Hyperbolic Geometry

Eric Lehman

February 2014 2

The course follows the chapter called "Non-Euclidean Geometry" in the book "Geo- metry" by David A. Brannan, Matthew F. Esplen and Jeremy J. Gray and gives alternative proofs of some theorems in a more geometric and less computational presentation. In the two usual models of Hyperbolic Geometry the lines are parts of Euclidean circles orthogonal to a border wich is a circle or a line. Therefore the course begins with some complements in Euclidean and Conformal Geometry about circles, inversion and circle bundles. These tools give a geometric insight in basic concepts in Hyperbolic Geometry such as Möbius transformations, parallelism and ultraparallelism, orthogonality, asymptotic triangles and area defect. Hyperbolic Geometry might be useful, but most of all it is beautiful ! II. Reading Hyperbolic Geometry 32

4 Non-Euclidean Geometry 33 § 1. chapter 6. Non-Euclidean Geometry : the two usual models of a hyperbolic plane ...... 33 § 2. 6.1.1 Non-Euclidean Geometry : the two usual models of a hyperbolic plane ...... 35 § 3. 6.1.2 Existence of hyperbolic lines ...... 36 Table of contents § 4. 6.1.3 Inversion preserves inversion points ...... 38 5 The group of Hyperbolic Geometry 39 § 1. 6.2.1 Non-Euclidean Transformations and Möbius Transformations ...... 39 § 2. 6.2.2 The Canonical Form of a Hyperbolic Transfor- I. Circles in Euclidean Geometry 1 mation ...... 42 § 3. 6.2.3 The Canonical Form of a Hyperbolic Transfor- 1 Power of a point with respect to a circle 3 mation ...... 43 § 1. Equations of a circle ...... 3 § 2. Orthogonal circles ...... 6 6 Distance 45 § 1. 6.3.1 The distance formula ...... 45 2 Reflection in a circle 11 § 2. 6.3.2 Midpoints and 6.3.4 Reflection ...... 47 § 1. Definition of inversions ...... 11 § 3. 6.3.2 Circles ...... 47 § 2. Images of lines and circles ...... 14 § 4. 6.3.2 Reflections ...... 47 § 3. Description of inversions using complex numbers . . . 17 § 5. 6.3.2 Midpoints ...... 47 § 6. 6.4.1 Triangles ...... 47 3 Circle bundles 19 § 7. 6.4.2 Orthogonal lines ...... 47 § 1. Different kinds of circle bundles ...... 19 § 2. Orthogonal bundles ...... 22 § 3. General definitions and properties ...... 25

3 4 TABLE OF CONTENTS Thème I

Circles in Euclidean Geometry

Chapitre 1

Power of a point with respect to a circle

§ 1. Equations of a circle § 2. Orthogonal circles

§ 1. Equations of a circle

1.1 Circle of center .a; b/ and radius R We suppose given an orthonormal frame .O; {; j / of a Euclidean plane P. E Let us denote by C.; R/ or simply by C theE circle with center .a; b/ and radius R. A point M.x; y/ belongs to that circle if M R or M 2 R2 0 or by Pytagoras theorem D D .x a/2 .y b/2 R2 0 (1.1) C D That characteristic relation is called an equation of C . We deﬁne the function fC by

2 2 2 fC .x; y/ .x a/ .y b/ R D C The equations of C are ˛fC .x; y/ 0 D where ˛ is any real number different from 0. When ˛ 1, we say that the equation is the normal equation of C . We may write the normal equationD of C as

x2 y2 2ax 2by c 0; where c a2 b2 R2 (1.2) C C D D C Interpretation of c. If c > 0, c is the square of the distance from the origine of coordinates O to a contactpoint T of a tangent to the circle C through O. This is also true if c 0, since then O is on the circle C . D

3 4 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY.CH.1. POWER OF A POINT WITH RESPECT TO A CIRCLE

T C R

pc .a; b/

If c < 0, c is the square of the length of the segment OB where B is a point common to the circle C and to the line through O orthogonal to the line O (what happens when O ?). D

C B R

...... p c ...... .a; b/ O

1.2 Power of a point with respect to a circle Deﬁnition. Let C be a circle with center .a; b/ and radius R and let M.x; y/ be any point. We denote by d the distance from M to the center of the circle. The power of the point M with respect to the circle C denoted by PC .M / is the number 2 2 PC .M / d R D Using the notations of the preceding paragraph, we have

Proposition.P C .M / fC .x; y/. D Remark. c PC .O/ fC .0; 0/. D D Theorem 1. (See proof on next page) Let C be a circle, M.x; y/ a point and d a line through M . If d intersects C in two points P and Q, then MP MQ if M is outside C PC .M / D MP MQ if M is inside C Remark 1. The important consequence of this theorem is that the product MP MQ is constant when turning the line d around the point M . § 1. EQUATIONS OF A CIRCLE 5

Remark 2. Using the scalar product,we have in all cases

PC .M / !MP !MQ D In particular PC .M / 0 if and only if M belongs to C . D

Q C C

P P M M Q

Theorem 2. Let C be a circle, M.x; y/ a point outside C and d a line through M tangent to C at a contact point T . Then 2 PC .M / MT D

C T R p PC .M / d

Proof of theorem 2. Let be the center of the circle C and R its radius. Denote the length M by d. The angle M T is a right angle and thus by Pythagoras’ theorem 2 2 2 2 2 MT M T d R PC .M / D D D Proof of theorem 1. Let P 0 be the point opposite to the point Q on the circle C . Since QP 0 is a diameter of C , the line PP 0 is orthogonal to the line PQ. Thus, the point P is the orthogonal projection of1P 0 on the line MQ, and

!MP !MQ MP!0 !MQ D Let us write !MQ and MP! as !MQ !M !Q and MP! !M P!, we have 0 D C 0 D C 0 P! !Q and and so 0 D 2 2 2 2 !MP !MQ .!M !Q/ .!M !Q/ !M !Q d R PC .M / D C D D D 6 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY.CH.1. POWER OF A POINT WITH RESPECT TO A CIRCLE

1.3 Use of complex numbers We use the same notations as above and put

z x iy and w a ib D C D C We denote the complex conjugate of a complex number z by z. Then z x iy and N N D x2 y2 zz ; .x a/2 .y b/2 .z w/.z w/ and a2 b2 ww C D C D C D The equation of C may be written

.z w/.z w/ R2 0 D or

zz .wz wz/ c 0 C C D

2 where c ww R PC .O/ R. D D 2

§ 2. Orthogonal circles

2.1 Angle of intersecting circles For any curves of class C1, intersecting in a point T , one deﬁnes the angle of these two curves as the angle of their tangents in T . If two circles are intersecting, there are two points of intersection, and by the reﬂection through the line joining the centers, one sees that the angles are the same if one looks at nonoriented angles and that oriented angles are opposite.

