Hyperbolic Geometry. Eric Lehman

Transcription

1 Hyperbolic Geometry Eric Lehman February 2014

2 2 The course follows the chapter called "Non-Euclidean Geometry" in the book "Geometry" 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!

3 II. Reading Hyperbolic Geometry 32 Table of contents I. Circles in Euclidean Geometry 1 1 Power of a point with respect to a circle 3 1. Equations of a circle Orthogonal circles Reflection in a circle Definition of inversions Images of lines and circles Description of inversions using complex numbers Circle bundles Different kinds of circle bundles Orthogonal bundles General definitions and properties Non-Euclidean Geometry chapter 6. Non-Euclidean Geometry : the two usual models of a hyperbolic plane Non-Euclidean Geometry : the two usual models of a hyperbolic plane Existence of hyperbolic lines Inversion preserves inversion points The group of Hyperbolic Geometry Non-Euclidean Transformations and Möbius Transformations The Canonical Form of a Hyperbolic Transformation The Canonical Form of a Hyperbolic Transformation Distance The distance formula Midpoints and Reflection Circles Reflections Midpoints Triangles Orthogonal lines

4 4 TABLE OF CONTENTS

5 Thème I Circles in Euclidean Geometry

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7 Chapitre 1 Power of a point with respect to a circle 1. Equations of a circle 2. Orthogonal circles 1.1 Circle of center.a; b/ and radius R 1. Equations of a circle We suppose given an orthonormal frame.o; E{; Ej / of a Euclidean plane P. Let us denote by C.; R/ or simply by C the circle with center.a; b/ and radius R. A point M.x; y/ belongs to that circle if M D R or M 2 R 2 D 0 or by Pytagoras theorem.x a/ 2 C.y b/ 2 R 2 D 0 (1.1) That characteristic relation is called an equation of C. We define the function f C by The equations of C are f C.x; y/ D.x a/ 2 C.y b/ 2 R 2 f C.x; y/ D 0 where is any real number different from 0. When D 1, we say that the equation is the normal equation of C. We may write the normal equation of C as x 2 C y 2 2ax 2by C c D 0; where c D a 2 C b 2 R 2 (1.2) 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 D 0, since then O is on the circle C. 3

8 4 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 1. POWER OF A POINT WITH RESPECT TO A CIRCLE p c T R C.a; b/ O 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?). p c.. B.... O R C.a; b/ 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 P C.M / is the number P C.M / D d 2 R 2 Using the notations of the preceding paragraph, we have Proposition. P C.M / D f C.x; y/. Remark. c D P C.O/ D f C.0; 0/. 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 P C.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.

9 1. EQUATIONS OF A CIRCLE 5 Remark 2. Using the scalar product,we have in all cases P C.M / D MP!! MQ In particular P C.M / D 0 if and only if M belongs to C. Q C C P M P 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 P C.M / D M T 2 p PC.M / T d R C M Proof of theorem 2. Let be the center of the circle C and R its radius. Denote the length M by d. The angle 1 M T is a right angle and thus by Pythagoras theorem M T 2 D M 2 T 2 D d 2 R 2 D P C.M / 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 of P 0 on the line MQ, and! MQ and Let us write! P 0 D Q! and and so! MP 0 as!!! MP MQ D. M C Q/!!. M!! MP MQ D MP! 0! MQ! MQ D M! C Q! and! MP 0 D! M C! P 0, we have! Q/ D! M 2! Q 2 D d 2 R 2 D P C.M /

10 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 D x C iy and w D a C ib We denote the complex conjugate of a complex number z by Nz. Then Nz D x iy and x 2 C y 2 D zz ;.x a/ 2 C.y b/ 2 D.z w/.z w/ and a 2 C b 2 D ww The equation of C may be written.z w/.z w/ R 2 D 0 or zz.wz C wz/ C c D 0 where c D ww R 2 D P C.O/ 2 R. 2.1 Angle of intersecting circles 2. Orthogonal circles For any curves of class C 1, 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.

