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

(Added Solution - Need help on asymptote)
(→‎Solution: condense + formatting)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
−
<asy>
+
<center><asy>
 
unitsize(12mm);
 
unitsize(12mm);
 
pair C=(0,0), B=(4 * dir(60)), A = (8,0), D=(2 * dir(60));
 
pair C=(0,0), B=(4 * dir(60)), A = (8,0), D=(2 * dir(60));
 
pair E=(1,0), F=(2,0);
 
pair E=(1,0), F=(2,0);
−
 
 
draw(C--B--A--C);
 
draw(C--B--A--C);
−
draw(A--D);
+
draw(A--D);draw(D--E);draw(B--F);
−
draw(D--E);
+
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
−
draw(B--F);
+
label("\(C\)",C,SW);
−
 
+
label("\(B\)",B,N);
−
dot(A);
+
label("\(A\)",A,SE);
−
dot(B);
+
label("\(D\)",D,NW);
−
dot(C);
+
label("\(E\)",E,S);
−
dot(D);
+
label("\(F\)",F,S);
−
dot(E);
+
label("\(60^\circ\)",C+(.1,.1),ENE);
−
dot(F);
+
label("\(2\)",1*dir(60),NW);
−
 
+
label("\(2\)",3*dir(60),NW);
−
label("C",C,SW);
+
label("\(\theta\)",(7,.4));
−
label("B",B,N);
+
label("\(1\)",(.5,0),S);
−
label("A",A,SE);
+
label("\(1\)",(1.5,0),S);
−
label("D",D,NW);
+
label("\(x-2\)",(5,0),S);
−
label("E",E,S);
+
</asy></center>
−
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>
+
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
  
−
<math>\frac{x}{x^2 - 3x + 8} \leq \frac{1}{4\sqrt{2}-3}</math>
+
<cmath>\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}</cmath>
  
−
<math>\frac{x\sqrt{3}}{x^2 - 3x + 8} \leq \frac{\sqrt{3}}{4\sqrt{2}-3}</math>
+
With calculus, taking the [[derivative]] and setting equal to zero will give the maximum value of <math>\tan \theta</math>. Otherwise, we can apply [[AM-GM]]:
  
−
<math>\tan\theta \leq \frac{\sqrt{3}}{4\sqrt{2}-3}</math>
+
<cmath>
 +
\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}</cmath>
  
−
Which means that the minimum is  
+
Thus, the minimum is at 
−
<math>\frac{\sqrt{3}}{4\sqrt{2}-3} \Rightarrow \mathbf{(D)}</math>
+
<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 12:17, 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

[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} (Error compiling LaTeX. Unknown error_msg)

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

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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