2004 AIME II Problems/Problem 13
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
.
![[asy] pointpen = black; pathpen = black+linewidth(0.7); pair D=(0,0), E=(15,0), F=IP(CR(D, 75/7), CR(E, 45/7)), A=D+ (5+(75/7))/(75/7) * (F-D), C = E+ (3+(45/7))/(45/7) * (F-E), B=IP(CR(A,3), CR(C,5)); D(MP("A",A,(1,0))--MP("B",B,N)--MP("C",C,NW)--MP("D",D)--MP("E",E)--cycle); D(D--A--C--E); D(MP("F",F)); MP("3",(B+C)/2,NW); MP("5",(A+B)/2,NE); MP("15",(D+E)/2); [/asy]](http://latex.artofproblemsolving.com/d/2/3/d23b2e8101396634464f7a743d499f001c678203.png)
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
.
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 • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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