11 2. ORTHOGONAL CIRCLES 7 T Orthogonality conditions Theorem. Let C and C 0 be two circles with respective centers.a; b/ and 0.a 0 ; b 0 / and respective radii R and R 0, 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 : D R 2 C R The triangle T 0 is rectangle in T 3. P C 0./ D R 2 4. P C. 0 / D R a a 0 / 2 C.b b 0 / 2 D R 2 C R 02 Theorem. Two circles C and C 0 with real equations x 2 C y 2 2ax 2by C c D 0 and x 2 C y 2 2a 0 x 2b 0 y C c 0 D 0 are orthogonal if and only if Proof. The condition 5 above may be written 2.aa 0 C bb 0 / D c C c 0 a 2 C b 2 R 2 C a 02 C b 02 R 02 D 2aa 0 C 2bb 0 We get the result from c D a 2 C b 2 R 2 and c 0 D a 02 C b 02 R 02. 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/.

12 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.x 2 C y 2 / 2Ax 2By C D 0 and 0.x 2 C y 2 / 2A 0 x 2B 0 y C 0 D 0 are orthogonal if and only if 2.AA 0 C BB 0 / D 0 C 0 Proof. We have to check the three possible situations : 1) The two curves are circles, that is 0 and 0 0. We have a D Ą, b D, c D B and similar relations with. Thus the condition of orthogonality may be written 2. Ą A0 C B 0 B 0 / D C 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 D 0. The line C 0 is orthogonal to C if and only if it goes through the center. Ą ; / of C, that is B Multiplying by, one gets 2A 0 Ą 2B 0 B C 0 D 0 2AA 0 C 2BB 0 D 0 which is the expected relation since 0 D 0. 3) Both curves are lines, that is D 0 D 0. These two lines are orthogonal if and only if the vectors.a; B/ and.a 0 ; B 0 / are orthogonal, that is AA 0 CBB 0 D 0. Since D 0 D 0, we still have the expected relation. 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 are orthogonal if and only if 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.

13 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 C 1 and C 2 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 A 0 and B 0 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, A 0 and B 0 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 crossratio of these four points in that order is by defition.a; BI C; D/ D c a d a d b d a ). ** 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.

14 10 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 1. POWER OF A POINT WITH RESPECT TO A CIRCLE

15 Chapitre 2 Reflection in a circle 1. Definition of inversions 2. Images of lines and circles 3. Description of inversions using complex numbers 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 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 transformation in the plane called reflection in the unit circle or inversion with center O and power 1. O N orth pole m 0 M m South pole M 0 1. Definition of inversions Proposition. In the picture in the introduction above, if OM D d, then OM 0 D 1 d and OM OM 0 D 1. First proof. Let the line OM be the r axis and ON the z axis. The equation of the line NM is z 1 C r d D 1 or z D 1 r=d. The intesection with the circle z 2 C r 2 D 1 has coordinates.r; z/ such that.1 r=d/ 2 C r 2 D 1 or.1 C 1=d 2 /r 2.2=d/r D 0. This equation in r has two solutions : r D 0 corresponding to the North Pole and r D corresponding to m. Then the coordinates of m are r D and z D 1 d 2 1Cd 2. The equation of the line N m 0 is 2d 1Cd 2 2d 1Cd 2 and z D d 2 1 1Cd 2 and the coordinates of m 0 are r D 2d 1Cd 2 z D 1 C 1 d 2 1 1Cd 2 2d r D 1 dr 1Cd 2 11

16 12 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 2. REFLECTION IN A CIRCLE Thus the point M 0, intersection of this line with the line z D 0, has coordinates.0; 1 d /. Finally OM OM 0 D d 1 d D 1. Second proof proof. Recall that if AB is a diameter of the circle ABC if and only if the angle 1 ACB is a right angle. C A B By symmetry the line Sm 0 goes through M. Since SN is a diameter, the angle Sm 1 0 N is a right angle, thus the triangular NOM and M 0 OM are similar. N orth pole m 0 O M M 0 m 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 m 0 P O M Q M 0 m South pole

