Timeline for 3-piece dissection of square to equilateral triangle?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Dec 6 at 13:41 | history | edited | Joseph O'Rourke | CC BY-SA 4.0 |
added 193 characters in body
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| Sep 27, 2023 at 12:49 | vote | accept | Joseph O'Rourke | ||
| Sep 27, 2023 at 12:28 | comment | added | Timothy Chow | @AntonLukyanenko I don't follow your argument. If we dissect the square with corners (0,0), (0,1), (1,0), (1,1) using the curves $y=x(1-x)$ and $y=x(x-1)+1$, then we get three pieces and no right angles. | |
| Sep 26, 2023 at 23:15 | answer | added | Timothy Chow | timeline score: 4 | |
| Aug 31, 2017 at 11:31 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Image links broken; now fixed.
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| Nov 28, 2013 at 6:04 | comment | added | Benjamin Dickman | I comment only to point out that the so-called Dudeney construction may not, in fact, be due to H. Dudeney. For more details, see G.N. Frederickson's "Hinged Dissections: Swinging and Twisting" (pp. 8-10). books.google.com/… | |
| Nov 6, 2011 at 18:34 | comment | added | Anton Lukyanenko | A 3-piece dissection of the equilateral triangle would have to create 4 right angles to serve as corners of the square, and there are just a few ways this can be done. At first glance, none of them recombine as a square (though you can get a rectangle). It should be pretty easy to run through the options and rule them all out. | |
| Nov 5, 2011 at 16:29 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Added 5-piece dissection link.
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| Nov 5, 2011 at 14:30 | comment | added | Todd Trimble | Wow, ask and ye shall receive! Thank you, Joseph! | |
| Nov 5, 2011 at 13:40 | comment | added | Joseph O'Rourke | @Todd: Added a stable image (in a different orientation). | |
| Nov 5, 2011 at 13:37 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
added 23 characters in body
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| Nov 5, 2011 at 13:07 | comment | added | Todd Trimble | I just want to tell that applet to "hold still, dammit"! | |
| Nov 5, 2011 at 12:57 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
added 6 characters in body
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| Nov 5, 2011 at 2:18 | comment | added | Ken Fan | 2 is not possible because the side length of the triangle is more than the diagonal of the square. | |
| Nov 5, 2011 at 1:58 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |