1997 IMO Problems/Problem 1
Problem
In the plane the points with integer coordinates are the vertices of unit squares. The squares are colored alternatively black and white (as on a chessboard).
For any pair of positive integers and , consider a right-angled triangle whose vertices have integer coordinates and whose legs, of lengths and , lie along edges of the squares.
Let be the total area of the black part of the triangle and be the total area of the white part.
Let
(a) Calculate for all positive integers and which are either both even or both odd.
(b) Prove that for all and .
(c) Show that there is no constant such that for all and .
Solution
Let , , , and , be the lower left vertex, lower right vertex, upper right vertex, and upper left vertex of rectangle
For any pair of positive integers and , consider a rectangle whose vertices have integer coordinates and whose legs, of lengths and , lie along edges of the squares.
Let be the total area of the black part of the rectangle and be the total area of the white part.
Let
Now we do part (a) case: and which are both even
Since and which are both even, the total area of the rectangle is
Since every row has an even number of squares there are equally as many white squares than black squares for each row.
Since every column has an even number of squares there are equally as many white squares than black squares for each column.
This means that in the rectangle there are equal number of white squares and black squares.
Therefore and
(a) Calculate for all positive integers and which are either both even or both odd.
(b) Prove that for all and .
(c) Show that there is no constant such that for all and .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.