Question

从区间 $[-20, 18]$ 中随机均匀选取一个实数 $a$ 。多项式 $\begin{aligned} &x^4 + 2ax^3 + (2a - 2)x^2 \\ &\quad {}+ (-4a + 3)x - 2 \end{aligned}$ 的根全为实数的概率可写成 $\frac{m}{n}$ ，其中 $m$ 与 $n$ 是互质正整数。求 $m + n$ 。

A real number $a$ is chosen randomly and uniformly from the interval $[-20, 18].$ The probability that the roots of the polynomial $\begin{aligned} &x^4 + 2ax^3 + (2a - 2)x^2 \\ &\quad {}+ (-4a + 3)x - 2 \end{aligned}$ are all real can be written in the form $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：37

知识点：多项式因式分解二次方程几何概率

难度评级：2510

解答：

按是否含有 $a$ ：分组： $\begin{aligned} &(x^4 - 2x^2 + 3x - 2) \\ &\quad {}+ 2a(x^3 + x^2 - 2x) \\ &= (x - 1)(x + 2)(x^2 - x + 1) \\ &\quad {}+ 2ax(x - 1)(x + 2), \end{aligned}$ 所以该多项式分解为 $(x - 1)(x + 2)$ $\cdot \left(x^2 + (2a - 1)x + 1\right)$ 。

四个根全为实数，恰好当二次因式有实根，即 $(2a - 1)^2 - 4 \ge 0$ ，也就是 $a \le -\frac{1}{2}$ 或 $a \ge \frac{3}{2}$ 。在 $[-20, 18]$ 中被排除的区间 $\left(-\frac{1}{2}, \frac{3}{2}\right)$ 长度为 $2$ ，整个区间长度为 $38$ ，所以概率为 $\frac{36}{38} = \frac{18}{19}$ 。所求和为 $18 + 19 = 37$ 。

Group the terms by whether they involve $a:$ $\begin{aligned} &(x^4 - 2x^2 + 3x - 2) \\ &\quad {}+ 2a(x^3 + x^2 - 2x) \\ &= (x - 1)(x + 2)(x^2 - x + 1) \\ &\quad {}+ 2ax(x - 1)(x + 2), \end{aligned}$ so the polynomial factors as $(x - 1)(x + 2)$ $\cdot \left(x^2 + (2a - 1)x + 1\right).$

All four roots are real exactly when the quadratic factor has real roots, i.e. when $(2a - 1)^2 - 4 \ge 0,$ which means $a \le -\frac{1}{2}$ or $a \ge \frac{3}{2}.$ The excluded interval $\left(-\frac{1}{2}, \frac{3}{2}\right)$ has length $2$ inside $[-20, 18],$ which has length $38,$ so the probability is $\frac{36}{38} = \frac{18}{19}.$ The requested sum is $18 + 19 = 37.$

2018 AIME II 第 6 题

答案：37

难度评级：2510

解答：

其他年份的第 6 题