On the Convex Central Configurations of the Symmetric (ℓ + 2)-body Problem

For the 4-body problem there is the following conjecture: Given arbitrary positive masses, the planar 4-body problem has a unique convex central configuration for each ordering of the masses on its convex hull. Until now this conjecture has remained open. Our aim is to prove that this conjecture cannot be extended to the (ℓ + 2)-body problem with ℓ ⩾ 3. In particular, we prove that the symmetric (2n + 1)-body problem with masses m1 = … = m2n−1 = 1 and m2n = m2n+1 = m sufficiently small has at least two classes of convex central configuration when n = 2, five when n = 3, and four when n = 4. We conjecture that the (2n + 1)-body problem has at least n classes of convex central configurations for n > 4 and we give some numerical evidence that the conjecture can be true. We also prove that the symmetric (2n + 2)-body problem with masses m1 = … = m2n = 1 and m2n+1 = m2n+2 = m sufficiently small has at least three classes of convex central configuration when n = 3, two when n = 4, and three when n = 5. We also conjecture that the (2n + 2)-body problem has at least [(n +1)/2] classes of convex central configurations for n > 5 and we give some numerical evidences that the conjecture can be true.


INTRODUCTION AND STATEMENT OF THE MAIN RESULTS
The equations for the classical Newtonian N -body problem are given by is the position vector of the point mass m i in an inertial coordinate system, |q j − q i | is the Euclidean distance between the masses m j and m i , and G is the gravitational constant which can be taken to be equal to one by choosing conveniently the unit of time. The configuration space of the planar N -body problem is E = {(q 1 , . . . , q N ) ∈ R 2N : q i = q j , for i = j}.
is the center of mass of the system. Thus, a central configuration (q 1 , . . . , q N ) ∈ E of the N -body problem with positive masses m 1 , . . . , m N is a solution of the system of 2N equations (1.1) with i = 1, . . . , N, for some λ.
A configuration is called convex if none of the masses is contained in the interior of the convex hull of the other remaining masses.
McMillan and Bartky in 1932 (see [10]) proved that the 4-body problem for any set of positive masses has a convex central configuration. This result was reproved in a simpler way by Xia in 2004 (see [13]).
In the paper of MacMillan and Bartky, the following conjecture is implicit: For the planar 4-body problem with positive masses there is a unique convex central configuration for each ordering of the masses on its convex hull. This conjecture also appears explicitly in the papers of Pérez -Chavela and Santoprete [12], and of Albouy and Fu [1]. Until now this conjecture has remained open and the opinion of the people that have worked on it is that its proof can be hard. Numerically it is known that the 5-body problem with equal masses has only one convex central configuration, see [9] and [11]. Chen and Hsiao [6] provided necessary conditions for a central configuration of the 5-body problem to be convex, and also show some numerical convex central configurations for different values of the masses.
Our results are the following. Theorem 1 is proved in Section 3.
In Section 3.5 we give numerical evidence that Conjecture 1 holds. In Section 4.6 we give numerical evidence that Conjecture 2 holds.  A key point in the proofs of this paper is the computation of the real roots of polynomials. The exact number of the real roots of a real polynomial is determined using the Sturm sequences associated to the polynomial. For details on the Sturm sequences, see Section 4.2 of [8]. These sequences have been computed with the algebraic manipulators Maple and Mathematica. After the roots are determined using the bisection method that allows one to control an exact sufficient small interval containing the real root, see for details on the bisection method again Section 4.2 of [8]. After locating the roots, some determinants or series coefficients need to be evaluated; these evaluations are done by controlling the errors of the computation using interval analysis, see, for instance, [2] and references therein. Again all these computations have been done with Maple and Mathematica.