17 1. DEFINITION OF INVERSIONS 13 By symmetry the arcs _ mq and m _ 0 Q are of the same length on the same circle, thus the angles 1 mnq and 2 QN m 0 are of equal measure. The lines NQ and NP are thus the bisectors of the angle mn m 0. Then the bundle of lines.n m; N m 0 I NP; NQ/ is harmonic and thus also the division.m; M 0 I P; Q/. Since O is the middle of the segment PQ, we have OM OM 0 D OP 2 D OQ 2 D 1 2 D 1. Definitions. 1 ) Let P denote an Euclidean plane, O a point and R a length. We call inversion with center O and power R 2 and denote Inv O;R 2 the transformation of P X fog that associates to each point M the point M 0 such that O, M and M 0 are on a same ray OM OM 0 D R 2 2 ) We extend the definition of the inversion to the set P [ f1g by defining Inv O;R 2.O/ D 1 Inv O;R 2.1/ D O 3 ) Let be the circle with center O and radius R. The inversion Inv O;R 2 is also called reflection in the circle and is denoted Refl. O M M 0 We have the following immediate consequences of the definition. Proposition. The transformation Inv O;R 2 or Refl is involutive (which means that it is equal to its own inverse) : Inv O;R 2 ı Inv O;R 2 D Identity or Refl ı Refl D Identity Proposition. The image of a point inside the circle by Inv O;R 2 or Refl is outside. The image of a point outside is inside. Proposition. The fixed points of Inv O;R 2 or Refl are the points of. Proposition. If a circle is orthogonal to, then it is globally invariant by Inv O;R 2 or Refl. Definition. Let be a real positive number ( > 0). We call dilation by a factor and a center O and denote by Dil O; the transformation of P (or P [ f1g) that associates to each point M the point M 0 such that O, M and M 0 are on a same ray OM 0 D OM

18 14 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 2. REFLECTION IN A CIRCLE In the case of P [ f1g, the point 1 is its own image. Proposition. For any point O and any positive numbers R and Dil O; ı Inv O;R 2 ı Dil O; D Inv O;R 2 or equivalently Dil O; ı Inv O;R 2 D Inv O;R 2 ı Dil O; 1 2. Images of lines and circles Let be a circle with center O and radius R. We denote by f the transformation Inv O;R 2 or Refl. In the case of P [ f1g, the point 1 is on all lines (in fact, later on, we ll imagine the lines as being the circles passing through the point 1). Proposition 1. The image of a line through O is itself. ` O M M 0 Proof. Let ` be a line through O. The image of O is 1 and the image of 1 is O. For any point M of ` different from O and from 1, the ray from O through M is included in `, thus f.`/ ` and since f is involutive f.`/ D `.

19 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 M 0 O H H 0 `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 `0. We shall show that f.m / D M 0. Indeed, since OH 0 is a diameter of `0 the angle 2 OM 0 H 0 is a right angle, we have OM 0 OH 0 D cos 2 H 0 OM 0 D OH OM thus OM OM 0 D OH OH 0 D R 2 and f.m / D M 0. Then f.`/ `0. By the same argument, we prove that f.`0/ `. Finally, f being involutive, we have f.`/ D `0. 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 opposite 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. Proposition 4. Let C be a circle such that O is outside C. Then C 0 D f.c / is a circle such that O is outside C 0. We have C 0 D C if and only if C is orthogonal to. T O T 0 C 0 S 0 T 1 S 1 C 1 S C

20 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 T 1 and S 1. There is a circle C 1 tangent to OT in T 1 and tangent to OS in S 1. There is a real number such that the dilation t D Dil O; transforms C into C 1. The circle C 1 is orthogonal to and thus invariant, that is f.c 1 / D C 1. Let us call C 0 the image of C 1 by t. We have C 0 D t.c 1 / D t.f.c 1 // D t.f.t.c /// D t ı f ı t.c /, and since t ıf ıt D f, we have C 0 D f.c /. But we have also C 0 D t 2.C / D Dil O; 2.C /, 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. O N M 0 M M 1 S Proof. The power of the point O with respect to the circle C is ON OS D R 2. Let M be a point of C. Let us call M 1 the intersection of the line OM with the circle C other than M. We have OM OM 1 D P C.O/ D R 2. Let M 0 be the image of M 1 in the halfturn of center O. Then M 0 is on the ray issued from O and going through M and we have OM OM 0 D R 2. Thus M 0 is the image of M through f. The image of C is thus the circle image of C 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 center O and radius R2 r. Proof. Immediate consequence of the central symmetry. Proposition 5. Let C be a circle such that O is inside C. Then C 0 D f.c / is a circle such that O is inside C 0. Proof. Let be the center of C. If D O, use Lemma 2. If O, we use the same trick as to prove the proposition 4. Let us draw the line orthognal to the line O, that cuts C in N and S. The rays ON and OS intersect in N 1 and S 1. The dilation t with center O and ratio ON 1 ON transforms C in a cercle C 1 such as the one in the lemma 1. Thus f.c 1 / D s.c 1 /, where s is the central symmetry with center O (s can as well be called halfturn or dilation with ratio 1). The symmetry s commutes with t. Finally, put C 0 D t.s.c 1 //, we have and O is inside C 0. C 0 D t ı s.c 1 / D t ı f.c 1 / D t ı f ı t.c / D f.c /

