IMO 1994 SL G3

Origin: RUS | Category: Geometry

Problem

A circle \omega is tangent to two parallel lines l1 and l2. A second circle \omega1 is tangent to l1 at A and to \omega externally at C. A third circle \omega2 is tangent to l2 at B, to \omega externally at D, and to \omega1 externally at E. AD intersects BC at Q. Prove that Q is the circumcenter of triangle CDE. 8 This problem is false. However, it is true if “not outside ABM” is replaced by “not outside ABCD”.

Solution

We shall prove that AD is a common tangent of \omega and \omega2. Denote by K, L the points of tangency of \omega with l1 and l2 respectively. Let r, r1, r2 be the radii of \omega, \omega1, \omega2 respectively, and set KA = x, LB = y. It will be enough if we show that xy = 2r2, since this will imply that \triangleKLB and \triangleAKO are similar, where O is the center of \omega, and consequently that OA \perpKD (because D \inKB). Now if O1 is the center of \omega1, we have x2 = KA2 = OO2 1 −(KO−AO1)2 = (r+r1)2−(r−r1)2 = 4rr1 and analogously y2 = 4rr2. But we also have (r1 +r2)2 = O1O2 2 = (x−y)2 +(2r−r1−r2)2, so x2 −2xy + y2 = 4r(r1 + r2 −r), from which we obtain xy = 2r2 as claimed. Hence AD is tangent to both \omega, \omega2, and similarly BC is tangent to \omega, \omega1. It follows that Q lies on the radical axes of pairs of circles (\omega, \omega1) and (\omega, \omega2). Therefore Q also lies on the radical axis of (\omega1, \omega2), i.e., on the common tangent at E of \omega1 and \omega2. Hence QC = QD = QE. Second solution. An inversion with center at D maps \omega and \omega2 to parallel lines, \omega1 and l2 to disjoint equal circles touching \omega, \omega2, and l1 to a circle externally tangent to \omega1, l2, and to \omega. It is easy to see that the obtained picture is symmetric (with respect to a diameter of l1), and that line AD is parallel to the lines \omega and \omega2. Going back to the initial picture, this means that AD is a common tangent of \omega and \omega2. The end is like that in the first solution.