1966 AMC 12 Problem 28

Question

Five points

O,

A,

B,

C,

D

are taken in order on a straight line with distances

OA=a,

OB=b,

OC=c,

and

OD=d.

P

is a point on the line between

B

and

C

and such that

AP:PD=BP:PC.

Then

OP

equals:

\dfrac{b^2-bc}{a-b+c-d}

\dfrac{ac-bd}{a-b+c-d}

-\dfrac{bd+ac}{a-b+c-d}

\dfrac{bc+ad}{a+b+c+d}

\dfrac{ac-bd}{a+b+c+d}

Write

p=OP.

Then

AP=p-a,

PD=d-p,

BP=p-b,

and

PC=c-p

Cross-multiply the two given ratios after making those substitutions

Let

p=OP.

The given ratio becomes

\frac{p-a}{d-p}=\frac{p-b}{c-p}.

Cross-multiplication cancels the

p^2

terms and yields

p(a-b+c-d)=ac-bd.

Therefore

p=\frac{ac-bd}{a-b+c-d}.

Thus, the correct answer is B.

Problem 28 in other years