IMO 2002 SL G4

Origin: RUS | Category: Geometry

Problem

Circles S1 and S2 intersect at points P and Q. Distinct points A1 and B1 (not at P or Q) are selected on S1. The lines A1P and B1P meet S2 again at A2 and B2 respectively, and the lines A1B1 and A2B2 meet at C. Prove that as A1 and B1 vary, the circumcenters of triangles A1A2C all lie on one fixed circle.

Solution

Let O be the circumcenter of A1A2C, and O1, O2 the centers of S1, S2 respectively. First, from \angleA1QA2 = 180◦−\anglePA1Q −\angleQA2P = 1 2(360◦−\anglePO1Q − \angleQO2P) = \angleO1QO2 we obtain \angleA1QA2 = \angleB1QB2 = \angleO1QO2. Therefore \angleA1QA2 = \angleB1QP + \anglePQB2 = \angleCA1P + \angleCA2P = 180◦−\angleA1CA2, from which we con- clude that Q lies on the circum- circle of \triangleA1A2C. Hence OA1 = OQ. However, we also have O1A2 = O1Q. Consequently, O, O1 both lie on the perpendicular bisector of A1Q, so OO1 \perpA1Q. Similarly, OO2 \perpA2Q, leading to \angleO2OO1 = O1 O2 P Q A1 A2 B1 B2 C O S1 S2 180◦−\angleA1QA2 = 180◦−\angleO1QO2. Hence, O lies on the circle through O1, O2, Q, which is fixed.