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?