2023 AMC 10A Problem 19

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 $\boxed{\textbf{(E)}\ 1}$.

The center of any rotation is the one point that does not move, and every other point keeps its distance from it. Two point-image pairs therefore give two perpendicular bisectors whose intersection is the center; no knowledge of the rotation angle is needed. The same idea locates the center of rotation in any figure-matching problem.

Write $P = (r, s)$ and set $PA^2 = PA'^2$ and $PB^2 = PB'^2$. Expanding, the squared terms cancel: $-2r - 4s + 5 = -6r - 2s + 10$ gives $4r - 2s = 5$, and $-6r - 6s + 18 = -8r - 6s + 25$ gives $r = \frac72$. Then $s = \frac92$.

• Using the midpoints of AA' or BB' as the center, or computing r - s with the coordinates swapped and forgetting the absolute value.
• Assuming the rotation is $90^\circ$ and applying the formula $(x, y) \to (y, -x)$ around some guessed center.