IMO 1987 SL 10

Origin: ICE

Problem

Let S1 and S2 be two spheres with distinct radii that touch externally. The spheres lie inside a cone C, and each sphere touches the cone in a full circle. Inside the cone there are n additional solid spheres arranged in a ring in such a way that each solid sphere touches the cone C, both of the spheres S1 and S2 externally, as well as the two neighboring solid spheres. What are the possible values of n?

Solution

Denote by r, R (take w.l.o.g. r < R) the radii and by A, B the centers of the spheres S1, S2 respectively. Let s be the common radius of the spheres in the ring, C the center of one of them, say S, and D the foot of the perpendicular from C to AB. The centers of the spheres in the ring form a regular n-gon with cen- ter D, and thus sin(\pi/n) = s/CD. Using Heron’s formula on the trian- gle ABC, we obtain (r+R)2CD2 = 4rRs(r + R + s), and hence A B E2 E1 C E D R r s v sin2 \pi n = s2 CD2 = (r + R)2s 4(r + R + s)rR. (1) Choosing the unit of length so that r + R = 2, for simplicity of writing, we write (1) as 1/sin2(\pi/n) = rR (1 + 2/s) . Let now v be half the angle at the top of the cone. Then clearly R −r = (R + r) sin v = 2 sin v, giving us R = 1 + sin v, r = 1 −sin v. It follows that sin2 \pi n

 1 + 2 s  cos2 v. (2)

We need to express s as a function of R and r. Let E1, E2, E be collinear points of tangency of S1, S2, and S with the cone. Obviously, E1E2 = E1E + E2E, i.e., 2\sqrtrs + 2 \sqrt Rs = 2 \sqrt Rr = (R + r) cos v = 2 cosv. Hence, cos2 v = s( \sqrt R + \sqrtr)2 = s(R + r + 2 \sqrt Rr) = s(2 + 2 cosv). Substituting this into (2), we obtain 2 + cosv = 1/sin(\pi/n). Therefore 1/3 < sin(\pi/n) < 1/2, and we conclude that the possible values for n are 7, 8, and 9.