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IMO Shortlist 2012, Geometry 6

Source: IMO Shortlist 2012, Geometry 6

July 29, 2013
geometrycircumcircleincenterreflectionIMO Shortlist

Problem Statement

Let ABCABC be a triangle with circumcenter OO and incenter II. The points D,ED,E and FF on the sides BC,CABC,CA and ABAB respectively are such that BD+BF=CABD+BF=CA and CD+CE=ABCD+CE=AB. The circumcircles of the triangles BFDBFD and CDECDE intersect at P≠DP \neq D. Prove that OP=OIOP=OI.
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