# 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 1

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 