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

[john562]

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)?

 

[Yuqing]

Those are smaller squares within the larger square ABCD. To be specific, they are the four quarters of ABCD

 

[john562]

I don't understand how adding them equals zero.

 

[mxbob468]

what is this? like the easiest putnam problem ever?

 

[CellCoree]

These equations are obtained by considering the square ABCD with center P and its four smaller squares (AEFP, BFGP, CGHP, DHPE) formed by connecting P to the midpoints of the sides of ABCD. By the given property of f, the sum of the values at the four corners of each of these squares must equal 0. Therefore, we can write the equations as shown above, where P is the shared corner of all four smaller squares. By adding and subtracting these equations as shown, we can eliminate the values at the corners and isolate f(P), which must equal 0.