Difference between revisions of "1997 IMO Problems/Problem 1"
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Revision as of 11:01, 16 November 2023
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
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 , , , and , be the lower left vertex, lower right vertex, upper right vertex, and upper left vertex of rectangle respectively.
Let be the total area of the black part of the rectangle and be the total area of the white part.
Let
Part (a):
case: and which are both even
Since and are both even, the total area of the rectangle is which is also even
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
Let be the midpoint of line . Them is at coordinate Since both and are even, then has integer coordinates.
Starting with vertex , because the length of is even, then the color for the square inside rectangle closest to is the opposite color of the square inside rectangle closest to , then starting with vertex , because the length of is even, then the color of the square inside rectangle closest to is the opposite color of the square inside rectangle closest to . this means that the color of the square inside rectangle closest to is the same as the color of the square inside rectangle closest to . Likewise, the color of the square inside rectangle closest to is the same as the color of the square inside rectangle closest to .
This color pattern and the fact that the midpoint has integer coordinates indicates that triangle has the same color pattern as triangle rotated 180 degrees.
Therefore, the white area in triangle is the same as the white area in triangle and the black area in triangle is the same as the black area in triangle .
Thus and , which gives
Therefore when both and are even.
case: and which are both odd
Since and are both odd, the total area of the rectangle is which is also odd
Since the total area is odd, then is not an integer but and are.
This means that in the rectangle there are squares of one color and squares of the other color
Let be the midpoint of line . Them is at coordinate Since both and are odd, then has non-integer coordinates coordinates but their decimal portions are both . This means that lies in the middle of the center square.
Starting with vertex , because the length of is odd, then the color for the square inside rectangle closest to is the same color of the square inside rectangle closest to , then starting with vertex , because the length of is odd, then the color of the square inside rectangle closest to is the same color of the square inside rectangle closest to . this means that the color of the square inside rectangle closest to is the same as the color of the square inside rectangle closest to . Likewise, the color of the square inside rectangle closest to is the same as the color of the square inside rectangle closest to .
This color pattern and the fact that the midpoint in in the center of the middle square that triangle has the same color pattern as triangle rotated 180 degrees.
Therefore, the white area in triangle is the same as the white area in triangle and the black area in triangle is the same as the black area in triangle .
Thus
Therefore when both and are odd.
To summarize part (a),
Part (b):
(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.