olymmath

OMM 2020 Problem 2

    Let ABCABC be a triangle with incenter II . The line BIBI meets ACAC at DD . Let PP be a point on CICI such that DI=DPDI=DP ( P≠IP\ne I ), EE the second intersection point of segment BCBC with the circumcircle of ABDABD and QQ the second intersection point of line EPEP with the circumcircle of AECAEC . Prove that ∠PDQ=90∘\angle PDQ=90^\circ .

    Proposed by Ariel García

    Dataset text AI-MO AoPS. Quoted with source.

    Solutions

    No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

    Add a solution

    Similar problems

    These problems have the most words in common with this statement.

    Locator mx-omm/2020/p2 · Tag H37EZ21N22 · Edit this page on GitHub · Keys: j next, k previous