A spherical bowl with radius $R$ is filled with 3 equal-sized spheres of radius $r$. How high above the table surface are these spheres?
The center of the large sphere and the centers of the three small spheres define a pyramid whose base is parallel to the table surface, each base edge being $2 r$, and each side edge being $R-r$. The height of this pyramid is:

$$
m=\sqrt{(R-r)^{2}-\rho^{2}}
$$

where $\rho$ is the radius of the circle circumscribed around the pyramid's base; thus $\rho=\frac{2 r}{\sqrt{3}} \mathrm{~s}$, so

$$
m=\sqrt{R^{2}-2 R r+r^{2}-\frac{4 r^{2}}{3}}=\sqrt{\frac{3 R^{2}-6 R r-r^{2}}{3}}
$$

Therefore, the distance of the pyramid's base - the centers of the small spheres - from the table surface is

$$
R-\sqrt{\frac{3 R^{2}-6 R r-r^{2}}{3}}
$$

(Weisz Pál, Nyitra.)

The problem was also solved by: Bartók I., Beck P., Kertész G., Klein A., Losonczy I., Mixich P., Schmidl I.