Shortest Path #3

We are trying to minimize $\overline{AP}+\overline{PM}+\overline{BQ}+\overline{BM}.$

Draw a square $AMXY$ like the image on the left.

We can prove that $\triangle AMP \equiv \triangle AYR,$ and thus $\overline{PM}=\overline{RY}.$

Also note that $\overline{RY}+\overline{RQ}+\overline{BQ}\ge\overline{BY}.$

Therefore

$\begin{aligned} \overline{AP}+\overline{PM}+\overline{BQ}+\overline{BM} &=\overline{RQ}+\overline{RY}+\overline{BQ}+3 \\ & \ge \overline{BY}+3 \\ &= \sqrt{\overline{AB}^2+\overline{AY}^2}+3=3+3\sqrt{5}.\end{aligned}$

with equality achieved when $Q$ and $R$ lies on $\overline{BY}.$

$\therefore~a+b+c=3+3+5=\boxed{11}.$