IMO 1992 SL 7

Origin: IND

Problem

Circles G, G1, G2 are three circles related to each other as follows: Circles G1 and G2 are externally tangent to one another at a point W and both these circles are internally tangent to the circle G. Points A, B, C are located on the circle G as follows: Line BC is a direct common tangent to the pair of circles G1 and G2, and line WA is the transverse common tangent at W to G1 and G2, with W and A lying on the same side of the line BC. Prove that W is the incenter of the triangle ABC.

Solution

Let G1, G2 touch the chord BC at P, Q and touch the circle G at R, S respectively. Let D be the midpoint of the complementary arc BC of G. The homothety centered at R mapping G1 onto G also maps the line BC onto a tangent of G parallel to BC. It follows that this line touches G at point D, which is therefore the image of P under the homothety. Hence R, P, and D are collinear. Since \angleDBP = \angleDCB = \angleDRB, it follows that \triangleDBP ∼\triangleDRB and consequently that DP \cdotDR = DB2. Similarly, points S, Q, D are collinear and satisfy DQ \cdot DS = DB2 = DP \cdot DR.

Hence D lies on the radical axis of the circles G1 and G2, i.e., on their common tangent AW, which also implies that AW bisects the angle BAD. Furthermore, since DB = DC = DW = \sqrt DP \cdot DR, it follows from the lemma of (SL99-14) that W is the incenter of \triangleABC. Remark. According to the third solution of (SL93-3), both PW and QW contain the incenter of \triangleABC, and the result is immediate. The problem can also be solved by inversion centered at W.