Question

Triangle $ABC$ has $\angle C = 60^\circ$ and $BC = 4.$ Point $D$ is the midpoint of $BC.$ What is the largest possible value of $\tan(\angle BAD)?$

$\dfrac{\sqrt{3}}{6}$

$\dfrac{\sqrt{3}}{3}$

$\dfrac{\sqrt{3}}{2\sqrt{2}}$

$\dfrac{\sqrt{3}}{4\sqrt{2} - 3}$

$1$

Solution:

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 lines $AD$ and $AB$ have slopes $\tfrac{\sqrt{3}}{1 - x}$ and $\tfrac{2\sqrt{3}}{2 - x}.$ Using the tangent-difference formula and simplifying, $\tan(\angle BAD) = \dfrac{\sqrt{3}\,x}{x^2 - 3x + 8}.$

Setting the derivative to zero gives $x^2 = 8,$ so $x = 2\sqrt{2}.$ Substituting, $\tan(\angle BAD) = \dfrac{2\sqrt{6}}{16 - 6\sqrt{2}} = \dfrac{\sqrt{6}}{8 - 3\sqrt{2}} = \dfrac{\sqrt{3}}{4\sqrt{2} - 3}.$

Thus, D is the correct answer.

2008 AMC 12A Problem 24

Solution:

Problem 24 in Other Years