olymmath

ISL 2021 G2

    Let ABCDA B C D be a convex quadrilateral circumscribed around a circle with centre II. Let ω\omega be the circumcircle of the triangle ACIA C I. The extensions of BAB A and BCB C beyond AA and CC meet ω\omega at XX and ZZ, respectively. The extensions of ADA D and CDC D beyond DD meet ω\omega at YY and TT, respectively. Prove that the perimeters of the (possibly self-intersecting) quadrilaterals ADTXA D T X and CDYZC D Y Z are equal.

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

    Official text: IMO 2021 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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