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2000 USAMO Problems/Problem 4

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Problem

Find the smallest positive integer n such that if n squares of a 1000 \times 1000 chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board.

Solution

We claim that n = 1999 is the smallest such number. For n \le 1998, we can simply color any of the 1998 squares forming the top row and the left column, but excluding the top left corner square.

for(int i = 0; i < 10; ++i){ for(int j = 0; j < 10; ++j){  if((i == 0 || j == 9) && !(j-i == 9)) fill(shift(i,j...

We now show that no configuration with no colored right triangles exists for n = 1999. We call a row or column filled if all 1000 of its squares are colored. Then any of the remaining 999 colored squares must share a column or row, respectively, with one of the colored squares in a filled row or column. These two squares, and any other square in the filled row or column, form a colored right triangle, giving us a contradiction. Hence, no filled row or column may exist.

Let m be the number of columns with 1 colored square. Then there are 1999-m colored squares in the remaining columns, and in each of these < 1999-m columns that have a colored square must have at least two colored squares in them. These two colored squares will form a triangle with any other colored square in either of the rows containing the colored squares. Hence, each of the 1999-m colored squares must be placed in different rows, but as there are only 1000 rows, the inequality 1999 - m \le 1000 \Longrightarrow m \ge 999 holds. If m = 1000, then each column only has 1 colored square, leaving no place for the remaining 999, contradiction. If m = 999, then each of the 1000 rows has 1 black square, leaving no place for the other 999, contradiction. Hence n = \boxed{1999} is the minimal value.

See also

2000 USAMO (Problems • Resources: AoPS | ML)
Preceded by
Problem 3
1 2 3 4 5 6 Followed by
Problem 5
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