IMO 2003 SL G2

Origin: GRE | Category: Geometry

Problem

Three distinct points A, B, C are fixed on a line in this order. Let \Gamma be a circle passing through A and C whose center does not lie on the line AC. Denote by P the intersection of the tangents to \Gamma at A and C. Suppose \Gamma meets the segment PB at Q. Prove that the intersection of the bisector of \angleAQC and the line AC does not depend on the choice of \Gamma.

Solution

Denote by R the intersection point of the bisector of \angleAQC and the line AC. From ∆ACQ we get AR RC = AQ QC = sin \angleQCA sin \angleQAC . By the sine version of Ceva’s theorem we have sin \angleAPB sin \angleBPC \cdot sin \angleQAC sin \anglePAQ \cdot sin \angleQCP sin \angleQCA = 1, which is equivalent to sin \angleAPB sin \angleBPC = sin \angleQCA sin \angleQAC 2 because \angleQCA = \anglePAQ and \angleQAC = \angleQCP. Denote by S(XY Z) the area of a triangle XY Z. Then sin \angleAPB sin \angleBPC = AP \cdot BP \cdot sin \angleAPB BP \cdot CP \cdot sin \angleBPC = S(∆ABP) S(∆BCP) = AB BC , which implies that
AR RC 2 = AB BC . Hence R does not depend on \Gamma.