2014 AIME II Problems/Problem 14


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 $m+n$.


[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

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 $75+2=77$.


-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

[asy]  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm);  real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  real xmin = -8.455641974276588, xmax = 26.731282460265, ymin = -10.92318356252699, ymax = 9.023689834456471;  /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882);   draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819)--cycle, linewidth(2) + rvwvcq);   /* draw figures */ draw((-1.4934334172297545,2.6953043701763835)--(1.1286284157632023,-6.954814372303504), linewidth(2) + wrwrwr);  draw((xmin, -0.9930079421029264*xmin + 1.2123131258653241)--(xmax, -0.9930079421029264*xmax + 1.2123131258653241), linewidth(2) + wrwrwr); /* line */ draw((xmin, 0.0035082940460819836*xmin-6.958773932654766)--(xmax, 0.0035082940460819836*xmax-6.958773932654766), linewidth(2) + wrwrwr); /* line */ draw((xmin, -285.03882139434313*xmin-422.9911967079192)--(xmax, -285.03882139434313*xmax-422.9911967079192), linewidth(2) + wrwrwr); /* line */ draw((xmin, -1.7181023895538718*xmin + 0.12943284739433739)--(xmax, -1.7181023895538718*xmax + 0.12943284739433739), linewidth(2) + wrwrwr); /* line */ draw(circle((4.642656870506668,-0.8187240845670819), 7.071067811865476), linewidth(2) + wrwrwr);  draw((xmin, -285.0388213943529*xmin + 1322.5187184230485)--(xmax, -285.0388213943529*xmax + 1322.5187184230485), linewidth(2) + wrwrwr); /* line */ draw((-1.4934334172297545,2.6953043701763835)--(4.617849638067675,6.252300211899359), linewidth(2) + wrwrwr);  draw((4.617849638067675,6.252300211899359)--(-1.459546107520503,-6.96389444957376), linewidth(2) + wrwrwr);  draw(circle((-0.18240250073327363,-2.12975500106356), 5), linewidth(2) + wrwrwr);  draw((xmin, -285.0388213943432*xmin + 449.7637608575419)--(xmax, -285.0388213943432*xmax + 449.7637608575419), linewidth(2) + wrwrwr); /* line */ draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + wrwrwr);  draw((-1.4934334172297545,2.6953043701763835)--(4.642656870506668,-0.8187240845670819), linewidth(2) + wrwrwr);  draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376), linewidth(2) + rvwvcq);  draw((-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504), linewidth(2) + rvwvcq);  draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + rvwvcq);  draw((4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819), linewidth(2) + rvwvcq);  draw((4.642656870506668,-0.8187240845670819)--(-1.4934334172297545,2.6953043701763835), linewidth(2) + rvwvcq);   /* dots and labels */ dot((-1.4934334172297545,2.6953043701763835),dotstyle);  label("$A$", (-1.3954084351380491,2.9230996889873015), NE * labelscalefactor);  dot((1.1286284157632023,-6.954814372303504),dotstyle);  label("$B$", (1.2093379191072373,-6.719031552166216), NE * labelscalefactor);  dot((8.199652712229643,-6.930007139864511),linewidth(4pt) + dotstyle);  label("$C$", (8.292420110475998,-6.741880204396438), NE * labelscalefactor);  dot((-1.459546107520503,-6.96389444957376),linewidth(4pt) + dotstyle);  label("$H$", (-1.3725597829078273,-6.787577508856881), NE * labelscalefactor);  dot((-1.4686261847907602,-4.375719926290057),linewidth(4pt) + dotstyle);  label("$E$", (-1.3725597829078273,-4.182831154611618), NE * labelscalefactor);  dot((4.617849638067675,6.252300211899359),linewidth(4pt) + dotstyle);  label("$L$", (4.705181710331174,6.441792132441429), NE * labelscalefactor);  dot((4.642656870506668,-0.8187240845670819),linewidth(4pt) + dotstyle);  label("$O$", (4.728030362561396,-0.6412900589272691), NE * labelscalefactor);  dot((4.117194931218359,-6.944329602191013),linewidth(4pt) + dotstyle);  label("$D$", (4.2025113612662945,-6.764728856626659), NE * labelscalefactor);  dot((4.6674641029456625,-7.889748381033524),linewidth(4pt) + dotstyle);  label("$F$", (4.750879014791618,-7.701523598065745), NE * labelscalefactor);  dot((4.651736947776926,-3.406898607850789),linewidth(4pt) + dotstyle);  label("$G$", (4.750879014791618,-3.2231877609423107), NE * labelscalefactor);  dot((1.5791517652735851,-0.3557971188372022),linewidth(4pt) + dotstyle);  label("$K$", (1.6663109637116735,-0.18431701432283698), NE * labelscalefactor);  dot((1.5870153428579534,-2.5972220054285695),linewidth(4pt) + dotstyle);  label("$P$", (1.6891596159418953,-2.4234849328845542), NE * labelscalefactor);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   /* end of picture */ [/asy] 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

2014 AIME II 14.png

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


See also

2014 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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