2009 AMC 12A Problem 20

Attempt Problem 20 of the 2009 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12A solutions, or check the answer key.

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20.

Convex quadrilateral

ABCD

has

AB = 9

and

CD = 12.

Diagonals

AC

and

BD

intersect at

E,

AC = 14,

and

\triangle AED

and

\triangle BEC

have equal areas. What is

AE?

\dfrac{9}{2}

\dfrac{50}{11}

\dfrac{21}{4}

\dfrac{17}{3}

6

Answer: E

Concepts:parallel lines similarity area ratio

Difficulty rating: 1930

Small hint:

Equal areas of

\triangle AED

and

\triangle BEC

imply

\triangle ACD

and

\triangle BCD

have equal areas

Big hint:

That forces

AB \parallel CD,

making

\triangle ABE \sim \triangle CDE

with ratio

\dfrac{9}{12}

Solution:

Adding

\triangle CED

to each of

\triangle AED

and

\triangle BEC

shows

\triangle ACD

and

\triangle BCD

have equal areas. They share base

CD,

so

A

and

B

are equidistant from line

CD,

meaning

AB \parallel CD.

Then

\triangle ABE \sim \triangle CDE

with ratio

\dfrac{AB}{CD} = \dfrac{9}{12} = \dfrac{3}{4},

so

\dfrac{AE}{EC} = \dfrac{3}{4}.

Writing

AE = 3x

and

EC = 4x,

we get

7x = AC = 14,

so

x = 2

and

AE = 6.

Thus, the correct answer is E.

Problem 20 in other years