2025 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 18 Easy

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, and 5 is

\frac { 1 } { \frac { 1 } { 3 } \left( \frac { 1 } { 4 } + \frac { 1 } { 4 } + \frac { 1 } { 5 } \right) } = \frac { 3 0 } { 7 } .

What is the harmonic mean of all the real roots of the 4050th degree polynomial

\prod _ { k = 1 } ^ { 2 0 2 5 } ( k x ^ { 2 } - 4 x - 3 ) = ( x ^ { 2 } - 4 x - 3 ) ( 2 x ^ { 2 } - 4 x - 3 ) ( 3 x ^ { 2 } - 4 x - 3 ) \cdots ( 2 0 2 5 x ^ { 2 } - 4 x - 3 ) ?

  • A.

    -\frac { 5 } { 3 }

  • B.

    -\frac { 3 } { 2 }

  • C.

    - \frac { 6 } { 5 }

  • D.

    - \frac { 5 } { 6 }

  • E.

    -\frac { 2 } { 3 }

Answer:B

Let x_k^{(1)} and x_k^{(2)} denote the roots of kx^2 - 4x - 3.

By Vieta's formulas: \frac{1}{x_k^{(1)}} + \frac{1}{x_k^{(2)}} = \frac{x_k^{(1)} + x_k^{(2)}}{x_k^{(1)} \cdot x_k^{(2)}} = \frac{\frac{4}{k}}{-\frac{3}{k}} = -\frac{4}{3}

Summing over all values: \sum_{k=1}^{2025}\left(\frac{1}{x_k^{(1)}} + \frac{1}{x_k^{(2)}}\right) = -\frac{4}{3} \times 2025

Computing the required expression: \frac{4050}{\sum_{k=1}^{2025}\left(\frac{1}{x_k^{(1)}} + \frac{1}{x_k^{(2)}}\right)} = \frac{4050}{-\frac{4}{3} \times 2025} = -\frac{3}{2}.

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