olymmath

ISL 2014 G1

The points PP and QQ are chosen on the side BCB C of an acute-angled triangle ABCA B C so that ∠PAB=∠ACB\angle P A B=\angle A C B and ∠QAC=∠CBA\angle Q A C=\angle C B A. The points MM and NN are taken on the rays APA P and AQA Q, respectively, so that AP=PMA P=P M and AQ=QNA Q=Q N. Prove that the lines BMB M and CNC N intersect on the circumcircle of the triangle ABCA B C. (Georgia)

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

Official text: IMO 2014 Shortlist booklet, page 9.

Solutions

Official solution Solution 1 by the IMO Problem Selection Committee

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