2022 AMC 8 第 23 题

先试着解答 2022 AMC 8 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 2022 AMC 8 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.

在九个小方格中,每格放入一个 △\bigtriangleup 或 ◯\bigcirc,这些方格组成一个 3×33 \times 3 网格。下面显示了一个有三个 △\bigtriangleup 连成一线的示例。
有多少种配置同时有三个 △\bigtriangleup 连成一线,并且有三个 ◯\bigcirc 连成一线?
A △\bigtriangleup or ◯\bigcirc is placed in each of the nine squares in a 3×33 \times 3 grid. Shown below is a sample configuration with three △\bigtriangleup’s in a line.
How many configurations will have three △\bigtriangleup’s in a line and three ◯\bigcirc’s in a line?
3939
4242
7878
8484
9696
答案:D
知识点:分类讨论对称性
小提示:
按网格中三角形的个数分类
Split by how many triangles are in the grid
大提示:
正好 33 个三角形时,对角线三角形线不行
With exactly 33 triangles, diagonal triangle lines do not work
解答:
设三角形个数为 kk。要同时有三角形线和圆形线,必须有 3≤k≤63\le k\le6。
若 k=3k=3,三角形必须成一条线。行或列有效,因为剩下的六个圆形中仍有完整的一行或一列;对角线无效,因为其余位置没有完整的圆形线。因此共有 66 种 k=3k=3 的配置。由对称性,也有 66 种 k=6k=6 的配置。
若 k=4k=4,先选一条三角形线。若这条线是行或列,额外的三角形有 66 个位置可选,每一种都可行,因为总有一条平行的行或列全是圆形,共 6⋅6=366\cdot6=36 种;若这条线是对角线,则无论额外的三角形放在哪里,都不能留下完整的圆形线。由对称性,k=5k=5 也有 3636 种。
总数为 6+36+36+6=846+36+36+6=84。
正确答案是 D。
Let kk be the number of triangles. To have both a triangle line and a circle line, we need 3≤k≤63\le k\le6.
If k=3k=3, the three triangles must form a line. A row or column works, because the remaining six circles contain a full circle row or column. A diagonal does not work, because its complement contains no full line of circles. Thus there are 66 configurations for k=3k=3. By symmetry, there are also 66 configurations for k=6k=6.
If k=4k=4, choose the triangle line and then the extra triangle. If the line is a row or column, all 66 possible extra positions work, because one parallel row or column remains all circles. This gives 6⋅6=366\cdot6=36 configurations. If the triangle line is a diagonal, no extra position leaves a full circle line. By symmetry, k=5k=5 also gives 3636 configurations.
The total number of configurations is 6+36+36+6=846+36+36+6=84.
Thus, the correct answer is D.
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