Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Diagram)
Line 16: Line 16:
 
</asy>
 
</asy>
  
−
==Solution==
+
==Solution 1==
 
As we can see,  
 
As we can see,  
  
Line 78: Line 78:
 
-Gamjawon
 
-Gamjawon
  
−
Solution 2.
+
==Solution 2==.
 
Here's a solution that doesn't need <math>\sin 15^\circ</math>.
 
Here's a solution that doesn't need <math>\sin 15^\circ</math>.
  
 
As above, get to <math>AP=HM</math>.  As in the figure, let <math>O</math> be the foot of the perpendicular from <math>B</math> to <math>AC</math>.  Then <math>BCO</math> is a 45-45-90 triangle, and <math>ABO</math> is a 30-60-90 triangle.  So <math>BO=5</math> and <math>AO=5\sqrt{3}</math>; also, <math>CO=5</math>, <math>BC=5\sqrt2</math>, and <math>MC=\dfrac{BC}{2}=5\dfrac{\sqrt2}{2}</math>.  But <math>MO</math> and <math>AH</math> are parallel, both being orthogonal to <math>BC</math>.  Therefore <math>MH:AO=MC:CO</math>, or <math>MH=\dfrac{5\sqrt3}{\sqrt2}</math>, and we're done.
 
As above, get to <math>AP=HM</math>.  As in the figure, let <math>O</math> be the foot of the perpendicular from <math>B</math> to <math>AC</math>.  Then <math>BCO</math> is a 45-45-90 triangle, and <math>ABO</math> is a 30-60-90 triangle.  So <math>BO=5</math> and <math>AO=5\sqrt{3}</math>; also, <math>CO=5</math>, <math>BC=5\sqrt2</math>, and <math>MC=\dfrac{BC}{2}=5\dfrac{\sqrt2}{2}</math>.  But <math>MO</math> and <math>AH</math> are parallel, both being orthogonal to <math>BC</math>.  Therefore <math>MH:AO=MC:CO</math>, or <math>MH=\dfrac{5\sqrt3}{\sqrt2}</math>, and we're done.
 +
 +
==Solution 3==
 +
Break our diagram into 2 special right triangle by dropping an altitude from <math>B</math> to <math>AC</math> we then get that <cmath>AC=5+5\sqrt{3}, BC=5\sqrt{2}.HC=\frac{5\sqrt2+5\sqrt6}{2}, HM=\frac{5\sqrt6}{2}, HN=\frac{5\sqrt6}{4}.</cmath> We know that <math>\triangle{AHD}\simeq \triangle{PND}</math> and are 30-60-90.  Thus, <cmath>AP=2 \cdot HN=\frac{5\sqrt6}{4}.</cmath>
 +
 +
Thus, <math>(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}</math>. So our final answer is <math>75+2=77</math>.
 +
 +
 +
<math>m+n=\boxed{077}</math>
  
 
== See also ==
 
== See also ==

Revision as of 02:42, 4 March 2018

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 \[AC=5+5\sqrt{3}, BC=5\sqrt{2}.HC=\frac{5\sqrt2+5\sqrt6}{2}, HM=\frac{5\sqrt6}{2}, HN=\frac{5\sqrt6}{4}.\] We know that $\triangle{AHD}\simeq \triangle{PND}$ and are 30-60-90. Thus, \[AP=2 \cdot HN=\frac{5\sqrt6}{4}.\]

Thus, $(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}$. So our final answer is $75+2=77$.


$m+n=\boxed{077}$

See also

2014 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_14&oldid=92683"