Deﬁnition. Two circles are orthogonal if they are intersecting and if their angles at the intersecting points are right angles. Remark. Since the tangent to a circle through a point T of a circle is orthogonal to the corresponding radius, we see that two circles with centers and 0 are orthogonal if the angle T 0 is a right angle. § 2. ORTHOGONAL CIRCLES 7

2.2 Orthogonality conditions

Theorem. Let C and C 0 be two 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 orthogonal if and only if one of the following equivalent conditions is fulﬁlled : 1. 2 R2 R 2 0 D C 0 2. The triangle T 0 is rectangle in T 2 3. PC ./ R 0 D 2 4. PC . / R 0 D 0 5. .a a /2 .b b /2 R2 R 2 0 C 0 D C 0 Theorem. Two circles C and C 0 with real equations

2 2 2 2 x y 2ax 2by c 0 and x y 2a0x 2b0y c0 0 C C D C C D are orthogonal if and only if

2.aa bb / c c 0 C 0 D C 0 Proof. The condition 5 above may be written

2 2 2 2 2 2 a b R a0 b0 R0 2aa0 2bb0 C C C D C We get the result from c a2 b2 R2 and c a 2 b 2 R 2. D C 0 D 0 C 0 0 Comment. Why is it convenient to use equations that are not necessarily normal ? The normal equation x2 y2 2ax 2by c 0 can only be used for genuine circles ; the equation ˛.x2 Cy2/ C2Ax C2By C D0 will describe a genuine circle if ˛ 0 AND will describeC a line if ˛ 0 and .A;C B/D .0; 0/. Thus .˛; A; B; / will describe¤ a circle-or-line iff .˛; A; B/ .0;D 0; 0/. ¤ ¤ 8 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY.CH.1. POWER OF A POINT WITH RESPECT TO A CIRCLE

Theorem. Two circle-or-lines C and C 0 with real equations

2 2 2 2 ˛.x y / 2Ax 2By 0 and ˛0.x y / 2A0x 2B0y 0 0 C C D C C D are orthogonal if and only if

2.AA BB / ˛ ˛ 0 C 0 D 0 C 0 Proof. We have to check the three possible situations : A B 1) The two curves are circles, that is ˛ 0 and ˛0 0. We have a ˛ , b ˛ , c ˛ and similar relations with ’. Thus the condition¤ of orthogonality¤ may beD writtenD D

A A B B 2. 0 0 / 0 ˛ ˛0 C ˛ ˛0 D ˛ C ˛0

Multiplying both sides by ˛˛0, one gets the expected relation. 2) C is a circle and C 0 is a line, that is ˛ 0 and ˛0 0. The line C 0 is orthogonal to ¤A B D C if and only if it goes through the center . ˛ ; ˛ / of C , that is A B 2A0 2B0 0 0 ˛ ˛ C D Multiplying by ˛, one gets 2AA0 2BB0 ˛ 0 C D which is the expected relation since ˛0 0. 3) Both curves are lines, that is ˛ D˛ 0. These two lines are orthogonal if and only D 0 D if the vectors .A; B/ and .A0;B0/ are orthogonal, that is AA0 BB0 0. Since ˛ ˛0 0, we still have the expected relation. C D D D

Theorem. Two circles C and C 0 with complex equations

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

ww ww c c 0 C 0 D C 0 Notation. We denote by C the circle with center O.0; 0/ and radius 1. Theorem. A circle C with equation x2 y2 2ax 2by c 0 is orthogonal to C if and only if c 1. A circle-or-line withC equation ˛.x2 yC2 2AxD 2By 0 is orthogonal to CDif and only if ˛ . C C D D § 2. ORTHOGONAL CIRCLES 9

Exercices

Exercice 1. Let C be a circle orthogonal to C. a) Show by algebraic computation(s) that the center of the circle C is outside the circle C. b) Show the same result geometrically.

Exercice 2. Let C1 and C2 be two genuine circles. Show that the set of points that have same power relatively to both circles is a line or is empty. When is it empty ? When it is a line, this line is called the radical axis of the two circles. Show that if one has three circles the radical axis of the circles two by two are concurrent or parallel.

* Exercice 3. Let C and C 0 be two circles with distinct centers and 0. Let ` be the line 0. We denote by A and B the intersections of C and ` and by A0 and B0 the intersections of C 0 and `. Show that the circles C and C 0 are orthogonal if and only if the four points A, B, A0 and B0 form a harmonic division, that is their cross ratio is equal to 1.(If the abscissae of four points on a line A, B, C and D are denoted a, b, c and d, then the c a d b crossratio of these four points in that order is by defition .A; B C;D/ da da ). I D ** Exercice 4. a) Is it possible to find three circles two by two orthogonal ? b) Is it possible to find four circles two by two orthogonal ? c) Is it possible to find five circles two by two orthogonal ? Hint. The answer will be easy when we’ll have got the concept of circle-bundles. 10 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY.CH.1. POWER OF A POINT WITH RESPECT TO A CIRCLE Chapitre 2

Reﬂection in a circle

§ 1. Definition of inversions § 2. Images of lines and circles § 3. Description of inversions using complex N orth pole numbers m0 The stereographic projection transforms a usual Euclidean plane into a sphere minus one point. Thus you have to take away a point from the sphere or to add one to the plane. We call that extra point the point at infinity. There is a natural transformation on the OM M 0 sphere : the reflection in the equatorial plane, exchanging points in the two hemispheres, the North hemisphere and the South hemisphere. By stereographic projection, this reflection becomes a trans- m formation in the plane called reflection in the unit circle or inversion with center O and power 1. South pole

§ 1. Deﬁnition of inversions 1 Proposition. In the picture in the introduction above, if OM d, then OM 0 d and OM OM 1. D D 0 D First proof. Let the line OM be the r axis and ON the z axis. The equation of the line NM is z r 1 1 C d D or z 1 r=d. The intesection with the circle z2 r2 1 has coordinates .r; z/ such that .1D r=d/ 2 r2 1 or .1 1=d 2/r2 .2=d/rC D0. This equation in r has two C D C D 2d solutions : r 0 corresponding to the North Pole and r 1 d 2 corresponding to m. Then D 2 D C the coordinates of m are r 2d and z d 1 and the coordinates of m are r 2d D 1 d 2 D 1 d 2 0 D 1 d 2 1 d 2 C C C and z 1d 2 . The equation of the line N m0 is D C 1 d 2 1d 2 1 z 1 C r 1 dr D C 2d D 1 d 2 C 11 12 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 2. REFLECTION IN A CIRCLE

Thus the point M , intersection of this line with the line z 0, has coordinates .0; 1 /. 0 D d Finally OM OM d 1 1. 0 D d D Second proof proof. Recall that if AB is a diameter of the circle ABC if and only if the angle ACB is a right angle.

1 AB

By symmetry the line Sm0 goes through M . Since SN is a diameter, the angle Sm0N is a right angle, thus the triangular NOM and M OM are similar. N orth pole 0

m0 1

OM M 0

South pole

Third proof. Denote the points of intersection of the sphere and the line OM by P and Q and join the lines NP and NQ :

N orth pole

PQ OM M 0

South pole § 1. DEFINITION OF INVERSIONS 13

_ _ By symmetry the arcs mQ and m0Q are of the same length on the same circle, thus the angles mNQ and QN m0 are of equal measure. The lines NQ and NP are thus the bisectors of the angle mN m . Then the bundle of lines .N m; N m NP;NQ/ is harmonic 0 0I and thus also the division .M; M 0 P;Q/. Since O is the middle of the segment PQ, we have OM OM OP 2 OQ2 I 12 1. 0 D D D D Definitions. 1°) Let P denote an Euclidean plane, O a point and R a length. We call inver- 2 sion with center1O and2 power R and denote Inv 2 the transformation of P O that O;R X f g associates to each point M the point M 0 such that – O, M and M 0 are on a same ray 2 – OM OM 0 R 2°) We extend theD definition of the inversion to the set P by defining [ f1g – Inv 2 .O/ O;R D 1 – Inv 2 . / O O;R 1 D 3°) Let be the circle with center O and radius R. The inversion InvO;R2 is also called reflection in the circle and is denoted Refl .

