Spatial bi stacked central configurations formed by two dual regular polyhedra

This is a preprint of: Spatial bi-stacked central configurations formed by two dual regular polyhedra, Montserrat Corbera, Jaume Llibre, Ernesto Pérez-Chavela, J. Math. Anal. Appl., vol. 43), 648 659, 04. DOI: [0.06/j.jmaa.03..05] Spatial bi stacked central configurations formed by two dual regular polyhedra Montserrat Corbera a,, Jaume Llibre b, Ernesto Pérez Chavela c a Departament de Tecnologies Digitals i de la Informació, Universitat de Vic, 08500 Vic, Barcelona, Catalonia, Spain. b Departament de Matemàtiques, Universitat Autònoma de Barcelona, 0893 Bellaterra, Barcelona, Catalonia, Spain. c Departamento de Matemáticas, Universidad Autónoma Metropolitana-I, Av. San Rafael Atlixco 86, 09340 México D.F., México. Abstract In this paper we prove the existence of two new families of spatial stacked central configurations, one consisting of eight equal masses on the vertices of a cube and six equal masses on the vertices of a regular octahedron, and the other one consisting of twenty masses at the vertices of a regular dodecahedron and twelve masses at the vertices of a regular icosahedron. The masses on the two different polyhedra are in general different. We note that the cube and the octahedron, the dodecahedron and the icosahedron are dual regular polyhedra. The tetrahedron is itself dual. There are also spatial stacked central configurations formed by two tetrahedra, one and its dual. Keywords: n body problem, spatial central configurations, dual regular polyhedra 000 MSC: 70F0, 70F5

. Introduction We consider the spatial N body problem m k q k = N j = j k G m k m j q k q j q k q j 3, k =,..., N, where q k R 3 is the position vector of the punctual mass m k in an inertial coordinate system, and G is the gravitational constant which can be taken equal to one by choosing conveniently the unit of time. We fix the center of mass N i= m i q i / N i= m i of the system at the origin of R 3N. The configuration space of the N body problem in R 3 is E = {q,..., q N ) R 3N : N m i q i = 0, q i q j for i j}. i= Given m,..., m N a configuration q,..., q N ) E is central if there exists a positive constant λ such that q k = λ q k, k =,..., N. Thus a central configuration q,..., q N ) E of the N body problem with positive masses m,..., m N is a solution of the system of the N vectorial equations with N unknowns q k for k =,..., N plus λ > 0) N j = j k m j q k q j q k q j 3 = λ q k. ) Central configurations play a main role in celestial mechanics, the transition of behaviour or bifurcations of a N body system happens at central configurations, total collapse and some kind of escapes are asymptotic to central configurations. We also must mention that they generate the homographic solutions, i.e. solutions where the configuration of the bodies is central for all time, in other words the configuration is similar with itself for all time, these are the unique explicitly solutions of the N body problem known until now see [4] for more details). The simplest spatial central configuration for the 4 body problem for 4 arbitrary masses is the tetrahedron, actually this is the unique spatial central configuration for N = 4 modulo homotheties and rotations). In

general for N 5, N equal masses at the vertices of a regular polyhedron of N vertices always form a central configuration see []). For N 5 many particular central configurations have been discovered, but still is interesting to find new classes of central configurations. In the literature the reader can find several papers studying spatial central configurations consisting of nested regular polyhedra, see for instance [, 3, 4] and the references therein. Another important class of central configurations are the so called stacked central configurations, where a proper subgroup of particles form by themselves a central configuration. They were introduced by Hampton in 005 in a planar five problem [6], since then they have been studied by many authors, see for instance [5, 3, 0] for some planar stacked central configurations and [7, 5,,, 7] for some spatial stacked central configurations. In this paper we are interested in a particular class of spatial stacked central configurations formed by two dual regular polyhedra; that is, the vertices of one polyhedron correspond to the faces of the other one and viceversa. More precisely, we will study central configurations formed by a cube and its dual, the octahedron. We will also analyze the central configurations formed by the dodecahedron and its dual, the icosahedron. Since both regular polyhedra are by itself a central configuration, the above classes form a bi stacked central configurations. As far as we know, this is the first time that this kind of central configurations are detected with two different dual polyhedra, because for the tetrahedron such kind of central configurations where studied in [4, 6]. Note that the dual of a regular tetrahedron is another regular tetrahedron facing in the opposite directions. Let P n denote a regular polyhedron with n vertices. The simplest way to create the dual polyhedron of P n is by finding the barycenter of each of its faces, and then connecting these barycenters so that they become the vertices of the new dual polyhedron, which we denote by Pn d. Definition. Given a central configuration formed by two dual regular polyhedra P n and P d n, we say that it is of see Figure ): Type A if P d n is inside P n. Type B if the faces of P d n intersect the faces of P n. Tipe C if P d n is outside P n. The main result of this paper is the following. 3

