Competition problem — Combinatorics

Combinatorics Difficulty 4.8 AIME Prove it Brazil

The edges of a cube are labeled from $1$ to $12$ in an arbitrary manner. Show that it is not possible to get the sum of the edges at each vertex the same. Show that we can get eight vertices with the same sum if one of the labels is changed to $13$ .

Solution

Each edge contributes to the total sum twice, one for each of its vertices. So if each vertex has sum $v$ , the sum of all numbers is $8v = 2(1+2+\cdots+12) = 4 \cdot 39 \Rightarrow v = \frac{39}{2}$ , which can't be possible.

The following diagram shows a solution for sums equal to $13$ .

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