AMC 10 · 2020 · #17

Grade 11 algebra

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

roots-of-unitycomplex-numberspolynomial-rootsvieta-formulas caseworksystematic-enumerationidentify-subproblems ↑ Prerequisites: complex-numberspolynomial-roots

📏 Long solution 💡 4 insights

Problem

Count the fifth-degree polynomials with real coefficients and constant term 2020 whose set of roots is closed under multiplication by a fixed cube root of one: whenever a number is a root, that number times the cube root is a root too.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Make a Systematic List

Tool #4 (Introduce a Variable): name ω and work out its arithmetic (ω³ = 1, 1 + ω + ω² = 0, ω² = ω) before touching the polynomial. Tool #5 (Pattern): applying the closure rule twice shows roots arrive in triples r, ω r, ω² r. Tool #14 (Extreme Principle): degree 5 is a hard ceiling — it kills the possibility of two separate triples and pins the polynomial down to one triple. Tool #2 (Systematic List): the only freedom left is how to split 5 into three multiplicities, so list those splits and keep the ones with real coefficients. That count is the answer.

1STEP 1

What the cube root does

Cubing it returns to one.

ω² = (-1-i√(3))/2 = ω, ω³ = 1, 1 + ω + ω² = 0

2STEP 2

Roots come in triples

The constant term means zero is not a root.

P(0) = 2020 ≠ 0; r → ω r → ω² r → ω³ r = r

3STEP 3

Only one triple fits

Degree five allows only one triple.

P(x) = (x-r)^m₁(x-ω r)^m₂(x-ω² r)^m₃, m₁ + m₂ + m₃ = 5, m_j ≥ 1

4STEP 4

One root is real

Real coefficients force one real root.

r ∈ {r, ω r, ω² r} → some root t is real; roots = t, ω t, ω² t

5STEP 5

List the multiplicity splits

List the possible splits.

m₁ + 2m = 5 → (m₁, m₂, m₃) = (3,1,1) or (1,2,2)

6STEP 6

Use the cubic identity

The triple product is a clean identity.

(x-t)(x-ω t)(x-ω² t) = x³ - t³, (x-ω t)(x-ω² t) = x² + tx + t²

7STEP 7

Pin the real root and count

The constant term fixes the real root, so 2.

-t⁵ = 2020 → t = -⁵√(2020); count = 2 → (C)

Answer

Check both winners against every requirement. With t = -⁵√(2020) ≈ -4.582, split (3,1,1) gives x⁵ + 9.164x⁴ + 20.996x³ + 96.208x² + 881.680x + 2020 and split (1,2,2) gives x⁵ - 4.582x⁴ + 20.996x³ + 96.208x² - 440.840x + 2020. Both are monic of degree 5, both have real coefficients, both end in 2020, and both root sets are {t, ω t, ω² t}, which multiplication by ω permutes. So 2 works and the answer is not 0 or 1. It cannot exceed 2 because the only free choice was the multiplicity split, and only two splits are conjugate-balanced — so 3 and 4 are out.

💡Key takeaway

Multiplying by ω = (-1+i√(3))/2 loops back to the start after three steps, so roots come in triples t, ω t, ω² t; degree 5 leaves room for only one triple, real coefficients force t real and the other two to share a multiplicity, and 5 = 3+1+1 = 1+2+2 leaves exactly 2 polynomials, choice (C).

Work out what ω does
Roots come in triples
Only one triple fits
One of the three roots is real
List the multiplicity splits
Use the identity x³ - t³
Pin down t and count