2022 AMC 12B 第 18 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

18.

5×55 \times 5 网格中的每个小方格或填充或为空,并且至多有八个相邻方格,相邻指共边或共顶点。 网格按如下规则变换:

任一填充方格若有两个或三个填充邻居,则保持填充。任一空方格若恰有三个填充邻居,则变为填充。 其他所有方格保持为空或变为空。

下图给出一个变换示例。

假设 5×55 \times 5 网格有一圈空方格包围一个 3×33 \times 3 子网格。有多少种初始构型会在一次变换后得到只有中心一个填充方格的网格?(旋转或反射得到的构型视为不同。)

Each square in a 5×55 \times 5 grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:

Any filled square with two or three filled neighbors remains filled. Any empty square with exactly three filled neighbors becomes a filled square. All other squares remain empty or become empty.

A sample transformation is shown in the figure below.

Suppose the 5×55 \times 5 grid has a border of empty squares surrounding a 3×33 \times 3 subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.)

1414

1818

2222

2626

3030

答案:C
知识点:过程模拟分类讨论
难度评级:2000
解答:

只有内部 3×33 \times 3 子网格可以初始填充。 要使中心在变换后填充,若它原为空,则需恰有 33 个填充邻居; 若它原已填充,则需有 2233 个填充邻居。

其他所有方格在变换后都必须为空。 3×33 \times 3

关键限制是,任何外圈方格都不能恰有三个填充邻居,这排除了将 33 子网格外边上的整排三格都填充的情况。 88 (0,0)(0,0)8,4,4,48,4,4,42020 {(1,1),(1,0),(1,1)},{(1,1),(1,1),(1,1)},{(1,1),(1,1),(1,0)},{(1,1),(0,1),(1,0)}. \begin{gathered} \{(-1,-1),(-1,0),(1,1)\},\\ \{(-1,-1),(-1,1),(1,-1)\},\\ \{(-1,-1),(-1,1),(1,0)\},\\ \{(-1,-1),(0,1),(1,0)\}. \end{gathered}

枚举满足这些条件的构型,可知每个有效构型都恰有三个填充格:中心初始为空的有 20+2=2220+2=22 个,中心初始填充的有 个,总数为 。

所以正确答案是 C

Only the inner 3×33 \times 3 squares can start filled. For the center to be filled afterward, if it began empty it needs exactly 33 filled neighbors, and if it began filled it needs 22 or 3.3.

Every other square must end empty. The key restriction is that no border square may acquire exactly three filled neighbors, which rules out filling all three squares along an outer edge of the 3×3.3 \times 3.

If the center starts empty, exactly 33 of its 88 neighbors must be filled. Checking these triples up to square symmetry leaves four types. With ring coordinates centered at (0,0),(0,0), representatives are {(1,1),(1,0),(1,1)},{(1,1),(1,1),(1,1)},{(1,1),(1,1),(1,0)},{(1,1),(0,1),(1,0)}. \begin{gathered} \{(-1,-1),(-1,0),(1,1)\},\\ \{(-1,-1),(-1,1),(1,-1)\},\\ \{(-1,-1),(-1,1),(1,0)\},\\ \{(-1,-1),(0,1),(1,0)\}. \end{gathered} Their symmetry-orbit sizes are 8,4,4,4,8,4,4,4, giving 2020 configurations.

If the center starts filled, the same neighbor check leaves only the two configurations in which the other filled cells are opposite corner neighbors. Hence the total is 20+2=22.20+2=22.

Thus, the correct answer is C.

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