Theorem. We consider n masses equal to M at the vertices of a regular polyhedron P n of n vertices. Without loss of generality we can assume that M =. We consider n additional masses equal to m at the vertices of the dual polyhedron Pn d scaled by a factor a, then the following statements hold. a) For each regular polyhedron P n we can find a function ma) and a non empty set D R such that for all a D the configuration formed by the vertices of P n and Pn d with m = ma) is central. Moreover the set D contains central configurations of the three Types A, B and C. b) For each regular polyhedron P n different from the dodecahedron and the icosahedron we can find two values µ < µ such that b.) If m 0, µ ), then there are three central configurations, two of Type A and one of Type B. b.) If m = µ, then there are two central configurations, one of Type A and one of Type B. b.3) If m = µ, µ ), then there is a unique central configurations of Type B. b.4) If m = µ, then there are two central configurations one of Type B and one of Type C. b.5) If m = µ, ), then there are three central configurations, one of Type B and two of Type C. c) For the dodecahedron and the icosahedron we can find three values µ < µ < µ 3 such that c.) If m 0, µ ), then there are three central configurations, one of Type A and two of Type B. c.) If m = µ, then there are two central configurations, one of Type A and one of Type B. c.3) If m = µ, µ ), then there is a unique central configurations of Type A. c.4) If m = µ, then there are two central configurations one of Type A and one of Type C. c.5) If m = µ, µ 3 ], then there are three central configurations, one of Type A and two of Type C. c.6) If m = µ 3, ), then there are three central configurations, one of Type A, one of Type B and one of Type C. 4

More information on the central configurations described in Theorem is provided in Theorems 4 and 6 of Section for the cube and the octahedron, in Theorems 8 and 0 of Section 3 for the dodecahedron and the icosahedron, and in Theorems and of Section 4 for the tetrahedron.. Central configurations formed by a cube and an octahedron We consider 8 equal masses m = = m 8 = at the vertices of a cube with positions q =,, ), q =,, ), q 3 =,, ), q 4 =,, ), q 5 =,, ), q 6 =,, ), q 7 =,, ), q 8 =,, ). We consider 6 additional equal masses m 9 = = m 4 = m at the vertices of a regular octahedron with positions q 9 = a, 0, 0), q 0 = a, 0, 0), q = 0, a, 0) q = 0, a, 0), q 3 = 0, 0, a), q 4 = 0, 0, a), for some a > 0. It is easy to check that the center of mass of these two polyhedra is located at the origin of coordinates. After some computations we can see that for this kind of configuration equations ) become a 3 m m R 3/ a 4a ) a 3 R 3/ ) 3 9 8 7 4a ) R 3/ 4a ) R 3/ λ = 0, a λ = 0, where R = a a 3 and R = a a 3. We compute λ from the first equation of ) and we substitute it into the second equation. The resulting equation can be written as m fa) ga) = 0 where fa) = ga) = a a 4a a 3 a R 3/ a 3 a R 3/, 3 9 ) 8 7 4a ) R 3/ 4a ) R 3/. ) 5

