olymmath

ISL 2018 G2

    Let ABCA B C be a triangle with AB=ACA B=A C, and let MM be the midpoint of BCB C. Let PP be a point such that PB<PCP B<P C and PAP A is parallel to BCB C. Let XX and YY be points on the lines PBP B and PCP C, respectively, so that BB lies on the segment PX,CP X, C lies on the segment PYP Y, and ∠PXM=∠PYM\angle P X M=\angle P Y M. Prove that the quadrilateral APXYA P X Y is cyclic. (Australia)

    Dataset text NuminaMath olympiads. Quoted with source.

    Official text: IMO 2018 Shortlist booklet, page 8.

    Solutions

    Official solution Solution 1 by the IMO Problem Selection Committee

    The terms of the source let olymmath link to this solution but not copy it. Read the solution (IMO official site).

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