10,11

In a rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$ ( $A A 1|| B B 1 \| C C 1|| D D 1$ ), it is known that $A B=B C=2 a$, $A A 1=a$. A plane section passes through points $B 1$ and $D$ parallel to the line $A C$. Find the radius of the sphere that touches this section and the three faces of the parallelepiped with a common vertex $B$.
Let $O$ be the center of the given sphere of radius $R, \angle B 1 B O=\angle C B O=\angle D B O=\alpha$ (Fig.1). Consider a regular pyramid, all plane angles at the vertex of which are right angles (Fig.2). Then the lateral edge of this pyramid forms an angle $\alpha$ with the height. Let the lateral edges of this pyramid be equal. Then the sides of the base are $\sqrt{2}$, and the projection of the lateral edge onto the plane of the base is $\frac{\sqrt{2}-\sqrt{3}}{3}$, therefore

$$
\sin \alpha=\frac{\frac{\sqrt{2}-\sqrt{3}}{3}}{1}=\frac{\sqrt{6}}{3}, \cos \alpha=\frac{\sqrt{3}}{3}, \operatorname{tg} \alpha=\sqrt{2}
$$

The given parallelepiped and its section, mentioned in the problem, are symmetric relative to the plane $B B 1 D 1 D$, so the sphere touches the cutting plane at point $P$, lying on the diagonal $D B 1$, and the planes of the face $A B C D$ - at point $Q$, lying on the diagonal $B D$ of the square $A B C D$. Consider the section of the parallelepiped by the plane $B B 1 D 1 D$. We get a rectangle $B B 1 D 1 D$ and a circle of radius $R$ with center at point $O$, touching the leg $B D$ of the right triangle $B D B 1$ at point $Q$, the hypotenuse $D B 1$ - at point $P$, and $\angle B O Q=\alpha$. Let $\angle B D B 1=2 \beta$. The center of the circle inscribed in an angle lies on the bisector of this angle, so $\angle B D O=\beta$, and since

$$
\operatorname{tg} 2 \beta=\frac{B H_{1}}{B D}=\frac{a}{2 a \sqrt{2}}=\frac{1}{2 \sqrt{2}}
$$

and $\operatorname{tg} 2 \beta=\frac{2 \operatorname{tg} \beta}{1-\operatorname{tg}^2 \beta}$, from the equation $\frac{1}{2 \sqrt{2}}=\frac{2 \operatorname{tg} \beta}{1-\operatorname{tg}^2 \beta}$ we find that $\operatorname{tg} \beta=3-2 \sqrt{2}$. Further, we find:

$$
D Q=\frac{O Q}{\operatorname{tg} \beta}=\frac{R}{3-2 \sqrt{2}}=R(3+2 \sqrt{2})
$$

$$
B Q=O Q \operatorname{tg} \angle B O Q=O Q \operatorname{tg} \angle O B B 1=R \operatorname{tg} \alpha=R \sqrt{2}
$$

and since $O Q=D Q \operatorname{tg} \beta$, we get the equation

$$
R=(2 a \sqrt{2}-R \sqrt{2})(3-2 \sqrt{2})
$$

from which we find that $R=\frac{4-2 \sqrt{2}}{3} a$.

## Answer

$\frac{4-2 \sqrt{2}}{3} a=\frac{2 a \sqrt{2}}{\sqrt{2}+\operatorname{sig} t\left(\frac{1}{2} \operatorname{arct} \lg \frac{\sqrt{2}}{1}\right)}$.