Problem

Source: Ukrainian Mathematical Olympiad 2023. Day 1, Problem 10.3

Tags: geometry, tangency



Let $I$ be the incenter of the triangle $ABC$, and $P$ be any point on the arc $BAC$ of its circumcircle. Points $K$ and $L$ are chosen on the tangent to the circumcircle $\omega$ of triangle $API$ at point $I$, so that $BK = KI$ and $CL = LI$. Show that the circumcircle of triangle $PKL$ is tangent to $\omega$. Proposed by Mykhailo Shtandenko