IMO 1997 Problem 1

In the plane the points with integer coordinates are the vertices of unit squares. The squares are coloured alternately black and white (as on a chessboard). For any pair of positive integers $ m$ and $ n$, consider a right-angled triangle whose vertices have integer coordinates and whose legs, of lengths $ m$ and $ n$, lie along edges of the squares. Let $ S_1$ be the total area of the black part of the triangle and $ S_2$ be the total area of the white part. Let $ f(m,n):=| S_1 - S_2 |$.

a) Calculate $ f(m,n)$ for all positive integers $ m$ and $ n$ which are either both even or both odd.

b) Prove that $ f(m,n) \leq \frac 12 \max \{m,n \}$ for all $ m$ and $ n$.

c) Show that there is no constant $ C\in\mathbb{R}$ such that $ f(m,n) < C$ for all $ m$ and $ n$.

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(a) If both $m$ and $n$ are even then $f(m,n)=0$, otherwise $f(m,n)=1/2$. The former follows from symmetry and the latter from $2f(m,n)=1$.

(b) Let $n\le m$ and consider $m$ columns and $n$ rows. The idea is to consider the $2i+1$-th and $2i+2$-th columns as a group. It could be seen that the area difference in each group is at most $1$. Hence if $m$ is even it is obvious. If $m$ is odd, then the total difference from all but the last column is at most $(m-1)/2$, and adding the last column the difference is bounded from above by $$(m-1)/2+1/2=m/2.$$

(c) Consider the right triangle of sides $n$ and $n+1$ where $n$ is even. Within whole squares, whose total area is $n^2/2$, from (a) both colors have the same area. For the rest, whose total area is $n/2$, it could be seen that one color has area $$\left\{\left(\frac{n}{n}\right)^2+\dots+\left(\frac{1}{n}\right)^2\right\}\frac{n}{2(n+1)}=\frac{2n+1}{12}$$ and hence the other color has area $\frac{4n-1}{12}$. Thus $$f(n,n+1)=\frac{n-1}{6},$$ which is unbounded.