AMC 8 · 2022 · #4

Grade 8 geometry-2d

Archetype Paper Fold-and-Cut Symmetry └ Sub-archetype Composition of Reflections of a Pictured Figure

reflection-symmetryspatial-visualizationcoordinate-geometry physical-representationreflection-unfolding ↑ Prerequisites: reflection-symmetryline-symmetry

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

An upright letter M sits in the first quadrant of a coordinate plane. Line p is the horizontal axis, and line q is the diagonal y = x. Reflect the M first across line q, then reflect the result across line p. Pick the choice whose picture shows the final image — both the location and the orientation of the letter must match.

Pick an answer.

(A)

(diagram) M rotated 90° clockwise (lying on its right side)

(B)

(diagram) M rotated 270° clockwise (lying on its left side)

(C)

(diagram) M rotated 90° clockwise (alternate orientation)

(D)

(diagram) M rotated 180° (upside-down W)

(E)

(diagram) M rotated 270° clockwise (alternate orientation)

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Create a Physical Representation

Two reflections are easy to mis-visualize in the head, especially over the diagonal y = x. Tool #10 (Physical) wins here: cut a small upright M out of paper, draw lines p and q on a sheet, and physically flip the cutout over q, then flip the result over p. That removes all guesswork about orientation. Tool #17 (Visualize) is the same move done mentally for the student who's ready. Tool #3 (Eliminate) is the multiple-choice safety net — track the final quadrant first and three of the five options drop out immediately. Tool #1 (Diagram) keeps a clean coordinate sketch of where each intermediate image lands. After the picture work, a one-line check via the "composition of two reflections through intersecting lines = single rotation" theorem confirms the answer.

1STEP 1

Sketch it: mark p as the x-axis, q as y = x, and draw the upright M in the first quadrant with its peaks pointing up (+y).

M₀ centered near (0.25, 0.6), peaks point +y

2STEP 2

Reflect over q (y = x): swapping coordinates sends the center from (0.25, 0.6) to (0.6, 0.25) — still first quadrant, now lower-right.

(x, y) → (y, x): (0.25, 0.6) → (0.6, 0.25)

3STEP 3

Track orientation: the up-arrow (0, +1) maps to (+1, 0), so the M lies on its side with peaks pointing right — that is rotate(270°) M.

peaks direction: (0, +1) → (+1, 0); orientation = rotate(270°) M

4STEP 4

Reflect over p (x-axis): only y flips sign, so (0.6, 0.25) moves to (0.6, -0.25) — leaving quadrant I for the fourth quadrant.

(x, y) → (x, -y): (0.6, 0.25) → (0.6, -0.25)

5STEP 5

The x-axis flip keeps x, so peaks still point right — orientation stays rotate(270°) M, now in quadrant IV; only (E) matches both.

Final: rotate(270°) M in Quadrant IV → (E)

Answer

(diagram) M rotated 270° clockwise (alternate orientation)

Quick orientation sanity check on the five choices: (A) rotate(90°) M in Q4 — wrong orientation. (B) rotate(270°) M in Q2 — right orientation, wrong quadrant. (C) rotate(90°) M in Q1 — wrong orientation and wrong quadrant. (D) rotate(180°) M in Q3 — wrong orientation. (E) rotate(270°) M in Q4 — matches both the orientation derived in Steps 3 and 5 and the Q4 position derived in Step 4. The answer is internally consistent: the original M in Q1 ends up in the diagonally-opposite-by-one-quadrant Q4, which is exactly where a single 90° clockwise rotation about the origin would send it.

💡Key takeaway

This AMC 8 problem only needs the Grade 8 rule for what reflections do to coordinates — flip the M over y=x (swap x and y), then over the x-axis (flip the sign of y), and you land on (E)!

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