11.6. A sphere passes through all the vertices of one face of a cube and is tangent to all the edges of the opposite face of the cube. Find the ratio of the volume of the sphere to the volume of the cube.
## Solution.

Let the sphere pass through the vertices $A, B, C, D$. We introduce a coordinate system such that the origin of coordinates coincides with point $A$, the $O x$ axis is the line $(A D)$, the $O y$ axis is the line $(A B)$, and the $\mathrm{Oz}$ axis is the line $\left(A A_{1}\right)$.

Let the edge of the cube be equal to $a$. Then in coordinates

$$
A=(0,0,0), \quad B=(0, a, 0), \quad D=(a, 0,0)
$$

The equation of the sphere from the conditions of the problem with the center at point $O=\left(x_{0}, y_{0}, z_{0}\right)$ and radius $R$ has the form

$$
\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}+\left(z-z_{0}\right)^{2}=R^{2}
$$

Since the sphere is tangent to all edges of the face $A_{1} B_{1} C_{1} D_{1}$ of the cube, the intersection of the sphere with this face is a circle inscribed in the square $A_{1} B_{1} C_{1} D_{1}$. This circle is tangent to all sides of the square $A_{1} B_{1} C_{1} D_{1}$ at the midpoints of the sides of the square - at points $P, Q, T, S$. In coordinates we have

$$
P=(a / 2,0, a), \quad Q=(0, a / 2, a), \quad T=(a / 2, a, a), \quad S=(a, a / 2, a) .
$$

Points A, B, D, Q belong to the sphere. Therefore, substituting the coordinates of these into the equation of the sphere, we obtain the system

$$
\left\{\begin{array}{l}
x_{0}^{2}+y_{0}^{2}+z_{0}^{2}=R^{2} \\
x_{0}^{2}+\left(y_{0}-a\right)^{2}+z_{0}^{2}=R^{2} \\
\left(x_{0}-a\right)^{2}+y_{0}^{2}+z_{0}^{2}=R^{2} \\
x_{0}^{2}+\left(y_{0}-a / 2\right)^{2}+\left(z_{0}-a\right)^{2}=R^{2}
\end{array}\right.
$$

Then we solve the system with respect to the unknown $R$ and obtain

$$
R=a \cdot \frac{\sqrt{41}}{8}
$$

Since

$$
V_{\text {cube }}=a^{3}, \quad V_{\text {sphere }}=\frac{4}{3} \pi R^{3}
$$

then

$$
\frac{V_{\text {sphere }}}{V_{\text {cube }}}=\frac{\frac{4}{3} \pi\left(a \cdot \frac{\sqrt{41}}{8}\right)}{a^{3}}=\frac{41 \pi \sqrt{41}}{384}
$$