M 0 M O

We have the following immediate consequences of the deﬁnition.

Proposition. The transformation InvO;R2 or Reﬂ is involutive (which means that it is equal to its own inverse) :

Inv 2 Inv 2 Identity or Reﬂ Reﬂ Identity O;R ı O;R D ı D

Proposition. The image of a point inside the circle by InvO;R2 or Reﬂ is outside . The image of a point outside is inside .

Proposition. The ﬁxed points of InvO;R2 or Reﬂ are the points of .

Proposition. If a circle is orthogonal to , then it is globally invariant by InvO;R2 or Reﬂ . Deﬁnition. Let be a real positive number ( > 0). We call dilation by a factor and a center O and denote by DilO; the transformation of P (or P ) that associates to each [ f1g point M the point M 0 such that – O, M and M 0 are on a same ray – OM OM 0 D 14 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 2. REFLECTION IN A CIRCLE

In the case of P , the point is its own image. [ f1g 1 Proposition. For any point O and any positive numbers R and

DilO; Inv 2 DilO; Inv 2 ı O;R ı D O;R or equivalently

DilO; InvO;R2 InvO;R2 DilO; 1 ı D ı

§ 2. Images of lines and circles

Let be a circle with center O and radius R. We denote by f the transformation InvO;R2 or Reﬂ . In the case of P , the point is on all lines (in fact, later on, we’ll imagine the lines as being the circles[ f1g passing through1 the point ). 1 Proposition 1. The image of a line through O is itself.

M 0 M O

Proof. Let ` be a line through O. The image of O is and the image of is O. For any point M of ` different from O and from , the ray1 from O through M 1is included in `, thus f .`/ ` and since f is involutive f .`/1 `. D § 2. IMAGES OF LINES AND CIRCLES 15

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. `

M 0 M

O HH 0

Proof. Let M be any point of ` and denote by M 0 the intersection of the ray originating in O and going through M and the circle ` . We shall show that f .M / M . Indeed, since 0 D 0 OH 0 is a diameter of `0 the angle OM 0H 0 is a right angle, we have

OM 0 OH cos H 0OM 0 OH 0 D D OM thus OM OM OH OH R2 and f .M / M . Then f .`/ ` . 0 D 0 D D 0 0 By the same argument, we prove2 that f .`0/ `. Finally, f being involutive, we have f .`/ ` . D 0 Remark that we have got at the same time2f .`0/ ` and so we have proved the following proposition. D Proposition 3. Let ` be a circle such that O ` . Denote by H the point opposite to O on 0 2 0 0 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 E0, then the line ` is just the line EE0. If the circle `0 intersects in exactly one point T , then the line ` is just the line tangent to (and to `0) in T .

Proposition 4. Let C be a circle such that O is outside C . Then C 0 f .C / is a circle such that O is outside C . We have C C if and only if C is orthogonalD to . 0 0 D T

T1 T 0 C C1 C 0 O S 0 S1 S 16 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 2. REFLECTION IN A CIRCLE

Proof. Let us draw the tangents from O to C and denote the contact points by T and S. The rays OT and OS intersect in T1 and S1. There is a circle C1 tangent to OT in T1 and tangent to OS in S1. There is a real number such that the dilation t DilO; transforms C D into C1. The circle C1 is orthogonal to and thus invariant, that is f .C1/ C1. Let us call D C 0 the image of C1 by t. We have C 0 t.C1/ t.f .C1// t.f .t.C /// t f t.C /, D D D 2 D ı ı and since t f t f , we have C f .C /. But we have also C t .C / Dil 2 .C /, ı ı D 0 D 0 D D O; which shows that C 0 is a circle such that O is outside C 0. Lemma 1. Let N and S be two opposite points of (NS is a diameter of ) and let C be a circle going through N and S. The image C 0 of C by f is the circle image of C by the reﬂection in the line NS.

N M

M 0 O

S Proof. The power of the point O with respect to the circle C is ON OS R2. Let D M be a point of C . Let us call M1 the intersection of the line OM with the circle C other 2 than M . We have OM OM1 PC .O/ R . Let M be the image of M1 in the D D 0 halfturn of center O. Then M 0 is on the ray issued from O and going through M and we 2 have OM OM 0 R . Thus M 0 is the image of M through f . The image of C is thus the circle image ofDC through the central symmetry through O ; but since the fole picture is invariant through a reﬂectio in the line through O orthogonal to NS, we may aswell say that C 0 is the image of C in the reﬂection in the line NS. Lemma 2. The image through f of a circle with center O and radius r is the circle with R2 center O and radius r . Proof. Immediate consequence of the central symmetry. Proposition 5. Let C be a circle such that O is inside C . Then C f .C / is a circle such 0 D that O is inside C 0. Proof. Let be the center of C . If O, use Lemma 2. If O, we use the same trick as to prove the proposition 4. Let us drawD the line orthognal to¤ the line O, that cuts C in N and S. The rays ON and OS intersect in N1 and S1. The dilation t with center O and ON1 ratio ON transforms C in a cercle C1 such as the one in the lemma 1. Thus f .C1/ s.C1/, where s is the central symmetry with center O (s can as well be called halfturn orD dilation with ratio 1). The symmetry s commutes with t. Finally, put C t.s.C1//, we have 0 D C 0 t s.C1/ t f .C1/ t f t.C / f .C / D ı D ı D ı ı D and O is inside C 0. § 3. DESCRIPTION OF INVERSIONS USING COMPLEX NUMBERS 17

Definition. We call Riemann sphere the set P P . The circles of P are the circles of P and the unions of a line and the one pointD set[ f1g. Thus from now on a Euclidean circle will be a usual circle or a line to which is addedf1g the point . f1g Proposition 5. The transformation f , inversion or reflection through a circle, transforms circles into circles, preserves the measure of angles and changes the orientation. Remark. The reflection in a line is a special case of the reflection in a circle, since lines are special circles. But we know allready that the properties above are true for lines. Proof. Look at the following drawing ;

M 0

M O

§ 3. Description of inversions using complex numbers

We are restraining ourselves to the inversion with center O and power 1, that is to reﬂection in the unit circle C.

1 Theorem. The inversion with center O and power 1 transforms z into z , where z is the comples conjugate of z. N N

i 1 1 i 1 1 i Proof. Write z in the polar form : z e . Then z e and z e . Thus O, 1 D D N D z and z are on a same ray with origin O. Moreover N

1 1 z 1 j j jz j D D N or OM OM R2, where R 1. 0 D D Equation of generalized circles. The equation of a circle may be written zz wz wz c 0 and the equation of a line !z !z c 0, where w and ! are complexN numbers N NC andDc is real. We put all these togetherN in oneN C formD

˛zz !z !z 0 N N N C D 18 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 2. REFLECTION IN A CIRCLE

where ˛ and are real numbers and ! C such that ˛ 0 or ! 0. The (generalized) circle is the same if you multiply all three2 coefﬁcients by¤ a common¤ real constant .

