2001 AMC 12 Problem 22

Problem 22 of 25HarderGeometry

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}

H

and

\overline{EG}

J.

The area of rectangle

ABCD

70.

Find the area of triangle

EHJ.

Show solution

Solution

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},

\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.