Question

A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A'(5,6).$ What distance does the origin $O(0,0),$ move under this transformation?

$\ 0$

$\ 3$

$\ \sqrt{13}$

$\ 4$

$\ 5$

Solution:

The dilation scale factor is $\frac32$ , since the radius changes from $2$ to $3$ . The center of dilation $C$ lies on the line through $A(2,2)$ and $A'(5,6)$ .

The vector from $A$ to $A'$ is $(3,4)$ . Since $CA:CA'=1:\frac32$ , the vector from $C$ to $A$ is twice the vector from $A'$ back to $A$ , so $C=(2,2)-2(3,4)=(-4,-6).$

Under a scale factor $\frac32$ dilation about $C$ , a point moves by half its distance from $C$ . Since $CO=\sqrt{(-4)^2+(-6)^2}=\sqrt{52}$ , the origin moves $\frac12\sqrt{52}=\sqrt{13}$ .

Thus, the correct answer is C.

2016 AMC 10B Problem 20

Solution:

Problem 20 in Other Years