IMO 1985 SL 22

Origin: USS

Problem

A circle with center O passes through points A and C and intersects the sides AB and BC of the triangle ABC at points K and N, respectively. The circumscribed circles of the triangles ABC and KBN intersect at two distinct points B and M. Prove that \angleOMB = 90◦.

Solution

Assume that \triangleABC is acute (the case of an obtuse \triangleABC is similar). Let S and R be the centers of the circumcircles of \triangleABC and \triangleKBN, respectively. Since \angleBNK = \angleBAC, the triangles BNK and BAC are similar. Now we have \angleCBR = \angleABS = 90◦−\angleACB, which gives us BR \perpAC and consequently BR \parallelOS. Similarly BS \perpKN implies that BS \parallelOR. Hence BROS is a parallelogram. Let L be the point symmetric to B with respect to R. Then RLOS is also a parallelogram, and since SR \perpBM, we obtain OL \perpBM. However, we also have LM \perpBM, from which we conclude that O, L, M are collinear and OM \perpBM. Second solution. The lines BM, NK, and CA are the radical axes of pairs of the three circles, and hence they intersect at a single point P. Also, the quadrilateral MNCP is cyclic. Let OA = OC = OK = ON = r. We then have BM \cdot BP = BN \cdot BC = OB2 −r2, PM \cdot PB = PN \cdot PK = OP 2 −r2. It follows that OB2 −OP 2

BP(BM −PM) = BM 2 −PM 2, which implies that OM \perpMB. A B C K L M N O P R S