IMO 1983 SL 23

Origin: USS

Problem

Let K be one of the two intersection points of the circles W1 and W2. Let O1 and O2 be the centers of W1 and W2. The two common tangents to the circles meet W1 and W2 respectively in P1 and P2, the first tangent, and Q1 and Q2, the second tangent. Let M1 and M2 be the midpoints of P1Q1 and P2Q2, respectively. Prove that \angleO1KO2 = \angleM1KM2.

Solution

We note that \angleO1KO2 = \angleM1KM2 is equivalent to \angleO1KM1 = \angleO2KM2. Let S be the intersection point of the common tangents, and let L be the second point of in- tersection of SK and W1. Since \triangleSO1P1 ∼\triangleSP1M1, we have SK \cdot SL = SP 2 = SO1 \cdot SM1 which implies that points O1, L, K, M1 lie on a circle. Hence \angleO1KM1

\angleO1LM1 = \angleO2KM2. P1 P2 Q1 Q2 O1 O2 M1 M2 K L S