hyperbolic geometry - uefcs.uef.fi/matematiikka/kurssit/hyperbolicgeometry/...chapitre 1 power of a...

Hyperbolic Geometry

Eric Lehman

Mars-April 2014

Table of contents

I. Summary 11 Power of a point with respect to a circle 3§ 1. Equations of a circle . . . . . . . . . . . . . . . . . . 3§ 2. Orthogonal circles . . . . . . . . . . . . . . . . . . . . 6§ 3. Circle bundles . . . . . . . . . . . . . . . . . . . . . . 8§ 4. Existence of circles . . . . . . . . . . . . . . . . . . . 11

2 Reflections 19§ 1. Definition of inversions and reflections . . . . . . . . . 19§ 2. Description of inversions using complex numbers . . . 20§ 3. Images of lines and circles . . . . . . . . . . . . . . . 21§ 4. Images of circle bundles . . . . . . . . . . . . . . . . 23

3 Non-Euclidean Geometry 25§ 1. chapter 6.1.1. Non-Euclidean Geometry : the two usual

models of a hyperbolic plane . . . . . . . . . . . . . . 25§ 2. 6.1.2 Existence of hyperbolic lines . . . . . . . . . . . 30§ 3. 6.1.3 Inversion preserves inversion points . . . . . . . 31§ 4. 6.2.1-6.2.3. The group of Hyperbolic Geometry . . . . 32

II. Reading Brennan 334 6.1-6.3 Non-Euclidean Geometry, Transformations, Dis-

tance 35§ 1. Page 263. What is Non-Euclidean Geometry ? . . . . . 35§ 2. Page 268-269 . . . . . . . . . . . . . . . . . . . . . . 36§ 3. Page 288 . . . . . . . . . . . . . . . . . . . . . . . . . 37§ 4. Page 289 . . . . . . . . . . . . . . . . . . . . . . . . . 38§ 5. Page 294. Reflection lemma . . . . . . . . . . . . . . 38

5 6.4 Non-Euclidean Geometry, Transformations, Dis-tance 41

§ 1. Euclidean Triangles . . . . . . . . . . . . . . . . . . . 41§ 2. Page 298 . . . . . . . . . . . . . . . . . . . . . . . . . 45§ 3. 03 . . . . . . . . . . . . . . . . . . . . . . . . . 45§ 4. Page 307-312 Pythagoras’ theorem and Lobachevski’s

formula . . . . . . . . . . . . . . . . . . . . . . . . . 45

6 6.5 Tessellation 47

3

4 TABLE OF CONTENTS

Thème I

Summary of February course and complements

Chapitre 1

Power of a point with respect to acircle

§ 1. Equations of a circle§ 2. Orthogonal circles§ 3. Circle bundles

§ 4. Existence of circles

§ 1. Equations of a circle

1.1 Circle of center �.a; b/ and radius R

We suppose given an orthonormal frame .O;E{; Ej / of a Euclidean plane P .Let C.�;R/ or simply C be a circle with center �.a; b/ and radius R and let M.x; y/

be a point. The power of the point M with respect to the circle C denoted by PC .M/ is thenumber

PC .M/ WD d2�R2 D .x�a/2C .y� b/2�R2 D x2Cy2� 2ax� 2byCa2C b2�R2Proposition. The point M belongs to the circle C if and only if PC .M/ D 0, or

.x � a/2 C .y � b/2 �R2 D 0which is the normal equation of C . We may write the normal equation of C using complexnumbers as

zz � .wz C wz/C c D 0

where z D x C iy ; z D x � iy and w D aC ib.

Theorem 1. Let C be a circle and M.x; y/ a point. For any line through M intersectingC let us call the two intersection points P and Q (distinct or the same, in case the line is

3

4 THÈME I. SUMMARY. CH. 1. POWER OF A POINT WITH RESPECT TO A CIRCLE

tangent to the circle). Then the product MP �MQ is constant. If M is outside the circleMP �MQ D PC .M/, if M is inside the circle MP �MQ D jPC .M/j D �PC .M/ and ifM is on the circle MP �MQ D 0.

Remark. Using the scalar product, we have in all cases��!MP � ��!MQ D PC .M/.

M

Q

P

C

MQ

P

C

1.2 Radical axis of two circles

Theorem and definition. Let C and C 0 be two circles with distinct centers �.a; b/ and�0.a0; b0/. The set of points M such that PC .M/ D PC 0.M/ is a line orthogonal to line��0. This line is called the radical axis of the two circles.

Proof. A point M.x; y/ is such that PC .M/ D PC 0.M/ if and only if :

x2 C y2 � 2ax � 2by C a2 C b2 �R2 D x2 C y2 � 2a0x � 2b0y C a02 C b02 �R02

or2.a0 � a/x C 2.b0 � b/y C a02 C b02 � a2 � b2 �R02 CR2 D 0

which is a line orthogonal to the vector��!��0. �

Remark 1. The radical axis of two point circles � and �0 is the bisector of the segment��0.

Remark 2. To find the radical axis of two circles it is enough to find one point K of thisradical axis. Then the radical axis is the line through K which is orthogonal to the linejoining the centers of the two circles.

Theorem. If two circles C and C 0 intersect in two points A and B , the radical axis of Cand C 0 is the line AB .

Proof. The points A and B are such that PC .A/ D 0 D PC 0.A/ and PC .B/ D 0 D PC 0.B/.These two points belong then to the radical axis of the two circles. �

Theorem. If two circles C and C 0 are tangent in a point T , the radical axis of C and C 0 isthe line through T orthogonal to the line joining the centers of the two given circles, that isthe common tangent to these two circles..

§ 1. EQUATIONS OF A CIRCLE 5

1.3 Radical center of three circles

Theorem and definition. LetC ,C 0 andC 00 be three circles with nonaligned centers�.a; b/,�0.a0; b0/ and �00.a00; b00/ (nonaligned means that the points do not belong to a commonline). There is one point S such that PC .S/ D PC 0.S/ D PC 00.S/. This point is called theradical center of the three circles.

Proof. Let us call � the radical axis of C 0 and C 00, �0 the radical axis of C 00 and C and�00 the radical axis of C and C 0. The lines � and �0 are not parallel since orthogonalrespectively to the distinct lines �0�00 and �00�. Thus � and �0 intersect in a point S suchthat PC 0.S/ D PC 00.S/ and PC 00.S/ D PC .S/ and thus PC .S/ D PC 0.S/, which meansthat S 2 �00. �How to construct the radical axis of two non intersecting circles with distinct centers.

1)

� �0

2) Draw a circle intersecting C and C 0

S

3) Draw a line through S orthogonal to ��0

� �0S

Remark 1. An other way to do it is to draw a common tangent. Let K be the middle of thethe segment T T 0 where T and T 0 are the points of contact. Then K belongs to the radicalaxis. But this method fails if one of the circles is inside the other one.

� �0

KTT 0

Remark 2. Let p D PC .S/ D PC 0.S/ D PC 00.S/ ; since these three quantities are equalthey have same sign : if p < 0, the point S is inside all three circles ; if p > 0, the pointS is outside all three circles ; if p D 0, the three circles have S as a common point. As aconsequence, we see that if two of the circles do not intersect each other, then S is outsideall three.


1.4 Generalized circles

The equation of the previous circle C may be written

˛.x2 C y2/ � 2Ax � 2By C D 0 (1)

where A D ˛a, B D ˛b and D ˛c D ˛.a2 C b2 � R2/. If ˛ ¤ 0, we have the equationof a genuine circle when R2 > 0, that is A2

˛2C B2

˛2� ˛> 0 or A2 C B2 � ˛ > 0. If we

choose ˛ D 0, we get the equation

�2Ax � 2By C D 0which is the equation of a line if .A;B/ ¤ .0; 0/. Thus the equation .1/ may be used todescribe as well circles as lines. If we add one point at infinity, denoted1, we may thinkthe lines as circles going through the point1.

Notice that the linear combination of the equations of two circles is a circle when circleis taken in the general sense. In particular, the radical axis of two genuine circles is obtainedusing coefficients 1 and �1 and the normalized equations of the circles.

§ 2. Orthogonal circles

2.1 Definition

Definition. Two circles C.�;R/ and C 0.�0; R0/ are orthogonal if they are intersecting intwo points T and T 0 such that the angles 1�T�0 and 2�T 0�0 are right angles (if C 0 is a pointcircle, it is orthogonal to C when the point is on C ).

� �0

T

T 0

A circle C.�.a; b/; R/ and a line with equation uxCvyCw D 0 are orthogonal if andonly if uaC vb C w D 0.

