In a game, all points in the (x, y) plane with coordinates obeying $ x, y \in \mathbb{Z} $ are labelled as
belonging to one of three players, i.e., either to Alice, Bob or Carol.
Show that one of the players will possess four points whose vertices form a rectangle.
Here is the problem I've been thinking for days. It seems easy since the coordinates are unlimited.
But it is also hard to find pigeonholes I want to divide.
I've encountered several problems like finding a parallelogram on a $n \times n$ chessboard with $2n$ pawns. It is relatively simpler to me.
Best Answer
HINT: Consider the points $\langle 0,n\rangle$, $\langle 1,n\rangle$, $\langle 2,n\rangle$, and $\langle 3,n\rangle$ for $n=0,1,2,\ldots\;$. Imagine that each point is labelled $A$, $B$, or $C$. Thus, one row might be labelled $ABCA$, another might be labelled $BCCB$, and so on.