Problem:
Each vertex of a cube is to be labeled with an integer from through , 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?
Answer Choices:
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Solution:
The sum of the four numbers on the vertices of each face must be . The only sets of four of the numbers that include and have a sum of are , and . Three of these sets contain both and . Because two specific vertices can belong to at most two faces, the vertices of one face must be labeled with the numbers , and two of the faces must include vertices labeled and . Thus and must mark two adjacent vertices. The cube can be rotated so that the vertex labeled is at the lower left front, and the vertex labeled is at the lower right front. The numbers , and must label vertices on the left face. There are ways to assign these three labels to the three remaining vertices of the left face. Then the numbers , and must label the vertices of the right face adjacent to the vertices labeled , and , respectively. Hence there are possible arrangements.
The problems on this page are the property of the MAA's American Mathematics Competitions