Solution:
Letting $x=\frac{a}{a+b}$, we have to solve
$$m \arcsin \frac{x}{2} + n \arcsin \frac{1-x}{2} = \pi$$
This is convex in $x$, so if it is to have a solution, we must find that the LHS exceeds $\pi$ at one of the endpoints. Thus $\max (m, n) \geq 7$. If $\min (m, n) \leq 5$ we can find a solution by the intermediate value theorem. Also if $\min (m, n) \geq 7$ then
$$m \arcsin \frac{x}{2} + n \arcsin \frac{1-x}{2} \geq 14 \arcsin (1 / 4) > \pi$$
The inequality $\arcsin (1 / 4) > \frac{\pi}{14}$ can be verified by noting that
$$\sin \frac{\pi}{14} < \frac{\pi}{14} < \frac{3.5}{14} = \frac{1}{4}$$
The final case is when $\min (m, n) = 6$. We claim that this doesn't actually work. If we assume that $n=6$, we may compute the derivative at 0 to be
$$\frac{m}{2} - 6 \cdot \frac{1}{\sqrt{3}} = \frac{m-\sqrt{48}}{2} > 0$$
so no solution exists.