7. On the edge $B C$ of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$, a point $M$ is chosen. A sphere constructed on the segment $C_{1} M$ as its diameter touches the planes of four faces of the parallelepiped, and one of them at a point lying on the edge $B_{1} B$. It is known that $B M=1, C M=3$. Find the length of the edge $A A_{1}$, the radius of the sphere, and the volume of the parallelepiped.
Answer: $A A_{1}=5, R=2, V=32$.

Solution. Since the center of the sphere is the midpoint of the segment $C_{1} M$ and lies in the face $B C C_{1} B_{1}$, the sphere does not touch this face. Note that if the segment $C_{1} M$ is not perpendicular to the planes $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$, then the sphere does not touch them, which is impossible since the sphere is supposed to touch the planes of four faces of the parallelepiped. Furthermore, the sphere cannot touch the face $C C_{1} D_{1} D$ (touching would imply that $M C_{1} \perp C C_{1} D_{1} D$, and from this it would follow that the planes $C C_{1} D_{1} D$ and $A_{1} B_{1} C_{1} D_{1}$ coincide). Therefore, the sphere touches the faces $A B C D, A_{1} B_{1} C_{1} D_{1}, A A_{1} D_{1} D$, and $A B B_{1} A_{1}$, and the latter at a point lying on the edge $B B_{1}$ (denote it as $T$).

Consider the parallelogram $B B_{1} C_{1} C$. Opposite sides of the parallelogram are equal, so $B_{1} C_{1}=B C=B M+M C=1+3=4$. Using the equality of tangents drawn from a point to a circle, we find that $B_{1} T=B_{1} C_{1}=4, B T=B M=1$. Therefore, $C C_{1}=B B_{1}=B T+T B_{1}=5$. By the Pythagorean theorem for triangle $C C_{1} M$, we find that $C_{1} M=\sqrt{C C_{1}^{2}-C M^{2}}=4$, from which the radius of the sphere is 2.

The volume of the parallelepiped $V$ is equal to the product of the area of its base and its height. We choose $B C C_{1} B_{1}$ as the base; then the height is equal to the radius of the sphere (since the center of the sphere lies in the face $B C C_{1} B_{1}$ and it touches the face $A A_{1} D_{1} D$). The area of the base is $C_{1} M \cdot B C=4 \cdot 4=16$. Therefore, $V=2 \cdot 16=32$.