AoPSWiki

Try our innovative online adaptive learning system, Alcumus.
Over 1100 problems and 60+ video lessons. FREE!

Personal tools

Log in

Search

Toolbox

2008 AMC 12A Problems/Problem 24

From AoPSWiki

Problem

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

$\mathrm{(A)}\ \frac{\sqrt{3}}{6}\qquad\mathrm{(B)}\ \frac{\sqrt{3}}{3}\qquad\mathrm{(C)}\ \frac{\sqrt{3}}{2\sqrt{2}}\qquad\ma...$

Solution

Let $x = CA$ . Then $\tan\theta = \tan(\angle BAF - \angle DAE)$ , and since $\tan\angle BAF = \frac{2\sqrt{3}}{x-2}$ and $\tan\angle DAE = \frac{\sqrt{3}}{x-1}$ , we have

$\tan\theta = \frac{\frac{2\sqrt{3}}{x-2} - \frac{\sqrt{3}}{x-1}}{1 + \frac{2\sqrt{3}}{x-2}\cdot\frac{\sqrt{3}}{x-1}}= \frac{x...$

With calculus, taking the derivative and setting equal to zero will give the maximum value of $\tan \theta$ . Otherwise, we can apply AM-GM:

$\begin{align*}\frac{x^2 - 3x + 8}{x} = \left(x + \frac{8}{x}\right) -3 &\geq 2\sqrt{x \cdot \frac 8x} - 3 = 4\sqrt{2} - 3...$

Thus, the maximum is at $\frac{\sqrt{3}}{4\sqrt{2}-3} \Rightarrow \mathbf{(D)}$ .

See also

2008 AMC 12A (Problems • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2008_AMC_12A_Problems/Problem_24"

Category: Intermediate Geometry Problems

Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us