15. Given that $A X$ is a ray inside $\angle N A M$, $\angle X A M=30^{\circ}, \angle X A N=45^{\circ}, P$ is a point on $A X$, $|A P|=1$, a line through $P$ intersects $A M, A N$ at $M, N$, find the size of $\angle A P N$ when the sum of the distances from $M, N$ to $A X$ is minimized, and find the length of $M N$ at this time.
15. As shown in the figure, draw $M C \perp O X$ at $C$, and $N D \perp O X$ at $D$. Let $<A P N=\theta .30^{\circ}<\theta<135^{\circ}$. Using the Law of Sines in $\triangle A P N$ and $\triangle A P M$, we get
$$
\begin{aligned}
\frac{A P}{\sin \left(\theta+45^{\circ}\right)} & =\frac{P N}{\frac{\sqrt{2}}{2}}, \frac{A P}{\sin \left(\theta-30^{\circ}\right)}=\frac{M P}{\frac{1}{2}} \\
M C+N D & =\frac{\sin \theta}{2 \sin \left(\theta-30^{\circ}\right)}+\frac{\sqrt{2} \sin \theta}{2 \sin \left(\theta+45^{\circ}\right)} \\
& =\frac{\sqrt{3}+1}{(\cot \theta+1)(\sqrt{3}-\cot \theta)} \geqslant \frac{\sqrt{3}+1}{\left(\frac{\sqrt{3}+1}{2}\right)^{2}}=\frac{4}{\sqrt{3}+1},
\end{aligned}
$$

At this point, $\cot \theta=\frac{\sqrt{3}-1}{2}$, i.e., $\theta=\operatorname{arccot} \frac{\sqrt{3}-1}{2}$, and $M N=\frac{M C+M D}{\sin \theta}=(\sqrt{3}-1)$
$$
\sqrt{8-2 \sqrt{3}} \text {. }
$$