AMC 12 B, 2024

Answer (B): Let $O$ be the center of $\triangle ABC$ and $\triangle DEF$, let $P$ be the foot of the perpendicular from $D$ to $\overline{AB}$, and let $M$ be the midpoint of $\overline{AB}$.

The condition $\theta \le 60^\circ$ implies that $P$ lies on $\overline{AM}$ (as opposed to lying on $\overline{BM}$). Furthermore, $O$ is the center of the circle containing points $A$, $B$, and $D$, so by the Inscribed Angle Theorem $\angle DOA = 2\angle DBA$. It suffices to compute $\tan \angle DBA$ and then use the Double Angle Formula to obtain $\tan \theta$.

The area of $\triangle ABC$ is $\frac{\sqrt{3}}{4} \cdot 14^2 = 49\sqrt{3}$. Because hexagon $ADBECF$ consists of $\triangle ABC$ plus three copies of $\triangle ADB$, the area of $\triangle ADB$ is
$$\frac{91\sqrt{3} - 49\sqrt{3}}{3} = 14\sqrt{3}.$$
This implies that $DP = 2\sqrt{3}$. Furthermore, $OM = \frac{7\sqrt{3}}{3}$ and $OD = OA = 2 \cdot OM = \frac{14\sqrt{3}}{3}$. The Pythagorean Theorem yields
$$\begin{aligned} MP &= \sqrt{OD^2 - (OM + DP)^2} \\ &= \sqrt{\left(\frac{14\sqrt{3}}{3}\right)^2 - \left(\frac{13\sqrt{3}}{3}\right)^2} \\ &= \sqrt{\frac{14^2 - 13^2}{3}} = \sqrt{\frac{27}{3}} = 3. \end{aligned}$$
Finally, $BP = 7 + 3 = 10$, so $\tan \angle DBA = \frac{2\sqrt{3}}{10} = \frac{\sqrt{3}}{5}$ and
$$\tan \theta = \tan(2\angle DOB) = \frac{\frac{2\sqrt{3}}{5}}{1 - \frac{3}{25}} = \frac{5\sqrt{3}}{11}.$$