Triangle $ABC$ has $\angle C = 60^{\circ}$ and $BC = 4$. Point $D$ is the midpoint of $BC$. What is the largest possible value of $\tan{\angle BAD}$? $\mathrm{(A)}\ \frac{\sqrt{3}}{6}\qquad$ $\mathrm{(B)}\ \frac{\sqrt{3}}{3}\qquad$ $\mathrm{(C)}\ \frac{\sqrt{3}}{2\sqrt{2}}\qquad$ $\mathrm{(D)}\ \frac{\sqrt{3}}{4\sqrt{2}-3}\qquad$ $\mathrm{(E)}\ 1$
Solution
Place $C=(0,0)$, $B=(4,0)$, $A=t(\cos60^\circ,\sin60^\circ)$, $D=(2,0)$ (midpoint of $BC$).
$\vec{AB}=(4-\frac t2,-\frac{\sqrt3}2t)$ and $\vec{AD}=(2-\frac t2,-\frac{\sqrt3}2t)$. The tangent of the angle between them is $\frac{|\vec{AB}\times\vec{AD}|}{\vec{AB}\cdot\vec{AD}}=\dfrac{\sqrt3\,t}{t^2-3t+8}=\dfrac{\sqrt3}{t+\frac8t-3}$.
By AM-GM, $t+\frac8t\ge2\sqrt8=4\sqrt2$, with equality at $t=2\sqrt2$, so the maximum is $\frac{\sqrt3}{4\sqrt2-3}$.
The maximum value simplifies to $\boxed{\dfrac{\sqrt3}{4\sqrt2-3}}$, answer (D).
Triangle $ABC$ has $\angle C=90^\circ$ and $BC=4$. Point $D$ is the midpoint of $\overline{BC}$. What is the largest possible value of $\tan\angle BAD$?
