Difference between revisions of "2014 AIME II Problems/Problem 14"
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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>. | ||
| + | |||
| + | <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> | ||
Revision as of 21:45, 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 
. 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∘)=
.
And we know that 
.
Finally if we calculate 
.
. So our final answer is 
.
