\section*{Problem \(5-341045=340945\)}

A regular tetrahedron \(A B C D\) has an inscribed sphere \(K\) (this is the sphere that touches all four triangular faces \(A B C, A B D, A C D, B C D\)).

A second regular tetrahedron \(P Q R S\) is inscribed in this sphere (i.e., its vertices \(P, Q, R, S\) all lie on the surface of the sphere \(K\)).

What is the ratio \(V_{2}: V_{1}\) of the volume \(V_{2}\) of such a tetrahedron \(P Q R S\) to the volume \(V_{1}\) of \(A B C D\)?
}

The incenter and circumcenter of a regular tetrahedron coincide with the intersection of its medians. Since these medians intersect in the ratio \(3: 1\) and are perpendicular to the respective faces, the circumradius of a tetrahedron is exactly three times its inradius.

Since the inradius \(r_{1}\) of \(A B C D\) is exactly the circumradius of \(P Q R S\), the inradius \(r_{2}\) of the latter is therefore exactly \(\frac{1}{3} r_{1}\).

By the intercept theorem, it quickly follows that the edge lengths and inradii of regular tetrahedra are in the same ratio, while their volumes are in the cube of this ratio. Therefore,

\[
V_{2}: V_{1}=\left(r_{2}: r_{1}\right)^{3}=(1: 3)^{3}=1^{3}: 3^{3}=1: 27
\]