6th Iberoamerican 1991 Problem 1 The number 1 or the number -1 is assigned to each vertex of a cube. Then each face is given the product of its four vertices. What are the possible totals for the resulting 14 numbers? Solution
: 14, 6, 2, -2, -6, -10. If every vertex is 1, we get 14 and that is clearly the highest possible total. The lowest possible total cannot be lower than -14, but we cannot even achieve that because if all the vertices are -1, then all the faces are 1. If we change a vertex, then we also change three faces. If the vertex and the three faces are all initially the same, then we make a change of ±8. If three are of one kind and one the opposite, then we make a change of ±4. If two are of one kind and two the opposite, then we make no change. Thus any sequence of changes must take us to 14 + 4n for some integer n. But we have already shown that the total is greater than -14 and at most 14, so the only possibilities are -10, -6, -2, 2, 6, 10 and 14. We show that 10 is not possible. If more than 2 vertices are -1, then the vertex total is at most 2, there are only 6 faces, so the total is less than 10. If all vertices are 1, then the total is 14. If all but one vertex is 1, then the total is 6. So the only possibility for 10 is just two vertices -1. But however we choose any two vertices, there is always a face containing only one of them, so at least one face is -1, so the face total is at most 4 and the vertex total is 4, so the total is less than 10. The other totals are possible, for example: 14: all vertices 1 6: one vertex -1, rest 1 2: three vertices of one face -1, rest 1 -2: all vertices -1 -6: all vertices but one -1 -10: two opposite corners 1, rest -1 6th Ibero 1991 © John Scholes jscholes@kalva.demon.co.uk 11 July 2002