9. The central angle of sector $OAB$ is $45^{\circ}$, with a radius of 1. Rectangle $PQMN$ is inscribed in this sector (as shown in the figure). The minimum length of the diagonal $PM$ of rectangle $PQMN$ is
$\begin{array}{l}\text { 9. In the figure } O M \text {, let } \angle N P M=\alpha, P M \\ =l \text {, then } P Q=N M=l \sin \alpha, Q M=P N \\ =l \cos \alpha . \quad \because \angle C=45^{\circ}, \quad \therefore O P=P N \\ =l \cos \alpha \text {. In } \triangle O M Q \text {, by the Pythagorean theorem we get } \\ 1=O M^{2}=O M^{2}+O Q^{2}=(l \cos \alpha)^{2} \\ +(l \cos \alpha+l \sin \alpha)^{2}=l^{2}(1+\sin 2 \alpha \\ +\cos ^{2} \alpha), \quad l^{2}=\frac{1}{1+\sin 2 \alpha+\cos ^{2} \alpha} \\ =-\frac{2}{2 \cos \alpha+\cos \alpha} \\ =\frac{2}{3+\sqrt{5} \sin (2 \alpha+\beta)} \text {, where } \\ \cos \varphi=-\sqrt{5}, \quad \sin \phi=\frac{1}{\sqrt{5}} . \quad \therefore \text { the minimum value of } l \\ \text { is } \sqrt{\frac{2}{3+\sqrt{5}}}=\sqrt{\frac{6-2 \sqrt{5}}{4}} \\ =\frac{\sqrt{5}-1}{2} \text {. } \\\end{array}$