2024 AMC 12B Problem 19

The six vertices lie on the circumcircle of radius $R = \dfrac{14}{\sqrt3},$ so $R^2 = \dfrac{196}{3}.$ Going around, the central angles alternate between $\theta$ (three times) and $120^\circ - \theta$ (three times). The cyclic-hexagon area is $$\begin{aligned} &\tfrac12 R^2\bigl(3\sin\theta + 3\sin(120^\circ - \theta)\bigr) \\ &= 98\bigl(\sin\theta + \sin(120^\circ - \theta)\bigr). \end{aligned}$$ By sum-to-product, $\sin\theta + \sin(120^\circ - \theta)$ $= 2\sin 60^\circ\cos(\theta - 60^\circ)$ $= \sqrt3\cos(60^\circ - \theta).$ Setting the area to $91\sqrt3$ gives $\sqrt3\cos(60^\circ - \theta)$ $= \dfrac{91\sqrt3}{98} = \dfrac{13\sqrt3}{14},$ so $\cos(60^\circ - \theta) = \dfrac{13}{14}$ and $\sin(60^\circ - \theta) = \dfrac{3\sqrt3}{14}.$ Then $\tan(60^\circ - \theta) = \dfrac{3\sqrt3}{13},$ and $$\begin{aligned} &\tan\theta = \tan\bigl(60^\circ - (60^\circ - \theta)\bigr) \\ &= \frac{\sqrt3 - \frac{3\sqrt3}{13}}{1 + \sqrt3\cdot\frac{3\sqrt3}{13}} \\ &= \frac{\frac{10\sqrt3}{13}}{\frac{22}{13}} \\ &= \frac{5\sqrt3}{11}. \end{aligned}$$ Thus, the correct answer is B.