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

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==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>.
 
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>.
  

Revision as of 00:18, 21 May 2014

Problem

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

Diagram

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

Solution

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 $\angle{HAB}=15^\circ$.


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


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


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


$PD=2ND$ and $AP=AD-PD=2HD-2ND=2HN$ or $AP=2HN=HM$


Now we know that $HM=AP$, we can find for $HM$ which is simpler to find.


We can use point $B$ to split it up as $HM=HB+BM$,


We can chase those lengths and we would get


$AB=10$, so $OB=5$, so $BC=5\sqrt{2}$, so $BM=\dfrac{1}{2} \cdot BC=\dfrac{5\sqrt{2}}{2}$


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


So HB=10 sin (15∘)=$\dfrac{5(\sqrt{6}-\sqrt{2})}{2}$.


And we know that $AP = HM = HB + BM = \frac{5(\sqrt6-\sqrt2)}{2} + \frac{5\sqrt2}{2} = \frac{5\sqrt6}{2}$.


Finally if we calculate $(AP)^2$.


$(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}$. So our final answer is $75+2=77$.


$m+n=\boxed{77}$


Thank you.


-Gamjawon

See also

2014 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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