A cube with edge $a$ has a sphere inscribed around each vertex with a radius of $\frac{a}{4} \sqrt{5}$, and we remove every part of the cube that belongs to at least one sphere. What is the volume of the remaining solid?
The first one takes a spherical segment around a corner of the cube, which is as large as one-eighth of the volume of the sphere, partly because each plane of the cube passing through the corner bisects the sphere, and the three planes together divide it into $2^{3}=8$ congruent parts, and partly because the radius of the sphere is smaller than the edge of the cube.

![](https://cdn.mathpix.com/cropped/2024_05_02_dff71ecb246e81b96cffg-1.jpg?height=602&width=585&top_left_y=320&top_left_x=759)

Figure 1

Taking one of the three adjacent vertices of the previously mentioned sphere as the center of our second sphere, this sphere already intrudes into the first one, because the sum of their radii, $2r$, is greater than the edge of the cube, since $\sqrt{5} / 2 > 1$. The common part of these two spheres is a lens-shaped body, and one-fourth of this belongs to the cube, because the two planes forming the edge divide the space into four congruent angular domains.

The spheres around two more distant vertices of the cube (the endpoints of a diagonal in a face or in space) do not overlap, because $2r$ is already smaller than even the $\sqrt{2}a$ diagonal of the face.

Accordingly, if we subtract the volume of the eighth sphere, multiplied by 8, from the volume of the cube, then we have subtracted twice the volume of the 12 lens quarters located along the edges, so we need to add it back once. Their total volume is equal to three whole lenses, or six half-lenses, noting that by half-lens here we mean the spherical segment that the plane perpendicular to the line segment connecting the centers of the spheres and bisecting it cuts from one of the spheres.

The height of the segment

$$
m=(r-a / 2)=a(\sqrt{5}-2) / 4,
$$

so its volume, according to the known formula 1, is

$$
V_{sz}=\frac{\pi}{3} m^{2}(3 r-m)=\frac{\pi a^{3}}{96}(5 \sqrt{5}-11)
$$

thus the volume of the remaining body is

$$
V_{m}=a^{3}-\frac{4 \pi}{3} r^{3}+6 V_{sz}=a^{3}\left(1-\frac{\pi}{48}(33-10 \sqrt{5})\right)=a^{3} \cdot 0.304
$$

Remark. Let us outline the thought process of another solution. Cut the cube and the remaining body in half with planes perpendicular to the edges of the cube and bisecting them. The 3 such symmetry planes divide the cube and the remaining body into $2^{3}=8$ parts, each small cube contains one vertex of the original cube, the parts of the remaining body are congruent, and the surface of each contains a part of the surface of only one sphere. Move the two parts of the cube that are intersected by each of the cube's body diagonals to each other's place. Thus, the 8 vertices of the original cube end up at the original center of the cube, and the parts of the spherical surface that bound the remaining body form parts of a single spherical surface (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_dff71ecb246e81b96cffg-2.jpg?height=605&width=569&top_left_y=131&top_left_x=759)

Figure 2

On the other hand, the vertices of the 8 parts that form at the original center of the cube spread out to the places of the original cube's vertices. We see a cube, the interior of which is missing, and circular holes of radius $a / 4$ have formed on its faces. Six spherical segments fall beyond the faces of the new cube (the above half-lenses), in which the "spherical excavator" does not find material, so the volume of the removed material is $V_{\text{sphere}}-6 V_{\text{segment}}$, and that of the remaining body is $V_{\text{cube}}-V_{\text{sphere}}+6 V_{\text{segment}}$.

[^0]: ${ }^{1} \mathrm{The}$ 344.2 formula in the school function tables.