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Minimum diameter for $n$ points, given that distance between any two of them is greater than or equal to 1.

2

1

$begingroup$

There are $n$ points on a plane, such that distance between any two of them $geq 1$. Question is, what is minimum possible diameter for such set of points, that is minimum of distance between two farthest points. At first I thought optimal configuration would be regular $n$-gon with side 1. For this configuration the answer is $$fracsin(fracpincdotlfloorfracn2rfloor)sin(fracpin)$$

For $n=2,3$, this is obviously true, answer is $1$.
however for $n=7$, There is a configuration with diameter $2$, regular hexagon with side $1$, and its center.

I can't prove that regular polygon is answer for $n=4,5,6$. But I still think that it is optimal, So how can I prove it for $n=4,5,6$ and what is the answer in general case?

geometry combinatorial-geometry

share|cite|improve this question

edited Mar 31 at 6:14

Alex Ravsky

43.1k32583

asked Mar 24 at 13:37

snowAuouesnowAuoue

334

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For four, look at the diagonals of the convex hull of the four points. Since all sides are $geq1$, if the diagonals are $<sqrt2$, then by the law of cosines the angles are $<pi/2$. Therefore, the interior angles wouldn't add up to $2pi$. Well, I assumed that the convex hull was a quadrilateral. The cases in which the convex hull is a triangle, or a segment are similar, but easier.
  $endgroup$
  – user647486
  Mar 24 at 14:03
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I also realized that for $n=6$, regular pentagon with side $1$ and its center has diameter of $2cos(fracpi10)=1.90211...$, so regular hexagon is not optimal
  $endgroup$
  – snowAuoue
  Mar 24 at 16:32
  

  
  
  
  

add a comment | 

2

1

$begingroup$

There are $n$ points on a plane, such that distance between any two of them $geq 1$. Question is, what is minimum possible diameter for such set of points, that is minimum of distance between two farthest points. At first I thought optimal configuration would be regular $n$-gon with side 1. For this configuration the answer is $$fracsin(fracpincdotlfloorfracn2rfloor)sin(fracpin)$$

For $n=2,3$, this is obviously true, answer is $1$.
however for $n=7$, There is a configuration with diameter $2$, regular hexagon with side $1$, and its center.

I can't prove that regular polygon is answer for $n=4,5,6$. But I still think that it is optimal, So how can I prove it for $n=4,5,6$ and what is the answer in general case?

geometry combinatorial-geometry

share|cite|improve this question

edited Mar 31 at 6:14

Alex Ravsky

43.1k32583

asked Mar 24 at 13:37

snowAuouesnowAuoue

334

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For four, look at the diagonals of the convex hull of the four points. Since all sides are $geq1$, if the diagonals are $<sqrt2$, then by the law of cosines the angles are $<pi/2$. Therefore, the interior angles wouldn't add up to $2pi$. Well, I assumed that the convex hull was a quadrilateral. The cases in which the convex hull is a triangle, or a segment are similar, but easier.
  $endgroup$
  – user647486
  Mar 24 at 14:03
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I also realized that for $n=6$, regular pentagon with side $1$ and its center has diameter of $2cos(fracpi10)=1.90211...$, so regular hexagon is not optimal
  $endgroup$
  – snowAuoue
  Mar 24 at 16:32
  

  
  
  
  

add a comment | 

$begingroup$

There are $n$ points on a plane, such that distance between any two of them $geq 1$. Question is, what is minimum possible diameter for such set of points, that is minimum of distance between two farthest points. At first I thought optimal configuration would be regular $n$-gon with side 1. For this configuration the answer is $$fracsin(fracpincdotlfloorfracn2rfloor)sin(fracpin)$$

For $n=2,3$, this is obviously true, answer is $1$.
however for $n=7$, There is a configuration with diameter $2$, regular hexagon with side $1$, and its center.

I can't prove that regular polygon is answer for $n=4,5,6$. But I still think that it is optimal, So how can I prove it for $n=4,5,6$ and what is the answer in general case?

geometry combinatorial-geometry

share|cite|improve this question

edited Mar 31 at 6:14

Alex Ravsky

43.1k32583

asked Mar 24 at 13:37

snowAuouesnowAuoue

334

$endgroup$

share|cite|improve this question

edited Mar 31 at 6:14

Alex Ravsky

43.1k32583

asked Mar 24 at 13:37

snowAuouesnowAuoue

334

share|cite|improve this question

edited Mar 31 at 6:14

Alex Ravsky

43.1k32583

asked Mar 24 at 13:37

snowAuouesnowAuoue

334

edited Mar 31 at 6:14

Alex Ravsky

43.1k32583

edited Mar 31 at 6:14

Alex Ravsky

43.1k32583

asked Mar 24 at 13:37

snowAuouesnowAuoue

334

asked Mar 24 at 13:37

snowAuouesnowAuoue

334