Timeline for Points on $k$ Circles
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 9, 2015 at 18:52 | comment | added | Morteza | @PietroMajer No I think, for example for $k=2$, we have $f(k)=9$, but your parameter is $6$. Your parameter is exactly $3k$, isn't it? | |
| Feb 9, 2015 at 15:09 | comment | added | Pietro Majer | (@Morteza: the quantity I defined above is larger than or equal to $f(k)-1$; I was asking if it coincides...) | |
| Feb 9, 2015 at 12:18 | comment | added | Morteza | @JosephO'Rourke The sketch of proof for f(2)=9 is using some cases for those nine points, 5+4 or 6+3 or..., then replacing one of them with the new point (10th point). The counterexample for 8 is three circles with their pairwise intersections (6 points) and one extra point on each of them. I think the answer for "unit circles" is very small. | |
| Feb 9, 2015 at 12:14 | comment | added | Morteza | @PietroMajer No, the number $f(k)−1$ is the maximum n such that there exist a set $S$ of points $P_1,P_2,...,P_n$ such that for each $i$, $S−P_i$ is on $k$ circles but the whole $S$ is not. | |
| Feb 9, 2015 at 0:58 | comment | added | Joseph O'Rourke | An interesting variant might be to replace "$k$ circles" by "$k$ unit-radius circles." | |
| Feb 9, 2015 at 0:57 | comment | added | Joseph O'Rourke | Although I don't doubt it, I would appreciate a proof sketch that $f(2)=9$. | |
| Feb 9, 2015 at 0:43 | comment | added | Pietro Majer | The number $f(k)-1$ is also the minimum $n$ such that for any set of $n$ points $S$ in the plane there is at most a union of $k$ circles $X=C_1\cup\dots\cup C_k$ such that $S\subset X$, isn't it? | |
| Feb 8, 2015 at 22:27 | comment | added | domotorp | I think providing some more related results and links would be useful. | |
| Feb 8, 2015 at 19:13 | history | asked | Morteza | CC BY-SA 3.0 |