A line with equation ux C vy C w D 0 is orthogonal to the vector uE{ C v E| . Twolines with equations ux C vy C w D 0 and u0x C v0y C w D 0 are orthogonal if.uE{ C v E|/ � .u0E{ C v0 E|/ D 0, or uu0 C vv0 D 0.

§ 2. ORTHOGONAL CIRCLES 7

2.2 Characteristic properties

Theorem. LetC andC 0 be two genuine circles with respective centers�.a; b/ and�0.a0; b0/and respective radii R and R0, intersecting in points T and T 0. The circles C and C 0 are or-thogonal if and only if one of the following equivalent conditions is fulfilled :

1. ��02 D R2 CR022. The triangle T��0 is rectangle in T

3. PC 0.�/ D R24. PC .�0/ D R025. .a � a0/2 C .b � b0/2 D R2 CR02

Theorem. Two (generalized) circles C and C 0 with real equations

˛.x2 C y2/ � 2Ax � 2By C D 0 and ˛0.x2 C y2/ � 2A0x � 2B 0y C 0 D 0are orthogonal if and only if

˛ 0 C ˛0 � 2AA0 � 2BB 0 D 0

Proof. 1°) If the two circles are genuine circles, the condition 5 above may be written

a2 C b2 �R2 C a02 C b02 �R02 D 2aa0 C 2bb0

or since a2 C b2 �R2 D c D ˛

and a02 C b02 �R02 D c0 D 0

˛0

˛C 0

˛0D 2A

˛

A0

˛0C 2B

˛

B 0

˛0

Multiplying by ˛˛0, we get the result.2°) If C is a genuine circle and C 0 a line, they are orthogonal if and only if the line is

a diameter of the circle, which means that the center of the circle belongs to the line. Let˛.x2Cy2/�2Ax�2ByC D 0 be the equation of the circle and �2A0x�2B 0yC 0 D 0the equation of the line. The center of the circle has coordinates .A

˛; B˛/ belongs to the line

if and only if �2A0 A˛� 2B 0 B

˛C 0 D 0. Multiplying by ˛ and taking into account that

˛0 D 0, we get the condition ˛ 0 C ˛0 � 2AA0 � 2BB 0 D 0.3°) IfC andC 0 are lines with equations�2Ax�2ByC D 0 and�2A0x � 2B 0y C 0 D 0.

These two lines are orthogonal if and only if the vectors �2AE{ � 2B E| and �2A0E{ � 2B 0 E|are orthogonal or AA0CBB 0 D 0, using ˛ D ˛0 D 0, we still get the characteristic relation˛ 0 C ˛0 � 2AA0 � 2BB 0 D 0. �

Comment. Why is it convenient to use equations that are not necessarily normal ? Thenormal equation x2 C y2 C 2ax C 2by C c D 0 can only be used for genuine circles ;the equation ˛.x2 C y2/ � 2Ax � 2By C D 0 will describe a genuine circle if ˛ ¤ 0

AND will describe a line if ˛ D 0 and .A;B/ ¤ .0; 0/. Thus .˛; A;B; / will describe acircle-or-line iff .˛; A;B/ ¤ .0; 0; 0/.


Theorem. Two circles C and C 0 with complex equations

zz � .wz C wz/C c D 0 and zz � .w0z C w0z/C c0 D 0are orthogonal if and only if

ww0 C ww0 D c C c0

Notation. We denote by C the circle with center O.0; 0/ and radius 1.

Theorem. A circle C with equation x2 C y2 � 2ax � 2by C c D 0 is orthogonal to C ifand only if c D 1. A circle-or-line with equation ˛.x2 C y2 � 2Ax � 2By C D 0 isorthogonal to C if and only if ˛ D .

§ 3. Circle bundles

From now on, "circle" will mean "general circle". Thus a circle is either a "genuinecircle" or a line. The lines are the circles that go through1.

3.1 General definitions and properties

Definition. Given two distinct circles C1 and C2, with equations

f1.x; y/ WD ˛1.x2Cy2/�2A1x�2B1yC 1 D 0 and f2.x; y/ WD ˛2.x2Cy2/�2A2x�2B2yC 2 D 0we call bundle B of circles determined by C1 and C2, the circles with equations

�f1.x; y/C �f2.x; y/ D 0 where .�; �/ 2 R2, but .�; �/ ¤ .0; 0/Proposition and definitions. Let B be a bundle determined by two distinct circles C1 andC2. If C1 and C2 have 2 common points P andQ, then all the circles in B go through these2 points. The bundle is called bundle with base points P and Q. If C1 and C2 have exactly1 common point P , they are tangent in that point and tangent to a line d , then all the circlesin B go through P and are tangent to d . The bundle is called tangent bundle . If C1 and C2have no common point, then all the circles in B are orthogonal to the circles in the bundleB0 with base points P and Q. The bundle is called bundle with base limit P and Q.

Proof. Let P.xP ; yP /. If f1.xP ; yP / D 0 and f2.xP ; yP / D 0, then �f1.xP ; yP / C�f2.xP ; yP / D 0 and the same holds forQ. Thus if C1 and C2 go through P andQ, it willbe true for all circles in the bundle B determined by C1 and C2.

The case of the tangent bundle can be viewed as a limiting case of the preceding one andthe bundle with limit points is the bundle orthogonal to the one with base points as definedbelow. �

Remark. If B is a bundle with base points P and Q D 1, then the circles belonging to Bare the lines through P . If B is a tangent bundle of circles tangent to the line d at the pointP D 1, then the circles belonging to B are the lines parallel to d . If B is a bundle withlimit points P and Q D 1, then the circles belonging to B are the genuine circles withcenter P .

§ 3. CIRCLE BUNDLES 9

Theorem and definition. Let B be a bundle of circles. The set of circles orthogonal to thecircles in B form a circle bundle B0. The bundles B and B0 are called orthogonal bundles.

Proof. Let C1 and C2 be two distinct circles in B. Write the equations of C1 and C2

˛1.x2Cy2/�2A1x�2B1yC 1 D 0 and ˛2.x

2Cy2/�2A2x�2B2yC 2 D 0A circle C 0 is orthogonal to all the circles in B if and only if it is orthogonal to C1 and C2.Thus we have to solve the homogeneous linear system in .˛0; A0; B 0; 0/�

˛1 0 C 1˛0 � 2A1A0 � 2B1B 0 D 0

˛2 0 C 2˛0 � 2A2A0 � 2B2B 0 D 0

This system of two equations in four unknowns is of rank 2, since the circles C1and C2are distinct (thus .˛1; A1; B1; 1/ and .˛2; A2; B2; 2/ are not proportional). The solutionis then a linear combination of any two independent solutions. �

3.2 Different kinds of circle bundles

Definition. Two bundles of lines or circles are orthogonal if every circle (or line) in onebundle is orthogonal to all the circles or lines in the other bundle.

If B is a bundle with base points P and Q, then the orthogonal bundle B0 is the bundlewith limit points P and Q and conversely. If B is a tangent bundle of circles tangent to theline d at the point P , then the orthogonal bundle B0 is the tangent bundle of circles tangentto the line d 0 at the point P , where d 0 is the line through P orthogonal to d .

Orthogonal bundles with base and limit points P and Q, where P ¤1 and P ¤1.

P Q


Orthogonal bundles with base and limit points P and Q D1.

P

Orthogonal tangent bundles tangent at point P ¤1 to orthogonal lines d and d 0.

P

d

d 0

§ 4. EXISTENCE OF CIRCLES 11

Orthogonal tangent bundles tangent at point P ¤1 to orthogonal lines d and d 0.

d

d 0P

§ 4. Existence of circles

Circle orthogonal to 2 circles and going through 1 point

Lemma. Given a circle bundle B with base points P and Q, then for any point M differentfrom P and Q, there is exactly one circle belonging to B and going through M .

Proof. If the point M is on the line PQ (including the case M D 1), then the circle is theline PQ. If not, PQM is a real triangle and there is exactly one circle going through thethree vertices. �

Lemma. Given a bundle B of circles tangent to a line d in P , for any point M differentfrom P , there is exactly one circle belonging to B and going through M .

Proof. If P D 1, the circles are parallel lines to d , and through each point M distinctfrom1 there is one and only one parallel to d . If P ¤ 1, we have two cases. First case :M 2 d , then d is the only circle through P andM tangent to d at P . Second case :M … d ,then there is only one circle tangent to d in P and going through M ; in fact you get thecenter as intersection of the bisector of segment PM and the line ortogonal to d in P . �

Lemma. Given a bundle B of circles with limit points P and Q, for any point M , there isexactly one circle belonging to B and going through M .

