Can we place $18$ points in a regular hexagon of side $2$ such that the minimal distance between points is $>1$?
This a follow-up of this question. In the answers provided for it there are shown solutions for $14$, $15$, $16$, and $17$ points. Also, we can place $19$ points so that the minimal distance between them is exactly $1$.
Thank you for your interest!
Assume that we placed $18$ points in a regular hexagon of side $2$ such that the minimal distance between points is $2r$. It follows that we can pack $18$ circles of radius $r$ into a regular hexagon of side $2+\tfrac{2r}{\sqrt{3}}$, or $18$ unit circles into a regular hexagon of side $\tfrac 2r+\tfrac{2}{\sqrt{3}}$. But the smallest known side of such a hexagon is $4+\tfrac{2}{\sqrt{3}}$. It follows $r\le \tfrac 12$.
I expect that an easy proof of the example optimality following from a partition of the hexagon into $17$ pieces of diameter at most $1$ is impossible. I guess that a proof of the example optimality is hard.
One of approaches to the proof, started by Hagen von Eitzen is to localize positions of points in a solution. This approach was inductively used to solve a similar problem below. I proposed it at the final stage of All-Ukrainian student mathematical olympiad in 2001. No participants achieved an advance solving the problem. Also I found this problem (without a solution) in a book “How nonstandard problems are solved” by A. Ya. Kanel'-Belov and A. K. Koval'dgi, (Moskow, MCNMO, 1997, in Russian), see Problem 15 at p. 49.
In a cube $Q$ with an edge $1$ are placed $8$ points. Whether always among them there exist two points placed at distance at most $1$?
First we remark that a maximal distance from a polyhedron to a point outside it can be reached in one of vertices of the polyhedron. Now let $x_1,\dots, x_8\in Q$. Suppose that all distances between points are greater than $1.$ Then in every of $8$ closed cubes at the picture can be placed at most one point $x_i.$ Without loss of generality we can suppose that $x_1\in M_1.$ If $x_1=(x_1^1,x_1^2,x_1^3),$ then $x_1\le a_1,$ where $|(1,1/2,1/2)-(a_1,0,0)|=1,$ thus $a_1=1-1/\sqrt{2}<1/2$ (otherwise for all $i\in\{1,\dots,4\}$ we have $|A_i-x_1|<1$ and therefore there exists $j\ne 1$ such that $|x_i-x_j|<1$). We can similarly prove that $x_1^2\le a_1,x_1^3\le a_1.$ Thus, a point $x_1$ is in cube $M'_1$ with an edge $a_1.$ Similar arguments can be used for all other $x_i$. Assume that it is already proved that all points $x_i$ have to be in small cubes with an edge $a_n.$ Similarly to the previous we can prove, that all of them must be in small cubes with edge $a_{n+1},$ where $|(1,a_n,a_n)-(a_{n+1},0,0)|=1,$ thus $2a_n^2+a^2_{n+1}-2a_{n+1}=0$. If $a_{n+1}>a_n,$ then $3a_{n+1}^2-2a_{n+1}>0$ and therefore $a_{n+1}>2/3,$ that is impossible, because $a_{n+1}\le a_1<1/2.$ Thus a sequence $\{a_n\}$ has a limit $a$, and $3a^2-2a=0.$ Hence $a=0.$ Thus all points are placed in vertices of the cube $S,$ a contradiction.
