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Brazil counting and probability
Problem
A real number with absolute value less than is written in each cell of an array, so that the sum of the numbers in each square is zero. Show that for odd the sum of all the numbers is less than .


Solution
Consider a -shaped piece formed by a square and two neighbouring squares. The sum of its numbers is the opposite of the number that completes the to a square, so the sum of numbers in any is less than . We proceed by induction on . For , divide the array in four regions: a square, an and two single cells. The sum is less than , because of the single cells and the . For bigger odd , divide the array in a square of side , several squares of side and a : The sum of the square of side is less than by induction hypothesis and there are only an and a single square adding to less than . So the sum is less than and the proof is complete.
Techniques
Induction / smoothingColoring schemes, extremal arguments