Proof. There is only one Apolonius circle with respect to the two points P and Q with theration MP

MQ. �

Theorem. Given a bundle B of circles, for any pointM , there is exactly one circle belongingto B and going through M .

Proof. The three lemmas (or lemmata) above show that the theorem is valid for any kind ofbundle. �


Theorem. Given two circles and a point M , there is exactly one circle orthogonal to thesetwo circles and going through M .

Proof. Let us call C1 and C2 the two given circles and B the bundle of circles determined byC1 and C2. Let B0 be the bundle orthogonal to B. We are looking for a circle C 0 belongingto the bundle B0 and going through M . The preceding theorem tells us that there is one andonly one such point. �

Circle orthogonal to 3 circles

Theorem. Let C1, C2 and C3 be three circles.1°) If all three circles belong to a common bundle B, then C 0 is any circle belonging to

the bundle B0 orthogonal to B.2°) If the three circles do not belong to one common bundle, then there is at most one

circle C 0 orthogonal to all three. If there is such a circle, then this circle is unique.3°) If the radical center of the three circles C1, C2 and C3 is inside these circles, there

is no circle orthogonal to these three circles. If the radical center of the three circles C1,C2 and C3 is outside (or on) these circles, then there is one circle orthogonal to these threecircles. The center of this circle is the radical center.

Proof. We have to solve the homogeneous linear system in .˛0; A0; B 0; 0/8<: ˛1 0 C 1˛0 � 2A1A0 � 2B1B 0 D 0

˛2 0 C 2˛0 � 2A2A0 � 2B2B 0 D 0

˛3 0 C 3˛0 � 2A3A0 � 2B3B 0 D 0

If the three circles belong to one bundle, the vectors .˛1; A1; B1; 1/, .˛2; A2; B2; 2/ and.˛3; A3; B3; 3/ are linearly dependent and we are back to the preceding theorem. If not,the rank of the system is 3, so the set of solutions are all proportional and all describe thesame solution.

BUT, we have to be careful : it is true that for every circle there are elements .˛; A;B; / 2R4 unique up to the product by a constant different from zero, but the converse is not true :given an element .˛; A;B; / 2 R4, there is a circle with corresponding equation if andonly if the square of the radius is greater or equal to zero, where : R2 D A2CB2�˛

˛2. Thus

we have proved that given three circles not belonging to a common bundle, there is at mostone circle orthogonal to the three circles.

3°) Let us call K the radical center of C1, C2 and C3 and denote by p D PC1.�/ DPC2.�/ D PC3.�/ the common power of K relatively to these three circles. If p > 0, thecircle C with center� and radiusR D pp is the unique circle orthogonal to C1, C2 and C3(when p D 0, this circle is a point circle). If p < 0, there is no circle orthogonal to all threecircles C1, C2 and C3, since the centers of two orthogonal circles are outside each other. �

Corollary 1. Let C1 be a circle, B be a circle bundle with base points P and Q and B0 thebundle orthogonal to B. If the circle C1 belongs to B0, then all the circles in B are orthogonalto C1. If not, then there is one and only one circle C belonging to B and orthogonal to C1.

Proof. Let C2 and C3 be two circles belonging to B0. The properties "C2B" and "C isorthogonal to C2 and to C3" are equivalent. Since B0 is a circle bundle with limit points,the radical axis of the circles C2 and to C3 is outside these two circles, and thus the radicalcenter of C1, C2 and C3 is outside these three circles and we may apply the theorem. �


The radical center is inside the three circles.

S

The radical center is outside the three circles.

C

C2 C3

C1


Corollary 2. Let C1 be a circle, B be a circle bundle with limit points P and Q and B0 thebundle orthogonal to B. If the circle C1 belongs to B0, then all the circles in B are orthogonalto C1. If not, then there at most one circle C belonging to B and orthogonal to C1. Moreprecisely, if C1 separates the points P and Q (that is : one point is inside or on the circleand the other one outside), then there is no circle in the bundle B orthogonal to C1. If C1does not separate the points P and Q, then there is one circle in the bundle B orthogonal toC1.

Proof. Let C2 and C3 be two circles belonging to B0. The properties C2B and C is ortho-gonal to C2 and to C3 are equivalent. Thus if C1 does not belong to B0, there is at most onecircle belonging to B and orthogonal to C1. To determine if there is one or zero such circle,let us look at the situation where Q D 1 : The bundle B0 is the set of circles centered atP ; there is a circle centered at P orthogonal to C if and only if P is outside or on C . Torecover the general case use an inversion (see next chapter). �

We are illustrating these two corollaries in the following paragraph.

Examples of circles belonging to one circle bundle and orthogonal to one circle

Let B be a circle bundle and C1 a circle. We call B0 the bundle orthogonal to B. Wesuppose that C1 … B0.

Situation 1. Let P be a point in the Euclidean plane. Let B be the bundle of lines goingthrough P and let C1 be any line or any genuine circle with center � different from P .Then there is exactly one element of B which is orthogonal to C1. (It is the line P� or theline through P ortogonal to the line C1).

Proof. The bundle B is the circle bundle with base points P and1. Use corllary 1. � �

P

C1

�

Situation 2. Let ` be a line in the Euclidean plane. Let B be the bundle of lines parallel to` and let C1 be any circle. If C1 is a genuine circle, there is one and only one line `1 in thebundle B orthogonal to C1 : `1 is the line parallel to ` that goes through the center �1 ofC1. If C1 is a line which does not belong to B0, then the only element of B orthogonal to C1is the point circle1.

Proof. There is one and only one parallel to a line through a point. �

`1

`C1

�1

Situation 3. Let P be a point in the Euclidean plane. Let B be the bundle of all circleshaving P as center and let C1 be any genuine circle. If P is outside C1 or on C1, one candraw a tangent from P to the circle C . Let us call T the contact point. There is one and onlyone circle C1 in the bundle B orthogonal to C : C1 is the circle with center P and radiusPT . If P is inside C , then no circle belonging to B is orthogonal to C .

Proof. If P is outside C1 or on C1, one can draw a tangent to C1 that goes through P . LetT be the contact point. The circle with center P and radius PT is a circle orthogonal to C1(if T D P , the point circle P is orthogonal to C1). We know that there is at most one circle,so then this circle is the one and only orthogonal to C1 and belonging to B.


T

C1P�1

If P is inside the circle C1, no circle with center P can be orthogonal to C . In fact, therelation d2 D R21 CR2 is impossible for all R when d < R1. �

P

C1

�1

�

An other proofOne easy construction of the circle C tangent to ` at P and orthogonal to C1 usinginversion (see next chapter).

C1

C

P

P �

`

`�Let P � be the image of Pin the reflection through C1.

Construction of the circle C belonging to the circle bundle with limit points P and Qand orthogonal to a circle C1 which does not separate P and Q.

C1

C

P Q

Q�

Let Q� be the image of Qin the reflection through C1.The circle C has to beorthogonal to the circle PQQ�.The point � is the radical centerof the circle C1 and the circlesgoing through P and Q.

�


Circle orthogonal to 3 circles which are not in a common bundle

Theorem. Let C1.�1; R1/, C2.�2; R2/ and C3.�3; R3/ be three circles which do not be-long to a common bundle and let S be their radical center. If S is inside the three circles,then there is no circle C orthogonal to C1, to C2 and to C3. If S is outside the three circles,then there is a circle C orthogonal to C1, to C2 and to C3.

Proof. Let C.�;R/ be a solution. For i D 1; 2 and 3, one has ��2i D R2i C R2. Thusdi D ��i > Ri and� is outside all three circles C1, C2 and C3. The circleC is orthogonalto the circles C1, C2 and C3, if the power of � with respect to each of the three circles isR2. And thus � has to be the point S . If S is outside or on the circles C1, C2 and C3, thenS is a center of a circle orthogonal to each of the three circles C1, C2 and C3. �

Corllary. Given a circle C and two distinct points P andQ, there is one circle � orthogonalto C and going through P and Q. If C does not belong to the bundle with limit points Pand Q, the circle � is unique. If C belongs to the bundle with limit points P and Q, thenall the circles going through P and Q are orthogonal to C .

Proof. The circle � has to be orthogonal to three circles. The radical center of these threecircles is outside them since P and Q are point circles. �

The radical center is inside the three circles.

S


The radical center is outside the three circles.

C

C1 C2

C3

Chapitre 2

Reflections

§ 1. Definition of inversions and reflections§ 2. Description of inversions using complex

numbers§ 3. Images of lines and circles§ 4. Images of circle bundles

In this chapter the plane is the usual plane union one point de-noted 1 and called point at infinity. A circle is either a genuinecircle or a straight line to which is added the point at infinity. Thus

a line is a circle which goes through1.Let C be circle. If C is a genuine circle with center� and ra-

dius R then the terms "reflection in C " and "inversion with center� and power R2" have the same meaning. If C is a line the term"reflection in C " have the usual meaning of a line reflection, withthe complement that the image of1 is1.

