2025 AMC 12A Problem 12

Below is the professionally curated solution for Problem 12 of the 2025 AMC 12A, 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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Concepts:Vieta’s Formulasquadraticharmonic mean

Difficulty rating: 1630

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)?\prod_{k=1}^{2025}(kx^2 - 4x - 3) = (x^2 - 4x - 3)(2x^2 - 4x - 3)(3x^2 - 4x - 3)\cdots(2025x^2 - 4x - 3)?

53-\dfrac{5}{3}

32-\dfrac{3}{2}

65-\dfrac{6}{5}

56-\dfrac{5}{6}

23-\dfrac{2}{3}

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.

Problem 12 in Other Years