Spatial Convex but Non-strictly Convex Double-Pyramidal Central Configurations of the (2n+2)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(2n+2)$$\end{document}-Body Problem

A configuration of the N bodies is convex if the convex hull of the positions of all the bodies in R3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {R}^3$$\end{document} does not contain in its interior any of these bodies. And a configuration is strictly convex if the convex hull of every subset of the N bodies is convex. Recently some authors have proved the existence of convex but non-strictly convex central configurations for some N-body problems. In this paper we prove the existence of a new family of spatial convex but non-strictly convex central configurations of the (2n+2)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(2n+2)$$\end{document}-body problem.


Introduction and Statement of the Main Result
The equations of motion of the spatial N -body problem are for k = 1, . . . , N , where G is the gravitational constant which will be taken equal to one by choosing conveniently the unit of time, q k ∈ R 3 is the position vector of the punctual mass m k in an inertial coordinate system, and the two dots denote the second derivative with respect to the time t.
The configuration space of the spatial N -body problem is E = {(q 1 , . . . , q N ) ∈ R 3N : q k = q j , for k = j}.
Given N positive masses m 1 , . . . , m N a configuration (q 1 , . . . , q N ) ∈ E is central if there exists a positive constant λ such thaẗ q k = −λ (q k − cm) , is the center of mass of the system. Consequently a central configuration (q 1 , . . . , q N ) ∈ E of the N -body problem with masses m 1 , . . . , m N is a solution of the system of equations for k = 1, . . . , N and for some λ.
We do not know the first author which shown that N equal masses located at the vertices of a regular N -gon is a central configuration of the N -body problem for all N ≥ 2.
If we put equal masses at the vertices of any regular polyhedron, then we obtain a spatial central configuration of the N -body problem with N equal to the number of vertices (see [2]).
Other spatial central configurations of the N -body problem are the pyramidal central configurations, which consist of N = n + 1 masses, with n equal masses at the vertices of a regular n-gon and the (n + 1)th mass in the orthogonal straight line to the plane containing the n-gon and passing through its barycenter (see for instance [5] and [17]).
Another simple spatial central configurations are the double pyramidal central configurations, which consist of N = n + 2 masses, n equal masses at the vertices of a regular n-gon and the other two equal masses symmetrically localized with respect to the plane containing the n-gon on the orthogonal straight line to this plane and passing through its barycenter (see [8][9][10][11][12][13][14]18,19]). In [15] the authors study these previous central configurations adding a new mass on the barycenter of the n-gon. In [4] we studied these double pyramidal central configurations but when the two masses on the orthogonal straight line to the plan containing the n-gon through its barycenter are not equal, we called such central configurations by bi-pyramidal central configurations.
We recall that a configuration of the N bodies is convex if the convex hull of the positions of all the bodies in R 3 does not contain in its interior any of these bodies. And a configuration is strictly convex if the convex hull of every subset of the N bodies is convex.
As far as we know the first example of a convex but non-strictly convex central configuration appeared in 2010 in the paper [7] of Gidea and Llibre for the planar 5-body problem. In 2012 Chen and Hsiao [3] reproved this previous result of the planar 5-body problem and provided a spatial convex but non-strictly convex central configuration for the 7-body problem. In these two first examples of convex but non-strictly convex central configurations three of the bodies are on a straight line. More recently in 2018 Fernandes, Garcia and Mello in [6] provided convex but non-strictly convex central configuration for the planar 8-body problem, and for the spatial 10-and 20-body problem. We think that these are all the know convex but non-strictly convex central configuration which appear until now in the literature. Inspired in the convex but non-strictly convex central configuration for the spatial 10-body problem found by Fernandes, Garcia and Mello in this paper we shall prove the existence of convex but non-strictly convex central configuration for the spatial (2n + 2)-body problem for all n ≥ 4. More precisely, we prove that for n ≥ 4 there are spatial convex but non-strictly convex central configurations of the (2n + 2)-body problem consisting of n masses equal to m 1 = 1 (we choose the unit of mass equal to m 1 ) at the vertices of a regular n-gon, n masses equal to m at the vertices of the n-gon whose vertices are the midpoints of the edges of the initial n-gon, and two masses equal to μ on the straight line orthogonal to the plane containing the two n-gons passing through their barycenters. Moreover, we show that for n = 3 does not exist such spatial convex but non-strictly convex central configurations. Note that the convex hull of such central configurations is formed by two equal pyramides with the base formed by the big n-gon, glued by their bases. So we call them spatial convex but non-strictly convex double-pyramidal central configurations of the 2n + 2-body problem. Thus our main result is the following one.

