Spending the weekend folding polyhedra and finding convex internal angles

So I’ve managed to resolve the circle construction question from a few posts back and determined that any number n can be constructed by a fairly small number of operations; more precisely the 2-logarithm the number but I’ve put off putting together a good description of the proof as I’ve been preoccupied with school, work, and another new mini-project: Polyhedra.

I originally got the idea that I should look back into Polyhedra after I overheard some students mentioning the concept of the Euler characteristic of a graph (V + F – E) and recognizing that I remembered very little of that particular theory I set to refreshing on the theory and making lots of paper polyhedra, drawing Hamiltonian paths, and colorings and so on as I stumbled over them.


As I reviewed the E + F – E = 2 proofs I also had to recognize that I had some gaps in my knowledge regarding the formal existence of triangulation of polygons and continued on the merry chain of ‘better check that’ untill I at the bottom of the chain ended up with

Every polygon has a strictly convex angle. (An internal angle less than \pi)

A statement for which I designed the following proof (which I unfortunately realized had some issues as I was typing it)

Let  P be the polygon A_1 A_2 \cdots A_{n-1}A_{n}spanned by the n points A_1, A_2, ..., A_n in the plane. Consider the line L formed by extending the segment A_1 A_2 and let A_k be a vertex on $P$ which is of maximal distance from $L$. Or more precisely has the following property

  • For any other vertex A on P the distance from A to the line L is shorter or equal to the distance between B and $L$.

Now (using the Paralell postulate) let L' be a line through B which is parallel to L. By construction all vertices on P either lie on L or on the same side of L' as L (A_1 and A_2). Let us for convenience say that P lies on the ‘south side’ of L'.

At most one of Bs neighbors in the polygon on P lie on L' (for otherwise it would be a redundant vertex) and so the angle formed by these three points contained on the south side of $L’$ is strictly convex and it is an internal angle of P. (It is internal for a ray from due north into B intersects the polygon 0 times (an even number). We have therefore found a strictly convex internal angle!

I will have to admit that I realized an issue with this proof as I finished it as I had somewhat embarrasingly not properly considered how to formulate (prove) that the ‘south’ angle at B was indeed the internal one and it sort of exposes that I haven’t collected enough formal devices to differentiate between angles of the form CDE and $latex EDC without coordinates or images and the proof isn’t as satisfactory as I’d like and I will see if I can fix it.

The image I had in my mind for why this was true was after all pretty straight forward.


Using higher maths the existence of a strictly convex internal angle can be straightforwardly deduced from the existence of the convex hull as the convex angles of the hull put upper bounds on at least 3 internal angles of any polygon. But it of course has the issue of just kicking the ball further down the corridor as you’d have to prove the existence of a convex hull.


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