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

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− | 14. In | + | 14. 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 AH⊥BC, <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 PN⊥BC. 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>. |

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

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Thank you. | Thank you. | ||

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+ | -Gamjawon |

## Revision as of 04:54, 31 March 2014

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

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

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