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2016 AMC 12A 第 14 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。

14.

一个立方体的每个顶点都要标上 11 到 88 中的一个整数，每个整数恰用一次，并且每个面的四个顶点数字之和都相同。通过旋转立方体可以互相得到的标法视为相同。有多少种不同的标法？

Each vertex of a cube is to be labeled with an integer from 11 through 8,8, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?

11

33

66

1212

2424

答案：C
知识点：正方体分类讨论对称性
难度评级：1730
小提示：

每个顶点属于 33 个面，所以六个面和的总和为 3(1+2+⋯+8)3(1+2+\cdots+8)。

Each vertex lies on 33 faces, so the six face-sums total 3(1+2+⋯+8)3(1+2+\cdots+8)

大提示：

每个面的和为 1818；证明 11 和 88 必须在相邻顶点上。

The face-sum is 18;18; show 11 and 88 must be on adjacent vertices

解答：

每个顶点属于 33 个面，所以 6S=3(1+2+⋯+8)=1086S=3(1+2+\cdots+8)=108，从而每个面的和为 S=18S=18。

包含 11 且和为 1818 的四元子集是 {1,2,7,8}\{1,2,7,8\}，{1,3,6,8}\{1,3,6,8\}，{1,4,5,8}\{1,4,5,8\}，和 {1,4,6,7}\{1,4,6,7\}。其中只有一个不含 88，所以经过 11 的三个不同面中至少有两个包含 88。立方体的两个顶点恰好在相邻时才共同位于两个面上；因此 11 和 88 相邻。

旋转立方体，使 11 位于左下前顶点，88 位于右下前顶点。经过 11 而不含 88 的唯一一个面必须放置 4,6,74,6,7，它们有 3!=63!=6 种排列。每种排列都会迫使 5,3,25,3,2 位于三个相对顶点，其余各面的和也都为 1818。因此共有 66 种标法。

因此，正确答案是 C。

Each vertex belongs to 33 faces, so 6S=3(1+2+⋯+8)=108,6S=3(1+2+\cdots+8)=108, giving each face-sum S=18.S=18.

The four-element subsets containing 11 with sum 1818 are {1,2,7,8},\{1,2,7,8\}, {1,3,6,8},\{1,3,6,8\}, {1,4,5,8},\{1,4,5,8\}, and {1,4,6,7}.\{1,4,6,7\}. Only one omits 8,8, so at least two of the three distinct faces through 11 contain 8.8. Two vertices of a cube lie on two common faces exactly when they are adjacent; hence 11 and 88 are adjacent.

Rotate the cube so that 11 is at the lower-left-front vertex and 88 at the lower-right-front vertex. The unique face through 11 that does not contain 88 must use 4,6,7,4,6,7, and these can be placed in 3!=63!=6 orders. Each order forces 5,3,25,3,2 at the three opposite vertices, and the remaining face sums are then 18.18. Hence there are 66 arrangements.

Thus, the correct answer is C.

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