Putnam (2009): Solution to problem A1

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:

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} }\ \$. $\quad \blacksquare$