Hyperbolic Geometry

Transcription

Hyperbolic Geometry

Hyperbolic Geometry
Eric Lehman
Mars-April 2014
2
3 Non-Euclidean Geometry
§ 1. chapter 6.1.1. Non-Euclidean Geometry : the two 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 . . . .
Table of contents
II. Reading Brennan
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I. Summary
1 Power of a point with respect to a circle
§ 1. Equations of a circle . . . . . . . . . .
§ 2. Orthogonal circles . . . . . . . . . . . .
§ 3. Circle bundles . . . . . . . . . . . . . .
§ 4. Existence of circles . . . . . . . . . . .
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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 . . . . . . . . . . . . .
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6.1-6.3 Non-Euclidean Geometry, Transformations, Distance
§ 1. Page 263. What is Non-Euclidean Geometry ? . . . . .
§ 2. Page 268-269 . . . . . . . . . . . . . . . . . . . . . .
§ 3. Page 288 . . . . . . . . . . . . . . . . . . . . . . . . .
§ 4. Page 289 . . . . . . . . . . . . . . . . . . . . . . . . .
§ 5. Page 294. Reflection lemma . . . . . . . . . . . . . .
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6.4 Non-Euclidean Geometry, Transformations, Distance
§ 1. Euclidean Triangles . . . . . . . . . . . . . . . . . . .
§ 2. Page 298 . . . . . . . . . . . . . . . . . . . . . . . . .
§ 3. Page 303 . . . . . . . . . . . . . . . . . . . . . . . . .
§ 4. Page 307-312 Pythagoras’ theorem and Lobachevski’s
formula . . . . . . . . . . . . . . . . . . . . . . . . .
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6.5 Tessellation
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TABLE OF CONTENTS
Thème I
Summary of February course and complements
Chapitre 1
Power of a point with respect to a
circle
§ 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{ ; jE/ 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 the
number
PC .M / WD d 2
R2 D .x
a/2 C .y
b/2
R2 D x 2 C y 2
2ax
2by C a2 C b 2
R2
Proposition. The point M belongs to the circle C if and only if PC .M / D 0, or
a/2 C .y
.x
b/2
R2 D 0
which is the normal equation of C . We may write the normal equation of C using complex
numbers as
zz
where z D x C iy
;z D x
.wz C wz/ C c D 0
iy
and
w D a C i b.
Theorem 1. Let C be a circle and M.x; y/ a point. For any line through M intersecting
C 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 circle
MP MQ D PC .M /, if M is inside the circle MP MQ D jPC .M /j D PC .M / and if
M is on the circle MP MQ D 0.
! !
Remark. Using the scalar product, we have in all cases MP MQ D PC .M /.
Q
C
C
P
P
M
M
Q
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 ; b 0 /. 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 b 2
R2 D x 2 C y 2
2a0 x
2b 0 y C a02 C b 02
R02
or
2.a0
a/x C 2.b 0
b/y C a02 C b 02
!
which is a line orthogonal to the vector 0 .
a2
b2
R02 C R2 D 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 this
radical axis. Then the radical axis is the line through K which is orthogonal to the line
joining 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 C
and 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 is
the line through T orthogonal to the line joining the centers of the two given circles, that is
the common tangent to these two circles..
§ 1. EQUATIONS OF A CIRCLE
5
1.3 Radical center of three circles
Theorem and definition. Let C , C 0 and C 00 be three circles with nonaligned centers .a; b/,
0 .a0 ; b 0 / and 00 .a00 ; b 00 / (nonaligned means that the points do not belong to a common
line). There is one point S such that PC .S / D PC 0 .S / D PC 00 .S /. This point is called the
radical 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 orthogonal
respectively to the distinct lines 0 00 and 00 . Thus and 0 intersect in a point S such
that PC 0 .S / D PC 00 .S / and PC 00 .S / D PC .S / and thus PC .S / D PC 0 .S /, which means
that S 2 00 .
How to construct the radical axis of two non intersecting circles with distinct centers.
2) Draw a circle intersecting C and C 0
1)
S

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

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

K
T0
0
Remark 2. Let p D PC .S / D PC 0 .S / D PC 00 .S / ; since these three quantities are equal
they have same sign : if p < 0, the point S is inside all three circles ; if p > 0, the point
S is outside all three circles ; if p D 0, the three circles have S as a common point. As a
consequence, we see that if two of the circles do not intersect each other, then S is outside
all three.
6
THÈME I. SUMMARY.
1.4 Generalized circles
The equation of the previous circle C may be written
˛.x 2 C y 2 /
2Ax
2By C D 0
where A D ˛a, B D ˛b and D ˛c D ˛.a2 C b 2
2
2
of a genuine circle when R2 > 0, that is A
C B˛2
˛2
choose ˛ D 0, we get the equation
2Ax
(1)
R2 /. If ˛ ¤ 0, we have the equation
> 0 or A2 C B 2 ˛ > 0. If we
˛
2By C D 0
which is the equation of a line if .A; B/ ¤ .0; 0/. Thus the equation .1/ may be used to
describe as well circles as lines. If we add one point at infinity, denoted 1, we may think
the lines as circles going through the point 1.
Notice that the linear combination of the equations of two circles is a circle when circle
is taken in the general sense. In particular, the radical axis of two genuine circles is obtained
using 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 in
two points T and T 0 such that the angles T 0 and T 0 0 are right angles (if C 0 is a point
circle, it is orthogonal to C when the point is on C ).
1
2
T
0

