Better butterfly theorem in the isotropic plane

Transcription

1 Mathematical Communications 2006, Better butterfly theorem in the isotropic plane Jelena Beban-Brkić Abstract. A real affine plane A 2 is called an isotropic plane I 2,if in A 2 a metric is induced by an absolute {f,f}, consisting of the line at infinity f of A 2 and a point F f. Better butterfly theorem is one of the generalisations of the wellknown butterfly theorem [],[4]. In this paper the better butterfly theorem has been adapted for the isotropic plane and its validity in I 2 has been proved. Key words: isotropic plane, better butterfly theorem AMS subject classifications: 5N25 Received March 3, 2006 Accepted March 27, Isotropic plane Let P 2 R be a real projective plane, f a real line in P 2, and A 2 = P 2 \f the associated affine plane. The isotropic plane I 2 R isarealaffineplanea 2 where the metric isintroduced with a real line f P 2 and a real point F incidental with it. The ordered pair {f,f}, F f iscalled the absolute figure of the isotropic plane I 2 R [2], [3]. In the affine model, where x = x /x 0, y = x 2 /x 0, the absolute figure is determined by the absolute line f x 0 =0,andtheabsolute point F 0:0:. We will first define some terms and point out some properties of triangles and circlesin I 2 that are going to be used further on. The geometry of I 2 could be seen for example in Sachs[2], or Strubecker [3]. All straight lines through the point F are called isotropic straight lines. A triangle in I 2 iscalled allowable if none of itssidesisisotropic. An isotropic circle parabolic circle or simply circle isaregular2 nd order curve in P 2 R which touchesthe absolute line f in the absolute point F. In I 2 there exists a three parametric family of circles, given by y = Rx 2 + αx + β, R 0, α, β R. Each circle can be reduced to the normal form y = Rx 2. Two circles k i y = R i x 2 + α i x + β i, i =, 2 are called congruent if R = R 2 ; they are called concentric if α = α 2. Department of Geomatics, Faculty of Geodesy, University of Zagreb, Kačićeva 26, HR Zagreb, Croatia, jbeban@geof.hr

2 34 J. Beban-Brkić 2. Better butterfly theorem Euclidean version Let there be 2 concentric circleswith the common centre O. Alinecrosses thetwo circlesat pointsp, Q and P, Q, M being the common midpoint of PQ and P Q. Through M, draw two lines AA B B and CC D D and connect AD, A D, BC, B C.LetX, Y, Z, W be the pointsof intersection of PP Q Q with AD, B C, A D, and BC, respectively. Then The proof isto be found in [4]. MX + MZ = MY + MW Figure. Better butterfly theorem Isotropic version This statement remains valid in the isotropic plane provided concentric circles are replaced by congruent and concentric circles and the corresponding equation for the signed lengths reads: dm,x + dm,z = dm,y dm,w 2 Figure 2. Better butterfly theorem in I 2

3 Better butterfly theorem in the isotropic plane 35 The proof dependson the following lemma: Lemma. In the allowable triangle RST let RU be a non-isotropic straight line connecting the vertex R with some point U on the opposite side ST of R. Let s introduce angles α = UR,RS, andβ = TR,RU. Then du, R = α dt,r β dr, S 3 Proof. Without loos of generality, we can assume that the vertex coordinates are asfollows: S0, 0, T t, 0, Rr,r 2, and Uu, 0, with t r u see Figure 3. For angles α, β and we have: Figure 3. α = UR,RS=uRS uur= s 2 r 2 s r r 2 u 2 r u, 4 β = TR,RU=uRU utr= u 2 r 2 r 2 t 2, 5 u r r t and = TR,RS=uRS utr= s 2 r 2 r 2 t 2. 6 s r r t Inserting 4, 5, and 6 in 3, together with du, R =r u, dt,r=r t, dr, S = r an equality isobtained. Proof of the theorem. Let k and k be two congruent and concentric circlesin I 2, k y = Rx 2, k y = Rx 2 + s, s 0andletM be the midpoint of the chord PQ of k. Let us choose the coordinate system as shown in the affine model in Figure 2, i.e. the tangent on the circle k parallel to the chord PQ asthe x-axis, and the isotropic straight line through M asthe y-axis. Choosing M0,m, for the chord PQ we have PQ y = m, andp p,m, Qq,m, P p,m, Q q,m. Note that p 2 = q 2 = m 2 2 R,andp = q = m s R.

