Timeline for Tiling the square with rectangles of small diagonals
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| May 11, 2015 at 11:25 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
deleted 43 characters in body
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| May 11, 2015 at 7:18 | comment | added | Aaron Meyerowitz | Note that for the square surrounded by $4$ rectangles it is better to have the square $\varepsilon \times \varepsilon$ and the rectangles of diagonal $\sqrt{\frac12+\varepsilon^2}.$ | |
| Nov 2, 2013 at 23:40 | comment | added | Gerry Myerson | That's OK, math is eternal. | |
| Nov 2, 2013 at 23:37 | comment | added | Joseph O'Rourke | @GerryMyerson: Oh, you are right! Cannot correct now... | |
| Nov 2, 2013 at 22:26 | comment | added | Gerry Myerson | The two partitions under the heading "For $k=8$...," isn't the one on the left actually $k=6$? | |
| Jul 23, 2013 at 15:46 | comment | added | Wlodek Kuperberg | This looks like the optimal tiling for $k=8$ - hence a counterexample to the conjecture. Very nice! Now, if the general solution looks "like this" (vertical blocks of congruent rectangles stacked forming strips of the square), then the general problem could be possibly reduced to a number theory question on expressing $k$ as the sum of a certain number, close to $\sqrt{k}$ I suppose, of positive integers that differ from each other by as little as possible. | |
| Jul 23, 2013 at 15:32 | vote | accept | Wlodek Kuperberg | ||
| Jul 23, 2013 at 10:03 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Calculations for three partitions into 8 rectangles.
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| Jul 23, 2013 at 2:06 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Removed that example for k=8: found a better one.
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| Jul 23, 2013 at 1:46 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
added 223 characters in body
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| Jul 23, 2013 at 0:18 | comment | added | Wlodek Kuperberg | Oops! I meant all sides of the rectangles are rational. Sorry. Correcting right away. | |
| Jul 22, 2013 at 23:58 | comment | added | Joseph O'Rourke | @WlodzimierzKuperberg: Did you mean that all diagonal lengths are square roots of rationals? | |
| Jul 22, 2013 at 23:56 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Neglected the squareroot! And added WK's better partition.
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| Jul 22, 2013 at 23:49 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Neglected the squareroot!
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| Jul 22, 2013 at 23:29 | comment | added | Wlodek Kuperberg | Cut the square with a vertical line into two rectangles, the left one of size $x\times1$ the right one $(1-x)\times1$, where $x=41/72$. Cut the left one into three congruent horizontal rectangles and the right one horizontally into two. The diagonals of all five rectangles come out equal and smaller than $0.66$. I believe this is the optimal tiling for $k=5$. (I don't know yet how to embed drawings here.) | |
| Jul 22, 2013 at 23:27 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
added 22 characters in body
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| Jul 22, 2013 at 21:39 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Replaced with a better quality image.
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| Jul 22, 2013 at 21:31 | history | answered | Joseph O'Rourke | CC BY-SA 3.0 |