G20 (7-3, Czechoslovakia) A tetrahedron $ABCD$ is divided into two parts by a plane parallel to $AB$ and $CD$. It is known that the ratio of the distance from $AB$ to this plane and the distance from $CD$ to the plane is equal to $k$. Find the ratio of the volumes of the two parts into which the tetrahedron is divided.
Let the plane $\alpha$ parallel to $AB, CD$ intersect the tetrahedron at a parallelogram $PQRS$, with $PQ \parallel SR \parallel CD$ and $QR \parallel PS \parallel AB$, as shown in Figure 4-20.

The lengths of the two adjacent sides of the median section of the frustum $ABRS PQ$ parallel to the face $PQRS$ are $\left(1+\frac{k}{2}\right) RQ$ and $\frac{1}{2} PQ$, which can be easily derived using the properties of parallel lines. Therefore, its area $M_{1} = \frac{1}{2} \left(1+\frac{k}{2}\right) S_{\square PQRS}$.

Similarly, the area of the median section of the frustum $PQRS CD$ parallel to the face $PQRS$ is $\frac{1}{2} \frac{2k+1}{2k} S_{\square PQRS}$.

Using the volume formula for the frustum and the given conditions, it is easy to find that the volume ratio of the two solids is
$$
\begin{aligned}
& k\left(S_{\text {■PRRS }} + 4 M_{1}\right) : \left(S_{\text {■FRRs }} + 4 M_{2}\right) \\
= & \frac{k(1+k+2)}{1+2 \frac{2k+1}{2k}} \\
= & \frac{k^2(k+3)}{3k+1} .
\end{aligned}
$$