10,11

The edge of a regular tetrahedron is $a$. A plane $P$ passes through vertex $B$ and the midpoints of edges $A C$ and $A D$. A sphere is tangent to the lines $A B, A C, A D$ and that part of the plane $P$ which is inside the tetrahedron. Find the radius of the sphere. (Find all solutions).
Let $M$ and $N$ be the midpoints of edges $A C$ and $A D$ respectively, $L$ and $T$ be the points of tangency of the specified sphere with edges $A D$, $A B$, and $A C$ respectively, $F$ be the point of tangency of the sphere with the plane $B M N$, $Q$ be the point of intersection of the midline $M N$ of triangle $A D C$ with the median $A A 1$ of triangle $A D C$, $H$ be the center of face $B C D$, and $E$ be the point of intersection of the altitude $A H$ with the segment $B Q$. Denote $\angle B A H=\alpha$. From the right triangle $B A H$ we find that

$$
\sin \alpha=\frac{B H}{A B}=\frac{\frac{a}{\sqrt{3}}}{a}=\frac{1}{\sqrt{3}}.
$$

The sphere is tangent to the edges $A D, A C$, and $A B$ of the trihedral angle of the regular tetrahedron $A B C D$, so its center $O$ lies on the height $A H$ of the tetrahedron. Let $r$ be the radius of the sphere. Then

$$
O A=\frac{\alpha^{2}}{\sin r_{2}}=\frac{r}{\frac{1}{\sqrt{3}}}=r \sqrt{3}.
$$

Consider the section of the tetrahedron and the sphere by the plane $A B A 1$. We obtain triangle $A B A 1$, where $A B=a$, the height $A H=a \sqrt{\frac{2}{3}}$, $B H=\frac{a}{\sqrt{3}}$, and a circle of radius $r$ with center $O$ on $A H$, tangent to side $A B$ at point $L$ and
to the median $B Q$ at point $F$. Draw a line through vertex $A$ parallel to $B A 1$, and extend segment $B Q$ until it intersects this line at point $G$. From the equality of triangles $A Q G$ and $A 1 Q B$ we get that $A G=B A 1=\frac{a \sqrt{3}}{2}$, and from the similarity of triangles $A E G$ and $H E B-$

$$
\frac{A H}{E H}=\frac{A G}{B H}=\frac{\frac{a \sqrt{3}}{2 \sqrt{3}}}{3}=\frac{3}{2}.
$$

Thus,

$$
A E=\frac{3}{5} A H=\frac{3}{5} a \sqrt{\frac{2}{3}}, E H=\frac{2}{5} A H=\frac{2}{5} a \sqrt{\frac{2}{3}}.
$$

Denote $\angle B E H=\beta$. From the right triangle $B E H$ we find that

$$
\operatorname{ctg} \beta=\frac{E H}{B H}=\frac{\frac{2}{k a} \sqrt{\frac{2}{3}}}{\frac{a \sqrt{3}}{3}}=\frac{2 \sqrt{2}}{5}, \sin \beta=\frac{1}{\sqrt{1++2 g^{2} \beta}}=\frac{5}{\sqrt{33}}.
$$

Then

$$
O E=\frac{O F^{\prime}}{\sin \beta}=\frac{r \sqrt{33}}{5}.
$$

Suppose the sphere and vertex $A$ are on the same side of plane $P$. Then $A E=O A+O E$, or

$$
\frac{3}{5} a \sqrt{\frac{2}{3}}=r \sqrt{3}+\frac{r \sqrt{33}}{5}
$$

from which we find that

$$
r=\frac{\frac{3}{a} \sqrt{\frac{2}{2}}}{\sqrt{3}+\frac{\sqrt{33}}{3}}=\frac{\alpha \sqrt{2}}{5+\sqrt{11}}.
$$

If the sphere and vertex $A$ are on opposite sides of plane $P$, then $A E=O A-O E$, or

$$
\frac{3}{5} a \sqrt{\frac{2}{3}}=r \sqrt{3}-\frac{r \sqrt[3]{33}}{3}
$$

from which we find that

$$
r=\frac{\frac{3 a}{a} \sqrt{\frac{2}{3}}}{\sqrt{3}-\frac{\sqrt[3]{33}}{3}}=\frac{a \sqrt{2}}{5-\sqrt{11}}.
$$

## Answer

$\frac{a \sqrt{2}}{5 \pm \sqrt{11}}$.