2006 AMC 12A Problem 22

Place the hexagon at the center of the circle. Corresponding to each vertex is an arc from which only the two sides meeting there are entirely visible. These six congruent arcs make up the complementary probability $\tfrac{1}{2},$ so each arc measures $30^\circ.$ Take the arc centered on the ray from the center $O$ through a vertex $A,$ and let $P$ be its upper endpoint. Then $\angle POA = 15^\circ,$ and at $P$ a third side is just becoming visible, so $P$ lies on that side’s supporting line. Its distance from $O$ is the apothem $\sqrt{3}.$ Hence $\sqrt{3} = r\sin 15^\circ = r \cdot \dfrac{\sqrt{6} - \sqrt{2}}{4},$ giving $$r = \frac{4\sqrt{3}}{\sqrt{6} - \sqrt{2}} = 3\sqrt{2} + \sqrt{6}.$$ Thus, the correct answer is D.