AMC 10 · 2004 · #23

Grade 11 algebra
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
complex-numberspolynomial-rootsvieta-formulas contradiction-elementaryeasier-related-problem ↑ Prerequisites: parity
📏 Long solution 💡 3 insights
Problem
A degree-2004 polynomial has real coefficients and 2004 distinct complex roots. One root is 0, and the real parts add up to the same total as the imaginary parts. Four of the five listed quantities are forced to equal 0. Decide which one is not forced.

Pick an answer.

(A)
$c_0$
(B)
$c_{2003}$
(C)
$b_2b_3...b_{2004}$
(D)
$\sum_{k = 1}^{2004}{a_k}$
(E)
$\sum_{k = 1}^{2004}{c_k}$
How to solve
Strategy Eliminate Possibilities

The question has two halves, and both need doing. First, show four of the choices are pinned to 0 — each follows from one structural fact (a root at 0, conjugate symmetry, Vieta's relation between the root sum and (c₂₀₀₃), and a parity count on the real roots). Second, and this is the half that is easy to skip: prove the survivor really can be nonzero by building an explicit polynomial that meets every condition and still has (P(1) ≠ 0). Eliminating four choices only tells you where to look; it does not prove the fifth is achievable. A small degree-4 model of the same setup gives a cheap way to sanity-check all five predictions at once.

1STEP 1

Conjugate pairs kill both sums

Real coefficients pair the roots by conjugation, so the imaginary parts cancel and both sums are 0.

Σ_k=1²⁰⁰⁴ b_k = 0 ⟹ Σ_k=1²⁰⁰⁴ a_k = 0 ⟹ Σ_k=1²⁰⁰⁴ z_k = 0
2STEP 2

A root at 0 kills the constant term

Substituting zero and one identifies two of the listed quantities as values of the polynomial.

c₀ = P(0) = 0, Σ_k=1²⁰⁰⁴ c_k = P(1)
3STEP 3

Vieta pins the next coefficient

Vieta ties the next coefficient to the root sum, which is zero, so it vanishes too.

P(x) = c₂₀₀₄Π_k=1²⁰⁰⁴(x - z_k) ⟹ c₂₀₀₃ = -c₂₀₀₄Σ_k=1²⁰⁰⁴ z_k = 0
4STEP 4

Parity forces a second real root

A parity count forces a second real root, so the product contains a zero factor.

#{nonreal roots} is even ⟹ #{real roots} is even ⟹ ∃ i ≥ 2, b_i = 0 ⟹ b₂b₃… b₂₀₀₄ = 0
5STEP 5

Build a polynomial with nonzero coefficient sum

An explicit polynomial makes the coefficient sum nonzero, so the answer is the sum of the coefficients, choice (E).

P(x) = x(x-2)(x-3)…(x-2003) (x + 2007005), P(1) = 2002! · 2007006 ≠ 0
Answer
Σ_k = 1²⁰⁰⁴{c_k}
Shrink the problem and test all five claims at once. Degree 4 with the same conditions: roots (0, 2, 3, -5) — distinct, real, containing 0, summing to 0. Then (P(x) = x(x-2)(x-3)(x+5) = x⁴ - 19x² + 30x). Read off the predictions: (c₀ = 0) (A holds), (c₃ = 0) (B holds), (Σ a_k = 0) and (Σ b_k = 0) (D holds), one of (b₂, b₃, b₄) is 0 so the product is 0 (C holds), and (Σ c_k = 1 + 0 - 19 + 30 + 0 = 12 = P(1) ≠ 0). Four forced zeros and one escapee, exactly as claimed. It is worth stating what the elimination alone does not establish: the fact that no proof of (Σ c_k = 0) came to hand is not a proof that it can be nonzero. Only the explicit construction settles that.
💡Key takeaway

Ruling out four choices only points at the fifth; to claim a quantity can be nonzero you have to build one example that actually is.

  • Conjugate pairs kill both sums
  • A root at 0 kills the constant term
  • Vieta pins the next coefficient
  • Parity forces a second real root
  • Build a polynomial with nonzero coefficient sum

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