Prove that a function defined on points in a plane is zero Let $n\ge3$ be an integer, and $f:P\to\mathbb R$ be a function defined on any point in the plane $P$, with the property that for any regular n-gon $<A_1A_2A_3\cdots A_n>$,
$$f(A_1)+f(A_2)+f(A_3)+\cdots+f(A_n) = 0 \quad \forall n\in\mathbb N/\{1,2\}$$
Prove that $f$ is the zero function, i.e, $f(t) = 0 \quad\forall t\in P$.

My initial idea was that if I could prove that $f$ is constant, my proof would be complete. I proceeded with my idea like shown in the diagram which I have attached because it is slightly hard to put it into words. Forgive my extremely poor paint skills.

I label the red points as $A_2$ and $A_3$(which is which doesn't matter), the blue as $A_1$ and the green as $A_4$. $A_1A_2A_3$ and  $A_2A_3A_4$ are supposed to equilateral triangles, (regular 3-gon feels weird)
Now given that $f(A_1)+f(A_2)+f(A_3)=0$ and $f(A_4)+f(A_2)+f(A_3)=0$, we can conclude that $f(A_1)=f(A_4)$. I continue my argument by asserting that this can be done for any two arbitrarily chosen points in the plane by taking $A_2$ and  $A_3$ on the perpendicular bisector of $A_i$ and $A_j$. Thus $f(A)$ is constant for all points on the plane. Thus $f$ is the zero function.
Are there any flaws in my argument? Is this rigorous? Anyone has any better ideas? By better I mean I know I could prove this algebraically, but this geometrical argument (if it is correct) is a lot more elegant than any algebraic argument I can think of, but maybe someone can do even better?
 A: The vertices of a regular $n$-gon in argand plane with center $z$ and one of the vertices $z+\omega$, are $\{z+\omega\alpha^j| j=1,\ldots,n \}$ where, $\alpha=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$.
So, the functon satisfies $\sum\limits_{j=1}^nf(z+\omega\alpha^j)=0$, for arbitrary $z,\omega \in \mathbb{C}$.
In particular replacing $z$ with $z-\omega\alpha^k$, for a fixed $k$ and choosing $\omega=1$ gives us, 
$\sum\limits_{j=1}^nf(z-\alpha^k+\alpha^j)=0$
Summing over all $k=1,\ldots ,n$ we have, $\sum\limits_{k=1}^n\sum\limits_{j=1}^nf(z-\alpha^k+\alpha^j)=0$
or, $\sum\limits_{k=1}^n\sum\limits_{l=1}^nf(z-\alpha^k+\alpha^{k+l})=0$ (where, $j \equiv k+l \mod n $)
Now, $\sum\limits_{k=1}^n\bigg(\sum\limits_{l=1}^n f(z-\alpha^k+\alpha^{k+l})\bigg) = nf(z) + \sum\limits_{k=1}^n\bigg(\sum\limits_{l=1}^{n-1} f(z-\alpha^k+\alpha^{k+l})\bigg)$
$= nf(z) + \sum\limits_{l=1}^{n-1}\bigg(\sum\limits_{k=1}^n f(z-\alpha^k+\alpha^{k+l})\bigg) = nf(z)$
[Since, when $l=n$, $f(z-\alpha^k+\alpha^{k+l})=f(z)$ 
and for $l\neq n$, $\bigg(\sum\limits_{l=1}^n f(z-\alpha^k+\alpha^{k+l})\bigg)= \bigg(\sum\limits_{l=1}^n f(z-(1-\alpha^{l})\alpha^k)\bigg)=0$]
Thus, $nf(z)=0$, or $f(z)=0$.
A: Suppose $f(O)=\varepsilon$, where O is the origin, and $\varepsilon\neq0$.
Consider the points $A(\cos( \pi/6), \sin( \pi/6))$ and $B(\cos( \pi/6), -\sin (\pi/6))$.  Then OAB is an equilateral triangle so $f(A)+f(B) = -\varepsilon$
Now consider the equilateral triangle ABC, where $C=(2\cos (\pi/6), 0)$.  Hence $f(C)=\varepsilon$.
By similar reasoning, we must have points $D(0, 2\cos (\pi/6)), E(-2\cos (\pi/6), 0), F(0, -2\cos (\pi/6))$ such that $f(D)=f(E)=f(F)=\varepsilon$.   
But CDEF is a square, so we have a contradiction.  QED