The unit circle centered at the origin will be denoted C.

§ 1. Definition of inversions and reflectionsDefinitions. Let P denote an Euclidean plane.

1°) LetO be a point and R a length. We call inversion with centerO and power R2 anddenote InvO;R2 the transformation of P[f1g that associates to each pointM the pointM 0such that

– if M … fO;1g, M and M 0 are on a same ray with origin O and OM �OM 0 D R2– if M D O , InvO;R2.O/ D1– if M D1, InvO;R2.1/ D OThe inversion InvO;R2 is also called reflection in the circle C and is denoted ReflC .2°) Let ` be a line. We call reflection in the line ` and denote Refl` the transformation

that associates to each point M the point M 0 such that– if M … `, then ` is the perpendicular bisector (jauan kestinormaali ?) of the segmentMM 0

– if M 2 `, then M 0 DM .

OM

M 0C

We have the following immediate consequences of the definition.

Proposition. Let � be either C or `, Refl� is involutive (which means that it is equal to itsown inverse) :

M M 0

`

Refl� ı Refl� D Identity

Proposition. The fixed points of Refl� are the points belonging to � .

19

20 THÈME I. SUMMARY. CH. 2. REFLECTIONS

A circle � is dividing the plane in three parts : the set of points belonging to � and thetwo sides of � . If � is a genuine circle, the two sides are the inside containing the center(unless the radius is 0) and the outside containing1. If � is a line1 belongs to � .

Proposition. The image of a point on one side of the circle � by Refl� is on the other sideof the circle � .

Proposition. If a circle is orthogonal to � , then it is globally invariant by Refl� .

§ 2. Description of inversions using complex numbers

We are restricting ourselves to the inversion with centerO and power 1, that is to reflec-tion in the unit circle C.

Theorem. The inversion with center O and power 1 transforms z into 1Nz , where Nz is the

complex conjugate of z.

Proof. Write z in the polar form : z D �ei� . Then 1zD ��1e�i� and 1

Nz D ��1ei� . Thus O ,z and 1

Nz are on a same ray with origin O . Moreover

jzj � j1Nz j D � � ��1 D 1

i.e. OM �OM 0 D R2, where R D 1. �

Equation of genuine and generalized circles. The equation of a circle C.�.a; b/; R/ is

x2 C y2 � 2ax � 2by C a2 C b2 �R2 D 0Introducing the normal form z D x C iy, we have 2x D z C Nz and 2y D �iz C i Nz. Thenz Nz D x2 C y2 and the equation of the circle may be written

z Nz � a.z C Nz/ � b.�iz C i Nz/C a2 C b2 �R2 D 0or

z Nz � .a � ib/z � .aC ib/ Nz C a2 C b2 �R2 D 0which may be written

z Nz � Nwz � w Nz C c D 0where w D aC ib is the center of the circle.

The equation of a generalized circle may then be written

˛z Nz � N!z � ! Nz C D 0where ˛ and are real numbers and ! 2 C such that .˛; !/ ¤ .0; 0/. The (generalized)circle is a genuine circle with center !

˛if ˛ ¤ 0. It is a line if ˛ D 0.

Theorem. Two circles C and C 0 with equations

˛z Nz � N!z � ! Nz C D 0 and ˛0z Nz � N!0z � !0 Nz C 0 D 0are orthogonal if and only if

§ 3. IMAGES OF LINES AND CIRCLES 21

!!0 C !!0 � ˛0 � ˛ 0 D 0

The equation of the circle C isz Nz � 1 D 0

thus ˛ D 1, ! D 0 and D �1. The circle C is orthogonal to C if and only if 0 D � ˛i.e.

˛ D In the case when C is a line, it is orthogonal to C if and only if D 0, that is, the line goesthrough O .

§ 3. Images of lines and circles

Theorem. The circle C with equation

˛z Nz � N!z � ! Nz C D 0

has as image through ReflC the circle C 0 with equation

z Nz � N!z � ! Nz C ˛ D 0Proof. the point M corresponding to the complex number z belongs to C 0 if and only ifReflC.M/ 2 C , that is

˛1

Nz1

Nz � N!1

Nz � !1

Nz C D 0

since the complex number correspnding to ReflC.M/ is 1Nz . Multiplying by z Nz we finally get z Nz � N!z � ! Nz C ˛ D 0: �

Theorem. A circle C with equation

˛z Nz � N!z � ! Nz C D 0

is globally invariant by ReflC if and only if

˛ D

that is, if and only if it is orthogonal to C.

Theorem. The reflection in the x-axis (the line y D 0) is described by z 7�! Nz. Theimage through ReflOx of the circle C with equation ˛z Nz � N!z � ! Nz C D 0 is thus˛z Nz � !z � N! Nz C D 0. This circle is globally invariant if and only if ! D N!, whichmeans :if C is a genuine circle, then it is centered onOx ; if C is a line, then it is orthogonal toOx.


Consequences

We denote ReflC by f .

Proposition 1. The image of a line ` through O is itself.

O

M

M 0

� `

Proof. ` is orthogonal to C. �

Proposition 2. Let ` be a line such that O … `. Denote by H the orthogonal projection ofO on ` and by H 0 the image by f of H . The image by f of ` is the circle `0 with diameterOH 0.

O

M

M 0� `

H H 0

`0

Proof. The equation of ` may be written

� N!z � ! Nz C D 0where ¤ 0. The equation of the image `0 is then

z Nz � N!z � ! Nz D 0which is the equation of a genuine circle passing through O . The point H is the point on `which is nearest O . Thus its image H 0 is the point on `0 wich is the farthest from O , that isthe point diametrically opposed to O . �

Remark that we have got at the same time f .`0/ D ` and so we have proved the follo-wing proposition.

Proposition 3. Let `0 be a circle such that O 2 `0. Denote by H 0 the point diametricallyopposed to O on that circle and by H the image by f of H 0. The image by f of `0 is theline ` through H and orthogonal to OH 0.Remark. If the circle `0 intersects � in two points E et E 0, then the line ` is just the lineEE 0. If the circle `0 intersects � in exactly one point T , then the line ` is just the line tangentto � (and to `0) in T .

§ 4. IMAGES OF CIRCLE BUNDLES 23

Proposition 4. The transformation f transforms circles into circles, preserves the measureof angles and changes the orientation.

Proof. Look at the following drawings :

O

M 0

M

C

`

M 0M

and get convinced ! �

§ 4. Images of circle bundlesTheorem. Let P andQ be two points and let f be any reflection (through a line or througha genuine circle). Let P 0 and Q0 be the images by f of P and Q. The image by f of thecircle bundle with base points P andQ is the circle bundle with base points P 0 andQ0. Theimage by f of the circle bundle with limit points P and Q is the circle bundle with limitpoints P 0 and Q0.Proof. The image by f of a circle C is a circle C 0 and if P 2 C , then P 0 2 C 0 and the samewith Q. Thus any circle in the bundle B with base points P and Q is a circle belonging tothe bundle B0 with base points P 0 and Q0. Conversely, any circle in B0 can be obtained inthis way.

The circles in the bundle B� with limit points P and Q are those that are orthogonal tothe circles belonging to B. Since f transforms circles into circles and preserves orthogona-lity, the image of B� is the bundle B�0 with limit points P 0 and Q0. �Remark. If Q D 1, then the bundle B with base points P and Q is the bundle of linesthrough P (and through1) and the bundle B0 with limit points P and Q is the bundle ofcircles with center P (including the point circles P and1).

Theorem. Let P be a point and ` a line through P and let f be any reflection (through aline or through a genuine circle). Let P 0 and `0 be the images by f of P and `. The imageby f of the tangent bundle B of circles through P and tangent to ` is the tangent bundle B0of circles through P 0 and tangent to `0.Proof. The image by f of a circle C is a circle C 0 and if P 2 C , then P 0 2 C 0. Thereflection preserves contacts, thus any circle through P and tangent to ` is a circle throughP 0 and tangent to `0. Conversely, any circle in B0 can be obtained in this way. �Remark. If P D1, then the bundle B is the bundle of lines parallel to `.

Theorem. Let C be a circle and P and Q two points such that ReflC .P / D Q. Then Cbelongs to the circle bundle with limit points P and Q.


