 107-122
MATH 6118-090
Spring 2006
192
MATH 6118-090
APPENDIX C. AN OVERVIEW OF EGYPTIAN MATHEMATICS
Spring 2006
List of Figures
1.1
Archimedes Argument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
2.1
2.2
2.3
Pythagorean Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A triangle with 3 right angles . . . . . . . . . . . . . . . . . . . . . . . . . .
Converse of Pythagorean Theorem . . . . . . . . . . . . . . . . . . . . . . .
23
24
25
3.1
3.2
3.3
Initial positions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Final isometries for SSS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Corollary 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
32
34
4.1
4.2
Star Trek Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Reason for the Name . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
36
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
5.10
5.11
5.12
5.13
5.14
5.15
Theorem 5.3 . . . . . . . . . . . . . . . . .
Chords for the Power of P . . . . . . . . .
Medians . . . . . . . . . . . . . . . . . . .
The triangle, its incircle, and its excircles
Circumcenter . . . . . . . . . . . . . . . .
Euler line . . . . . . . . . . . . . . . . . .
9 Point Circle . . . . . . . . . . . . . . . .
Pedal Triangle . . . . . . . . . . . . . . .
Simson Line . . . . . . . . . . . . . . . . .
Nagel Point . . . . . . . . . . . . . . . . .
Nagel Segment . . . . . . . . . . . . . . .
Miquel Point . . . . . . . . . . . . . . . .
Morley’s Theorem . . . . . . . . . . . . .
Fermat point . . . . . . . . . . . . . . . .
Gergonne point . . . . . . . . . . . . . . .
6.1
6.2
6.3
6.4
Quadratrix of Hippias .
Archimedean spiral . . .
Doubling the Square . .
Richmond’s construction
7.1
7.2
7.3
7.4
Alternate Interior Angles .
Exterior Angle Theorem . .
SAA Congruence . . . . . .
Uniqueness of the midpoint
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of the regular
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41
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pentagon
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194
LIST OF FIGURES
7.5
7.6
7.7
7.8
7.9
7.10
First step in Saccheri-Legendre Theorem
Additivity of Defect . . . . . . . . . . .
Right triangle defect . . . . . . . . . . .
Existence of Rectangles . . . . . . . . .
Step 1 . . . . . . . . . . . . . . . . . . .
Poincaré Disk Model . . . . . . . . . . .
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84
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9.1
9.2
9.3
9.4
9.5
9.6
9.7
Poincaré line . . . . . . . . . . .
Poincaré lines through A . . . . .
Poincaré line in Case III . . . . .
Multiple parallels through A . . .
Limiting Parallel Poincaré Lines
Lobachevskii’s Theorem . . . . .
After the first translation . . . .
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108
108
109
111
111
112
112
10.1 The Saccheri quadrilateral associated with 4ABC . . . . . . . . . . . . . .
116
11.1 Poincaré Disk Saccheri quadrilateral . . . . . . . . . . . . . . . . . . . . . .
11.2 Acute Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.3 Obtuse Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
121
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122
12.1
12.2
12.3
12.4
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130
141
141
144
13.1 Horocyclic Sector in the Poincaré Disk model . . . . . . . . . . . . . . . . .
13.2 Tractrix curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
147
148
14.1 H1 and H2 are horocycles and C1 is a circle in the Poincaré Disk model . .
153
15.1 Right triangles in D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15.2 Hyperbolic Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
158
162
16.1
16.2
16.3
16.4
16.5
16.6
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166
167
168
169
170
172
B.1 Yale Tablet diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
B.2 Susa tablet diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
184
187
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Isometries in H . . . . . . . . . . . . . . . . . .
Doubly Asymptotic Triangle . . . . . . . . . . . .
Trebly Asymptotic Triangle . . . . . . . . . . . .
Angle of Parallelism: left in D, right in H . . .
Saccheri Quadrilateral . . . . .
Complementary Segments . . .
Engel’s Theorem . . . . . . . .
Coordinates in Poincaré Plane
P in the Poincaré Disk . . . . .
Hyperbolic Graph Paper . . . .
MATH 6118-090
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Spring 2006
List of Tables
1.1
1.2
1.3
Early Calculations of π . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Calculation of decimal value of π . . . . . . . . . . . . . . . . . . . . . . . .
