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# A chessboard

Best Answer googon97, 17 March 2013 - 04:28 AM

Spoiler for Assuming it's not a trick question

Go to the full post

7 replies to this topic

### #1 Utkrisht123

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Posted 16 March 2013 - 11:15 AM

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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### #2 dark_magician_92

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Posted 16 March 2013 - 04:40 PM

Spoiler for

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### #3 TimeSpaceLightForce

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Posted 16 March 2013 - 05:23 PM

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?

Spoiler for

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### #4 TimeSpaceLightForce

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Posted 16 March 2013 - 05:47 PM

Spoiler for

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### #5 googon97

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Posted 17 March 2013 - 04:28 AM   Best Answer

Spoiler for Assuming it's not a trick question

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### #6 Utkrisht123

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Posted 17 March 2013 - 10:32 AM

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?

Spoiler for

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.

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### #7 Prime

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Posted 17 March 2013 - 10:01 PM

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 by Prime, 17 March 2013 - 10:05 PM.

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Past prime, actually.

### #8 Utkrisht123

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Posted 18 March 2013 - 08:20 AM

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

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