AIME 2014 II · 第 14 题

Solution 1

Let us just drop the perpendicular from $B$ to $AC$ and label the point of intersection $O$. We will use this point later in the problem. 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 $75+2=77$. $m+n=\boxed{077}$

-Gamjawon -edited by srisainandan6 to clarify and correct a small mistake

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,

$$HC=\frac{5\sqrt2+5\sqrt6}{2}$$ $MC=\frac{BC}{2},$

$$HM=\frac{5\sqrt6}{2}$$ $$HN=\frac{5\sqrt6}{4}$$ 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 $75+2=\boxed{077}$.

Solution 4

Draw the $45-45-90 \triangle AHC$. Now, take the perpendicular bisector of $BC$ to intersect the circumcircle of $\triangle ABC$ and $AC$ at $F, L, G$ as shown, and denote $O$ to be the circumcenter of $\triangle ABC$. It is not difficult to see by angle chasing that $AHBGO$ is cyclic, namely with diameter $AB$. Then, by symmetry, $EH = HB$ and as $HB, OG$ are both subtended by equal arcs they are equal. Hence, $EH = GO$. Now, draw line $HL$ and intersect it at $AC$ at point $K$ in the diagram. It is not hard to use angle chase to arrive at $AEOL$ a parallelogram, and from our length condition derived earlier, $AL \parallel HG$. From here, it is clear that $AK = KG$; that is, $P$ is just the intersection of the perpendicular from $K$ down to $BC$ and $AD$! After this point, note that $AP = PF$. It is easily derived that the circumradius of $\triangle ABC$ is $\frac{10}{\sqrt{2}}$. Now, $APO$ is a $30-60-90$ triangle, and from here it is easy to arrive at the final answer of $\boxed{077}$. ~awang11's sol

Solution 5

Let $BO \perp AC, O \in AC.$

Let $ME \perp BC, E \in AD.$

$MB = MC, \angle C = 45^\circ \implies$ points $M, E, O$ are collinear.

$HN = NM, AH||NP||ME \implies AP = PE.$

In $\triangle ABO \hspace{10mm} \angle A = 30^\circ \implies AO = AB \cos 60^\circ = 5 \sqrt{3}.$

In $\triangle AEO \hspace{10mm} \angle A = 15^\circ, \angle O = 90^\circ + 45^\circ = 135^\circ \implies$

$$\angle AEO = 30^\circ \implies$$ $$AE = AO \frac {\sin 135^\circ}{\sin 30^\circ} = 5 \sqrt{3} \cdot \sqrt{2} = 5 \sqrt{6} \implies$$ $$AP = 5 \sqrt {\frac {3}{2}} \implies AP^2 = \frac {75}{2} \implies \boxed{\textbf{077}}.$$ vladimir.shelomovskii@gmail.com, vvsss

Video solution

https://www.youtube.com/watch?v=SvJ0wDJphdU