IMO 2002 SL G1

Origin: FRA | Category: Geometry

Problem

Let B be a point on a circle S1, and let A be a point distinct from B on the tangent at B to S1. Let C be a point not on S1 such that the line segment AC meets S1 at two distinct points. Let S2 be the circle touching AC at C and touching S1 at a point D on the opposite side of AC from B. Prove that the circumcenter of triangle BCD lies on the circumcircle of triangle ABC.

Solution

To avoid working with cases, we use oriented angles modulo 180◦. Let K be the circumcenter of \triangleBCD, and X any point on the common tangent to the circles at D. Since the tangents at the ends of a chord are equally inclined to the chord, we have \angleBAC = \angleABD + \angleBDC + \angleDCA = \angleBDX + \angleBDC + \angleXDC = 2\angleBDC = \angleBKC. It follows that B, C, A, K are concyclic, as required.