ISL 2010 G1

Let $A B C$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $B C, C A, A B$ respectively. One of the intersection points of the line $E F$ and the circumcircle is $P$ . The lines $B P$ and $D F$ meet at point $Q$ . Prove that $A P=A Q$ . (United Kingdom)

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

Official text: IMO 2010 Shortlist booklet, page 45.

Solutions

Official solution Solution 1 by the IMO Problem Selection Committee

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