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

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<math>M</math> is the midpoint of <math>BC</math> and <math>N</math> is the midpoint of <math>HM</math> | <math>M</math> is the midpoint of <math>BC</math> and <math>N</math> is the midpoint of <math>HM</math> | ||

− | <math>AHC</math> is a <math>45-45-90</math> triangle, so | + | <math>AHC</math> is a <math>45-45-90</math> triangle, so <math>\angle{HAB}=15^\circ</math>. |

− | <math>AHD</math> is <math>30-60-90</math>. | + | <math>AHD</math> is <math>30-60-90</math> triangle. |

− | <math>AH</math> and <math>PN</math> are parallel lines so <math>PND</math> is <math>30-60-90</math> also. | + | <math>AH</math> and <math>PN</math> are parallel lines so <math>PND</math> is <math>30-60-90</math> triangle also. |

Then if we use those informations we get <math>AD=2HD</math> and | Then if we use those informations we get <math>AD=2HD</math> and | ||

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<math>PD=2ND</math> and <math>AP=AD-PD=2HD-2ND=2HN</math> or <math>AP=2HN=HM</math> | <math>PD=2ND</math> and <math>AP=AD-PD=2HD-2ND=2HN</math> or <math>AP=2HN=HM</math> | ||

− | Now we know that HM=AP, we can find for HM which is simpler to find. | + | Now we know that <math>HM=AP</math>, we can find for <math>HM</math> which is simpler to find. |

− | We can use point B to split it up as HM=HB+BM, | + | We can use point <math>B</math> to split it up as <math>HM=HB+BM</math>, |

We can chase those lengths and we would get | We can chase those lengths and we would get | ||

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Thank you. | Thank you. | ||

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## Revision as of 04:42, 31 March 2014

14. In △ABC, AB=10, ∠A=30∘, and ∠C=45∘. Let H, D, and M be points on the line BC such that AH⊥BC, ∠BAD=∠CAD, 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.