2001 AMC 12 Problem 22

Question

In rectangle

ABCD,

points

F

and

G

lie on

\overline{AB}

so that

AF = FG = GB

and

E

is the midpoint of

\overline{DC}.

Also,

\overline{AC}

intersects

\overline{EF}

at

H

and

\overline{EG}

at

J.

The area of rectangle

ABCD

is

70.

Find the area of triangle

EHJ.

\dfrac{5}{2}

\dfrac{35}{12}

3

\dfrac{7}{2}

\dfrac{35}{8}

Triangle

EFG

has base

\dfrac{1}{3}AB

and the full height, so its area is

\dfrac{1}{6}

of the rectangle

Use

EC \parallel AB

to get

\dfrac{EH}{EF} = \dfrac{3}{5}

and

\dfrac{EJ}{EG} = \dfrac{3}{7}

from similar triangles

Triangle

EFG

has base

FG = \dfrac{1}{3}AB

and height equal to the rectangle’s height, so its area is

\dfrac{1}{6}(70) = \dfrac{35}{3}.

Because

EC \parallel AF,

triangles

AFH

and

CEH

are similar with ratio

\dfrac{EC}{AF} = \dfrac{3}{2},

so

\dfrac{EH}{EF} = \dfrac{3}{5}.

Likewise

\dfrac{EJ}{EG} = \dfrac{3}{7}.

Then

\dfrac{[EHJ]}{[EFG]}

= \dfrac{EH}{EF}\cdot\dfrac{EJ}{EG}

= \dfrac{3}{5}\cdot\dfrac{3}{7}

= \dfrac{9}{35},

giving

[EHJ] = \dfrac{9}{35}\cdot\dfrac{35}{3} = 3.

Thus, the correct answer is C.

Problem 22 in other years