T0
A circle C..a; b/; R/ and a line with equation ux C vy C w D 0 are orthogonal if and
only if ua C 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. Two
lines with equations ux C vy C w D 0 and u0 x C v 0 y C w D 0 are orthogonal if
.uE{ C v |E/ .u0E{ C v 0 |E/ D 0, or uu0 C vv 0 D 0.
§ 2. ORTHOGONAL CIRCLES
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2.2 Characteristic properties
Theorem. Let C and C 0 be two genuine circles with respective centers .a; b/ and 0 .a0 ; b 0 /
and respective radii R and R0 , intersecting in points T and T 0 . The circles C and C 0 are orthogonal if and only if one of the following equivalent conditions is fulfilled :
1. 02 D R2 C R02
2. The triangle T 0 is rectangle in T
3. PC 0 ./ D R2
4. PC .0 / D R02
5. .a
a0 /2 C .b
b 0 /2 D R2 C R02
Theorem. Two (generalized) circles C and C 0 with real equations
˛.x 2 C y 2 /
2Ax
2By C D 0
and
˛ 0 .x 2 C y 2 /
2AA0
2BB 0 D 0
2A0 x
2B 0 y C 0 D 0
are orthogonal if and only if
˛ 0 C ˛ 0
Proof. 1°) If the two circles are genuine circles, the condition 5 above may be written
a2 C b 2
or since a2 C b 2
R2 D c D
R2 C a02 C b 02
˛
and a02 C b 02
R02 D 2aa0 C 2bb 0
R02 D c 0 D
0
˛0
A A0
B B0
0
C2
C 0 D2
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
˛.x 2 C y 2 / 2Ax 2By C D 0 be the equation of the circle and 2A0 x 2B 0 y C 0 D 0
the equation of the line. The center of the circle has coordinates . A
; B / belongs to the line
˛ ˛
0A
0B
0
if and only if 2A ˛ 2B ˛ C 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°) If C and C 0 are lines with equations 2Ax 2ByC D 0 and 2A0 x 2B 0 y 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 AA0 C BB 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 ? The
normal equation x 2 C y 2 C 2ax C 2by C c D 0 can only be used for genuine circles ;
the equation ˛.x 2 C y 2 / 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 a
circle-or-line iff .˛; A; B/ ¤ .0; 0; 0/.
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THÈME I. SUMMARY.
Theorem. Two circles C and C 0 with complex equations
zz
.wz C wz/ C c D 0
and
zz
.w 0 z C w 0 z/ C c 0 D 0
ww 0 C ww 0 D c C c 0
Notation. We denote by C the circle with center O.0; 0/ and radius 1.
Theorem. A circle C with equation x 2 C y 2 2ax 2by C c D 0 is orthogonal to C if
and only if c D 1. A circle-or-line with equation ˛.x 2 C y 2 2Ax 2By C D 0 is
orthogonal to C if and only if ˛ D .
§ 3. Circle bundles
From now on, "circle" will mean "general circle". Thus a circle is either a "genuine
circle" or a line. The lines are the circles that go through 1.
3.1 General definitions and properties
Definition. Given two distinct circles C1 and C2 , with equations
2
f1 .x; y/ WD ˛1 .x Cy 2 / 2A1 x 2B1 yC1 D 0
and
f2 .x; y/ WD ˛2 .x 2 Cy 2 / 2A2 x 2B2 yC2 D 0
we 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 and
C2 . If C1 and C2 have 2 common points P and Q, then all the circles in B go through these
2 points. The bundle is called bundle with base points P and Q. If C1 and C2 have exactly
1 common point P , they are tangent in that point and tangent to a line d , then all the circles
in B go through P and are tangent to d . The bundle is called tangent bundle . If C1 and C2
have no common point, then all the circles in B are orthogonal to the circles in the bundle
B 0 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 for Q. Thus if C1 and C2 go through P and Q, it will
be 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 and
the bundle with limit points is the bundle orthogonal to the one with base points as defined
below.
Remark. If B is a bundle with base points P and Q D 1, then the circles belonging to B
are the lines through P . If B is a tangent bundle of circles tangent to the line d at the point
P D 1, then the circles belonging to B are the lines parallel to d . If B is a bundle with
limit points P and Q D 1, then the circles belonging to B are the genuine circles with
center P .
§ 3. CIRCLE BUNDLES
9
Theorem and definition. Let B be a bundle of circles. The set of circles orthogonal to the
circles in B form a circle bundle B 0 . The bundles B and B 0 are called orthogonal bundles.
Proof. Let C1 and C2 be two distinct circles in B. Write the equations of C1 and C2
˛1 .x 2 C y 2 / 2A1 x
2B1 y C 1 D 0
and
˛2 .x 2 C y 2 / 2A2 x
2B2 y C 2 D 0
A 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 2A1 A0 2B1 B 0 D 0
˛2 0 C 2 ˛ 0 2A2 A0 2B2 B 0 D 0
This system of two equations in four unknowns is of rank 2, since the circles C1 and C2
are distinct (thus .˛1 ; A1 ; B1 ; 1 / and .˛2 ; A2 ; B2 ; 2 / are not proportional). The solution
is 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 one
bundle 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 B 0 is the bundle
with limit points P and Q and conversely. If B is a tangent bundle of circles tangent to the
line d at the point P , then the orthogonal bundle B 0 is the tangent bundle of circles tangent
to 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
10
THÈME I. SUMMARY.
Orthogonal bundles with base and limit points P and Q D 1.
P
Orthogonal tangent bundles tangent at point P ¤ 1 to orthogonal lines d and d 0 .
d
P
d0
§ 4. EXISTENCE OF CIRCLES
11
Orthogonal tangent bundles tangent at point P ¤ 1 to orthogonal lines d and d 0 .
d
P
d0
§ 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 different
from 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 the
line PQ. If not, PQM is a real triangle and there is exactly one circle going through the
three vertices.
Lemma. Given a bundle B of circles tangent to a line d in P , for any point M different
from 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 distinct
from 1 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 and M 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 the
center 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 is
exactly 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 the
MP
.
ration MQ
Theorem. Given a bundle B of circles, for any point M , there is exactly one circle belonging
to B and going through M .
Proof. The three lemmas (or lemmata) above show that the theorem is valid for any kind of
bundle.
12
THÈME I. SUMMARY.
Theorem. Given two circles and a point M , there is exactly one circle orthogonal to these
two circles and going through M .
Proof. Let us call C1 and C2 the two given circles and B the bundle of circles determined by
C1 and C2 . Let B 0 be the bundle orthogonal to B. We are looking for a circle C 0 belonging
to the bundle B 0 and going through M . The preceding theorem tells us that there is one and
only 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 B 0 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 three
circles. 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 2A1 A0 2B1 B 0 D 0
˛2 0 C 2 ˛ 0 2A2 A0 2B2 B 0 D 0
:
˛3 0 C 3 ˛ 0 2A3 A0 2B3 B 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 the
same solution.
BUT, we have to be careful : it is true that for every circle there are elements .˛; A; B; / 2
R4 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 and
2
2 ˛
only if the square of the radius is greater or equal to zero, where : R2 D A CB
. Thus
˛2
we have proved that given three circles not belonging to a common bundle, there is at most
one 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 ./ D
PC2 ./ D PC3 ./ the common power of K relatively to these three circles. If p > 0, the
p
circle C with center and radius R D p 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 three
circles 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 B 0 the
bundle orthogonal to B. If the circle C1 belongs to B 0 , then all the circles in B are orthogonal
to 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 B 0 . The properties "C 2 B" and "C is
orthogonal to C2 and to C3 " are equivalent. Since B 0 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 radical
center of C1 , C2 and C3 is outside these three circles and we may apply the theorem.
13
The radical center is inside the three circles.
S
The radical center is outside the three circles.
C2
C3
C
C1
14
THÈME I. SUMMARY.
Corollary 2. Let C1 be a circle, B be a circle bundle with limit points P and Q and B 0 the
bundle orthogonal to B. If the circle C1 belongs to B 0 , then all the circles in B are orthogonal
to C1 . If not, then there at most one circle C belonging to B and orthogonal to C1 . More
precisely, if C1 separates the points P and Q (that is : one point is inside or on the circle
and the other one outside), then there is no circle in the bundle B orthogonal to C1 . If C1
does not separate the points P and Q, then there is one circle in the bundle B orthogonal to
C1 .
Proof. Let C2 and C3 be two circles belonging to B 0 . The properties C 2 B and C is orthogonal to C2 and to C3 are equivalent. Thus if C1 does not belong to B 0 , there is at most one
circle 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 B 0 is the set of circles centered at
P ; there is a circle centered at P orthogonal to C if and only if P is outside or on C . To
recover 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 B 0 the bundle orthogonal to B. We
suppose that C1 … B 0 .
Situation 1. Let P be a point in the Euclidean plane. Let B be the bundle of lines going
through 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 the
line through P ortogonal to the line C1 ).
C1

