IMO 1997 SL 25

Origin: POL

Problem

The bisectors of angles A, B, C of a triangle ABC meet its cir- cumcircle again at the points K, L, M, respectively. Let R be an internal point on the side AB. The points P and Q are defined by the following conditions: RP is parallel to AK, and BP is perpendicular to BL; RQ is parallel to BL, and AQ is perpendicular to AK. Show that the lines KP, LQ, MR have a point in common.

Solution

Let MR meet the circumcircle of triangle ABC again at a point X. We claim that X is the common point of the lines KP, LQ, MR. By symmetry, it will be enough to show that X lies on KP. It is easy to see that X and P lie on the same side of AB as K. Let Ia = AK \capBP be the excenter of \triangleABC corresponding to A. It is easy to calculate that \angleAIaB = \gamma/2, from which we get \angleRPB = \angleAIaB = \angleMCB = \angleRXB. Therefore R, B, P, X are concyclic. Now if P and K are on distinct sides of BX (the

other case is similar), we have \angleRXP = 180◦−\angleRBP = 90◦− \beta/2 = \angleMAK = 180◦−\angleRXK, from which it follows that K, X, P are collinear, as claimed. Remark. It is not essential for the statement of the problem that R be an internal point of AB. Work with cases can be avoided using oriented angles. A B C R M K L X P Ia Q