Type A Type B Type C n Pn d the faces of P inside P n d and n P n intersect P d n outside P n 8 a 0, ) a [, 3] a 3, ) 0 a 0, ) a [, 5 6 5] a 5 6 5, ) 4 a 0, ) a [, 9] a 9, ) Figure : Plots of central configurations formed by two dual regular polyhedra. 6

f 0.3 0. 0. 0. 0. 0.3 3 4 5 6 7 8 a 0.5 0.5 g 3 4 5 6 7 8 a Figure : The graphs of f and g Therefore m = ma) = ga) fa). 3) The solution of ) given by 3) provides a central configuration if and only if m > 0. Next we analyze the sign of the functions f and g for a > 0. Lemma 3. The functions f and g satisfy the following properties for a > 0 see the graphs of f and g in Figure ). a) The functions f and g are defined for all a > 0. b) Let α =.7875... and α = 3.68586.... Then fα ) = 0, fα ) = 0, fa) > 0 when a 0, α ) α, ) and fa) < 0 when a α, α ). c) Let β = 0.893884... and β =.08366.... Then gβ ) = 0, gβ ) = 0, ga) > 0 when a β, β ) and ga) < 0 when a 0, β ) β, ). Proof. It is easy to check that equations R = 0 and R = 0 have no real solutions, so R > 0 and R > 0 for all a R. Therefore f and g are defined for all a > 0. This proves statement a). By solving equation fa) = 0 see Appendix A) we get two real solutions with a > 0, a = α and a = α. Then by analyzing the sign of f for a > 0 we get statement b). Finally we solve equation ga) = 0 and we get also two real roots with a > 0, a = β and a = β see Appendix B). By analyzing the sign of g for a > 0 we get statement c). 7

From Lemma 3 we see that ma) > 0 when a 0, β ) α, β ) α, ). In short we have proved the following result. Theorem 4. For each a 0, β ) α, β ) α, ) we can find a unique value of m = ma) = ga)/fa) such that the configuration formed by the cube nested with an octahedron is central. The central configuration is of Type A when a 0, β ), it is of Type B when a α, β ), and it is of Type C when a α, ) see Figure ). We note that Theorem 4 corresponds to statement a) of Theorem for the cube and its dual polyhedron, the octahedron. If in Theorem 4 we replace m by /m and a by /a, then we obtain statement a) of Theorem for the octahedron and its dual polyhedron, the cube. Now we are interested in the number of central configurations for each value of m > 0. Lemma 5. The function ma) = ga)/fa) when a > 0 satisfies the following properties see Figure 3). a) mβ ) = mβ ) = 0, b) lim a 0 ma) = 0, c) lim a α ma) = and lim a α ma) =, d) Let γ = 0.7049775... and γ = 4.645077.... Then a = γ is a local maximum of m with mγ ) = µ = 0.03840360..., a = γ is a local minimum with mγ ) = µ = 74.48633..., m is increasing in 0, γ ) γ, ), and it is decreasing in γ, α ) α, α ) α, γ ). Proof. Statement a) follows from Lemma 3c), and statements b) and c) follow immediately from the computation of the corresponding limits. The derivative of m becomes m a) = ga)/fa) where ga) = 3a a 3 ) R 90 8 3 8 6 R 96a 3a a 4 0a 9 ) ) 8 9 3 a 5 7 4 ) R 5/ R 5/ 36a 3 a 5a 3 ) a 5a 3 ) ) R 5/ R 5/ 8.

µ 0.03 0.0 0.0 m 0. 0.4 0.6 γ 0.8 β a 0, β ) a 0 8 6 4 m α.4.6.8 β a α, β ) a 500 400 300 00 00 µ m α γ a α, ) 4 5 6 7 8 9 0 a Figure 3: The graphic of the function m = ma) = ga)/fa). 9

