1992 AMC 12 Problem 24

Question

Let

ABCD

be a parallelogram of area

10

with

AB=3

and

BC=5.

Locate

E,

F,

and

G

on segments

\overline{AB},

\overline{BC},

and

\overline{AD},

respectively, with

AE=BF=AG=2.

Let the line through

G

parallel to

\overline{EF}

intersect

\overline{CD}

at

H.

The area of the quadrilateral

EFHG

is

4

4.5

5

5.5

6

Use

AB

and

AD

as an affine coordinate basis

In that basis,

E=(\frac{2}{3},0),

F=(1,\frac{2}{5}),

and

G=(0,\frac{2}{5})

; use the parallel condition to locate

H

Use

AB

and

AD

as basis vectors. Then

\begin{aligned} E&=\left(\frac23,0\right),\\ F&=\left(1,\frac25\right),\\ G&=\left(0,\frac25\right). \end{aligned}

A line from

G

parallel to

EF

meets

CD

at

H=(\frac{1}{2},1).

The shoelace formula gives the coordinate area of

EFHG

as

\frac{1}{2}.

Affine coordinates scale all areas by the area

10

of

ABCD,

so

[EFHG]=\frac12\cdot10=5.

Thus the correct answer is C.

Problem 24 in other years