$23 \cdot 31$ In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $P, M, N$ are the midpoints of edges $A A_{1}$, $B C, C C_{1}$ respectively. The figure formed by the intersection lines of the plane determined by these three points with the surface of the cube is
(A) a rhombus.
(B) a regular pentagon.
(C) a regular hexagon.
(D) a hexagon with only equal sides.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] As shown in the figure, connect $M N, B C_{1}$. It is easy to prove that $M N \parallel B C_{1}$, and $M N=\frac{1}{2} B C_{1}$.

Since $A D_{1} \cong B C_{1} \cong 2 M N$, then
$P E \cong M N$, which means $E$ lies on the cross-section.

Now, take the midpoints $F, G$ of $A B, D_{1} C_{1}$, and connect $P N$. Similarly, we have
$F M \perp E G \geqq \frac{1}{2} P N$, which means points $F, G$ also lie on the cross-section.

In summary, the cross-section is a hexagon $PFMNGE$, and it is a regular hexagon.
Therefore, the answer is $(C)$.