2005 AMC 10B Problem 23

Question

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

The midsegment

\overline{FE}

has length

\dfrac{AB+DC}{2}

The two smaller trapezoids share a height, so their areas compare like the averages of their parallel sides

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.

Problem 23 in other years