10. (ICE 3) Let $S_{1}$ and $S_{2}$ 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 $S_{1}$ and $S_{2}$ externally, as well as the two neighboring solid spheres. What are the possible values of $n$ ?
10. Denote by $\mathrm{r}, R$ (take w.lo.g. $r<R$ ) the radii and by $A, B$ the centers of the spheres $S_{1}, S_{2}$ 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 $A B$. The centers of the spheres in the ring form a regular $n$-gon with center $D$, and thus $\sin (\pi / n)=s / C D$. Using Heron's formula on the triangle $A B C$, we obtain $(r+R)^{2} C D^{2}=$ $4 r R s(r+R+s)$, and hence
$$
\sin ^{2} \frac{\pi}{n}=\frac{s^{2}}{C D^{2}}=\frac{(r+R)^{2} s}{4(r+R+s) r R} .
$$

Choosing the unit of length so that $r+R=2$, for simplicity of writing, we write (1) as $1 / \sin ^{2}(\pi / n)=r R(1+2 / s)$. 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
$$
\frac{1}{\sin ^{2} \frac{\pi}{n}}=\left(1+\frac{2}{8}\right) \cos ^{2} v
$$

We need to express $s$ as a function of $R$ and $r$. Let $E_{1}, E_{2}, E$ be collinear points of tangency of $S_{1}, S_{2}$, and $S$ with the cone. Obviously, $E_{1} E_{2}=$ $E_{1} E+E_{2} E$, i.e., $2 \sqrt{r s}+2 \sqrt{R s}=2 \sqrt{R r}=(R+r) \cos v=2 \cos v$. Hence,
$$
\cos ^{2} v=s(\sqrt{R}+\sqrt{r})^{2}=s(R+r+2 \sqrt{R r})=s(2+2 \cos v) .
$$

Substituting this into (2), we obtain $2+\cos v=1 / \sin (\pi / n)$. Therefore $1 / 3<\sin (\pi / n)<1 / 2$, and we conclude that the possible values for $n$ are 7. 8. and 9 .