![]() ![]() This means that we will again have a subboard all of whose corners are the same color, and we re done. There are exactly 6 ways to color in a column in this way here they are: Since we have 7 columns, but only 6 ways of coloring them, two columns will be colored in the exact same fashion. (This is because permuting rows and permuting columns doesn t change anything.) Therefore, for this case we may assume that 6ħ every single column has exactly 2 blue squares and 2 white squares. First, note that the argument above would hold even if we assumed that another column had 3 blue squares. Case 2: 2 blue squares in the first column. This means that we ll have a subboard all of whose corners are white, as required. ![]() Indeed, there are only 4 ways to color in the first 3 rows of a column by using at most 1 blue square here they are: By the pigeonhole principle, since there are only 4 potential colorings, and 6 columns to color in, some two columns will agree on the first 3 rows. As noted above, if any of the remaining columns have blue squares in 2 of those rows, we d have a subboard with matching corners for example, if column 5 has 2 blue squares in the first 3 rows, here s what can happen: Therefore, we need to assume that each of the remaining 6 columns each have at least 2 white squares in the first 3 rows here s an example of such a coloring: (note that for now, we re ignoring the fourth row, hence the light blue squares) As should be clear, we re now having an issue with having a subboard all of whose corners are white. Since permuting the rows doesn t change anything, let s assume that the first column has blue squares in rows 1, 2 and 3. Case 1: 3 blue squares in the first column. Either the first column has 2 of each color, or we have at least 3 squares of 5Ħ one color (either blue or white.) Since there s no difference between blue and white in the context of the problem, let s start with the case where we have at least 3 blue squares in the first column. Let s consider the first column of the checkerboard. ![]() Let us proceed by contradiction: assume that there exists a coloring such that no subboard has all four corners of the same color. Note that we get a subboard whose corners are the same color if two columns agree on two squares: for example, if column 2 and column 5 both have blue squares in the 1nd and 4th rows, then we d get the following (the squares for which we don t know the colors are colored in light blue): As another example, in the coloring given in the question, columns 1 and 6 both have blue squares in the 3rd and 4th row, which similarly leads to a desired subboard. Here s an example of a coloring: see if you can find a subboard with four corners of the same color! Solution: Let us first discuss how to approach this. ![]() Show that there is a subboard all of whose corners are blue or all of whose corners are white. Here s an example of a subboard, with squares shaded in red: Now, suppose that each of the 28 squares is colored either blue or white. A subboard of a checkerboard is a board you can cut-out of the checkerboard by only taking the squares which are between a specified pair of rows and a specified pair of columns. Show that if we take n + 1 numbers from the set then a 1 + a 3 + a 4 + a 5 = a 3 + a 5 + a 7 + a 8 a 1 + a 4 = a 7 + a 8 Thus, this will allow us to find two disjoint subsets of the 10 numbers with the same sum. ![]()
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