Let the parallelepiped $A B C D A 1 B 1 C 1 D 1$ be circumscribed around a cylinder (Fig.1). Then the parallelepiped is a right one, the circle of one base of the cylinder is inscribed in the base $A B C D$ of the parallelepiped, and the circle of the second base is inscribed in the base $A 1 B 1 C 1 D 1$. Since circles are inscribed in the parallelograms $A B C D$ and $A 1 B 1 C 1 D 1$, these parallelograms are rhombuses. If $A C$ and $A 1 C 1$ are the major diagonals of these rhombuses, then $A C 1$ and $A 1 C$ are the major diagonals of the parallelepiped. Since the heights of the cylinder and the parallelepiped are equal, the areas of the bases of the parallelepiped and the cylinder are in the same ratio as their volumes, and since the radius of the circle inscribed in the rhombus is $r$, the area of the rhombus is equal to its semiperimeter multiplied by $r$. If $a$ is the side of the rhombus, then

$$
\frac{2 \pi y}{\pi r^{2}}=\frac{2 \alpha}{\pi r}=\frac{\dot{5}}{\pi}
$$

from which we find that $a=\frac{5}{2}$. Let the circle with center $O$, inscribed in the rhombus $A B C D$, touch the side $A B$ at point $F$, and $K$ be the midpoint of $A B$ (Fig.2). Denote $\angle O K B=\alpha$. Then

$$
\begin{aligned}
& \sin \alpha=\frac{\partial F}{\partial K}=\frac{\stackrel{r}{4}}{4}=\frac{1}{b}, \cos \alpha=\frac{3}{b}, \\
& \sin \angle O A B=\sin \frac{\alpha}{2}=\sqrt{\frac{1-\operatorname{mb} \beta}{2}}=\frac{1}{\sqrt{5}},
\end{aligned}
$$

therefore

$$
\begin{gathered}
O A=\frac{O F}{d n \angle T A B}=r \sqrt{5}, A C=2 O A=2 \sqrt{5} \\
\operatorname{tg} \angle C A C 1=\frac{\sigma C_{1}}{A C}=\frac{5+}{2 r \sqrt{A}}=\frac{\sqrt{5}}{2}, \cos \angle C A C 1=\frac{2}{3}
\end{gathered}
$$

Consider the plane $A C C 1 A 1$ (Fig.3). The centers $O$ and $O 1$ of the circles inscribed in the rhombuses $A B C D$ and $A 1 B 1 C 1 D 1$ are the midpoints of the sides $A C$ and $A 1 C 1$ of the rectangle $A C C 1 A 1$. Let $M$ and $N$ be the points of intersection of the first circle with the segment $A C$, and $M 1$ and $N 1$ be the points of intersection of the second circle with the segment $A 1 C 1$, with $M M 1 \| N N 1$. Let the diagonal $A C 1$ intersect the segments $M M 1$ and $N N 1$ at points $P$ and $Q$ respectively. Drop the perpendicular $P H$ from point $P$ to $N N 1$. Then

$$
P Q=\frac{P H}{m \Delta M P Q}=\frac{M A N}{m \Delta B A C_{1}}=\frac{2 r}{\sqrt{n}}=3 r .
$$

## Answer

$3 r$.
10,11

The radius of the base of the cylinder is equal to $r$, and the height is equal to $5 r$. A parallelepiped is circumscribed around the cylinder, the ratio of the volume of which to the volume of the cylinder is $\frac{\tilde{5}}{\pi}$. Find the length of the segment of the larger diagonal of the parallelepiped lying inside the cylinder.