AMC 10 · 2022 · #18

Grade 8 geometry-2d

Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Periodic Placement Formula

rotation-isometryreflection-symmetrypattern-recognition easier-related-problempattern-recognitionconvert-to-algebra ↑ Prerequisites: rotation-isometry

📏 Long solution 💡 3 insights

Problem

A transformation rotates the plane some number of degrees counterclockwise about the origin and then reflects across the vertical axis. Such transformations are applied in order, with the angle growing one degree at a time. Find the least positive index at which a point returns to itself.

Pick an answer.

(A)

359

(B)

360

(C)

719

(D)

720

(E)

721

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 the one-step rule

The angle follows one rule.

θ_n = 180° - θ_n-1 - n°

2STEP 2

Compute the first steps

Compute the first six steps.

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

3STEP 3

Separate odd and even

The angle alternates between two families.

θ₂k-1 = 180° - k°, θ₂k = -k°

4STEP 4

Confirm the pattern

Check the pattern one step further.

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

5STEP 5

Solve the odd family

The odd family gives one answer.

n_odd = 359

6STEP 6

Solve the even family

The even family's answer is far larger.

n_even = 720

7STEP 7

Take the smaller

The smaller is 359.

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.