Theorem 1
The following statements hold for the central configurations of the (2n + 2)-body problem consisting of n masses equal to 1 at the vertices of a regular n-gon inscribed in a circle of radius 1, n masses equal to m at the vertices of the n-gon whose vertices are the midpoints of the edges of the initial n-gon, and two masses equal to μ on the straight line orthogonal to the plane containing the two n-gons passing through their barycenters and at a distance a of it. We note that the spatial convex but non-strictly convex central configuration found by Fernandes, Garcia and Mello for the 10-body problem is the central configuration of Theorem 1 for n = 4.
In Sect. 2 we provide the explicit equations for the spatial convex but non-strictly convex double-pyramidal central configurations of the 2n + 2-body problem.

Equations of the Central Configurations
We consider N = 2n + 2 with n ∈ N, n ≥ 3, n equal masses m 1 = · · · = m n at the vertices of a (regular) n-gon, m n+1 = · · · = m 2n = m at the midpoint (regular) n-gon (the n-gon whose vertices are the midpoints of the edges of the initial n-gon), and 2 masses m 2n+1 = m 2n+2 = μ located symmetrically on the straight line orthogonal to the plane containing the n-gons passing through their center. Without loss of generality we can choose the unit of mass so that m 1 = · · · = m n = 1, and we can take the unit of length so that the radius of the circle containing the initial n-gon be one. By using complex coordinates in the plane that contains the regular n-gons, the positions of the vertices of the initial n-gon can be written as q j = (e iϕ j , 0) ∈ C × R with ϕ j = 2π j/n for j = 1, . . . , n and the vertices of the midpoint n-gon can be written as q j+n = α(e i(ϕ j −π/n) , 0) ∈ C × R with α = cos(π/n) for j = 1, · · · , n. Let the positions of the masses m 2n+1 and m 2n+2 be q 2n+1 = (0, a) and q 2n+2 = (0, −a), respectively, with a > 0.

Proof of Statement (b) of Theorem 1
We start with some auxiliary lemmas. We define

Lemma 1 Let
then the following statements hold.
(a) a * is not defined for n ≥ 473, Proof In [16] it is proved that β n /n is increasing with n and that β n /n > 1 for n ≥ 473. Therefore statement (a) follows directly from the results in [16]. Moreover for n ≥ 473 because β n /n > 1 (see again [16]). This proves statement (c). Statements (b), (d) and (e) follow from simple computations. Finally, computing the values of a * for n = 3, . . . , 472 we get statement (f).

Lemma 2 The function H
Independently of us and almost simultaneously Barrabés and Cors in [1] have proved that H (n) > 0 for all n ≥ 5. Here we give our proof for n ≥ 473 which is different from their proof.
Proof We define Then H (n) can be written as Since α = cos(π/n), we have that F (2πn/n ± π/n) = 0. Therefore On the other hand, since [ n 2 ] denotes the integer part of n 2 and Next we want to see which terms in the sum A are positive and which are negative when n ≥ 473. The sum A can be written as Notice that the first term in the sum A corresponds to x = 2π/n and the last term corresponds to x = π − 3π/n when n is odd, and x = π − 2π/n when n is even. Let I x = [2π/n, π − 3π/n] when n is odd, and I x = [2π/n, π − 2π/n] when n is even. We claim that the function G has a unique zero for x ∈ I x and this zero belongs to the interval (π − 5π/n, π − 4π/n). Next we prove the claim.
To simplify the computations we introduce the change of variables z = x + π/n and consider the equation Let g(z) be the numerator of the factorization of G(z), after doing the substitutions cos π n = α, Squaring both sides of this equation to drop off the square roots and doing the substitution c = cos z we get an equation equivalent to g(z) = 0 given by Note that all solutions of equation G(x) = G(z − π/n) = 0 are solutions of g(c) = 0 with c = cos z.
First we prove that the number of solutions of g(c) = 0 does not change for α ∈ D. The number of solutions of g(c) = 0 can change when g(c) = 0 and g (c) = 0 simultaneously. We compute Res[ g(c), g (c), c] and we get the following polynomial in the variable α Applying Sturm's algorithm we see that there are no zeroes of T for α ∈ D. Therefore, using the properties of resultants, we get that the number of real solutions of g(c) = 0 does not change for α ∈ D.
where B 4 is a polynomial of degree 14 in γ with positive coefficients, so they are positive for α ∈ [cos(π/473), 1) because γ is positive in [cos(π/473), 1). In short if n is even, H (n) > 0 for all n ≥ 473. This completes the proof.
Then the proof follows from the fact that α b 1 − b 2 > 0 for all n ≥ 473 (see Lemma 2).

