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

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Then using right triangle <math>AHB</math>, we have HB=10 sin (15∘)
 
Then using right triangle <math>AHB</math>, we have HB=10 sin (15∘)
  
So HB=10 sin (15∘)=<math>\dfrac{5(\sqrt{6}\sqrt{2})}{2}</math>.
+
So HB=10 sin (15∘)=<math>\dfrac{5(\sqrt{6}-\sqrt{2})}{2}</math>.

Revision as of 22:42, 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 $AP=AD-PD=2HD-2ND=2HN$ 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∘)=$\dfrac{5(\sqrt{6}-\sqrt{2})}{2}$.

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