IMO 2012 solutions
Problem 1 (Evangelos Psychas, Greece) It is obvious that \[\angle JFL=\angle JBM-\angle FMB=\frac12(\angle BAC+\angle BCA)-\frac12\angle BCA=\frac12\angle BAC,\] therefore \(J,L,A\) and \(F\) belong to a circle which implies that \(\angle JFS=90^{\circ}\). But \(\angle …
IMO 2012 solutions Read More