42nd IMO 2001 shortlist Problem G5 ABC is an acute-angled triangle. B' is a point on the perpendicular bisector of AC on the opposite side of AC to B such that angle AB'C = 2A. A' and C' are defined similarly (with ∠CA'B = 2∠C, ∠BC'A = 2∠B). The lines AA' and B'C' meet at A". The points B" and C" are defined similarly. Find AA'/A"A' + BB'/B"B' + CC'/C"C'. Solution