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45 degrees everywhere

Source: India IMOTC 2024 Day 2 Problem 2

May 31, 2024
geometry

Problem Statement

Let ABCABC be an acute angled triangle with AC>ABAC>AB and incircle ω\omega. Let ω\omega touch the sides BC,CA,BC, CA, and ABAB at D,E,D, E, and FF respectively. Let XX and YY be points outside △ABC\triangle ABC satisfying ∠BDX=∠XEA=∠YDC=∠AFY=45∘.\angle BDX = \angle XEA = \angle YDC = \angle AFY = 45^{\circ}. Prove that the circumcircles of △AXY,△AEF\triangle AXY, \triangle AEF and △ABC\triangle ABC meet at a point Z≠AZ\ne A.
Proposed by Atul Shatavart Nadig and Shantanu Nene
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