How to place 14 dots on the plane
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.
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)
geometry
asked Dec 21 at 5:49
Makina
1,118115
add a comment |
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.
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)
geometry
asked Dec 21 at 5:49
Makina
1,118115
2
Compare: math.stackexchange.com/questions/2575268/…
– Chris Culter
Dec 21 at 6:15
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?
– kasperd
Dec 21 at 12:11
Yeah, no matter the orientation, 31 remains the max here.
– Makina
Dec 21 at 14:09
add a comment |
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.
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)
geometry
asked Dec 21 at 5:49
Makina
1,118115
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.
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)
geometry
geometry
asked Dec 21 at 5:49
Makina
1,118115
asked Dec 21 at 5:49
Makina
1,118115
edited Dec 21 at 5:55
asked Dec 21 at 5:49
Makina
1,118115
asked Dec 21 at 5:49
Makina
1,118115
asked Dec 21 at 5:49
Makina
1,118115
1,118115
2
Compare: math.stackexchange.com/questions/2575268/…
– Chris Culter
Dec 21 at 6:15
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?
– kasperd
Dec 21 at 12:11
Yeah, no matter the orientation, 31 remains the max here.
– Makina
Dec 21 at 14:09
add a comment |
2
Compare: math.stackexchange.com/questions/2575268/…
– Chris Culter
Dec 21 at 6:15
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?
– kasperd
Dec 21 at 12:11
Yeah, no matter the orientation, 31 remains the max here.
– Makina
Dec 21 at 14:09
2
2
Compare: math.stackexchange.com/questions/2575268/…
– Chris Culter
Dec 21 at 6:15
Compare: math.stackexchange.com/questions/2575268/…
– Chris Culter
Dec 21 at 6:15
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?
– kasperd
Dec 21 at 12:11
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?
– kasperd
Dec 21 at 12:11
Yeah, no matter the orientation, 31 remains the max here.
– Makina
Dec 21 at 14:09
Yeah, no matter the orientation, 31 remains the max here.
– Makina
Dec 21 at 14:09
add a comment |
1 Answer
1
active
oldest
votes
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:
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.
answered Dec 21 at 6:40
Chris Culter
20k43482
Thank you for your answer. Amazing, i wonder how do people even come up with these...
– Makina
Dec 21 at 14:09
add a comment |
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1 Answer
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1 Answer
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active
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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:
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.
answered Dec 21 at 6:40
Chris Culter
20k43482
Thank you for your answer. Amazing, i wonder how do people even come up with these...
– Makina
Dec 21 at 14:09
add a comment |
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:
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.
answered Dec 21 at 6:40
Chris Culter
20k43482
Thank you for your answer. Amazing, i wonder how do people even come up with these...
– Makina
Dec 21 at 14:09
add a comment |
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:
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.
answered Dec 21 at 6:40
Chris Culter
20k43482
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:
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.
answered Dec 21 at 6:40
Chris Culter
20k43482
answered Dec 21 at 6:40
Chris Culter
20k43482
answered Dec 21 at 6:40
Chris Culter
20k43482
answered Dec 21 at 6:40
Chris Culter
20k43482
20k43482
Thank you for your answer. Amazing, i wonder how do people even come up with these...
– Makina
Dec 21 at 14:09
add a comment |
Thank you for your answer. Amazing, i wonder how do people even come up with these...
– Makina
Dec 21 at 14:09
Thank you for your answer. Amazing, i wonder how do people even come up with these...
– Makina
Dec 21 at 14:09
Thank you for your answer. Amazing, i wonder how do people even come up with these...
– Makina
Dec 21 at 14:09
add a comment |
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2
Compare: math.stackexchange.com/questions/2575268/…
– Chris Culter
Dec 21 at 6:15
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?
– kasperd
Dec 21 at 12:11
Yeah, no matter the orientation, 31 remains the max here.
– Makina
Dec 21 at 14:09