P
Proof. The bundle B is the circle bundle with base points P and 1. Use corllary 1. `
`1
1
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 the
bundle B orthogonal to C1 : `1 is the line parallel to ` that goes through the center 1 of
C1 . If C1 is a line which does not belong to B 0 , then the only element of B orthogonal to C1
is the point circle 1.
Proof. There is one and only one parallel to a line through a point.
Situation 3. Let P be a point in the Euclidean plane. Let B be the bundle of all circles
having P as center and let C1 be any genuine circle. If P is outside C1 or on C1 , one can
draw a tangent from P to the circle C . Let us call T the contact point. There is one and only
one circle C1 in the bundle B orthogonal to C : C1 is the circle with center P and radius
P T . 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 . Let
T be the contact point. The circle with center P and radius P T 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.
15
T
P
C1
1

P
If P is inside the circle C1 , no circle with center P can be orthogonal to C . In fact, the
relation d 2 D R12 C R2 is impossible for all R when d < R1 .
One easy construction of the circle C tangent to ` at P and orthogonal to C1 using
inversion (see next chapter).
`
C1
P
Let P be the image of P
in the reflection through C1 .
`
P
C
Construction of the circle C belonging to the circle bundle with limit points P and Q
and orthogonal to a circle C1 which does not separate P and Q.
C1
Let Q be the image of Q
in the reflection through C1 .
Q
The circle C has to be
orthogonal to the circle PQQ .
The point is the radical center
of the circle C1 and the circles
going through P and Q.
P
Q

