Question

一组数的调和平均数，定义为这些数的倒数的算术平均数的倒数。例如， $4$ 、 $4$ 和 $5$ 的调和平均数为

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

下面这个 $4050$ 次多项式所有实根的调和平均数是多少？

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

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

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

$-\dfrac{5}{3}$

$-\dfrac{3}{2}$

$-\dfrac{6}{5}$

$-\dfrac{5}{6}$

$-\dfrac{2}{3}$

答案：B

知识点：韦达定理二次方程调和平均数

难度评级：1840

小提示：

对每个因式 $kx^2 - 4x - 3$ ，由韦达定理可得根和为 $\tfrac{4}{k}$ ，根积为 $-\tfrac{3}{k}$ ，所以倒数和是二者的商。

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

大提示：

调和平均数等于根的个数除以所有倒数之和。

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

解答：

看其中一个因式 $kx^2 - 4x - 3$ 。它的判别式 $16 + 12k$ 为正，所以它有两个不同的实根。来自不同因式的根也互不相同：若 $x$ 是下标为 $k_1$ 和 $k_2$ 的两个因式的公共根，就会有 $(k_1-k_2)x^2=0$ ，但 $x\ne0$ 。由韦达定理，同一个因式的两个根的倒数之和为 $\frac{x_1 + x_2}{x_1 x_2} = \frac{\frac{4}{k}}{-\frac{3}{k}} = -\tfrac{4}{3}$ 。对全部 $2025$ 个因式求和，这些倒数的总和为 $2025 \cdot \left(-\tfrac{4}{3}\right) = -2700$ 。实根共有 $4050$ 个，所以调和平均数为 $\frac{4050}{-2700} = -\tfrac{3}{2}$ 。因此正确答案是 B。

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.

2025 AMC 10A 第 18 题

答案：B

难度评级：1840

小提示：

大提示：

解答：

其他年份的第 18 题