By proceeding in a similar way than in Appendix A and Appendix B we can transform equation ga) into a polynomial equation of degree 44. By solving numerically this polynomial equation we get the following real solutions satisfying equation ga) = 0 a =.70448..., a = 0.64853..., a = γ = 0.7049775..., a = γ = 4.645077.... Then by analyzing the sign of ga) for a > 0 remember that we only are interested in positive values of a) we get that ga) > 0 when a 0, γ ) γ, ) and ga) < 0 when a γ, α ) α, α ) α, γ ). This proves statement d). From Lemma 5 and Figure 3 we have that m increases from 0 to µ when a 0, γ ), it decreases from µ to 0 when a γ, β ), it decreases form to 0 when a α, β ), it decreases from to µ when a α, γ ). Finally m increases from µ to when a γ, ). In short we have proved the following result. Theorem 6. Let α, α, β and β be defined as in Lemma 3, let µ and µ be defined as in Lemma 5d), and let D = 0, γ ), D = γ, β ), D 3 = α, β ), D 4 = α, γ ) and D 5 = γ, ). a) For each value of m 0, µ ) there are three different central configurations, one with a D Type A), one with a D Type A) and one with a D 3 Type B). b) For m = µ there are two different central configurations, one with a = γ Type A), and other with a D 3 Type B). c) For m µ, µ ) there is a unique central configuration with a D 3 T ypeb). c) For m = µ there are two central configurations, one with a D 3 Type B) and other with a = γ Type C). d) For m µ, ) there are three different central configurations, one with a D 3 Type B), one with a D 4 Type C), and one with a D 5 Type C). We note that Theorem 6 corresponds to statement b) of Theorem for the cube and its dual polyhedron, the octahedron. As above if in Theorem 6 we replace m by /m and a by /a, then we obtain statement b) of Theorem for the octahedron and its dual polyhedron, the cube. 0

3. Central configurations formed by a dodecahedron and an icosahedron We consider the dodecahedron with vertices q =,, ), q =,, ), q 3 =,, ), q 4 =,, ), q 5 =,, ), q 6 =,, ), q 7 =,, ), q 8 =,, ), q 9 = 0, /φ, φ), q 0 = 0, /φ, φ) q = 0, /φ, φ), q = 0, /φ, φ), q 3 = /φ, φ, 0), q 4 = /φ, φ, 0), q 5 = /φ, φ, 0), q 6 = /φ, φ, 0), q 7 = φ, 0, /φ), q 8 = φ, 0, /φ), q 9 = φ, 0, /φ), q 0 = φ, 0, /φ), where φ = 5)/ is the golden ratio. Now we construct its dual icosahedron. The dodecahedron is composed by regular pentagonal faces. Let F {i, j, k, l, m} denote the face of the dodecahedron corresponding two the pentagon with vertices i, j, k, l, m and let K i be the barycenter of this pentagon. The faces of the dodecahedron and its corresponding barycenters are given in Table. We consider 0 equal masses m = = m 0 = at the vertices of the dodecahedron with vertices q i for i =,..., 0, and we consider additional equal masses m = = m 3 = m at the positions q i = a K i 0 for i =,..., 3 and for some a > 0. It is easy to check that the center of mass of these two polyhedra is located at the origin of coordinates. After some computations we can see that the equations of the central configurations ) for the configuration formed by the q i with i =,..., 3, become [ ) ) 5 3 5 5 a 5 m 85 38 5 R 3/ 3 5 a R 3/ 3 8 9 3 9 5 a 3 5 5 ) R)] 3/ a 3 5 R 3/ 4 ) λ = 0, 36 65 9 5 ) 5 3 5 5 ) m 8a 5 a R 3/ a R 3/ a 5 R 3/ a 5 3 R 3/ 4 0 5 5 ) aλ = 0, 65 9 ) 5 ) 4)

Face Barycenter F {,, 9, 3, 4} K = 0, φ φ, φ ) 5φ 5 ) φ F {, 3, 9, 0, 7} K =, 0, φ φ 5 5φ φ φ F {, 4, 3, 7, 9} K 3 =, φ ), 0 5φ 5 ) F {, 5, 9, 0, 8} K 4 = 5 φ ), 0, φ φ 5φ F {, 6, 4, 8, 0} K 5 = φ φ, φ ), 0 5φ 5 F {3, 5, 0, 5, 6} K 6 = 0, φ φ, φ ) 5φ 5 φ φ F {3, 7, 5, 7, 9} K 7 =, ) 5φ 5 φ ), 0 F {4, 6,, 3, 4} K 8 = 0, φ φ, ) 5φ 5 φ ) ) φ F {4, 7,,, 9} K 9 =, 0, φ φ 5 5φ F {5, 8, 6, 8, 0} K 0 = φ φ, ) 5φ 5 φ ), 0 ) F {6, 8,,, 0} K = 5 φ ), 0, φ φ 5φ F {7, 8,, 5, 6} K = 0, φ φ, ) 5φ 5 φ ) Table : The faces of the dodecahedron and its corresponding barycenters.

