Three, let the height of the regular tetrahedron $P-ABC$ be $PO$, and $M$ be the midpoint of $PO$. A plane passing through $AM$ is parallel to the edge $BC$, dividing the tetrahedron into upper and lower parts. Try to find the ratio of the volumes of these two parts.
Three, let the plane through $A M$ parallel to $B C$ intersect the edges $P B$ and $P C$ at $E$ and $F$, respectively, then $E F \parallel B C$.
Let $D$ be the midpoint of $B C$, and connect $P D$ to intersect $E F$ at $G$, then
$$
\frac{E F}{B C}=\frac{P G}{P D} .
$$

Since the distance from $A$ to the plane $P B C$ is the same as the distance from $A$ to the plane $P E F$, we have
$$
\frac{V_{P-A E F}}{V_{P-A B C}}=\frac{S_{\triangle P E F}}{S_{\triangle P B C}}=\left(\frac{P G}{P D}\right)^{2} .
$$

In $\triangle P D A$, draw a line through $O$ parallel to $P D$ intersecting $A G$ at $N$, then
$$
\begin{array}{r}
\frac{P G}{G D}=\frac{O N}{G D}=\frac{A O}{A D}=\frac{2}{3}, \\
\frac{P G}{P D}=\frac{2}{5}, \quad \text { hence } \frac{V_{P-A E F}}{V_{P-A B C}}=\frac{4}{25} .
\end{array}
$$

Therefore, the ratio of the volumes of the upper and lower parts is $\frac{4}{21}$ (as shown in the left lower figure).