Problem

Source: ELMO Shortlist 2024 G5

Tags: geometry



Let $ABC$ be a triangle with circumcenter $O$ and circumcircle $\omega$. Let $D$ be the foot of the altitude from $A$ to $\overline{BC}$. Let $P$ and $Q$ be points on the circumcircles of triangles $AOB$ and $AOC$, respectively, such that $A$, $P$, and $Q$ are collinear. Prove that if the circumcircle of triangle $OPQ$ is tangent to $\omega$ at $T$, then $\angle BTD=\angle CAP$. Tiger Zhang