2000 AMC 12 第 8 题

先试着解答 2000 AMC 12 第 8 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。

所有题目均经美国数学协会(MAA)官方合法授权使用。

8.

图形 00,11,22 和 33 分别由 11,55,1313 和 2525 个互不重叠的单位正方形组成。如果继续这个规律,图形 100100 中会有多少个互不重叠的单位正方形?
Figures 0,0, 1,1, 2,2, and 33 consist of 1,1, 5,5, 13,13, and 2525 nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?100?
1040110401
1980119801
2020120201
3980139801
4080140801
答案:C
知识点:找规律完全平方数
小提示:
每个图形都是一个菱形,各行长度按奇数 1,3,5,…1, 3, 5, \ldots 增大后再减小。
Each figure is a diamond whose rows are the odd numbers 1,3,5,…1, 3, 5, \ldots up and back down
大提示:
图形 nn 有 n2+(n+1)2n^2 + (n + 1)^2 个单位正方形。
Figure nn has n2+(n+1)2n^2 + (n + 1)^2 unit squares
解答:
图形 nn 是一个菱形,各行长度按奇数增大后再减小,所以共有 n2+(n+1)2n^2 + (n + 1)^2 个单位正方形。这与 n=0,1,2,3n = 0, 1, 2, 3 时的 1,5,13,251, 5, 13, 25 相符。
因此图形 100100 中单位正方形的个数为 1002+1012=10000+10201=20201 \begin{aligned} 100^2 + 101^2 &= 10000 + 10201 \\ &= 20201 \end{aligned} 个。
因此,正确答案是 C。
Figure nn is a diamond whose row lengths increase through the odd numbers and back down, giving a total of n2+(n+1)2n^2 + (n + 1)^2 unit squares. This matches 1,5,13,251, 5, 13, 25 for n=0,1,2,3.n = 0, 1, 2, 3.
Therefore figure 100100 has 1002+1012=10000+10201=20201 \begin{aligned} 100^2 + 101^2 &= 10000 + 10201 \\ &= 20201 \end{aligned} unit squares.
Thus, the correct answer is C.
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