What is the solution to a problem involving integers on an icosahedron?

[Ackbach]

Here is this week's POTW:

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In honor of Opalg's http://mathhelpboards.com/geometry-11/ask-height-icosahedron-18151.html, here is another problem involving icosahedrons:

On each face of a regular icosahedron is written a nonnegative integer such that the sum of all $20$ integers is $39$. Show that there are two faces that share a vertex and have the same integer written on them. (Recall that an icosahedron has $12$ vertices.)

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

Congratulations to kiwi for his correct answer, which follows:

Spoiler

Assume that no two faces with a common vertex have the same number written on them.

Consider any vertex. If we sum the integers on the 5 faces that involve this vertex the sum must be at least 0+1+2+3+4=10.

Doing this for all 12 vertices and summing (recognizing that every face will be counted more than once) we get a total of 120.

Now each face is triangular so it's integer was included for each of three vertices. Since no two faces with a common vertex have the same number written on them the total must be at least 120/3 = 40.

We are told that the total is 39 so we have a contradiction so two faces with a common vertex must have the same number on them.