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 $B$s 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.