2010 AMC 10B Problem 20

Consider the following diagram: Assume the regular hexagon has side length $1.$ The smaller circle is inscribed in an equilateral triangle of side length $1.$ The inradius of this equilateral triangle is $\dfrac{\sqrt3}{6}.$ The area of the circle is then $$\pi \cdot \left(\dfrac{\sqrt3}{6}\right)^2 = \dfrac{\pi}{12}.$$ Let $O$ be the center of the larger circle. Drop the perpendicular from $O$ to $\overline{GH}$ at $J.$ Draw $\overline{OG}.$ We have that $\triangle OJG$ is right. Since $\angle HGI = 60^{\circ},$ we also have that $\triangle OJG$ is a $30-60-90$ triangle. Let $OJ = r.$ Then $OG = 2r.$ We also have that $OG$ is the sum of the height of the hexagon, equilateral triangle, and radius of the circle. Then $$OG = \dfrac{\sqrt3}{2} + \sqrt3 + r.$$ Substituting in $OG,$ we get $$2r = \dfrac{\sqrt3}{2} + \sqrt3 + r.$$ Simplifying gives us $$r = \dfrac{3\sqrt3}{2}.$$ The area of the larger circle is then $$\pi \cdot \left(\dfrac{3\sqrt3}{2}\right)^2 = \dfrac{27}{4}\pi.$$ The desired ratio is then $$\dfrac{\frac{27\pi}{4}}{\frac{\pi}{12}} = 81.$$ Thus, D is the correct answer.