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2021 AMC 12A Fall 第 23 题

先试着解答 2021 AMC 12A Fall 第 23 题，然后核对你的答案与精心整理的解答，解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2021 AMC 12A Fall 解答，或核对答案。

所有题目均经美国数学协会（MAA）官方合法授权使用。

23.

首项系数为 11、且系数为实数的二次多项式称为“无礼”的，如果方程 p(p(x))=0p(p(x)) = 0 恰好有三个实数解。在所有无礼二次多项式中，有唯一一个多项式 p~(x)\tilde{p}(x)，使得其根的和最大。p~(1)\tilde{p}(1) 是多少？

A quadratic polynomial with real coefficients and leading coefficient 11 is called disrespectful if the equation p(p(x))=0p(p(x)) = 0 is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial p~(x)\tilde{p}(x) for which the sum of the roots is maximized. What is p~(1)?\tilde{p}(1)?

516\dfrac{5}{16}

12\dfrac{1}{2}

58\dfrac{5}{8}

11

98\dfrac{9}{8}

答案：A
知识点：多项式二次方程最优化
难度评级：2380
小提示：

若 pp 的根为 r,sr, s 则 p(p(x))=0p(p(x)) = 0 表示 p(x)=rp(x) = r 或 p(x)=sp(x) = s；恰有三个根要求其中一个方程有重根

If pp has roots r,s,r, s, then p(p(x))=0p(p(x)) = 0 means p(x)=rp(x) = r or p(x)=s;p(x) = s; exactly three roots requires one of these to have a double root

大提示：

在 (r−s)2+4s=0(r - s)^2 + 4s = 0 下令 u=r−su = r - s，最大化 r+s=−u22+ur + s = -\tfrac{u^2}{2} + u。

With (r−s)2+4s=0(r - s)^2 + 4s = 0 and letting u=r−s,u = r - s, maximize r+s=−u22+ur + s = -\tfrac{u^2}{2} + u

解答：

设 pp 的根为 rr 和 ss。那么 p(p(x))=0p(p(x)) = 0 分解成 p(x)=rp(x) = r 与 p(x)=sp(x) = s，两个判别式分别为 (r−s)2+4r(r - s)^2 + 4r 和 (r−s)2+4s(r - s)^2 + 4s。恰有三个实根意味着一个判别式为 00，另一个为正。

取 (r−s)2+4s=0(r - s)^2 + 4s = 0，并令 u=r−su = r - s。则 s=−u24s = -\tfrac{u^2}{4}，另一个判别式为 4u4u，所以 u>0u>0。此外，r+s=−u22+ur+s=-\tfrac{u^2}{2}+u，它在 u=1u=1 时达到最大值，此时 r=34r=\tfrac34 且 s=−14s=-\tfrac14。

所以 p~(x)=(x−34)(x+14)\tilde{p}(x) = \left(x - \tfrac34\right)\left(x + \tfrac14\right)，并且 p~(1)=14⋅54=516\tilde{p}(1) = \tfrac14\cdot\tfrac54 = \tfrac{5}{16}。

因此，正确答案是 A。

Let pp have roots rr and s.s. Then p(p(x))=0p(p(x)) = 0 splits into p(x)=rp(x) = r and p(x)=s,p(x) = s, with discriminants (r−s)2+4r(r - s)^2 + 4r and (r−s)2+4s.(r - s)^2 + 4s. Exactly three real roots means one discriminant is 00 and the other positive.

Take (r−s)2+4s=0(r - s)^2 + 4s = 0 and set u=r−s.u = r - s. Then s=−u24s = -\tfrac{u^2}{4} and the other discriminant is 4u,4u, so u>0.u>0. Also r+s=−u22+u,r+s=-\tfrac{u^2}{2}+u, maximized at u=1,u=1, giving r=34r=\tfrac34 and s=−14.s=-\tfrac14.

So p~(x)=(x−34)(x+14),\tilde{p}(x) = \left(x - \tfrac34\right)\left(x + \tfrac14\right), and p~(1)=14⋅54=516.\tilde{p}(1) = \tfrac14\cdot\tfrac54 = \tfrac{5}{16}.

Thus, the correct answer is A.

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