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Let $n>2$ be an integer. Let $f$ be a real-value function on a plane such that for every regular n-gon with vertices $A_1,A_2,...,A_n$, $f(A_1)+f(A_2)+...+f(A_n)=0$. Prove that $f$ is zero function.

Putnam 2009 asked to prove this for the case of a square which is much easier for me. However I don't have a clue on how to prove the general case.

Rikka
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    The AOPS forums usually have good solutions. – Sophie Nov 21 '16 at 08:12
  • Could you show your solution for the case of a square? – Student Nov 21 '16 at 08:28
  • BTW this is a 1996 Romanian team selection IMO question. – Dair Nov 21 '16 at 08:31
  • @Supermario: You can find a solution here: http://kskedlaya.org/putnam-archive/2009s.pdf (first question) – Dair Nov 21 '16 at 08:34
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    @Supermario: Let $\langle x,y\rangle$ be any point in the plane. Consider the nine points $\langle x+i,y+j\rangle$ with $i,j\in{-1,0,1}$. Add the values of $A$ at the four corner points, twice the values of $A$ at the four side points, and four times the value at the centre point. The total is $0$, since it's the sum of the totals for the four unit squares. The total for the corner points is $0$, and so is the total for the side points, which also form a square. Thus, the value at the centre must also be $0$. – Brian M. Scott Nov 21 '16 at 08:38
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    There is a solution here: http://www.artofproblemsolving.com/community/c6h51933p327816 – Dair Nov 21 '16 at 08:40
  • @Dair: Thank you – Rikka Nov 21 '16 at 08:45

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