olymmath

CN-LEAGUE 2014 Problem 2

    Let ABCABC be an acute triangle such that ∠BAC≠60∘\angle BAC \neq 60^\circ . Let D,ED,E be points such that BD,CEBD,CE are tangent to the circumcircle of ABCABC and BD=CE=BCBD=CE=BC ( AA is on one side of line BCBC and D,ED,E are on the other side). Let F,GF,G be intersections of line DEDE and lines AB,ACAB,AC . Let MM be intersection of CFCF and BDBD , and NN be intersection of CECE and BGBG . Prove that AM=ANAM=AN .

    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 cn-league/2014/p2 · Tag B3QVM6PB4E · Edit this page on GitHub · Keys: j next, k previous