2008 AMC 12A Problem 24

Place $C = (0, 0),$ $B = (2, 2\sqrt{3})$ so that $\angle C = 60^\circ$ and $BC = 4,$ and let $A = (x, 0)$ with $x \gt 0.$ Then $D = (1, \sqrt{3})$ is the midpoint of $BC.$ The vectors $\overrightarrow{AB} = (2-x, 2\sqrt{3})$ and $\overrightarrow{AD} = (1-x, \sqrt{3})$ have cross-product magnitude $\sqrt{3}\,x$ and dot product $x^2 - 3x + 8,$ which is always positive. Hence $$\tan(\angle BAD) = \dfrac{\sqrt{3}\,x}{x^2 - 3x + 8}.$$ The derivative has the sign of $8-x^2,$ so the unique maximum occurs at $x = 2\sqrt{2}.$ Substituting, $$\begin{aligned} \tan(\angle BAD) &= \dfrac{2\sqrt{6}}{16 - 6\sqrt{2}} \\ &= \dfrac{\sqrt{6}}{8 - 3\sqrt{2}} \\ &= \dfrac{\sqrt{3}}{4\sqrt{2} - 3}. \end{aligned}$$ Thus, D is the correct answer.