where R = a a 6 5 5, R = a a 6 5 5, R 3 = a 5 ) a 6 5 5, and R 4 = a 5 ) a 6 5 5. We compute λ from the first equation of 4) and we substitute it into the second equation. The resulting equation can be written as m fa) ga) = 0 where 65 9 5 ) 5 3 5 5 ) 5 5 fa) = 8a 0 ) ) )) ) 3 5 5 a a a 3 5 5 a ga) = 360 Therefore R 3/ 65 9 5 0 5 5 5 65 9 5 R 3/ ) ) ) 3 5 a a a 3 5 a, R 3/ 3 R 3/ 4 ) 8 9 3 9 ) 5 a a R 3/ a R 3/ a 5 R 3/ 3 a ) 5 R 3/. 4 m = ma) = ga) fa). 5) The solution of 4) given by 5) provides a central configuration if and only if m > 0. Next we analyze the sign of the functions f and g for a > 0. Lemma 7. The functions f and g satisfy the following properties for a > 0 see the graphs of f and g in Figure 4). a) The functions f and g are defined for all a > 0. b) Let α = 0.955686... and α =.5890.... Then fα ) = 0, fα ) = 0, fa) > 0 when a 0, α ) α, ) and fa) < 0 when a α, α ). c) Let β =.0645... and β =.560908.... Then gβ ) = 0, gβ ) = 0, ga) > 0 when a β, β ) and ga) < 0 when a 0, β ) β, ). 3

f 0.3 0. 0. 0. 0. 0.3 3 4 5 6 7 8 a g 0.5 0.5 3 4 5 6 7 8 a Figure 4: The graphs of f and g Proof. Equations R = 0, R = 0, R 3 = 0 and R 4 = 0 have no real solutions, so R > 0, R > 0, R 3 > 0 and R 4 > 0 for all a > 0. Therefore f and g are defined for all a > 0. This proves statement a). Using the notion of resultants we find all the real solutions of equation fa) = 0 with a > 0 with the help of an algebraic manipulator as Mathematica. These reals solutions are a = α and a = α see Appendix C for details). To complete the prove of statement b) we analyze the sign of f on the intervals 0, α ), α, α ), and α, ). By doing the same for the function g we get statement c) see also Appendix C). From Lemma 7 we see that ma) > 0 when a 0, α ) β, β ) α, ). In short we have proved the following result. Theorem 8. For each a 0, α ) β, β ) α, ) we can find a unique value of m = ma) = ga)/fa) such that the configuration formed by the dodecahedron nested with an icosahedron is central. The central configuration is of Type A when a 0, α ), it is of Type B when a β, β ) and when a α, 5 6 5), and it is of Type C when a 5 6 5, ) see Figure ). We note that Theorem 8 corresponds to statement a) of Theorem for the dodecahedron and its dual polyhedron, the icosahedron. If in Theorem 8 we replace m by /m and a by /a, then we obtain statement a) of Theorem for the icosahedron and its dual polyhedron, the dodecahedron. Now we are interested in the number of central configurations for each value of m > 0. 4