Computer calculations of π . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
11
12
15.1 Hyperbolic Equilateral Triangles . . . . . . . . . . . . . . . . . . . . . . . .
161
195
Index
cosh, 155
π
irrational, 64
transcendental, 65
Archimedes, 8
Biblical, 7
computation, 9
Egyptian, 8
Euler, 11
Gregory, 9
Mesopotamian, 8
Oughtred, 11
Shanks, 10
Wallis, 9
sinh, 155
tanh, 155
morbus cyclometricus, 65
9 Point Circle, 48
AAA congruence, 88
Alternate Interior Angle Theorem
Neutral Geometry, 74
alternate interior angles, 74
angle
exterior, 76
of parallelism, 90
parallelism, 119
remote interior, 76
fan, 97
parallelism, 101, 119
angle of parallelism
Poincaré disk model, 111
angle,right, 30
angles
alternate interior, 74
corresponding, 74
Apollonius, 18, 63
Archimedes, 17, 67
area
elliptic geometry, 148
function, 118
hyperbolic, 116–119
hyperbolic geometry, 118, 149
Axiom
Hyperbolic, 84
Hyperbolic, 73
Axioms
Birkhoff’s
Neutral Geometry, 27
Euclidean, 29
Hilbert
Archimedes’ Axiom, 26
Betweeness Axioms, 26
Circular Continuity Principle, 27
Congruence Axioms, 26
Continuity Axioms, 26
Dedekind’s Axiom, 26
Elementary Continuity Principle, 27
Incidence Axioms, 25
transformations, 30
axioms
complete, 105
consistent, 105
independent, 105
base
triangle, 115
Bernoulli
Johann, 64
bisector
exterior angle, 43
Bolyai, 20
János, 84
Bolyai-Lobachevskii Theorem, 144
centroid, 42
circle, 29
Spieker, 50, 51
unit, 107
inversion, 173
196
INDEX
circle of inversion, 173
circles
hyperbolic
Poincaré disk model, 113
orthogonal, 175
circumcenter, 46
circumference
hyperbolic circle, 162
circumradius, 46
complex numbers, 69
conchoid, 68
congruence
AAA, 88
SAA, 77
Side-Angle-Angle, 77
congruent, 30
constructible, 59
regular pentagon, 61
constructible length, 59
coordinates, Beltrami168, axial168
Lobachevskii, 168
Poincaré, 168
polar, 168
Weierstrass, 168
Beltrami, 170
corresponding angles, 74
corresponding points, 151
cross ratio, 134
Crossbar Theorem
singly asymptotic triangle, 91
curvature, 148
Cusa, 63
cyclic quadrilateral, 49
da Vinci, 64
De Morgan, 64
defect
additivity, 81
hyperbolic geometry, 86
of a triangle, 81
DeMoivre’s Theorem, 69
dilation, 175
Dinostratus, 62
distance, 29
Poincaré disk model, 109
equiangular, 151
equidistant
MATH 6118-090
197
points, 100
equidistant curve, 154
equidistant line, 151
Eratosthenes, 65
Euclid, 16
Euclid V
equivalent statements, 20
Euclid’s form, 33
Euclidean
definitions, 17
postulates, 17, 25
Eudoxus of Cnidus, 16
Euler line, 48
excenter, 43
excircle, 43
exradius, 43
exterior angle, 76
Exterior Angle Theorem, 76
singly asymptotic triangle, 91
fan angle, see angle, fan
Fermat point, 53
Feuerbach’s Theorem, 48
finite decomposition, 115
foot of the perpendicular, 75
Formula
Bolyai-Lobachevsky, 119
fractional linear transformation, 130
frame, 168
functions
hyperbolic trigonometric, 155
Gauss, 20, 69
Gauss
Karl Friedreich, 84
geodesic, 147
Gergonne point, 53
Gregory
James, 64
group
projective special linear, 132
special linear, 132
Heron of Alexandria, 18
Heron’s formula, 45
hyperbolic, 161
Hipparchus, 18
Hippias, 62
Spring 2006
198
INDEX
Hippocrates of Chios, 15, 65, 67
horizontal translation, 127
