10,11

The edge of a regular tetrahedron is $4 \sqrt{6}$. Find the radius of the sphere that touches the lateral faces of the tetrahedron at points lying on the sides of the base.
Let the sphere with center $O$ touch the planes of the faces $A B D, B C D$, and $A C D$ of the given regular tetrahedron $A B C D$ at points $N, M$, and $K$, respectively, which lie on the sides $A B, B C$, and $A C$ of the base $A B C$. Then the section of the sphere by the plane of the base $A B C$ is a circle inscribed in the equilateral triangle $A B C$. Therefore, points $N$, $M$, and $K$ are the midpoints of the sides of the base $A B C$ of the tetrahedron $A B C D$. The radii $O N, O M$, and $O K$ of the sphere are perpendicular to the planes of the faces $A B D, B C D$, and $A C D$, respectively, and pass through the midpoints of the segments $A B, B C$, and $A C$. Thus, point $O$ is equidistant from points $A, B$, and $C$. Therefore, point $O$ lies on the line $D Q$, where $Q$ is the center of the equilateral triangle $A B C$. Let $\angle M D O=\phi, A B=A D=a$. Then

$$
\begin{aligned}
& D M=A M=\frac{a \sqrt{3}}{2}, Q M=\frac{1}{3} A M=\frac{a \sqrt{3}}{6}, \sin \phi=\frac{Q M}{D M}=\frac{\frac{\alpha \sqrt{3}}{\frac{\pi}{2}}}{2}=\frac{1}{3},
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_acc13965d0370bfc03dcg-04.jpg?height=94&width=712&top_left_y=1554&top_left_x=679)

Therefore,

$$
O M=D M \operatorname{tg} \angle M D O=\frac{a \sqrt{3}}{2} \cdot \operatorname{tg} \phi=\frac{a \sqrt{3}}{2} \cdot \frac{1}{2 \sqrt{2}}=\frac{a \sqrt{6}}{8}=4 \sqrt{6} \cdot \frac{\sqrt{6}}{8}=3
$$

## Answer

3.00