ISL 2015 G1

Answer

Let $A B C$ be an acute triangle with orthocenter $H$ . Let $G$ be the point such that the quadrilateral $A B G H$ is a parallelogram. Let $I$ be the point on the line $G H$ such that $A C$ bisects $H I$ . Suppose that the line $A C$ intersects the circumcircle of the triangle $G C I$ at $C$ and $J$ . Prove that $I J=A H$ . (Australia)

Official text Transcription from NuminaMath olympiads, checked word for word against the official paper. Quoted with source.

Official text: IMO 2015 Shortlist booklet, page 7.

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I J=A H

Solutions

Official solution Solution 1 by the IMO Problem Selection Committee

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