C
C1
1
An other proof
16
THÈME I. SUMMARY.
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 belong 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 Ri2 C R2 . Thus
di D i > Ri and is outside all three circles C1 , C2 and C3 . The circle C is orthogonal
to the circles C1 , C2 and C3 , if the power of with respect to each of the three circles is
R2 . And thus has to be the point S . If S is outside or on the circles C1 , C2 and C3 , then
S 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 and Q, there is one circle orthogonal
to C and going through P and Q. If C does not belong to the bundle with limit points P
and Q, the circle is unique. If C belongs to the bundle with limit points P and Q, then
all 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 three
circles is outside them since P and Q are point circles.
The radical center is inside the three circles.
S
17
The radical center is outside the three circles.
C1
C2
C
C3
18
THÈME I. SUMMARY.
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 denoted 1 and called point at infinity. A circle is either a genuine
circle or a straight line to which is added the point at infinity. Thus
a line is a circle which goes through 1.
Let C be circle. If C is a genuine circle with center and radius 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, with
the complement that the image of 1 is 1.
The unit circle centered at the origin will be denoted C.
§ 1. Definition of inversions and reflections
Definitions. Let P denote an Euclidean plane.
1°) Let O be a point and R a length. We call inversion with center O and power R2 and
denote InvO;R2 the transformation of P [ f1g that associates to each point M the point M 0
such 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/ D 1
– if M D 1, InvO;R2 .1/ D O
The 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 segment
MM 0
– if M 2 `, then M 0 D M .
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 its
own inverse) :
Refl ı Refl D Identity
Proposition. The fixed points of Refl are the points belonging to .
19
C
O
M0
M
`
M
M0
20
THÈME I. SUMMARY.
CH. 2. REFLECTIONS
A circle is dividing the plane in three parts : the set of points belonging to and the
two sides of . If is a genuine circle, the two sides are the inside containing the center
(unless the radius is 0) and the outside containing 1. If is a line 1 belongs to .
Proposition. The image of a point on one side of the circle by Refl is on the other side
of 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 center O and power 1, that is to reflection in the unit circle C.
Theorem. The inversion with center O and power 1 transforms z into
complex conjugate of z.
Proof. Write z in the polar form : z D e i . Then z1 D z and z1N are on a same ray with origin O. Moreover
1
jzj j j D zN
1
1
e
i
and
1
zN
1
,
zN
D
where zN is the
1 i
e . Thus O,
D1
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 b 2
R2 D 0
Introducing the normal form z D x C iy, we have 2x D z C zN and 2y D
z zN D x 2 C y 2 and the equation of the circle may be written
z zN
a.z C zN /
b. iz C i zN / C a2 C b 2
or
z zN
.a
i b/z
.a C i b/zN C a2 C b 2
iz C i zN . Then
R2 D 0
R2 D 0
which may be written
z zN
wz
N
w zN C c D 0
where w D a C i b is the center of the circle.
The equation of a generalized circle may then be written
˛z zN
!z
N
! zN C D 0
where ˛ 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 zN
!z
N
! zN C D 0
and
˛ 0 z zN
!N 0 z
! 0 zN C 0 D 0
§ 3. IMAGES OF LINES AND CIRCLES
21
!! 0 C !! 0
˛0
˛ 0 D 0
The equation of the circle C is
z zN
thus ˛ D 1, ! D 0 and D
i.e.
1D0
1. The circle C is orthogonal to C if and only if 0 D ˛
˛D
In the case when C is a line, it is orthogonal to C if and only if D 0, that is, the line goes
through O.
§ 3. Images of lines and circles
Theorem. The circle C with equation
˛z zN
!z
N
! zN C D 0
has as image through ReflC the circle C 0 with equation
z zN
!z
N
! zN C ˛ D 0
Proof. the point M corresponding to the complex number z belongs to C 0 if and only if
ReflC .M / 2 C , that is
˛
11
zN zN
!N
1
zN
!
1
C D0
zN
since the complex number correspnding to ReflC .M / is z1N . Multiplying by z zN we finally get
z zN !z
N
! zN C ˛ D 0:
Theorem. A circle C with equation
˛z zN
!z
N
! zN 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 ! z.
N The
image through ReflOx of the circle C with equation ˛z zN !z
N
! zN C D 0 is thus
˛z zN !z !N zN C D 0. This circle is globally invariant if and only if ! D !,
N which
means :
if C is a genuine circle, then it is centered on Ox ; if C is a line, then it is orthogonal to Ox.
22
THÈME I. SUMMARY.
CH. 2. REFLECTIONS
Consequences
We denote ReflC by f .
Proposition 1. The image of a line ` through O is itself.

`
M0
M
O
Proof. ` is orthogonal to C.
Proposition 2. Let ` be a line such that O … `. Denote by H the orthogonal projection of
O on ` and by H 0 the image by f of H . The image by f of ` is the circle `0 with diameter
OH 0 .

`
M0
M
H0
H
O
`0
Proof. The equation of ` may be written
!z
N
! zN C D 0
where ¤ 0. The equation of the image `0 is then
z zN
!z
N
! zN D 0
which 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 is
the point diametrically opposed to O.
Remark that we have got at the same time f .`0 / D ` and so we have proved the following proposition.
Proposition 3. Let `0 be a circle such that O 2 `0 . Denote by H 0 the point diametrically
opposed to O on that circle and by H the image by f of H 0 . The image by f of `0 is the
line ` 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 line
EE 0 . If the circle `0 intersects in exactly one point T , then the line ` is just the line tangent
to (and to `0 ) in T .
§ 4. IMAGES OF CIRCLE BUNDLES
23
Proposition 4. The transformation f transforms circles into circles, preserves the measure
of angles and changes the orientation.
Proof. Look at the following drawings :
C
M0
M
M
M0
O
`
and get convinced !
§ 4. Images of circle bundles
Theorem. Let P and Q be two points and let f be any reflection (through a line or through
a genuine circle). Let P 0 and Q0 be the images by f of P and Q. The image by f of the
circle bundle with base points P and Q is the circle bundle with base points P 0 and Q0 . The
image by f of the circle bundle with limit points P and Q is the circle bundle with limit
points 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 same
with Q. Thus any circle in the bundle B with base points P and Q is a circle belonging to
the bundle B 0 with base points P 0 and Q0 . Conversely, any circle in B 0 can be obtained in
this way.
The circles in the bundle B with limit points P and Q are those that are orthogonal to
the 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 lines
through P (and through 1) and the bundle B 0 with limit points P and Q is the bundle of
circles with center P (including the point circles P and 1).
Theorem. Let P be a point and ` a line through P and let f be any reflection (through a
line or through a genuine circle). Let P 0 and `0 be the images by f of P and `. The image
by f of the tangent bundle B of circles through P and tangent to ` is the tangent bundle B 0
of 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 . The
reflection preserves contacts, thus any circle through P and tangent to ` is a circle through
P 0 and tangent to `0 . Conversely, any circle in B 0 can be obtained in this way.
Remark. If P D 1, 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 C
belongs to the circle bundle with limit points P and Q.
24
THÈME I. SUMMARY.
CH. 2. REFLECTIONS
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 through
P and Q. Then the circle C and the point circles P and Q all belong to the circle bundle
with limit points P and Q.
C
P
Q