Lemma 9. The function ma) = ga)/fa) when a > 0 satisfies the following properties see Figure 5). a) mβ ) = mβ ) = 0, b) lim a 0 ma) = 0, c) lim a α ma) = and lim a α ma) =, d) Let γ =.478858... and γ =.6746.... Then a = γ is a local maximum of m with mγ ) = µ =.73553..., a = γ is a local minimum with mγ ) = µ = 4.93060..., m increases in a 0, α ) β, γ ) γ, ) and decreases when a γ, β ) α, γ ). Proof. Statement a) of Lemma 9 follows from Lemma 7c), and statements b) and c) follow immediately from the computation of the corresponding limits. The derivative of m is m a) = ga)/fa) where ga) = g a) fa) ga) f a). Working in a similar way as in Appendix C we solve equation ga) = 0, and we get two real solutions with a > 0, a = γ and a = γ. Then by analyzing the sign of ga) for a > 0 we get that ga) > 0 when a γ, β ) α, γ ) and that ga) < 0 when a 0, α ) β, γ ) γ, ). This proves statement d). From Lemma 9 and Figure 5 we have that m increases from 0 to when a 0, α ), it increases from 0 to µ when a β, γ ), it decreases form µ to 0 when a γ, β ), it decreases from to µ when a α, γ ). Finally m increases from µ to when a γ, ). In short we have proved the following result. Theorem 0. Let α, α, β and β be defined as in Lemma 7, let ξ = 5 6 5, let µ and µ be defined as in Lemma 9d), let µ 3 = 48.843... and let D = 0, α ), D = β, γ ), D 3 = γ, β ), D 4 = α, ξ], D 5 = ξ, γ ) and D 6 = γ, ). a) For each value of m 0, µ ) there are three different central configurations, one with a D Type A), one with a D Type B) and one with a D 3 Type B). b) For m = µ there are two different central configurations, one with a = γ Type B), and other with a D Type A). 5

m m 0 8 6 4 0. 0.4 0.6 0.8 a 0, α ) α a µ.5 0.5..4 β γ β a β, β ) a 00 m 50 00 50 µ α γ 4 6 8 a α, ) a Figure 5: The graphic of the function m = ma) = ga)/fa). 6

c) For m µ, µ ) there is a unique central configuration with a D Type A). c) For m = µ there are two central configurations, one with a D Type A) and other with a = γ Type C). d) For m µ, µ 3 ] there are three different central configurations, one with a D Type A), one with a D 5 Type C), and one with a D 6 Type C). e) For m µ 3, ) there are three different central configurations, one with a D Type A), one with a D 4 Type B), and one with a D 6 Type C). We note that Theorem 0 corresponds to statement c) of Theorem for the dodecahedron and its dual polyhedron, the icosahedron. If in Theorem 0 we replace m by /m and a by /a, then we obtain statement c) of Theorem for the icosahedron and its dual polyhedron, the dodecahedron. 4. Central configurations formed by two dual tetrahedra We consider the regular tetrahedron with vertices q = 0, 0, 3/), q = 0, / 3, / 6), q 3 =, / 3, / 6), q 4 =, / 3, / 6). In order to construct its dual tetrahedron we compute the barycenter of each one of its triangular faces. As in the previous section, we denote by F {i, j, k} the face of the tetrahedron corresponding to the triangle with vertices i, j, k and we denote by K i its barycenter. The faces of the tetrahedron and its corresponding barycenters are given in Table. We consider 4 equal masses m = = m 4 = at the vertices of the tetrahedron with vertices q i for i =,..., 4 and we consider 4 additional equal masses m 5 = = m 8 = m at the positions q i = a K i 4 for i = 5,..., 8 and for some a > 0. We note that the barycenters K, K, K 3 and K 4 are the vertices of a new regular tetrahedron which corresponds to the initial one scaled by a factor /3 and rotated with the rotation matrix R = 0 0 0 0 0 0. 7

Face Barycenter ) F {,, 3} K = 3, 3 3, 3 6 F {,, 4} K = ) 3, 3 3, 3 6 F {, 3, 4} K 3 = 0, ) 3 3, 3 6 F {, 3, 4} K 4 = 0, 0, ) 6 Table : The faces of the tetrahedron and its corresponding barycenters. Therefore this configuration corresponds to the configuration of Type II studied in Theorem of [4], and studied also in [6]. By writing the results of [4] and [6] but using the notation of this paper we get the following two theorems. Theorem. Let be fa) = ga) = 3 a 9)a a a 9) 3/ 9 a 3 a a 3), 6 6 6a ) a a 9) 3/ a 6 a 3), and let α =.45669... and α = 9.6083... be the zeros of f with a > 0, and β = 0.4589907... and β = 4.94495... be the zeros of g with a > 0. For each a 0, β ) α, β ) α, ) we can find a unique value of m = ma) = ga)/fa) such that the configuration formed by the vertices of the two dual tetrahedra is central. The central configuration is of Type A when a 0, β ), it is of Type B when a α, β ), and it is of Type C when a α, ) see Figure ). We note that Theorem corresponds to statement a) of Theorem for two dual tetrahedra. Theorem. Let γ = 0.348567... and γ = 6.368... be the zeros of ga)/fa)) with a > 0, let be mγ ) = µ = 0.00034783... and mγ ) = µ = 880.330..., and let be D = 0, γ ), D = γ, β ), D 3 = α, β ), D 4 = α, γ ) and D 5 = γ, ). 8

