2008 AMC 12A Problems/Problem 24
Problem
Triangle 
has 
and 
. Point 
is the midpoint of 
. What is the largest possible value of 
?
Solution
![[asy] 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); [/asy]](http://latex.artofproblemsolving.com/1/1/2/1128545a032f8a76ab532ab0d377c56d9ca11993.png)
Let 
. Then 
, and since 
and 
, we have
![\[\tan\theta = \frac{\frac{2\sqrt{3}}{x-2} - \frac{\sqrt{3}}{x-1}}{1 + \frac{2\sqrt{3}}{x-2}\cdot\frac{\sqrt{3}}{x-1}}= \frac{x\sqrt{3}}{x^2-3x+8}\]](http://latex.artofproblemsolving.com/0/e/7/0e7e40fad497d77cfc1c1273257c72565ec82619.png)
With calculus, taking the derivative and setting equal to zero will give the maximum value of 
. 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} (Error compiling LaTeX. Unknown error_msg)
Thus, the minimum is at
.
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 | |
