Convexity and symmetry of central configurations in the five-body problem: Lagrange plus two

We study convexity and symmetry of central configurations in the five body problem when three of the masses ara located at the vertices of an equilateral triangle, that we call Lagrange plus two central configurations. First, we prove that the two bodies out of the vertices of the triangle cannot be placed on certain lines. Next, we give a geometrical characterization of such configurations in the sense as that of Dziobek, and we describe the admissible regions where the two remaining bodies can be placed. Furthermore, we prove that any Lagrange plus two central configuration is concave. Finally, we show numerically the existence of non-symmetric central configurations of the five body problem.


Introduction and main results
Central configurations of the planar n-body problem correspond to configurations r = (r 1 , . . . , r n ), r i ∈ R 2 , i = 1, . . . , n, for positive masses m 1 , . . . , m n moving under the Newtonian gravitational attraction such that the acceleration at each mass point is a constant multiple of the relative position with respect to the center of mass C: for all i = 1, . . . , n (in suitable units so the gravitational constant G = 1), where r i j = |r i − r j | is the Euclidean distance between the bodies at r i and r j . One of the reasons for the importance of the central configurations is that they lead to the so called homographic solutions of the planar n-body problem: the initial shape of the configuration is preserved as time varies. The first homographic solutions were found in the three-body problem, the collinear solution of Euler (1767, consisting in three bodies aligned) and the equilateral triangle solution of Lagrange (1772, consisting in three bodies of arbitrary mass located at the vertices of an equilateral triangle).
Counting up rotations and translations in the plane there are exactly five classes of central configurations for each choice of positive masses when n = 3. The finiteness of the number of classes of central configurations was a question posed by Chazy in [4], Wintner in [27], and reformulated by Smale in [25] as a challenge question for the 21st century. In a computer assisted proof, Hampton and Moeckel, in [16], gave an affirmative answer when n = 4 and any choice of the masses. This result was obtained analytically by Albouy and Kaloshin in [2], and extended the result to n = 5 for almost all choice of the masses. The question about the finiteness of the number of classes of central configurations is open for n > 5. We refer the reader to [19,22,24], and the references therein, for a wider introduction on central configurations, their usefulness and open problems.
The study of the geometry of central configurations is a complementary matter to the finiteness conjecture. An often studied geometric property is convexity. A configuration is convex if the polygon defined by the configuration have all interior angles less than or equal to 180 degrees, while is strictly convex if all interior angles are strictly less than 180 degrees. Otherwise the configuration is concave.
MacMillan and Bartky in 1932 ( [20]) showed that for any four positive masses and any ordering of the bodies there exist a convex central configuration, and using the Perpendicular Bisector Theorem (see Theorem 3) it must be strictly convex. MacMillan and Bartky also derived a geometrical condition which is necessary and sufficient to achieve non trivial configurations in the four-body problem, remaining to check the positivity of the masses. This geometrical condition is named sometimes as Dziobek's condition, see [10] and references therein. Recently, Corbera, Cors and Roberts [7] have classified the full set of convex central configurations in the Newtonian planar four-body problem.
In the five-body problem Williams ( [26]) and Chen and Hsiao ( [6]) study the existence of convex central configurations and give some geometric properties. Moreover, Llibre and Gidea ( [13]) and Chen and Hsiao ( [5]) show that in the five-body problem it is possible to achieve a convex but non-strictly convex central configurations, in contrast with the four-body problem. But questions concerning the characterization or classification of the set of convex central configurations in the five-body problem still remain open. See for example [1,6] and references therein.
Hampton in [15] introduces the concept of stacked central configurations in which a proper subset is also a central configuration. We follow the notation (n, k)-stacked central configuration introduced in [11], where n represents the total number of bodies and k, with k = 1, . . . , n − 3, the bodies that can be removed maintaining the configuration central. Hampton shows a family of concave central configurations where three bodies are at the vertices of an equilateral triangle and the other two placed symmetrically respect to one perpendicular bisector line of the triangle. Since then, several authors have paid attention to this kind of central configurations in the five-body problem, see for instance, [8,9,[11][12][13]17,18,21,23]. Several questions arise. One inquiry posted recently by Cornelio, Álvarez-Ramírez and Cors in [9] is the following: For a given central configuration of n-body problem, what is the number of (n, k)-stacked central configurations, for all k = 1, . . . , n − 3? When n = 5 and excluding collinear configurations, the authors give the answer: there can be none, one or two. Another question posed is: are there convex central configurations of the n-body problem for which is possible to add a body and yet to have a convex central configuration? For n = 4 the authors of [12] gave a negative answer. A different proof is given in [6].
In all the discussed studies about stacked central configurations of the five-body problem we observe two main characteristics: all of them are symmetric and none of them are strictly convex. From that observation, one interesting question arise: are those two conditions necessary to have a stacked central configuration? In the case of the (5, 1)-stacked central configurations, the answer is affirmative. Fernandes and Mello, in [12], show that there exists only one (5, 1)-stacked central configuration consisting in a square of equal masses at the vertices plus a body with arbitrary mass at the center of square. Therefore, such a central configuration is symmetric and concave. In the case of the (5, 2)-stacked central configurations, Gidea and Llibre in [13] prove the existence of a symmetric non-strictly convex stacked central configuration where three of the bodies are in Euler configuration.
In the present work, we study the existence of central configurations in the fivebody problem assuming that three bodies are located at the vertices of an equilateral triangle, that we call Lagrange plus two central configurations. In particular, all of these configurations are (5, 2)-stacked central configurations. Let r = (r 1 , r 2 , . . . , r 5 ) be a configuration out of the collision set, i.e., we assume r i j > 0, for all i = j. An equivalent set of equations for central configurations (1) is given by the Andoyer's equations (see the references [3] and [14]) is twice times the oriented area defined by the triangle with vertices at r i , r j and r k . Notice, for a given configuration r = (r 1 , r 2 , . . . , r 5 ), system (2) is a system of linear equations A(r) · M = 0 for the mass vector M = (m 1 . . . m 5 ) t .
Our first goal is to give a geometric characterization of Lagrange plus two central configurations in the five-body problem in the sense of Dziobek's condition.
Theorem 1 Consider a five-body configuration r = (r 1 , r 2 , . . . , r 5 ) so that r 1 , r 2 , r 3 , form an equilateral triangle, there are no three bodies lined-up, and none of the two remaining bodies r 4 , r 5 is located on a bisector line of the triangle.
(i) If the system of Andoyer's equations (2) has solution for m i , i = 1, . . . , 5 with m 4 , m 5 = 0, then the system is satisfied. (ii) If the configuration r satisfy the equations (3), then the system of Andoyer's equations (2) has non-trivial solution.
Notice that geometric condition (3) only depends on the relative positions of the bodies. After that, positivity in the masses must be checked from Andoyer's equations to ensure the existence of Lagrange plus two central configurations. The proof of Theorem 1 is given in Section 3. Also, in that section we include a characterization of the admissible regions in the plane where the two bodies out of the equilateral triangle can be placed (Proposition 4).
The second goal is to give an answer to the questions posted above about the convexity and symmetry of Lagrange plus two central configurations. More concretely: Next result is then a straightforward consequence of Theorem 2 and the result about (5,1) stacked central configurations of Fernandes and Mello [12].

