# Difference between revisions of "2014 AIME II Problems/Problem 14"

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==PROBLEM== | ==PROBLEM== | ||

− | + | In <math>\triangle{ABC}, AB=10, \angle{A}=30^\circ</math> , and <math>\angle{C=45^\circ}</math>. Let <math>H, D,</math> and <math>M</math> be points on the line <math>BC</math> such that <math>AH\perp{BC}</math>, <math>\angle{BAD}=\angle{CAD}</math>, and <math>BM=CM</math>. Point <math>N</math> is the midpoint of the segment <math>HM</math>, and point <math>P</math> is on ray <math>AD</math> such that <math>PN\perp{BC}</math>. Then <math>AP^2=\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | |

==DIAGRAM== | ==DIAGRAM== |

## Revision as of 01:17, 21 May 2014

## PROBLEM

In , and . Let and be points on the line such that , , and . Point is the midpoint of the segment , and point is on ray such that . Then , where and are relatively prime positive integers. Find .

## DIAGRAM

http://www.artofproblemsolving.com/Wiki/images/5/59/AOPS_wiki.PNG ( This is the diagram.)

## SOLUTION

As we can see,

is the midpoint of and is the midpoint of

is a triangle, so .

is triangle.

and are parallel lines so is triangle also.

Then if we use those informations we get and

and or

Now we know that , we can find for which is simpler to find.

We can use point to split it up as ,

We can chase those lengths and we would get

, so , so , so

Then using right triangle , we have HB=10 sin (15∘)

So HB=10 sin (15∘)=.

And we know that .

Finally if we calculate .

. So our final answer is .

Thank you.

-Gamjawon