2. Given are spheres with radii $R$ and $r$ ( $R<r$ ). The distance between the centers of these spheres is $c$, where $R-r<c<R+r$. Express the volume of the cone circumscribed around these spheres as a function of $R, d=R-r+c$. Prove that the ratio of the volume of this cone to the larger sphere is greater than or equal to 2. Determine the surface area of the cone for which this ratio is equal to 2.
Solution. Let $ABC$ be the fixed cross-section of the given cone ($A$ is the vertex of the cone). Let $O$ and $O_{1}$ be the centers of the given spheres, and $D$ and $D_{1}$ be their points of tangency that lie on the considered cross-section of the cone. Let $E$ be the midpoint of the base of the cross-section, and let $\alpha$ be the angle between the slant height and the height of the cone (see diagram on the right). From the right triangle $\triangle AOD$, we get $\frac{r}{AO} = \sin \alpha$, and from the right trapezoid $O_{1}D_{1}DO$ we get $\frac{d}{c} = \sin \alpha$. Therefore, $\frac{r}{AO} = \frac{d}{c}$, i.e., $AO = \frac{cr}{d}$. Hence, the height of the cone is

$$
h = AO + OO_{1} + O_{1}E = \frac{cr}{d} + c + R = \frac{c(r+d) + Rd}{d} = R \frac{c+d}{d}
$$

If we denote the radius of the base of the cone by $\rho$, from the triangle $AEC$ we get

$$
\rho = \tan \alpha = R \frac{c+d}{d} \frac{\frac{d}{c}}{\sqrt{1 - \frac{d^2}{c^2}}} = R \frac{c+d}{\sqrt{c^2 - d^2}}
$$

Therefore, the volume of the cone is

$$
V = \frac{1}{3} \pi \rho^2 h = \frac{\pi R^3}{3} \frac{(c+d)^2}{d(c-d)}
$$

and its ratio to the volume of the larger sphere is

$$
V : \frac{4}{3} \pi R^3 = \frac{(c+d)^2}{4d(c-d)}
$$

For this ratio to be greater than or equal to 2, it is necessary and sufficient that $(c+d)^2 \geq 8d(c-d)$, i.e., $(c-3d)^2 \geq 0$. The equality holds if and only if $c = 3d$. In this case, $\rho = R \sqrt{2}$, $h = 4R$, so the slant height of such a cone is $\sqrt{h^2 + \rho^2} = 3R \sqrt{2}$, and the surface area is

$$
P = \pi R \sqrt{2} (R \sqrt{2} + 3R \sqrt{2}) = 8 \pi R^2
$$