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 Refl2 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 inversion
exchanging C and `.
Chapitre 3
Non-Euclidean Geometry
§ 1. chapter 6.1.1. Non-Euclidean Geometry : the
two 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 at
http ://aleph0.clarku.edu/ djoyce/java/elements/elements.html
§ 1. chapter 6.1.1. Non-Euclidean Geometry : the two usual models of a hyperbolic
plane
The unit disc of Poincaré
Set of points. The points z D x C iy D .x; y/ such that x 2 C y 2 < 1 i.e.
D D fz 2 C j jzj < 1 g
The set D is called hyperbolic plane.
Boundary points. The points z D x C iy D .x; y/ such that x 2 C y 2 D 1 or C D fz 2
C j jzj D 1 g are not points of the hyperbolic plane, but the set C of these points plays an
important role. The points of C are called boundary points.
Set of lines. The lines are the intersection of D with the (generalized) circles orthogonal to
the 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 the
reflection through C. Let ˛.x 2 C y 2 / 2Ax 2By C D 0 be the equation of . The circle
is orthogonal to C if and only if ˛ D .
Theorem 4. Given two distinct points P and Q in D, there is a unique line (or more preciesely d line) going through P and Q.
Proof. Since P … C and Q … C, there is a unique circle going through P and Q and
orthogonal to C. In fact has to belong to the circle bundle B with base points P and Q
and 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 length
jdzj by the formula
jdzj
ds D
1 jzj2
The geodesics computed from that are the d lines. One immediate consequence of that
formula is that the angles in the hyperbolic plane D are the same as those in the Euclidean
plane. 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 Euclidean
geometry it is since it is the center of C, but in hyperbolic geometry it is just a point like the
others 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 by
compositions 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 g
The 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 set
L D fz 2 C j =.z/D0 g is not included in the hyperbolic plane, but plays an important
role. The points of L are called boundary points.
Set of lines. The lines are the intersections of H with the (generalized) circles orthogonal to
the line L. We call these lines h lines or simply lines.
Theorem 4. Given two distinct points P and Q in H, there is a unique h line going through
P and Q.
Proof. Since the line L is a circle and since P … L and Q … L, there is a unique circle
through P and Q and orthogonal to L (look at the picture below in the case the line PQ is
not parallel to the y-axis).
Remark 1. Let be a circle with equation ˛.x 2 C y 2 /
is such that \ H is a h line if and only if
BD0
and
2Ax
.˛; A/ ¤ .0; 0/
2By C D 0. The circle
§ 1. CHAPTER 6.1.1. NON-EUCLIDEAN GEOMETRY : THE TWO USUAL MODELS OF A HYPERBOLIC PLANE
Examples of h lines.
y
H
Q
P
L
O
x
Remark 2. The infinitesimal length ds in H is related to the infinitesimal euclidean length
jdzj by the simple formula
jdzj
ds D 2
y
The geodesics computed from that are the h lines. One immediate consequence of that
formula is that the angles in the hyperbolic plane H are the same as those in the Euclidean
plane. 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 in
hyperbolic geometry it is just a h-line like all the other h-lines and the space is completely
homogeneous : 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 by
compositions 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 such
azCb
a transformation which is a homography z 7! czCd
, with ad bc ¤ 0 ? There are several
answers but we just want one, eventually as simple as possible The choice has to be such
that the image of the circle z D e i becomes the real line y D 0, i.e. z z D 0, thus for all
ae i C b
ae
N i C bN
D
i
ce C d
ce
N i C dN
or
that is
.ae i C b/.ce
N
.adN
N i C .acN
c b/e
i
C dN / D .ae
N
ac
N C b dN
i
i
N
C b/.ce
C d/
N / C .b cN
bd
ad
N /e
i
D0
27
28
THÈME I. SUMMARY.
The condition becomes
acN
adN
ac
N C b dN
c bN
Nbd
D
D
0
0
We want also that 0 7 ! i , i.e. b D id . Let us choose d D 1, then b D i and
a C ic D 0
acN ac
N Ci Ci D 0
thus c D i a and i aaN ai
N a C 2i D 0 and finally aaN D 1. Thus a D e i' , b D i , c D i e i'
and d D 1. Thus the transformation is
z7 !
Let us choose ' such that i e
i'
e i' z C i
D
i e i' z C 1
i
z C ie
z ie
i'
i'
D 1, we get
z7 !i
1Cz
1 z
2
/
This transformation transforms 1 into 1, i into i 1Ci
D i .1Ci
D
1 i
2
into C1.
The inverse transformation from H to D is
z7 !
1,
1 into 0 and
i
z i
zCi
the image of 1 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 e i where is the measure in radians of the oriented angle in the
following picture
xCi
x
i
H
i
x
L
i
§ 1. CHAPTER 6.1.1. NON-EUCLIDEAN GEOMETRY : THE TWO USUAL MODELS OF A HYPERBOLIC PLANE
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
`6
`2
H
`3
`1
`4
L
`5
O
x
Group of transformations
Theorem 1. Let ` be a d-line (respectively h-line) which is part of a genuine circle orthogonal to C. The reflection with repect to , denoted r is such that
r .D/ D D
r .C/ D C
and
More precisely, the regions D1 and D2 are exchanged.
C
D2
D1
`

