Timeline for What is the maximum number of points on a plane such that some other point A is the nearest neighbor for all of them? [closed]
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 23, 2021 at 8:15 | history | closed |
Yoav Kallus Mark Wildon Andreas Blass Willie Wong Aaron Meyerowitz |
Not suitable for this site | |
| Apr 19, 2021 at 13:53 | review | Close votes | |||
| Apr 23, 2021 at 8:15 | |||||
| Apr 19, 2021 at 12:51 | comment | added | Stefan Kohl♦ | @EpsilonNet In your construction, the nearest point to a point $p$ on the outer circle is the point $p'$ on the inner circle in between $p$ and $A$. | |
| Apr 19, 2021 at 11:22 | comment | added | Epsilon Net | In the comments, it is stated that 5 points are the maximal number of such points. But what about the following configuration: let A be the center of the unit circle, take 4 points on the circle at the north, south, and east and west 'poles'. The next 4 points are the same on the circle of radius 1/3 around A. These are already 8 points, and of course, it can be continued to infinity. Or I just didn't understand the question | |
| Jul 26, 2020 at 20:49 | comment | added | Alexander Piperski | Thank you very much for your responses! It was really helpful. | |
| Jul 25, 2020 at 18:17 | comment | added | M. Winter | This seems like a reformulation of the kissing number problem. | |
| Jul 25, 2020 at 4:09 | comment | added | Alapan Das | You may prove that there may maximum 5 points by drawing circles around the points passing through $A$, which doesn't include any other point. | |
| Jul 25, 2020 at 3:40 | comment | added | Gerry Myerson | If you weaken the nearest neighbor to a nearest neighbor, you can use a regular hexagon with $A$ at its center. | |
| Jul 25, 2020 at 1:19 | comment | added | mathworker21 | i misread the problem, but thanks | |
| Jul 25, 2020 at 1:17 | comment | added | David E Speyer | To add a little more detail, $|AX| < |XY|$ and $|AY| < |XY|$ since $A$ is nearest neighbor to $X$ and $Y$, so $XY$ is the longest edge in triangle $AXY$, so $\angle XAY$ is the smallest angle. Since the sum of the angles in a triangle is $180$, this means $\angle XAY < 60$. | |
| Jul 25, 2020 at 1:08 | comment | added | LSpice | @mathworker21, if it were less than that, then $X$ and $Y$ would be closer to each other than to $A$. | |
| Jul 25, 2020 at 1:00 | review | Close votes | |||
| Jul 29, 2020 at 20:34 | |||||
| Jul 25, 2020 at 0:39 | comment | added | Gjergji Zaimi | Five is the maximum because for any two points X and Y from the configuration, the angle XAY has to exceed 60. This question is more suitable for math.stackexchange | |
| Jul 25, 2020 at 0:36 | review | First posts | |||
| Jul 25, 2020 at 0:50 | |||||
| Jul 25, 2020 at 0:35 | history | asked | Alexander Piperski | CC BY-SA 4.0 |