Graph Theory proof via induction

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Homework Statement



Prove by induction that the graph of any triangulation of a polygon will have at least 2 vertices of degree 2

Hint: Split the triangulation graph into 2 triangulation graphs at some chord e


The Attempt at a Solution



Ok I am pretty terrible at induction proofs, so bare with me.

This is trivial for the case when we only have a triangle.

Suppose this is true for n vertices. then we want to show that it is true for n+1 vertices.

Basically i have no clue how to do this problem.

My guess is that we have to make e the smallest triangle possible, but that only proves that there is one edge of degree 2.

Any help is appreciated.
 
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any takers?
 
First I absolutely refuse to "bare" with you! I don't know you that well.

Assume that the graph of any triangulation of any polygon with k sides has at least two vertices of degree 2. Now look at a polygon with k+1 sides.drawing a chord between two "almost adjacent" vertices (i.e. they have exactly one other vertex between them) divides the polygon into a triangle and a polygon with k sides.
 
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wow it took me a few read overs but then i finally got it, and it blew my mind.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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