AMC 10 · 2020 · #23

Grade 12 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

complex-numberscomplex-polar-formroots-of-unityequal-spacing caseworkextremal-constructioneasier-related-problem ↑ Prerequisites: complex-numberscomplex-polar-form

📏 Long solution 💡 4 insights

Problem

For which integers at least two is this statement always true: if that many complex numbers all have modulus one and add to zero, then they must be equally spaced around the unit circle? Count how many such integers there are.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Eliminate Possibilities

Tool #1 (Draw a Diagram): unit modulus means "on the circle" and zero sum means "the arrows close into a loop" — the picture drives everything. Tool #16 (Change Focus): the statement is a "for all" claim, so the two directions cost wildly different amounts of work — proving n works needs a real argument, but killing n needs only ONE bad configuration. Hunt for the bad configuration first. Tool #3 (Eliminate Possibilities): walk n = 2, 3, 4, … and sort each one into works / fails. Tool #4 (Introduce a Variable): for the small cases, name the angles and compute |z₁ + z₂| exactly. Tool #6 (Guess and Check): build one explicit four-point counterexample by hand. Tool #5 (Look for a Pattern): notice that any antipodal pair w, -w contributes 0 to the sum, which turns the single counterexample into a recipe for every larger n.

1STEP 1

Turn both conditions into pictures

Modulus one means a point on the unit circle.

z_k = e^iθ_k for each k, and Σ_k=1ⁿ z_k = 0 ⇔ (z₁ + … + z_n)/n = 0

2STEP 2

Pin down equally spaced

Equally spaced means all gaps equal.

equally spaced ⇔ {z₁, …, z_n} = {ζ e^(2π i k)/n}_k=0ⁿ⁻¹, all neighbour gaps = 2π/n

3STEP 3

The case of two

Two must be opposite, so it holds.

z₁ + z₂ = 0 → z₂ = -z₁, gaps π and π

4STEP 4

Measure an angle for three

For three, each pair's angle is forced.

2 + 2 Re (z₁z₂) = 1 ⟹ Re (z₁z₂) = -1/2

5STEP 5

Conclude for three

All three angles agree, so it holds.

cos(θ₁ - θ₂) = -1/2 → θ₁ - θ₂ = ±2π/3, and likewise for the other two pairs

6STEP 6

A counterexample for four

Two opposite pairs are not equally spaced.

1 + (-1) + e^iπ/4 + (-e^iπ/4) = 0, gaps 45°, 135°, 45°, 135° ≠ 90°

7STEP 7

Every even case breaks

The same trick breaks every even case.

Σ_j=1^m(e^iθ_j + (-e^iθ_j)) = 0, with θ_j = jπ/100n, so a gap of π/100n ＜ 2π/n appears

8STEP 8

Every odd case breaks

Add opposite pairs to the triple.

(1 + ω + ω²) + Σ_j=1^k(w_j - w_j) = 0; for odd n, e^(2π i d)/n = -1 is impossible

9STEP 9

Count the survivors

Only 2 survive.

good n = {2, 3} → count = 2 → (B)

Answer

Check the pieces numerically. The n = 3 claim is consistent with the obvious example 1 + ω + ω² = 0 for ω = e^(2π i)/3, and the step-4 identity predicts Re (1 · ω) = cos (-2π/3) = -1/2, which matches. The n = 4 counterexample checks out digit by digit: 1 + (-1) + 0.7071(1 + i) - 0.7071(1 + i) = 0, all four moduli are 1, and the angle gaps 45°, 135°, 45°, 135° are visibly unequal. The n = 5 counterexample 1, ω, ω², i, -i sums to 0 + 0 = 0 and sits at 0°, 120°, 240°, 90°, 270°, gaps 90°, 30°, 120°, 30°, 90° — not the 72° a regular pentagon needs. The count 2 also has to be at least 2, since n = 2 and n = 3 both pass, and the only way to reach the tempting choice (C) = 3 would be to believe n = 4 forces a square, which the explicit counterexample refutes. Finally, the argument is safe under either reading of the problem: the counterexamples use distinct points, so they work whether or not the z_k are required to be distinct.

💡Key takeaway

Two opposite points on the circle always cancel, so once n ≥ 4 you can build a lopsided set that still sums to zero; only n = 2 and n = 3 are too cramped to cheat, so the count is 2.

Turn both conditions into pictures
Pin down "equally spaced" and the cheap direction
n = 2 works
n = 3: measure one pair exactly
n = 3: the angle must be 120°
n = 4 breaks
Every even n ≥ 4 breaks
Every odd n ≥ 5 breaks
Count the survivors