IMO 1989 SL 1

Origin: AUS

Problem

Let ABC be a triangle. The bisector of angle A meets the circumcircle of triangle ABC in A1. Points B1 and C1 are defined similarly. Let AA1 meet the lines that bisect the two external angles at B and C in point A0. Define B0 and C0 similarly. If SX1X2...Xn denotes the area of the polygon X1X2 . . . Xn, prove that SA0B0C0 = 2SAC1BA1CB1 \geq4SABC.

Solution

Let I denote the intersection of the three internal bisectors. Then IA1 = A1A0. One way proving this is to realize that the circumcircle of ABC is the nine-point circle of A0B0C0, hence it bisects IA0, since I is the orthocenter of A0B0C0. An- other way is through noting that IA1 = A1B, which follows from \angleA1IB = \angleIBA1 = (\angleA + \angleB)/2, and A1B = A1A0 which follows from \angleA1A0B = \angleA1BA0 = 90◦− A B C A1 B1 C1 A0 B0 C0 I \angleIBA1. Hence, we obtain SIA1B = SA0A1B. Repeating this argument for the six triangles that have a vertex at I and adding them up gives us SA0B0C0 = 2SAC1BA1CB1. To prove SAC1BA1CB1 \geq2SABC, draw the three altitudes in triangle ABC inter- secting in H. Let X, Y , and Z be the symmetric points of H with respect the sides BC, CA, and AB, respectively. Then X, Y, Z are points on the circumcircle of \triangleABC (because \angleBXC = \angleBHC = 180◦−\angleA). Since A1 is the midpoint of the arc BC, we have SBA1C \geqSBXC. Hence SAC1BA1CB1 \geqSAZBXCY = 2(SBHC + SCHA + SAHB) = 2SABC.