2023 AMC 10A Problem 19

1. What does a rotation keep? When you spin a shape around a point $P$, every point stays the same distance from $P$. So $P$ is as far from $A$ as from $A'$, and as far from $B$ as from $B'$.
2. Where are such points? The points that are the same distance from two given points form a line: the perpendicular bisector (it goes through the midpoint, at a right angle). So $P$ is on the perpendicular bisector of $BB'$ and on the perpendicular bisector of $AA'$.
3. Start with the easy pair, $B$ and $B'$. $B = (3, 3)$ and $B' = (4, 3)$ are side by side on the line $y = 3$. The midpoint is $\left(\frac72, 3\right)$, and the line crossing at a right angle is straight up and down:
$$x = \tfrac72.$$
4. Now $A = (1, 2)$ and $A' = (3, 1)$. The midpoint is $\left(2, \frac32\right)$. Going from $A$ to $A'$ is $2$ right and $1$ down, so the slope is $-\frac12$. A perpendicular line has slope $2$ (flip it and change the sign). So the bisector is
$$y = \tfrac32 + 2(x - 2).$$
5. Where do they cross? Put $x = \frac72$ in: $y = \frac32 + 2 \times \frac32 = \frac32 + 3 = \frac92$. So $P = \left(\frac72, \frac92\right)$.
6. Check one pair of distances (squared): from $P$ to $A$, $\left(\frac52\right)^2 + \left(\frac52\right)^2 = \frac{25}{2}$; from $P$ to $A'$, $\left(\frac12\right)^2 + \left(\frac72\right)^2 = \frac{1}{4} + \frac{49}{4} = \frac{25}{2}$. ✓
7. Finally $|r - s| = \left|\frac72 - \frac92\right| = 1$.

The answer is $\boxed{\textbf{(E)}\ 1}$.

The center of a rotation is the one point that does not move, and every other point keeps its distance from it. So each point and its image give one line the center must lie on, and two such lines pin the center down. You never need to know the angle of the turn.

Write the equal distances as equations with $P = (r, s)$. From $PB^2 = PB'^2$:
$$(r-3)^2 + (s-3)^2 = (r-4)^2 + (s-3)^2.$$
The $s$ parts are the same on both sides, and the $r^2$ terms cancel: $-6r + 9 = -8r + 16$, so $r = \frac72$. Doing the same with $PA^2 = PA'^2$ gives $4r - 2s = 5$, so $s = \frac92$, and $|r - s| = 1$.

A rotation about $P$ keeps every point at the same distance from $P$. So $PA = PA'$, meaning $P$ lies on the perpendicular bisector of $AA'$, and likewise $P$ lies on the perpendicular bisector of $BB'$.

Bisector of $BB'$: $B(3,3)$ and $B'(4,3)$ are on a horizontal line with midpoint $\left(\frac72, 3\right)$, so the bisector is the vertical line $x = \frac72$.

Bisector of $AA'$: $A(1,2)$ and $A'(3,1)$ have midpoint $\left(2, \frac32\right)$ and slope $-\frac12$, so the bisector has slope $2$:
$$y - \tfrac32 = 2(x - 2).$$
At $x = \frac72$ this gives $y = \frac32 + 3 = \frac92$.

So $P = \left(\frac72, \frac92\right)$ and $|r - s| = \left|\frac72 - \frac92\right| = 1$.

Check: $PA^2 = \left(\frac52\right)^2 + \left(\frac52\right)^2 = \frac{25}{2}$ and $PA'^2 = \left(\frac12\right)^2 + \left(\frac72\right)^2 = \frac{25}{2}$.

The answer is $\boxed{\textbf{(E)}\ 1}$.