[ $\left[\begin{array}{ll}\text { Cylinder } & \\ \text { Cone } & \end{array}\right]$ [ Tangent spheres $]$
A cylinder with a base radius of 1 and height $\frac{12}{3+2 \sqrt{3}}$ has three identical spheres inscribed in it such that the spheres touch the upper base of the cylinder, its lateral surface, and each other. Find the volume of the cone whose base coincides with the lower base of the cylinder and which touches all three spheres.
Note that

$$
\frac{12}{3+2 \sqrt{3}}=\frac{12(2 \sqrt{3}-3)}{12-9}=4(2 \sqrt{3}-3)=8 \sqrt{3}-12 .
$$

Let $O^{\prime} 1, O^{\prime} 2, O^{\prime} 1$ be the orthogonal projections of the centers $O 1, O 2, O 3$ of the spheres onto the plane of the lower base of the cylinder (Fig.2), $O$ be the center of this base, $r$ be the radius of the spheres, $V$ be the desired volume of the cone. The point $O$ is the center of the circle circumscribed around the equilateral triangle $O^{\prime} 1 O^{\prime} 2 O^{\prime} 1$ with side $2 r$, so $O O^{\prime} 1=\frac{2 r}{\sqrt{3}}$, and since the radius of the base of the cylinder is 1, then $\frac{2 r}{\sqrt{3}}+r=1$, from which

$$
r=\frac{\sqrt{3}}{2+\sqrt{3}}=\sqrt{3}(2-\sqrt{3})=2 \sqrt{3}-3 .
$$

Consider the axial section of the cylinder passing through the point $O 1$ (Fig.3). We get a rectangle $A B C D$, an isosceles triangle $B C E$ with base $B C=2$ and height $E O$ ( $E$ is the vertex of the cone) and a circle with center $O 1$ and radius $r=2 \sqrt{3}-3$, touching $A B, A D$ and $B E$ at points $F, G$ and $H$ respectively. Denote $\angle O 1 B F=\alpha$. Then

$$
\begin{aligned}
& \operatorname{tg} \alpha=\frac{O_{1} F}{B F}=\frac{O_{1} F}{A B-A F}=\frac{r}{A B-r}=\frac{2 \sqrt{3}-3}{9 \sqrt{3}-12-2 \sqrt{3}+3}=\frac{2 \sqrt{3}-3}{3 \sqrt{3}-9}=\frac{1}{3}, \\
& \angle B E O=\angle A B E=2 \cdot \angle O 1 B F=2 \alpha \\
& \operatorname{tg} \angle B E O=\operatorname{tg} 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}=\frac{2 \cdot \frac{1}{3}}{1-\frac{1}{3}}=\frac{3}{4}, O E=\frac{B O}{\operatorname{tg} \angle B E O}=\frac{1}{\operatorname{tg} 2 \alpha}=\frac{4}{3} . \\
& V=\frac{1}{3} \pi \cdot O B^2 \cdot O E=\frac{1}{3} \pi \cdot 1^2 \cdot \frac{4}{3}=\frac{4}{9} \pi
\end{aligned}
$$

Therefore,

## Answer

$\frac{4}{9} \pi$