29
30
THÈME I. SUMMARY.
Questions. How is the preceding theorem if ` is part of a Euclidean line ? Draw the pictures
corresponding to the H-plane.
Definitions. A hyperbolic reflection is a restriction to D of a reflection in a (generalized
euclidean) circle, whose intersection with D (respectively H) is a d -line (respectively hline).
The group of the hyperbolic plane is the set of transformations that can be written as
compositions of reflections, called hyperbolic transformations.
Now recall that in the two models, the hyperbolic angles are the same as the Euclidean
angles. 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 magnitudes of angles.
§ 2. 6.1.2 Existence of hyperbolic lines
Origin Lemma. Let A be a point of D distinct from O. There is a (hyperbolic) reflection r
such that r.A/ D O and r.O/ D A.
Comment. In the hyperbolic plane D, all the points play the same role. The point O is special
in the representation we have chosen, but any other point could have been chosen to be the
center 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 that
r.P / D Q (and then r.Q/ D P ).
Comment. In Euclidean geometry, given two points P and Q there is one line ` such that
the reflection in ` exchanges P and Q. The line is the bisector of the segment PQ. In
the 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 the
hyperbolic segment PQ (which is the arc of the circle going through P and Q, which is
orthogonal to C and included in D).
Proof. We are looking for a circle belonging to the circle bundle with limit points P and
Q and orthogonal to C. Since C does not separate the points P and Q, there is one and only
one such circle and ` D \ D.
Geometrical construction of the line `.
In the disc model
C
`
Q
P
B bundle with base points P and Q
B 0 bundle with limit points P and Q
§ 3. 6.1.3 INVERSION PRESERVES INVERSION POINTS
31
and in the half plane model
`
Q
P
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 a
unique line AB wich is part of a Euclidean circle C . Thus C cannot have more than two
points 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 and Q in H, there is a unique hyperbolic line going
through 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 . Then
there 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 that
given 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 a
line `0 that goes through A2 . Then by a rotation r transform `0 into `2 . The transformation
f 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 ` be
any 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 .
32
THÈME I. SUMMARY.
§ 4. 6.2.1-6.2.3. The group of Hyperbolic Geometry
The inversions in C [ f1g
In the complex plane, the inversion with center ˛ and radius R is described by
z7 !˛C
R2
z ˛
If the circle with center ˛ is orthogonal to C, then j˛j2 D R2 C 1 and the inversion or
reflection through ` D \ D becomes
.z/ D
˛z 1
z ˛
What happens if ` is a line ? The formula becomes .z/ D e i z
Theorem 1 and 2. The direct hyperbolic transformation can always be described by a Möbius transformation of the form
M.z/ D
az C b
bz C a
where jbj < jaj
The reflection B W z 7 ! zN is an indirect hyperbolic transformation ; all indirect transformations 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 in
2 lines `1 and `2 .
If `1 and `2 intersect, the transformation is a "rotation". If `1 and `2 are parallel the
transformation 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 :
z7 !K
z m
1 mz
where jKj D 1
and
jmj < 1
z7 !K
z m
1 mz
where jKj D 1
and
jmj < 1
Indirect :
4.2 6.3.1 The distance formula
Properties of a distance
1. 8.z1 ; z2 / 2 D2 W
8.z1 ; z2 / 2 D2 W
d.z1 ; z2 / > 0
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. 8M 2 Direct subgroup of D; 8.z1 ; z2 / 2 D2
x D th
1
.y/ D
d.M.z1 /; M.z2 // D d.z1 ; z2 /
1 1Cy
ln
2 1 y
The formula
d.z1 ; z2 / D th
1
ˇ
ˇ z
ˇ 1 z2 ˇ
ˇ
ˇ
1 z1 z2
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 points C group of transformations C invariant objects.
2. Start from Euclid.
Let us look at the first answer : there are several models to describe the hyperbolic
plane : D, H, the upperhalf of the hyperboloide z 2 x 2 y 2 D 1. In this last model
the 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 Klein
model where the line are Euclidean segements in a unit disc. The central projection of the
half-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 postulates
1,2,3 and 4 we are going first to "Ordered Geometry", then to "Absolute Geometry" (the
choices of adjectives are those of Coxeter) and finally to elliptic, Euclidean or hyperbolic
Geometry. In Brennan Non-Euclidean means hyperbolic.
1.2 Ordered Geometry
The basic concept is intermediacy or betweenness, denoted ŒABC to say that the points
A, B and C are on a same line and B is between A and C . One starts from the following
axioms
1. 92 points
2. 8A; B; 9C such that ŒABC
3. ŒABC H) A ¤ C
4. ŒABC H) ŒCBA and not ŒBCA ; from here you may define the words : ray, segment and line.
35
36
THÈME II. READING BRENNAN.
CH. 4. 6.1-6.3 NON-EUCLIDEAN GEOMETRY, TRANSFORMATIONS, DISTANCE
9
C ¤D =
5. C 2 line AB
H) A 2 line CD
;
D 2 line AB
6. 9 line AB; 9Dsuch that D … line AB
7. (Better than Pasch’s axiom) Given triangle ABC and points D and E such that
ŒBCD and ŒCEA, the line DE intersects the open interval AB.
A
E
B
C
D
Sylvester’s problem of collinear points
Sylvester’s conjecture (1893) : It is not possible to arrange any finite number of points
so that a line through every two of them passes through a third, unless they all lie in the
same 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 3n
).
7
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 can
be summerized as follows : consider all the couples (point, line) where "point" is one of
the n given points P and "line" is a line passing through exactly two of the n given points
other 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 is
finite and thus has a smallest element d . Then you prove that you can build a new couple
for 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 statement 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 to
C : 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 circles
p and q.
As a consequence we get the Lemma 1, page 268, since O is a point like all the others
in the hyperbolic geometry. We get also some methods to construct the line ` such that
Refl` .p/ D q.
§ 3. Page 288
3.1 Example 3 : Midpoint of 41 i and 43 i
r
1
We have to find X such that
.X
X2
a/.X b/
D 1 C r2
D r2
Eliminating r we get
X2
.a C b/X C ab D X 2
i.e.
XD
and
rD
p
X2
ab C 1
D
aCb
1p 2
1D
19
16
thus
mDX
rD
1 3
4 4
1
C 43
4
19
p
16
105
D
162
1
19
16
p
105
D
16
D 0; 547 065 577 13 : : :
38
§ 4. Page 289
Use origin lemma and preservation of circles through reflections : all Euclidean circles
INCLUDED 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 zN
w zN
wz
N C1D0
i.e.
w zN 1
zN wN
The reflection through this circle is given by
zD
z0 D
w zN 1
zN wN
We just change z into z 0 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 disc
model D, 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 and
make a conjecture. Prove your conjecture.
C
B
Hints :
Introduce A0 D ReflC .A/
The bisector of a hyperbolic
angle has to be the bisector of the
angle made by the tangents to the
circles.
A
O
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 :
`
C
P0
O
P
`?
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 the
point of BC such that the lines IA0 and BC are orthogonal. Let B 0 be the point of CA such
that the lines IB 0 and CA are orthogonal. Let C 0 be the point of AB such that the lines IC 0
and AB are orthogonal. Construct the circle going through A0 , B 0 and C 0 . What can you
say about the circle ?
Exercise 4 to 7. Are the medians, the segment bisectors, the altitudes, the outer angle bisectors concurrent ?
40
Chapitre 5
6.4 Non-Euclidean Geometry,
Transformations, Distance
§ 1. Euclidean Triangles
A
1.1 Médianes et centre de gravité
Définition. Soit ABC un triangle. On note A0 le milieu du segment BC . On appelle médiane du triangle issue de A la droite AA0 ou le segment AA0 ou encore la longueur de ce
segment.
Théorème. Les médianes d’un triangle sont concourantes.
Démonstration. Soit ABC un triangle. On note A0 le milieu du segment BC , B 0 le milieu
de CA et C 0 le milieu de AB. Les droites BB 0 et C C 0 sont sécantes, car si elles étaient
parallèles, on aurait par le théorème de Thalès BA D BC 0 D 12 BA, soit B D A ce
qui est impossible. Notons G le point d’intersection de BB 0 et de C C 0 . La réciproque du
théorème de Thalès nous donne B 0 C 0 ==BC et B 0 C 0 D 12 CB. Appliquons alors le théorème
de Thalès aux droites BB 0 et C C 0 coupées par les sécantes BC et B 0 C 0 . Il vient
GB 0
GB
D
1
2
ou encore
GB 0
BB 0
D
Définition. Le point de concours des médianes d’un triangle s’appelle le centre de gravité
du triangle.
Remarque. Le centre de gravité G d’un triangle ABC 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
C
A
1
3
ce qui détermine le point G de façon unique (« au tiers de la médiane à partir de la base »).
Appliquons notre résultat au triangle CBA : les médianes BB 0 et AA0 se coupent en un
point G 0 situé au tiers de la médiane BB 0 . Le point G 0 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 A C B C C D A C 2A0 D B C 2B 0 D C C 2C 0 .
A0
B
B0
C
G
B
A0
C
42
CH. 5. 6.4 NON-EUCLIDEAN GEOMETRY, TRANSFORMATIONS, DISTANCE
sommets. Le point G est le centre de gravité au sens physique d’une plaque homogène
infiniment mince de forme triangulaire de sommets A, B et C .
C 1. Démontrer l’affirmation de la remarque précédente.
C1
B1
A
C0
B0
G
B
A0
C
A1
1.2 Médiatrices et cercle circonscrit

