2025 AMC 12A Problem 24

The centers of the $12$ outer circles lie on a circle of radius $r + 1,$ forming a regular $12$-gon. Adjacent centers are $2$ apart (both circles have radius $1$), and the central angle between them is $30^\circ.$ Thus $2(r + 1)\sin 15^\circ = 2,$ so $r + 1 = \dfrac{1}{\sin 15^\circ}.$ Since $\sin 15^\circ = \dfrac{\sqrt{6} - \sqrt{2}}{4},$ $$r + 1 = \frac{4}{\sqrt{6} - \sqrt{2}} = \sqrt{6} + \sqrt{2}.$$ Then $r = \sqrt{6} + \sqrt{2} - 1,$ so $a + b + c = 6 + 2 - 1 = 7.$ Thus, the correct answer is C.