2025 AMC 12A Problem 12

Attempt Problem 12 of the 2025 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 12A solutions, or check the answer key.

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12.

The harmonic mean of a collection of numbers is the reciprocal of the arithmetic mean of the reciprocals of the numbers in the collection. For example, the harmonic mean of 4,4,4, 4, and 55 is 113(14+14+15)=307.\frac{1}{\frac{1}{3}\left(\frac{1}{4} + \frac{1}{4} + \frac{1}{5}\right)} = \frac{30}{7}. What is the harmonic mean of all the real roots of the 40504050th degree polynomial k=12025(kx24x3)=(x24x3)(2x24x3)(3x24x3)(2025x24x3)? \begin{aligned} &\small \prod_{k=1}^{2025}(kx^2 - 4x - 3) \\ &= (x^2 - 4x - 3) \\ &\quad {}\cdot (2x^2 - 4x - 3) \\ &\quad {}\cdot (3x^2 - 4x - 3)\cdots \\ &\quad (2025x^2 - 4x - 3)? \end{aligned}

53-\dfrac{5}{3}

32-\dfrac{3}{2}

65-\dfrac{6}{5}

56-\dfrac{5}{6}

23-\dfrac{2}{3}

Answer: B
Concepts:Vieta’s Formulasquadraticharmonic mean
Difficulty rating: 1630
Solution:

Each factor kx24x3kx^2 - 4x - 3 has discriminant 16+12k>0,16 + 12k \gt 0, so it has two real roots; there are 40504050 roots in all.

For the roots of kx24x3,kx^2 - 4x - 3, the sum of reciprocals is sumproduct=4/k3/k=43,\dfrac{\text{sum}}{\text{product}} = \dfrac{4/k}{-3/k} = -\dfrac{4}{3}, independent of k.k.

Summing over all 20252025 factors, 1r=2025(43)=2700.\displaystyle\sum \frac{1}{r} = 2025\left(-\frac{4}{3}\right) = -2700. The harmonic mean is 40502700=32.\frac{4050}{-2700} = -\frac{3}{2}.

Thus, the correct answer is B.

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