2014 AIME II Problems/Problem 14

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 . Point is the midpoint of the segment , and point is on ray such that PN⊥BC. 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 ∠HAB=15∘.

is .

and are parallel lines so is also.

Then if we use those informations we get and

and or

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

, so , so , so

Then using right triangle , we have HB=10 sin (15∘)

So HB=10 sin (15∘)=$\dfrac{5(\sqrt{6}−\sqrt{2})}{2}$ (Error compiling LaTeX. Unknown error_msg).