2000 AMC 12 第 12 题

先试着解答 2000 AMC 12 第 12 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 2000 AMC 12 解答,或核对答案。

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12.

设 AA、MM 和 CC 是非负整数,且 A+M+C=12A + M + C = 12。求下式的最大值:A⋅M⋅C+A⋅M+M⋅C+C⋅A? \begin{aligned} &A \cdot M \cdot C + A \cdot M \\ &\quad {}+ M \cdot C + C \cdot A \end{aligned}\text{?}
Let A,A, M,M, and CC be nonnegative integers such that A+M+C=12.A + M + C = 12. What is the maximum value of A⋅M⋅C+A⋅M+M⋅C+C⋅A? \begin{aligned} &A \cdot M \cdot C + A \cdot M \\ &\quad {}+ M \cdot C + C \cdot A? \end{aligned}
6262
7272
9292
102102
112112
答案:E
知识点:因式分解算术-几何平均不等式最优化
小提示:
AMC+AM+MC+CAAMC + AM + MC + CA =(A+1)(M+1)(C+1)= (A + 1)(M + 1)(C + 1) −(A+M+C)−1- (A + M + C) - 1。
AMC+AM+MC+CAAMC + AM + MC + CA =(A+1)(M+1)(C+1)= (A + 1)(M + 1)(C + 1) −(A+M+C)−1- (A + M + C) - 1
大提示:
三个因数 A+1,M+1,C+1A+1, M+1, C+1 的和为 1515,所以它们越接近相等,乘积越大。
The three factors A+1,M+1,C+1A+1, M+1, C+1 sum to 15,15, so their product is largest when they are equal
解答:
注意 AMC+AM+MC+CA=(A+1)(M+1)(C+1)−(A+M+C)−1。 \begin{aligned} &AMC + AM + MC + CA \\ &\quad = (A + 1)(M + 1)(C + 1) \\ &\quad {}- (A + M + C) - 1 \end{aligned}\text{。}
因为 A+M+C=12A + M + C = 12,原式等于 (A+1)(M+1)(C+1)−13(A + 1)(M + 1)(C + 1) - 13。这三个因数的和为 1515,乘积在它们都等于 55 时最大,等于 53=1255^3 = 125。
最大值为 125−13=112125 - 13 = 112。
因此,正确答案是 E。
Observe that AMC+AM+MC+CA=(A+1)(M+1)(C+1)−(A+M+C)−1. \begin{aligned} &AMC + AM + MC + CA \\ &\quad = (A + 1)(M + 1)(C + 1) \\ &\quad {}- (A + M + C) - 1. \end{aligned}
Since A+M+C=12,A + M + C = 12, this equals (A+1)(M+1)(C+1)−13.(A + 1)(M + 1)(C + 1) - 13. The three factors sum to 15,15, so their product is maximized when each equals 5,5, giving 53=125.5^3 = 125.
The maximum value is 125−13=112.125 - 13 = 112.
Thus, the correct answer is E.
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