Competition · AMC preparation · step 4 of 4

AMC 10 · 2022A · #18

Grade 8 arithmetic
rotation-isometryreflection-symmetrypattern-recognition easier-related-problempattern-recognitionconvert-to-algebra ↑ Prerequisites: rotation-isometry
📏 Long solution 💡 3 insights
Problem
Let T_k rotate the plane k° counterclockwise about the origin and then reflect across the y-axis. Find the least positive integer n such that applying T₁, T₂, T₃, …, T_n in order returns the point (1, 0) to itself.

Pick an answer.

(A)
359
(B)
360
(C)
719
(D)
720
(E)
721

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Solve an Easier Related Problem

Since only the angle matters (radius is preserved), track θ_n alone. Tool #9 (Easier Problem): compute θ₁, θ₂, θ₃, θ₄ explicitly using the rule that reflection across the y-axis sends angle φ to 180° - φ. Tool #5 (Look for a Pattern): the odd-index and even-index angles fall into two clean arithmetic sequences. Tool #13 (Convert to Algebra): set the closed-form expressions equal to a multiple of 360° and solve for the smallest valid n. Tool #3 (Eliminate): compare the two cases and pick the smaller.

1STEP 1

Write one step as angles

Rotation adds k° to the angle; reflection across the y-axis sends φ to 180°-φ. Composing gives T_k: θ → 180°-θ-k°.

θ_n = 180° - θ_n-1 - n°
2STEP 2

Compute the first few angles

Run the recurrence from θ₀=0°: θ₁=179°, θ₂=-1°, θ₃=178°, θ₄=-2°. Compute first, then guess.

θ₁ = 179°, θ₂ = -1°, θ₃ = 178°, θ₄ = -2°, θ₅ = 177°, θ₆ = -3°
3STEP 3

Split odd and even indices

Split by parity: odd steps give θ₂k-1=180°-k, even steps give θ₂k=-k.

θ₂k-1 = 180° - k°, θ₂k = -k°
4STEP 4

Verify the pattern

Induction check: assuming θ₂k=-k, the recurrence reproduces both formulas at 2k+1 and 2k+2, so they hold.

θ₂k+1 = 180 - (k+1), θ₂k+2 = -(k+1) — confirmed.
5STEP 5

Solve the odd case

Odd case n=2k-1: need 180°-k ≡ 0 (mod 360°), so k=180, giving n=359.

n_odd = 359
6STEP 6

Solve the even case

Even case n=2k: need -k ≡ 0 (mod 360°), so k=360, giving n=720.

n_even = 720
7STEP 7

Take the smaller answer

Compare the candidates: 359 ＜ 720, so the least n is 359, choice (A).

min(359, 720) = 359 → (A)
Answer
359
Direct check at n = 359: this is odd with k = 180, so θ₃₅₉ = 180° - 180° = 0° — point lands exactly back on (1, 0). Direct check that no smaller n works: for any odd n = 2k - 1 with 1 ≤ k ≤ 179 we have θ_n = 180 - k ∈ {1°, …, 179°}, none of which is a multiple of 360°. For any even n = 2k with 1 ≤ k ≤ 359 we have θ_n = -k° ∈ {-1°, …, -359°}, none of which is a multiple of 360°. So n = 359 is indeed the minimum.
💡Key takeaway

Only the angle moves — the radius stays 1. After running T_k a few times you see two clean rules: odd steps land at 180° - k, even steps land at -k. Solving "when is this a multiple of 360°?" gives n = 359 (odd) and n = 720 (even), so the smallest is (A) 359.

  • Write one step as angles
  • Compute the first few angles
  • Split odd and even indices
  • Verify the pattern
  • Solve the odd case
  • Solve the even case
  • Take the smaller answer

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