Theorem. Two circles C and C 0 with equations

˛zz !z !z 0 and ˛0zz ! z !0z 0 0 N N N C D N N0 N C D are orthogonal if and only if

!! !! ˛ ˛ 0 C 0 D 0 C 0 and C is orthogonal to C if and only if 0 ˛ ; in the case when C is a line it is orthogonal to C if and only if 0, that is theD line goes through O. D Theorem. The circle C with equation

˛zz !z !z 0 N N N C D

has as inverse the circle C 0 with equation

zz !z !z ˛ 0 N N N C D Proof. M(z) belongs to C 0 if and only if

1 1 1 1 ˛ ! ! 0 z z N z z C D N N N N or zz !z !z ˛ 0: N N N C D Chapitre 3

Circle bundles

§ 1. Different kinds of circle bundles In this chapter lines are circles, more precisely the lines are § 2. Orthogonal bundles the circles going through the point at inﬁnity . Given any two 1 distinct (generalized) circles with equations f1.x; y/ 0 and § 3. General deﬁnitions and properties D f2.x; y/ 0, they determine a unique bundle of circles : the set of all circlesD with an equation

1f1.x; y/ 2f2.x; y/ 0 C D

§ 1. Different kinds of circle bundles

Deﬁnition. Let A and B be two points different from . We call circle bundle with base points A and B, the set of all circles passing through A1and B.

Proposition. Let B be a circle bundle with base points A and B. For any point C , there is one and only one circle in B passing through C .

19 20 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 3. CIRCLE BUNDLES

Proof. If C is not on the line AB, there is a unique ("usual") circle passing through A, B and C (circle circumscribed to the triangle ABC ). If C is on the line AB (including the case when C ), the circle will be the whole line AB. D 1 Proposition and deﬁnition. Let A and B be two points (different from each other and MA different from ) and let Œ0; . The set of points M such that MB is a circle called Apolonius1 circle relatively2 toC1 the segment AB with ratio . D Remark. If 0, the Apolonius circle is the point circle A (circle with center A and radius 0). If ,D the Apolonius circle is the point circle B (circle with center B and radius 0). If 1D, the 1 Apolonius circle is the line, bisector of the segment AB. D Proof. Choose the middle of the segment AB as origine of the frame and choose the x- axis such that the coordinates of A are . a; 0/. Then B.a; 0/ and the relation characterising M.x; y/, MA2 2MB2 may be written D .x a/2 y2 2..x a/2 y2/ C C D C or .1 2/.x2 y2/ 2.1 2/ax .1 2/a2 0 C C C C D which is indeed the equation of a circle. Deﬁnition. Let A and B be two points different from . We call circle bundle with limit points A and B, the set of all Apolonius circles relatively1 to the segment AB.

Proposition. Let A and B be two points different from . For any point C , there is exactly one circle passing through C and belonging to the circle1 bundle with limit points A and B. CA Proof. Choose CB . The Apolonius circle with ratio goes through C and it is the only one. D Theorem. Let A and B be two points different from . The circles belonging to the circle bundle with base points A and B are all orthogonal to1 all the circles belonging to the circle bundle with base points A and B. We say that these two bundles are orthogonal. § 1. DIFFERENT KINDS OF CIRCLE BUNDLES 21

Proof. We may prove it using the equations. Indeed we have already the equations of the circles belonging to the bundle with limit points :

1 2 .x2 y2/ 2 C ax a2 0 C C 1 2 C D . Now, let us look at the equation of a circle belonging to the bundle with base points : x2 y2 2ux 2vy c 0. This circle goes through A and B, thus a2 2ua c 0 andCa2 2ua c 0C, whichD implies u 0 and c a2. The equationsC of theseC circlesD are then C D D D x2 y2 2vy a2 0 C D The relation charasterizing orthogonality (2.aa bb / c c ) 0 C 0 D C 0 1 2 2.2 C a 0 0 . 2v// a2 a2 1 2 C D becomes 0 0, which is a true relation. We shallD see a much simpler proof later on. Deﬁnition. Let A be a point different from . We call concentric circle bundle with center A, the set of all circles with center A. 1

Deﬁnition. Let A be a point different from . We call bundle of lines through A the set of all the lines going through A. 1

A 22 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 3. CIRCLE BUNDLES

Deﬁnition. Let A be a point different from and d a line through A. We call tangential bundle of circles through A and tangent to d1the set containing d and all the circles tangent to d at the point A.

Deﬁnition. Let d be a line through. We call bundle of lines parallel to d the set of all the lines (including d) that are parallel to d.

§ 2. Orthogonal 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. § 2. ORTHOGONAL BUNDLES 23

Generic example.

Let A be the center of the picture above. Let B be any point different from A and different from . Choose the inversion with center B and power BA2. The image of the point A is itself.1 The image of the line BA is itself. The image of a line through A is a circle passing through A and B. We know that inversion preserves the angles. Thus, the image of a circle with center A is a circle orthogonal to the preceding one, unless the circle goes through B, in which case the image is the bisector of the segment AB. We see that we obtain two orthogonal bundles. We recognize the ﬁrst two examples above.

AB 24 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 3. CIRCLE BUNDLES

Reciprocally, if we have a circle bundle with base points, by an inversion in one of the base points, we get a bundle of concurrent lines. The bundles of circles with limit points becomes a bundle of circles orthogonal to all the lines of the line bundle. Thus we have orthogonal bundles. What happens if we start with a bundle B of tangent circles to a line d at a point A ? Let us denote by d 0 the line orthogonal to d going through A and by B0 the bundle of circles tangent to d 0 in A. Let C B and C 0 B0. These two circles meet at point A and the tangents at A to these circles2 are orthogonal2 ; thus these two circles are orthogonal and the bundles B and B0 are orthogonal.

A d 0

What happens if we transform the above picture by an inversion with center A ? All the circles are transformed into lines. The circles in B become lines parallel to d and those in B0 become lines parallel to d 0, that is orthogonal to d. Thus we get

A d 0

which is indeed much simpler ! § 3. GENERAL DEFINITIONS AND PROPERTIES 25

§ 3. General deﬁnitions and properties

From now on we use the word "circle" to mean "circle or line". We may still use the word line in the usual meaning, that is "a circle that goes through the point ". When we need to talk about a circle that is not a line we’ll say "genuine circle" or "circle1 such that does not belong to it". The equation of a circle is thus 1

˛.x2 y2/ 2Ax 2By 0 C C D (if ˛ 0, it is a line, if ˛ 0, it is a genuine circle). D ¤ 3.1 Deﬁnition of bundles

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

2 2 2 2 f1.x; y/ ˛1.x y / 2A1x 2B1y 1 0 and f2.x; y/ ˛2.x y / 2A2x 2B2y 2 0 WD C C D WD C C D we call bundle B of circles determined by C1 and C2, the circles with equations

2 f1.x; y/ f2.x; y/ 0 where .; / R , but .; / .0; 0/ C D 2 ¤ 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 B0 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 / 0 and f2.xP ; yP / 0, then f1.xP ; yP / D D C f2.xP ; yP / 0 and the same holds for Q. Thus if C1 and C2 go through P and Q, it will D 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. Recall. Two circles with equations

2 2 2 2 ˛.x y / 2Ax 2By 0 and ˛0.x y / 2A0x 2B0y 0 0 C C D C C D are orthogonal if and only if

˛ 0 ˛0 2AA0 2BB0 0 C D Theorem and deﬁnition. Let B be a bundle of circles. The set of circles orthogonal to the circles 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

2 2 2 2 ˛1.x y / 2A1x 2B1y 1 0 and ˛2.x y / 2A2x 2B2y 2 0 C C D C C D 26 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 3. CIRCLE BUNDLES

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;B0; 0/ ˛1 1˛ 2A1A 2B1B 0 0 C 0 0 0 D ˛2 2˛ 2A2A 2B2B 0 0 C 0 0 0 D

This system of two equations in four unknowns is of rank 2, since the circles C1and 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 Existence of circles Circle orthogonal to 2 circles and going through 1 point Theorem. 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 ), then the circle is the line PQ. If not, PQM is a real triangle and there is exactly oneD 1 circle going through the three vertices. Theorem. 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 , the circles are parallel lines to d, and through each point M distinct from thereD is 1 one and only one parallel to d. If P , we have two cases. First case : M 1d, then d is the only circle through P and M tangent¤ 1 to d at P . Second case : M d, then2 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 . Theorem. 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 .

