We inscribe a sphere around a unit-edged cube, then we cut the sphere with the plane of each of the cube's faces. Let's calculate the area of the parts of the spherical surface divided by the 6 circles cut out.
The diameter of the circumscribed sphere of the cube is equal to the body diagonal, which is $\sqrt{3}$ units, so its radius is $r=\sqrt{3} / 2$, and its surface area is $F=4 \pi r^{2}=3 \pi$ square units.

![](https://cdn.mathpix.com/cropped/2024_05_02_05373bca4ce8eaeafdaeg-1.jpg?height=580&width=553&top_left_y=263&top_left_x=775)

A plane of one face of the cube intersects the surface of the sphere in a (non-great) circle and divides it into two spherical caps. The height $m$ of the smaller spherical cap is given by the difference between the radius and the distance from the center to the plane. This distance is equal to half the edge length of the cube, so the surface area of the smaller spherical cap is:

$$
F_{\mathrm{s}}=2 \pi r m=\pi \sqrt{3}\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)=\pi \frac{3-\sqrt{3}}{2}
$$

Considering the small circles cut by all 6 face planes, these planes are pairwise parallel, and the spherical caps cut by these planes do not overlap. The small circles cut by two adjacent face planes intersect at two vertices of the cube, which are the endpoints of the edge formed by the planes, thus dividing each circle into 4 quarter-circles. The smaller spherical cap is also divided into 5 parts by the 4 quarter-arcs. Among the surface parts, 4 have 2 vertices and 2 quarter-arcs as boundaries, and 1 has 4 vertices and 4 quarter-arcs. The latter type of part appears 6 times on the entire sphere, while the former type appears $6 \cdot 4 / 2$ times, because in the product $6 \cdot 4$, each such part is counted twice. No other type of surface part is formed. By symmetry, the 4-sided parts are congruent to each other, as are the 2-sided parts.

Thus, denoting the surface area of the 4-sided and 2-sided parts by $F_{4}$ and $F_{2}$ respectively, where the index refers to the number of sides of the part, we get the following system of equations:

$$
F_{4}+4 F_{2}=F_{\mathrm{s}}, \quad 6 F_{4}+12 F_{2}=F
$$

From this,

$$
\begin{aligned}
& F_{4}=\frac{F}{3}-F_{\mathrm{s}}=\frac{\pi}{2}(\sqrt{3}-1) \approx 1.15 \text { square units } \\
& F_{2}=\frac{F_{\mathrm{s}}}{2}-\frac{F}{12}=\frac{\pi}{4}(2-\sqrt{3}) \approx 0.21 \text { square units. }
\end{aligned}
$$

Rózsa Buda (Dombóvár, Gőgös I. g. III. o. t.)

Note. The four-sided spherical surface part under consideration is called a hanging dome in architecture, or a regular Czech vault.

It is not appropriate to call the surface parts spherical quadrilaterals or triangles, as is often read in papers, because these terms are used for parts of the sphere's surface bounded by 4 or 2 great circle arcs, respectively.