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"
Line 15: Line 15:
 
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>

Revision as of 22:35, 29 March 2014

14. In △ABC, AB=10, ∠A=30∘, and ∠C=45∘. Let H, D, and M be points on the line BC such that AH⊥BC, ∠BAD=∠CAD, and $BM=CM$. Point $N$ is the midpoint of the segment $HM$, and point $P$ is on ray $AD$ such that PN⊥BC. Then $AP^2=\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. 

http://www.artofproblemsolving.com/Wiki/images/5/59/AOPS_wiki.PNG ( This is the diagram.)

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 ∠HAB=15∘.

$AHD$ is $30-60-90$.

$AH$ and $PN$ are parallel lines so $PND$ is $30-60-90$ also.

Then if we use those informations we get $AD=2HD$ and

$PD=2ND$ and $AP=AD-PD=2HD-2ND=2HN$ or $AP=2HN=HM$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_14&oldid=61283"