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 $\boxed{113}$ .

Link for anti-xray TI-Nspire

www.aimetinspirecheats.com/antixray/purchase/sale_127462836

2013 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2013 AIME I Problems/Problem 9

Problem 9

Solution

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