2025 AMC 10A Problem 18

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

Small hint:

For each factor

kx^2 - 4x - 3,

Vieta gives root sum

\tfrac{4}{k}

and product

-\tfrac{3}{k},

so the reciprocal-sum is their quotient

Big hint:

The harmonic mean is the number of roots divided by the total sum of reciprocals

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{\frac{4}{k}}{-\frac{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.