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Point $A$ on the plane is given. How many other points can be placed on this plane under the condition that $A$ is the nearest neighbor for any of these points?

I can think of a regular pentagon with $A$ as its center, its vertices being the $5$ points in question, but are there any configurations that include more points?

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    $\begingroup$ 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 $\endgroup$
    – Gjergji Zaimi
    Commented Jul 25, 2020 at 0:39
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    $\begingroup$ @mathworker21, if it were less than that, then $X$ and $Y$ would be closer to each other than to $A$. $\endgroup$
    – LSpice
    Commented Jul 25, 2020 at 1:08
  • $\begingroup$ 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 $\angle XAY < 60$. $\endgroup$
    – David E Speyer
    Commented Jul 25, 2020 at 1:17
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    $\begingroup$ This seems like a reformulation of the kissing number problem. $\endgroup$
    – M. Winter
    Commented Jul 25, 2020 at 18:17
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    $\begingroup$ @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 $A$. $\endgroup$
    – Stefan Kohl ♦
    Commented Apr 19, 2021 at 12:51

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