IMO 2000 SL G1

Origin: NET | Category: Geometry

Problem

In the plane we are given two circles intersecting at X and Y . Prove that there exist four points A, B, C, D with the following property: For every circle touching the two given circles at A and B, and meeting the line XY at C and D, each of the lines AC, AD, BC, BD passes through one of these points.

Solution

Denote by k1, k2 the given circles and by k3 the circle through A, B, C, D. We shall consider the case that k3 is inside k1 and k2, since the other case is analogous. Let AC and AD meet k1 at points P and R, and BC and BD meet k2 at Q and S respectively. We claim that PQ and RS are the common tangents to k1 and k2, and therefore P, Q, R, S are the desired points. The circles k1 and k3 are tangent to each other, so we have DC \parallelRP. P A X B Q C D Y S R k1 k2 k3 Since AC \cdot CP = XC \cdot CY = BC \cdot CQ, the quadrilateral ABQP is cyclic, implying that \angleAPQ = \angleABQ = \angleADC = \angleARP. It follows that PQ is tangent to k1. Similarly, PQ is tangent to k2.