# 2005 AMC 10B Problems/Problem 23

## Problem

In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC}$, $E$ as the midpoint of $\overline{BC}$, and $F$ as the midpoint of $\overline{DA}$. The area of $ABEF$ is twice the area of $FECD$. What is $AB/DC$? $\mathrm{(A)} 2 \qquad \mathrm{(B)} 3 \qquad \mathrm{(C)} 5 \qquad \mathrm{(D)} 6 \qquad \mathrm{(E)} 8$

## Solution 1

Since the height of both trapezoids are equal, and the area of $ABEF$ is twice the area of $FECD$, $\frac{AB+EF}{2}=2\left(\frac{DC+EF}{2}\right)$. $\frac{AB+EF}{2}=DC+EF$, so $AB+EF=2DC+2EF$. $EF$ is exactly halfway between $AB$ and $DC$, so $EF=\frac{AB+DC}{2}$. $AB+\frac{AB+DC}{2}=2DC+AB+DC$, so $\frac{3}{2}AB+\frac{1}{2}DC=3DC+AB$, and $\frac{1}{2}AB=\frac{5}{2}DC$. $\frac{AB}{DC} = \boxed{5}$.

## Solution 2

Mark $DC=z$, $AB=x$, and $FE=y.$ Note that the heights of trapezoids $ABEF$ & $FECD$ are the same. Mark the height to be $h$.

Then, we have that $\dfrac{x+y}{2}\cdot h=2(\dfrac{y+z}{2} \cdot h)$.

From this, we get that $x=2z+y$.

We also get that $\dfrac{x+z}{2} \cdot h= 3(\dfrac{y+z}{2} \cdot h)$.

Simplifying, we get that $2x=z+3y$

Notice that we want $\dfrac{AB}{DC}=\frac{x}{z}$.

Dividing the first equation by $z$, we get that $\dfrac{x}{z}=2+\dfrac{y}{z}\implies 3(\dfrac{x}{z})=6+3(\dfrac{y}{z})$.

Dividing the second equation by $z$, we get that $2(\dfrac{x}{z})=1+3(\dfrac{y}{z})$.

Now, when we subtract the top equation from the bottom, we get that $\dfrac{x}{z}=5$

Hence, the answer is $\boxed{5}$

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