olymmath

INMO 2024 Problem 1

    In triangle ABCA B C with CA=CBC A=C B, point EE lies on the circumcircle of ABCA B C such that ∠ECB=90∘\angle E C B=90^{\circ}. The line through EE parallel to CBC B intersects CAC A in FF and ABA B in GG. Prove that the centre of the circumcircle of triangle EGBE G B lies on the circumcircle of triangle ECFE C F.

    Dataset text NuminaMath olympiads. 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

    Locator in-inmo/2024/p1 · Tag 1NY2V8M0D8 · Edit this page on GitHub · Keys: j next, k previous