THE SYMMETRIC RESTRICTED + 2-BODY PROBLEM WITH EQUAL MASSES AT THE VERTICES OF A REGULAR -GON
Now for k = 1, 2, 3 we consider the S k -symmetric restricted ( + 2)-body problem with masses equal to one at the vertices of a regular -gon with positions x i = cos α i , y i = sin α i , 1 i where α i = 2π(i − 1)/ + α and α = 0 when k = 1, 2 and α = π/ when k = 3, and two infinitesimal masses with m = 0 located at the points (x +1 , y +1 ) and (x +1 , −y +1 ). Here (x +1 , y +1 ) corresponds to the position of the infinitesimal mass in a central configuration of the restricted + 1-body problem with equal masses at the vertices of a regular -gon.
The restricted ( + 1)-body problem with equal masses at the vertices of a regular -gon has been studied by several authors. Arenstorf in [3] proved that for the central configurations of the restricted (3 + 1)-body problem with 3 equal masses at the vertices of an equilateral triangle, the infinitesimal mass must be on one of the three straight lines passing through the barycenter and a vertex of the triangle. He also proved that on each straight line there are exactly four positions for the infinitesimal mass in a central configuration. Bang and Elmabsout in [4] and [5] and later on Fernandes et al. in [7] studied the problem for 3. The results of [4,5,7] are summarized in the following lemma.

