AMC12 2005 A

AMC12 2005 A · Q24

AMC12 2005 A · Q24. It mainly tests Polynomials, Functions basics.

Let $P(x)=(x-1)(x-2)(x-3)$. For how many polynomials $Q(x)$ does there exist a polynomial $R(x)$ of degree $3$ such that $P(Q(x)) = P(x) \cdot R(x)$?

设 $P(x)=(x-1)(x-2)(x-3)$。有多少个多项式 $Q(x)$ 存在，使得存在次数为 $3$ 的多项式 $R(x)$ 满足 $P(Q(x)) = P(x) \cdot R(x)$？

(A) 19 19

(B) 22 22

(D) 27 27

(E) 32 32

Answer

Correct choice: (B)

正确答案：（B）

Solution

We can write the problem as Since $\deg P(x) = 3$ and $\deg R(x) = 3$, $\deg P(x)\cdot R(x) = 6$. Thus, $\deg P(Q(x)) = 6$, so $\deg Q(x) = 2$. Hence, we conclude $Q(1)$, $Q(2)$, and $Q(3)$ must each be $1$, $2$, or $3$. Since a quadratic is uniquely determined by three points, there can be $3*3*3 = 27$ different quadratics $Q(x)$ after each of the values of $Q(1)$, $Q(2)$, and $Q(3)$ are chosen. However, we have included polynomials $Q(x)$ which are linear, rather than quadratic. Namely, Clearly, we could not have included any other constant functions. For any linear function, we have $2\cdot Q(2) = Q(1) + Q(3)$ because $Q(2)$ is the $y$-coordinate of the midpoint of $(1, Q(1))$ and $(3, Q(3))$. So we have not included any other linear functions. Therefore, the desired answer is $27 - 5 = \boxed{\textbf{(B) }22}$.

我们可以将题目写成：由于 $\deg P(x) = 3$ 且 $\deg R(x) = 3$，所以 $\deg(P(x)\cdot R(x)) = 6$。因此 $\deg(P(Q(x))) = 6$，从而 $\deg Q(x) = 2$。于是可知 $Q(1)$、$Q(2)$、$Q(3)$ 必须各自等于 $1$、$2$ 或 $3$。由于一个二次多项式由三个点唯一确定，当选定 $Q(1)$、$Q(2)$、$Q(3)$ 的值后，就有 $3*3*3 = 27$ 个不同的二次多项式 $Q(x)$。然而，其中包含了一些实际上是一次而非二次的多项式 $Q(x)$。具体为：显然不可能包含其他常数函数。对于任意一次函数，都有 $2\cdot Q(2) = Q(1) + Q(3)$，因为 $Q(2)$ 是点 $(1, Q(1))$ 与 $(3, Q(3))$ 的中点的 $y$ 坐标。因此也不会包含其他一次函数。所以所求答案为 $27 - 5 = \boxed{\textbf{(B) }22}$。

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