2014 AIME II Problems/Problem 14

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

2014 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2014 AIME II Problems/Problem 14

Contents

PROBLEM

Diagram

Solution

See also