10,11

The bases of three equal cones are located in the same plane and touch each other. The axial section of each cone is an equilateral triangle with side $a$. Find the radius of the sphere that touches the lateral surface of each cone and the plane in which their bases are located.
Let $Q$ be the point of tangency of the indicated sphere with center $O$ with the plane of the bases of the cones, and $O_1, O_2, O_3$ be the centers of the bases of the cones. Then $O_1 O_2 O_3$ is an equilateral triangle with side $a$, so

$$
Q O_1 = Q O_2 = Q O_3 = \frac{a}{\sqrt{3}}
$$

Draw a plane through the height $C O_1$ of one of the cones and parallel to the radius $Q O$ of the sphere. We obtain an equilateral triangle $A B C$ (the axial section of the cone) and a circle tangent to the lateral side, for example, $A C$, at point $D$, and the extension of the base $A B$ at point $Q$. By the problem's condition, $O_1 A = \frac{a}{2}, C O_1 = \frac{a \sqrt{3}}{2}$. Let $Q O = r, \angle O A Q = \phi$. Then

$$
\phi = \frac{1}{2} \left(180^{\circ} - \angle C A O_1 \right) = \frac{1}{2} \left(180^{\circ} - 60^{\circ} \right) = 60^{\circ}
$$

From the right triangle $O A Q$, we find that

$$
A Q = O Q \operatorname{ctg} \phi = r \operatorname{ctg} 60^{\circ} = \frac{r}{\sqrt{3}}.
$$

Since $Q O_1 = A Q + A O_1$, we have the equation

$$
\frac{a}{\sqrt{3}} = \frac{r}{\sqrt{3}} + \frac{a}{2}
$$

from which we find that $r = \frac{a (2 - \sqrt{3})}{2}$.

## Answer

$\frac{a (2 - \sqrt{3})}{2}$