1974 AMC 12 Problem 25

Question

In parallelogram

ABCD

of the accompanying diagram, line

DP

is drawn bisecting

BC

at

N

and meeting

AB

(extended) at

P.

From vertex

C,

line

CQ

is drawn bisecting side

AD

at

M

and meeting

AB

(extended) at

Q.

Lines

DP

and

CQ

meet at

O.

If the area of parallelogram

ABCD

is

k,

then the area of triangle

QPO

is equal to

k

\frac{6k}{5}

\frac{9k}{8}

\frac{5k}{4}

2k

Answer: C

Concepts:parallelogram coordinate geometry triangle area

Difficulty rating: 2100

Small hint:

Use

A=(0,0),

B=(1,0),

D=(u,v),

and

C=(u+1,v)

Big hint:

The midpoint lines meet the baseline at

Q=(-1,0)

and

P=(2,0)

; then find

O

’s height

Solution:

Choose coordinates

A=(0,0),

B=(1,0),

D=(u,v),

and

C=(u+1,v),

so

k=v.

The midpoints are

\begin{aligned} M&=(\frac{u}{2},\frac{v}{2}),\\ N&=(1+\frac{u}{2},\frac{v}{2}). \end{aligned}

Extending

CM

and

DN

to

y=0

gives

Q=(-1,0)

and

P=(2,0).

Parameterizing

DP

and

CQ

shows their intersection is

O=(\frac{3u}{4}+\frac{1}{2},\frac{3v}{4}).

Thus

QP=3

and the height from

O

is

\frac{3v}{4},

so

\begin{aligned} [QPO]&=\frac12(3)\left(\frac{3v}{4}\right)\\ &=\frac{9v}{8}=\frac{9k}{8}. \end{aligned}

Therefore, the correct answer is C.