Problem

Source: India IMOTC 2024 Day 2 Problem 2

Tags: geometry



Let $ABC$ be an acute angled triangle with $AC>AB$ and incircle $\omega$. Let $\omega$ touch the sides $BC, CA,$ and $AB$ at $D, E,$ and $F$ respectively. Let $X$ and $Y$ be points outside $\triangle ABC$ satisfying \[\angle BDX = \angle XEA = \angle YDC = \angle AFY = 45^{\circ}.\]Prove that the circumcircles of $\triangle AXY, \triangle AEF$ and $\triangle ABC$ meet at a point $Z\ne A$. Proposed by Atul Shatavart Nadig and Shantanu Nene