4.21. On a sphere of radius 2, three pairwise tangent circles of radius 1 are arranged. Find the radius of the smallest circle located on the given sphere and tangent to all three given circles.
4.21. Let $O$ be the center of the sphere; $O_{1}, O_{2}$ and $O_{3}$ be the centers of the given circles; $O_{4}$ be the center of the desired circle. Considering the section of the sphere by the plane $O O_{1} O_{2}$, it is easy to prove that $O O_{1} O_{2}$ is an equilateral triangle with side $\sqrt{3}$. The line $O O_{4}$ passes through the center of the triangle $O_{1} O_{2} O_{3}$ perpendicular to its plane, and therefore, the distances from the vertices of this triangle to the line $O O_{4}$ are equal to 1. Let $K$ be the point of tangency of the circles with centers $O_{1}$ and $O_{4}$; $L$ be the foot of the perpendicular dropped from $O_{1}$ to $O O_{4}$; $N$ be the foot of the perpendicular dropped from $R$ to $O_{1} L$. Since $\triangle O_{1} K N$ is similar to $\triangle O O_{1} L$, then $O_{1} N = O L \cdot O_{1} K / O O_{1} = \sqrt{2 / 3}$, and therefore, the desired radius $O_{4} K$ is equal to $L N = 1 - \sqrt{2 / 3}$.