2008 AMC 12A Problem 24

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

Place

C = (0, 0),

B = (2, 2\sqrt{3}),

and

A = (x, 0)

with

x \gt 0

Using slopes gives

\tan(\angle BAD) = \dfrac{\sqrt{3}\,x}{x^2 - 3x + 8};

maximize over

x

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.

Problem 24 in other years