79. 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 the circle, smaller than the given ones, which is also located on the given sphere and touches each of the given circles.
79. Notations: $O$ - center of the sphere, $O_{1}, O_{2}, O_{3}$ - centers of the given circles, $O_{4}$ - center of the required circle. Clearly, $\triangle O_{1} O_{2} O_{3}$ is equilateral. Let's find its sides ( $M$ - the point of tangency of the circles with centers $O_{1}$ and $\left.O_{2}\right) \cdot\left|O_{1} M\right|=\left|O_{2} M\right|=$

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$=\sqrt{3},\left|O_{1} O_{2}\right|=\sqrt{3} . O O_{4}$ is perpendicular to the plane $O_{1} O_{2} O_{3}$ and passes through the center of $\triangle O_{1} O_{2} O_{3}$, the distances from $O_{1}, O_{2}$ and $O_{3}$ to $O O_{4}$ are 1. Let $K$ be the point of tangency of the circles $O_{1}$ and $O_{4}, L$ be the foot of the perpendicular dropped from $O_{1}$ to $O O_{4} . K N \perp L O_{1}$, $\left|O_{1} L\right|=\left|O_{1} K\right|=1,\left|0 O_{1}\right|=\sqrt{3}$. From the similarity of the right triangles $O_{1} K N$ and $O O_{1} L$ we find $\left|O_{1} N\right|=\sqrt{\frac{2}{3}}$. Thus, the required radius $\left|O_{4} K\right|=|L N|=1-\sqrt{\frac{2}{3}}$.