21 3. DESCRIPTION OF INVERSIONS USING COMPLEX NUMBERS 17 Definition. We call Riemann sphere the set P D P [ f1g. The circles of P are the circles of P and the unions of a line and the one point set f1g. Thus from now on a Euclidean circle will be a usual circle or a line to which is added 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 O M 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. Theorem. The inversion with center O and power 1 transforms z into 1, where Nz is the Nz comples conjugate of z. Proof. Write z in the polar form : z D e i. Then 1 z D 1 e i and 1 Nz D 1 e i. Thus O, z and 1 are on a same ray with origin O. Moreover Nz or OM OM 0 D R 2, where R D 1. jzj j 1 Nz j D 1 D 1 Equation of generalized circles. The equation of a circle may be written z Nz Nwz w NzCc D 0 and the equation of a line N!z! Nz C c D 0, where w and! are complex numbers and c is real. We put all these together in one form z Nz N!z! Nz C D 0

22 18 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 2. REFLECTION IN A CIRCLE where and are real numbers and! 2 C such that 0 or! 0. The (generalized) circle is the same if you multiply all three coefficients by a common real constant. Theorem. Two circles C and C 0 with equations z Nz N!z! Nz C D 0 and 0z Nz N! 0 z! 0 Nz C 0 D 0 are orthogonal if and only if!! 0 C!! 0 D 0 C 0 and C is orthogonal to C if and only if 0 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. Theorem. The circle C with equation has as inverse the circle C 0 with equation Proof. M(z) belongs to C 0 if and only if z Nz N!z! Nz C D 0 z Nz N!z! Nz C D Nz Nz N! 1 Nz! 1 Nz C D 0 or z Nz N!z! Nz C D 0:

23 Chapitre 3 Circle bundles 1. Different kinds of circle bundles 2. Orthogonal bundles 3. General definitions and properties In this chapter lines are circles, more precisely the lines are the circles going through the point at infinity 1. Given any two distinct (generalized) circles with equations f 1.x; y/ D 0 and f 2.x; y/ D 0, they determine a unique bundle of circles : the set of all circles with an equation 1 f 1.x; y/ C 2 f 2.x; y/ D 0 1. Different kinds of circle bundles Definition. Let A and B be two points different from 1. We call circle bundle with base points A and B, the set of all circles passing through A and B. A 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

24 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 D 1), the circle will be the whole line AB. Proposition and definition. Let A and B be two points (different from each other and different from 1) and let 2 Œ0; C1. The set of points M such that MA D is a circle MB called Apolonius circle relatively to the segment AB with ratio. Remark. If D 0, the Apolonius circle is the point circle A (circle with center A and radius 0). If D 1, the Apolonius circle is the point circle B (circle with center B and radius 0). If D 1, the Apolonius circle is the line, bisector of the segment AB. 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/, MA 2 D 2 MB 2 may be written or.x C a/ 2 C y 2 D 2..x a/ 2 C y 2 /.1 2 /.x 2 C y 2 / C 2.1 C 2 /ax C.1 2 /a 2 D 0 which is indeed the equation of a circle. Definition. Let A and B be two points different from 1. We call circle bundle with limit points A and B, the set of all Apolonius circles relatively to the segment AB. A B Proposition. Let A and B be two points different from 1. For any point C, there is exactly one circle passing through C and belonging to the circle bundle with limit points A and B. Proof. Choose D CA. The Apolonius circle with ratio goes through C and it is the only CB one. Theorem. Let A and B be two points different from 1. The circles belonging to the circle bundle with base points A and B are all orthogonal to all the circles belonging to the circle bundle with base points A and B. We say that these two bundles are orthogonal.

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

26 22 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 3. CIRCLE BUNDLES Definition. Let A be a point different from 1 and d a line through A. We call tangential bundle of circles through A and tangent to d the set containing d and all the circles tangent to d at the point A. d 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. 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.

27 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 1. Choose the inversion with center B and power BA 2. The image of the point A is itself. 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. A B

28 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 B 0 the bundle of circles tangent to d 0 in A. Let C 2 B and C 0 2 B 0. These two circles meet at point A and the tangents at A to these circles are orthogonal ; thus these two circles are orthogonal and the bundles B and B 0 are orthogonal. d 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 B 0 become lines parallel to d 0, that is orthogonal to d. Thus we get d A d 0 which is indeed much simpler!