Lemma 1. For
3 consider the central configurations of the restricted ( + 1)-body problem with equal masses at the vertices of a regular -gon of radius 1.
(a) The infinitesimal mass is located on an axis of symmetry of the -gon (see Theorem 2 in [4]).
(b) Assuming that the bodies are arranged so that the positive x-axis is a semiaxis of symmetry containing one of the primaries, the possible positions for the infinitesimal mass are (0, 0) and (ρ 1 , 0) with ρ 1 > 1.
(c) Assuming that the bodies are arranged so that the positive x-axis is a semiaxis of symmetry that does not contain any of the primaries, the possible positions for the infinitesimal mass are (0, 0), (ρ 2 , 0) and (ρ 3 , 0) with 0 < ρ 2 < 1 < ρ 3 .
Neither [3,5] nor [7] provide the explicit values of ρ 1 , ρ 2 and ρ 3 . In this paper these values are necessary and they will be given when we need them.
3. THE S 1 -SYMMETRIC (2n + 1)-BODY PROBLEM 3.1. Equations of the S 1 -symmetric (2n + 1)-body Problem Since the set of central configurations is invariant under rotations and dilations and we have the first integral of the center of mass, without loss of generality we can assume that the center of masses is at the origin of coordinates and that q 1 = (1, 0). Since the center of masses is at the origin, we have (cm x , cm y ) = (0, 0), so 2n+1 i=1 m i x i = 0, and x 2 = −(1/2 + n i=3 x i + mx 2n ). Notice that due to the symmetry 2n+1 i=1 m i y i is identically zero. Taking into account these conditions together with the S 1 -symmetry, the 4n + 2 equations e 1 = 0, . . . , e 4n+2 = 0 given by (1.1) with N = 2n + 1 for the central configurations of the S 1symmetric (2n + 1)-body problem reduce to the following 2n equations: Indeed, the equations e 1 and e 2n+2 are omitted because And from the S 1 -symmetry, we have e i = e 2n+1−i , e i+2n+1 = −e 4n+2−i , for 2 i n, e 2n+1 = e 2n and e 4n+2 = −e 4n+1 . From the first equation e 2 = 0 we isolate λ and we substitute it into the other 2n − 1 equations, which we denote by (3.1) In short, we have 2n − 1 equations and 2n − 1 unknowns y 2 , x i , y i , 3 i n, x 2n , y 2n .
We are interested in the central configurations of the S 1 -symmetric restricted 2n + 1-body problem, see Section 2. Since the number of primaries is odd, without loss of generality we can assume that the line of symmetry is the x-axis and that one of the primaries is in the positive xaxis. We also assume that the infinitesimal mass m 2n is in the x-axis, i. e., y 2n = 0. Straightforward computations show that, when m = 0, x i = cos α i , y i = sin α i , with α i = 2π(i − 1)/(2n − 1), 1 i 2n − 1 and y 2n = 0, all the equations in (3.1) are identically zero except the equation E 2n = 0, which becomes f n (x 2n ) = 0 with
In short, we have three equations and three unknowns y 2 , x 4 and y 4 .
In the next result we provide all the convex central configurations of the S 1 -symmetric restricted (3 + 2)-body problem having the three primaries with masses equal to one located at the vertices of the equilateral triangle (x 1 , y 1 ) = (1, 0), (x 2 , y 2 ) = (−1/2,  Here a = −1.6197896088. . . is a root of the polynomial Proof. First we look for the central configurations of the S 1 -symmetric restricted (3 + 2)-body problem with m 4 on the axis of symmetry containing m 1 , i. e., with y 4 = 0. In the previous subsection we have seen that under this condition x 4 must satisfy the equation f 2 (x 4 ) = 0.
Since we are only interested in convex central configurations, we can restrict to x 4 < −1/2. Under this hypothesis the equation f 2 (x 4 ) = 0 becomes 3(2x 4 + 1) Squaring both sides of this equation and dropping the denominators, we get the polynomial equation 3x 4 P (x) = 0. Solving numerically this polynomial equation, we get exactly two solutions with x 4 < −1/2, but only x 4 = a is a solution for the initial Eq. (3.2). This proves statement (b).
Taking the axis of symmetry through the mass m 3 , or equivalently rotating the solution of statement (b) through angle −2π/3, we get statement (a).
Clearly, the configuration with (x 4 , y 4 ) = (x 5 , y 5 ) = (a, 0) and the one with (x 4 , y 4 ) = (−a/2, − √ 3a/2) and (x 5 , y 5 ) = (−a/2, √ 3a/2) are both convex. Moreover, if we take the axis of symmetry through the mass m 2 , by symmetry we get again the configuration given in statement (a). Thus, these are the unique convex central configurations of the restricted (3 + 2)-body problem.  Proof. Using the Implicit Function Theorem, we will see that the solutions of system E 4 = 0, E 7 = 0 and E 9 = 0 with m = 0 given in Proposition 1 can be continued to solutions of this system with m > 0 sufficiently small. We start with the solution (x 4 , y 4 ) = (−a/2, − √ 3a/2). The determinant of the Jacobian of the system E 4 = 0, E 7 = 0 and E 9 = 0 (with respect to the variables y 2 , x 4 , y 4 ), evaluated at m = 0, Since this determinant is different from zero by the Implicit Function Theorem there exists an analytic family y 1 2 (m), x 1 4 (m) and y 1 4 (m) of solutions of the system E 4 = 0, E 7 = 0 and E 9 = 0 with , defined for m > 0 sufficiently small. Notice that the system E 4 = 0, E 7 = 0 and E 9 = 0 is not analytic in all its variables in a neighborhood of m = 0, y 2 = √ 3/2 and (x 4 , y 4 ) = (a, 0) because the equation E 9 = 0 contains the term m/y 2 4 . After making the change of variables y 4 = μY 4 /2 with μ = m 1/3 , we obtain a new system of equations which is analytic in all its variables in a neighborhood of μ = 0, y 2 = √ 3/2, x 4 = a and Y 4 = 0 and it is given by

Consider now the system of equations
which is analytic in all its variables in a neighborhood of μ = 0, y 2 = √ 3/2, x 4 = a and Y 4 = 0. Substituting μ = 0, y 2 = √ 3/2, x 4 = a into the systemĒ 4 = 0,Ē 7 = 0 andĒ 9 = 0, we find that E 4 andĒ 7 are identically zero and that the equationĒ 9 = 0 is equivalent to The solutions of this equation are Y 4 = 0 and Clearly, the systemĒ 4 = 0,Ē 7 = 0 andĒ 9 = 0 is analytic in all its variables in a neighborhood of the solution μ = 0, Moreover, the determinant of the Jacobian of the system (with respect to the variables y 2 , x 4 , Y 4 ) evaluated at this solution is Therefore, applying the Implicit Function Theorem, there exists a unique analytic family of solutions . Since for m = 0, they are two different convex central configurations, it follows that for m > 0 sufficiently small they continue being two different convex central configurations. This completes the proof of the theorem.

