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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>
  
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<math>AHC</math> is a <math>45-45-90</math> triangle, so ∠HAB=15∘.
+
<math>AHC</math> is a <math>45-45-90</math> triangle, so <math>\angle{HAB}=15^\circ</math>.
  
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<math>AHD</math> is <math>30-60-90</math>.
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<math>AHD</math> is <math>30-60-90</math> triangle.
  
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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.
+
<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  
Line 17: Line 17:
 
<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.
  
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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
Line 38: Line 38:
  
 
Thank you.
 
Thank you.
−
 
−
--[[User:Gamjawon|Gamjawon]] 22:47, 29 March 2014 (EDT)
 

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 $BM=CM$. 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.

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