2025 AMC 12A 第 12 题

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{。}

求下面这个

4050

次多项式所有实根的调和平均数：

\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}\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{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}

答案：B

知识点：韦达定理二次方程调和平均数

难度评级：1630

小提示：

调和平均数是

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

，其中

N

是根的个数

The harmonic mean is

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

where

N

is the number of roots

大提示：

对

kx^2 - 4x - 3

，两个根的倒数和为

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

，与

k

无关

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

解答：

每个因式

kx^2 - 4x - 3

的判别式为

16 + 12k \gt 0

，所以它有两个实根；总共有

4050

个根。

对

kx^2 - 4x - 3

的根，倒数之和为

\dfrac{\text{和}}{\text{积}} = \dfrac{\frac{4}{k}}{-\frac{3}{k}} = -\dfrac{4}{3}

，与

k

无关。

对全部

2025

个因式求和，

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

。调和平均数为

\frac{4050}{-2700} = -\frac{3}{2}\text{。}

因此，正确答案是 B。

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.