Proof of Theorem 1 for = 5
For the symmetric 7-body problem (that is, n = 3), the equations of central configuration (3.1) are reduced to 5 equations: (by the fist integral of the center of masses), and the ones for the symmetric restricted (5 + 2)-body problem are reduced to the equation f 3 (x 6 ) = 0, see Section 3.1.
In order to prove the convexity of our central configurations, we will use the following result. Lemma 2. Let q i for i = 1, . . . , N be the vertices of an n-gon ordered sequentially counterclockwise. If the signed areas of all the triangles formed by three consecutive vertices of the n-gon are positive, then the n-gon is convex.
REGULAR AND CHAOTIC DYNAMICS Vol. 25 No. 3 2020 Recall that the signed area of a triangle formed by the consecutive vertices of an n-gon (not necessarily regular) in the counterclockwise order, p 1 = (x 1 , y 1 ), p 2 = (x 2 , y 2 ) and p 3 = (x 3 , y 3 ), is given by Squaring both sides of this equation and dropping the denominators, we get the polynomial equation −8192x 6 P (x 6 ) = 0 where P (x 6 ) is a polynomial of degree 35 in the variable x 6 . This polynomial equation has exactly four real roots with x 6 < cos(4π/5) of which only x 6 = a 1 and x 6 = a 2 are solutions of the initial Eq. (3.3).
It is easy to see that the configurations with (x 6 , y 6 ) = (a 1 , 0) and the one with (x 6 , y 6 ) = (a 2 , 0) are both convex because they satisfy the conditions of Lemma 2. This proves statements (d) and (e).
We can also take the mass m 6 on an axis of symmetry passing through m 4 and m 5 . This corresponds to rotating the solutions (x 6 , y 6 ) = (a 1 , 0) and (x 6 , y 6 ) = (a 2 , 0) through angles −4π/5 and −2π/5, respectively. Note that by symmetry, if we take the mass m 6 on the axis of symmetry through m 2 (respectively, m 3 ), we obtain the same configuration as if we take the axis of symmetry through m 5 (respectively, m 4 ).
The system E 3 = 0, E 6 = 0, E 9 = 0, E 10 = 0, E 13 = 0 is analytic in all its variables in a neighborhood of the solutions m = 0, U =Ũ and V =Ṽ i for i = 1, 2, 3. Let J (U, V, m) denote the Jacobian of the system E 3 = 0, E 6 = 0, E 9 = 0, E 10 = 0, E 13 = 0 with respect to the variables U , V . Straightforward computations show that Since these determinants are different from zero by the Implicit Function Theorem there exist three analytic families of solutions of the system E 3 = 0, E 6 = 0, E 9 = 0, E 10 = 0, E 13 = 0 defined for m > 0 sufficiently small. They are given by The system E 3 = 0, E 6 = 0, E 9 = 0, E 10 = 0, E 13 = 0 is not analytic in all its variables in a neighborhood of the solution m = 0, X =X and Y = (x 6 , y 6 ) = (a, 0) with either a = a 1 or a = a 2 because the equation E 13 = 0 contains the term m/y 2 6 . After making the change of variables y 6 = μY 6 /2 with μ = m 1/3 , we obtain a new system of equations which is analytic in all its variables in a neighborhood of μ = 0, U =Ũ , x 6 = a and Y 6 = 0 and it can be written as Now we consider the system of equations which is also analytic in all its variables in a neighborhood of μ = 0, U =Ũ , x 6 = a, and Y 6 = 0. Substituting μ = 0, U =Ũ and x 6 = a into the systemĒ 3 = 0,Ē 6 = 0,Ē 9 = 0,Ē 10 = 0,Ē 13 = 0, we find thatĒ 3 ,Ē 6 ,Ē 9 ,Ē 10 are identically zero and that the equationĒ 13 = 0 is equivalent tō The solutions of this equation are If a = a 1 , then Y 6 = ±b 1 = ±1.1302790764. . ., and if a = a 2 , then Y 6 = ±b 2 = ±0.4564622776. . . . By symmetry, we are only interested in the positive values of Y 6 . Clearly, the systemĒ 3 = 0,Ē 6 = 0,Ē 9 = 0,Ē 10 = 0,Ē 13 = 0 is analytic in all its variables in a neighborhood of the solutions μ = 0, U =Ũ , x 6 = a i , Y 6 = b i with i = 1, 2. Moreover, the Jacobian of the system (with respect to the variables U , x 6 , Y 6 ) evaluated at these solutions takes the values 92.9486133. . . when i = 1 and −2892.8337318417. . . when i = 2. Therefore, applying the Implicit Function Theorem, there exist two analytic families of solutions of the system E 3 = 0,Ē 6 = 0,Ē 9 = 0,Ē 10 = 0,Ē 13 = 0 defined for μ > 0 sufficiently small. They are given by 6 (μ)/2, then these two families provide the two families of solutions of the system E 3 = 0, E 6 = 0, E 9 = 0, E 10 = 0, E 13 = 0 given by Since for m = 0 we have five different central configurations that are convex (see Fig. 2), it follows that for m > 0 sufficiently small they continue being five different convex central configurations. This completes the proof of the theorem.

