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

(Created page with "==Problem 12== == Solution == == See also == {{AIME box|year=2012|n=I|num-b=11|num-a=13}}")
 
(Problem 12)
Line 1: Line 1:
 
==Problem 12==
 
==Problem 12==
 +
Let <math>\triangle ABC</math> be a right triangle with right angle at <math>C.</math> Let <math>D</math> and <math>E</math> be points on <math>\overline{AB}</math> with <math>D</math> between <math>A</math> and <math>E</math> such that <math>\overline{CD}</math> and <math>\overline{CE}</math> trisect <math>\angle C.</math> If <math>\frac{DE}{BE} = \frac{8}{15},</math> then <math>\tan B</math> can be written as <math>\frac{m \sqrt{p}}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, and <math>p</math> is a positive integer not divisible by the square of any prime. Find <math>m+n+p.</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 01:32, 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

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