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2012 AIME I Problems/Problem 12

Revision as of 11:14, 18 March 2012 by Flamewire (talk | contribs) (→‎Solution)

Problem 12

Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$

Solution

Solution 1

Without loss of generality, set $CB = 1$. Then, by the Angle Bisector Theorem on triangle $DCB$, we have $CD = \frac{8}{15}$. We apply the Law of Cosines to triangle $DCB$ to get $1 + \frac{64}{225} - \frac{8}{15} = BD^{2}$, which we can simplify to get $BD = \frac{13}{15}$.

Now, we have $\cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}$ by another application of the Law of Cosines to triangle $DCB$, so $\cos \angle B = \frac{11}{13}$. In addition, $\sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}$, so $\tan \angle B = \frac{4\sqrt{3}}{11}$.

Our final answer is $4+3+11 = \boxed{018}$.

Solution 2

(This solution does not use the Angle Bisector Theorem but uses more trig.)

Find values for all angles in terms of $\angle B$. $\angle CEB = 150-B$, $\angle CED = 30+B$, $\angle CDE = 120-B$, $\angle CDA = 60+B$, and $\angle A = 90-B$.

Use the law of sines on $\triangle CED$ and $\triangle CEB$:

In $\triangle CED$, $\frac{8}{\sin 30} = \frac{CE}{\sin (120-B)}$. This simplifies to $16 = \frac{CE}{\sin (120-B)}$.

In $\triangle CEB$, $\frac{15}{\sin 30} = \frac{CE}{\sin B}$. This simplifies to $30 = \frac{CE}{\sin B}$.

Solve for $CE$ and equate them so that you get $16\sin (120-B) = 30\sin B$.

From this, $\frac{8}{15} = \frac{\sin B}{\sin (120-B)}$.

Use a trig identity on the denominator on the right to obtain: $\frac{8}{15} = \frac{\sin B}{\sin 120 \cos B - \cos 120 \sin B}$

This simplifies to $\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \cosB + \sin B}{2}} = \frac{2\sin B}{\sqrt{3}\cos B + \sin B}$ (Error compiling LaTeX. Unknown error_msg)

This gives $\frac{15}{8} = \frac{\sqrt{3}\cos B + \sin B}{2\sin B} = \frac{\sqrt{3}\cos B}{2\sinB} + \frac{1}{2} = \frac{\sqrt{3}\cot B}{2} + \frac{1}{2}.$ (Error compiling LaTeX. Unknown error_msg)

Thus: $\frac{11}{7} = \frac{\sqrt{3}\cot B}{2}$ and $\cot B = \frac{11*2}{8\sqrt{3}}$.

Since $\cot B = \frac{1}{\tan B}$, $tan{B} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}$. Our final answer is $4 + 3 + 11 = 18$.

See also

2012 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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