PROBLEM OF THE MONTH

OCTOBER 2001


 






Let X be a region in the plane with area strictly greater then one. Show that X contains two at least two distinct points with coordinates (a,b) and (c,d) such that a-c and b-d are both integers.

Cover the plane with non overlapping squares of edge one, whose lower left corners are points with integer coordinates. More precisely, use the squares $ S_{m,n}$ with lower left corners $ (m,n)\in{\bf Z}\times {\bf Z}$ where

$\displaystyle S_{m,n}=\{(x,y)\mid m\le x< m+1, n\le y< n+1\}.$

The region $ X$ is then decomposed into pieces $ X\cap S_{m,n}$ by the squares $ S_{m,n}$. Denote by $ O=(0,0)$ the origin and translate each square $ S_{m,n}$ along the segment joing $ O$ and $ (m,n)$ so that all squares are superimposed onto $ S_{0,0}$. The pieces $ X\cap S_{m,n}$ are therefore translated onto subsets $ X_{m,n}$ of the square $ S_{0,0}$. The total area of all these pieces is the area of $ X$ which is assumed to be greater than one ($ =$ the area of the square $ S_{0,0}$), hence at least two such translates must meet. Let $ p=(x,y)$ be a point which is both in $ X_{m,n}$ and $ X_{r,s}$ with $ (m,n)\ne (r,s)$. It follows that the point $ (a,b)=(x+m,y+n)$ lies in $ X\cap S_{m,n}$, while $ (c,d)=(x+r,y+r)$ lies in $ X\cap S_{r,s}$. Thus both $ (a,b)$ and $ (c,d)$ lie in $ X$, and furthermore $ a-c=m-r$ and $ b-d=n-s$ are both integers.

The simple way to phrase what we have just proved is that if $ X$ is a region in the plane with area greater than one, then there exist a point $ P$ with integer coordinates such that $ X$ and its translate along the segment $ OP$ have a point in common.