Proof of Theorem 1 for = 7
For the symmetric 9-body problem (that is, n = 4), the equations of central configuration are reduced to 7 equations:  The system E 3 = 0, E 4 = 0, E 8 = 0, E 11 = 0, E 12 = 0, E 13 = 0, E 17 = 0 is not analytic in all its variables in a neighborhood of the solution m = 0, U =Ũ and (x 8 , y 8 ) = (a 3 , 0) because the equation E 17 = 0 contains the term m/y 2 8 . After making the change of variables y 8 = μY 8 /2 with μ = m 1/3 , we obtain a new equivalent system of equations which is analytic in all its variables in a neighborhood of μ = 0, U =Ũ , x 8 = a 3 and Y 8 = 0 and it can be written as Now we consider the system of equations . . , 4. Since for m = 0 we have four different central configurations that are convex (see Fig. 3), it follows that for m > 0 sufficiently small they continue being four different convex central configurations. This completes the proof of the theorem.

Numerical Evidence that Conjecture 1 is True
From Lemma 1 we know that for all n 3 the equation f n (x 2n ) = 0 has two solutions with x 2n < 0, which we denote by a 1 (n) and a 2 (n), where a 1 (n) < a 2 (n). We have computed numerically these two solutions for n = 4, . . . , 150 (see Fig. 4 for the plot of these solutions as a function of n). Each solution provides n central configurations of the S 1 -symmetric restricted (2n + 1)-body problem that are given by with j = 1, 2 and i = 1, . . . , n − 1. Applying Lemma 2, we see that the central configurations (x 2n , y 2n ) i,1 are concave for all i = 1, . . . , n − 1, whereas the central configurations (x 2n , y 2n ) i,2 are convex for all i = 1, . . . , n − 1. In short, for n = 4, . . . , 150 the restricted (2n + 1)-body problem with 2n − 1 masses at the vertices of a regular n-gon has exactly n classes of convex central configurations. We have applied the Implicit function Theorem as in the proof of Theorem 1 for = 3, 5, 7 to continue these central configurations to the symmetric (2n + 1)-body problem with m > 0 sufficiently small for n = 4, . . . , 20, and in all cases we have seen that the corresponding determinants are different from zero. In particular, we see that the absolute value of these determinants increases as n increases. This gives numerical evidence that Conjecture 1 is true.

