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Posers and Puzzles

Posers and Puzzles

  1. 14 Apr '08 14:05
    Can you cut a chess board into four rectangles so they have the ratios 1:1, 1:2, 1:3 and 1:4 respectively?
    You cannot cut squares in parts.
  2. 14 Apr '08 15:44
    I think you cannot do that. Maybe my calculations are wrong:
    I suggest letting x be the largest part of the board. Then we can rite the following equasion:
    x + 1/2x + 1/3x + 1/4x = 64
    Multiply the equasion by 12:
    12x + 6x + 4x + 3x = 768
    x = 30.72
  3. 14 Apr '08 15:45
    Originally posted by FabianFnas
    Can you cut a chess board into four rectangles so they have the ratios 1:1, 1:2, 1:3 and 1:4 respectively?
    You cannot cut squares in parts.
    Lets see...

    To cut the board into 4 rectangles, we must make 3 paralle cuts or 2 perpendicular cuts.

    The first option is impossible since the 1:1 rectangle is a square.

    The second option is impossible since the rectangle having no common edge with the 1:1 rectangle is a square, but 1:2, 1:3, 1:4 are not squares.

    So there is no solution, even if cutting squares in parts is allowed.
  4. 14 Apr '08 16:02 / 1 edit
    Well, a square is a special kind of rectangle with the ratio 1:1. We do agree on this, don't we?
  5. 14 Apr '08 16:12
    Oh, I misunderstood the question. I thought you meant that the rectangles relate to each other as 1 to 1, 1 to 2 and so on by their square.
  6. 14 Apr '08 16:21
    Easier to say with a diagram, but split into the following blocks:

    a1->d8 (2:1)
    e1->f8 (4:1)
    g1->h6 (3:1)
    g7->h8 (1:1)

    I got there by looking for +ve integer solutions to a^2 + 2b^2 + 3c^2 + 4d^2 = 64.
  7. Subscriber deriver69
    Keeps
    14 Apr '08 16:38
    * *|* * * * * *
    * *|* * * * * *
    -----------------
    * * * * * * * *
    * * * * * * * *
    -----------------
    * * * * * * * *
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  8. Subscriber deriver69
    Keeps
    14 Apr '08 16:41
    I think I used a simlar method, the areas of the simpler ratios were 1,2,3 and 4. I then multiplied these by 4 to make a total of 64 (doubling the sizes). This was a slight hunch because 64=2^6 so I assumed trebling would not help.
  9. 14 Apr '08 16:56
    Totally correct! Well done.
    (and nice graphics, deriver69 !)
  10. 14 Apr '08 18:21
    Can you cut the chessboard into four congruent shapes, each of which contains exactly one of the following squares: a1, b2, c3, d4?
  11. 14 Apr '08 18:31
    Originally posted by Jirakon
    Can you cut the chessboard into four congruent shapes, each of which contains exactly one of the following squares: a1, b2, c3, d4?
  12. 14 Apr '08 20:08
    Originally posted by David113
    Lets see...

    To cut the board into 4 rectangles, we must make 3 paralle cuts or 2 perpendicular cuts.

    The first option is impossible since the 1:1 rectangle is a square.

    The second option is impossible since the rectangle having no common edge with the 1:1 rectangle is a square, but 1:2, 1:3, 1:4 are not squares.

    So there is no solution, even if cutting squares in parts is allowed.
    Shame:'(
    I forgot the third way, which gives a solution
  13. 14 Apr '08 21:45
    Originally posted by mtthw
    [fen]pRRRRRRR/pRrrrrrr/pRrPPPPr/pRrPppPr/pRrrRpPr/pRRRRpPr/ppppppPr/PPPPPPPr[/fen]
    I like mine better..

  14. Subscriber AThousandYoung
    Proud Boys Beware
    18 Apr '08 05:00
    Originally posted by mtthw
    [fen]pRRRRRRR/pRrrrrrr/pRrPPPPr/pRrPppPr/pRrrRpPr/pRRRRpPr/ppppppPr/PPPPPPPr[/fen]
    Great. A swastika.
  15. Subscriber AThousandYoung
    Proud Boys Beware
    18 Apr '08 05:01
    Originally posted by Guych
    I think you cannot do that. Maybe my calculations are wrong:
    I suggest letting x be the largest part of the board. Then we can rite the following equasion:
    x + 1/2x + 1/3x + 1/4x = 64
    Multiply the equasion by 12:
    12x + 6x + 4x + 3x = 768
    x = 30.72
    You're calculating areas. The problem is about ratios.