Shortest Path #2

We are trying to minimize $\overline{AP}+\overline{PQ}+\overline{QM}.$

Rotate the rectangle about point $C$ for $180^{\circ}$ to get rectangle $A'B'CD'.$

Note that $\overline{PQ}=\overline{PQ'}$ and $\overline{QM}=\overline{Q'M'}.$

Also, it is clear that $\overline{AP}+\overline{PQ'}+\overline{Q'M'}\ge\overline{AM'}.$

Therefore $\overline{AP}+\overline{PQ}+\overline{QM}=\overline{AP}+\overline{PQ'}+\overline{Q'M'}\ge\overline{AM'}=3\sqrt{5}.$

$\therefore~m^2=(3\sqrt{5})^2=\boxed{45}.$

The shortest distance traveled from $A$ to $M$ is that traveled by light. Therefore, the sides $\overline{BC}$ and $\overline{CD}$ acts like mirrors where line $\overline{AP}$ incidents at an angle of $\theta$ and reflects at the same angle of $\theta$ . If we fold the reflected beams at the mirrors, we find that the image $\overline{PR}$ is colinear with incident beam $\overline{AP}$ making $\overline{AR}$ a straight line as expected of how light should travel.

We note that the light beam travels vertically a distance of $\overline{RS} =$ $\overline{AB}+\overline{CD} =$ $3+3= 6$ and horizontally $\overline{SM} =$ $\overline{BC}+\overline{DM} =$ $2+1 = 3$ . By Pythagorean theorem , $\overline{AR}^2 =$ $\overline{RS}^2 +\overline{SM}^2$ , $\implies m^2 =$ $6^2+3^2 = \boxed{45}$ .