29 3. GENERAL DEFINITIONS AND PROPERTIES 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 1". When we need to talk about a circle that is not a line we ll say "genuine circle" or "circle such that 1 does not belong to it". The equation of a circle is thus.x 2 C y 2 / 2Ax 2By C D 0 (if D 0, it is a line, if 0, it is a genuine circle). 3.1 Definition of bundles Definition. Given two distinct circles C 1 and C 2, with equations f 1.x; y/ WD 1.x 2 Cy 2 / 2A 1 x 2B 1 yc 1 D 0 and f 2.x; y/ WD 2.x 2 Cy 2 / 2A 2 x 2B 2 yc2 D 0 we call bundle B of circles determined by C 1 and C 2, the circles with equations f 1.x; y/ C f 2.x; y/ D 0 where.; / 2 R 2, but.; /.0; 0/ Proposition and definitions. Let B be a bundle determined by two distinct circles C 1 and C 2. If C 1 and C 2 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 C 1 and C 2 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 C 1 and C 2 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.x P ; y P /. If f 1.x P ; y P / D 0 and f 2.x P ; y P / D 0, then f 1.x P ; y P / C f 2.x P ; y P / D 0 and the same holds for Q. Thus if C 1 and C 2 go through P and Q, it will be true for all circles in the bundle B determined by C 1 and C 2. 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.x 2 C y 2 / 2Ax 2By C D 0 and 0.x 2 C y 2 / 2A 0 x 2B 0 y C 0 D 0 are orthogonal if and only if 0 C 0 2AA 0 2BB 0 D 0 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 C 1 and C 2 be two distinct circles in B. Write the equations of C 1 and C 2 1.x 2 Cy 2 / 2A 1 x 2B 1 y C 1 D 0 and 2.x 2 Cy 2 / 2A 2 x 2B 2 y C2 D 0

30 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 C 1 and C 2. Thus we have to solve the homogeneous linear system in. 0; A 0 ; B 0 ; 0 / 1 0 C 1 0 2A 1 A 0 2B 1 B 0 D C 2 0 2A 2 A 0 2B 2 B 0 D 0 This system of two equations in four unknowns is of rank 2, since the circles C 1 and C 2 are distinct (thus. 1; A 1 ; B 1 ; 1 / and. 2; A 2 ; B 2 ; 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 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. 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 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. 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 ration MP MQ. Circle orthogonal to 3 circles Theorem. Let C 1, C 2 and C 3 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 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; A 0 ; B 0 ; 0 / 8 < 1 0 C 1 0 2A 1 A 0 2B 1 B 0 D C 2 0 : 3 0 C 3 0 2A 2 A 0 2A 3 A 0 2B 2 B 0 D 0 2B 3 B 0 D 0 If the three circles belong to one bundle, the vectors. 1; A 1 ; B 1 ; 1 /,. 2; A 2 ; B 2 ; 2 / and. 3; A 3 ; B 3 ; 3 / are linearly dependent and be we are back to the preceding theorem. If not,

31 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; / 2 R 4 unique up to the product by a constant different from zero, but the converse is not true : given an element. ; A; B; / 2 R 4, there is a circle with corresponding equation if and only if the square of the radius is greater or equal to zero, where : R 2 D A2 CB 2. 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. 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 2 B, then all the circles in B are orthogonal to C. In this paragraph we ll suppose that C 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 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 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. P C Lemma 2. Let ` be a line in the Euclidean plane. Let B be the bundle of lines parallel to ` and let C be any genuine circle. There is one and only one line `1 in the bundle B orthogonal to C : `1 is the line parallel to ` that goes through the center of C. ` `1 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 C 1 in the bundle B orthogonal to C : C 1 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 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 P T is a circle orthogonal to C (if T D P, the point circle P is orthogonal to C ). We know that there is at most one circle, so it is this one. T P C

