Problem

Source: JBMO 2024

Tags: geometry, circumcircle, collinear, Balkan, JBMO



Let ABC be a triangle such that AB<AC. Let the excircle opposite to A be tangent to the lines AB,AC, and BC at points D,E, and F, respectively, and let J be its centre. Let P be a point on the side BC. The circumcircles of the triangles BDP and CEP intersect for the second time at Q. Let R be the foot of the perpendicular from A to the line FJ. Prove that the points P,Q, and R are collinear. (The excircle of a triangle ABC opposite to A is the circle that is tangent to the line segment BC, to the ray AB beyond B, and to the ray AC beyond C.) Proposed by Bozhidar Dimitrov, Bulgaria