1966 AMC 12 Problem 32

Question

Let

M

be the midpoint of side

AB

of triangle

ABC.

Let

P

be a point on

AB

between

A

and

M,

and let

MD

be drawn parallel to

PC

and intersecting

BC

at

D.

If the ratio of the area of triangle

BPD

to that of triangle

ABC

is denoted by

r,

then:

\dfrac12\lt r\lt1

depending upon the position of

P

r=\dfrac12

independent of the position of

P

\dfrac12\leq r\lt1

depending upon the position of

P

\dfrac13\lt r\lt\dfrac23

depending upon the position of

P

r=\dfrac13

independent of the position of

P

Answer: B

Concepts:area ratio median (geometry)parallel lines

Difficulty rating: 1650

Small hint:

Triangles

MDP

and

MDC

have equal bases on parallel lines

Big hint:

Add their areas to

[BMD]

and use that

CM

is a median

Solution:

Since

MD\parallel PC,

triangles

MDP

and

MDC

have the same base

MD

and equal altitudes, so their areas are equal. Hence

\begin{aligned} [BPD] &=[BMD]+[MDP]\\ &=[BMD]+[MDC]\\ &=[BMC]. \end{aligned}

Because

CM

is a median,

[BMC]=\frac{[ABC]}{2}.

Thus

r=\frac{1}{2}

for every allowed

P.

Therefore, the correct answer is B.

1966 AMC 12 Problem 32

Answer: B

Difficulty rating: 1650

Small hint:

Big hint:

Solution:

Problem 32 in other years