2010 AIME I 第 6 题

Question

设

P(x)

是一个实系数二次多项式，满足对所有实数

x

都有

\begin{aligned} x^2 - 2x + 2 &\le P(x) \\ &\le 2x^2 - 4x + 3 \end{aligned}\text{，}

并且

P(11) = 181

。求

P(16)

。

Let

P(x)

be a quadratic polynomial with real coefficients satisfying

\begin{aligned} x^2 - 2x + 2 &\le P(x) \\ &\le 2x^2 - 4x + 3 \end{aligned}

for all real numbers

x,

and suppose

P(11) = 181.

Find

P(16).

答案

答案：406

知识点：二次方程配方法多项式

难度评级：2390

小提示：

配方：两个界限二次式分别等于

(x-1)^2 + 1

和

2(x-1)^2 + 1

，它们有共同顶点

(1, 1)

Complete the square: both bounding quadratics equal

(x-1)^2 + 1

and

2(x-1)^2 + 1,

with common vertex

(1, 1)

大提示：

P(x) - (x-1)^2 - 1

处处非负且在

x = 1

处为零，所以它是

c(x-1)^2

；用

P(11) = 181

求

c

P(x) - (x-1)^2 - 1

is nonnegative everywhere and zero at

x = 1,

so it is

c(x-1)^2;

use

P(11) = 181

to find

c

解答：

配方后，条件为

\begin{aligned} (x-1)^2 + 1 &\le P(x) \\ &\le 2(x-1)^2 + 1 \end{aligned}\text{。}

当

x = 1

时，两个界都等于

1

，所以

P(1) = 1

。二次式

P(x) - \left((x-1)^2 + 1\right)

对所有

x

非负，并在

x = 1

处为零，所以

x = 1

是二重根：

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

，其中

a

为常数。

由

P(11) = 100a + 1 = 181

得

a = \frac{9}{5}

。于是

\begin{aligned} P(16) &= \frac{9}{5} \cdot 225 + 1 \\ &= 405 + 1 = 406 \end{aligned}\text{。}

Completing the square, the condition reads

\begin{aligned} (x-1)^2 + 1 &\le P(x) \\ &\le 2(x-1)^2 + 1. \end{aligned}

At

x = 1

both bounds equal

1,

so

P(1) = 1.

The quadratic

P(x) - \left((x-1)^2 + 1\right)

is nonnegative for all

x

and vanishes at

x = 1,

so

x = 1

is a double root:

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

for some constant

a.

From

P(11) = 100a + 1 = 181

we get

a = \frac{9}{5}.

Then

\begin{aligned} P(16) &= \frac{9}{5} \cdot 225 + 1 \\ &= 405 + 1 = 406. \end{aligned}

2010 AIME I 第 6 题

答案：406

难度评级：2390

小提示：

大提示：

解答：

其他年份的第 6 题