MathLabs

Problem 2

Into each box of a 2012 by 2012 square grid, insert a real number between 0 and 1 inclusive. Split the grid into two non-empty rectangles of boxes by a line parallel to a side of the grid. Suppose that for every such split at least one resulting rectangle has sum at most 1. Determine the maximum possible sum of all inserted numbers.
Step 3 of 5: Use the split condition
a<c⟹R(1,a)>1 and R(a+1,n)>1.a<c\Longrightarrow R(1,a)>1\text{ and }R(a+1,n)>1.
Detailed analysis

If a<c, maximality of a makes the upper block through row a exceed 1, while minimality of c makes the lower block from row a+1 exceed 1. Splitting between these rows would then have both sums greater than 1, impossible. Thus a=c.