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2025 AMC 12A Problem 12

Problem 12 of 25IntermediateAlgebraProbability & Statistics

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(kx2−4x−3)=(x2−4x−3)⋅(2x2−4x−3)⋅(3x2−4x−3)⋯(2025x2−4x−3)? \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}

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Solution

Each factor kx2−4x−3kx^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 kx2−4x−3,kx^2 - 4x - 3, the sum of reciprocals is sumproduct=4k−3k=−43,\dfrac{\text{sum}}{\text{product}} = \dfrac{\frac{4}{k}}{-\frac{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 4050−2700=−32.\frac{4050}{-2700} = -\frac{3}{2}. Thus, the correct answer is B.
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Tagged: Vieta’s Formulas · quadratic · harmonic mean

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