# Function in the Plane That Vanishes

Here is problem A1 from *70th Annual William Lowell Putnam Mathematical Competition* (2010).

Let f be a real-valued function on the plane such that for every square ABCD in the plane,

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Copyright © 1996-2018 Alexander Bogomolny

Let f be a real-valued function on the plane such that for every square ABCD 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:

(*) |
f(A) + f(X) + f(P) + f(W) = 0, f(B) + f(Y) + f(P) + f(X) = 0, f(C) + f(Z) + f(P) + f(Y) = 0, f(D) + f(W) + f(P) + f(Z) = 0. |

Now observe that, since ABCD and XYZW are squares,

f(A) + f(B) + f(C) + f(D) = 0, and f(X) + f(Y) + f(Z) + f(W) = 0. |

So adding up the four identities in (*) gives 4f(P) = 0, implying

#### Note

The conditions of the problem are too strong. It would suffice to select a couple of the axes and require the identity

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,

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Copyright © 1996-2018 Alexander Bogomolny

Let f be a real-valued function on the plane such that for every equilateral triangle ABC 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:

f(A) + f(B) + f(P) = 0, f(B) + f(C) + f(P) = 0, f(C) + f(D) + f(P) = 0, f(D) + f(E) + f(P) = 0, f(E) + f(F) + f(P) = 0, f(F) + f(A) + f(P) = 0. |

Summing all up gives

2(f(A) + f(C) + f(E)) + 2(f(B) + f(D) + f(F)) + 6f(P) = 0. |

But triangles ADE and BDF are also equilateral, implying

f(A) + f(C) + f(E) = f(B) + f(D) + f(F) = 0, |

and consequently 6f(P) = 0.

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Copyright © 1996-2018 Alexander Bogomolny