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In every polyhedron there is at least one pair of faces with the same number of sides.

Solution:

Let $N$ be the greatest number of sides in a face of a given polyhedron. Then the number of edges of each of the N adjacent faces each has the number of sides between $3$ and $N$. There bound to be $2$ with the same number of sides.

I don't understand why "the number of edges of each of the N adjacent faces each has the number of sides between $3$ and $N$"? Anyone helps?

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  • $\begingroup$ What's the least number of sides of a polygon? The maximum is $N$ by hypothesis. $\endgroup$ – Ian Mateus Jan 13 '14 at 19:56
  • $\begingroup$ The argument seems to depend on the fact that two faces cannot have more than one edge in common, so $N$ must be less than the number of faces. $\endgroup$ – Henning Makholm Jan 13 '14 at 20:06
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At least 3 since the face with the least amount of sides is a triangle.

At most N since you are using N to name the maximum number of sides any face in the polyhedron has. Hope that helps.

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  • $\begingroup$ Also the argument is considering any face $F$ that has the maximum of N edges, and then thinking about the N faces that touch $F$. That is what "the N adjacent faces" means. $\endgroup$ – user40919 Jan 13 '14 at 20:14
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Let $F$ be a face with the maximum number of sides of any face let $N$ denote that number. Then there are $N$ adjacent faces to $F$. Each face has any number of sides between $3$ and $N$( because $N$ is maximal). Now count the number of faces so far. We have the $N$ faces and $N-3$ options for number of sides. Use pigeonhole principle.

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Well it cannot be more than N because this is the max by definition. Obviously it cannot be 0. It also cannot be 1 because it already has 1 joining the original face with N.

So try and rule out 2. Think about the minimum number of faces a polyhedron has.

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