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

m (typo)
(Added Solution - Need help on asymptote)
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<math>\textbf{(A)} \ \frac {\sqrt {3}}{6} \qquad \textbf{(B)} \ \frac {\sqrt {3}}{3} \qquad \textbf{(C)} \ \frac {\sqrt {3}}{2\sqrt {2}} \qquad \textbf{(D)} \ \frac {\sqrt {3}}{4\sqrt {2} - 3} \qquad \textbf{(E)}\ 1</math>
 
<math>\textbf{(A)} \ \frac {\sqrt {3}}{6} \qquad \textbf{(B)} \ \frac {\sqrt {3}}{3} \qquad \textbf{(C)} \ \frac {\sqrt {3}}{2\sqrt {2}} \qquad \textbf{(D)} \ \frac {\sqrt {3}}{4\sqrt {2} - 3} \qquad \textbf{(E)}\ 1</math>
  
−
==Solution==  
+
==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("<math>60^\circ</math>",C,NE);
 +
label("2",1*dir(60),NW);
 +
label("2",3*dir(60),NW);
 +
label("<math>\theta</math>",(7,.4));
 +
label("1",(.5,0),S);
 +
label("1",(1.5,0),S);
 +
label("x-2",(5,0),S);
 +
</asy>
 +
 
 +
 
 +
 
 +
Where <math>CA = x</math>
 +
 
 +
<math>\tan\theta = \tan(\angle BAF - \angle DAE)</math>
 +
 
 +
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
 +
 
 +
<math>\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}}</math>
 +
 
 +
Multiplying numerator and denominator by <math>(x-2)(x-1)</math>
 +
 
 +
<math>\tan\theta = \frac{x\sqrt{3}}{x^2-3x+8}</math>
 +
 
 +
If you know calculus, you can use that right here to max <math>\tan\theta</math>, but if you don't:
 +
 
 +
By AM-GM
 +
 
 +
<math>\frac{x + \frac{8}{x}}{2} \geq \sqrt{8}</math>
 +
 
 +
<math>x + \frac{8}{x} \geq 4\sqrt{2}</math>
 +
 
 +
<math>x + \frac{8}{x} -3 \geq 4\sqrt{2} - 3</math>
 +
 
 +
<math>\frac{x^2 - 3x + 8}{x} \geq 4\sqrt{2} - 3</math>
 +
 
 +
<math>\frac{x}{x^2 - 3x + 8} \leq \frac{1}{4\sqrt{2}-3}</math>
 +
 
 +
<math>\frac{x\sqrt{3}}{x^2 - 3x + 8} \leq \frac{\sqrt{3}}{4\sqrt{2}-3}</math>
 +
 
 +
<math>\tan\theta \leq \frac{\sqrt{3}}{4\sqrt{2}-3}</math>
 +
 
 +
Which means that the minimum is
 +
<math>\frac{\sqrt{3}}{4\sqrt{2}-3} \Rightarrow \mathbf{(D)}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|ab=A|num-b=23|num-a=25}}
 
{{AMC12 box|year=2008|ab=A|num-b=23|num-a=25}}

Revision as of 11:05, 24 February 2008

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 $\tan{\angle BAD}$?

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

Solution

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("<math>60^\circ</math>",C,NE);
label("2",1*dir(60),NW);
label("2",3*dir(60),NW);
label("<math>\theta</math>",(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)


Where $CA = x$

$\tan\theta = \tan(\angle BAF - \angle DAE)$

Since $\tan\angle BAF = \frac{2\sqrt{3}}{x-2}$ and $\tan\angle DAE = \frac{\sqrt{3}}{x-1}$, we have

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

Multiplying numerator and denominator by $(x-2)(x-1)$

$\tan\theta = \frac{x\sqrt{3}}{x^2-3x+8}$

If you know calculus, you can use that right here to max $\tan\theta$, but if you don't:

By AM-GM

$\frac{x + \frac{8}{x}}{2} \geq \sqrt{8}$

$x + \frac{8}{x} \geq 4\sqrt{2}$

$x + \frac{8}{x} -3 \geq 4\sqrt{2} - 3$

$\frac{x^2 - 3x + 8}{x} \geq 4\sqrt{2} - 3$

$\frac{x}{x^2 - 3x + 8} \leq \frac{1}{4\sqrt{2}-3}$

$\frac{x\sqrt{3}}{x^2 - 3x + 8} \leq \frac{\sqrt{3}}{4\sqrt{2}-3}$

$\tan\theta \leq \frac{\sqrt{3}}{4\sqrt{2}-3}$

Which means that the minimum is $\frac{\sqrt{3}}{4\sqrt{2}-3} \Rightarrow \mathbf{(D)}$

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
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