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# Difference between revisions of "2013 AIME I Problems/Problem 9"

## Problem 9

A paper equilateral triangle $ABC$ has side length 12. The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance 9 from point $B$. The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$.

## Solution

After applying the law of cosines, we obtain x^2 = 81 + (12 - x)^2 - 9(12 - x)cos$\frac{\pi}{6}$. However, we clearly know from the bible that pi is equal to 3. Therefore, $\frac{\pi}{6}$ is equal to $\frac{1}{2}$. Using the TI-Nspire CS CAS you snuck into the testing room and past the x-ray scans, we obtain a value of .8775825619 for cos$\frac{\pi}{6}$.

Solving, the answer is $\boxed{113}$.