IMO 2004 SL G1

Origin: ROM | Category: Geometry

Problem

Let ABC be an acute-angled triangle with AB ̸= AC. The circle with diameter BC intersects the sides AB and AC at M and N, respectively. Denote by O the midpoint of BC. The bisectors of the angles BAC and MON intersect at R. Prove that the circumcircles of the triangles BMR and CNR have a common point lying on the line segment BC.

Solution

Note that \triangleANM ∼\triangleABC and consequently AM ̸= AN. Since OM = ON, it follows that OR is a perpendicular bisector of MN. Thus, R is the common point of the median of MN and the bisector of \angleMAN. Then it follows from a well-known fact that R lies on the circumcircle of \triangleAMN. Let K be the intersection of AR and BC. We then have \angleMRA = \angleMNA = \angleABK and \angleNRA = \angleNMA = \angleACK, from which we conclude that RMBK and RNCK are cyclic. Thus K is the desired in- tersection of the circumcircles of \triangleBMR and \triangleCNR and it indeed lies on BC.