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Team Selection Test for JBMO

Geometry Difficulty 5.1 AIME, harder Prove it Turkey

In an equilateral triangle $ABC$ , let $D$ be a point on the side $[BC]$ other than the vertices. Let $I$ be the excenter of the triangle $ABD$ opposite to the side $[AB]$ and $J$ be the excenter of the triangle $ACD$ opposite to the side $[AC]$ . Let the circumcircles of the triangles $AIB$ and $AJC$ intersect at the point $E$ for the second time. Prove that $A$ is the incenter of the triangle $IEJ$ .

Solution

First note that $\angle IBA = \angle JCA = 60^\circ$ . Then $\angle IEA = \angle JEA = 60^\circ$ and $\angle IEJ = 120^\circ$ . Therefore A is on the angle bisector of the angle IEJ. Next observe that $\angle IAD = 90^\circ + \angle BAD/2$ and $\angle JAD = 90^\circ + \angle CAD/2$ . Hence $\angle IAJ = 360^\circ - 180^\circ - (\angle BAD + \angle DAC)/2 = 150^\circ$ and therefore A is inside the triangle IEJ.

Let X be the incenter of the triangle IEJ. We know that X is inside the triangle IEJ and $\angle IXJ = 90^\circ + \angle IEJ/2 = 150^\circ$ . Thus I, X, A, J are cyclic. Since E, X, A are collinear we get $X = A$ as desired.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.