2005 AMC 10B Problem 23

Below is the professionally curated solution for Problem 23 of the 2005 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 10B solutions, or check the answer key.

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Concepts:trapezoid area ratio midpoint

Difficulty rating: 1630

23.

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 $\dfrac{AB}{DC}?$

$2$

$3$

$5$

$6$

$8$

Solution:

Let $AB = a$ and $DC = c.$ The midsegment $\overline{FE}$ has length $\dfrac{a + c}{2},$ and $ABEF$ and $FECD$ have the same height.

Their areas are proportional to the averages of their parallel sides, so $\dfrac{a + \frac{a+c}{2}}{\frac{a+c}{2} + c} = \dfrac{3a + c}{a + 3c} = 2.$

Then $3a + c = 2a + 6c,$ so $a = 5c$ and $\dfrac{AB}{DC} = 5.$

Thus, C is the correct answer.

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