olymmath

RU-ALLRUSSIAN 1984 Problem 11

    ABC is a triangle and P is any point. The lines PA, PB, PC cut the circumcircle of ABC again at A'B'C' respectively. Show that there are at most eight points P such that A'B'C' is congruent to ABC.

    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 ru-allrussian/1984/p11 · Tag PNGP9M24DZ · Edit this page on GitHub · Keys: j next, k previous