At the base of the right prism $A B C D A 1 B 1 C 1 D 1$ lies a rhombus $A B C D$ with angle $B A D$ equal to $2 \arccos \frac{1}{3}$. A sphere touches all segments of the broken line ABCC1A1 and intersects the edge BB1 at points B1 and M. Find the volume of the prism and the radius of the sphere, if $B 1 M=1$.

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Let a sphere of radius $R$ touch the segments $A B, B C, C C 1, C 1 A 1$ of the broken line $A B C C 1 A 1$ at points $P, Q, K$ and $L$ respectively. Points $P$ and $Q$ lie on the sphere, so the center $O$ of the sphere lies in the plane perpendicular to segment $P Q$ and passing through its midpoint, and since the prism is straight and its base is a rhombus, this plane coincides with the plane $B B 1 D 1 D$. At the same time, since the sphere passes through points $B 1$ and $M$, its center lies in the plane perpendicular to segment $B 1 M$ and passing through its midpoint $N$. These two planes intersect along the line $O N$, parallel to $B D$. Let $L^{\prime}$ be the center of the rhombus $A 1 B 1 C 1 D 1$. The line $A 1 C 1$ is perpendicular to the plane $B B 1 D 1 D$, so $O L^{\prime} \perp A 1 C 1$. At the same time, the radius of the sphere drawn to the point $L$ of tangency with the line $A 1 C 1$ is also perpendicular to $A 1 C 1$, so the points $L^{\prime}$ and $L$ coincide. Let the line passing through point $O$ parallel to the lateral edge of the prism intersect the diagonals $B 1 D 1$ and $B D$ of the bases at points $H$ and $G$ respectively. Since $O K \perp C C 1$ (as the radius of the sphere drawn to the point of tangency of the sphere with the segment $C C 1$), the quadrilateral $O H C 1 K$ is a rectangle, so

$$
C 1 K=O H=B 1 N=\frac{1}{2} B 1 M=\frac{1}{2}
$$

Then $C 1 L = C 1 K = \frac{\mathbf{1}}{\mathbf{2}}$ as segments of tangents drawn from the same point $C 1$ to the sphere. Let the height of the prism be $h$. Then

$$
C Q=C K=C C 1-C 1 K=h-\frac{1}{2}, B M=B B 1-B 1 M=h-1 .
$$

By the tangent-secant theorem

$$
B Q=\sqrt{B B_{1} \cdot B M}=\sqrt{h(h-1)}
$$

Then

$$
B 1 C 1=B C=B Q+Q C=\sqrt{h(h-1)}+h-\frac{1}{2}
$$

On the other hand, from the right triangle $C 1 L B 1$ we find that

$$
B 1 C 1=\frac{C_{1} L}{\operatorname{cose} \angle B_{1} C_{1} L}=\frac{C_{1} L}{\operatorname{cose} \frac{1}{2} \angle B_{1} C_{1} D_{1}}=\frac{\frac{1}{2}}{\frac{1}{3}}=\frac{3}{2}
$$

Thus, we obtain the equation

![](https://cdn.mathpix.com/cropped/2024_05_06_6ba7643fcbc3b9c222e2g-52.jpg?height=89&width=377&top_left_y=2345&top_left_x=841)

from which we find that $h=\frac{4}{3}$. Then

$$
B P=B Q=\sqrt{h(h-1)}=\sqrt{\frac{4}{3} \cdot \frac{1}{3}}=\frac{2}{3}, C K=h-\frac{1}{2}=\frac{4}{3}-\frac{1}{2}=\frac{5}{6}
$$

Let $\angle B A D=\alpha$. By the condition $\cos \frac{\frac{\mathbf{2}}{2}}{2}=\frac{1}{3}$, so

$$
\sin \frac{\frac{1}{2}}{2} \sqrt{1-\frac{1}{9}}=\frac{2 \sqrt{2}}{3}, \sin \alpha=2 \sin \frac{\frac{2}{2}}{2} \cos \frac{\frac{2}{2}}{2}=2 \cdot \frac{2 \sqrt{2}}{3} \cdot \frac{1}{3}=\frac{4 \sqrt{2}}{9}
$$

Therefore,

$$
V_{A B C D A} 1 B 1 C 1 D 1=S_{A B C D} \cdot h=A B \cdot A D \sin \alpha \cdot h=\left(\frac{3}{2}\right)^2 \cdot \frac{4 \sqrt{2}}{9} \cdot \frac{4}{3}=\frac{4}{3} \sqrt{2} .
$$

Finally, from the right triangles $B P G$ and $O P G$ we find that

$$
\begin{gathered}
P G=B P \operatorname{ctg} \angle P G B=B Q \operatorname{ctg} \frac{\frac{2}{2}}{2} \frac{2}{3} \cdot \frac{1}{2 \sqrt{2}}=\frac{1}{3 \sqrt{2}} \\
R 2=O P 2=\sqrt{P G^{2}+O G^{2}}=\sqrt{P G^{2}+C K^{2}}=\sqrt{\left(\frac{1}{3 \sqrt{2}}\right)^{2}+\left(\frac{5}{6}\right)^{2}}=\frac{\sqrt{3}}{2}
\end{gathered}
$$

## Answer

$V=\frac{4}{3} \sqrt{2}, R=\frac{\sqrt{3}}{2}$.