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

 untranslated text remains in its original form.
4. Find the radius of the sphere that touches all the edges of a regular tetrahedron with edge length \(a\).

ANSWER. \(r = \frac{a \sqrt{2}}{4}\)

SOLUTION.

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
\[ KM = \frac{CK}{3} = \frac{a \sqrt{3}}{6}, \quad AM = 2KM = \frac{a \sqrt{3}}{3}, \quad DM = \sqrt{AD^2 - AM^2} = \sqrt{a^2 - \frac{a^2}{3}} = a \sqrt{\frac{2}{3}}. \]

Consider the

composite

parts

of the

segment

DM:

\[ OM = \sqrt{OK^2 - KM^2} = \sqrt{r^2 - \frac{a^2}{12}} \]
\[ OD = \sqrt{ON^2 + DN^2} = \sqrt{r^2 + \frac{a^2}{4}}. \]

Since \(DM = OM + OD\), we arrive at the equation for \(r\):

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

Squaring both sides of the equation, we get \(r = \frac{a \sqrt{2}}{4}\).