A friend asked me a question to ponder over:

You got $14$ dots which you need to place on a plane in such a way so that you get the maximum amount of similar distances between each $2$ points. I managed to get $31$ ($12$ first hexagon $+ 12$ second hexagon$ + 7$ distance between each point of $2$ hexagons) by drawing $2$ hexagons one below the other with the distance between the centers equal to the side length

The answer is incorrect though according to him.

enter image description here

Any insight towards the solution would be helpful.

PS: don't mind the black-red, I made such distinction just for the sake of pointing out the position of $14$ dots ($2$ hexagons + $2$ centers)

  • 2
    $\begingroup$ Compare: math.stackexchange.com/questions/2575268/… $\endgroup$ – Chris Culter Dec 21 '18 at 6:15
  • $\begingroup$ As far as I can tell your construction has one degree of freedom as you can rotate the direction in which the two hexagons are offset from each other. Have you verified if any such rotation could give you an extra pair of points with the same distance making it 32? $\endgroup$ – kasperd Dec 21 '18 at 12:11
  • $\begingroup$ Yeah, no matter the orientation, 31 remains the max here. $\endgroup$ – Makina Dec 21 '18 at 14:09

Apparently the maximum is $33$, as was worked out in C. Schade: Exakte Maximalzahlen gleicher Abstände, Diploma thesis directed by H. Harborth, Techn. Univ. Braunschweig 1993. Sadly, I couldn't find a description of the proof. This maximum can be realized in two ways, up to graph isomorphism:

enter image description here

Image source: Jean-Paul Delahaye, Les graphes-allumettes, (in French), Pour la Science no. 445, November 2014, pages 108-113. http://www.lifl.fr/~jdelahay/pls/2014/252.pdf

More resources are listed at https://oeis.org/A186705.

  • $\begingroup$ Thank you for your answer. Amazing, i wonder how do people even come up with these... $\endgroup$ – Makina Dec 21 '18 at 14:09

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