3. Four balls with equal radii $r$ are given, and they touch each other pairwise. A fifth ball is described around them. Find the radius of the fifth ball.
Solution. The centers of the four balls are vertices of a regular tetrahedron with edge $2 r$. The center of the fifth ball is at the intersection of the heights of this tetrahedron, and since the distance from the intersection of the heights of a regular tetrahedron with edge $a$ to any of its vertices is $\frac{a \sqrt{6}}{4}$, the radius of the fifth ball will be

$$
\frac{2 r \sqrt{6}}{2}+r=\frac{r}{2}(2+\sqrt{6})
$$