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

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

(wrong problem)
m (spacing)
 
(11 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 +
== Problem ==
 +
[[Triangle]] <math>ABC</math> has <math>\angle C = 60^{\circ}</math> and <math>BC = 4</math>.  Point <math>D</math> is the [[midpoint]] of <math>BC</math>.  What is the largest possible value of <math>\tan{\angle BAD}</math>?
  
 +
<math>\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</math>
 +
 +
== 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 <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
 +
 +
<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>
 +
 +
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]]:
 +
 +
<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}\end{align*}</cmath>
 +
 +
Thus, the maximum is at 
 +
<math>\frac{\sqrt{3}}{4\sqrt{2}-3} \Rightarrow \mathbf{(D)}</math>.
 +
 +
== Solution 2 ==
 +
We notice that <math>\tan(x)</math> is strictly increasing on the interval <math>[0, \frac{\pi}{2})</math> (if <math>\angle BAD\ge 90^\circ</math>, then it is impossible for <math>\angle C=60^\circ</math>), so we want to maximize <math>\angle BAD</math>.
 +
 +
Consider the circumcircle of <math>BAD</math> and let it meet <math>AC</math> again at <math>F</math>. Any point <math>P</math> between <math>A</math> and <math>F</math> on line <math>AC</math> is inside this circle, so it follows that <math>\angle BPD>\angle BAD</math>. Therefore to maximize <math>\angle BAD</math>, the circumcircle of <math>BAD</math> must be tangent to <math>AC</math> at <math>A</math>. By PoP we find that <math>CA^2=CD\cdot CB \Rightarrow AC = 2\sqrt{2}</math>.
 +
 +
Now our computations are straightforward:
 +
<cmath>\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}}</cmath>
 +
<cmath>=\frac{4\sin \angle ABD\cdot AB}{AB^2+AD^2-4} = \frac{4 AC \sin \angle ACB}{AB^2 + AD^2 - 4}</cmath>
 +
<cmath>=\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}}</cmath>
 +
<cmath>=\boxed{\frac{\sqrt{3}}{4\sqrt{2}-3}}</cmath>
 +
 +
== See also ==
 +
{{AMC12 box|year=2008|ab=A|num-b=23|num-a=25}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 13:35, 7 June 2018

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

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=94982"