a) For each value of m 0, µ ) there are three different central configurations, one with a D Type A), one with a D Type A) and one with a D 3 Type B). b) For m = µ there are two different central configurations, one with a = γ Type A), and other with a D 3 Type B). c) For m µ, µ ) there is a unique central configuration with a D 3 Type B). c) For m = µ there are two central configurations, one with a D 3 Type B) and other with a = γ Type C). d) For m µ, ) there are three different central configurations, one with a D 3 Type B), one with a D 4 Type C), and one with a D 5 Type C). We note that Theorem corresponds to statement b) of Theorem for two dual tetrahedra. Acknowledgements The first two authors are supported by a MINECO/FEDER grant number MTM008 03437. The second author is also supported by an AGAUR grant number 009SGR 40, by ICREA Academia and by FP7-PEOPLE- 0-IRSES 36338 and 38999. The three authors are partially supported by CONACYT, México, Grant 8790, and by Red de Cuerpos Académicos del PROMEP-México. Appendix A. In this appendix we solve the equation fa) = 0 defined in Section. This equation can be written as a 4a a 3 a R 3/ = a 3 a R 3/. By squaring booth sides of this equation and isolating the terms containing R 3/ we get ) 4 a 3) 33 a 4 6 a 4 a4 6a 3 9a R 3 a4 6a 3 9a R 3 9 = a R 3/.

We square again both sides of this equation and we drop the denominators by multiplying both sides of the resulting equation by the least common denominator. Then we get the polynomial equation 64 33 8 ) a 6 7 58 ) a 4 83648 0944 ) a 93 ) 458990 a 0 0 73837 8400 ) a 8 337887 68699 ) a 6 5679 009780 ) a 4 4 448939 8079888 ) a 6 309799 79376 ) a 0 43 70667 73 ) a 8 96 8467 696 ) a 6 747954 7 58 ) a 4 708588 7 58 ) a 5344 7 58 ) = 0. We find numerically for instance with the help of Mathematica) all real solutions of this polynomial equation and we get a = ±.7875..., a = ±.84837..., a = ±3.68586.... Finally we check which of these solutions are also solutions of the equation fa) = 0, and we get that the real solutions of fa) = 0 are exactly a =.7875..., and a = 3.68586.... Appendix B. In this appendix we solve the equation ga) = 0 defined in Section. By proceeding as in Appendix we transform equation ga) = 0 into the 0

polynomial equation 3369 9396 388 3 9 ) 6 a 8 3369 9396 388 3 9 ) 6 a 6 4 3369 9396 388 3 9 ) 6 a 4 4 983 897804 4095 3 3856 ) 6 a 3 836805 4603068 43096 3 99388 ) 6 a 0 4 0393 64769 676776 3 45876 ) 6 a 8 4 37547449 33330868 374404 3 090964 ) 6 a 6 6 37480969 039948 963944 3 597708 ) 6 a 4 7 448334005 899644 3787303 3 908956 ) 6 a 4860 6736799 948780 84840 3 560 ) 6 a 0 4374 787409 356 34568 3 809 ) 6 a 8 96 739 4576 96648 3 805788 ) 6 a 6 5344 3369 9396 388 3 9 ) 6 a 4 0883968 83 54 3 6 ) 6 a = 0 We find numerically all real solutions of this polynomial equation obtaining a = 0, a = ±0.893884..., a = ±.0368..., a = ±.0836.... Finally we check which of these solutions are also solutions of the equation ga) = 0, and we get that the real solutions of ga) = 0 are a = 0, a = ±0.893884..., a = ±.0836.... Appendix C. In this appendix we solve equations fa) = 0 and ga) = 0 defined in Section 3. Let r = R /, r = R /, r 3 = R / 3 and r 4 = R / 4. Then f can

