olymmath

SHARYGIN 2012 Problem 5

    On side ACAC of triangle ABCABC an arbitrary point is selected DD. The tangent in DD to the circumcircle of triangle BDCBDC meets ABAB in point C1C_1; point A1A_1 is defined similarly. Prove that A1C1∥ACA_1C_1 \parallel AC.

    Official text Sharygin Geometry Olympiad 2012 official paper, page 2. Quoted with source.

    A model read this text from the scan of the paper by OCR, with vision/local (GLM-OCR-8bit). No person has checked it against the paper yet.

    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 sharygin/2012/p5 · Tag GAN6T7KGTW · Edit this page on GitHub · Keys: j next, k previous