IMO 1991 SL 5

Origin: SPA

Problem

In the triangle ABC, with ∡A = 60◦, a parallel IF to AC is drawn through the incenter I of the triangle, where F lies on the side AB. The point P on the side BC is such that 3BP = BC. Show that ∡BFP = ∡B/2.

Solution

Let P1 be the point on the side BC such that \angleBFP1 = \beta/2. Then \angleBP1F = 180o −3\beta/2, and the sine law gives us BF BP1 = sin(3\beta/2) sin(\beta/2)

3 −4 sin2(\beta/2) = 1 + 2 cos \beta. Now we calculate BF BP . We have \angleBIF = 120o −\beta/2, \angleBFI = 60o and \angleBIC = 120o, \angleBCI = \gamma/2 = 60o −\beta/2. By the sine law, BF = BI sin(120o −\beta/2) sin 60o , BP = 1 3BC = BI sin 120o 3 sin(60o −\beta/2). It follows that BF BP = 3 sin(60o−\beta/2) sin(60o+\beta/2) sin2 60o = 4 sin(60o −\beta/2) sin(60o + \beta/2) = 2(cos\beta −cos 120o) = 2 cos\beta + 1 = BF BP1 . Therefore P \equivP1.