Putnam 2009, Problem A-1

Let $f$ be a real-valued function on the plane such that for every square $ABCD$ in the plane, $f(A) + f(B) + f(C) + f(D) = 0$. Does it follow that $f(P) = 0$ for all points $P$ in the plane?

Solution:
Yes, it does follow. Let \( P \) be any point in the plane. Let \( ABCD \) be any square with center \( P \). Let \( E,F,G,H \) be the midpoints of the segments \( AB,BC,CD,DA \), respectively. The function \( f \) must satisfy the equations
\[
\begin{aligned}
0 &= f(A) + f(B) + f(C) + f(D) \\
0 &= f(E) + f(F) + f(G) + f(H) \\
0 &= f(A) + f(E) + f(P) + f(H) \\
0 &= f(B) + f(F) + f(P) + f(E) \\
0 &= f(C) + f(G) + f(P) + f(F) \\
0 &= f(D) + f(H) + f(P) + f(G).
\end{aligned}
\]
If we add the last four equations, then subtract the first equation and twice the second equation, we obtain \( 0 = 4f(P) \), whence \( f(P) = 0 \).

**Remark.** Problem 1 of the 1996 Romanian IMO team selection exam asks the same question with squares replaced by regular polygons of any (fixed) number of vertices.