SHARYGIN 2012 Problem 3

A circle with center $I$ touches sides $AB, BC, CA$ of triangle $ABC$ in points $C_1, A_1, B_1$ . Lines $AI, CI, B_1I$ meet $A_1C_1$ in points $X, Y, Z$ respectively. Prove that $\angle YB_1Z = \angle XB_1Z$ .

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

Solutions

Similar problems