# Find structure geometry of $A_1, A_2,…,A_n$ such that $\prod_{i<j} A_iA_j$ is maximum

In any triangle we have the well-known inequality:

$$\sin{A}\sin{B}\sin{C} \le \frac{3\sqrt{3}}{8} (1)$$

Signification of inequality (1): Let three points $A, B, C$ lie on a circle then $AB.BC.CA$ maximum when $ABC$ is an equilateral triangle.

Similarly (4 points lie on circle):

Let four points $A_1, A_2, A_3, A_4$ lie on a circle then $A_1A_2.A_1A_3.A_1A_4.A_2A_3.A_2A_4.A_3A_4$ maximum when $A_1A_2A_3A_4$ is a square.

Similarly (n points lie on circle):

Let $n$ points $A_1, A_2,....,A_n$ lie on a circle then $\prod_{i<j} A_iA_j$ maximum when $A_1A_2...A_n$ is a n-regular polygon

In Three-dimensional space: Let four points $A_1, A_2, A_3, A_4$ in a sphere then $A_1A_2.A_2A_3.A_3A_4.A_4A_1$ maximum when $A_1A_2A_3A_4$ is a regular tetrahedron

In Three-dimensional space: Let eight points $A_1, A_2, A_3,...,A_8$ in a sphere then $\prod_{i<j} A_iA_j$ maximum when $A_1A_2\cdots A_8$ is a cube

The question: Let $n$ points $A_1, A_2,...,A_n$ lie on a sphere (in a Euclidean space with $m$ dimentions, $m < n$) find structure geometry of $A_1, A_2,...,A_n$ such that $\prod_{i<j} A_iA_j$ is maximum

Note that: The question ask by me and Ngo Quang Duong.

• The function $\log$ is "completely monotone" in the sense of Cohn-Kumar-Minton arxiv.org/abs/1308.3188 so their results show that the octahedron and the icosahedron are also optimal. – DES-SupportsMonicaAndTransfolk Sep 6 '16 at 16:03
• Are you sure about the cube, though? Often in such problems some square antiprism does better than a cube. – Noam D. Elkies Sep 9 '16 at 4:33
• Yes, I am sure, Thank Dear Dr. @NoamD.Elkies – Oai Thanh Đào Sep 9 '16 at 7:15
• I conjecture that: Let $n$ points $A_1, A_2,...,A_n$ lie on a sphere (in a Euclidean space with $m$ dimentions, $m < n$) then $\prod_{i<j} A_iA_j$ is maximum when structure geometry of $A_1, A_2,...,A_n$ satisfying the conditions if $i-j=l-m$ then $d(A_i, A_j) = d(A_l, A_m)$ for all $i, j, l, m=1, 2, \cdots, n$. – Oai Thanh Đào Sep 11 '16 at 12:01

The question of distributing $N$ points on the sphere so that the product of their mutual distances is maximized is asked in

L.L. Whyte, Unique arrangements of points on a sphere, Amer. Math. Monthly, 59 (1952), 606-611.

Such points are sometimes referred to as logarithmic points because they minimize the discrete logarithmic energy $$E_{0}(\omega_{N})=\sum_{i<j}\log\frac{1}{|x_{i}-x_{j}|}$$ among all point configurations $\omega_{N}=\{x_{1},x_{2},\ldots,x_{N}\}$ on the unit sphere $S^{2}$. They are also known as elliptic Fekete points.

Instead of the logarithmic energy, the minimization can be performed with respect to the Riesz energy $$E_{s}(\omega_{N})=\sum_{i<j}\frac{1}{|x_{i}-x_{j}|^{s}},$$ where $s>0$. The Thompson problem corresponds to the minimization of the electrostatic potential energy when $s=1$.

According to

P.D. Dragnev, Log-optimal configurations on the sphere, Contemporary Mathematics Vol. 661 (2016)

the Whyte problem is rigorously solved for the following 7 values of $n$ only, namely,

• $n=1,2,3,4$ : solutions are the trivial'' ones
• $n=5$ : two points at the Poles and three points forming an equilateral triangle on the equator (P.D. Dragnev, D.A. Legg, and D.W. Townsend 2002)
• $n=6$ : regular octahedron (A.V. Kolushov and V.A. Yudin 1997)
• $n=12$ :regular icosahedron (N.N. Andreev 1996)

For large values of $n$, the numerical determination of optimal points is difficult as the number of local minima grows exponentially with $n$. Designing a fast algorithm to compute in polynomial time “nearly optimal” logarithmic energy points is Problem #7 in Smale's list of problems for this century.

Configuration of 1600 points and Voronoi cells, near optimal with respect to the $s$-energy for $s=1$ (Image by R. Womersley from this paper)

• I conjecture that: Let $n$ points $A_1, A_2,...,A_n$ lie on a sphere (in a Euclidean space with $m$ dimentions, $m < n$) then $\prod_{i<j} A_iA_j$ is maximum when structure geometry of $A_1, A_2,...,A_n$ satisfying the conditions if $i-j=l-m$ then $d(A_i, A_j) = d(A_l, A_m)$ for all $i, j, l, m=1, 2, \cdots, n$. – Oai Thanh Đào Sep 11 '16 at 12:00