Utkrisht123 3 Report post Posted March 16, 2013 Two diagonally opposite corner squares are removed from a regular chessboard. Now is it possible to cover all 62 squares with exactly 31 rectangles ( no rectangle should overlap each other ). If yes then how? If no then why? Share this post Link to post Share on other sites

0 googon97 1 Report post Posted March 17, 2013 NO it's impossible. Imagine the chess board pattern. The opposite corners are the same color. One rectangle can cover on tile of each color. That leaves two of the same tile that can't be covered with the last rectangle. Share this post Link to post Share on other sites

0 dark_magician_92 4 Report post Posted March 16, 2013 Pretty sure it cant be done. trying to pinpoint the problem in it. Share this post Link to post Share on other sites

0 TimeSpaceLightForce 11 Report post Posted March 16, 2013 Two diagonally opposite corner squares are removed from a regular chessboard. Now is it possible to cover all 62 squares with exactly 31 rectangles ( no rectangle should overlap each other ). If yes then how? If no then why? yes,because you didnt say the sides of rectagles should be aligned with the sides of the squares Share this post Link to post Share on other sites

0 TimeSpaceLightForce 11 Report post Posted March 16, 2013 squares are considered rectangles too ..some may be 2 x 1 , 3 x 1 , 1 x 1 or 2 x 2 ? Share this post Link to post Share on other sites

0 Utkrisht123 3 Report post Posted March 17, 2013 Two diagonally opposite corner squares are removed from a regular chessboard. Now is it possible to cover all 62 squares with exactly 31 rectangles ( no rectangle should overlap each other ). If yes then how? If no then why? yes,because you didnt say the sides of rectagles should be aligned with the sides of the squares Good point. i didnt thought that way But if I say that all rectangles should be completely aligned and should be of same size then what would you say. Share this post Link to post Share on other sites

0 Prime 15 Report post Posted March 17, 2013 (edited) The stipulation that all 31 rectangles are “of the same size” is still a bit ambiguous. The size could be interpreted as area. If rectangles were equal that would imply they are equal in area (2 squares each) and dimensions. However, do we need a stipulation that rectangles must be alined on square boundaries? If rectangle's dimensions were specified as 1x2, then the problem would be solved by googon97 in post #5. But those rectangles could be 1/3 x 6, or 1/2 x 4. Still, it is impossible to cover up the board with 31 of those rectangles. Furthermore, can we prove that we could or could not cover the board with 31 equal area (2 squares each) rectangles of any dimensions? Edited March 17, 2013 by Prime Share this post Link to post Share on other sites

0 Utkrisht123 3 Report post Posted March 18, 2013 The stipulation that all 31 rectangles are “of the same size” is still a bit ambiguous. The size could be interpreted as area. If rectangles were equal that would imply they are equal in area (2 squares each) and dimensions. However, do we need a stipulation that rectangles must be alined on square boundaries? If rectangle's dimensions were specified as 1x2, then the problem would be solved by googon97 in post #5. But those rectangles could be 1/3 x 6, or 1/2 x 4. Still, it is impossible to cover up the board with 31 of those rectangles. Furthermore, can we prove that we could or could not cover the board with 31 equal area (2 squares each) rectangles of any dimensions? If you want to then You are welcome to do so Share this post Link to post Share on other sites

Two diagonally opposite corner squares are removed from a regular chessboard. Now is it possible to cover all 62 squares with exactly 31 rectangles ( no rectangle should overlap each other ).

If yes then how? If no then why?

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