B
O
A0
G
A0
A1
Théorème. Les médiatrices d’un triangle sont concourantes en un point qui est le centre de
l’unique cercle qui passe par les trois sommets du triangle.
Démonstration. Soit ABC un triangle. Les médiatrices des segments AB et CA ne sont
pas parallèles car sinon les sommets du triangle seraient alignés. Soit O le point commun à
ces médiatrices. Comme O appartient à la médiatrice de AB, on a OB D OA et comme O
appartient à la médiatrice de AC , on a OA D OC . On en déduit OB D OC , ce qui prouve
que la médiatrice de BC passe par O.
Le cercle de centre O et de rayon OA passe donc par B et par C . Il existe donc un cercle
passant par les trois sommets du triangle. Enfin si un cercle passe par les trois sommets du
triangle, son centre est sur les médiatrices des trois côtés et est donc confondu avec O et
son rayon est OA. A
B
A
Rappel. (admis) La médiatrice d’un segment AB est la droite , ensemble des points équidistants de A et de B. La droite est la perpendiculaire à AB passant par le milieu M du
segment AB. On appelle médiatrices d’un triangle les médiatrices des côtés du triangle.
M
A
Corollaire. Soit ABC un triangle. On note A0 ,
B 0 et C 0 les milieux des côtés et G le centre
de gravité. On note A1 , B1 et C1 les points
tels que ABA1 C , BCB1 A et CAC1 B soient
des parallélogrammes. L’homothétie de centre G
et de rapport 2 transforme le triangle A0 B 0 C 0
en ABC et ABC en A1 B1 C1 ; l’homothétie de
centre G et de rapport 12 transforme A1 B1 C1
en ABC et ABC en A0 B 0 C 0 . Ces trois triangles ont les mêmes médianes, les divisions
.A; A0 I A1 ; G/, .B; B 0 I B1 ; G/ et .C; C 0 I C1 ; G/
sont harmoniques et A0 est le milieu de A1 A, B 0 le
milieu de B1 B et C 0 le milieu de C1 C .
C
Définitions. L’unique cercle qui passe par les trois sommets d’un triangle est appelé le
cercle circonscrit au triangle et son centre O est appelé le centre du cercle circonscrit au
triangle.
§ 1. EUCLIDEAN TRIANGLES
43
Définition. Soit ABC un triangle. On note A0 , B 0 et C 0
les milieux des côtés. Le cercle circonscrit au triangle
A0 B 0 C 0 est appelé le cercle d’Euler ou cercle des neuf
points ou cercle de Feuerbach du triangle ABC .
Nous noterons O 0 le centre du cercle circonscrit à
A0 B 0 C 0 et H le centre du cercle circonscrit à A1 B1 C1 .
L’homothétie de centre G et de rapport 2 transforme
O 0 en O et O en H ; l’homothétie de centre G et de rapport 12 transforme H en O et O en O 0 . On remarque
que les points O, G, O 0 et H sont alignés, que la division .O; O 0 I H; G/ est harmonique et que O 0 est le
milieu de HO.
La droite HO 0 GO s’appelle la droite d’Euler du triangle ABC .
A
C1
C H
O0
B
B1
B0
G
O
C
A0
H O 0G O
1.3 Quadrangle orthocentrique
Lemme 1. Soit uE , vE et w
E trois vecteurs, alors uE .w
E
vE/ C vE .E
u
w/
E Cw
E .E
v
uE / D 0.
A1
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 uE D AB, vE D AC et w
E D AD. Définition. Soit ABC un triangle. On appelle hauteur du triangle ABC issue de A la droite
passant par A et perpendiculaire à BC . L’intersection P de la hauteur issue de A avec la
droite 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.
Première démonstration. Les hauteurs du triangle ABC sont les médiatrices du triangle
A1 B1 C1 . Elles sont donc concourantes en le point H , centre du cercle circonscrit au triangle
A1 B1 C1 . Deuxième démonstration. Les hauteurs CR et BQ ne sont pas parallèles car sinon les
sommets A, B et C seraient alignés. Soit H le point d’intersection de CR et de BQ. On
! !
a 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.
Proposition. Si H est l’orthocentre d’un triangle ABC , alors A est l’orthocentre du triangle
BCH , B est l’orthocentre du triangle ACH et C est l’orthocentre du triangle ABH .
A
Q
R
B
C
P
C1
A
R
B
B
Q
H
P
C
A1
44
Définition. Un quadrangle est un ensemble de quatre points, appelés ses sommets. Il est
orthocentrique si chacun de ses sommets est l’orthocentre du triangle formé par les trois
autres sommets du quadrangle.
Proposition. Un quadrangle est orthocentrique si un sommet du quadrangle est l’orthocentre 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 triangle
orthique commun aux quatre triangles BCD, ACD, ABD et ABC .
A
H O0
O
B
P A0
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 milieu
du 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 donc
O 0 P D O 0 A0 , ce qui montre que P appartient au cercle d’Euler de centre O 0 et de rayon
O 0 A0 . Ce résultat appliqué aux triangles BCA et CAB montre que les pieds des hauteurs
C
