2021 AMC 10B Problem 9

We know where $P$ ends up, so work backward: undo the last move first, then the first move.

Step 1: undo the reflection over $y = -x$.

1. Reflecting over the line $y = -x$ swaps the two coordinates and changes both signs. For example, $(1, 0)$ goes to $(0, -1)$.
2. Doing the same reflection twice brings a point back, so to undo it, just do it again. $(-6, 3)$ goes to $(-3, 6)$.
3. So after the rotation, $P$ was at $(-3, 6)$.

Step 2: undo the rotation about $(1, 5)$.

1. To undo a quarter turn counterclockwise, make a quarter turn clockwise about the same center $(1, 5)$.
2. Measure $(-3, 6)$ from the center: it is $4$ to the left and $1$ up, an offset of $(-4, 1)$.
3. A quarter turn clockwise changes an offset $(u, v)$ into $(v, -u)$. Check on a simple case: the offset $(0, 1)$ (straight up) turns clockwise to $(1, 0)$ (straight right). So $(-4, 1)$ becomes $(1, 4)$.
4. Add the center back: $P = (1 + 1, \ 5 + 4) = (2, 9)$.

Step 3: check forward.

1. $(2, 9)$ is $(1, 4)$ from the center. A counterclockwise quarter turn changes $(u, v)$ into $(-v, u)$, giving $(-4, 1)$, which is the point $(-3, 6)$.
2. Reflecting over $y = -x$ gives $(-6, 3)$. Correct.

So $a = 2$, $b = 9$, and $b - a = 7$.

The answer is $\boxed{\textbf{(D)}\ 7}$.

To undo several moves, undo them one by one in the opposite order, like taking off shoes before socks. To rotate around a point that is not the origin, measure from that point, rotate the offset, and add the point back. A quick forward check at the end catches sign mistakes.

Work backwards from the final image, undoing the last transformation first.

Undo the reflection. Reflection over $y = -x$ sends $(x, y) \mapsto (-y, -x)$ and is its own inverse. The point that reflects to $(-6, 3)$ is $(-3, 6)$. This is the rotated image of $P$.

Undo the rotation. To undo a $90^\circ$ counterclockwise rotation about $(1,5)$, rotate $90^\circ$ clockwise about $(1,5)$. The offset of $(-3,6)$ from the center is $(-4, 1)$. A clockwise quarter turn sends an offset $(u, v)$ to $(v, -u)$, so $(-4, 1) \mapsto (1, 4)$. Adding the center back: $P = (1 + 1, 5 + 4) = (2, 9)$.

So $a = 2$, $b = 9$, and $b - a = 7$.

Check forward: offset of $(2,9)$ is $(1,4)$; counterclockwise gives $(-4, 1)$, i.e. the point $(-3, 6)$; reflecting over $y=-x$ gives $(-6, 3)$. Correct.

The answer is $\boxed{\textbf{(D)}\ 7}$.