2018 AIME II 第 6 题

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

小提示：

寻找对所有

a

都成立的根：

x = 1

与

x = -2

始终都是根

Look for roots that work for every

a:

both

x = 1

and

x = -2

are always roots

大提示：

提出因式

(x - 1)(x + 2)

后，二次式

x^2 + (2a - 1)x + 1

有实根恰好当

(2a - 1)^2 \ge 4

After factoring out

(x - 1)(x + 2),

the quadratic

x^2 + (2a - 1)x + 1

has real roots exactly when

(2a - 1)^2 \ge 4

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}\text{，}

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

小提示：

大提示：

解答：

其他年份的第 6 题