AMC 10 · 2025 · #12

Grade 11 algebra
Archetype Pair-Sum Invariant └ Sub-archetype Uniform Pair-Sum Recovery
harmonic-meanvieta-formulasquadratic-equationspolynomial-roots identify-subproblems ↑ Prerequisites: vieta-formulasquadratic-equations
📏 Medium solution 💡 2 insights
Problem
A degree-4050 polynomial is written as a product of 2025 quadratics, one for each k from 1 to 2025: the k-th factor is k x² - 4x - 3. Every factor shares the same middle coefficient -4 and the same constant -3; only the leading coefficient k changes. Find the harmonic mean of all the real roots, where the harmonic mean of n numbers is n divided by the sum of their reciprocals.

Pick an answer.

(A)
$-\frac{5}{3}$
(B)
$-\frac{3}{2}$
(C)
$-\frac{6}{5}$
(D)
$-\frac{5}{6}$
(E)
$-\frac{2}{3}$
How to solve
Strategy Identify Subproblems

Expanding a degree-4050 polynomial is impossible by hand, so the winning move is to notice it is already factored: a product is zero exactly when one factor is zero, so its roots are just the roots of the 2025 separate quadratics. That breaks one monstrous problem into 2025 tiny identical-looking ones. Because the harmonic mean only needs the sum of reciprocals of the roots, and Vieta's formulas give that sum for each quadratic without solving it, every quadratic turns out to contribute the same fixed amount. Handling one quadratic then handles all of them at once.

1STEP 1

Split the product into its quadratic factors

A product is zero when one factor is zero, so the roots are just the roots of the 2025 quadratics k x² - 4x - 3 — no expanding needed.

Π_k=1²⁰²⁵(k x² - 4x - 3) = 0 ⇔ k x² - 4x - 3 = 0 for some k
2STEP 2

Check each quadratic has two real roots

The discriminant 16 + 12k stays positive, so every quadratic has 2 real roots: 2025 times 2 = 4050 real roots, exactly the degree.

Δ_k = (-4)² - 4(k)(-3) = 16 + 12k ＞ 0 (k ≥ 1)
3STEP 3

Aim at the sum of reciprocals, not the roots

Finding all 4050 roots is needless: the harmonic mean only wants the total sum of reciprocals, so aim straight at that single sum.

HM = 4050/(Σ_i=1⁴⁰⁵⁰ 1/r_i)
4STEP 4

Each quadratic contributes the same -43\frac{4}{3}

Vieta gives 1p\frac{1}{p} + 1q\frac{1}{q} = p+qpq\frac{p+q}{pq} = -bc\frac{b}{c}; the leading coefficient cancels, so each quadratic contributes -bc\frac{b}{c} = -43\frac{4}{3}.

1/p + 1/q = (p+q)/pq = -b/a/c/a = -b/c = --4/-3 = -4/3
5STEP 5

Add up the pieces and divide

All 2025 pieces give -43\frac{4}{3}, so the reciprocal sum is 2025 times -43\frac{4}{3} = -2700; then 4050 divided by -2700 reduces to -32\frac{3}{2}, choice (B).

Σ_i=1⁴⁰⁵⁰ 1/r_i = 2025 · (-4/3) = -2700, HM = 4050/-2700 = -3/2
Answer
-3/2
The harmonic mean landed at exactly -32\frac{3}{2}, matching the value a single quadratic k x² - 4x - 3 gives on its own, which is a strong consistency check: whenever every group in a collection has the same harmonic mean, the whole collection has that harmonic mean too. The sign is also sensible. Each pair of roots satisfies 1p\frac{1}{p} + 1q\frac{1}{q} = -43\frac{4}{3} ＜ 0, so the reciprocals lean negative overall, forcing a negative harmonic mean, and only choices with negative values appear. The magnitude 32\frac{3}{2} sits comfortably among the offered options.
💡Key takeaway

When a huge polynomial is already a product, its roots are just the roots of the factors, and the harmonic mean only needs the sum of reciprocals, which Vieta's formulas hand you without ever solving for a single root.

  • Split the product into its quadratic factors
  • Check each quadratic has two real roots
  • Aim at the sum of reciprocals, not the roots
  • Each quadratic contributes the same -43\frac{4}{3}
  • Add up the pieces and divide

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