Proof that the sum of exterior angles of a complex polygon or a non-planar polygon is greater than 2 Pi

3danglecomputational geometrygeometrypolygons

In the context of writing a ray-tracing program, I must be able to determine if a polygon in three-dimensional space is a planar simple convex polygon. My conjecture is that if I take the sum of the exterior angles (absolute values, always in the range $[0, \pi]$) then this sum is equal to $2\pi$ if and only if it is a planar simple convex polygon.

I firmly believe this to be true, but I seek somewhat rigorous proof. I already managed to show that the sum equals $2\pi$ for convex polygons and that it exceeds $2\pi$ for concave ones. However, I do not have any ideas on how to continue to prove that complex and non-planar polygons have a sum of exterior angles greater than $2\pi$.

For complex planar polygons, I managed to show (w.l.o.g.) that the exterior angles which "turn to the right" minus those that "turn to the left" must be a $2\pi n$ for some natural $n$. From there it's trivial to show that if $n>1$ then the total must be greater than $2\pi$, but for $n=0$ and $n=1$ I do not know how to proceed.

For non-planar polygons, I thought about what would happen if we project it to 2-dimensions then somehow show that the projection has an exterior angle sum strictly between $2\pi$ and that of the non-planar version, but I once again cannot fathom how to show that. (Edit: this idea does not work since there are counterexamples of non-planar polygons whose projections onto two-dimensional space have a larger exterior angle sum.)

Best Answer

The answer lies in Lemma 2.1 of Sullivan, Curves of Finite Total Curvature.

The paper refers to exterior angles as turning angles, and for a polygon $P$ designates the sum of turning angles (total curvature) as $TC(P)$

Here's a paraphrased version of the key lemma:

Lemma 2.1. Suppose $P$ is a polygon in $\mathbb R^3$. If $P'$ is obtained from $P$ by deleting one vertex $v_n$ then $TC(P') ≤ TC(P)$. We have equality here if $v_{n−1}v_n v_{n+1}$ are collinear in that order, or if $v_{n−2}v_{n−1}v_{n}v_{n+1}v_{n+2}$ lie convexly in some plane, but never otherwise.

Please refer to the paper for a proof, along with background and definitions. It's a short simple argument, using a mapping that takes 3D polygons $P$ to spherical polygons $P_S$ on the unit sphere $\mathbb S^2$. The total curvature of $P$ is the same as the boundary length of $P_S$.

Lemma 2.1 nicely takes care of both cases.

For a non-planar polygon $P$, removing vertices until only a triangle $T$ is left defines a sequence of polygons whose total curvature is strictly decreasing in at least one pair of the sequence (corresponding to where some $v_{n−2}v_{n−1}v_{n}v_{n+1}v_{n+2}$ is not planar). Since $TC(T)=2\pi$ we conclude that $TC(P)>2\pi$.

Similarly, for complex planar polygons $P$ there will be vertex sequences $v_{n−2}v_{n−1}v_{n}v_{n+1}v_{n+2}$ that are not convex and again $TC(P)>2\pi$.

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