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.)
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  http://www.artofproblemsolving.com/Wiki/images/5/59/AOPS_wiki.PNG ( This is the diagram.)  
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As we can see,
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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>
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<math>AHC</math> is a <math>45-45-90</math> triangle, so ∠HAB=15∘.
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<math>AHD</math> is <math>30-60-90</math>.
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<math>AH</math> and <math>PN</math> are parallel lines so <math>PND</math> is <math>30-60-90</math> also.
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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>

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 $BM=CM$. Point $N$ is the midpoint of the segment $HM$, and point $P$ is on ray $AD$ such that PN⊥BC. Then $AP^2=\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

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

As we can see,

$M$ is the midpoint of $BC$ and $N$ is the midpoint of $HM$

$AHC$ is a $45-45-90$ triangle, so ∠HAB=15∘.

$AHD$ is $30-60-90$.

$AH$ and $PN$ are parallel lines so $PND$ is $30-60-90$ also.

Then if we use those informations we get $AD=2HD$ and

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

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