text: # Difference between revisions of "2014 AIME II Problems/Problem 14"
  
  14. 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⊥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⊥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 $\angle{HAB}=15^\circ$.
  
  $AHD$ 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