2021 Fall AMC 12B Problem 15

Because the three squares are rotated by $0^\circ,$ $30^\circ,$ and $60^\circ,$ the figure has $12$-fold symmetry. Its $24$ vertices alternate every $15^\circ$: outer vertices are the square corners at distance $3\sqrt2$ from the center, and inner vertices are edge crossings at distance $2\sqrt3.$ Connecting the center to all $24$ vertices splits the polygon into $24$ triangles, each with sides $3\sqrt2$ and $2\sqrt3$ and included angle $15^\circ.$ The total area is $$\begin{aligned} &24 \cdot \tfrac12 (3\sqrt2)(2\sqrt3)\sin 15^\circ \\ &= 72\sqrt6 \cdot \dfrac{\sqrt6 - \sqrt2}{4}. \end{aligned}$$ This simplifies to $18\sqrt6(\sqrt6 - \sqrt2) = 108 - 36\sqrt3,$ so $a + b + c = 108 + 36 + 3 = 147.$ Thus, the correct answer is E.