# 2004 AIME II Problems/Problem 13

## Problem

Let $ABCDE$ be a convex pentagon with $AB \parallel CE, BC \parallel AD, AC \parallel DE, \angle ABC=120^\circ, AB=3, BC=5,$ and $DE = 15.$ Given that the ratio between the area of triangle $ABC$ and the area of triangle $EBD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

## Solution

Let the intersection of $\overline{AD}$ and $\overline{CE}$ be $F$. Since $AB \parallel CE, BC \parallel AD,$ it follows that $ABCF$ is a parallelogram, and so $\triangle ABC \cong \triangle CFA$. Also, as $AC \parallel DE$, it follows that $\triangle ABC \sim \triangle EFD$.

$[asy] pointpen = black; pathpen = black+linewidth(0.7); pair D=(0,0), E=(15,0), F=IP(CR(D, 75/7), CR(E, 45/7)), A=D+ (5+(75/7))/(75/7) * (F-D), C = E+ (3+(45/7))/(45/7) * (F-E), B=IP(CR(A,3), CR(C,5)); D(MP("A",A,(1,0))--MP("B",B,N)--MP("C",C,NW)--MP("D",D)--MP("E",E)--cycle); D(D--A--C--E); D(MP("F",F)); MP("5",(B+C)/2,NW); MP("3",(A+B)/2,NE); MP("15",(D+E)/2); [/asy]$

By the Law of Cosines, $AC^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cos 120^{\circ} = 49 \Longrightarrow AC = 7$. Thus the length similarity ratio between $\triangle ABC$ and $\triangle EFD$ is $\frac{AC}{ED} = \frac{7}{15}$.

Let $h_{ABC}$ and $h_{BDE}$ be the lengths of the altitudes in $\triangle ABC, \triangle BDE$ to $AC, DE$ respectively. Then, the ratio of the areas $\frac{[ABC]}{[BDE]} = \frac{\frac 12 \cdot h_{ABC} \cdot AC}{\frac 12 \cdot h_{BDE} \cdot DE} = \frac{7}{15} \cdot \frac{h_{ABC}}{h_{BDE}}$.

However, $h_{BDE} = h_{ABC} + h_{CAF} + h_{EFD}$, with all three heights oriented in the same direction. Since $\triangle ABC \cong \triangle CFA$, it follows that $h_{ABC} = h_{CAF}$, and from the similarity ratio, $h_{EFD} = \frac{15}{7}h_{ABC}$. Hence $\frac{h_{ABC}}{h_{BDE}} = \frac{h_{ABC}}{2h_{ABC} + \frac {15}7h_{ABC}} = \frac{7}{29}$, and the ratio of the areas is $\frac{7}{15} \cdot \frac 7{29} = \frac{49}{435}$. The answer is $m+n = \boxed{484}$.