A compact cube, to make it more suitable as a dice for games of chance, had its corners rounded with the sphere whose center is the center of the cube and which touches all the edges of the cube. What is the surface area and volume of the remaining solid?
Let the edge length of the cube be denoted by $e$. The diameter of the sphere in question is equal to the distance $e \sqrt{2}$ between two opposite edges of the cube. Indeed, the diagonal section passing through these two edges forms a rectangle whose sides are the edge and the diagonal of the face.

![](https://cdn.mathpix.com/cropped/2024_05_02_4bfe3e2f731c438105a0g-1.jpg?height=493&width=461&top_left_y=239&top_left_x=824)

Thus, the radius of the sphere is $r = e \sqrt{2} / 2$. Consequently, part of the sphere's surface is outside the cube, and part is inside, because the distance $d = e / 2$ from the center to the faces is smaller than $r$, while the distance from the center to the vertices is $e \sqrt{3} / 2$, which is larger than $r$. The sphere intersects the faces in circles that touch the edges bounding the faces, so their radius is $\varrho = e / 2$. Therefore, the remaining solid is bounded by 6 circular disks of radius $\varrho$ and 8 triangular segments of the sphere's surface.

It is more convenient to say that the faces of the cube cut 6 congruent spherical caps and 6 congruent spherical segments from the sphere. The height of the cap and the segment is $m = r - d = e(\sqrt{2} - 1) / 2$, so the total surface area and volume of the removed parts are:

$$
\begin{aligned}
f & = 6 \cdot 2 \pi r m = 3(2 - \sqrt{2}) \pi e^{2}, \\
v & = 6 \frac{\pi m}{6} \left(3 \varrho^{2} + m^{2}\right) = \frac{\pi e(\sqrt{2} - 1)}{2} \left[\frac{3 e^{2}}{4} + \frac{e^{2}(3 - 2 \sqrt{2})}{4}\right] = \\
& = \frac{\pi e^{3}}{4}(\sqrt{2} - 1)(3 - \sqrt{2}) = \frac{\pi e^{3}}{4}(4 \sqrt{2} - 5)
\end{aligned}
$$

With these, the remaining surface area and volume of the sphere are:

$$
\begin{gathered}
F_{1} = 4 \pi r^{2} - f = 2 \pi e^{2} - 3(2 - \sqrt{2}) \pi e^{2} = (3 \sqrt{2} - 4) \pi e^{2} \\
V = \frac{4 \pi}{3} r^{3} - v = \frac{4 \pi}{3} \frac{e^{3} \sqrt{2}}{4} - \frac{\pi e^{3}}{4}(4 \sqrt{2} - 5) = \frac{\pi e^{3}}{12}(15 - 8 \sqrt{2})
\end{gathered}
$$

This gives us the volume of the die, and its surface area is given by the sum of $F_{1}$ and the areas of the 6 circular disks:

$$
F = 6 \pi \varrho^{2} + F_{1} = \frac{3 \pi}{2} e^{2} + (3 \sqrt{2} - 4) \pi e^{2} = \frac{\pi}{2}(6 \sqrt{2} - 5) e^{2}
$$

Approximating the constants to four significant decimal places, we get $V \approx 0.9651 e^{3}$ and $F \approx 5.475 e^{2}$, which are 96.5% and 91.2% of the cube's volume and surface area, respectively.