2004 AMC 12A 第 23 题

Question

多项式

\begin{aligned} &P(x) = c_{2004} x^{2004} + c_{2003} x^{2003} \\ &\quad {}+ \cdots + c_1 x + c_0 \end{aligned}

的系数均为实数，且

c_{2004} \ne 0

。它有

2004

个不同的复数零点

z_k = a_k + b_k i

，其中

1 \le k \le 2004

，

a_k

和

b_k

均为实数，

a_1 = b_1 = 0

，并且

\sum_{k=1}^{2004} a_k = \sum_{k=1}^{2004} b_k\text{。}

下列哪个量可能是非零数？

A polynomial

\begin{aligned} &P(x) = c_{2004} x^{2004} + c_{2003} x^{2003} \\ &\quad {}+ \cdots + c_1 x + c_0 \end{aligned}

has real coefficients with

c_{2004} \ne 0

and

2004

distinct complex zeros

z_k = a_k + b_k i,

1 \le k \le 2004

with

a_k

and

b_k

real,

a_1 = b_1 = 0,

and

\sum_{k=1}^{2004} a_k = \sum_{k=1}^{2004} b_k.

Which of the following quantities can be a nonzero number?

c_0

c_{2003}

b_2 b_3 \ldots b_{2004}

\displaystyle\sum_{k=1}^{2004} a_k

\displaystyle\sum_{k=1}^{2004} c_k

答案：E

知识点：复数韦达定理多项式

难度评级：2350

小提示：

因为

z_1 = 0

是根，所以

c_0 = P(0) = 0

；非实根成共轭对出现。

Since

z_1 = 0

is a root,

c_0 = P(0) = 0;

nonreal roots come in conjugate pairs

大提示：

\sum c_k = P(1)

，它不一定为零；检查其他每个选项都必须为零。

\sum c_k = P(1),

which is not forced to be zero; check each other option must vanish

解答：

因为

z_1 = a_1 + b_1 i = 0

是根，所以

c_0 = P(0) = 0

。

非实零点成共轭对出现，所以

\sum b_k = 0

而题设于是强制

\sum a_k = 0

。系数

c_{2003}

等于

-c_{2004}

乘以根之和

\sum a_k + i \sum b_k = 0

，所以

c_{2003} = 0

。

因为次数为偶数，

z_2, \ldots, z_{2004}

中至少有一个实根，使得某个

b_k = 0

，所以

b_2 b_3 \cdots b_{2004} = 0

。因此 (A) 到 (D) 都必须为

0

。

另一方面，

\sum_{k=1}^{2004} c_k = P(1)

。一个有效多项式例如

P(x) = x(x - 2)(x - 3) \cdots

\cdot (x - 2003)

\cdot \left(x + \sum_{k=2}^{2003} k\right)

满足

P(1) \ne 0

，所以只有

\sum c_k

可能非零。

所以正确答案是 E。

Since

z_1 = a_1 + b_1 i = 0

is a root,

c_0 = P(0) = 0.

The nonreal zeros occur in conjugate pairs, so

\sum b_k = 0,

and the hypothesis then forces

\sum a_k = 0.

The coefficient

c_{2003}

equals

-c_{2004}

times the sum of the roots

\sum a_k + i \sum b_k = 0,

so

c_{2003} = 0.

Because the degree is even, at least one of

z_2, \ldots, z_{2004}

is real, making one

b_k = 0,

so

b_2 b_3 \cdots b_{2004} = 0.

Thus (A) through (D) all must be

0.

On the other hand,

\sum_{k=1}^{2004} c_k = P(1),

and a valid polynomial such as

P(x) = x(x - 2)(x - 3) \cdots

\cdot (x - 2003)

\cdot \left(x + \sum_{k=2}^{2003} k\right)

has

P(1) \ne 0.

So only

\sum c_k

can be nonzero.

Thus, the correct answer is E.

2004 AMC 12A 第 23 题

答案：E

难度评级：2350

小提示：

大提示：

解答：

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