32 28 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 3. CIRCLE BUNDLES P C 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 C 1 with center P were orthogonal, there would be a point T common to both circles such that the angle 1 P T would be a right angle. The set of points T such that 1 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 C 1 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 not belong to the bundle B orthogonal to B. Then there is exactly one element C 1 of B which is orthogonal to C. (A way to construct C 1 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 B 0 of lines going through the point P 0. By lemma 1, there is a line C1 0 through P 0 orthogonal to C 0. Using the inversion again the image C 1 of belongs to the bundle B and is orthogonal to C. C 0 1 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 C 1 of B which is orthogonal to C (see one possible construction of C 1 after the proof). Proof. Do an inversion with center P. The bundle B is transformed into a bundle B 0 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 B 0 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 B 0 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 C 1. ` P C Let P be the image of P in the reflection through C. P ` C 1

33 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 D then there is exactly one element C 1 of B which is orthogonal to C (see one possible construction of C 1 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 0 of C will separate P 0 and Q 0 D 1, that is, 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 C 1. C Let Q be the image of Q in the reflection through C. Q P Q C 1 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.

34 30 THÈME I. CIRCLES IN EUCLIDEAN GEOMETRY. CH. 3. CIRCLE BUNDLES Activities with Geogebra We have the Euclidean plane eventually with one extra point 1 and the two models of Hyperbolic plane : the disc model, where D D fz 2 Ckzj < 1 g and C D fz 2 Cjjzj D 1 g and the half-plane model where H D fz 2 C j =.z/ > 0 g and L D fz 2 C j =.z/ D 0 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 T 1 on C and draw the tangent t 1 to C through T 1. Let K 1 be the intersection of the lines AB and t 1. Draw the circle 1 with center K 1 and radius K 1 T 1. 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 D C. Given two points C and D in D draw a line ` such that the reflection in ` exchanges the points C and D. Make 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 D CD. 2) Choose a point E on 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 D EF. 6) Draw the point of 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 D ` D IJ _ which the intersection of g with D. 12) Hide the construction lines and circle and keep only c, C, D and h D `. 13) Move around the points C and 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?

35 3. GENERAL DEFINITIONS AND PROPERTIES 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 Given three circles. Draw a circle orthogonal to these three circles. When is it possible? When is it not possible?

36 Thème II Reading Brennan and all

37 Chapitre 4 Non-Euclidean Geometry 1. chapter 6. Non-Euclidean Geometry : the two usual models of a hyperbolic plane Non-Euclidean Geometry : the two usual models of a hyperbolic plane Existence of hyperbolic lines Inversion preserves inversion points For Euclid s Elements, please look at http ://aleph0.clarku.edu/ djoyce/java/elements/elements.html In this chapter we would like to use the two usual models of a hyperbolic plane : the disc D and the half-plane H of Poincaré. Indeed there are holomorphic functions whose restrictions to D (or 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. Exercice 1. Let D and H be subsets of the complex plane C defined by D D fz 2 C j jzj < 1 g and H D fz 2 C j =.z/ > 0 g and let f be the inversion with center i and power 2. Show that f.d/ D H and f.h/ 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 D.x; y/ such that x 2 C y 2 < 1 or The set D is called hyperbolic plane. D D fz 2 C j jzj < 1 g Boundary points. The points z 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. Theorem 4. Given two distinct points P and Q in D, there is a unique d P and Q. line going through 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

38 34 THÈME II. READING HYPERBOLIC GEOMETRY. CH. 4. NON-EUCLIDEAN GEOMETRY called reflection in the circle C). Let P 0 and Q 0 be the images of P and Q in this inversion. The circle going through P, P 0 and Q (it will also go through Q 0 ) is orthogonal to C. Remark. A circle is such that its intersection with D is a d.x 2 C y 2 / 2Ax 2By C D 0 is such that The half-plane of Poincaré D Set of points. The points z D.x; y/ such that y > 0 or The set H is called hyperbolic plane. H D fz 2 C j =.z/ > 0 g line if and only if its equation Boundary points. The points z D.x; y/ such that y D 0 or L D fz 2 C j =.z/ D 0 g are not points of the hyperbolic plane, but the set L of these points plays an important role. 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.x 2 C y 2 / 2Ax 2By C D 0 is such that Examples of h lines. B D 0 and. ; A/.0; 0/ y H P Q L O x

39 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 jdzj by the simple formula ds D jdzj y 2 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 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 H `2 `6 `1 `3 L O `4 `5 x

40 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 r C is such that r C.D/ D D and r C.C/ D C More precisely, the regions D 1 and D 2 are exchanged. C D 1 D 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 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 f such that f.p / D Q (and then f.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

41 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 B 0 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 P ` Q B bundle with base points P and Q B 0 bundle with limit points P and Q or in the half plane model ` Q L P 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.