AMC 8 2026 Problem 15
AMC8 2026 年第 15 题
立体几何★★★☆☆
Elijah has a large collection of identical wooden cubes which are white on 4 faces and gray 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 gray faces visible. What is the fewest number of cubes that he could glue together to ensure that no gray faces are visible, no matter how he rotates the figure?
Elijah 有很多相同的木制小立方体，每个立方体有 4 个白色面和 2 个灰色面，且这两个灰色面共享一条棱。他将一些立方体面对面粘在一起。下图展示了两个立方体粘合后，露出 3 个灰色面的情形。他至少需要粘合多少个小立方体，才能确保无论怎样旋转整个图形，都看不到任何灰色面？
- A.4
- B.6
- C.8
- D.9
- E.27
正确答案 · Correct Answer
A
解析 · Solution
解法1
将 4 个方块排成 2×2 田字格，每块的 2 个灰面朝内，各与相邻方块的灰面对接。8 个灰面全部隐藏，外露的全是白面。少于 4 块则无法完成所有对接，必有灰面外露。
Solution 1
Each cube has two gray faces. Place 4 cubes in a 2×2 square so that each cube's two gray faces point inward, each meeting a neighbor's gray face. All 8 gray faces are hidden; only white faces remain visible on the outside. Fewer than 4 cubes cannot pair up all gray faces, leaving at least one exposed.
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