# The second and the third smallest distances on the sphere

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### Abstract

Let s1 (n) denote the largest possible minimal distance among n distinct points on the unit sphere {Mathematical expression}. In general, let sk(n) denote the supremum of the k-th minimal distance. In this paper we prove and disprove the following conjecture of A. Bezdek and K. Bezdek: s2(n) = s1([n/3]). This equality holds for n > n0 however s2(12) > s1(4). We set up a conjecture for sk(n), that one can always reduce the problem of the k-th minimum distance to the function s1. We prove this conjecture in the case k=3 as well, obtaining that s3(n) = s1([n/5]) for sufficiently large n. The optimal construction for the largest second distance is obtained from a point set of size [n/3] with the largest possible minimal distance by replacing each point by three vertices of an equilateral triangle of the same size e{open}. If e{open} → 0, then s2 tends to s1([n/3]). In the case of the third minimal distance, we start with a point set of size [n/5] and replace each point by a regular pentagon.

Original language English 55-65 11 Journal of Geometry 46 1-2 https://doi.org/10.1007/BF01231000 Published - Mar 1993

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Set of points
Denote
Equilateral triangle
Pentagon
Disprove
Unit Sphere
Minimum Distance
Supremum
Equality
Tend
Distinct

### ASJC Scopus subject areas

• Geometry and Topology

### Cite this

In: Journal of Geometry, Vol. 46, No. 1-2, 03.1993, p. 55-65.

Research output: Contribution to journalArticle

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abstract = "Let s1 (n) denote the largest possible minimal distance among n distinct points on the unit sphere {Mathematical expression}. In general, let sk(n) denote the supremum of the k-th minimal distance. In this paper we prove and disprove the following conjecture of A. Bezdek and K. Bezdek: s2(n) = s1([n/3]). This equality holds for n > n0 however s2(12) > s1(4). We set up a conjecture for sk(n), that one can always reduce the problem of the k-th minimum distance to the function s1. We prove this conjecture in the case k=3 as well, obtaining that s3(n) = s1([n/5]) for sufficiently large n. The optimal construction for the largest second distance is obtained from a point set of size [n/3] with the largest possible minimal distance by replacing each point by three vertices of an equilateral triangle of the same size e{open}. If e{open} → 0, then s2 tends to s1([n/3]). In the case of the third minimal distance, we start with a point set of size [n/5] and replace each point by a regular pentagon.",
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