2005 AMC 10B Problem 23

Let $AB = a$ (the longer base) and $DC = c$ (the shorter base). We need two facts about the segment $EF$ that joins the midpoints of the two legs:
- $EF$ is parallel to the bases and halfway between them, so it cuts the height $h$ into two halves of $\frac{h}{2}$.
- Its length is the average of the bases: $EF = \frac{a + c}{2}$. (For example, with bases $10$ and $2$, the middle segment is $6$.)

1. Areas. The area of a trapezoid is $\text{height} \times \frac{\text{sum of bases}}{2}$. Both pieces have height $\frac{h}{2}$:
$$\text{Area}(ABEF) = \frac{h}{2} \cdot \frac{a + EF}{2}, \qquad \text{Area}(FECD) = \frac{h}{2} \cdot \frac{EF + c}{2}.$$
2. Use the condition. The top area is twice the bottom area. Both have the same factor $\frac{h}{2} \cdot \frac{1}{2}$ in front, so it cancels:
$$a + EF = 2(EF + c), \quad\text{so}\quad a = EF + 2c.$$
3. Put in $EF$.
$$a = \frac{a + c}{2} + 2c.$$
Multiply both sides by $2$: $2a = a + c + 4c$, so $a = 5c$.
4. Ratio. $\frac{AB}{DC} = \frac{a}{c} = 5$.

Check with $c = 1$, $a = 5$: then $EF = 3$, the top piece has base sum $5 + 3 = 8$ and the bottom has $3 + 1 = 4$. Since they share a height, the top area is twice the bottom. It works.

The answer is $\boxed{\textbf{(C)}\ 5}$.

The midpoint segment of a trapezoid is the one fact that turns this picture into algebra: it gives the new base length and splits the height evenly. With equal heights, comparing areas is the same as comparing base sums. Checking with actual numbers at the end confirms the answer quickly.

Test the answer choices. Let $DC = 1$ and try $AB = x$ from the choices. Then $EF = \frac{x+1}{2}$ and we need the top base sum to be twice the bottom one. For $x = 2$: $EF = 1.5$, sums $3.5$ and $2.5$, not double. For $x = 5$: $EF = 3$, sums $8$ and $4$, exactly double. So the answer is $5$.