Question

设 $f(x)$ 是一个实系数三次多项式，满足 $\begin{aligned} |f(1)| = |f(2)| &= |f(3)| \\ &= |f(5)| = |f(6)| \\ &= |f(7)| = 12. \end{aligned}$ 求 $|f(0)|$ 。

Let $f(x)$ be a third-degree polynomial with real coefficients satisfying $\begin{aligned} |f(1)| = |f(2)| &= |f(3)| \\ &= |f(5)| = |f(6)| \\ &= |f(7)| = 12. \end{aligned}$ Find $|f(0)|.$

答案

答案：72

知识点：多项式韦达定理方程组

难度评级：2930

解答：

$f(x) - 12$ 与 $f(x) + 12$ 都是三次多项式，所以它们各自在 $1, 2, 3, 5, 6, 7$ 中恰好三个点取零。把它们写成 $c(x - r_1)(x - r_2)(x - r_3)$ 与 $c(x - s_1)(x - s_2)(x - s_3)$ ，这两个三次多项式相差常数 $24$ ，所以它们的 $x^2$ 系数和 $x$ 系数相同：两组三个根有相同的和与相同的两两乘积和。把 $\{1,2,3,5,6,7\}$ 分成两个和相等的三元组，唯一方式是 $\{2,3,7\}$ 与 $\{1,5,6\}$ （和都为 $12$ ），且两者的两两乘积和确实都为 $41$ 。

若有必要，把 $f$ 换成 $-f$ （这不改变 $|f(0)|$ ），则 $f(x) = c(x-2)(x-3)(x-7)$ $+ 12$ $= c(x-1)(x-5)(x-6) - 12$ 。令 $x = 0$ ，得到 $-42c + 12 = -30c - 12$ ，所以 $c = 2$ ，且 $f(0) = -42 \cdot 2 + 12 = -72$ 。因此 $|f(0)| = 72$ 。

Each of $f(x) - 12$ and $f(x) + 12$ is a cubic, so each vanishes at exactly three of $1, 2, 3, 5, 6, 7.$ Writing them as $c(x - r_1)(x - r_2)(x - r_3)$ and $c(x - s_1)(x - s_2)(x - s_3),$ the two cubics differ by the constant $24,$ so their $x^2$ and $x$ coefficients agree: the root triples have equal sums and equal sums of pairwise products. The only partition of $\{1,2,3,5,6,7\}$ into two triples of equal sum is $\{2,3,7\}$ and $\{1,5,6\}$ (each summing to $12$ ), and indeed both have pairwise-product sum $41.$

Replacing $f$ by $-f$ if necessary (which does not change $|f(0)|$ ), we have $f(x) = c(x-2)(x-3)(x-7)$ $+ 12$ $= c(x-1)(x-5)(x-6) - 12.$ Setting $x = 0$ gives $-42c + 12 = -30c - 12,$ so $c = 2$ and $f(0) = -42 \cdot 2 + 12 = -72.$ Thus $|f(0)| = 72.$

2015 AIME I 第 10 题

答案：72

难度评级：2930

解答：

其他年份的第 10 题