Function in the Plane That Vanishes

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. Let P be a point in the plane and ABCD any square with the center at P. The midpoints X, Y, Z, W of the sides of the square form another square with the same center P. Moreover, joining the opposite points of that square divides ABCD into four smaller squares, say, AXPW, BYPX, CZPY, DWPZ, to which of each we may apply the given property of function f:

Now observe that, since ABCD and XYZW are squares,

So adding up the four identities in (*) gives 4f(P) = 0, implying f(P) = 0. As P was an arbitrary point in the plane, f is identically 0.

Note

The conditions of the problem are too strong. It would suffice to select a couple of the axes and require the identity f(A) + f(B) + f(C) + f(D) = 0 for the corners of any parallelogram (or square) with sides parallel to the axes. The problem would make sense also if restricted to the grid points of a regular grid. For the problem on the grid, ABCD must be a grid square (or parallelogram) with the center also a grid point.

I think that a more interesting problem comes up if one considers vertices of a triangle:

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

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

Solution

The answer again is Yes. Let P be an arbitrary point in the plane and ABCDEF a regular hexagon with the center P. All triangles ABP, BCP, CDP, DEP, EFP, FAP are regular. We have six equalities:

Summing all up gives

But triangles ADE and BDF are also equilateral, implying

and consequently 6f(P) = 0.