0

I am reading Aigner & Ziegler's Proofs from the BOOK Chapter 35 and came across the following statement:

The astute reader may have noticed a subtle point in our reasoning. Does a triangulation [of a polygon] always exist? Probably everybody’s first reaction is: Obviously, yes! Well, it does exist, but this is not completely obvious, and, in fact, the natural generalization to three dimensions (partitioning into tetrahedra) is false!

The authors then proceed to give a proof of the existence of a triangulation of a polygon. However, they use the "fact" that the sum of the interior angles of a polygon with $n$ vertices is $(n-2) 180^\circ$. However, the most obvious proof of this fact that comes to mind is to add up the interior angles of the $n - 2$ triangles is that would triangulate the polygon.

To avoid such circular reasoning in the proof, what are other ways of seeing this fact?

Earthliŋ
  • 2,490
  • 2
    Non-rigorously: Walk around the polygon clockwise. At each point where you turn, you turn $180-x,$ clockwise, where $x$ is the interior angle at that point. The total turning across all $n$ corners is $180n-s,$ where $s$ is the sum of the interior angles. But the total turning is $360$ degrees. So $s=180n-360.$ – Thomas Andrews Jun 08 '21 at 19:53
  • You compute the sum of their suplementary angles (exterior), which on one hand is $2\pi$ and on the other is equal $n\pi$ minus the sum of the interior angles. That the sum of the exterior angles is $2\pi$ you can do by picking a vertex and drawing parallel lines to the sides through that vertex and using angles between parallels. – plop Jun 08 '21 at 19:54
  • 1
    In theory, the total turn could be any multiple of $360$ degrees, which is why this is non-rigorous - it takes some work to prove that a polygonal path in the plane is not simple if it rotates you $720$ degrees or more. – Thomas Andrews Jun 08 '21 at 19:56

0 Answers0