TITLE: Are functions that sum to zero over vertices of similar polygons identically zero?
QUESTION [12 upvotes]: This is a generalization of problem A1 from the 2009 Putnam competition. The original problem asks for proof that any function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ that sums to zero over the vertices of any square is necessarily zero everywhere. That is, if for any square with vertices $A$,$B$,$C$, and $D$,
\begin{equation}
f(A)+f(B)+f(C)+f(D)=0
\end{equation}
then $f(P)=0$ for any point $P\in \mathbb{R}^2.$
You can find the solution to this question here: Putnam 2009 A1 Points in a plane.
My question is whether this result can be generalized to any family of similar polygons. If $\Sigma_{i}f(A_i)=0$ whenever $A_1,...,A_n$ are the vertices of a polygon similar to a given polygon, does $f$ need to be $0$ everywhere?
This can be proven for regular $n$-gons by a method I describe in the comments. The proof for equilateral triangles amounts to adding the vertices of the three blue equilateral triangles in the picture and then subtracting the vertices of the the two red triangles leaving $3f(A)=0$ and, since $A$ is arbitrary, proving $f$ is zero everywhere. Essentially the same argument applies to any $n$-gons if you draw them evenly arranged around the center point. It doesn't matter if they overlap.

REPLY [4 votes]: Let me reformulate M. Winter's answer as follows.
Embed $\mathbb R^2$ to $\mathbb C$, the original question can be restated as below:

Let $f : \mathbb C \to \mathbb R$ be a function. Suppose there exists pairwise distinct $v_0 = 0$ and $v_1 \cdots, v_n \in \mathbb C$ such that $$\sum_{i=0}^n f(z+\alpha v_i) = 0$$ for all $\alpha \neq 0 \in \mathbb C$ and $z \in \mathbb C$. Show that $f \equiv 0$.

Without loss of generality, suppose none of $v_i$ is $-1$. For $0 \leq i,j \leq n$ Let $w_{ij} = v_i(v_j + 1)$. From the definition of $f$, for $0 \leq j \leq n$,
$$
\sum_{i=0}^n f(w_{ij}) = \sum_{i=0}^n f(0+(v_j + 1)v_i) = 0.
$$
Similarly, for $1 \leq i \leq n$,
$$
\sum_{j=0}^n f(w_{ij}) = \sum_{j=0}^n f(v_i + v_iv_j) = 0.
$$
Then
$$
\begin{aligned}
(n+1)f(0) &= \text{(the sum of the $n+1$ equations in the top)}\\
&- \text{(the sum of the $n$ equations in the bottom)} = 0.
\end{aligned}
$$
So $f(0) = 0$, in the same manner $f(z) = 0$ for all $z \in \mathbb C$.