2001年第42屆國際數學奧林匹亞競賽試題解答評析（續）

∠BAC= α = 60°∠CBA = β ∠ACB= γBP ' = BPAP " = AP' AB ' P AQ P" ∆BP'P β 2 AQ + QP" = AB + BP = AQ + QB QP = QB" " ∆AP' PAP BAC ∠PP ' = PP" BP"BP"" P ∆BPP" ∠PBQ = ∠PP'B = ∠PP"Q = β / 2 * * * * ∆BPP"BP = PP" = PP'' BPP ° β / 2 = 60 α + β = 60°++ 120°= 180° " P P"BCQ ∆ ** **** ** *** 120° − β= γ = β 2 ° β = 80 γ = 40° ∆ABC" P P"

ABQ ∆ABP ∆ AB + BP = AQ + QB ∆ABP " P