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Disclaimer: these are my personal solutions; they have not been reviewed, so be careful relying on them as there might be errors. Some competitions may have official solutions from the contest publishers which you may wish to refer to instead if you’re looking for verified solutions.

Problem A1

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

Open to see the solution; try solving it first!

Let’s consider a $3 \times 3$ grid of points $A, B, C, D, E, F, G, H, I$ arranged in alphabetical order, from left to right and top to bottom, such that point $E$ has position $(x, y)$ and the other points are displaced from $E$ by an arbitrary real value $t$ in a given direction, namely:

  • $A$ is at position $(x - t, y + t)$
  • $B$ is at position $(x , y + t)$
  • $C$ is at position $(x + t, y + t)$
  • $D$ is at position $(x - t, y )$
  • $E$ is at position $(x , y )$
  • $F$ is at position $(x + t, y )$
  • $G$ is at position $(x - t, y - t)$
  • $H$ is at position $(x , y - t)$
  • $I$ is at position $(x + t, y - t)$

We can visualize this1 as follows:

We can see that this arrangement includes the following (small) squares:

  • $ABED$
  • $BCFE$
  • $DEHG$
  • $EFIH$

As well as the square at a $45\degree$ angle $BDHF$ and the grand square $ACIG$:

Starting with the small squares, we know that the following hold:

$$ \begin{align*} f(A) + f(B) + f(D) + f(E) & = 0 \\ f(B) + f(C) + f(E) + f(F) & = 0 \\ f(D) + f(E) + f(G) + f(H) & = 0 \\ f(E) + f(F) + f(H) + f(I) & = 0 \\ \end{align*} $$

We can add all of these equations together to form the following:

$$ f(A) + 2 f(B) + f(C) + 2 f(D) + 4 f(E) + 2 f(F) + f(G) + 2 f(H) + f(I) = 0 $$

Rearranging this to group by similar factors and grouping accordingly, we have:

$$ (\underbrace{f(A) + f(C) + f(I) + f(G)}_{\textrm{the grand square}}) + 2( \overbrace{f(B) + f(D) + f(H) + f(F)}^{\textrm{the } 45\degree \textrm{ square}} ) + 4 f(E) = 0 $$

Since one of the terms is now the grand square $ACIG$, the value of that term is zero by problem definition. Similarly, the second term is the $45\degree$ square $BDHF$, so that term is also zero for the same reason.

Thus, we are just left with

$$ 4f(E) = 0 $$

which means that $f(E) = 0$. However, recall that $E$ represents an arbitrary point $(x, y)$ in the plane, so this shows that $\boxed{ \textrm{yes, } f(P) = 0 \textrm{ for all points } P \textrm{ in the plane} }\ \ $. $\quad \blacksquare$

  1. The diagrams on the page were programmatically defined via g9, an open source JavaScript toolkit for interactive visualizations. ↩︎