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Topic: Covering square with 2x1 rectangles (Read 395 times) |
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rmsgrey
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Covering square with 2x1 rectangles
« on: Aug 9th, 2021, 4:41am » |
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"Borrowed" from a recent Numberphile video:
Given a square, it's possible to divide it into some number of similar rectangles, each with sides in ratio 2:1. What numbers of rectangles are possible, and what numbers are impossible?
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Grimbal
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You can do any value except 1, 3 and 4.
2 is obvious.
For every N, you can make a new square with N+4 rectangles by surrounding the square by 4 rectangles. See 2 -> 6 for example.
For every N, you can make a new square with N+3 rectangles by splitting a rectangle in four. See 2 -> 5 for example.
7 is actually a 2 with one rectangle split in 9, giving 10, then four rectangles joined to obtain 7.
From 5, 6 and 7 you can construct any N > 7.
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| « Last Edit: Aug 11th, 2021, 1:03pm by Grimbal » |
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rmsgrey
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Re: Covering square with 2x1 rectangles
« Reply #2 on: Aug 11th, 2021, 6:52am » |
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Can you prove the impossibility of 1, 3 and 4?
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Grimbal
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Re: Covering square with 2x1 rectangles
« Reply #3 on: Aug 11th, 2021, 12:54pm » |
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1 rectangle: obvious.
I.e. the square must be covered by one rectangle. That rectangle must have area 1 and ratio 2:1. The rectangle must have dimensions sqrt(2) by sqrt(0.5). The diagonal of the rectangle is sqrt(2+0.5) = sqrt(2.5). This diagonal doesn't fit in the square with diagonal sqrt(2). So a rectangle with sufficient area doesn't fit in the square.
(Or more simpler: the square isn't a 2:1 rectangle.)
3 rectangles:
There are 4 corners and only 3 rectangles. One rectangle must fill two corners. That rectangle must have size is 1 x 0.5. The remaining area is also a 1 x 0.5 rectangle. And it must be filled by wo rectangles. Again, one rectangle must fill 2 corners of that area. This leaves only 2 ways, 1 x 0.5 or 0.5 x 0.25. One way doesn't give space for a third rectangle, the other leaves a 3:2 space which is not the right ratio.
4 rectangles:
If one rectangle fills 2 corners, there is always one rectangle that must fill a whole side. That leaves few possibilities.
The first rectangle would be 1 x 1/2. The second can only be 1/2 x 1/4. The third can be 1/2 x 1/4 leaving a 1/2 x 1/2 space or 3/4 x 3/8 leaving a narrow 3/4 x 1/8 rectangle. Both are the wrong ratio.
So remains the case where each corner of the square is filled by a different rectangle.
In that case, consider a rectangle that touches the center of the square. There must be one. That rectangle must reach from a corner to the center. It must have at least 1/2 in both dimensions, which actually means it must have dimensions 1 x 1/2. We are back to the previous case which didn't work. (Or in fact we contradict the hypothesis that each corner is filled by a different rectangle).
I think that covers it. No pun intended.
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| « Last Edit: Aug 11th, 2021, 1:01pm by Grimbal » |
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