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