2026 AMC 8 第 15 题
先试着解答 2026 AMC 8 第 15 题，然后核对你的答案与视频与精心整理的解答，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 2026 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。
15.
Elijah 有一些立方体，每个立方体有 个面未涂色，另外 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 个这样的立方体粘在一起后，仍有 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 faces and shaded on faces that share an edge. He glues some cubes together face-to-face. The figure below shows cubes being glued together, leaving 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?
小提示：
每个阴影面都必须粘到另一个立方体的一个面上。
Every shaded face must be glued to another cube
大提示：
如果只有 个立方体，总会有某个端点立方体最多只和一个立方体相邻。
With only cubes, some cube has at most one face-neighbor
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 的方阵时，每个立方体都可以把两个阴影面朝向内部，所以 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 square can each be oriented with its two shaded faces pointing inward, so cubes suffice.
其他年份的第 15 题
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