Proof. Let � be the center of the circle C and R its radius. The relation ReflC .P / D Q

means that�P ��Q D R2. The circle C is thus orthogonal to all the circles going throughP and Q. Then the circle C and the point circles P and Q all belong to the circle bundlewith limit points P and Q. �

P Q

C

�

Exercices

Exercice 1. Let C be a circle and P and Q two points inside C .a) Show that there is a circle �1 going through P and Q and orthogonal to C . Give a

construction of �1.b) Show that there is a circle �2 orthogonal to C and such that Refl�2 exchanges the

points P and Q. Give a construction of �2.c) Show that �1 and �2 are orthogonal.

Exercice 2. Let C be a circle and ` a line going through the center of C . Find an inversionexchanging C and `.

Chapitre 3

Non-Euclidean Geometry

§ 1. chapter 6.1.1. Non-Euclidean Geometry : thetwo usual models of a hyperbolic plane

§ 2. 6.1.2 Existence of hyperbolic lines§ 3. 6.1.3 Inversion preserves inversion points

§ 4. 6.2.1-6.2.3. The group of Hyperbolic Geometry

For Euclid’s Elements, please look athttp ://aleph0.clarku.edu/ djoyce/java/elements/elements.html

§ 1. chapter 6.1.1. Non-Euclidean Geometry : the two usual models of a hyperbolicplane

The unit disc of Poincaré

Set of points. The points z D x C iy D .x; y/ such that x2 C y2 < 1 i.e.

D D fz 2 C j jzj < 1 gThe set D is called hyperbolic plane.

Boundary points. The points z D x C iy D .x; y/ such that x2 C y2 D 1 or C D fz 2C j jzj D 1 g are not points of the hyperbolic plane, but the set C of these points plays animportant role. The points of C are called boundary points.

Set of lines. The lines are the intersection of D with the (generalized) circles orthogonal tothe unit circle C. We call these lines d�lines or simply lines.

Remark 1. Remember that a circle � is orthogonal to C if and only if it is invariant in thereflection through C. Let ˛.x2Cy2/�2Ax�2ByC D 0 be the equation of � . The circle� is orthogonal to C if and only if ˛ D .

Theorem 4. Given two distinct points P andQ in D, there is a unique line (or more precie-sely d�line) going through P and Q.

Proof. Since P … C and Q … C, there is a unique circle � going through P and Q andorthogonal to C. In fact � has to belong to the circle bundle B with base points P and Qand has to be orthogonal to the circle C which does not belong to B. �

25

26 THÈME I. SUMMARY. CH. 3. NON-EUCLIDEAN GEOMETRY

Remark 2. The infinitesimal length ds in D is related to the infinitesimal euclidean lengthjdzj by the formula

ds D jdzj1 � jzj2

The geodesics computed from that are the d�lines. One immediate consequence of thatformula is that the angles in the hyperbolic plane D are the same as those in the Euclideanplane. This is of course also valid for orthogonality.

Remark 3. In the disc model, the point O.0; 0/ seems to be a special point. In Euclideangeometry it is since it is the center of C, but in hyperbolic geometry it is just a point like theothers and the space is completely homogeneous : all the points play the same role.

Group GD. The group of this geometry is the set of all the transformations obtained bycompositions of reflections (through d -lines).

The half-plane of Poincaré

Set of points. The points z D x C iy D .x; y/ such that y > 0 i.e.

H D fz 2 C j =.z/ > 0 gThe set H is called hyperbolic plane.

Boundary points. The set of points z D x C iy D .x; y/ such that y D 0, that is, the setL D fz 2 C j =.z/D0 g is not included in the hyperbolic plane, but plays an importantrole. The points of L are called boundary points.

Set of lines. The lines are the intersections ofH with the (generalized) circles orthogonal tothe line L. We call these lines h�lines or simply lines.

Theorem 4. Given two distinct pointsP andQ inH, there is a unique h�line going throughP and Q.

Proof. Since the line L is a circle and since P … L and Q … L, there is a unique circlethrough P and Q and orthogonal to L (look at the picture below in the case the line PQ isnot parallel to the y-axis). �

Remark 1. Let � be a circle with equation ˛.x2 C y2/ � 2Ax � 2By C D 0. The circle� is such that � \H is a h�line if and only if

B D 0 and .˛; A/ ¤ .0; 0/

§ 1. CHAPTER 6.1.1. NON-EUCLIDEAN GEOMETRY : THE TWO USUAL MODELS OF A HYPERBOLIC PLANE 27

Examples of h�lines.y

O x

H

L

P

Q

Remark 2. The infinitesimal length ds in H is related to the infinitesimal euclidean lengthjdzj by the simple formula

ds D jdzjy2

The geodesics computed from that are the h�lines. One immediate consequence of thatformula is that the angles in the hyperbolic plane H are the same as those in the Euclideanplane. This is of course also valid for orthogonality.

Remark 3. In the half-plane model, the line x D 0 seems to be a special point, but inhyperbolic geometry it is just a h-line like all the other h-lines and the space is completelyhomogeneous : all the points play the same role and all the lines play the same role.

Group GH. The group of this geometry is the set of all the transformations obtained bycompositions of reflections (through h-lines).

Holomorphic transformations exchanging D and H

We are looking for a conformal transformation that transforms D into H. Is there sucha transformation which is a homography z 7! azCb

czCd , with ad � bc ¤ 0 ? There are severalanswers but we just want one, eventually as simple as possible The choice has to be suchthat the image of the circle z D ei� becomes the real line y D 0, i.e. z � z D 0, thus for all�

aei� C bcei� C d D

Nae�i� C NbNce�i� C Nd

or.aei� C b/. Nce�i� C Nd/ D . Nae�i� C Nb/.cei� C d/

that is.a Nd � c Nb/ei� C .a Nc � Nac C b Nd � Nbd/C .b Nc � Nad/e�i� D 0


The condition becomes �a Nd � c Nb D 0

a Nc � Nac C b Nd � Nbd D 0

We want also that 0 7�! i , i.e. b D id . Let us choose d D 1, then b D i and�aC ic D 0

a Nc � Nac C i C i D 0

thus c D ia and �ia Na � NaiaC 2i D 0 and finally a Na D 1. Thus a D ei' , b D i , c D iei'and d D 1. Thus the transformation is

z 7�! ei'z C iiei'z C 1 D �i

z C ie�i'z � ie�i'

Let us choose ' such that ie�i' D 1, we get

z 7�! i1C z1 � z

This transformation transforms 1 into1, i into i 1Ci1�i D i .1Ci/

2

2D �1, �1 into 0 and �i

intoC1.The inverse transformation from H to D is

z 7�! z � iz C i

the image of1 is 1, the image of �1 is i , the image of 0 is �1 and the image of 1 is �i .The image of a real x is ei� where � is the measure in radians of the oriented angle in thefollowing picture

i

�i

x

x � i

x C i H

L�

§ 1. CHAPTER 6.1.1. NON-EUCLIDEAN GEOMETRY : THE TWO USUAL MODELS OF A HYPERBOLIC PLANE 29

Parallelisme and ultraparallelisme

Definition. Two d�lines (respectively h�lines) are– intersecting if they have exactly one common point ;– parallel if the circles of which they are parts have a common point on the boundary ;– ultraparallel if the circles of which they are parts have no common point in D [ C

(respectively H [ L).

Examples. `1 and `2 are intersecting, `3 and `4 are parallel, `4 and `5 are parallel, `3 and`5 are parallel, `3 and `6 are parallel,`5 and `6 are parallel, `1 and `3 are ultraparallel,. . . ,`4 and `6 are ultraparallel.

y

O x

H

L

`1

`2

`3

`4 `5

`6

Group of transformations

Theorem 1. Let ` be a d-line (respectively h-line) which is part of a genuine circle � ortho-gonal to C. The reflection with repect to � , denoted r� is such that

r�.D/ D D and r�.C/ D CMore precisely, the regions D1 and D2 are exchanged.

�D1 D2

C

`


Questions. How is the preceding theorem if ` is part of a Euclidean line ? Draw the picturescorresponding to the H-plane.

Definitions. A hyperbolic reflection is a restriction to D of a reflection in a (generalizedeuclidean) circle, whose intersection with D (respectively H) is a d -line (respectively h-line).

The group of the hyperbolic plane is the set of transformations that can be written ascompositions of reflections, called hyperbolic transformations.

Now recall that in the two models, the hyperbolic angles are the same as the Euclideanangles. Since inversions transform circles into circles and preserve the magnitudes of angles,we have the following theorem.

Theorem 2. The hyperbolic transformations transform lines into lines and preserve the ma-gnitudes of angles.

§ 2. 6.1.2 Existence of hyperbolic linesOrigin Lemma. Let A be a point of D distinct from O . There is a (hyperbolic) reflection rsuch that r.A/ D O and r.O/ D A.