4 36 J. Beban-Brkić Let Aa,Ra 2, Bb,Rb 2, with a b, and Cc,Rc 2, Dd,Rd 2, with c d, be the four pointson the circle k, anda a 2,Ra + s, B b 2,Rb + s, with a b, C c 2,Rc + s, D d 2,Rd + s, with c d, the four points on the circle k. Let usintroduce anglesα = PM,MA= QM, MB and β = DM, MP = CM,MQ. Applying Lemma 2 to allowable triangles AD M, A DM, B CM,and BC M successively one gets dx, M = α dd,m β dm,a β dm,b dy,m = α dc, M From 7 and 7 2 we obtain: 7, 7 3, dx, M + dz, M Analogously, 7 3 and 7 4 yield that dy,m + dw, M dz, M = α dd, M β dm,a dw, M = α dc,m β dm,b = α dd,m + dd, M β = α dm,a + dm,a dc,m + dc, M β dm,b + dm,b 7 2, Using dy, M = dm, Y anddw, M = dm, W, the latter becomes dm,y + = α dm,w dc,m + dc, M +β dm,b +. 0 dm,b Showing that the right-hand sides in 8 and 0 are equal, i.e. α dd,m + β dd, M dm,a + dm,a = α dc,m + dc, M + β dm,b + dm,b the theorem will be proved. Using the point coordinates we can rewrite the identity given in to the following form β a + a which isequivalent to a + b a β + β + b a b a b + β + = α + α +, b b c c d d c + d c = α α c d + d c d. 2

5 Better butterfly theorem in the isotropic plane 37 Besides, knowing that AB isa chord through M, the following relationsare obtained: M,A,B collinear points det 0 m a Ra 2 =0 b Rb 2 ma b Ra b a b =0 a b = m R. 3 Analogously, for CD being a chord through M, wegetthat c d = m R. 4 Relationsgiven in 3 and 4 can be reached using the following lemma: Lemma 2. Let k be a circle in I 2,apointP I 2, P / k, ands, S 2 two points of intersection of a non-isotropic straight line g through P with k. The product fp :=dp, S dp, S 2 does not depend on the line g, but only on k and P. The proof isgiven in [2, p. 32]. So, a b = dm,a dm,b =dm,p dm,q =p q = p p = p 2 = m R, and c d = dm,c dm,d =dm,p dm,q =p q = p p = p 2 = m R. Analogously, a b = d dm,b =dm,p dm,q M,A = p q = 2 m s p p = p = R, 5 c d = dm,c dm,d =dm,p dm,q = p q = 2 m s p p = p = R. 6 Since A, A,andM aswell asa, M, andb are collinear pointsthe relations ma a a a Ra a +a s =0, ma b a b Ra b +a s =0 respectively, are valid. Subtracting these relations we get ma b +a Ra 2 b 2 a 2 Ra b =0. The chord A B being a non-isotropic line allows us to rewrite the latter equation as m + a Ra + b a2 R = 0, wherefrom, using 3, we finally obtain that a + b = a + b. 7

6 38 J. Beban-Brkić Following the similar procedure, it can be shown that c + d = c + d 8 holdsaswell. For the oriented anglesα, β, introduced at the beginning, we have asfollows: α = PM,MA=uMA upm= a 2 m 2 m 2 p 2 = Ra2 m, 9 a m m p a β = CM,MQ=uMQ ucm= q 2 m 2 m 2 c 2 = m Rc2. 20 q m m c c Finally, using the relations given in 3, 4,..., and 20 in 2 one getsthat 2 βa + b = αc + d m Rc 2 a m Ra 2 = m c m c Ra a Rc m Rc2 Ra2 m = m Rc2 Ra2 m. Ra c Ra c Acknowledgments. The author isgrateful to the refereesfor their valuable suggestions. References [] H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, The Mathematical Association of America, Washington D. C., 967. [2] H. Sachs, Ebene isotrope Geometrie, Vieweg-Verlag, Braunschweig; Wiesbaden, 987. [3] K. Strubecker, Geometrie in einer isotropen Ebene, Math.-naturwiss. Unterricht, 5962, , [4]