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:

$13a$

$17a-51$

$17-3a$

$\dfrac{17-3a}{4}$

$12a-34$

Answer: C

Concepts:similarity linear equation transformation

Difficulty rating: 1830

Small Hint:

The perspective lines in the diagram associate $x=3$ with $y=5$ and $x=4$ with $y=1$

Big Hint:

Because the two numbered segments are parallel, the induced relation between $x$ and $y$ is linear

Solution:

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.

1958 AMC 12 Problem 50

Answer: C

Difficulty rating: 1830

Small Hint:

Big Hint:

Solution:

Problem 50 in Other Years