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2025 AMC 10A Problem 18

Problem 18 of 25IntermediateAlgebra

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{gathered} \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 {}\cdot (2025x^2 - 4x - 3)? \end{gathered}

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Solution

Look at one factor kx24x3.kx^2 - 4x - 3. Its discriminant 16+12k16 + 12k is positive, so it has two distinct real roots. Roots from different factors are also distinct: a common root xx for indices k1k_1 and k2k_2 would satisfy (k1k2)x2=0,(k_1-k_2)x^2=0, but x0.x\ne0. By Vieta, the two reciprocals from one factor sum to x1+x2x1x2=4k3k=43.\frac{x_1 + x_2}{x_1 x_2} = \frac{\frac{4}{k}}{-\frac{3}{k}} = -\tfrac{4}{3}. Summing over all 20252025 factors, the reciprocals total 2025(43)=2700.2025 \cdot \left(-\tfrac{4}{3}\right) = -2700. There are 40504050 real roots in all, so the harmonic mean is 40502700=32.\frac{4050}{-2700} = -\tfrac{3}{2}. Therefore, the answer is B.

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Concepts: Vieta’s Formulas · quadratic · harmonic mean

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.