Comment. In the hyperbolic planeD, all the points play the same role. The pointO is specialin the representation we have chosen, but any other point could have been chosen to be thecenter of the boundary. Thus the above lemma is in fact equivalent to the following theorem.

Theorem. Given two points P and Q in D, there is one hyperbolic reflection r such thatr.P / D Q (and then r.Q/ D P ).

Comment. In Euclidean geometry, given two points P and Q there is one line ` such thatthe reflection in ` exchanges P and Q. The line is the bisector of the segment PQ. Inthe hyperbolic context, we shall see later (Theorem 3, page 293) that the hyperbolic line `(such that f , the reflection in `, exchanges P and Q) will be the hyperbolic bisector of thehyperbolic segment PQ (which is the arc of the circle going through P and Q, which isorthogonal to C and included in D).

Proof. We are looking for a circle � belonging to the circle bundle with limit points P andQ and orthogonal to C. Since C does not separate the points P andQ, there is one and onlyone such circle � and ` D � \D. �Geometrical construction of the line `.

In the disc model

P

B bundle with base points P and QB0 bundle with limit points P and Q

Q

C`

§ 3. 6.1.3 INVERSION PRESERVES INVERSION POINTS 31

and in the half plane model

P

Q

L

`

Theorem 3. Let A be a point in D. There exists infinite many lines through A.

Proof. Let � be any circle around A and included in D. For any point B on � there is aunique line AB wich is part of a Euclidean circle C . Thus C cannot have more than twopoints common with � . If C had 3 or more common points with � , it would be the circle �which is not orthogonal to C. �

The theorem 4 has already been proven.

Theorem 4. Given two distinct points P andQ inH, there is a unique hyperbolic line goingthrough P and Q.

Theorem 5. Let `1 and `2 be two lines, let A1 be a point in `1 and A2 a point in `2. Thenthere exists a transformation f such that f .A1/ D A2 and f .`1/ D f .`2/.

Partial proof. We have to know more about rotations to fulfil the proof, but let us admit thatgiven two lines through a point there is a rotation transforming one into the other.

By any suitable reflection s transform A1 into A2, then the line `1 is transformed into aline `0 that goes through A2. Then by a rotation r transform `0 into `2. The transformationf D r ı s suits the conditions of the theorem. �

§ 3. 6.1.3 Inversion preserves inversion points

Recall that "line" means hyperbolic line, if nothing else is specified.

Theorem 6. Let A, B be images of each other in a reflection f through a line `. Let `� beany line and let f � be the reflection through `�. Call A0, B 0 and `0 the images by f � of A,B and `. Then A0 and B 0 are images of each other in the reflection f 0 through the line `0.


§ 4. 6.2.1-6.2.3. The group of Hyperbolic Geometry

The inversions in C [ f1gIn the complex plane, the inversion with center ˛ and radius R is described by

z 7�! ˛ C R2

z � ˛If the circle � with center ˛ is orthogonal to C, then j˛j2 D R2C 1 and the inversion or

reflection through ` D � \D becomes

�.z/ D ˛z � 1z � ˛

What happens if ` is a line ? The formula becomes �.z/ D ei�zTheorem 1 and 2. The direct hyperbolic transformation can always be described by a Mö-bius transformation of the form

M.z/ D az C bbz C a where jbj < jaj

The reflection B W z 7�! Nz is an indirect hyperbolic transformation ; all indirect trans-formations may be written as z 7�!M.z/ where M is as above.

Rotations and translations

The direct transformation can always be described as the composition of 2 reflections in2 lines `1 and `2.

If `1 and `2 intersect, the transformation is a "rotation". If `1 and `2 are parallel thetransformation is a "limit rotation". If `1 and `2 are ultraparallel, the transformation is a"translation".

But be careful : translations do not commute in general.

4.1 The Canonical Form of a Hyperbolic Transformation

Direct :z 7�! K

z �m1 �mz where jKj D 1 and jmj < 1

Indirect :

z 7�! Kz �m1 �mz where jKj D 1 and jmj < 1

4.2 6.3.1 The distance formula

Properties of a distance

1. 8.z1; z2/ 2 D2 W d.z1; z2/ > 08.z1; z2/ 2 D2 W d.z1; z2/ D 0” z1 D z2

2. 8.z1; z2/ 2 D2 W d.z1; z2/ D d.z2; z1/

3. 8.z1; z2; z3/ 2 D3 d.z1; z2/C d.z2; z3/ > d.z1; z3/4. 8`; line 8.z1; z2; z3/2`3 where z2 between z1 and z3 W d.z1; z2/Cd.z2; z3/ D d.z1; z3/5. 8.z1; z2/ 2 D2 W d.z1; z2/ D d.z1; z2/6. 8M2 Direct subgroup of D;8.z1; z2/ 2 D2 d.M.z1/;M.z2// D d.z1; z2/

x D th�1.y/ D 1

2ln1C y1 � y

The formula

d.z1; z2/ D th�1�ˇ̌̌ z1 � z21 � z1z2

ˇ̌̌�

Thème II

Reading Brennan & All

Chapitre 4

6.1-6.3 Non-Euclidean Geometry,Transformations, Distance

§ 1. Page 263. What is Non-Euclidean Geometry ?

1.1 Two answers :

1. Set of pointsC group of transformationsC invariant objects.2. Start from Euclid.

Let us look at the first answer : there are several models to describe the hyperbolicplane : D, H, the upperhalf of the hyperboloide z2 � x2 � y2 D 1. In this last modelthe lines are the intersections of that surface with the planes going through the origine O .The central projection through O of that surface onto the plane z D 1 gives you the Kleinmodel where the line are Euclidean segements in a unit disc. The central projection of thehalf-hyperboloid through .0; 0;�1/ onto the plane z D 0 gives you the D model.

Let us look at the second answer :If we keep among the five postulates, the postulates 1, 2 and 5 we are going to "Affine

Geometry" where there is no difference between circle and ellipse. If we keep the postulates1,2,3 and 4 we are going first to "Ordered Geometry", then to "Absolute Geometry" (thechoices of adjectives are those of Coxeter) and finally to elliptic, Euclidean or hyperbolicGeometry. In Brennan Non-Euclidean means hyperbolic.

1.2 Ordered Geometry

The basic concept is intermediacy or betweenness, denoted ŒABC � to say that the pointsA, B and C are on a same line and B is between A and C . One starts from the followingaxioms

1. 92 points2. 8A;B; 9C such that ŒABC �

3. ŒABC � H) A ¤ C4. ŒABC � H) ŒCBA� and not ŒBCA� ; from here you may define the words : ray, seg-

ment and line.

35

36 THÈME II. READING BRENNAN.CH. 4. 6.1-6.3 NON-EUCLIDEAN GEOMETRY, TRANSFORMATIONS, DISTANCE

5.C ¤ D

C 2 line ABD 2 line AB

9=; H) A 2 line CD

6. 9 line AB; 9Dsuch that D … line AB7. (Better than Pasch’s axiom) Given triangle ABC and points D and E such thatŒBCD� and ŒCEA�, the line DE intersects the open interval AB .

A

B C D

E

Sylvester’s problem of collinear points

Sylvester’s conjecture (1893) : It is not possible to arrange any finite number of pointsso that a line through every two of them passes through a third, unless they all lie in thesame line.

It was solved only 40 years later !Equivalent statement : If n points are not all collinear, then there is at least one line

containing exactly two of them (in fact at least 3n7

).The theorem can be proved in ordered geometry and then it is valid in all the geometries

in which the axioms of ordered geometry are valid. The proof in usual Euclidean plane canbe summerized as follows : consider all the couples (point, line) where "point" is one ofthe n given points P and "line" is a line passing through exactly two of the n given pointsother than P . To each such couple (point, line) associate the distance of the "point" to the"line" and call S the set of distances obtained in such a way. If the set S is not empty, it isfinite and thus has a smallest element d . Then you prove that you can build a new couplefor which the distance is less than d , which is contradictory. Thus S D ¿.

This proof was not easy to find because everyone could feel that the truth of the state-ment had nothing to do with the metric of the plane...

1.3 Absolute Geometry

Absolute Geometry=Ordered geometry C Congruence.

§ 2. Page 268-269

2.1 Theorem 4 : Through two points there is a unique line

Proof. We are looking for a circle passing through to points p and q and orthogonal toC : look at the bundle B with limit points p and q. The circle C does not belong to the bundle

§ 3. PAGE 288 37

B, thus there is exactly one circle orthogonal to the three circles C and the two point circlesp and q. �

As a consequence we get the Lemma 1, page 268, since O is a point like all the othersin the hyperbolic geometry. We get also some methods to construct the line ` such thatRefl`.p/ D q.

