# Understand the problem

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?

Putnam 2009 A1

Geometry
Easy
##### Suggested Book
Mathematical Olympiad Challenges by Titu Andreescu

# Start with hints

Do you really need a hint? Try it first!

If the function is indeed identically 0, then you would need a geometrical argument to prove it. If not, you would need to find an example of a function which is not identically zero and satisfies the hypothesis of the question. In principle, you should start by playing around with some examples.
The answer to the question is YES.
Think about squares that have $P$ as a vertex. Consider the squares in picture. Clearly, $f(D)+f(C)+f(P)+f(E)=0 \\ f(C)+f(B)+f(I)+f(P)=0 \\ f(E)+f(P)+f(G)+f(F)=0 \\ f(P)+f(I)+f(H)+f(G)=0$ Adding, we get $4f(P)+2(f(C)+f(E)+f(G)+f(I))+ (f(D)+f(B)+f(H)+f(F))=0$.   As both $CEGI$ and $DBHF$ are squares, the second and the third terms are zero. Thus, $f(P)=0$.

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