$\mathrm{Az}$ asztalon fekvő (az asztalt érintő) négy golyó mindegyike érinti a másik hármat. Hármuknak ugyanakkora $R$ sugara van. Mekkora a negyedik golyó sugara.

Four spheres lying on a table (touching the table) each touch the other three. Three of them have the same radius $R$. What is the radius of the fourth sphere?
The centers of three spheres with radius $R$ form an equilateral triangle, whose side is $2 R$ and lies in a plane parallel to the table surface, at a distance of $R$ from it. The center of a fourth sphere with radius $x$ forms a regular tetrahedron with the centers of the previous three spheres, whose edge length is $R+x$, and the radius of the circumscribed circle of the base (as the $\frac{2}{3}$ of the altitude of the $2 R$ side equilateral triangle) is $\frac{2 R \sqrt{3}}{3}$. The height of the tetrahedron is $R-x$. In a regular tetrahedron, the radius of the circumscribed circle of the base, the height of the tetrahedron, and the edge length form a right triangle, with the edge length as the hypotenuse. Therefore, according to the Pythagorean theorem,

$$
(R+x)^{2}=\left(\frac{2 R \sqrt{3}}{3}\right)^{2}+(R-x)^{2}
$$

which gives

$$
4 R x=\frac{4 R^{2}}{3}, \quad \text { i.e. } \quad x=\frac{R}{3}
$$