# 2008 AMC 8 Problems/Problem 23

## Problem

In square $ABCE$, $AF=2FE$ and $CD=2DE$. What is the ratio of the area of $\triangle BFD$ to the area of square $ABCE$? $[asy] size((100)); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((3,0)--(9,9)--(0,3)--cycle); dot((3,0)); dot((0,3)); dot((9,9)); dot((0,0)); dot((9,0)); dot((0,9)); label("A", (0,9), NW); label("B", (9,9), NE); label("C", (9,0), SE); label("D", (3,0), S); label("E", (0,0), SW); label("F", (0,3), W); [/asy]$

$\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{2}{9}\qquad\textbf{(C)}\ \frac{5}{18}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{7}{20}$

## Solution

The area of $\triangle BFD$ is the area of square $ABCE$ subtracted by the the area of the three triangles around it. Arbitrarily assign the side length of the square to be $6$.

$[asy] size((100)); pair A=(0,9), B=(9,9), C=(9,0), D=(3,0), E=(0,0), F=(0,3); pair[] ps={A,B,C,D,E,F}; dot(ps); draw(A--B--C--E--cycle); draw(B--F--D--cycle); label("A",A, NW); label("B",B, NE); label("C",C, SE); label("D",D, S); label("E",E, SW); label("F",F, W); label("6",A--B,N); label("6",(10,4.5),E); label("4",D--C,S); label("2",E--D,S); label("2",E--F,W); label("4",F--A,W); [/asy]$

The ratio of the area of $\triangle BFD$ to the area of $ABCE$ is

$$\frac{36-12-12-2}{36} = \frac{10}{36} = \boxed{\textbf{(C)}\ \frac{5}{18}}$$

## Solution 2~Mr.BigBrain_AoPS

Say that $\overline{FE}$ has length $x$, and that from there we can infer that $\overline{AF} = 2x$. We also know that $\overline{ED} = x$, and that $\overline{DC} = 2x$. The area of triangle $BFD$ is the square's area subtracted from the area of the excess triangles, which is simply these equations: $$9x^2 - (3x^2 + \dfrac{x}{2}^2 + 3x^2)$$ $$9x^2 - 6.5x^2$$ $$2.5x^2$$ Thus, the area of the triangle is $2.5x^2$. We can now put the ratio of triangle $BFD$'s area to the area of the square $ABCE$ as a fraction. We have: $$\dfrac{2.5x^2}{9x^2}$$ $$\dfrac{2.5\cancel{x^2}}{9\cancel{x^2}}$$ $$\dfrac{2.5}{9}$$ $$\dfrac{5}{18}$$ Thus, our answer is $\boxed{C}$, $\boxed{\dfrac{5}{18}}$.