be written as fa) = 65 9 5 ) 5 3 5 5 ) 5 5 8a 0 ) ) )) ) 3 5 5 a a a 3 5 5 a r 3 65 9 5 0 r 3 ) ) ) 3 5 a a a 3 5 a. r 3 3 r 3 4 First we eliminate the denominators of the fractions which appear in fa) by multiplying equation fa) = 0 by the least common multiple of all the denominators. We obtain a new equation F 0 = 0, which is a polynomial in the variables a, r, r, r 3 and r 4. We consider four additional polynomials F = r R, F = r R, F 3 = r 3 R 3, F 4 = r 4 R 4. Clearly, the solutions of equation fa) = 0 are also solutions of system F 0 = 0, F = 0, F = 0, F 3 = 0, F 4 = 0. C.) We solve the system of polynomial equations C.) by means of resultants. The resultant of two polynomials P x) and Qx) with leading coefficient one, Res[P, Q], is the expression formed by the product of all the differences a i b j for i =,,..., n, j =,,..., m where a i with i =,,..., n are the roots of P and b j with j =,,..., m are the roots of Q. In order to see how to compute Res[P, Q], see for instance [8] and [9]. The main property of the resultant is that if P and Q have a common root then necessarily Res[P, Q] = 0. Consider now two multivariable polynomials, say P x, y) and Qx, y). These polynomials can be considered as polynomials in x with polynomial coefficients in y. The resultant with respect to x, which is denoted by Res[P, Q, x], is a polynomial in the variable y satisfying the following property: if x 0, y 0 ) is a solution of system P x, y) = 0, Qx, y) = 0 then Res[P, Q, x]y 0 ) = 0. In order to solve system C.), we eliminate the variable r by doing the resultant Res[F 0, F, r ]. We obtain a new polynomial equation F 0 = 0 with the variables a, r, r 3 and r 4. Next we eliminate the variable r by doing the resultant F = Res[ F 0, F, r ], we eliminate the variable r 3 by doing the resultant F = Res[ F, F 3, r 3 ], and finally we eliminate the variable r 4 by

doing the resultant F 3 = Res[ F, F 4, r 4 ]. We obtain in this way a polynomial equation F 3 = 0 of degree 04 in the variable a. We find numerically all real solutions of this polynomial equation obtaining 8 different solutions of which only two are real solutions of the initial equation fa) = 0 with a > 0. These two solutions are a = 0.955686..., a =.5890.... We repeat the same steps for the equation ga) = 0 obtaining in this case a polynomial equation F 3 = C a 4 F = 0, where F is a polynomial of degree 04 in the variable a, and C is a constant. As above we find numerically all real solutions of this polynomial equation obtaining 9 different solutions of which only two are real solutions of the initial equation ga) = 0 with a > 0. These two solutions are a =.0645..., a =.560908.... References [] F. Cedó and J. Llibre, Symmetric central configurations of the spatial n body problem, J. of Geometry and Physics 6 989) 367 394. [] M. Corbera and J. Llibre, Central configurations for the spatial n body problem of nested regular polyhedra, J. of Geometry and Physics 58 008) 4 5. [3] M. Corbera and J. Llibre, Central configurations of three nested regular polyhedra for the spatial 3n body problem, J. of Geometry and Physics 59 009) 3 339. [4] M. Corbera and J. Llibre, Central configurations of nested rotated regular tetrahedra, J. of Geometry and Physics 59 009) 379-394. [5] A. C. Fernandes and L.F. Mello, On stacked central configurations with n bodies when one body is removed, J. Math. Anal. Appl. 405 03) 30 35. [6] M. Hampton, Stacked central configurations: new examples in the planar five body problem, Nonlinearity 8 005) 99 304. [7] M. Hampton and M. Santoprete, Seven-body central configurations: a family of central configurations in the spatial seven-body problem, Celestial Mech. Dynam. Astronom. 99 007) 93-305. 3

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