Prove f(P) = 0 for Any Point P in the Plane | Putnam A1 Question Help

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The discussion centers on proving that if a real-valued function f satisfies the condition f(A) + f(B) + f(C) + f(D) = 0 for every square ABCD in the plane, then f(P) must equal 0 for any point P. The proof involves considering a square centered at point P and examining the midpoints of its sides, leading to a series of equations that ultimately show 0 = 4f(P). A participant expresses confusion about the reasoning behind the equations involving smaller squares within ABCD, questioning how their sums also equal zero. The conversation highlights the clarity of the proof while addressing the challenge some may face in understanding the logic behind it. The problem is noted as potentially one of the easier questions from the Putnam competition.
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Question:
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?

Answer:
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
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):
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.

Comments:
I don't understand why
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)?
 
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Those are smaller squares within the larger square ABCD. To be specific, they are the four quarters of ABCD
 
I don't understand how adding them equals zero.
 
what is this? like the easiest putnam problem ever?
 
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