Equations of the S 2 -symmetric (2n + 2)-body Problem
We assume that q 1 = (1, 0). Since the center of masses is at the origin, we have (cm x , cm y ) = (0, 0), so 2n+2 i=1 m i x i = 0, and x 2 = −( n i=3 x i + x n+1 /2 + 1/2 + mx 2n+1 ). Notice that due to the symmetry 2n+2 i=1 m i y i is identically zero. Taking into account all these conditions, the 4n + 4 equations e 1 = 0, . . . , e 4n+4 = 0 given by (1.1) with N = 2n + 2 for the central configurations of the planar S 2 -symmetric (2n + 2)-body problem reduce to the following 2n + 1 equations: From the first equation e 1 = 0 we isolate λ and substitute it into the other 2n equations, which we denote by In short, we have 2n equations and 2n unknowns y 2 , x i , y i , 3 i n, x n+1 , x 2n+1 , y 2n+1 . Now we consider the symmetric restricted (2n + 2)-body problem. From the results in Section 2 we know that the infinitesimal masses must be on one of the axes of symmetry of the 2ngon. Clearly, when the infinitesimal masses are on an axis of symmetry containing one of the primaries, the resulting central configuration cannot be convex, so we only consider the case where the primaries are on an axis of symmetry that does not contain any of the primaries. Without loss of generality we can assume that the 2n primaries with masses equal to one are at the vertices of the regular 2n-gon x i = cos α i , y i = sin α i , 1 i 2n with α i = 2π(i − 1)/2n and the two infinitesimal masses with m = 0 are at the points (x 2n+1 , y 2n+1 ) = r cos (π/2n), r sin (π/2n) and (x 2n+2 , y 2n+2 ) = r cos (π/2n), −r sin (π/2n) for some r > 0.

Proof of Theorem 2 for = 4
Next, we see that there are no convex central configurations of the S 2 -symmetric restricted (4 + 2)-body problem.
The proof of Theorem 2 for = 10 follows from the following result.

Some Numerical Evidence that Conjecture 2 is True
Proceeding as in Section 3.5 for the S 2 -symmetric (2n + 2)-body problem we have computed the two solutions of f n (r) for n = 6, . . . , 151. We denote these solutions by r = a 1 (n) and r = a 2 (n) with a 1 (n) < a 2 (n). Then applying Lemma 2, we have seen that the solution a 2 (n) does not lead to convex central configurations of the S 2 -symmetric rectricted (2n + 2)-body problem, whereas the solution a 1 (n) provides exactly n+1 2 convex central configurations. We have also applied the Implicit Function Theorem as in the proof of Theorem 2 for = 6, 8, 10 to continue these central configurations to the symmetric (2n + 2)-body problem with m > 0 sufficiently small for n = 6, . . . , 20 and in all cases we have seen that the corresponding determinants are different form zero. Moreover, the absolute value of these determinants increases as n increases. This gives numerical evidence that Conjecture 2 is true.

Equations of the S 3 -symmetric (2n + 2)-body Problem
Without loss of generality we can assume that the position of m 1 is fixed at the point q 1 = cos π/(2n) , sin π/(2n) and that the center of masses is at the origin of coordinates, so 2n+2 i=1 m i x i = 0, and using the S 3 -symmetry, we have x 2 = − cos π/(2n) + n i=3 x i + mx 2n+1 . Notice that using the S 2 -symmetry again 2n+2 i=1 m i y i is identically zero.
Taking into account all these conditions, the 4n + 4 equations e 1 = 0, . . . , e 4n+4 = 0 given by (1.1) with N = 2n + 2 for the central configurations of the planar symmetric (2n + 2)-body problem reduce to the following 2n + 1 equations: From the equation e 2n+3 = 0 we isolate λ and substitute it into the other 2n equations, which we denote by where E k is e k after the substitution. In short, we have 2n equations and 2n − 1 unknowns y 2 , x i , y i , 3 i n, x 2n+1 , y 2n+1 . Since we have more equations than unknowns, this system could have no solution. Now we consider the S 3 -symmetric restricted (2n + 2)-body problem where the 2n primaries with masses equal to one are at the vertices of the regular 2n-gon x i = cos(α i + π/(2n)), y i = sin(α i + π/(2n)), 1 i 2n with α i = π(i − 1)/n and the two infinitesimal masses with m = 0 are at the points (x 2n+1 , y 2n+1 ) and (x 2n+1 , −y 2n+1 ). Here (x 2n+1 , y 2n+1 ) is the position of the infinitesimal mass in a central configuration of the restricted (2n + 1)-body problem with 2n equal masses at the vertices of a regular 2n-gon. From Lemma 1, the position of the infinitesimal mass is on an axis of symmetry of the 2n-gon.

