Art of Problem Solving
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

2014 AIME II Problems/Problem 14: Difference between revisions

← Older editNewer edit →
Swamih (talk | contribs)
30 edits
No edit summary
Swamih (talk | contribs)
30 edits
No edit summary
Line 7: Line 7:
==SOLUTION==
==SOLUTION==
As we can see,  
As we can see,  


<math>M</math> is the midpoint of <math>BC</math> and <math>N</math> is the midpoint of <math>HM</math>
<math>M</math> is the midpoint of <math>BC</math> and <math>N</math> is the midpoint of <math>HM</math>


<math>AHC</math> is a <math>45-45-90</math> triangle, so <math>\angle{HAB}=15^\circ</math>.
<math>AHC</math> is a <math>45-45-90</math> triangle, so <math>\angle{HAB}=15^\circ</math>.


<math>AHD</math> is <math>30-60-90</math> triangle.
<math>AHD</math> is <math>30-60-90</math> triangle.


<math>AH</math> and <math>PN</math> are parallel lines so <math>PND</math> is <math>30-60-90</math> triangle also.
<math>AH</math> and <math>PN</math> are parallel lines so <math>PND</math> is <math>30-60-90</math> triangle also.


Then if we use those informations we get <math>AD=2HD</math> and  
Then if we use those informations we get <math>AD=2HD</math> and  


<math>PD=2ND</math> and <math>AP=AD-PD=2HD-2ND=2HN</math>  or <math>AP=2HN=HM</math>
<math>PD=2ND</math> and <math>AP=AD-PD=2HD-2ND=2HN</math>  or <math>AP=2HN=HM</math>


Now we know  that <math>HM=AP</math>, we can find for <math>HM</math> which is simpler to find.
Now we know  that <math>HM=AP</math>, we can find for <math>HM</math> which is simpler to find.


We can use point <math>B</math> to split it up as <math>HM=HB+BM</math>,
We can use point <math>B</math> to split it up as <math>HM=HB+BM</math>,


We can chase those lengths and we would get
We can chase those lengths and we would get


<math>AB=10</math>, so <math>OB=5</math>, so <math>BC=5\sqrt{2}</math>, so <math>BM=\dfrac{1}{2} \cdot BC=\dfrac{5\sqrt{2}}{2}</math>
<math>AB=10</math>, so <math>OB=5</math>, so <math>BC=5\sqrt{2}</math>, so <math>BM=\dfrac{1}{2} \cdot BC=\dfrac{5\sqrt{2}}{2}</math>


Then using right triangle <math>AHB</math>, we have HB=10 sin (15∘)
Then using right triangle <math>AHB</math>, we have HB=10 sin (15∘)


So HB=10 sin (15∘)=<math>\dfrac{5(\sqrt{6}-\sqrt{2})}{2}</math>.
So HB=10 sin (15∘)=<math>\dfrac{5(\sqrt{6}-\sqrt{2})}{2}</math>.


And we know that <math>AP = HM = HB + BM = \frac{5(\sqrt6-\sqrt2)}{2} + \frac{5\sqrt2}{2} = \frac{5\sqrt6}{2}</math>.
And we know that <math>AP = HM = HB + BM = \frac{5(\sqrt6-\sqrt2)}{2} + \frac{5\sqrt2}{2} = \frac{5\sqrt6}{2}</math>.


Finally if we calculate <math>(AP)^2</math>.
Finally if we calculate <math>(AP)^2</math>.


<math>(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}</math>. So our final answer is <math>75+2=77</math>.  
<math>(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}</math>. So our final answer is <math>75+2=77</math>.  


<math>m+n=\boxed{77}</math>
<math>m+n=\boxed{77}</math>


Thank you.
Thank you.


-Gamjawon
-Gamjawon

Revision as of 05:48, 31 March 2014

PROBLEM

$\boxed{14}$ 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 $m+n$.

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

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_14&oldid=61302"