Difference between revisions of "2008 AMC 12A Problems/Problem 24"

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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

$[asy]unitsize(12mm); pair C=(0,0), B=(4 * dir(60)), A = (8,0), D=(2 * dir(60)); pair E=(1,0), F=(2,0); draw(C--B--A--C); draw(A--D);draw(D--E);draw(B--F); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F); label("$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]$

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: $\tan\angle BAD = \frac{\sin \angle BAD}{\cos \angle BAD} = \frac{\frac{2\sin\angle ABD}{AD}}{\frac{AB^2+AD^2-BD^2}{2AB\cdot AD}}$ $=\frac{4\sin \angle ABD\cdot AB}{AB^2+AD^2-4} = \frac{4 AC \sin \angle ACB}{AB^2 + AD^2 - 4}$ $=\frac{4\sqrt{6}}{(4^2+(2\sqrt{2})^2-4\cdot 2\sqrt{2}) + (2^2+(2\sqrt{2})^2 - 2\cdot 2\sqrt{2}) - 4} = \frac{4\sqrt{6}}{32-12\sqrt{2}}$ $=\boxed{\frac{\sqrt{3}}{4\sqrt{2}-3}}$

@@ Line 24: / Line 24: @@
 label("\(1\)",(.5,0),S);
 label("\(1\)",(1.5,0),S);
-label("\(x-2)",(5,0),S);</asy>
+label("\(x-2\)",(5,0),S);</asy>
 Let <math>x = CA</math>. Then <math>\tan\theta = \tan(\angle BAF - \angle DAE)</math>, and since <math>\tan\angle BAF = \frac{2\sqrt{3}}{x-2}</math> and <math>\tan\angle DAE = \frac{\sqrt{3}}{x-1}</math>, we have

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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