IMO 1974 SL 5

Problem

I 5 (GBR 3) Let Ar, Br, Cr be points on the circumference of a given circle S. From the triangle ArBrCr, called \triangler, the triangle \triangler+1 is ob- tained by constructing the points Ar+1, Br+1, Cr+1 on S such that Ar+1Ar is parallel to BrCr, Br+1Br is parallel to CrAr, and Cr+1Cr is parallel to ArBr. Each angle of \triangle1 is an integer number of degrees and those integers are not multiples of 45. Prove that at least two of the triangles \triangle1, \triangle2, . . . , \triangle15 are congruent.

Solution

All the angles are assumed to be oriented and measured modulo 180◦. Denote by \alphai, \betai, \gammai the angles of triangle \trianglei, at Ai, Bi, Ci respectively. Let us determine the angles of \trianglei+1. If Di is the intersection of lines BiBi+1 and CiCi+1, we have \angleBi+1Ai+1Ci+1 = \angleDiBiCi+1 = \angleBiDiCi+1 + \angleDiCi+1Bi = \angleBiDiCi −\angleBiCi+1Ci = −2\angleBiAiCi. We conclude that \alphai+1 = −2\alphai, and analogously \betai+1 = −2\betai, \gammai+1 = −2\gammai. Therefore \alphar+t = (−2)t\alphar. However, since (−2)12 \equiv1 (mod 45) and consequently (−2)14 \equiv(−2)2 (mod 180), it follows that \alpha15 = \alpha3, since all values are modulo 180◦. Analogously, \beta15 = \beta3 and \gamma15 = \gamma3, and moreover, \triangle3 and \triangle15 are inscribed in the same circle; hence \triangle3 ∼= \triangle15.