13. In a regular triangular prism $A B C-A_{1} B_{1} C_{1}$, $A B=3, A A_{1}=4, M$ is the midpoint of $A A_{1}$, and $P$ is a point on $B C$. The shortest path from $P$ along the lateral surface of the prism, passing through the edge $C C_{1}$ to $M$, is $\sqrt{29}$. Let the intersection point of this shortest path with $O C_{1}$ be $N$. Find the lengths of $P C$ and $N C$.
13. Solution As shown in the figure, rotate the side $B B_{1} C_{1} C$ around the edge $C C_{1}$ by $120^{\circ}$, so that it lies in the same plane as the side $A A_{1} C_{1} C$. Point $P$ moves to the position of point $P_{1}$, and connect $M P_{1}$ intersecting $O C_{1}$ at $N$.
Then $M P_{1}$ is the shortest path from point $P$ along the lateral surface of the prism, passing through the edge $C C_{1}$, to point $M$.
Let $P C=x$, then $P_{1} C=x$.
In the right triangle $\triangle M A P_{1}$, by the Pythagorean theorem, we have $(3+x)^{2}+2^{2}=29$,
Solving for $x$ gives $x=2$. Thus, $P C=P_{1} C=2$.
Since $\frac{N C}{M A}=\frac{P_{1} C}{P_{1} A}=\frac{2}{5}$, it follows that $N C=\frac{4}{5}$.