We create a "die" from a spherical object by cutting off six equal spherical caps in such a way that the circles formed where the spherical caps were removed each touch four of their neighbors. What percentage of the total surface area of the die is the combined area of the six circles?

How many percent of the total surface area of the die is the combined area of the six circles?
Solution. Let the radius of the sphere be denoted by $R$, its center by $O$, and the radius of the circular faces by $r$. Consider one of the circular faces, and let the two opposite points of tangency on this face be $A$ and $B$.
![](https://cdn.mathpix.com/cropped/2024_05_02_3ca64263a3ef05510432g-1.jpg?height=338&width=982&top_left_y=264&top_left_x=558)

The plane $OAB$, due to the symmetry of the sphere and the cube, cuts a square out of the cube, with the side length of the square being $2r$. The length of the diagonal $AC$ is $AC = 2r\sqrt{2}$, hence $AO = \frac{AC}{2} = r\sqrt{2} = R$. The total area of the six circular faces is:

$$
t = 6r^2\pi = 6\left(\frac{R\sqrt{2}}{2}\right)^2\pi = 3R^2\pi
$$

Now, let's calculate the total surface area $F$ of the cube. The surface area is obtained by subtracting the surface area of the six spherical caps from the surface area of the sphere $F_{\text{g}} = 4R^2\pi$ and adding the area of the six circular faces. The surface area of one spherical cap is $F_1 = 2\pi Rm$, where $R$ is the radius of the sphere and $m$ is the height of the spherical cap: $m = R - r$, where $r = \frac{R\sqrt{2}}{2}$,

$$
\begin{aligned}
& m = \left(R - \frac{R\sqrt{2}}{2}\right) = R\left(1 - \frac{\sqrt{2}}{2}\right), \quad \text{and} \\
& F_1 = 2\pi R^2\left(1 - \frac{\sqrt{2}}{2}\right) = R^2\pi(2 - \sqrt{2})
\end{aligned}
$$

Thus, the surface area of the cube is:

$$
\begin{aligned}
F & = 4R^2\pi - 6R^2\pi(2 - \sqrt{2}) + 3R^2\pi = \\
& = R^2\pi(4 - 12 + 6\sqrt{2} + 3) = R^2\pi(6\sqrt{2} - 5)
\end{aligned}
$$

Finally, let's calculate what percentage of the total surface area of the cube is covered by the six circular faces:

$$
\frac{3R^2\pi}{R^2\pi(6\sqrt{2} - 5)} = \frac{3(6\sqrt{2} + 5)}{47} \approx 86.08\%
$$