## Task 10/75

Given is a regular pyramid with a square base. Each side face forms an angle $\alpha$ with the base. Two spheres are inscribed in this pyramid such that one sphere touches all five faces of the pyramid, while the other touches the four side faces of the pyramid and the surface of the first sphere.

In what ratio do the volumes of the two spheres stand to each other?
If the radius of the larger sphere is denoted by $\mathrm{R}$ and the radius of the smaller sphere by $r$, then the desired ratio is

$$
V_{2}: V_{1}=\frac{4}{3} r^{3} \pi: \frac{4}{3} R^{3} \pi=r^{3}: R^{3}
$$

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-261.jpg?height=668&width=511&top_left_y=237&top_left_x=287)

In the figure (cross-sectional view), the following relationships are valid:

$$
\frac{R}{h-R}=\frac{r}{h-2 R-r}=\cos \alpha
$$

(it is $\triangle S M_{1} B \sim \triangle \quad S M_{2} C \sim \triangle \quad S A Z$ according to the main similarity theorem; each of these triangles contains the angle $Z S A$ and a right angle; this also results in the equality of the third angle $\alpha$. From these relationships, one obtains after a short calculation

$$
R=h \frac{\cos \alpha}{1+\cos \alpha} \quad ; \quad r=h \frac{\cos \alpha(1-\cos \alpha)}{(1+\cos \alpha)^{2}}
$$

Thus, the desired ratio is

$$
\left(\frac{r}{R}\right)^{3}=\left(\frac{1-\cos \alpha}{1+\cos \alpha}\right)^{3}=\tan ^{6} \frac{\alpha}{2}
$$