2. Square $A B C D$ and square $A B E F$ are in planes that form a $120^{\circ}$ angle, $M, N$ are points on the diagonals $A C$, $B F$ respectively, and $A M=F N$. If $A B=1$, then the maximum value of $M N$ is $\qquad$ .
2. 1 Analysis: As shown in the figure, draw $M P \perp A B$ at $P$, connect $P N$, it can be proven that $P N \perp A B$, thus $\angle M P N=120^{\circ}$. Let $A M=A N=x$, then, $\frac{M P}{1}=\frac{A M}{\sqrt{2}}$, so $M P=\frac{x}{\sqrt{2}}$, similarly, $P N=\frac{\sqrt{2}-x}{\sqrt{2}}$, in $\triangle M N P$, using the cosine rule, $M N^{2}=P M^{2}+P N^{2}-2 P M \cdot P N \cos 120^{\circ}$ $=\frac{1}{2}\left(x^{2}-\sqrt{2} x+2\right)=\frac{1}{2}\left(x-\frac{\sqrt{2}}{2}\right)^{2}+\frac{3}{4}(0 \leqslant x \leqslant \sqrt{2})$, when $x=0$ or $x=\sqrt{2}$, $M N_{\text {max }}=1$.