On a sphere with a radius of 2, there are three circles with a radius of 1, each touching the other two. Find the radius of a smaller circle, also located on the given sphere, that touches each of the given circles.

#
Let $O$ be the center of the sphere (Fig.1); $O_1, O_2, O_3$ be the centers of three given circles located on the sphere; $Q$ be the center of the desired circle with radius $x$; $A$ be the point of tangency of the circles with centers $O_1$ and $O_2$. Consider the section of the sphere by a plane passing through point $A$ perpendicular to the common tangent $l$ to the circles with centers $O_1$ and $O_2$, passing through point $A$ (Fig.2). Points $O, O_1$, and $O_2$ lie in this plane. Thus, we have a circle of radius 2 with center $O$ and two chords $AB$ and $AC$, the midpoints of which are points $O_1$ and $O_2$ respectively, and since $AB$ and $AC$ are diameters of the given circles with centers $O_1$ and $O_2$, then $AB = AC = 2$, so $AOB$ and $AOC$ are equilateral triangles with side 2, hence,

$$
O_1O_2 = \frac{1}{2} BC = \sqrt{3}, \quad OO_1 = OO_2 = \sqrt{3}
$$

Similarly,

$$
O_1O_3 = O_2O_3 = \sqrt{3}, \quad OO_3 = \sqrt{3}
$$

Consider the regular tetrahedron $OO_1O_2O_3$. Let $\alpha$ be the angle that its height forms with a lateral edge. Then

$$
\sin \alpha = \frac{1}{\sqrt{3}}, \quad \cos \alpha = \sqrt{\frac{2}{3}}
$$

Note that the extension of the height of the tetrahedron, drawn from vertex $O$, passes through point $Q$. Consider the section of the sphere by a plane passing through point $D$ perpendicular to the common tangent $m$ to the circles with centers $O_1$ and $Q$, passing through their point of tangency $D$ (Fig.3). Points $O, O_1$, and $Q$ lie in this plane. Thus, we have a circle of radius 2 with center $O$ and two chords $DE$ and $DF$, the midpoints of which are points $O_1$ and $Q$ respectively, and since $DE$ and $DF$ are diameters of the circles with centers $O_1$ and $Q$, then $O_1D = 1$ and $QD = x$. Moreover,

$$
\angle DOO_1 = 30^\circ, \quad \angle O_1OQ = \alpha, \quad \angle DOQ = \angle O_1OQ - \angle DOO_1 = \alpha - 30^\circ
$$

Therefore,

$$
\begin{gathered}
x = QD = OD \sin \angle DOQ = 2 \sin (\alpha - 30^\circ) = 2 \left( \sin \alpha \cos 30^\circ - \cos \alpha \sin 30^\circ \right) = \\
= 2 \left( \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} - \sqrt{\frac{2}{3}} \cdot \frac{1}{2} \right) = 2 \left( \frac{1}{2} - \frac{\sqrt{\frac{2}{3}}}{2} \right) = 1 - \sqrt{\frac{2}{3}}
\end{gathered}
$$

## Answer

$1 - \sqrt{\frac{2}{3}}$