2021 AMC 10B Problem 9

P1-10~3 minPrint

Coordinate Geometry Transformations & Symmetry Start from the end state / desired conclusion and reverse

The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$

Show solution without answering

Key insight

Work backward and undo the last move first: reflecting (−6, 3) over y = −x gives (−3, 6). Then turn that point a quarter turn clockwise around (1, 5).

Solution

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

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

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

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

To undo a quarter turn counterclockwise, make a quarter turn clockwise about the same center $(1, 5)$ .
Measure $(-3, 6)$ from the center: it is $4$ to the left and $1$ up, an offset of $(-4, 1)$ .
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 $(1, 4)$ .
Add the center back: $P = (1 + 1, \ 5 + 4) = (2, 9)$ .

Step 3: check forward.

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

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

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(D)}\ 7}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Geometry, Chapter 16: The More Things Change. . . (§16.1, §16.2, §16.3, §16.4)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.4)
Introduction to Algebra, Chapter 8: Graphing Lines