§ 3. Page 288

3.1 Example 3 : Midpoint of 14i and 3

4i

1

r

We have to find X such that�X2 D 1C r2

.X � a/.X � b/ D r2

Eliminating r we getX2 � .aC b/X C ab D X2 � 1

i.e.

X D ab C 1aC b D

14� 34

14C 3

4

D 19

16

and

r DpX2 � 1 D 1

16

p192 � 162 D

p105

16

thus

m D X � r D 19 �p10516

D 0; 547 065 577 13 : : :


§ 4. Page 289

Use origin lemma and preservation of circles through reflections : all Euclidean circlesINCLUDED IN D are hyperbolic circles (but the centers are not the same !).

§ 5. Page 294. Reflection lemma

Remark. The book’s ˛ is my w.A circle which is orthogonal to C has the equation

z Nz � w Nz � Nwz C 1 D 0i.e.

z D w Nz � 1Nz � Nw

The reflection through this circle is given by

z0 D w Nz � 1Nz � Nw

We just change z into z0 in the former equation ! I do not understand why it is so simple !but it is.

§ 5. PAGE 294. REFLECTION LEMMA 39

TUESDAY, APRIL 1

Exercise 1. Choose three points A, B and C in a hyperbolic plane in the Poincaré discmodelD, such that they are not (hyperbolic-)colinear. Construct the three internal bisectors.Do they always have a common point ? Check your idea by moving your initial points andmake a conjecture. Prove your conjecture.

A

BC

O

Hints :� Introduce A0 D ReflC.A/� The bisector of a hyperbolicangle has to be the bisector of theangle made by the tangents to thecircles.

Exercise 2. Let ` be a hyperbolic line in D and let P be a point in D. Construct the(hyperbolic-)line through P orthogonal to `. Hint :

P

P 0C

O

`

`?

Exercise 3. Using again exercise 1, let I be the point of intersection of two angle bisectors.Construct the three lines orthogonal to the three sides of the triangle ABC . Let A0 be thepoint of BC such that the lines IA0 and BC are orthogonal. Let B 0 be the point of CA suchthat the lines IB 0 and CA are orthogonal. Let C 0 be the point of AB such that the lines IC 0and AB are orthogonal. Construct the circle � going through A0, B 0 and C 0. What can yousay about the circle � ?Exercise 4 to 7. Are the medians, the segment bisectors, the altitudes, the outer angle bisec-tors concurrent ?

Chapitre 5

6.4 Non-Euclidean Geometry,Transformations, Distance

§ 1. Euclidean Triangles

1.1 Médianes et centre de gravité

Définition. Soit ABC un triangle. On note A0 le milieu du segment BC . On appelle mé-diane du triangle issue de A la droite AA0 ou le segment AA0 ou encore la longueur de cesegment.

A

B CA0Théorème. Les médianes d’un triangle sont concourantes.Démonstration. Soit ABC un triangle. On note A0 le milieu du segment BC , B 0 le milieude CA et C 0 le milieu de AB . Les droites BB 0 et CC 0 sont sécantes, car si elles étaientparallèles, on aurait par le théorème de Thalès BA D �BC 0 D �1

2BA, soit B D A ce

qui est impossible. Notons G le point d’intersection de BB 0 et de CC 0. La réciproque duthéorème de Thalès nous donneB 0C 0==BC etB 0C 0 D 1

2CB . Appliquons alors le théorème

de Thalès aux droites BB 0 et CC 0 coupées par les sécantes BC et B 0C 0. Il vient

GB 0

GBD �1

2ou encore

GB 0

BB 0D 1

3

ce qui détermine le point G de façon unique (« au tiers de la médiane à partir de la base »). G

C

A

B CA0

B 0

Appliquons notre résultat au triangle CBA : les médianes BB 0 et AA0 se coupent en unpoint G0 situé au tiers de la médiane BB 0. Le point G0 est donc égal à (en géométrie on dit« confondu avec ») le point G, c’est-à-dire que la médiane AA0 passe par G.�Remarque. La démonstration utilisant le calcul barycentrique est beaucoup plus simple :3G D AC B C C D AC 2A0 D B C 2B 0 D C C 2C 0.Définition. Le point de concours des médianes d’un triangle s’appelle le centre de gravitédu triangle.Remarque. Le centre de gravitéG d’un triangleABC est l’équibarycentre de ses sommets.Mais le nom vient de ce que G est aussi l’équibarycentre de l’enveloppe convexe de ses

41

42 THÈME II. READING BRENNAN.CH. 5. 6.4 NON-EUCLIDEAN GEOMETRY, TRANSFORMATIONS, DISTANCE

sommets. Le point G est le centre de gravité au sens physique d’une plaque homogèneinfiniment mince de forme triangulaire de sommets A, B et C .

C 1. Démontrer l’affirmation de la remarque précédente.

B C

G

C1 B1

A1

A

A0

B 0C 0

Corollaire. Soit ABC un triangle. On note A0,B 0 et C 0 les milieux des côtés et G le centrede gravité. On note A1, B1 et C1 les pointstels que ABA1C , BCB1A et CAC1B soientdes parallélogrammes. L’homothétie de centre Get de rapport �2 transforme le triangle A0B 0C 0en ABC et ABC en A1B1C1 ; l’homothétie decentre G et de rapport �1

2transforme A1B1C1

en ABC et ABC en A0B 0C 0. Ces trois tri-angles ont les mêmes médianes, les divisions.A;A0IA1; G/, .B;B 0IB1; G/ et .C; C 0IC1; G/sont harmoniques et A0 est le milieu de A1A, B 0 lemilieu de B1B et C 0 le milieu de C1C .

A0GA A1

1.2 Médiatrices et cercle circonscrit

Rappel. (admis) La médiatrice d’un segment AB est la droite�, ensemble des points équi-distants de A et de B . La droite � est la perpendiculaire à AB passant par le milieu M dusegment AB . On appelle médiatrices d’un triangle les médiatrices des côtés du triangle.

Théorème. Les médiatrices d’un triangle sont concourantes en un point qui est le centre del’unique cercle qui passe par les trois sommets du triangle.

A B

M

� Démonstration. Soit ABC un triangle. Les médiatrices des segments AB et CA ne sontpas parallèles car sinon les sommets du triangle seraient alignés. Soit O le point commun àces médiatrices. Comme O appartient à la médiatrice de AB , on a OB D OA et comme Oappartient à la médiatrice de AC , on a OA D OC . On en déduit OB D OC , ce qui prouveque la médiatrice de BC passe par O .

Le cercle de centreO et de rayonOA passe donc par B et par C . Il existe donc un cerclepassant par les trois sommets du triangle. Enfin si un cercle passe par les trois sommets dutriangle, son centre est sur les médiatrices des trois côtés et est donc confondu avec O etson rayon est OA.�

Définitions. L’unique cercle qui passe par les trois sommets d’un triangle est appelé lecercle circonscrit au triangle et son centre O est appelé le centre du cercle circonscrit autriangle.

A

B C

O

A0

§ 1. EUCLIDEAN TRIANGLES 43

Définition. Soit ABC un triangle. On note A0, B 0 et C 0les milieux des côtés. Le cercle circonscrit au triangleA0B 0C 0 est appelé le cercle d’Euler ou cercle des neufpoints ou cercle de Feuerbach du triangle ABC .Nous noterons O 0 le centre du cercle circonscrit àA0B 0C 0 et H le centre du cercle circonscrit à A1B1C1.L’homothétie de centre G et de rapport �2 transformeO 0 enO etO enH ; l’homothétie de centreG et de rap-port �1

2transforme H en O et O en O 0. On remarque

que les points O , G, O 0 et H sont alignés, que la di-vision .O;O 0IH;G/ est harmonique et que O 0 est lemilieu de HO .La droite HO 0GO s’appelle la droite d’Euler du tri-angle ABC .

H G OO 0

HG

O

A

B C

A1

B1C1

C

A0

B 0

O 0

1.3 Quadrangle orthocentrique

Lemme 1. Soit Eu, Ev et Ew trois vecteurs, alors Eu � . Ew � Ev/C Ev � .Eu � Ew/C Ew � .Ev � Eu/ D 0.

Lemme 2. Soit A, B , C et D quatre points, alors

��!AB � ��!CD C��!AC � ��!DB C��!AD � ��!BC D 0 (5)

Démonstrations. Lemme 1. Développer et simplifier.Lemme 2. Appliquer le lemme 1 en posant Eu D ��!AB , Ev D ��!AC et Ew D ��!AD.�

Définition. SoitABC un triangle. On appelle hauteur du triangleABC issue deA la droitepassant par A et perpendiculaire à BC . L’intersection P de la hauteur issue de A avec ladroite BC s’appelle le pied de la hauteur issue de A.