horocycle, 151, 152
Poincaré disk model, 153
tangent, 152
horocycles
codirectional, 152
horoparallel, 89, 99
Hypatia, 19
hyperbolic plane, 84
Hyperbolic Pythagorean Theorem, 123
hypercycle, 154
axis, 154
base line, 154
distance, 154
Poincaré models, 154
radius, 154
tangent line, 154
hyperparallel, 89, 101
Law of Cosines, 43
hyperbolic, 161
for angles, 161
Law of Sines, 46
hyperbolic, 161
Lemoine point, 55
Lindemann, 65
line, 29
equidistant, 151
of enclosure, 174
line segment, 29
lines
parallel, 74
Poincaré disk model, 107
Lobachevskii, 20
Lobachevskii
Nikolai, 84
Lobachevskii’s formula, 112
Lobachevskii’s Theorem, 112
Iamblichus, 63
ideal point, 107
incircle, 43
Inequality
Triangle, 79
inradius, 43
Interior Angles
Alternate, 74
inverse, 173
inversion
in the unit circle, 127
isogon, 54
isogonal, 54
reflection, 54
isogonal
points, 54
isometric, 147
isometry, 30
direct, 32
orientation preserving, 32
orientation reversing, 32
reflection, 32
rotation, 32
translation, 32
median, 41
Menaechmus, 16
Menelaus, 18
metric, 29
Miquel point, 51
Model
Poincaré Disk, 107
Kant, 17
Lambert, 64
MATH 6118-090
Nagel, 50
Nagel point, 50
Nagel segment, 50
Nicomedes, 63, 68
conchoid of, 68
Nine Point Circle, 48
non-transversal side, 74
Nunes, 64
omega point, 107
orientation, 32
orthocenter, 47
Pappus, 18
parallel, limiting89
boundary, 89
divergently, see hyperparallel
limiting, 99
parallel lines, 74
Pasch’s Theorem
Spring 2006
INDEX
singly asymptotic triangle, 91
pedal triangle, 49
pencil of lines, 95
plane
hyperbolic, 84
Plato, 15
Playfair’s Postulate, 19
Poincaré Disk Model, 107
point
Fermat, 53
Gergonne, 53
Miquel, 51
Nagel, 50
Spieker, 50
point at infinity, 107
points
corresponding, 151
pole, 173
polygons
constructible, 72
Posidonius, 19
Postulate
Wallis’, 88
power, 175
of a point, 41
Proclus, 19
pseudosphere, 147
Ptolemy, 18, 19
Pythagoras of Samos, 15
Pythagorean Theorem, 23
Pythagoren Theorem
Converse, 24
quadratrix, 62
quadrilateral
cyclic, 49
Lambert, 87, 165
Saccheri, 86, 121
Saccheri, 115
associated, 115
ratio
cross, 134
rays
limiting parallel, 98
rectangle, 82
reflection, 32, 127
region
MATH 6118-090
199
polygonal, 115
remote interior angle, 76
Rhind Papyrus, 8
right angle, 30
rotation, 32
SAA congruence, 77
Saccheri, 86
Saccheri-Legendre Theorem, 80
SAT, 90
segment, 29
Nagel, 50
segments
complementary, 167
semiperimeter, 43
Side-Side-Side Theorem, 31
similar triangles, 37
similarity, 37
Simson line, 50
Speiker, 50
Spieker circle, 51
spiral
Archimedean, 63
SSS, 31
St. Vitus, 65
Star Trek Lemma, 35
Star Trek Logo, 36
symmedian, 55
point, 55
Thales of Miletus, 15
Thætetus of Athens, 15
tractrix, 147, 148
transformation, 29
fractional linear, 130
translation, 32
horizontal, 127
hyperbolic, 137
transversal side, 74
triangle
contact, 55
doubly asymptotic, 92
pedal, 49
similar, 37
singly asymptotic, 90
singly asymptotic
congruent, 91
Triangle Inequality, 79
Spring 2006
200
INDEX
triangulation, 115
trilateral, 90
ultraideal point, 107
ultraparallel, 89
unit circle, 107
Universal Hyperbolic Theorem, 86
vanishing point, 107
Vulcan Mind Meld, 36
Wantzel, 69
Weak Exterior Angle Theorem, 76
MATH 6118-090
Spring 2006