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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?

Below is my solution:

Create a 3x3 set of boxes (where each box has equal lengths as all of the other boxes) like so:

. . . .

. . . .

. . . .

. . . .

The periods represent vertices. Now, denote the value of the function at each vertice by a(1),...,a(16).

Now, since the sum of any four vertices making a square is zero, we get 18 unique equations (each containing a linear relationship between a unique quadruple of vertices):

9 (each representing a unit box)

+

4 (each representing a box of side length 2 units)

+

1 (this equation relates the vertices of the entire 3x3 square)

+

4 (each representing a box of side length sqrt(2) units (the diagonal squares)

Writing these into a matrix, we get an 18x16 matrix mapping a vector whose coordinates are the values of the function at each vertex) to the zero vector. Since we have 18 homogeneous equations and 16 variables, all of the variables must be zero. From this we can clearly conclude that the value of the function is zero for everywhere on the plane.

Is this solution correct? I can't see any errors, I just thought it was an interesting answer (it was different from the official answer, which I've posted below) and wasn't sure if I was missing anything.

OFFICIAL 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.

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# Problem A1 from the 2009 Putnam Exam

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