2004 AIME II Problems/Problem 13
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Problem
Let 
be a convex pentagon with 
and 
Given that the ratio between the area of triangle 
and the area of triangle 
is 
where 
and 
are relatively prime positive integers, find 
Solution
Let the intersection of 
and 
be 
. Since 
it follows that 
is a parallelogram, and so 
. Also, as 
, it follows that 
.
By the Law of Cosines, 
. Thus the length similarity ratio between 
and 
is 
.
Let 
and 
be the lengths of the altitudes in 
to 
respectively. Then, the ratio of the areas ![\frac{[ABC]}{[BDE]} = \frac{\frac 12 \cdot h_{ABC} \cdot AC}{\frac 12 \cdot h_{BDE} \cdot DE} = \frac{7}{15} \cdot \frac{h_{A...](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/0/d/6/0d6ce90073589d4df08e47c12af744b44d03a222.gif)
.
However, 
, with all three heights oriented in the same direction. Since , it follows that 
, and from the similarity ratio, 
. Hence 
, and the ratio of the areas is 
. The answer is 
.
See also
| 2004 AIME II (Problems • Resources) | ||
| Preceded by Problem 12 | Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