Nous noterons Q le pied de la hauteur issue de B et R le pied de la hauteur issue de C .Le triangle PQR est appelé le triangle orthique du triangle ABC .

Théorème. Les hauteurs d’un triangle sont concourantes.

P

A

B C

QR

Première démonstration. Les hauteurs du triangle ABC sont les médiatrices du triangleA1B1C1. Elles sont donc concourantes en le pointH , centre du cercle circonscrit au triangleA1B1C1.�Deuxième démonstration. Les hauteurs CR et BQ ne sont pas parallèles car sinon lessommets A, B et C seraient alignés. Soit H le point d’intersection de CR et de BQ. Ona alors AB perpendiculaire à CH , soit

��!AB � ��!CH D 0, et AC perpendiculaire à BH , soit��!

AC � ��!HB D 0. De la relation (5) du lemme 2, on déduit alors��!AH � ��!BC D 0, ce qui montre

que AH est la hauteur issue de A.�

Définition. On appelle orthocentre d’un triangle le point de concours de ses hauteurs. P

A

B C

QR

A1

B1C1

H

Proposition. SiH est l’orthocentre d’un triangleABC , alorsA est l’orthocentre du triangleBCH , B est l’orthocentre du triangle ACH et C est l’orthocentre du triangle ABH .


Définition. Un quadrangle est un ensemble de quatre points, appelés ses sommets. Il estorthocentrique si chacun de ses sommets est l’orthocentre du triangle formé par les troisautres sommets du quadrangle.

Proposition. Un quadrangle est orthocentrique si un sommet du quadrangle est l’ortho-centre du triangle formé par les trois autres sommets.

Proposition. Soit ABCD un quadrangle orthocentrique. Les quatre triangles BCD, ACD,ABD et ABC ont même triangle orthique.

Définition. On appelle triangle orthique d’un quadrangle orthocentrique ABCD le triangleorthique commun aux quatre triangles BCD, ACD, ABD et ABC .

Théorème. Le cercle d’Euler est le cercle circonscrit au triangle orthique.Démonstration. Soit ABC un triangle, P le pied de la hauteur issue de A et A0 le milieudu segment BC . On a vu que le centre O 0 du cercle d’Euler est le milieu du segment HO .Il résulte alors du théorème de Thalès que O 0 est sur la médiatrice de PA0. On a doncO 0P D O 0A0, ce qui montre que P appartient au cercle d’Euler de centre O 0 et de rayonO 0A0. Ce résultat appliqué aux triangles BCA et CAB montre que les pieds des hauteursQ et R appartiennent aussi au cercle d’Euler.�

H

O

A

B CP A0

O 0

Le cercle circonscrit au triangle orthique PQR passe donc par les milieux des côtés dutriangle ABC , mais pour la même raison par les milieux des côtés du triangle BCH , etc. Ilen résulte que le cercle d’Euler passe par les 9 points suivants :

P; Q; R; A0; B 0; C 0; Ma milieu de AH; Mb milieu de BH; Mc milieu de CH

C’est pourquoi ce cercle est aussi appelé le cercle des 9 points. Remarquer que comme lesangles en P , Q et R sont droits, les segments A0Ma, B 0Mb et C 0Mc sont des diamètres ducercle d’Euler.C 2. Il y a une lacune dans le raisonnement précédent dans lequel on a supposé que PQR est un triangle. Il estdemandé au lecteur de la combler.

Exercice 1. Soit H l’orthocentre d’un triangle ABC . Montrerque les cercles circonscrits aux triangles BCH , ACH , ABH etABC ont même rayon.

Exercice 2. Soit H l’orthocentre d’un triangle ABC . Montrerque les symétriques de H par rapport aux côtés du triangle appar-tiennent au cercle circonscrit à ABC .

Exercice 3. SoitH l’orthocentre d’un triangleABC ,A0 le milieudu segmentBC ,B 0 le milieu deCA,Ma le milieu deAH etMble milieu de BH . Montrer que A0B 0MaMb est un rectangle.

Exercice 4. Montrer que la médiane d’un triangle rectangle est undiamètre de son cercle d’Euler.

Exercice 5. SoitH l’orthocentre d’un triangleABC etPQR sontriangle orthique. Montrer que les quadrangles BCQR, CARP ,ABPQ, ARHQ, BPHR et CQHP sont inscriptibles. Jus-tifier les égalités suivantes : .PQ;PA/ D .PQ;PH/ D.CQ;CH/ D .CQ;CR/ D .BQ;BR/ D .BH;BR/ D.PH;PR/ D .PA;PR/. En déduire que les hauteurs et les cô-tés du triangle ABC sont les bissectrices de son triangle orthiquePQR.

1.4 Bissectrices, cercle inscrit et cercles exinscrits

Définition. Soit ABC un triangle. On appelle bissectrices de l’angle A les deux bissec-trices des deux droites AB et AC . On appelle bissectrice intérieure celle qui est l’axe desymétrie des demi-droites ŒAB/ et ŒAC / issues de A et contenant respectivement les pointsB et C . L’autre bissectrice s’appelle la bissectrice extérieure (orthogonale à la bissectriceintérieure).

§ 2. PAGE 298 45

Théorème. Les bissectrices intérieures d’un triangle sont concourantes en un point I , inté-rieur au triangle. La bissectrice intérieure issue d’un sommet A passe par le point Ia, pointd’intersection des bissectrices extérieures aux deux autres sommets.Définition. Soit ABC un triangle. On appellecercle inscrit au triangle ABC le cercle decentre I tangent aux trois côtés du triangle etintérieur au triangle. Le point I est désignécomme centre du cercle inscrit. On appellecercles exinscrits au triangleABC les cerclesde centre Ia, Ib et Ic tangents aux trois côtésdu triangle et extérieurs au triangle. Les pointsIa, Ib et Ic sont désignés comme centres descercles exinscrits.Remarque. Il est facile de démontrer queles bissectrices de deux sommets B et C nepeuvent pas être parallèles et se coupent doncen 4 points équidistants des trois côtés du tri-angle. Les bissectrices de A passent donc parces 4 points. La partie difficile est de montrerque les bissectrices intérieures sont concou-rantes. Une méthode simple consiste à utili-ser le calcul barycentrique. Nous renvoyonsle lecteur au chapitre 4, §4 pour avoir lescoordonnées barycentriques des centres descercles inscrit et exinscrits, ce qui permet unedémonstration simple du théorème ci-dessus.

A

CB

I

Ic

Ib

Ia

C 3. Le quadrangle IaIbIcI est orthocentrique et son triangle orthique est le triangle ABC . Comparer cetteproposition au résultat de l’exercice 5.

§ 2. Page 298

§ 3. Page 303

§ 4. Page 307-312 Pythagoras’ theorem and Lobachevski’s formula

Chapitre 6

6.5 Tessellation

THURSDAY, APRIL 3Exercise 1. Construct a hyperbolic segment that does not disappear when you move around its ends inside the discD.

B

A

A0

O

MN

Hints : � Introduce A0 D ReflC.A/.� The point M is the Euclidean

midpoint of the Euclidean segment AB .�N is at the intersection of the seg-

ment OM and the circle A0AB .

Move around the points A and B to check the rigidity of the construction.Exercise 2. Construct a regular octogone centered atO .

O

Exercise 3. In the following drawing in Euclidean plane, show that x D cos˛qcos2 ˛ � sin2 ˇ

, where x D OA,

47

48 THÈME II. READING BRENNAN. CH. 6. 6.5 TESSELLATION

˛ D 1OMA0 and ˇ D1AOM .

AA0O

T

M

49

Exercise 4. We intend to do a tessellation of the hyperbolic plane with regular polygons all of same size, notoverlapping (but two by two with common boundaries) and covering the whole plane. We want

– to have regular polygons with n edges ;– that each vertex is common to p such regular polygons.Let us call a a hyperbolic segment which is a side of a regular polygon with n sides centered atO . Let A be

the center of the circle � such that a � � .For instance, with n D 4 and p D 6, we want something beginning like (but covering the whole disc D).

A

:O

a

M

Show that the distanceOA has to be

OA D cos.�p/q

cos2.�p/� sin2.�

n/

Choose some values for n and p such that 1nC 1p< 12

, n > 3 and p > 3, compute the length OA andmake the drawing, beginning by drawing the circle C with radius 1 and putting the point A such that the distanceOA is the one you have computed.

hyperbolic geometry - uefcs.uef.fi/matematiikka/kurssit/hyperbolicgeometry/...chapitre 1 power of a...

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