Corollary 1 Consider a central configuration of the five-body problem. If the configuration is strictly convex, then the configuration is not stacked.
We notice that the claims of Theorem 1 and 2 are valid not only for the Newtonian potential, but also for any homogeneous potential 1/r α .
Finally, with respect to the question whether the symmetry is a necessary condition for the existence of stacked central configurations, the answer is negative. In Section 5 we show numerically the existence of non-symmetric central configurations containing an equilateral triangle. We perform a massive numerical exploration in the admissible

Preliminary results
Given five bodies with masses m 1 , . . . , m 5 and located at r 1 , . . . , r 5 (r i ∈ R 2 , i = 1, . . . , 5), we denote by c i j the line that passes through r i , r j , and by l i j the bisector line of the segment defined by the points r i , r j . See Fig. 1.
We recall the Perpendicular Bisector Theorem, see [22], which is a geometric tool to check if a configuration can be central (see Fig. 1 for an example of a configuration not allowed by this property).

Theorem 3 (Perpendicular Bisector) Let r i and r j be two positions of a planar central configuration, and consider the two open double cones determined by the lines c i j and l i j . If one of the double cone has nonempty intersection with the central configuration, then so does the other open double cone.
We introduce a technical lemma that gives a relation between the areas of the triangles i jk given by any configuration of five points, see for instance [26].

