Timeline for Equipartition of the circle [closed]
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 24, 2013 at 3:29 | history | closed |
Ricardo Andrade Andrés E. Caicedo Andrey Rekalo Stefan Kohl♦ Alexandre Eremenko |
Not suitable for this site | |
| Dec 23, 2013 at 21:04 | comment | added | The Masked Avenger | @Joce, sounds like some good problems to study. I imagine it will be exact for finitely many, but I don't know. Also root 2 is algebraic, not transcendental, and I guess also not transcendent. | |
| Dec 23, 2013 at 21:01 | answer | added | The Masked Avenger | timeline score: 1 | |
| Dec 23, 2013 at 20:55 | comment | added | Joce | I also incline to believe it's a good approximation, if not it should be possible to write $OX'_i\cdot OX'_{i+1}$ independently of $i$ as a combination of the other vectors in the drawing, but if the postulate is true then transcendent numbers such as $\sqrt{2}$ could be obtained with rationals. The school book displays an $N=17$ example which looks pretty good though, I'd be happy to bound the approximation error. Also, could it be exact for (some sequence of) constructible polygons? | |
| Dec 23, 2013 at 20:39 | comment | added | The Masked Avenger | More precisely, my reading of the described construction (which is supported by Joseph's picture) is so equivalent... . | |
| Dec 23, 2013 at 20:34 | comment | added | The Masked Avenger | Except I can I can mark off with a compass 7 equal length segments on another line, then use parallel lines to divide the diameter correspondingly. I maintain that the construction described in the question is equivalent to a compass and straightedge construction. | |
| Dec 23, 2013 at 20:21 | comment | added | Joce | The marked ruler is needed for dividing AB into 7 equal segments, I believe. | |
| Dec 23, 2013 at 19:46 | review | Close votes | |||
| Dec 24, 2013 at 3:29 | |||||
| Dec 23, 2013 at 19:37 | comment | added | Carlo Beenakker | @TheMaskedAvenger --- I presume the construction of the OP allows one to use a compass with a marked ruler (not just an unmarked straightedge); then heptagons can be constructed (and I would think all regular polygons as well) --- en.wikipedia.org/wiki/Heptagon | |
| Dec 23, 2013 at 19:23 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
replaced deprecated tag 'geometry'; replaced inappropriate tag 'equitable-partition'
|
| Dec 23, 2013 at 19:21 | comment | added | The Masked Avenger | This would imply constructibility of a heptagon by compass and straightedge, a contradiction. I think the method produces just a good approximation. | |
| Dec 23, 2013 at 19:17 | answer | added | Joseph O'Rourke | timeline score: 2 | |
| Dec 23, 2013 at 18:35 | review | First posts | |||
| Dec 23, 2013 at 18:36 | |||||
| Dec 23, 2013 at 18:20 | history | asked | Joce | CC BY-SA 3.0 |