IMO 1982 SL 12

Problem

B6 (FIN 3) Four distinct circles C, C1, C2, C3 and a line L are given in the plane such that C and L are disjoint and each of the circles C1, C2, C3 touches the other two, as well as C and L. Assuming the radius of C to be 1, determine the distance between its center and L.

Solution

Let y be the line perpendicular to L passing through the center of C. It can be shown by a continuity argument that there exists a point Y \iny such that an inversion \Psi centered at Y maps C and L onto two concentric circles <C and <L. Let < X denote the image of an object X under \Psi. Then the circles @ Ci touch <C externally and <L internally, and all have the same radius. Let us now rotate the picture around the common center Z of <C and <L so that @ C3 passes through Y . Applying the inversion \Psi again on the picture thus obtained, <C and <L go back to C and L, but @ C3 goes to a line C′ 3 parallel to L, while the images of @ C1 and @ C2 go to two equal circles C′ 1 and C′ 2 touching L, C′ 3, and C. This way we have achieved that C3 becomes a line. Denote by O1, O2, O respectively the centers of the circles C′ 1, C′ 2, C and by T the point of tangency of the circles C′ 1 and C′ 2. If x is the common radius of the circles C′ 1 and C′ 2, then from \triangleO1TO we obtain that (x −1)2 + x2 = (x + 1)2, and L C′ C′ C′ T x x O1 O O2 thus x = 4. Hence the distance of O from L equals 2x −1 = 7.