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 same bundle, then there is at most one circle C 0 orthogonal to all three. This circle is unique .

Proof. We have to solve the homogeneous linear system in .˛0;A0;B0; 0/ 8 ˛1 1˛ 2A1A 2B1B 0 < 0 C 0 0 0 D ˛2 2˛ 2A2A 2B2B 0 0 C 0 0 0 D : ˛3 3˛ 2A3A 2B3B 0 0 C 0 0 0 D

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 be we are back to the preceding theorem. If not, § 3. GENERAL DEFINITIONS AND PROPERTIES 27 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; / R4 unique up to the product by a constant different from zero, but the converse is not true :2 given an element .˛; A; B; / R4, there is a circle with corresponding equation if and 2 2 A2 B2 ˛ only if the square of the radius is greater or equal to zero, where : R C˛2 . Thus we have proved that given three circles not belonging to a common bundle,D there is at most one circle orthogonal to the three circles. To be more precise we are going to begin with three lemmas, where the bundles are very simple.

Circle belonging to one circle bundle and orthogonal to one circle

Let B be a circle bundle and C a circle. We call B the bundle orthogonal to B. If C B, then all the circles in B are orthogonal to C . In this paragraph we’ll suppose that C 2 B . … Lemma 1. Let P be a point in the Euclidean plane. Let B be the bundle of lines going through P and let C be any line or any genuine circle with center different from P . Then there is exactly one element of B which is orthogonal to C . (It is the line P or the line C through P ortogonal to the line C ). Proof. If C is a genuine circle, the lines orthogonal to C are the diameters, that is, the lines P going through the center of C . If C is a line, there is always one unique line through a point P orthogonal to C . The only line going through that center and through the point P is the line P . Lemma 2. Let ` be a line in the Euclidean plane. Let B be the bundle of lines parallel to ` ` C and let C be any genuine circle. There is one and only one line `1 in the bundle B orthogonal `1 to C : `1 is the line parallel to ` that goes through the center of C . Proof. There is one and only one parallel to a line through a point. Lemma 3. Let P be a point in the Euclidean plane. Let B be the bundle of all circles having P as center and let C be any genuine circle. If P is outside C or on C , 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 PT . If P is inside C , then no circle belonging to B is orthogonal to C . Proof. If P is outside C or on C , one can draw a tangent to C that goes through P . Let T be the contact point. The circle with center P and radius PT is a circle orthogonal to C (if T P , the point circle P is orthogonal to C ). We know that there is at most one circle, so it isD this one. T

P C 28 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 3. CIRCLE BUNDLES

If P is inside the circle C , no circle with center P can be orthogonal to C . In fact, let us call the center of C . If the circle C and a circle C1 with center P were orthogonal, C there would be a point T common to both circles such that the angle P T would be a right P angle. The set of points T such that P T is a right angle is the circle with diameter P . Since P is inside C , the circle does not intersect C ; thus the circles C and C1 cannot be orthogonal.

Theorem 1. Let P and Q be two points in the Euclidean plane. Let B be the bundle of circles going through P and Q. Let C be any circle whitch do not1 belong to the bundle B orthogonal to B. Then there is exactly1 one element C1 of B which is orthogonal to C . (A way to construct C1 is the following : construct the image of P (or Q) in the reﬂection through circle C and then the circle through P , Q and that image.)

Proof. Make an inversion with center Q. Call P 0 the image of P and C 0 the image of C . The bundle B is transformed into the bundle B0 of lines going through the point P 0. By lemma 1, there is a line C10 through P 0 orthogonal to C 0. Using the inversion again the image C1 of C10 belongs to the bundle B and is orthogonal to C .

Theorem 2. Let P be a point in the Euclidean plane and ` a line through P . Let ` be the line going through P and orthogonal to `. Let B be the bundle of circles going through P and tangent to `. Let C be any circle which is not tangent to ` at the point P . Then there is exactly one element C1 of B which is orthogonal to C (see one possible construction of C1 after the proof).

Proof. Do an inversion with center P . The bundle B is transformed into a bundle B0 of lines parallel to ` and C is transformed to C 0. If C 0 is a genuine circle, lemma 2 proves that there is a unique line in B0 orthogonal to C 0 and thus if C does not go through P , there is a unique element in B orthogonal to C .

If C 0 is a line, that line is not parallel to ` and thus there is no line in B0 orthogonal to C 0 and thus if C goes through P there is no circle with strictly positive radius in B orthogonal to C . But the point circle P is on C and thus it is orthogonal to C .

Construction of the circle C1.

Let P be the image of P in the reﬂection through C . P `

C1 § 3. GENERAL DEFINITIONS AND PROPERTIES 29

Theorem 3. Let P and Q be two points in the Euclidean plane. Let B be the bundle of circles with limit points P and Q and let C be a circle. Denote the set of points of the line PQ which are strictly between P and Q by PQŒ. If C PQŒ ¿ then there is exactly \ D one element C1 of B which is orthogonal to C (see one possible construction of C1 after the proof). If C PQŒ ¿ then there is no circle in B orthogonal to C . \ ¤ Comment. If C PQŒ ¿, that is if C cuts PQŒ then C separates the points P and Q, thus in an inversion\ with¤ center Q the image C of C will separate P and Q , that is, 0 0 0 D 1 P 0 will be in the circle C 0. In the other case it will be outside C 0. Proof. Make an inversion with center Q and apply Lemma 3.

Construction of the circle C1.

Let Q be the image of Q in the reﬂection through C . Q

PQ C1

Circle orthogonal to 1 circle and going through 2 points Theorem. 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 C 0 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. Use theorem 1 above. 30 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 3. CIRCLE BUNDLES

Activities with Geogebra

We have the Euclidean plane eventually with one extra point and the two models of Hyperbolic plane : the disc model, where 1

D z C z < 1 and C z C z 1 D f 2 k j g D f 2 jj j D g and the half-plane model where

H z C .z/ > 0 and L z C .z/ 0 D f 2 j = g D f 2 j = D g 1. In the Euclidean plane. Draw a picture showing two orthogonal circles.

