Problem
COM-B1-M07-P024 Two Diagonals on \(10\times10\)
All cells on both diagonals of a \(10\times10\) board are removed. The remaining area is \(80\). Prove that it cannot be tiled by straight \(1\times4\) tetrominoes.
Chessboard coloring gives no contradiction. Try the color \(i+j\pmod4\).
Color cell \((i,j)\) by the residue \(i+j\pmod4\). Any straight tetromino covers one cell of each of the four colors. On the full \(10\times10\) board, the counts of colors \(0,1,2,3\) are \(25,24,25,26\). On the main diagonal \((i,i)\), colors alternate between \(2\) and \(0\), so \(5\) cells of color \(0\) and \(5\) cells of color \(2\) are removed. On the other diagonal \((i,11-i)\), the sum of indices is \(11\), so all these \(10\) cells have color \(3\). After removal the counts are \(20,24,20,16\). They are not equal. But a tiling by straight tetrominoes would require equal numbers of all four colors. Contradiction.
Strong challenge: area and usual chessboard coloring pass, so hidden four-coloring is needed.