Timeline for union of regular polygons
Current License: CC BY-SA 2.5
19 events
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Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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Nov 22, 2010 at 9:20 | comment | added | Fedor Petrov | @Gerry: yes, also I did not say that subgroups are different, but said it about polygons:) | |
Nov 21, 2010 at 23:32 | comment | added | Gerry Myerson | Strictly speaking, there's a difference between the polygon and group formulations of the problem, in that you're forbidding a one-gon but permitting a one-element subgroup. Of course it makes no difference to the solution of the problem. | |
Nov 18, 2010 at 20:16 | vote | accept | Fedor Petrov | ||
Nov 18, 2010 at 3:11 | answer | added | Sergey Norin | timeline score: 4 | |
Nov 17, 2010 at 13:09 | comment | added | HenrikRüping | Let $G\cong \mathbb{Z}/n$, $G_i:=<n_i>$, $n=lcm(n/n_1,..,n/n_k)$. If all $n/n_i$'s were pairwise coprime, then every way of inscribing the regular gons looks exactly the same (and is hence minimal and has especially a common point) (Chinese remainder theorem). So I am wondering, whether one can restrict to that case ... | |
Nov 17, 2010 at 12:19 | comment | added | Fedor Petrov | @Gerry: thanks, I must think about this. Hopefully now it is more clear, then originally. | |
Nov 17, 2010 at 12:17 | history | edited | Fedor Petrov | CC BY-SA 2.5 |
added 216 characters in body
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Nov 17, 2010 at 10:55 | comment | added | Gerry Myerson | Where the "elements" are the vertices; note Fedor says, "here polygon is just the set of its vertices." So a 3-gon contributes 3 vertices, a 4-gon contributes 4, making 7 in all, but if they have one in common that brings it down to 6. Fedor, if you ever want to hire an interpreter, I'm available. | |
Nov 17, 2010 at 8:47 | comment | added | Fedor Petrov | @Elizabeth: no, cardinality is just cardinality (number of distinct elements) | |
Nov 17, 2010 at 8:36 | comment | added | Elizabeth S. Q. Goodman | Does minimal cardinality in this context refer to the smallest possible size of a regular polygon containing such a union? If so, I'm with Gjergji: each $b_i$-gon, each translate on the unit circle of the $b_i$th roots of unity if you want to be algebraic, is only contained in $c$-gons where $c$ is a multiple of $b_i$. For this the common-vertex arrangement works, all at 1. I don't understand where the minimal cardinality of 6 for $b_1=3, b_2=4$ comes from. It's not my field but I'm curious to know what is going on. | |
Nov 17, 2010 at 7:24 | comment | added | Fedor Petrov | @Joseph: yes. @Gerry: you understand right, I mean that (at least) one of minima is achieved for polygons having one and the same vertex. | |
Nov 17, 2010 at 7:23 | history | edited | Fedor Petrov | CC BY-SA 2.5 |
added 6 characters in body
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Nov 17, 2010 at 5:01 | comment | added | Gerry Myerson | @Gjergi, if $k=2$, $b_1=3$, $b_2=4$, then (provided, as ever, that I understand the problem) minimal cardinality is 6 whereas the lcm of the $b_i$ is 12. | |
Nov 17, 2010 at 2:53 | comment | added | Gjergji Zaimi | @Gerry, but isn't the minimal cardinality just LCM(b_i)? | |
Nov 17, 2010 at 2:08 | comment | added | Gerry Myerson | @Gjergi, if I understand the question, your example shows the minimal cardinality for 2, 3, 12 is 12 - but that cardinality can (also) be achieved by a configuration in which all three polygons share a vertex. A true counterexample would be one where there's a configuration strictly better than any in which all polygons share a vertex. But I, too, confess to some uncertainty about the meaning of the question. | |
Nov 17, 2010 at 0:36 | comment | added | Gjergji Zaimi | A 2-gon and a regular triangle can fit inside a 12-gon without having any vertex in common. Maybe I don't understand the question. | |
Nov 17, 2010 at 0:27 | comment | added | Joseph O'Rourke | "they share a vertex" = "they all share one vertex"? | |
Nov 16, 2010 at 23:30 | history | asked | Fedor Petrov | CC BY-SA 2.5 |