[ Regular Tetrahedron ]

Find the radius of the sphere that touches all the edges of a regular tetrahedron with edge $a$.

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## First method.

Extend the given regular tetrahedron to a parallelepiped by drawing three pairs of parallel planes through opposite edges (Fig.1). Since the diagonals of each face of the resulting parallelepiped are equal and perpendicular to each other, this parallelepiped is a cube. Its edge is $\frac{a \sqrt{2}}{2}$. The sphere inscribed in this cube touches all the edges of the original regular tetrahedron. Its radius is half the edge of the cube, i.e., $\frac{a \sqrt{2}}{4}$.

## Second method.

Let $M$ be the center of the face $ABC$ of the regular tetrahedron $ABCD$, and $O$ be the center of the given sphere. Since point $O$ is equidistant from the sides of triangle $ABC$, its projection onto the plane of face $ABC$ coincides with point $M$, i.e., the center of the sphere lies on the height of the tetrahedron. Let $r$ be the desired radius, $K$ be the midpoint of $AB$, and $N$ be the midpoint of $AD$. Then

\[
\begin{aligned}
& K M=\frac{1}{3} C K=\frac{a \sqrt{3}}{6}, A M=\frac{a \sqrt{3}}{3}, D M=\sqrt{A D^{2}-A M^{2}}=\sqrt{a^{2}-\frac{a^{2}}{3}}=a \sqrt{\frac{2}{3}} \\
& O M=\sqrt{O K^{2}-K M^{2}}=\sqrt{r^{2}-\frac{a^{2}}{12}}, O D=\sqrt{O N^{2}+D N^{2}}=\sqrt{r^{2}+\frac{a^{2}}{4}}
\end{aligned}
\]

Since $O M + O D = D M$, we have the equation

\[
\sqrt{r^{2}-\frac{a^{2}}{12}}+\sqrt{r^{2}+\frac{a^{2}}{4}}=a \sqrt{\frac{2}{3}}
\]

## Problem