1958 AMC 12 Problem 50

Question

In this diagram a scheme is indicated for associating all the points of segment

AB

with those of segment

A'B',

and reciprocally. To describe this association scheme analytically, let

x

be the distance from a point

P

on

AB

to

D

and let

y

be the distance from the associated point

P'

of

A'B'

to

D'.

Then for any pair of associated points, if

x=a,

x+y

equals:

13a

17a-51

17-3a

\dfrac{17-3a}{4}

12a-34

The perspective lines in the diagram associate

x=3

with

y=5

and

x=4

with

y=1

Because the two numbered segments are parallel, the induced relation between

x

and

y

is linear

The two numbered segments are parallel, so projection through the fixed intersection point gives a linear relation between

x

and

y.

The endpoint associations shown are

(x,y)=(3,5)

and

(x,y)=(4,1).

The slope is

\frac{1-5}{4-3}=-4,

so

y=-4x+17.

If

x=a,

then

\begin{aligned} x+y &=a+(-4a+17)\\ &=17-3a. \end{aligned}

Therefore, the correct answer is C.

Problem 50 in other years