Call the $(2n+1)$-gon $P$, and label an arbitrary vertex $A$. We assume without loss of generality that $A$ is one of the chosen vertices. Let the other two chosen vertices in clockwise order be $B$ and $C$, respectively, and suppose there are $k$ edges of $P$ contained within minor arc $BAC$. Because $\triangle ABC$ contains the center of $P$, we have $\angle BAC < 90^\circ$. This implies $k\cdot \frac{180}{2n+1} < 90$; hence $1\le k \le n$.
For a fixed $k$ between $1$ and $n$ inclusive, it is easy to verify that there are $k$ choices of $B$ and $C$ such that exactly $k$ edges of $P$ are contained within minor arc $BAC$ . This gives us a total of $\sum_{k=1}^n k = n(n+1)/2$ valid choices for $B$ and $C$. Because we have $\binom{2n}{2}$ ways to choose $B$ and $C$ from all the remaining vertices of $P$, the probability that a randomly chosen triangle contains the center of $P$ is $$ \frac{\frac{n(n+1)}{2}}{\binom{2n}{2}} = \frac{n+1}{2(2n-1)}. $$
Note: This probability implies there are a total of $\frac{n(n+1)(2n+1)}{6} = \sum_{k=1}^n k^2$ triangles that contain the center of $P$. Is there also a bijective proof that counts this directly?
It isn't a necessary and sufficient condition.
- The implication : "points $z_k$s are the vertices of a regular polygon" (assumed indexed in the trigonometric order) implies "sum = $0$" is true.
Indeed the vertices of a regular $n$-gon are the images of the "standard" regular $n$-gon using this "similitude"
$$s(z)=A(z-z_G) \ \ \text{with} \ \ A=re^{i \theta } \ \ \text{and} \ \ z_G=\frac1n \sum z_k \tag{1}$$
where $\theta$ is the rotation angle, $r$ is the enlargment/shrinking factor and $z_G$ is the barycenter of points $z_k$.
Relationship (1) can be expressed in a slightly different way:
$$s(z)=Az+B$$
Therefore
$$\sum_{k=1}^n z_k \omega_n^k=\sum_{k=1}^n s(\omega_n^k) \omega_n^k=\sum_{k=1}^n (A\omega_n^k +B) \omega_n^k$$
$$=A\underbrace{\sum_{k=1}^n \omega_n^{2k}}_0 +B \underbrace{\sum_{k=1}^n \omega_n^k}_0=0\tag{2}$$
The fact that the first summation in (2) is zero is a consequence of a more general result one can find here. A specific proof could also be given by distinguishing the cases $n$ odd/$n$ even.
- Whereas, in the other direction, "sum=$0$" doesn't imply "points $z_k$ constitute a regular polygon".
Here is a counterexample for $n=4$ with $w_4=i$ :
$$z_1=0+0i,\ \ z_2=3-2i, \ \ z_3=-1+2i, \ \ z_4=1-3i$$
their sum
$$\sum_{k=1}^4 z_k \omega_4^k=(0)i+(3-2i)i^2+(-1+2i)i^3+(1-3i)i^4$$
$$=-(3-2i)-i(-1+2i)+(1-3i)=0$$
whereas the $z_k$ aren't the vertices of a square.
Fig 1: Case $n=8$: A "continuous" mapping of the regular octagon made by the $w_8^k$s onto the regular octagon of the $z_k$s. The "continuous" aspect is of course unimportant ; I have used it because of its aesthetical appeal.
Matlab program of the figure:
n=8;
z=exp((i*2*pi/n)*(0:n));
A=@(t)(t);B=@(t)(1-exp(i*(t-1)));
for t=1:0.1:4
plot(A(t)*z+B(t));
end;
Best Answer
If the lines from the center of the polygon to the points on the sides have the same angle between consecutive lines ($2\pi/n$), the resulting polygon will again be regular so its center will be the same as the original one.
I will leave it to others to work out what happens when the angles between consecutive lines are arbitrarily specified (the only restrictions being that they are positive and total $2\pi$).