Proof of Theorem 1 for n = 4
We are interested in the solutions of system (14) with m, μ > 0. When n = 4 we claim that we can find a solution of system (14) with μ = 0 and m > 0 that can be continued to a family of solutions of (14) with μ > 0 small and m > 0. Now we prove the claim. When n = 4 the coefficients of system (14) are We assume that a 11 = 0 and we look for solutions of (14) with μ = 0. Under these assumptions, from (14) we get m = −b 1 /a 11 . Straightforward computations show that a 11 > 0 for a > a ,1 = 5 58 a 11 < 0 for a < a ,1 , and a 11 = 0 for a = a ,1 ; and b 1 > 0 for a > a ,2 = 4 2 7 2/3 3 0 for a < a ,2 , and b 1 = 0 for a = a ,2 . Therefore m > 0 when a ∈ (a ,1 , a ,2 ). On the other hand, from (14) again, μ = 0 when N μ (a) = a 21 b 1 − a 11 b 2 = 0. We see that N μ (a ,1 ) > 0 and N μ (a ,2 ) < 0. Therefore, since N μ is defined for all a ∈ R + , there exists at least a zero a = a μ of N μ with a μ ∈ (a ,1 , a ,2 ). If this zero does not satisfy condition D(a μ ) = a 11 a 22 − a 21 a 12 = 0, then this solution can be continued to a family of solutions of (14) with m > 0 and μ > 0 small, which is given by If the value a μ is such that D(a μ ) = 0, then either system (14) has no solutions when N m (a μ ) = a 12 b 2 − a 22 b 1 = 0, or it has infinitely many solutions when N m (a μ ) = 0. We note that the first case is not possible because, since a 11 = 0, from condition D(a μ ) = 0 we get a 22 = a 21 a 12 /a 11 and from condition N μ (a μ ) = 0 we get b 2 = b 1 a 21 /a 11 , then N m (a μ ) = 0. The infinitely many solutions that we can have when N μ (a μ ) = 0, D(a μ ) = 0 and N m (a μ ) = 0 are In this last case, since −b 1 /a 11 > 0 there exists infinitely many families of solutions of (14) with m > 0 and μ > 0 small. Notice that if a 11 = 0, that is a = a ,1 , then the solution is m = m(a ,1 ) ≈ 2.2217548449 and μ = μ(a ,1 ) ≈ 2.4304510854 = 0 (see (15)). This completes the proof of the claim, so Theorem 1 is proved for n = 4.
Remark: We can eliminate the square roots of equation N μ (a) = 0 by squaring conveniently as many times as we need, then dropping off the denominators we get a polynomial equation of degree 24 in the variable a whose set of solutions contains at least all solutions of equation N μ (a) = 0. This polynomial equation has two positive real solutions that can be computed numerically but only a = a μ ≈ 1.1778268479 is a solution of N μ (a) = 0. Proceeding in the same way with the equation D(a) = 0 we arrive to a polynomial equation of degree 36 in the variable a that has also two positive real solutions but only a = a d ≈ 0.7111336024 is a solution of D(a) = 0. Therefore N μ and D are not zero simultaneously, so for all positive a with a = a d system (14) has a unique solution m = m(a), μ = μ(a) which is given by (15), and for a = a d system (14) has no solution. The solution m = m(a), μ = μ(a) provides a central configurations when m(a) > 0 and μ(a) > 0. Now we find the set where these conditions are satisfied.
As above we eliminate the square roots of equation N m (a) = 0, we drop off the denominators we get a polynomial equation of degree 36 in the variable a that has 6 real solutions but only the solutions a = a m,1 ≈ 0.1092983002, a = a m,2 ≈ 1.3587461717 and a = a m,3 = 3.7016660737 are solutions of N m (a) = 0. Analyzing the signs of N μ , D and N m we get that N μ (a) > 0 for a ∈ (0, a μ ) and N μ (a) < 0 for a ∈ (a μ , +∞); D(a) > 0 for a ∈ (a d , +∞) and D(a) < 0 for a ∈ (0, a d ); and N m (a) > 0 for a ∈ (a m,1 , a m,2 )∪(a m,3 , +∞) and N m (a) < 0 for a ∈ (0, a m,1 )∪(a m,2 , a m,3 ). So m(a) > 0  and μ(a) > 0 for a ∈ (a d , a μ ) = (0.7111336024, 1.1778268479) and the configuration is central. Notice that m(a) = 0 for all a ∈ (a d , a μ ), therefore when n = 4 there are no solutions of (14) with m = 0 and μ > 0.

