Timeline for Conway's lesser-known results
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Oct 2, 2020 at 7:32 | comment | added | Zach Teitler | @EmilJeřábek Thank you. All five of the triangles are similar, but one of them is a different size than the other four. Yes, it is obvious. I feel very stupid; I should have noticed that myself. | |
| Oct 2, 2020 at 7:20 | history | edited | Emil Jeřábek | CC BY-SA 4.0 |
restore information lost in an edit
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| Oct 2, 2020 at 7:18 | comment | added | Emil Jeřábek | @ZachTeitler If you actually draw it, you will quickly find out that if you start with a right triangle not similar to the 1–2–$\sqrt5$ one, you’ll end up with 4 congruent triangles, and a fifth noncongruent triangle. | |
| Jun 16, 2020 at 17:06 | comment | added | Zach Teitler | I am curious what is special about the $1$-$2$-$\sqrt{5}$ right triangle? Any right triangle yields a decomposition into $5$ congruent pieces in this way, giving a tiling of the plane. I suppose for some special ones such as isosceles right triangle it will be a periodic tiling, but presumably for "most" right triangles it will be aperiodic. Has this been studied? | |
| Apr 24, 2020 at 21:34 | comment | added | Michael Chaney | I regret learning about this. This is another tiling to add to my maze project at github.com/mdchaney/jsmaze | |
| Apr 21, 2020 at 22:15 | history | edited | Ian Agol | CC BY-SA 4.0 |
added 2 characters in body
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| Apr 19, 2020 at 6:51 | history | edited | Robin Houston | CC BY-SA 4.0 |
At least one reader of this answer has gone on to credit the tessellation to Radin, possibly as a result of lack of clarity on Wikipedia. I’ll try to be clear here.
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| Apr 13, 2020 at 22:30 | comment | added | Dan Rust | Just for completeness, and for those looking to read further on this tiling substitution, this is Conway and Radin's 'pinwheel tiling', for which there is a huge literature and still many open questions. | |
| Apr 13, 2020 at 15:51 | comment | added | Andreas Blass | @probably_someone Since each triangle has nonempty interior and therefore contains a point with rational coordinates, one doesn't need to construct a counting scheme; counting schemes for the points with rational coordinates are available "off the shelf". | |
| Apr 12, 2020 at 20:59 | comment | added | Nick S | As a side note, if one replace the diagonal in the rectangle with the other diagonal, the tiling becomes periodic. | |
| Apr 12, 2020 at 20:57 | comment | added | Nick S | @BlueRaja The set of orientations is countable, but it does distribute uniformly on the unit circle. | |
| Apr 12, 2020 at 20:55 | comment | added | Emil Jeřábek | Well, it’s easy to see from the relative orientation of the large triangle to the five constituents that all the angles belong to the abelian group generated by $\pi/2$ and $\arctan 2$. | |
| Apr 12, 2020 at 19:17 | comment | added | Kevin Casto | I would also wager that, say the cosine of the orientation angles are all algebraic. Maybe there's a nice way to describe them. | |
| Apr 12, 2020 at 17:23 | comment | added | probably_someone | @BlueRaja There are only countably infinitely many triangles in this tiling (it's fairly straightforward to construct a counting scheme), so the set of orientations must have measure zero. | |
| Apr 12, 2020 at 17:14 | comment | added | BlueRaja | I wonder, does it appear in every orientation? | |
| S Apr 12, 2020 at 13:42 | history | answered | Robin Houston | CC BY-SA 4.0 | |
| S Apr 12, 2020 at 13:42 | history | made wiki | Post Made Community Wiki by Robin Houston |