2. In the Euclidean plane. Let C and C 0 be two orthogonal circles with centers A and B. a) Is it possible to have A inside C 0 and B outside C ? b) Is it possible to have A inside C 0 and B inside C ? c) Is it possible to have A outside C 0 and B outside C ? 3. In the Euclidean plane. Choose two points A and B. Draw 10 circles belonging to the circle bundle with base points A and B. Where are the centers of these circles ? 4. In the Euclidean plane. Choose two points A and B. Draw the line AB and one circle C going through A and B. Choose a point T1 on C and draw the tangent t1 to C through T1. Let K1 be the intersection of the lines AB and t1. Draw the circle 1 with center K1 and radius K1T1. Draw other circles 2, 3 ::: by the same procedure. Hide the lines, points and circles which are not elements of the circle bundle with limit points A and B. 5. In the Euclidean plane. Draw a bundle of Euclidean lines going through one point and the orthogonal bundle of concentric circles. Transform these two bundles by an inversion (that is reflection in a circle). Drag the inversion circle and observe the result. 6. In the hyperbolic plane D. We put c C. Given two points C and D in D draw a line ` such that the reflection in ` exchanges the points C and D. MakeD the picture in such a way that you can drag around the points C and D. When is the line ` a Euclidean line ? You may do it this way (with automatic labelling) : 1) Draw the line a CD. 2) Choose a point EDon c. 3) Draw the circle d that goes through the points C , D and E. 4) Draw the point of intersection of d and c other than E and call it F . 5) Draw the line b EF . 6) Draw the point ofD intersection of CD and EF and call it G. 7) Draw the tangents to the circle d from G. 8) Choose one of the two contact points of the tangents with d and call it H . 9) Draw the circle g with center G and radius GH. 10) Let I and J be the intersection points of g and c. _ 11) Draw the arc h ` IJ which the intersection of g with D. 12) Hide the constructionD D lines and circle and keep only c, C , D and h `. 13) Move around the points C and D. D The line ` is the bisector of the hyperbolic segment CD. 7. In the Euclidean plane. Given a line d and a circle C in a Euclidean plane, find an inversion that transforms d into C . What is the image of C ? § 3. GENERAL DEFINITIONS AND PROPERTIES 31

8. In the Euclidean plane. Find an inversion that transforms D into H. Draw any picture in H and draw its image in D. Draw any picture in D and draw its image in H. 9. In the hyperbolic plane D. Given two points P and Q in D draw a line ` that goes through the points P and Q. Make the picture in such a way that you can drag around the points P and Q. Draw the hyperbolic segment PQ. 10. In the hyperbolic plane D. Draw a triangle PQR. Add the magnitude of the three angles. Move the picture. What can you say ? What happens if the vertices are near or even on C ? 11. In the hyperbolic plane H. Do the same exercises as 6., 9. and 10. 12. Given three circles. Draw a circle orthogonal to these three circles. When is it possible ? When is it not possible ? Thème II

Reading Brennan and all Chapitre 4

Non-Euclidean Geometry

§ 1. chapter 6. Non-Euclidean Geometry : the two In this chapter we would like to use the two usual models of usual models of a hyperbolic plane a hyperbolic plane : the disc D and the half-plane H of Poincaré. § 2. 6.1.1 Non-Euclidean Geometry : the two Indeed there are holomorphic functions whose restrictions to D (or usual models of a hyperbolic plane H) are bijections of one on the other. We can also deduce the properties of one model from those of the other one by an inversion. § 3. 6.1.2 Existence of hyperbolic lines § 4. 6.1.3 Inversion preserves inversion points Exercice 1. Let D and H be subsets of the complex plane C deﬁ- ned by D z C z < 1 and H z C .z/ > 0 D f 2 j j j g D f 2 j = g For Euclid’s Elements, please look at and let f be the inversion with center i and power 2. Show that http ://aleph0.clarku.edu/ djoyce/java/elements/elements.html f .D/ H and f .H/ D. D D

§ 1. chapter 6. Non-Euclidean Geometry : the two usual models of a hyperbolic plane

The unit disc of Poincaré Set of points. The points z .x; y/ such that x2 y2 < 1 or D C D z C z < 1 D f 2 j j j g The set D is called hyperbolic plane. Boundary points. The points z .x; y/ such that x2 y2 1 or C z C z 1 are not points of the hyperbolicD plane, but the set C ofC theseD points playsD f an2 importantj j j D role.g 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. Theorem 4. Given two distinct points P and Q in D, there is a unique 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… be invariant in the inversion with center O and power 1 (also

33 34 THÈME II. READING HYPERBOLIC GEOMETRY. CH. 4. NON-EUCLIDEAN GEOMETRY

called reﬂection in the circle C). Let P 0 and Q0 be the images of P and Q in this inversion. The circle going through P , P 0 and Q (it will also go through Q0) is orthogonal to C. Remark. A circle is such that its intersection with D is a d line if and only if its equation ˛.x2 y2/ 2Ax 2By 0 is such that C C D ˛ D The half-plane of Poincaré Set of points. The points z .x; y/ such that y > 0 or D H z C .z/ > 0 D f 2 j = g The set H is called hyperbolic plane. Boundary points. The points z .x; y/ such that y 0 or L z C .z/ 0 are not points of the hyperbolic plane,D but the set L of theseD points playsD f an2 importantj = D role.g The points of L are called boundary points. Set of lines. The lines are the intersection of H with the (generalized) circles orthogonal to the line L. We call these lines h 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. A circle is such that its intersection with H is a h line if and only if its equation ˛.x2 y2/ 2Ax 2By 0 is such that C C D B 0 and .˛; A/ .0; 0/ D ¤ Examples of h lines. y

H Q P

L O x § 2. 6.1.1 NON-EUCLIDEAN GEOMETRY : THE TWO USUAL MODELS OF A HYPERBOLIC PLANE 35

Remark 2. The inﬁnitesimal length ds in H is related to the inﬁnitesimal euclidean length dz by the simple formula j j dz ds j j D y2 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. There is a conformal transformation that transforms H into D (see page 322). Thus the remarks above are also valid for the hyperbolic plane D. Remark 4. 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 a point as the others and the space is completely homogeneous : all the points play the same role. Same remark is valid in the halfplane model.

§ 2. 6.1.1 Non-Euclidean Geometry : the two usual models of a hyperbolic plane

The unit disc of Poincaré 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 H `2

`3 `1

L `4 `5 O x 36 THÈME II. READING HYPERBOLIC GEOMETRY. CH. 4. NON-EUCLIDEAN GEOMETRY

Group of transformations Theorem 1. Let ` be a d-line which is part of a genuine circle C . The reﬂection with repect to C , denoted rC is such that

rC .D/ D and rC .C/ C D D

More precisely, the regions D1 and D2 are exchanged.

D D1 2 `

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 h- line). 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.

