The Chicago Math Meetup Group Message Board › Zero sum in a cube

Zero sum in a cube

Miguel A
Miguel_A.
Chicago, IL
Post #: 1,130
We assign a +1 or -1 to each vertex of a cube. Then to each face of the cube we assign the product of the numbers of its vertices. Is it possible to make an assignation so that the sum of the 14 numbers (8 vertices and 6 faces) is zero? Find such assignation or prove that it is impossible. (Don't use brute force.)

Miguel
Miguel A
Miguel_A.
Chicago, IL
Post #: 1,140
We assign a +1 or -1 to each vertex of a cube. Then to each face of the cube we assign the product of the numbers of its vertices. Is it possible to make an assignation so that the sum of the 14 numbers (8 vertices and 6 faces) is zero? Find such assignation or prove that it is impossible. (Don't use brute force.)

It is impossible.

We can imagine any such assignation as starting by assigning +1 to all the vertices (so all the numbers are +1 and the sum will be 14), and then replacing successively the number assigned to some of the vertices from +1 to -1. Note that if we change the number assigned to vertex 'v' from +1 to -1, the numbers on the three faces adjacent to 'v' will change their signs, i.e., they will change from +1 to -1 or from -1 to +1. So, if we represent the numbers assigned to a vertex and the three faces adjacent to it as a 4-tuple, the possible changes are

(+1,+1,+1,+1) -> (-1,-1,-1,-1)

(+1,+1,+1,-1) -> (-1,-1,-1,+1)

(+1,+1,-1,-1) -> (-1,-1,+1,+1)

(+1,-1,-1,-1) -> (-1,+1,+1,+1)

and permutations of those. In each case the total sum changes by -8, -4, 0 and +4 respectively, always a multiple of 4. But by changing the initial value of the sum, which is 14, by multiples of 4, we can never get 0, because 14 is not a multiple of 4.

Miguel






Powered by mvnForum

People in this
Meetup are also in:

Sign up

Meetup members, Log in

By clicking "Sign up" or "Sign up using Facebook", you confirm that you accept our Terms of Service & Privacy Policy