Question

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

$\frac{1}{\frac{1}{3}\left(\frac{1}{4} + \frac{1}{4} + \frac{1}{5}\right)} = \frac{30}{7}.$

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

$\begin{gathered} \prod_{k=1}^{2025}(kx^2 - 4x - 3) \\ {}= (x^2 - 4x - 3) \\ \quad {}\cdot (2x^2 - 4x - 3) \\ \quad {}\cdot (3x^2 - 4x - 3)\cdots \\ \quad {}\cdot (2025x^2 - 4x - 3)? \end{gathered}$

$-\dfrac{5}{3}$

$-\dfrac{3}{2}$

$-\dfrac{6}{5}$

$-\dfrac{5}{6}$

$-\dfrac{2}{3}$

Answer: B

Concepts:Vieta’s Formulas quadratic harmonic mean

Difficulty rating: 1840

Solution:

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

2025 AMC 10A Problem 18

Answer: B

Difficulty rating: 1840

Solution:

Problem 18 in Other Years