3. In a tetrahedron with edge $a$, a sphere is inscribed and another sphere is circumscribed around it. Find the corresponding radii and the volume of the pyramid with vertices at the points where the inscribed sphere touches the tetrahedron.
Solution. The centers of the inscribed and circumscribed spheres within and around the regular tetrahedron $ABCS$ (see diagram) coincide and are located at the intersection of the heights of the tetrahedron. Let this point be $O$.

It is clear that:

$$
\overline{SO} + \overline{OK} = \frac{a \sqrt{6}}{3}
$$

(height of a regular tetrahedron with edge $a$),

$$
\overline{OK}^2 = \overline{OA}^2 - \overline{AK}^2
$$

Since $\overline{AK} = \frac{a \sqrt{3}}{3}$, from equation (2) we get

$$
\overline{OK}^2 = \overline{SO}^2 - \frac{a^2}{3}
$$

If we use equation (1), we find that the radius of the inscribed sphere is
$\overline{OK} = \frac{a \sqrt{6}}{12}$, and the radius of the circumscribed sphere is $\overline{OS} = \frac{a \sqrt{6}}{4}$.

The height of the pyramid, with vertices at the points of tangency of the inscribed sphere with the tetrahedron with edge $a$ and that tetrahedron, is $\frac{a}{3}$, and therefore the volume of the pyramid is

$$
V = \frac{1}{3} \cdot \frac{1}{4} \cdot \left(\frac{a}{3}\right)^2 \cdot \sqrt{3} \cdot \frac{1}{3} \frac{a}{3} \sqrt{6} = \frac{a^2 \sqrt{3}}{324}
$$