Question

Elijah 有一些立方体，每个立方体有 $4$ 个面未涂色，另外 $2$ 个相邻的面涂了阴影。他把这些立方体面贴面地粘在一起。图中显示 $2$ 个这样的立方体粘在一起后，仍有 $3$ 个阴影面可见。为了保证无论怎样旋转整个图形，都没有阴影面可见，至少需要多少个立方体？

Elijah has a large collection of identical wooden cubes which are plain on $4$ faces and shaded on $2$ faces that share an edge. He glues some cubes together face-to-face. The figure below shows $2$ cubes being glued together, leaving $3$ shaded faces visible. What is the fewest number of cubes that he could glue together to ensure that no shaded faces are visible, no matter how he rotates the figure?

$4$

$6$

$8$

$9$

$27$

答案：A

知识点：正方体立体几何图论

难度评级：1450

视频讲解：

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文字解答：

如果没有阴影面可见，那么每个立方体的两个阴影面都必须粘到相邻立方体上，所以每个立方体需要两个面相邻的邻居。若只有三个立方体，面相邻关系是一条链，因此端点立方体只有一个面相邻的邻居。四个立方体排成 $2\times2$ 的方阵时，每个立方体都可以把两个阴影面朝向内部，所以 $4$ 个立方体足够。

For no shaded face to be visible, each cube must have both of its shaded faces glued to neighboring cubes, so each cube needs two face-neighbors. With only three cubes, the face-adjacency graph is a path, so an end cube has only one face-neighbor. Four cubes arranged as a $2\times2$ square can each be oriented with its two shaded faces pointing inward, so $4$ cubes suffice.

2026 AMC 8 第 15 题

答案：A

难度评级：1450

视频讲解：

文字解答：

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