10,11

The base of the prism is an equilateral triangle $ABC$, and the lateral edges of the prism $AA_1, BB_1$, and $CC_1$ are perpendicular to the base. A sphere, whose radius is equal to the side length of the base of the prism, touches the plane $A_1B_1C_1$ and the extensions of the segments $AB_1, BC_1$, and $CA_1$ beyond points $B_1, C_1$, and $A_1$ respectively. Find the side length of the base of the prism, given that the lateral edges are equal to 1.
Let a sphere of radius $a$ with center at point $O$ touch the base plane $A 1 B 1 C 1$ of a given prism at point $P 1$, and the lines $A B 1, B C 1$, and $C A 1$ at points $K, L$, and $M$ respectively, with $A 1 B 1=a$. Denote $P 1 A 1=x, P 1 B 1=y$, and $P 1 C 1=z$. If the extension of the radius $O P 1$ intersects the base $A B C$ at point $P$, then $P A=x, P B=y, P C=z$. By the theorem of the equality of tangent segments drawn from a single point to a sphere,

$$
A 1 M=A 1 P 1=x, B 1 K=B 1 P 1=y, C 1 L=C 1 P 1=z
$$

From the right triangles $OMC$ and $OPC$, we find that

$$
\begin{gathered}
O C 2=O M 2+C M 2=O M 2+(C A 1+A 1 M) 2=a 2+\left(\sqrt{1+a^{2}}+x\right) 2 \\
O C 2=O P 2+P C 2=(1+a) 2+z 2
\end{gathered}
$$

Therefore,

$$
(1+a) 2+z 2=a 2+\left(\sqrt{1+a^{2}}+x\right) 2
$$

Similarly,

$$
\begin{aligned}
& (1+a) 2+x 2=a 2+\left(\sqrt{1+a^{2}}+y\right) 2 \\
& (1+a) 2+y 2=a 2+\left(\sqrt{1+a^{2}}+z\right) 2
\end{aligned}
$$

Subtracting the second equation from the first, the third from the second, and the third from the first, we get three equations:

$$
\begin{aligned}
& z 2-x 2=\left(\sqrt{1+a^{2}}+x\right) 2-\left(\sqrt{1+a^{2}}+y\right) 2=(x-y)\left(2 \sqrt{1+a^{2}}+x+y\right) \\
& x 2-y 2=\left(\sqrt{1+a^{2}}+y\right) 2-\left(\sqrt{1+a^{2}}+z\right) 2=(y-z)\left(2 \sqrt{1+a^{2}}+y+z\right) \\
& z 2-y 2=\left(\sqrt{1+a^{2}}+x\right) 2-\left(\sqrt{1+a^{2}}+z\right) 2=(x-z)\left(2 \sqrt{1+a^{2}}+x+z\right) .
\end{aligned}
$$

Assume that $x>y$. Since $2 \sqrt{1+a^{2}}+x+y>0$, it follows from the first equation that $z>x$, so $z>y$.

At the same time, from the third equation, it follows that $z<y$, which is impossible. Assuming that $x<y$ also leads to a contradiction. Therefore, $x=y$. Similarly, $y=z$. Thus, $P 1 A 1=P 1 B 1=P 1 C 1$, i.e., $P 1$ is the center of the equilateral triangle $A 1 B 1 C 1$ with side $a$. Therefore,

$$
x=P 1 A 1=P 1 B 1=P 1 C 1=\frac{a \sqrt{3}}{3}
$$

Substituting $z=x=\frac{a \sqrt{3}}{3}$ into the equation

$$
(1+a) 2+z 2=a 2+\left(\sqrt{1+a^{2}}+x\right) 2
$$

we get the equation

$$
(1+a) 2+\frac{a^{2}}{3}=a 2+\left(\sqrt{1+a^{2}}+\frac{a \sqrt{3}}{3}\right) 2
$$

from which we find that $a=\sqrt{44}-6$.

## Answer

$\sqrt{44}-6$.