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

(Problem 9)
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==Problem 9==
 
==Problem 9==
 
A paper equilateral triangle <math>ABC</math> has side length 12. The paper triangle is folded so that vertex <math>A</math> touches a point on side <math>\overline{BC}</math> a distance 9 from point <math>B</math>. The length of the line segment along which the triangle is folded can be written as <math>\frac{m\sqrt{p}}{n}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>n</math> are relatively prime, and <math>p</math> is not divisible by the square of any prime. Find <math>m+n+p</math>.
 
A paper equilateral triangle <math>ABC</math> has side length 12. The paper triangle is folded so that vertex <math>A</math> touches a point on side <math>\overline{BC}</math> a distance 9 from point <math>B</math>. The length of the line segment along which the triangle is folded can be written as <math>\frac{m\sqrt{p}}{n}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>n</math> are relatively prime, and <math>p</math> is not divisible by the square of any prime. Find <math>m+n+p</math>.
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== Solution ==
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<math>\boxed{113}</math>
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== See also ==
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{{AIME box|year=2013|n=I|num-b=8|num-a=10}}

Revision as of 20:45, 16 March 2013

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

$\boxed{113}$

See also

2013 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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