IMO 1999 SL G6

Origin: RUS | Category: Geometry

Problem

Two circles Ω1 and Ω2 touch internally the circle Ωin M and N, and the center of Ω2 is on Ω1. The common chord of the circles Ω1 and Ω2 intersects Ωin A and B. MA and MB intersect Ω1 in C and D. Prove that Ω2 is tangent to CD.

Solution

We first introduce the following lemmas.

Lemma 1. Let ABC be a triangle, I its inenter and Ia the center of the excircle touching BC. Let A′ be the center of the arc : BC of the circumcircle not containing A. Then A′B = A′C = A′I = A′Ia. Proof. The result follows from a straightforward calculation of the relevant angles. Lemma 2. Let two circles k1 and k2 meet each other at points X and Y and touch a circle k internally in points M and N, respectively. Let A be one of the intersections of the line XY with k. Let AM and AN intersect k1 and k2 respectively at C and E. Then CE is a common tangent of k1 and k2. Proof. Since AC\cdotAM = AX\cdotAY = AE\cdotAN, the points M, N, E, C lie on a circle. Let MN meet k1 again at Z. If M ′ is any point on the common tangent at M, then \angleMCZ = \angleM ′MZ = \angleM ′MN = \angleMAN (as oriented angles), implying that CZ \parallelAN. It follows that \angleACE = \angleANM = \angleCZM. Hence CE is tangent to k1 and analogously to k2. In the main problem, let us define E and F respectively as intersections of NA and NB with Ω2. Then applying Lemma 2 we get that CE and DF are the common tangents of Ω1 and Ω2. If the circles have the same radii, the result trivially holds. Otherwise, let G be the intersection of CE and DF. Let O1 and O2 be the centers of Ω1 and Ω2. Since O1D = O1C and \angleO1DG = \angleO1CG = 90◦, it follows that O1 is the midpoint of the shorter arc of the circumcircle of \triangleCDG. The center O2 is located on the bisector of \angleCGD, since Ω2 touches both GC and GD. O O1 O2 A B C D E F G M N X Y However, it also sits on Ω1, and using Lemma 1 we obtain that O2 is either at the incenter or at the excenter of \triangleCDG opposite G. Hence, Ω2 is either the incircle or the excircle of CDG and thus in both cases touches CD. Second solution. Let O be the center of \Gamma, and r, r1, r2 the radii of \Gamma, \Gamma1, \Gamma2. It suffices to show that the distance d(O2, CD) is equal to r2. The homothety with center M and ratio r/r1 takes \Gamma1, C, D into \Gamma, A, B, respectively; hence CD \parallelAB and d(C, AB) = r−r1 r d(M, AB). Let O1O2 meet XY at R. Then d(O2, CD) = O2R + r−r1 r d(M, AB), i.e., d(O2, CD) = O2R + r −r1 r [O1O2 −O2R + r1 cos \angleOO1O2], (1) since O, O1, and M are collinear. We have O1X = O1O2 = r1, OO1 = r −r1, OO2 = r −r2, and O2X = r2. Using the cosine law in the triangles OO1O2 and XO1O2, we obtain that cos \angleOO1O2 = 2r2 1−2rr1+2rr2−r2 2r1(r−r1) and O2R = r2 2r1 . Substituting these values in (1) we get d(O2, CD) = r2.