Q et R appartiennent aussi au cercle d’Euler. Le cercle circonscrit au triangle orthique PQR passe donc par les milieux des côtés du
triangle ABC , mais pour la même raison par les milieux des côtés du triangle BCH , etc. Il
en 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 les
angles en P , Q et R sont droits, les segments A0 Ma , B 0 Mb et C 0 Mc sont des diamètres du
cercle d’Euler.
C 2. Il y a une lacune dans le raisonnement précédent dans lequel on a supposé que PQR est un triangle. Il est
demandé au lecteur de la combler.
Exercice 1. Soit H l’orthocentre d’un triangle ABC . Montrer
que les cercles circonscrits aux triangles BCH , ACH , ABH et
ABC ont même rayon.
Exercice 2. Soit H l’orthocentre d’un triangle ABC . Montrer
que les symétriques de H par rapport aux côtés du triangle appartiennent au cercle circonscrit à ABC .
Exercice 3. Soit H l’orthocentre d’un triangle ABC , A0 le milieu
du segment BC , B 0 le milieu de CA, Ma le milieu de AH et Mb
le milieu de BH . Montrer que A0 B 0 Ma Mb est un rectangle.
Exercice 4. Montrer que la médiane d’un triangle rectangle est un
diamètre de son cercle d’Euler.
Exercice 5. Soit H l’orthocentre d’un triangle ABC et PQR son
triangle orthique. Montrer que les quadrangles BCQR, CARP ,
ABPQ, ARHQ, BPHR et CQHP sont inscriptibles. Justifier 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 côtés du triangle ABC sont les bissectrices de son triangle orthique
PQR.
1.4 Bissectrices, cercle inscrit et cercles exinscrits
Définition. Soit ABC un triangle. On appelle bissectrices de l’angle A les deux bissectrices des deux droites AB et AC . On appelle bissectrice intérieure celle qui est l’axe de
symétrie des demi-droites ŒAB/ et ŒAC / issues de A et contenant respectivement les points
B et C . L’autre bissectrice s’appelle la bissectrice extérieure (orthogonale à la bissectrice
intérieure).
§ 2. PAGE 298
45
Théorème. Les bissectrices intérieures d’un triangle sont concourantes en un point I , intérieur au triangle. La bissectrice intérieure issue d’un sommet A passe par le point Ia , point
d’intersection des bissectrices extérieures aux deux autres sommets.
Définition. Soit ABC un triangle. On appelle
cercle inscrit au triangle ABC le cercle de
centre I tangent aux trois côtés du triangle et
intérieur au triangle. Le point I est désigné
comme centre du cercle inscrit. On appelle
cercles exinscrits au triangle ABC les cercles
Ib
de centre Ia , Ib et Ic tangents aux trois côtés
A
du triangle et extérieurs au triangle. Les points
Ic
Ia , Ib et Ic sont désignés comme centres des
cercles exinscrits.
Remarque. Il est facile de démontrer que
les bissectrices de deux sommets B et C ne
I
peuvent pas être parallèles et se coupent donc
C
B
en 4 points équidistants des trois côtés du triangle. Les bissectrices de A passent donc par
ces 4 points. La partie difficile est de montrer
que les bissectrices intérieures sont concouIa
rantes. Une méthode simple consiste à utiliser le calcul barycentrique. Nous renvoyons
le lecteur au chapitre 4, §4 pour avoir les
coordonnées barycentriques des centres des
cercles inscrit et exinscrits, ce qui permet une
démonstration simple du théorème ci-dessus.
C 3. Le quadrangle Ia Ib Ic I est orthocentrique et son triangle orthique est le triangle ABC . Comparer cette
proposition au résultat de l’exercice 5.
§ 2. Page 298
§ 3. Page 303
§ 4. Page 307-312 Pythagoras’ theorem and Lobachevski’s formula
46
Chapitre 6
6.5 Tessellation
THURSDAY, APRIL 3
Exercise 1. Construct a hyperbolic segment that does not disappear when you move around its ends inside the disc
D.
A0
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 segment OM and the circle A0 AB.
A
N
O
M
B
Move around the points A and B to check the rigidity of the construction.
Exercise 2. Construct a regular octogone centered at O.
O
Exercise 3. In the following drawing in Euclidean plane, show that x D q
47
cos ˛
cos2
˛
sin2 ˇ
, where x D OA,
48
1
CH. 6. 6.5 TESSELLATION
1
˛ D OMA0 and ˇ D AOM .
T
M
O
A0
A
49
Exercise 4. We intend to do a tessellation of the hyperbolic plane with regular polygons all of same size, not
overlapping (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 at O. 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).
:
M
a
O
A
Show that the distance OA has to be
OA D q
cos. p/
cos2 . p/
sin2 . n/
1
1
Choose some values for n and p such that n
C p
< 12 , n > 3 and p > 3, compute the length OA and
make the drawing, beginning by drawing the circle C with radius 1 and putting the point A such that the distance
OA is the one you have computed.