# Difference between revisions of "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 $[asy] unitsize(20); pair A = MP("A",(-5sqrt(3),0)), B = MP("B",(0,5),N), C = MP("C",(5,0)), M = D(MP("M",0.5(B+C),NE)), D = MP("D",IP(L(A,incenter(A,B,C),0,2),B--C),N), H = MP("H",foot(A,B,C),N), N = MP("N",0.5(H+M),NE), P = MP("P",IP(A--D,L(N,N-(1,1),0,10))); D(A--B--C--cycle); D(B--H--A,blue+dashed); D(A--D); D(P--N); markscalefactor = 0.05; D(rightanglemark(A,H,B)); D(rightanglemark(P,N,D)); MP("10",0.5(A+B)-(-0.1,0.1),NW); [/asy]$

## Solution 1

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 $BM=\dfrac{1}{2} \cdot BC=\dfrac{5\sqrt{2}}{2}$

We can also use Law of Sines: $$\frac{BC}{AB}=\frac{\sin\angle A}{\sin\angle C}$$ $$\frac{BC}{10}=\frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}}\implies BC=5\sqrt{2}$$

Then using right triangle $AHB$, we have $HB=10 \sin 15^\circ$

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

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 $75+2=77$. $m+n=\boxed{077}$

Thank you.

-Gamjawon

==Solution 2==. Here's a solution that doesn't need $\sin 15^\circ$.

As above, get to $AP=HM$. As in the figure, let $O$ be the foot of the perpendicular from $B$ to $AC$. Then $BCO$ is a 45-45-90 triangle, and $ABO$ is a 30-60-90 triangle. So $BO=5$ and $AO=5\sqrt{3}$; also, $CO=5$, $BC=5\sqrt2$, and $MC=\dfrac{BC}{2}=5\dfrac{\sqrt2}{2}$. But $MO$ and $AH$ are parallel, both being orthogonal to $BC$. Therefore $MH:AO=MC:CO$, or $MH=\dfrac{5\sqrt3}{\sqrt2}$, and we're done.

## Solution 3

Break our diagram into 2 special right triangle by dropping an altitude from $B$ to $AC$ we then get that $$AC=5+5\sqrt{3}, BC=5\sqrt{2}.$$ Since $\triangle{HCA}$ is a 45-45-90, $$HC=\frac{5\sqrt2+5\sqrt6}{2}$$ $MC=\frac{BM}{2},$ $$HM=\frac{5\sqrt6}{2}$$ $$HN=\frac{5\sqrt6}{4}$$ We know that $\triangle{AHD}\simeq \triangle{PND}$ and are 30-60-90. Thus, $$AP=2 \cdot HN=\frac{5\sqrt6}{2}.$$ $(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}$. So our final answer is $75+2=77$. $m+n=\boxed{077}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 