2005 AMC 10B Problem 23

Problem 23 of 25HarderGeometry

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}?

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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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Concepts: trapezoid · area ratio · midpoint

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.