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.

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

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.

Well, a square is a special kind of rectangle with the ratio 1:1. We do agree on this, don't we?

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.

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.

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* * * * * * * *
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* * * * * * * *
* * * * * * * *
* * * * * * * *
* * * * * * * *

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.

Totally correct! Well done.
(and nice graphics, deriver69 !)

Can you cut the chessboard into four congruent shapes, each of which contains exactly one of the following squares: a1, b2, c3, d4?

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?

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.

Originally posted by mtthw
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Originally posted by mtthw
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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