AMC12 2025 A

AMC12 2025 A · Q12

AMC12 2025 A · Q12. It mainly tests Vieta / quadratic relationships (basic), Fractions.

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 \[\frac{1}{\frac{1}{3}(\frac{1}{4}+\frac{1}{4}+\frac{1}{5})}=\frac{30}{7}.\] What is the harmonic mean of all the real roots of the $4050$th degree polynomial \[\prod_{k=1}^{2025} (kx^2-4x-3)=(x^2-4x-3)(2x^2-4x-3)(3x^2-4x-3)...(2025x^2-4x-3)?\]

一个数列的调和平均数是该数列倒数的算术平均数的倒数。例如，$4,4$ 和 $5$ 的调和平均数为 \[\frac{1}{\frac{1}{3}(\frac{1}{4}+\frac{1}{4}+\frac{1}{5})}=\frac{30}{7}.\] 下列 $4050$ 次多项式 \[\prod_{k=1}^{2025} (kx^2-4x-3)=(x^2-4x-3)(2x^2-4x-3)(3x^2-4x-3)\dots(2025x^2-4x-3)?\] 的所有实根的调和平均数是多少？

(A) -\frac{5}{3} -\frac{5}{3}

(B) -\frac{3}{2} -\frac{3}{2}

(D) -\frac{5}{6} -\frac{5}{6}

(E) -\frac{2}{3} -\frac{2}{3}

Answer

Correct choice: (B)

正确答案：（B）

Solution

We will need to determine the sum of the reciprocals of the roots. To find the sum of the reciprocals of the roots $p,q$ of the quadratic $ax^2+bx+c$, we use Vieta's formulas. Recall that $p+q = -b/a$ and $pq = c/a$. Therefore, \[\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{\frac{-b}{a}}{\frac{c}{a}} = \frac{-b}{a} \cdot \frac{a}{c} = \frac{-b}{c},\] which doesn't depend on $a$. The sum of the reciprocals of the roots of the quadratic $x^2-4x-3$ is $\frac{-(-4)}{-3} = -4/3.$ The same is true for every quadratic in the form $ax^2-4x-3$. The sum of all the reciprocals of the roots of $ax^2+bx+c$ is $2025 \cdot \left(-\frac{4}{3}\right).$ Because we have $2025$ quadratics, there are $2 \cdot 2025 = 4050$ total roots. Our answer is $\frac{1}{\frac{1}{4050}\cdot \frac{-4\cdot 2025}{3}} =\boxed{(B)\ -\dfrac{3}{2}}$.

我们需要确定根的倒数之和。对于二次方程 $ax^2+bx+c$ 的根 $p,q$，利用 Vieta 公式，$p+q = -b/a$，$pq = c/a$。因此，\[\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{\frac{-b}{a}}{\frac{c}{a}} = \frac{-b}{c},\] 不依赖于 $a$。对于 $x^2-4x-3$，根的倒数之和为 $\frac{-(-4)}{-3} = -4/3$。对于形式 $ax^2-4x-3$ 的每个二次方程均如此。所有根的倒数之和为 $2025 \cdot \left(-\frac{4}{3}\right)$。有 $2025$ 个二次方程，总共有 $2 \cdot 2025 = 4050$ 个根。调和平均数为 $\frac{1}{\frac{1}{4050}\cdot \frac{-4\cdot 2025}{3}} =\boxed{(B)\ -\dfrac{3}{2}}$。

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