USAMO 2014 — Days 1 and 2 Question 5

Let $ABC$ have orthocenter $H$. Let $P$ be the second intersection of the circumcircle of $AHC$ with the internal bisector of $\angle BAC$. Let $X$ be the circumcenter of $APB$ and $Y$ the orthocenter of $APC$. Prove that $XY$ equals the circumradius of $ABC$.

Proposed by Titu Andreescu, Cosmin Pohoata (as credited in the supplied compilation).