13. (1991 National High School Mathematics League) Let the height of the regular tetrahedron $P-ABC$ be $PO$, and let $M$ be the midpoint of $PO$. A plane passing through $AM$ and parallel to the edge $BC$ divides the tetrahedron into upper and lower parts. Find the ratio of the volumes of these two parts.
13. Let the plane through $AM$ parallel to $BC$ intersect edges $PB$ and $PC$ at $E$ and $F$, respectively, then $EF \parallel BC$. Let $D$ be the midpoint of $BC$. Connecting $PD$ intersects $EF$ at $G$, then $\frac{EF}{BC} = \frac{PG}{PD}$. Since the distance from $A$ to the plane $PBC$ is the same as the distance from $A$ to the plane $PEF$, we have
$$
\frac{V_{P-AEF}}{V_{P-ABC}} = \frac{V_{A-PEF}}{V_{A-PBC}} = \frac{S_{\triangle PEF}}{S_{\triangle PBC}} = \left(\frac{PG}{PD}\right)^2.
$$

In $\triangle PDA$, draw a line through $O$ parallel to $PD$ intersecting $AG$ at $N$, then $\triangle PGM$
$$
\begin{array}{l}
\text { and } \triangle ANO \sim \triangle AGD, \text { so } \frac{PG}{GD} = \frac{ON}{GD} \\
\frac{PG}{PD} = \frac{PG}{PG + GD} = \frac{1}{1 + \frac{3}{-2}} = \frac{2}{5}. \\
\text { Therefore, } \frac{V_{P-AEF}}{V_{P-ABC}} = \left(\frac{2}{5}\right)^2 = \frac{4}{25}.
\end{array}
$$

Thus, the ratio of the volumes of the two parts is $\frac{4}{21}$.
Note: This problem can be appropriately modified to be the 7th IMO problem: Given a tetrahedron $ABCD$, the length of edge $AB$ is $a$, and the length of edge $CD$ is $b$. The distance between the skew lines $AB$ and $CD$ is $d$, and the angle between them is $\delta$. The tetrahedron is divided into two parts by a plane $\gamma$ parallel to edges $AB$ and $CD$. If the distances from $AB$ and $CD$ to the plane $\gamma$ are in the ratio $k$, find the ratio of the volumes of the two parts. $\left(\frac{k^2(k+3)}{3k+1}\right)$