ehsan2004 wrote: Prove that $f(x)=0$ for all reals $x$. You wanted to say "$f(A)=0$ for all points $A$", right?

perfect_radio wrote: ehsan2004 wrote: Prove that $f(x)=0$ for all reals $x$. You wanted to say "$f(A)=0$ for all points $A$", right? excuse me, my meant was $f(x)\equiv 0$

Take $A \neq B$. Let $\ell$ be the perpendicular bisector of $AB$. Construct a rhombus $ACBD$, with $C,D \in \ell$ and $\measuredangle DAC = \measuredangle DBC = \dfrac{\pi}{3}$. This yields $f(A)+f(C)+f(D)=0=f(B)+f(C)+f(D)$, so $f(A)=f(B)=t$, $\forall A \neq B$. Therefore $nt=0$, so $t=0$. Have I done something wrong? It looks too good to be true (because I used the condition given only for $n=3$)

Actually, $n$ is fixed in the original statement.

enescu wrote: Actually, $n$ is fixed in the original statement. Oops... sorry . do you know the solution for $n \geq 4$?

Yes. Let $A$ be an arbitrary point. Consider a regular $n-$gon $AA_{1}A_{2}\ldots A_{n-1}.$ Let $k$ be an integer, $0\leq k\leq n-1.$ A rotation with center $A$ of angle $\dfrac{2k\pi}{n}$ sends the polygon $AA_{1}A_{2}\ldots A_{n-1}$ to $A_{k0}A_{k1}\ldots A_{k,n-1},$ where $A_{k0}=A$ and $A_{ki}$ is the image of $A_{i}$, for all $i=1,2,\ldots,n-1.$ From the condition of the statement, we have \[ \sum_{k=0}^{n-1} \sum_{i=0}^{n-1}{f(A_{ki})}=0. \] Observe that in the sum the number $f(A)$ appears $n$ times, therefore \[ nf(A)+\sum_{k=0}^{n-1} \sum_{i=1}^{n-1}{f(A_{ki})}=0. \] On the other hand, we have \[ \sum_{k=0}^{n-1} \sum_{i=1}^{n-1}{f(A_{ki})}=\sum_{i=1}^{n-1} \sum_{k=0}^{n-1}{f(A_{ki})}=0, \] since the polygons $A_{0i}A_{1i}\ldots A_{n-1,i}$ are all regular $n-$gons. From the two equalities above we deduce $f(A)=0,$ hence $f$ is the zero function.

We may assume \(n\) is even, since for \(n\) odd, the sum of the vertices in any \(2n\)-gon is zero. Now let \(A_1\cdots A_n\) be a regular \(n\)-gon. For each \(i\) and \(j\), let \(M_{ij}\) be the midpoint of \(\overline{A_iA_j}\) (so in particular, \(M_iM_i=A_i\)), and let \(O\) be the center of the \(n\)-gon. We know since \(M_{i1}M_{i2}\cdots M_{in}\) and \(M_{1,1+i}M_{2,2+i}\cdots M_{n,n+i}\) are regular \(n\)-gons that \begin{align*} 0=\sum_i\sum_jf(M_{ij}) =n\cdot f(O)+\sum_j\sum_{\substack{i<n\\ i\ne n/2}}f(M_{j,j+i}) &=n\cdot f(O) \end{align*}

ehsan2004 wrote: Let $ f: \mathbb{R}^2 \rightarrow \mathbb{R} $ be a function such that for every regular $ n $-gon $ A_1A_2 \ldots A_n $ we have $ f(A_1)+f(A_2)+\cdots +f(A_n)=0 $. Prove that $ f(x)=0 $ for all reals $ x $. The claim for just $n=4$: https://aops.com/community/p1703551