7. Edge $A_{1} A$ of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ is perpendicular to its face $A B C D$. Sphere $\Omega$ touches the edges $B B_{1}, B_{1} C_{1}, C_{1} C, C B, C_{1} D_{1}$, and touches the edge $C_{1} D_{1}$ at a point $K$ such that $C_{1} K=16, K D_{1}=1$.

a) Find the length of edge $A_{1} A$.

b) Suppose it is additionally known that sphere $\Omega$ touches the edge $A D$. Find the volume of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ and the radius of sphere $\Omega$.
Answer: $A_{1} A=32, V=8192, R=4 \sqrt{17}$.

Solution. a) From the condition, it follows that $B B_{1} \perp A B C D$, therefore $B B_{1} \perp B C$ and the face $B C C_{1} B_{1}$ is a rectangle. A circle can be inscribed in this rectangle (the section of the sphere by the plane $B C C_{1} B_{1}$), which means this face is a square. Let $F$ be the point of tangency of the given sphere with the edge $B_{1} C_{1}$. The segments $C_{1} K$ and $C_{1} F$ are equal as tangents to the sphere drawn from the same point. Moreover, a circle inscribed in a square touches its sides at their midpoints. Therefore, $F$ is the midpoint of $B_{1} C_{1}$ and $B_{1} C_{1}=2 \cdot C_{1} F=2 \cdot C_{1} K=32$. The lateral edges of the parallelepiped are equal to each other, and $B C C_{1} B_{1}$ is a square, so $A A_{1}=C C_{1}=B_{1} C_{1}=32$.

b) The center of the sphere is located on a line perpendicular to the plane of the square $B C C_{1} B_{1}$ and passing through its center, so it is equidistant from the lines $A D$ and $A_{1} D_{1}$. Therefore, if the sphere touches one of these edges, it touches the other as well. Let $M$ be the point of tangency of the sphere with the edge $A_{1} D_{1}$, and the circle obtained by the section of the sphere by the plane $A_{1} B_{1} C_{1} D_{1}$ be denoted as $\omega$. Let $O$ be the center of $\omega$.

The center of a circle inscribed in an angle lies on the bisector of this angle, so $\angle O C_{1} D_{1} + \angle O D_{1} C_{1} = \frac{1}{2} (\angle F C_{1} D_{1} + \angle M D_{1} C_{1}) = \frac{1}{2} \cdot 180^{\circ} = 90^{\circ}$. Thus, the triangle $O C_{1} D_{1}$ is a right triangle, $O K = \sqrt{C_{1} K \cdot D_{1} K} = 4$.

The area of the parallelogram $A_{1} B_{1} C_{1} D_{1}$ is $B_{1} C_{1} \cdot F M = 2 \cdot C_{1} F \cdot 2 \cdot O K = 2 \cdot 16 \cdot 8 = 256$. The volume of the parallelepiped is $A A_{1} \cdot S_{A_{1} B_{1} C_{1} D_{1}} = 32 \cdot 256 = 8192$.

Let $Q$ be the center of the sphere. Then $Q K$ is its radius; and the triangle $O K Q$ is a right triangle. Therefore, $Q K = \sqrt{O K^{2} + O Q^{2}} = \sqrt{O K^{2} + \frac{1}{4} A A_{1}^{2}} = 4 \sqrt{17}$.