2025 AMC 12A Problem 12

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{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}

-\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: 1630

Small hint:

The harmonic mean is

\dfrac{N}{\sum \frac{1}{r}},

where

N

is the number of roots

Big hint:

For

kx^2 - 4x - 3,

the two roots have reciprocal-sum

\dfrac{\frac{4}{k}}{-\frac{3}{k}} = -\dfrac{4}{3},

the same for every

k

Solution:

Each factor

kx^2 - 4x - 3

has discriminant

16 + 12k \gt 0,

so it has two real roots; there are

4050

roots in all.

For the roots of

kx^2 - 4x - 3,

the sum of reciprocals is

\dfrac{\text{sum}}{\text{product}} = \dfrac{\frac{4}{k}}{-\frac{3}{k}} = -\dfrac{4}{3},

independent of

k.

Summing over all

2025

factors,

\displaystyle\sum \frac{1}{r} = 2025\left(-\frac{4}{3}\right) = -2700.