Proof of Theorem 1 for n ≥ 5
Proceeding in a similar way than in the case n = 4, we claim that for all n ≥ 5 we can find at least a value of a for which there exist a solution of (14) with m = 0 and μ > 0 that can be continued to a family of solutions of (14) with m > 0 and μ > 0. This family of solutions will provide a family of central configurations of our problem, this proves Theorem 1 for n ≥ 5. We note that when n ≥ 5 there are no solutions of (14) with μ = 0 and m > 0 (see the remark at the end of this section) is for that reason that the case n ≥ 5 cannot be treated as the cas n = 4.
We assume that a 12 = 0 and we look for solutions of (14) with m = 0. If m = 0 and a 12 = 0, then from (14) we get μ = −b 1 /a 12 . We analyze the sign of −b 1 /a 12 . From Lemma 1, if 5 ≤ n < 53, then −b 1 /a 12 > 0 for a ∈ (1/ √ 3, a * ), if 53 ≤ n ≤ 472, then −b 1 /a 12 > 0 for a ∈ (a * , 1/ √ 3); and if n ≥ 473, then −b 1 /a 12 > 0 for a ∈ (0, 1/ √ 3). Now we prove that there exists at least a value of a such that m = 0 in the region where μ > 0. If m = 0, then N m (a, n) must be zero. From Lemma 3 we have that if 5 ≤ n ≤ 472, then N m (1/ √ 3, n) and N m (a * , n) have different signs, and if n ≥ 473 then lim a→0 + N m (a, n) = +∞ and N m (1/ √ 3, n) < 0. Since N m (a, n) is well defined for all a ∈ R + we get that for all n ≥ 5 there exist at least a value of a, a = a m , such that N m (a m , n) = 0 and μ > 0. Notice that if a 12 = 0 (i.e. a = 1/ √ 3, see Lemma 1) then N m (1/ √ 3, n) is not zero, so the assumption a 12 = 0 is not restrictive. If D(a m , n) = 0, then the solution of (14) wit a = a m is unique and it can be continued to a family of solutions of (14) in a with m > 0 and μ > 0. As in (15), this family is given by m = N m (a, n)/D(a, n) and μ = N μ (a, n)/D(a, n). If the value a m is such that D(a m , n) = 0, then either system (14) has no solutions when N μ (a m , n) = 0, or it has infinitely many solutions when N μ (a m , n) = 0. We note that the first case is not possible. This is proved following the same arguments than in the case n = 4.
When N m (a m , n) = 0, D(a m , n) = 0 and N μ (a m , n) = 0 we have infinitely many solutions that are given by In this last case since −b 1 /a 12 > 0, for all n ≥ 5 there exists infinitely many families of solutions of (14) with m > 0 and μ > 0. This proves the claim, so Theorem 1 is proved for n ≥ 5. Remark: We have computed numerically the values of a for which N m (a, n) = 0 and the values of a for which D(a, n) = 0 for 5 ≤ n ≤ 600 and they never coincide. Moreover for 5 ≤ n ≤ 600 we have found that both functions, N m (a, n) and D(a, n), have a unique positive zero. Therefore we have strong numerical evidences that for each n ≥ 5 system (14) cannot have infinitely many solutions and that there exists a unique value of a such that the solution of system (14) satisfies that m = 0 and μ > 0.
Proceeding as in the case n = 4 we have computed numerically for 5 ≤ n ≤ 600 the set S where the solution of system (14) provides a central configuration; that is, m > 0 and μ > 0. Analyzing the functions N m (a, n), N μ (a, n) and D(a, n) for 5 ≤ n ≤ 600 we see that these functions have a unique positive zero denoted by a m (n), a μ (n) and a d (n) respectively. Moreover N m (a, n) > 0 for a ∈ (0, a m (n)) and N m (a, n) < 0 for a ∈ (a m (n), +∞); N μ (a, n) > 0 for a ∈ (0, a μ (n)) and N μ (a, n) < 0 for a ∈ (a μ (n), +∞); and D(a, n) > 0 for a ∈ (a d (n), +∞) and D(a, n) < 0 for a ∈ (0, a d (n)). After computing numerically the values of a m (n), a μ (n) and a d (n) for 5 ≤ n ≤ 600 we get that a d (n) < a m (n) < a μ (n) for all 5 ≤ n ≤ 128 and a μ (n) < a m (n) < a d (n) for all 128 < n ≤ 600. In short, S = (a d (n), a m (n)) when n ≤ 128 and S = (a m (n), a d (n)) when n > 128. Notice that μ = 0 for all a ∈ S, therefore there are no solutions of (14) with μ = 0 and m > 0. In Figures 1 (a) and (b) we plot the functions a μ (n), a m (n) and a d (n) for 5 ≤ n ≤ 600. In Figure 2 we plot the length of the interval S for 5 ≤ n ≤ 600. In Table 1 we give the numerical value of S for some values of n. , a 12 = 2  In a similar way, if μ = 0 then

