| Math puzzle |
Wed, 01 June 2005 02:01  |
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This was brought to me by one of my maths students a few weeks ago, I thought you guys might like a crack at it.
Take a cube.
Label each of the 8 vertices(corners) either 1 or -1.
Label each of the 6 faces(sides) with the product of the 4 vertices that define it.
Q: Prove that the sum of all 14 numbers is never 0.
image showing the labelling of the visible vertices and two faces.
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| Re: Math puzzle |
Thu, 02 June 2005 17:14  |
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 wizard |
 | Officer Cadet 3rd Year |
Messages: 279
Registered: January 2004
Location: Aachen, Germany | |
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Hi gible,
as a starting point, let's assume that we have 0 vertices with value -1, thus 8 vertices with value +1. We'd have a total value of 14 then (8 vertices + 6 faces, all value +1).
We now have the possibility to change vertices from +1 to -1. We can do that repeatedly up to 8 times.
Regarding one single change, there are a few possibilities how the total value will change. As each vertex is adjacent to three faces, all possibilities have in common that the value of exactly three (all adjacent) faces will be changed - either from +1 to -1 or vice versa. Each change of a face changes the total face sum by +2 or -2 (i.e. change from +1 to -1: face sum is decreased by 2). If we have three changes, the total value can change only by -6, -2, +2 or +6, other values are not possible.
Each vertex change causes a change in the vertex sum too - by -2. One more negative vertex, one less positive vertex.
Thus, if we add the face sum and the vertex sum, we can only change the total sum by -8, -4, 0 or +4. All those values are dividable by 4, and if we add 1 to 8 of those values, the result will still be dividable by 4 (positive or negative). If we add any of those numbers to the starting value of 14, we'll never get a value of 0. +2 or -2 are the closest values possible.
All in all, the possible total sums are -10, -6, -2, +2, +6, +10, +14.
q.e.d.
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| Re: Math puzzle |
Fri, 03 June 2005 03:19 |
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Well done to Mazda and wizard, with both essentially identical proofs. I'm assuming the proof I received by email also belongs to wizard...its a perfect match 
For those who're interested here is the alternative proof I already had:
Labelling the vertices ABCDEFGH and considering the product of the 14 numbers gives (ABCDEFGH)^4 which must equal 1.
Hence there must be an even number of -1's in the forteen numbers. A zero sum requires 7 -1's which is not even.
Hence sum != 0
qed.
[Updated on: Fri, 03 June 2005 03:26] Report message to a moderator
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