IMO 2003 SL G5

Origin: KOR | Category: Geometry

Problem

Let ABC be an isosceles triangle with AC = BC, whose incenter is I. Let P be a point on the circumcircle of the triangle AIB lying inside the triangle ABC. The lines through P parallel to CA and CB meet AB at D and E, respectively. The line through P parallel to AB meets CA and CB at F and G, respectively. Prove that the lines DF and EG intersect on the circumcircle of the triangle ABC.

Solution

The triangles PDE and CFG are homothetic; hence lines FD, GE, and CP intersect at one point. Let Q be the intersection point of the line CP and the circumcircle of \triangleABC. The required statement will follow if we show that Q lies on the lines GE and FD. Since \angleCFG = \angleCBA = \angleCQA, the quadrilateral AQPF is cyclic. Analogously, BQPG is cyclic. However, the isosceles trapezoid BDPG is also cyclic; it follows that B, Q, D, P, G lie on a circle. Therefore we get \anglePQF = \anglePAC, \anglePQD = \anglePBA. (1) Since I is the incenter of \triangleABC, we have \angleCAI

2\angleCAB

2\angleCBA = \angleIBA; hence CA is the tangent at A to the circumcircle of \triangleABI. This implies that \anglePAC = \anglePBA, and it follows from (1) that \anglePQF = \anglePQD, i.e., that F, D, Q are also collinear. Similarly, G, E, Q are collinear and the claim is thus proved. A B C I P D E G F Q