Difference between revisions of "2012 AIME I Problems/Problem 12"

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Now, we have <math>cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}</math> by another application of the Law of Cosines to triangle <math>DCB</math>, so <math>cos \angle B = \frac{11}{13}</math>. In addition, <math>sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}</math>, so <math>tan \angle B = \frac{4\sqrt{3}}{11}</math>.  
 
Now, we have <math>cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}</math> by another application of the Law of Cosines to triangle <math>DCB</math>, so <math>cos \angle B = \frac{11}{13}</math>. In addition, <math>sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}</math>, so <math>tan \angle B = \frac{4\sqrt{3}}{11}</math>.  
  
Our final answer is <math>4+3+11 = 18</math>.
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Our final answer is <math>4+3+11 = \boxed{018}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=11|num-a=13}}
 
{{AIME box|year=2012|n=I|num-b=11|num-a=13}}

Revision as of 11:54, 17 March 2012

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

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 $4+3+11 = \boxed{018}$.

See also

2012 AIME I (ProblemsAnswer KeyResources)
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