Polygons!!

$120 + (\dfrac{n - 2}{n})180 + (\dfrac{m - 2}{m})180 = 360 \implies \dfrac{n - 2}{n} + \dfrac{m - 2}{m} = \dfrac{4}{3} \implies \dfrac{m + n}{mn} = \dfrac{1}{3} \implies m(n) = \dfrac{3n}{n - 3}$ , where $n > 3$ . Since $m(n)$ is monotonic decreasing for $n > 3 \implies n = 4$ maximizes $m(n)$ and $m(4) = 12$ . That is; $m(n) \leq m(4) = 12$ for $n > 3$ .

For area $A$ of the dodecagon: Let $x$ be a side of dodecagon.

$\dfrac{x}{2} = r\sin(\dfrac{\pi}{12}) \implies r = \dfrac{x}{2\sin(\dfrac{\pi}{12})} \implies h = \dfrac{x}{2} \cot(\dfrac{\pi}{12}) \implies A = 3\cot(\dfrac{\pi}{12})x^2$ .

$\cos(\dfrac{\pi}{12}) = \cos(\dfrac{\pi}{4} - \dfrac{\pi}{6}) = \dfrac{1}{2\sqrt{2}}(\sqrt{3} + 1)$

$\sin(\dfrac{\pi}{12}) = \sin(\dfrac{\pi}{4} - \dfrac{\pi}{6}) = \dfrac{1}{2\sqrt{2}}(\sqrt{3} - 1)$

$\implies A = 3(\dfrac{\sqrt{3} + 1}{\sqrt{3} - 1})x^2 = 3(2 + \sqrt{3})x^2$

$A = 3(2 + \sqrt{3})x^2 = 2 + \sqrt{3} \implies x = \dfrac{1}{\sqrt{3}}$ .

The area of the square $A^{*} = y^2 = 2 \implies y = \sqrt{2}$

$\implies \dfrac{y}{x} = \sqrt{\dfrac{2}{3}} = \sqrt{\dfrac{a}{b}} \implies a + b = \boxed{5}$ .