Question 156, As shown in the figure, the edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is $4, O$ is the center of the square $A D D_{1} A_{1}$, $E$ and $F$ are the midpoints of $B_{1} C_{1}$ and $C C_{1}$ respectively, $P$ is a moving point on the line segment $O B$, $M$ is a moving point on the base $A B C D$, and $N$ is a moving point on the line segment $E F$. Then the minimum value of $P M+P N$ is $\qquad$ -
Question 156, Solution: As shown in the figure, take the center of face $\mathrm{BCC}_{1} \mathrm{~B}_{1}$ as $Q$, then $O Q \perp$ face $\mathrm{BCC}_{1} \mathrm{~B}_{1}$, so $O Q \perp E F$. Also, $\mathrm{EF} \perp \mathrm{BC}_{1}$, hence $E F \perp$ face $O B Q$, so the foot of the perpendicular from point $\mathrm{P}$ to line segment $E F$ is the intersection of $E F$ and $B C_{1}$, which is also the midpoint of $E F$. Therefore, let $N$ be the midpoint of $EF$.

Take $G$ as the midpoint of $A D$, then $G$ is the projection of $O$ onto the plane $A B C D$. Hence, let $M$ be the projection of $P$ onto line segment $B G$.
Now rotate the plane $O B N$ around $O B$ so that point $N$ falls onto the plane $B O G$, and let the new position of point $N$ be $N'$. Then $\mathrm{PM}+\mathrm{PN}=\mathrm{PM}+\mathrm{PN}'$
$\geq \mathrm{d}\left(\mathrm{N}', \mathrm{BG}\right)=\mathrm{BN} \cdot \sin \angle \mathrm{N}' \mathrm{BG}=\mathrm{BN} \cdot \sin (\angle \mathrm{OBQ}+\angle \mathrm{OBG})$
It is easy to calculate that $\mathrm{BN}=3 \sqrt{2}, \sin \angle \mathrm{OBQ}=\frac{2}{\sqrt{6}}, \cos \angle \mathrm{OBQ}=\frac{\sqrt{2}}{\sqrt{6}}, \sin \angle \mathrm{OBG}=\frac{1}{\sqrt{6}}, \cos \angle \mathrm{OBQ}=$ $\frac{\sqrt{5}}{\sqrt{6}}$, substituting into (*) we get:
$\mathrm{PM}+\mathrm{PN} \geq 3 \sqrt{2}\left(\frac{2}{\sqrt{6}} \times \frac{\sqrt{5}}{\sqrt{6}}+\frac{\sqrt{2}}{\sqrt{6}} \times \frac{1}{\sqrt{6}}\right)=\sqrt{10}+1$
In summary, the minimum value of $\mathrm{PM}+\mathrm{PN}$ is $\sqrt{10}+1$.