IMO 1986 SL 15

Origin: NET

Problem

Let ABCD be a convex quadrilateral whose vertices do not lie on a circle. Let A′B′C′D′ be a quadrangle such that A′, B′, C′, D′ are the centers of the circumcircles of triangles BCD, ACD, ABD, and ABC. We write T (ABCD) = A′B′C′D′. Let us define A′′B′′C′′D′′ = T (A′B′C′D′) = T (T (ABCD)). (a) Prove that ABCD and A′′B′′C′′D′′ are similar. (b) The ratio of similitude depends on the size of the angles of ABCD. Determine this ratio.

Solution

(a) This is the same problem as SL82-14. (b) If S is the midpoint of AC, we have B′S = AC cos \angleD 2 sin \angleD , D′S = AC cos \angleB 2 sin \angleB , B′D′ = AC sin(\angleB+\angleD) 2 sin \angleB sin \angleD . These formulas are true also if \angleB > 90◦or \angleD > 90◦. We similarly obtain that A′′C′′ = B′D′

sin(\angleA′+\angleC′) 2 sin \angleA′ sin \angleC′ . Therefore A′′C′′ = AC sin2(\angleA + \angleC) 4 sin \angleA sin \angleB sin \angleC sin \angleD.