Let the spheres with centers $O_1, O_2$, and $O_3$ touch the base of a cone with vertex $A$, and the center $O_4$ of the fourth sphere lies on the height of the cone, which is equal to $H$. Consider a regular tetrahedron $O_1 O_2 O_3 O_4$, with an edge length of 2. If $h$ is its height, and $\alpha$ is the angle between the height $O_4 M$ and the lateral edge, then

$$
h=2 \sqrt{\frac{2}{3}}, \sin \alpha=\frac{O_{2} M}{O_{1} O_{2}}=\frac{\frac{2 \sqrt{3}}{3}}{2}=\frac{\sqrt{3}}{3}
$$

Suppose the sphere with center $O_4$ touches the lateral surface of the cone at point $B$. Then $O_4 B \perp A B$ and

$$
A O_4=\frac{O_{1} B}{\sin r_{2}}=\frac{\frac{1}{\sqrt{3}}}{3}=\sqrt{3}
$$

Therefore,

$$
H=1+h+O_4 A=1+2 \sqrt{\frac{2}{3}}+\sqrt{3}
$$

## Answer

$1+2 \sqrt{\frac{2}{3}}+\sqrt{3}$
8,9 |

Four spheres of radius 1 touch each other pairwise. Find the height of the cone that contains these spheres such that all of them touch the lateral surface and three of them touch the base of the cone.