Are functions that sum to zero over vertices of similar polygons identically zero?

Question

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.

Answer 1

To visualize what subrosar probably meant with his "regular $n$-gon"-technique, I created the following picture for the case of the 7-gon

In his post, subrosar used the same technique for the triangle.

You create $n$ copies of the $n$-gone, so that they all meat in a single vertex, and you arrange them in a symmetric pattern (the black 7-gons in the picture). You then "cancel out" all vertices but the central one by inserting further concentric $n$-gons. This works similar for all regular $n$-gons, but it is best seen for odd $n$, as then no additional vertices coincide.

This technique can be generalized to polygons in which all vertices lie on a common circle, e.g. for this irregular 5-gon:

And in fact, there is an even more general way to do this that works for all polygons, convex and non-convex:

Let $P$ be a polygon with vertices $v_1,...,v_n$. Choose a point $x\in\Bbb R^2$ that is not a vertex of $P$. Now, for each vertex $v_i$ of $P$, place a polygon $P_i$ in $\Bbb R^2$, that is congruent to $P$ (by rotation and scaling, no mirroring), and has its vertex $v_1^i$ (its version of the vertex $v_1$) at $x$, and its vertex $v_2^i$ at $v_i$.

The points $\{v_k^i\mid i=2,...,n\}$ now form the vertices of a polygon $\bar P_k$, that is congruent to $P$.

If $x$ is choosen to be the circumcenter of the polygon and all vertices of the polygon are on that circumcircle, we just obtain the same technique as above.

So, indeed, the task can be stated with any polygon (or if you want so, with any set of points of size at least three).

Answer 2

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 $z \in \mathbb C$.