Three balls of the same radius touch each other pairwise and a certain plane. The base of the cone is located in this plane. All three spheres touch the lateral surface of the cone externally. Find the angle at the vertex of the axial section of the cone, if the height of the cone is equal to the diameter of the ball.

#
Let $O^{\prime} 1$, O'2, O'3 be the orthogonal projections of the centers $O 1, O 2$, O3 of the given spheres onto the base plane of the cone (Fig.2), $R$ be the radius of the spheres, $A$ be the vertex of the cone, $O$ be the center of the base of the cone, $r$ be its radius, $\phi$ be the angle in the axial section of the cone. The point $O$ is the center of the circle circumscribed around the equilateral triangle O'1O'2O'3 with side $2 R$, so $O O^{\prime} 1=\frac{2 R}{\sqrt{3}}$. Consider the axial section of the cone passing through the point $O 1$ (Fig.3). We obtain an isosceles triangle $A B C$ with base $B C=2 r$, height $A O=2 R$, and a circle tangent to the lateral side $A C$ at some point $M$, and the extension of the base $B C$ beyond point $C$ at point $O^{\prime} 1$. Let the line passing through point $A$ be tangent to this circle at point $P$, not lying on $A C$. Since $A O=P O^{\prime}=2 R$, the line $A P$ is parallel to $B C$. Denote $\angle P A O 1=\alpha$. Then

$$
\begin{gathered}
\operatorname{tg} \alpha=\frac{O_{1} P}{A P}=\frac{O_{P} P}{O O_{1}}=\frac{R \sqrt{3}}{2 R}=\frac{\sqrt{3}}{2} \\
\angle C A O 1=\angle P A O 1=\alpha, \angle C A P=2 \alpha
\end{gathered}
$$

Therefore,
$\operatorname{tg} \frac{\varphi}{2}=\operatorname{tg}\left(90^{\circ}-2 \alpha\right)=\operatorname{ctg} 2 \alpha=\frac{1}{\operatorname{tg}^{2 \alpha}}=\frac{1-\operatorname{tg}^{2} \alpha}{2 \operatorname{ran}_{\beta 2}}=$

$=1-\frac{3}{1 \sqrt{3}}=\frac{1}{1 \sqrt{3}}=\frac{\sqrt{3}}{12}$.

## Answer

$2 \operatorname{arctg} \frac{\sqrt{3}}{12}=4 \operatorname{arctg} \frac{2-\sqrt{3}}{2+\sqrt{3}}=\pi-4 \operatorname{arctg} \frac{\sqrt{3}}{2}$.