2014 AIME II Problems/Problem 14
In , and . Let and be points on the line such that , , and . Point is the midpoint of the segment , and point is on ray such that . Then , where and are relatively prime positive integers. Find .
http://www.artofproblemsolving.com/Wiki/images/5/59/AOPS_wiki.PNG ( This is the diagram.)
As we can see,
is the midpoint of and is the midpoint of
is a triangle, so .
and are parallel lines so is triangle also.
Then if we use those informations we get and
Now we know that , we can find for which is simpler to find.
We can use point to split it up as ,
We can chase those lengths and we would get
, so , so , so
Then using right triangle , we have HB=10 sin (15∘)
So HB=10 sin (15∘)=.
And we know that .
Finally if we calculate .
. So our final answer is .
|2014 AIME II (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|