AMC 10 · 2021 · #9
Grade 8 arithmeticProblem
The point P(a,b) in the xy-plane is first rotated counterclockwise by 90∘ 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 ?
Pick an answer.
AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Start with an unknown point $P(a, b)$. First rotate it counterclockwise by $90^\circ$ about the point $(1, 5)$; then reflect the result about the line $y = -x$. The final image is $(-6, 3)$. Find $b - a$.
Givens: Original point $P = (a, b)$ — unknown; Transformation 1: rotate $90^\circ$ counterclockwise about $C = (1, 5)$; Transformation 2: reflect about line $y = -x$ (sends $(x, y) \to (-y, -x)$); Final image after both transformations: $(-6, 3)$; Answer choices: (A) $1$, (B) $3$, (C) $5$, (D) $7$, (E) $9$
Unknowns: The value $b - a$
Understand
Restated: Start with an unknown point $P(a, b)$. First rotate it counterclockwise by $90^\circ$ about the point $(1, 5)$; then reflect the result about the line $y = -x$. The final image is $(-6, 3)$. Find $b - a$.
Givens: Original point $P = (a, b)$ — unknown; Transformation 1: rotate $90^\circ$ counterclockwise about $C = (1, 5)$; Transformation 2: reflect about line $y = -x$ (sends $(x, y) \to (-y, -x)$); Final image after both transformations: $(-6, 3)$; Answer choices: (A) $1$, (B) $3$, (C) $5$, (D) $7$, (E) $9$
Plan
Primary tool: #11 Work Backwards
Secondary: #1 Draw a Diagram, #7 Identify Subproblems, #3 Eliminate Possibilities
Tool #11 (Work Backwards): the problem hands us the end state and asks for the start — undo the reflection first, then undo the rotation. Tool #7 (Subproblems): each transformation is its own self-contained inverse. Tool #1 (Diagram): sketch the four key points (final image, post-rotation image, rotation center $C$, and $P$) on a quick coordinate grid to keep signs straight. Tool #3 (Eliminate): the answer is one of $\{1, 3, 5, 7, 9\}$ — all odd — so a parity sanity-check (if $a, b$ are integers, $b - a$ inherits the right parity) is built in.
Execute — Answer: D
8.G.A.1 Step 1 - Undo the reflection about $y = -x$.
- Reflection sends $(x, y) \mapsto (-y, -x)$; this reflection is its own inverse, so applying it again to $(-6, 3)$ undoes it.
- Apply the rule: $(-6, 3) \mapsto (-3, 6)$.
- So just after the rotation (before the reflection) the point was $(-3, 6)$.
💡 Grade 8 reflections: reflecting twice over the same line returns the original.
8.G.A.3 Step 2 - Now undo the $90^\circ$ CCW rotation about $C = (1, 5)$ — that is, rotate $(-3, 6)$ by $90^\circ$ CW about $C$.
- Step 1: translate so $C$ sits at the origin.
- Subtract $C$ from the point: $(-3, 6) - (1, 5) = (-4, 1)$.
💡 Grade 8 translation: shift the center to the origin so a simple rotation rule applies.
8.G.A.1 Step 3 - Apply the $90^\circ$ CW rotation about the origin.
- The rule for CW rotation is $(x, y) \mapsto (y, -x)$.
- Apply to $(-4, 1)$: $(-4, 1) \mapsto (1, 4)$.
💡 Grade 8 rotation: $90^\circ$ CW sends $(x, y) \to (y, -x)$.
8.G.A.3 Step 4 - Translate back: add $C = (1, 5)$ to $(1, 4)$.
- Result: $(2, 9)$.
- So the original point is $P = (a, b) = (2, 9)$, giving $a = 2$ and $b = 9$.
💡 Grade 8 translation: undo the shift to land back in the original coordinate system.
4.NBT.B.4 Step 5 - Compute $b - a = 9 - 2 = 7$.
- Match to the choices: $7$ is choice (D).
💡 Grade 4 subtraction: $9 - 2 = 7$, exactly choice (D).
8.G.A.1 Undo the reflection about $y = -x$. Reflection sends $(x, y) \mapsto (-y, -x)$; 8.G.A.3 Now undo the $90^\circ$ CCW rotation about $C = (1, 5)$ — that is, rotate $(-3, 8.G.A.1 Apply the $90^\circ$ CW rotation about the origin. The rule for CW rotation is $ 8.G.A.3 Translate back: add $C = (1, 5)$ to $(1, 4)$. Result: $(2, 9)$. So the original 4.NBT.B.4 Compute $b - a = 9 - 2 = 7$. Match to the choices: $7$ is choice (D). Review
Reasonableness: Run the transformations forward on $P = (2, 9)$ to confirm. Translate by $-C$: $(2, 9) - (1, 5) = (1, 4)$. Rotate $90^\circ$ CCW about origin (rule $(x, y) \to (-y, x)$): $(1, 4) \to (-4, 1)$. Translate back by $+C$: $(-4, 1) + (1, 5) = (-3, 6)$. Reflect over $y = -x$: $(-3, 6) \to (-6, 3)$. Final image $(-6, 3)$ matches the given. $b - a = 7$ is the right answer.
Alternative: Tool #6 (Guess and Check) on the five answer choices: the answer is one of $\{1, 3, 5, 7, 9\}$. The two transformations preserve distance from the rotation center and then swap signs/coordinates — both are isometries — so the final image $(-6, 3)$ has the same distance from $C' = (-5, -1)$ (the image of $C$ under the reflection) as $P$ has from $C$. A coordinate-bashing forward check on each candidate's $(a, b)$ pair eliminates four of them and confirms (D). Less elegant than working backwards but completely mechanical.
CCSS standards used (min grade 8)
8.G.A.1Verify experimentally the properties of rotations, reflections, and translations (Knowing reflection-over-$y=-x$ rule $(x,y)\to(-y,-x)$ and $90^\circ$ CW rotation rule $(x,y)\to(y,-x)$.)8.G.A.3Describe the effect of dilations, translations, rotations, and reflections on coordinates (Translating to the rotation center, rotating, then translating back to recover the pre-image.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Computing $9 - 2 = 7$ and the coordinate shifts $\pm(1, 5)$.)
⭐ This AMC 10 problem only needs Grade 8 transformation rules you already know — reflect over $y = -x$ (a self-inverse) to undo step 2, then translate to center, $90^\circ$ CW rotate, translate back to undo step 1. Starting from $(-6, 3)$, the chain gives $P = (2, 9)$, so $b - a = 9 - 2 = 7$, choice (D).
⭐ This AMC 10 problem only needs Grade 8 transformation rules you already know — reflect over $y = -x$ (a self-inverse) to undo step 2, then translate to center, $90^\circ$ CW rotate, translate back to undo step 1. Starting from $(-6, 3)$, the chain gives $P = (2, 9)$, so $b - a = 9 - 2 = 7$, choice (D).
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