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

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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 <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>. | 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 <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.) |

+ | |||

+ | As we can see, | ||

+ | |||

+ | <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 ∠HAB=15∘. | ||

+ | |||

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

+ | |||

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

+ | |||

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

+ | |||

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

## Revision as of 22:34, 29 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 ∠HAB=15∘.

is .

and are parallel lines so is also.

Then if we use those informations we get and

and $AP=AD−PD=2HD−2ND=2HN$ (Error compiling LaTeX. ! Package inputenc Error: Unicode char \u8:− not set up for use with LaTeX.) or