2016 AMC 10B Problem 20

A dilation is set by two things: its scale factor and its center. Find both.

1. Scale factor. The radius grows from $2$ to $3$, so every distance from the center is multiplied by $k = \frac32$.
2. How a point moves. The center $Z$, a point $P$ and its image $P'$ lie on one line, and $P'$ is $\frac32$ times as far from $Z$ as $P$ is. So the point moves away from the center by half its distance from the center. In coordinates:
$$\text{movement of } P = \tfrac12 \times (P - Z).$$
3. Find the center. $A = (2,2)$ moves to $(5,6)$, a movement of $(3, 4)$. That must be half of $A - Z$, so $A - Z = (6, 8)$. Then
$$Z = (2 - 6,\ 2 - 8) = (-4, -6).$$
4. Move the origin. $O - Z = (0 - (-4),\ 0 - (-6)) = (4, 6)$. Half of that is $(2, 3)$. So $O$ moves to $(2, 3)$.
5. The distance moved is
$$\sqrt{2^2 + 3^2} = \sqrt{13}.$$

Check: $Z + \frac32(A - Z) = (-4, -6) + \frac32(6, 8) = (-4 + 9,\ -6 + 12) = (5, 6)$, which is $A'$.

The answer is $\boxed{\textbf{(C)}\ \sqrt{13}}$.

The two radii give the scale factor for free, and one point with its image then locates the center. Once you know both, every point's movement follows one rule, which is why the origin moves a definite distance even though it is not on either circle.

Skip the center. The movement of a point $P$ is $\frac12 P$ plus some fixed shift $c$ (the same $c$ for every point). For $A$: $(3, 4) = \frac12(2, 2) + c = (1, 1) + c$, so $c = (2, 3)$. For the origin, $\frac12 O = (0, 0)$, so it moves by exactly $c = (2, 3)$, a distance of $\sqrt{13}$.

The radii grow from $2$ to $3$, so the scale factor is $k = \tfrac32$. A dilation with center $Z$ sends any point $P$ to
$$P' = Z + k(P - Z), \qquad\text{so}\qquad P' - P = (k - 1)(P - Z) = \tfrac12 (P - Z).$$

Find $Z$ from the known pair $A \mapsto A'$: the displacement of $A$ is $A' - A = (3, 4)$, and it must equal $\tfrac12(A - Z)$. Hence $A - Z = (6, 8)$ and $Z = (2,2) - (6,8) = (-4, -6)$.

Now the origin: its displacement is $\tfrac12(O - Z) = \tfrac12(4, 6) = (2, 3)$, so $O$ moves to $(2, 3)$, a distance of
$$\sqrt{2^2 + 3^2} = \sqrt{13}.$$

The answer is $\boxed{\textbf{(C)}\ \sqrt{13}}$.