§ 3. 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/ O and r.O/ A. D D 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 f such that f .P / Q (and then f .Q/ P ). D D 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 § 3. 6.1.2 EXISTENCE OF HYPERBOLIC LINES 37 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. Recall that a circle C such that the reflection in C transforms P into Q is orthogonal to all circles going through P and Q. Thus we have to find a circle C orthogonal to C and belonging to the circle bundle B defined by the point circles P and Q. Thus C has to be orthogonal to any circle in the budle B0 of circles through P and Q and orthogonal to C. Choose a circle through P and Q that intersects C. Geometrical construction of the line `. In the disc model

C ` Q B bundle with base points P and Q

P B0 bundle with limit points P and Q

or in the half plane model

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. 38 THÈME II. READING HYPERBOLIC GEOMETRY. CH. 4. NON-EUCLIDEAN GEOMETRY

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/ A2 and f .`1/ f .`2/. D D 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 r s suits the conditions of the theorem. D ı

§ 4. 6.1.3 Inversion preserves inversion points

From now on "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, B0 and `0 the images by f of A, B and `. Then A0 and B0 are images of each other in the reflection f 0 through the line `0. Proof. Recall that A and B are images of each other in a reflection f through a line `, if ` belongs to the circle bundle wich is orthogonal to the bundle with base points A and B. Since a reflection is an inversion (more precisely the restriction to D of an inversion), the bundles are transformed to bundles and orthogonality is preserved (and the image of D is D), `0 is a part of a circle orthogonal to C and orthogonal to the bundle with base points A0 and B0. Thus A0 and B0 are images of each other in the reflection f 0 through the line `0. Chapitre 5

The group of Hyperbolic Geometry

§ 1. 6.2.1 Non-Euclidean Transformations and It can be showed that any isometry of the plane in Euclidean Möbius Transformations Geometry can be written as the composition of at most three reflec- § 2. 6.2.2 The Canonical Form of a Hyperbolic tions. Transformation In Hyperbolic Plane Geometry, the group is also the group of transformations that can be written as compositions of reflections. § 3. 6.2.3 The Canonical Form of a Hyperbolic We have here also direct and indirect hyperbolic transformations Transformation depending on the parity of the number of reflections. Here also any element of the group may be written as the composition of at most 3 In Euclidean Plane Geometry a transformation that preserves reflections. The composition of two reflections is the restriction to D distances (and thus angles) is called an isometry. The group of the of a Möbius transformation. A general Möbius transformation is a Euclidean Geometry is the group of isometries. Every isometry can bijection from C onto itself which can be written as a homographic be obtained as a composition of reflections. A composition of an function even number of reflections is a direct isometry preserving the orientations of angles. A composition of an odd number of reflections is a az b z C where .a; b; c; d/ C4 and ad bc 0 indirect isometry changing the orientations of angles into their op- 7! cz d 2 ¤ C posites. The composition of two reflections through two lines d and But in the disc model we have to restrain ourselves to the Möbius d d d 0 is a translation if 0 and a rotation if not. The composition transformations M such that M.D/ D which are those such of three reflections is eitherk a reflection or a "glide reflection" with that c b and d a D no fixed point : D D az b z C where .a; b/ C2 and b < a 7! bz a 2 j j j j C In the half-plane model we have to restrain ourselves to the Möbius transformations M such that M.H/ H which are those such that .a; b; c; d/ R4 and ad bc >D 0. 2

§ 1. 6.2.1 Non-Euclidean Transformations and Möbius Transformations

The inversions in C [ f1g 1 Let us look at the transformation z z . It is not an inversion since the points 0, z 1 7! and z are not on a common ray (unless z is real). Lat us write z in the polar form, we have

39 40 THÈME II. READING HYPERBOLIC GEOMETRY. CH. 5. THE GROUP OF HYPERBOLIC GEOMETRY

z rei and thus 1 1 e i . But 1 1 ei ; thus 0, z and 1 are on a same ray and D z D r z D r z z 1 1. The inversion with center O and power 1 transforms z into 1 . j j j z j D z

Proposition. The inversion with center O and power R2 may be desribed by the transformation R2 z 7! z

i R2 R2 i Proof. If z re , we get z0 z r e . The points z and z0 are on same ray with origine O andD the product of theirD modulesD is R2.

Proposition. The inversion with center ˛ and power R2 may be desribed by the transformation R2 z ˛ 7! C z ˛ Proof. Call t the translation such that t.0/ ˛, f the inversion with center ˛ and power 2 D 2 1 R and call i the inversion with center 0 and power R . To compute f .z/, we ﬁrst do t , R2 R2 then i and then t. Thus f .z/ is obtained by z z ˛ z ˛ z ˛ ˛. 7! 7! 7! C

The complex form of reﬂections in a hyperbolic plane

Lemma 1. Let ` be a hyperbolic line intersection of D with the circle with center ˛ orthogonal to C. The (hyperbolic) reﬂection in `, denoted , is given by

˛z 1 .z/ D z ˛ Proof. The reﬂection is the restriction to D of the inversion with center ˛ and power the square of the radius R of the circle . Since is orhogonal to C we have d 2 R2 12 and since the square of the distance from O to ˛ is ˛˛, we have R2 ˛˛ 1. TheD formulaC of last proposition then becomes D

˛˛ 1 ˛z 1 .z/ ˛ D C z ˛ D z ˛

What happens if ` is a line ? Use the following sketch to show that the limit of the preceding formula is just euclidean reflection through the line ` through O and orthogonal to the line that joins O and ˛ when ˛ goes to infinity in a specified direction. ˛ Let ˛ Kei , where K . The formula becomes .z/ z e2i. 2 /z D ! 1 D ˛ D C § 1. 6.2.1 NON-EUCLIDEAN TRANSFORMATIONS AND MÖBIUS TRANSFORMATIONS 41

2 e2i z Direction in which ˛ goes to inﬁnty z z

2i. / e C 2 z

This describes a symmetry relatively to the line through O orthogonal to the direction of ˛. How to look at homographic functions as linear functions ? A homographic function is a function that has the form

ax b x0 C .1/ D cx d C

Numerator N N 0 If we look at x and x0 as fractions x Denominator D and x0 D , then we’ll get back the formula .1/ by writing D D D 0 Â Ã Â ÃÂ Ã N 0 a b N D0 D c d D The nice thing with this trick is that when we compose two homographic functions we get a new homographic function with coefﬁcients that are elements of the product of two matrices. Usually the function has to be non constant, that is ad bc 0, which may be translated by saying that the matrix has determinant different from zero and¤ thus is invertible. Möbius transformations are complex homographic functions.

Composition of two hyperbolic reﬂections Let ˛ and ˇ be two complex numbers outside the unit disc C D. Considering the reﬂections described by [

˛z 1 ˇz 1 .z/ and .z/ D z ˛ D z ˇ We want to show that is a Möbius function and we want to ﬁnd which Möbius function it is. ı Let us call B the complex conjugation (B.z/ z) and introduce the Möbius functions D ˇz 1 ˛z 1 ˛z 1 Mˇ .z/ ;M˛.z/ and M˛.z/ D z ˇ D z ˛ D z ˛ 42 THÈME II. READING HYPERBOLIC GEOMETRY. CH. 5. THE GROUP OF HYPERBOLIC GEOMETRY

we have Mˇ B and M˛ B B M˛ D ı D ı D ı and thus, since B B idC, we have ı D Mˇ B B M˛ Mˇ M˛ ı D ı ı ı D ı and the matrix associated to the Möbius function is ı Âˇ 1ÃÂ˛ 1Ã Âˇ˛ 1 ˛ ˇ Ã 1 ˇ 1 ˛ D ˛ ˇ ˛ˇ 1 Finally az b . /.z/ C where a ˇ˛ 1 and b ˛ ˇ ı D bz a D D C Thus the composition of two hyperbolic reﬂections is the restriction to D of a Möbius transformation M that preserves C. Notice that M has also to map D on D and not on C D. For that we look at the image of 0, that is b and we must have M.0/ < 1, or X a j j b < a j j j j Exercice 1. Verify that b < a , when a ˇ˛ 1, b ˛ ˇ, ˛ > 1 and ˇ > 1. In the book the conversej j isj proven.j ThusD D j j j j Theorem 1 and 2. The direct hyperbolic transformation can always be desribed by a Möbius transformation of the form az b M.z/ C where b < a D bz a j j j j C Since B is an indirect hyperbolic transformation, all indirect transformation may be written as z M.z/ where M is as above. 7! 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.

