1963 AMC 12 Problem 38

Question

Point

F

is taken on the extension of side

AD

of parallelogram

ABCD.

BF

intersects diagonal

AC

at

E

and side

DC

at

G.

If

EF=32

and

GF=24,

then

BE

equals:

4

8

10

12

16

Parameterize points on

FB

by their fraction of the distance from

F

to

B

If

F=tD

in affine coordinates based at

A,

the parameters of

G

and

E

are

\frac{t-1}{t}

and

\frac{t}{t+1}

Use affine coordinates

A=0,

B=u,

D=v,

C=u+v,

and

F=tv.

A point on

FB

is

su+t(1-s)v.

Intersecting

DC

gives

s_G=\frac{t-1}{t},

while intersecting

AC

gives

s_E=\frac{t}{t+1}.

Therefore

\begin{aligned} \frac{GF}{EF} &=\frac{s_G}{s_E} =\frac{t^2-1}{t^2}\\ &=\frac{24}{32}=\frac34, \end{aligned}

so

t=2

and

s_E=\frac{2}{3}.

Thus

EF=(\frac{2}{3})FB=32,

making

FB=48

and

BE=16.

Therefore, the correct answer is E.

Problem 38 in other years