Prove that the vector sum of the vertices of an n-sided regular polytope whose center is at the orgin is zero

Question

I need to prove this (assuming it's true):

The vector sum of the vectors pointing to the vertices of an n-sided regular polytope whose center is at the origin of a Euclidean space is zero.

If it has an even number of vertices, it's clearly zero by symmetry (at least where my imagination works). I can't think of a way to prove this in the general case, though.

Answer 1

The Chebyshev center of a regular polytope is unique. Assume that the Chebyshev center $C$ of a regular polytope with centroid in the origin is not the origin. Then we must have $\varphi(C)=C$ for any $\varphi$ in the symmetry group of the polytope, or $C\in\operatorname{Fix}(\varphi)$. If we have two different elements in the symmetry group of the polytope such that the corresponding sets of fixed points are lines through the origin, we have a contradiction. But the last condition is met by any polytope with more than two faces, hence it is trivial.

Answer 2

For odd number n, let P(k) be the vector from center to vertex k. Let vector S = sum of all such vectors = P(1) + P(2) + ... + P(n). Let side(k) of the polygon be the side clock-wise adjacent to vertex k.

Consider vector P(1), and that the side((n+1)/2) is directly opposite to vertex 1. By symmetry, the line that coincide with vector P(1) passes through and bisects the side directly opposite to vertex 1, which is side((n+1)/2), and also bisects the angle between vectors P((n+1)/2) and P((n+1/2)+1). The size of the angle is 2Pi/n.

By symmetry, vector sum P((n+1)/2) + P((n+1/2)+1) = - 2 x cos ((2Pi/n)/2) P(1). Note the vector sum is in opposite direction of P(1). ( you can draw a picture to visualise this)

Rewrite this as P(1a) + P(1b) = - 2 cos (Pi/n) P(1), i.e. P(ka) and P(kb) are the 2 vectors from center of the regular polygon to the vertices adjacent to the side that is directly opposite to vertex k.

Then by symmetry, sum vectors P(ka) + P(kb) = - 2 cos (Pi/n) P(k). Repeat such equation for k = 1 to n.

Adding these n equations. By symmetry, left hand side = 2 S, and right hand size = - 2 cos (Pi/n) S. (Remember vector sum S = P(1) + P(2) +... P(k) + ...+ P(n))

Thus 2 S = - 2 cos (Pi/n) S => 2 [ 1 + cos (Pi/n) ] S = zero vector.

Since [ 1 + cos (Pi/n) ] is never zero, that means vector S = zero vector, i.e. P(1) + P(2) + ... + P(n) = zero vector

QED.