§ 2. 6.2.2 The Canonical Form of a Hyperbolic Transformation

Direct : z m z K where K 1 and m < 1 7! 1 mz j j D j j Indirect : z m z K where K 1 and m < 1 7! 1 mz j j D j j § 3. 6.2.3 THE CANONICAL FORM OF A HYPERBOLIC TRANSFORMATION 43

§ 3. 6.2.3 The Canonical Form of a Hyperbolic Transformation Theorem 4. The equation of a line can be written

˛.x2 y2/ 2Ax 2By ˛ 0 C C D Proof. The equation of C is x2 y2 1 0 C D Thus a circle with equation ˛.x2 y2/ 2Ax 2By 0 is orthogonal to C if and only if C C D ˛ . 1/ 1 0 C D or simply ˛. D 44 THÈME II. READING HYPERBOLIC GEOMETRY. CH. 5. THE GROUP OF HYPERBOLIC GEOMETRY Chapitre 6

§3 and 4. Distance and geometrical problems

§ 1. 6.3.1 The distance formula § 6. 6.4.1 Triangles § 2. 6.3.2 Midpoints and 6.3.4 Reﬂection § 7. 6.4.2 Orthogonal lines § 3. 6.3.2 Circles § 4. 6.3.2 Reﬂections § 5. 6.3.2 Midpoints

§ 1. 6.3.1 The distance formula

Properties of a distance

2 1. .z1; z2/ D d.z1; z2/ > 0 8 2 2 W .z1; z2/ D d.z1; z2/ 0 z1 z2 8 2 W D ” D 2 2. .z1; z2/ D d.z1; z2/ d.z2; z1/ 8 2 W D 3 3. .z1; z2; z3/ D d.z1; z2/ d.z2; z3/ > d.z1; z3/ 8 2 C 3 4. `; line .z1; z2; z3/ ` where z2 between z1 and z3 d.z1; z2/ d.z2; z3/ d.z1; z3/ 8 8 2 W C D 2 5. .z1; z2/ D d.z1; z2/ d.z1; z2/ 8 2 W D 2 6. M Direct subgroup of D; .z1; z2/ D d.M.z1/; M.z2// d.z1; z2/ 8 2 8 2 D Comment. The group of the hyperbolic plane is generated by the reﬂections. The properties 5. and 6. together are infact equivalent to

2 f reflection of D; .z1; z2/ D d.f .z1/; f .z2// d.z1; z2/ 8 8 2 D (notice that complex conjugation is a specific reflection).

45 46 THÈME II. READING HYPERBOLIC GEOMETRY. CH. 6. DISTANCE

Hyperbolic trigonometry We’ll use the notations ch, sh and th for cosh, sinh and tanh, more convenient when you realy compute by hand ! Deﬁnition. ex e x ex e x sh x ex e x ch x C sh x th x D 2 D 2 D ch x D ex e x C Addition formulae.

ch .a b/ ch a ch b sh a sh b C D C sh .a b/ ch a sh b sh a ch b C D C ch a sh b sh a ch b th .a b/ C C D ch a ch b sh a sh b C Divide numerator and denominator by ch a ch b, you get

th a th b th.a b/ C C D 1 th a th b C 1 The functions ch, sh, th and th . ch ch sh

1 th

1 sh th § 2. 6.3.2 MIDPOINTS AND 6.3.4 REFLECTION 47

1 Explicit expression of the function th . x x 2x e e e 1 2x 2x From the expression y thx ex e x we get y 2x and y.e 1/ e 1 D D D e 1 C D or 1 y e2x.1 y/. Thus C C C D 1 y e2x C D 1 y and 1 1 1 y x th .y/ ln C D D 2 1 y Speed=velocity=rapidity in Special Relativity Let us take c 1, where c is the speed of light. The formula for adding velocities D v1 v2 V C D 1 v1v2 C If you deﬁne the rapidity r in terms of the velocity v by r th 1.v/, we have D th.v1/ th.v2/ th.R/ C th.v1 v2/ D 1 th.v1/th.v2/ D C C and thus R r1 r2 D C Thus the rapidity is additive, and if you had a velocity bigger than c, then the coresponding rapidity would be bigger than inﬁnity !

The formula

1ˇ z1 z2 ˇÁ d.z1; z2/ th ˇ ˇ D ˇ1 z1z2 ˇ

§ 2. 6.3.2 Midpoints and 6.3.4 Reﬂection

Use exercise 6 and 9, page 28 and 29.

§ 3. 6.3.2 Circles § 4. 6.3.2 Reﬂections § 5. 6.3.2 Midpoints § 6. 6.4.1 Triangles § 7. 6.4.2 Orthogonal lines 48 THÈME II. READING HYPERBOLIC GEOMETRY. CH. 6. DISTANCE

Exercises

Exercise 1. Let two point circles P and Q and a circle C be three circles in a common circle bundle. Show that the reﬂection in C transforms P in Q. Draw a picture that illustrates your proof. Exercise 2. 1°) Draw the picture of two tangential circle bundles B and B0. Call T the point common to all the circles belonging to the two bundles. 2°) Draw the images of B and B in an inversion with center O where O T . 0 ¤ 2°) Draw the images of B and B0 in an inversion with center T . Exercise 3. Let PQR be a triply asymptotic triangle, where P , Q and R are boundary points (belonging to C). Let S be a point outside C D. The rays with origin S and through [ P , Q and R intersect C in P 0, Q0 and R0. Find, if it exists, a hyperbolic reﬂection that transform the asymptotic triangle PQR and P 0Q0R0 one into the other. Make a drawing and prove your results. Exercise 4. 1°) Let C and C 0 be two circles with different centers, show that the set of points M having the same power relative to both circles ( that is PC .M / PC .M /) is a line. This D 0 line is called the radical axis of the circles C and C 0. 2°) Let and be the centers of two circles C and C . We suppose . Verify 0 0 ¤ 0 that the radical axis of the circles C and C 0 is orthogonal to the line 0. 3°) Show that if the circles C and C 0 intersect in two points A and B, then the line AB is the radical axis of C and C 0. 4°) Show that if C and C have a common tangent TT where T C et T C , then 0 0 2 0 2 0 the midpoint of the segment TT 0 belongs to the radical axis of the circles C and C 0. 5°) Show that if ` is the radical axis of two circles C and C 0, then `, C and C 0 belong to a common bundle of circles. 6°) Let C , C 0 and C 00 be three circles with centers which are not on a common line. 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. Show that `, `0 and `00 are concurrent, that is that there is a point belonging to these three lines. This point is called the radical center of the three circles C , C 0 and C 00. 7°) Use the notion of radical center to give a method of construction of the radical axis of two circles, where one is included in the other. 8°) a) Draw the circles C1, C2 and C3 with equations

x2 y2 1 0 C D .x 4/2 y2 4 0 C D .x 1/2 .y 3/2 2 0 C D

Use geometric methods to construct the circle orthogonal to C1, C2 and C3. b) Compute the equation of and check that it is orthogonal to the circles C1, C2 and C3. Check that it is the same as the one you constructed by geometric methods.