We divided the radius of a sphere ($r=10 \mathrm{dm}$) into two parts such that the part falling towards the center is the mean proportional between the whole radius and the other part. We placed a plane perpendicular to the radius through the dividing point, and in the larger of the spherical segments thus formed, we constructed a regular quadrilateral pyramid whose apex lies on the surface of the sphere. Find the volume, surface area, and slant angle of the pyramid.
The part of the ray falling towards the center is

$$
x=\frac{r}{2}(\sqrt{5}-1)
$$

Let $\rho$ denote the radius of the base circle of the spherical cap; then, by the Pythagorean theorem,

$$
\rho=\frac{r}{\sqrt{2}} \sqrt{\sqrt{5}-1}
$$

The base of our pyramid is a square inscribed in a circle of radius $\rho$; its side is

$$
a=\rho \sqrt{2}=r \sqrt{\sqrt{5}-1}
$$

The height of the pyramid's side face is

$$
m=\sqrt{(r+x)^{2}+\left(\frac{a}{2}\right)^{2}}=\frac{r}{2} \sqrt{3 \sqrt{5}+5}
$$

Thus, the surface area of the pyramid is

$$
F=a^{2}+4 a \cdot \frac{m}{2}=r^{2}(\sqrt{2 \sqrt{5}+10}+\sqrt{5}-1)
$$

Its volume is

$$
V=\frac{a^{2}}{3}(r+x)=\frac{2}{3} r^{3}
$$

The tangent of the slant angle is:

$$
\operatorname{tg} \alpha=\frac{r+x}{\frac{a}{2}}=\frac{\sqrt{5}+1}{\sqrt{\sqrt{5}-1}}=\frac{1}{2}(\sqrt{\sqrt{5}+1})^{3}
$$

(Henrik Kolos.)

Number of solutions: 24.