$\left[\begin{array}{l}{[\text { Tangent Spheres }}\end{array}\right]$

The angle at the vertex of the axial section of a cone is $60^{\circ}$. Inside the cone, there are three spheres of radius 1. Each sphere touches two others, the base of the cone, and its lateral surface. Find the radius of the base of the cone.
Let $O_1, O_2$ and $O_3$ (Fig.1) be the centers of the spheres, $O$ be the center of the base circle of the cone, $R$ be its radius, and $A, B$ and $C$ be the orthogonal projections of points $O_1, O_2$ and $O_3$ onto the base plane of the cone (Fig.2). Then $O$ is the center of the equilateral triangle $ABC$ with side length 2. Therefore,

$$
O A=O B=O C=\frac{2 \sqrt{3}}{3}
$$

Draw a plane through the axis of the cone and parallel to the radius $O_1 A$ of one of the spheres (Fig.3). We obtain a circle of radius 1 inscribed in an angle with vertex $D$, equal to $60^{\circ}$. Then

$$
A D=A O_1 \operatorname{ctg} 30^{\circ}=\sqrt{3} .
$$

Therefore,

$$
R=O D=O A+A D=\frac{2 \sqrt{3}}{3}+\sqrt{3}=\frac{5 \sqrt{3}}{3}
$$

## Answer

$\frac{5}{\sqrt{3}}$.