Proof of Theorem 3 for = 4
For the central configurations of the S 3 -symmetric 6-body problem (that is, n = 2) Eqs. (3.1) reduce to the four equations and the three unknowns y 2 , x 5 and y 5 , plus the parameter m. Here The equation E 5 = 0 for the S 2 -symmetric restricted (4 + 2)-body problem when y 5 = 0 becomes Now we see that the central configuration of the S 3 -symmetric restricted (4 + 2)-body problem with (x 5 , y 5 ) = (0, ρ 3 ) cannot be continued to a family of central configurations of the S 3 -symmetric 6-body problem with m > 0 small. To do that we consider system (5.2) as a system of the four unknowns y 2 , x 5 , y 5 and m and apply the Inverse Function Theorem. The Jacobian of the system with respect to the variables y 2 , x 5 , y 5 and m evaluated at the solution y 2 = sin(3π/4), x 5 = 0, y 5 = ρ 3 and m = 0 becomes 3.9841276914. . . = 0. Therefore, the solution is isolated and cannot be continued to a family of solutions of system (5.2) for m > 0 sufficiently small.
We consider now the central configuration of the S 3 -symmetric restricted (4 + 2)-body problem with (x 5 , y 5 ) = (ρ 3 , 0). We will also see that this central configuration cannot be continued to a family of central configurations of the S 3 -symmetric restricted 6-body problem with μ > 0 sufficiently small. Notice that in this case m 5 and m 6 collide, so system (5.2) is not analytic in all its variables in a neighborhood of y 2 = sin(3π/4), x 5 = ρ 3 , y 5 = 0 and m = 0. Proceeding as in the proofs of Theorems 1 and 2, after making the change of variables y 5 = μY 5 /2 with μ = m 1/3 we obtain a new system of equations HereẼ 20 ,Ẽ 50 ,Ẽ 80 andẼ 110 are functions that do not depend on μ. Finally, we consider the system of equationsĒ which is also analytic in its variables in a neighborhood of (5.4). Substituting (5.4) into system (5.5), we find thatĒ 2 ,Ē 5 andĒ 8 are identically zero, and that the equationĒ 11 = 0 is equivalent to the equation Clearly, the derivatives ofĒ 2 ,Ē 5 ,Ē 8 , andĒ 11 with respect to μ when μ = 0 are zero, so the Jacobian of system (5.5) with respect to the variables y 2 , x 5 , Y 5 and μ evaluated at the solution (5.4) with Y 5 = b is zero. In this case the Inverse Function Theorem is not sufficient to prove that the solution (5.4) with Y 5 = b cannot be continued analytically to a solution with μ > 0 sufficiently small. We assume that it can be continued, that is, that y 2 = sin(3π/4) + y 21 μ + y 22 μ 2 + y 23 μ 3 + O(μ 4 ), is a solution of (5.5) and we shall arrive at a contradiction. Indeed, we substitute this solution into (5.5), then expanding in power series of μ we get We can see thatĒ 83 (0, 0, y * 23 ) = −23.3594334957. . . = 0. Therefore, the solution (5.4) with Y 5 = b cannot be continued analytically to a solution of system (5.2) with μ > 0 sufficiently small.