Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2008 AMC 12A Problems/Problem 24

Revision as of 17:06, 27 January 2017 by Sw993 (talk | contribs) (→‎Solution)

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 $\tan{\angle BAD}$?

$\mathrm{(A)}\ \frac{\sqrt{3}}{6}\qquad\mathrm{(B)}\ \frac{\sqrt{3}}{3}\qquad\mathrm{(C)}\ \frac{\sqrt{3}}{2\sqrt{2}}\qquad\mathrm{(D)}\ \frac{\sqrt{3}}{4\sqrt{2}-3}\qquad\mathrm{(E)}\ 1$

Solution

Solution 1

unitsize(12mm);
pair C=(0,0), B=(4 * dir(60)), A = (8,0), D=(2 * dir(60));
pair E=(1,0), F=(2,0);
draw(C--B--A--C);
draw(A--D);draw(D--E);draw(B--F);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("\(C\)",C,SW);
label("\(B\)",B,N);
label("\(A\)",A,SE);
label("\(D\)",D,NW);
label("\(E\)",E,S);
label("\(F\)",F,S);
label("\(60^\circ\)",C+(.1,.1),ENE);
label("\(2\)",1*dir(60),NW);
label("\(2\)",3*dir(60),NW);
label("\(\theta\)",(7,.4));
label("\(1\)",(.5,0),S);
label("\(1\)",(1.5,0),S);
label("\(x-2)",(5,0),S); (Error making remote request. Unknown error_msg)

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

\[\tan\theta = \frac{\frac{2\sqrt{3}}{x-2} - \frac{\sqrt{3}}{x-1}}{1 + \frac{2\sqrt{3}}{x-2}\cdot\frac{\sqrt{3}}{x-1}}= \frac{x\sqrt{3}}{x^2-3x+8}\]

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\\ \frac{x}{x^2 - 3x + 8} &\leq \frac{1}{4\sqrt{2}-3}\\ \frac{x\sqrt{3}}{x^2 - 3x + 8} = \tan \theta &\leq \frac{\sqrt{3}}{4\sqrt{2}-3}\end{align*}

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

Solution 2

We notice that $\tan(x)$ is strictly increasing on the interval $[0, \frac{\pi}{2})$ (if $\angle BAD\ge 90^\circ$, then it is impossible for $\angle C=60^\circ$), so we want to maximize $\angle BAD$.

Consider the circumcircle of $BAD$ and let it meet $AC$ again at $F$. Any point $P$ between $A$ and $F$ on line $AC$ is inside this circle, so it follows that $\angle BPD>\angle BAD$. Therefore to maximize $\angle BAD$, the circumcircle of $BAD$ must be tangent to $AC$ at $A$. By PoP we find that $CA^2=CD\cdot CB \Rightarrow AC = 2\sqrt{2}$.

Now our computations are straightforward: \[\tan\angle BAD = \frac{\sin \angle BAD}{\cos \angle BAD} = \frac{\frac{2\sin\angle ABD}{AD}}{\frac{AB^2+AD^2-BD^2}{2AB\cdot AD}}\] \[=\frac{4\sin \angle ABD\cdot AB}{AB^2+AD^2-4} = \frac{4 AC \sin \angle ACB}{AB^2 + AD^2 - 4}\] \[=\frac{4\sqrt{6}}{(4^2+(2\sqrt{2})^2-4\cdot 2\sqrt{2}) + (2^2+(2\sqrt{2})^2 - 2\cdot 2\sqrt{2}) - 4} = \frac{4\sqrt{6}}{32-12\sqrt{2}}\] \[=\boxed{\frac{\sqrt{3}}{4\sqrt{2}-3}}\]

See also

2008 AMC 12A (Problems • Answer Key • 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
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_24&oldid=82594"