Recall that
Straightforward computations show that a 11 < 0 for a < a with a = 1 2 9 2 a 11 = 0 for a = a, and a 11 < 0 for a < a; b 2 > 0 for a > √ 5/2, b 2 = 0 for a = √ 5/2, and b 2 < 0 for a < √ 5/2; and finally a 21 > 0 for a > 1/ √ 2, a 22 = 0 for a = 1/ √ 2, and a 22 < 0 for a < 1/ √ 2. Therefore m > 0 (that is, −b 1 /a 11 > 0 and −b 2 /a 21 > 0 simultaneously) when a < a < √ 5/2. By eliminating the square roots of equation N μ (a) = 0 and dropping off the denominators we get a polynomial equation P μ (a) = 0 of degree 24 in the variable a. Doing the change of variables γ = (4 − 5a)/(5a − 6) in the polynomial P μ (a) (notice that γ > 0 in a < a < √ 5/2) as above we see that this polynomial equation has no solutions in the interval a < a < √ 5/2. Therefore there is no solution of system (14) with μ = 0 and m > 0. Now we prove that D and N μ (respectively, N m ) are not zero simultaneously. We eliminate the square roots of equation N μ (a) = 0 (respectively, N m (a) = 0) by squaring conveniently as many times as we need, then dropping off the denominators we get a polynomial equation P μ (a) = 0 of degree 24 (respectively, P m (a) = 0 of degree 36). Doing the same with equation D(a) = 0 we get a polynomial equation P D (a) = 0 of degree 36. We do the resultant of the polynomials P D and P μ , and of P D and P m with respect to a, and we get two numbers different from zero. Then by the properties of the resultants these two pairs of polynomials cannot have a common zero.