olymmath

SHARYGIN 2010 Problem 21

    A given convex quadrilateral ABCDABCD is such that ∠ABD+∠ACD>∠BAC+∠BDC\angle ABD + \angle ACD > \angle BAC + \angle BDC. Prove that SABD+SACD>SBAC+SBDCS_{ABD} + S_{ACD} > S_{BAC} + S_{BDC}.

    Official text Sharygin Geometry Olympiad 2010 official paper, page 3. 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/2010/p21 · Tag 7F52BNMGDS · Edit this page on GitHub · Keys: j next, k previous