Lemma 1 Consider five points in the plane
Proof It is a straightforward computation from the definition of i jk .
Consider now that three of the five bodies are located at the vertices of an equilateral triangle, that is, in Lagrange plus two configuration, see Proof Without loss of generality we can suppose that r 1 , r 2 and r 3 are at the vertices of the equilateral triangle ordered counterclockwise, and that r 4 and r 5 are aligned with r 3 , see Fig. 2. Consider the Andoyer's equations (2) corresponding to f 12 , f 13 and f 23 . Using that R 12 = R 23 = R 13 we have that  Proof Without loss of generality, we can consider that r 1 , r 2 and r 3 are at the vertices of an equilateral triangle ordered counterclockwise, and that r 4 ∈ c 12 . Applying the Perpendicular Bisector Theorem with respect to r 1 and r 2 , we have that r 5 ∈ c 12 or r 5 ∈ l 12 . The first case is not possible, because there are no central configurations of the five body problem with four bodies in a line. Thus r 5 ∈ l 12 . We use again the Perpendicular Bisector Theorem twice: first, with respect r 1 and r 4 , we have that r 5 must be in a different half plane defined by the line c 12 than r 3 ; second, with respect r 1 and r 3 , r 4 cannot be located between r 1 and r 2 . See Fig. 3.
Introducing 124 = 0, R 15 = R 25 , 235 = − 135 into equations f 13 and f 23 of Andoyer's equations, we have that As m 4 = 0, we have that the second factor, which depends only on r 4 , must vanish. Without loss of generality, we consider that the side of the triangle equals 1 and r 4 is at the right hand side of r 2 at a distance a > 0. Therefore, Squaring both sides one gets a polynomial with all the coefficients positive, so the equation has no solution for a > 0, which proves the claim.

Proposition 3
Consider a five-body central configuration r = (r 1 , r 2 , . . . , r 5 ) so that three of them form an equilateral triangle. If one of the two remaining bodies is on a bisector line of the triangle, then the other one is on the same line.
Proof The proof is straightforward using the Perpendicular Bisector Theorem with respect the bisector line that contains two bodies.

Geometric condition for Lagrange plus two central configurations
In this section we give proof of Theorem 1 that states a geometric condition on the configuration of five bodies with three at the vertices of an equilateral triangle, in order to have non-trivial solutions for the masses of Andoyer's equations. We avoid two particular configurations: when three of the bodies are aligned or when any of bodies, excluding the vertices of the triangle, is on a bisector l i j of the triangle. Notice that if a configuration is central, by Propositions 1, 2 and 3, these two particular configurations are excluded except in the case that three of the bodies are on the same bisector line l i j , studied by Llibre and Mello [18]. Therefore, we can exclude them from our study.
Moreover, after the proof of Theorem 1, we study the admissible regions where the two bodies, that are not at the vertices of the equilateral triangle, can be placed.

Proof of Theorem 1
We recall the hypothesis of the theorem. Consider a five-body configuration r = (r 1 , r 2 , . . . , r 5 ) so that r 1 , r 2 , r 3 , form an equilateral triangle, there are no three bodies lined-up, and none of the two remaining bodies r 4 , r 5 is located on a bisector line of the triangle. Without loss of generality, we suppose r 1 , r 2 , r 3 ordered counterclockwise.
First, we want to see that if the system of Andoyer's equations have non-trivial solution for m i , i = 1, . . . , 5, with m 4 , m 5 = 0, then the system (3) is satisfied. We consider the system of Andoyer's equations given in (2) as a linear system for the masses where A ∈ M 10×5 and M = (m 1 . . . m 5 ) t . Notice that, from the hypothesis, all i jk = 0, for all i, j, k. We introduce in the equations the fact that the mutual distances between r 1 , r 2 and r 3 are equal. In particular, the subsystem given by the equations corresponding to f 12 , f 13 and f 23 : must have non-trivial solution. Notice that all the coefficients of m 4 and m 5 in (6) cannot vanish because r 4 , r 5 / ∈ l i j , i, j = 1, 2, 3. Then, for a fixed configuration, system (6) must have rank one, that is, which is the system of the statement (3). Second, we want to see that if system (3) is satisfied, then Andoyer's equations have non-trivial solution for m i , i = 1, . . . , 5. More concretely, we will prove that under the hypothesis the Andoyer's linear system (5) has kernel of dimension at least one. Equations (3) ensure that system (6) has rank one, so we can remove two equations (for example the ones corresponding to f 13 and f 23 ) from (5) to obtain the linear system with A ∈ M 8×5 . We want to prove that there exist non trivial solutions of (8).
For simplicity in what follows we write 234 , 235 , 145 , 245 and 345 instead of the corresponding expressions.
Second, we can obtain R 14 and R 15 from equations F 1 = 0, F 2 = 0, as follows. Using the third relation in (9), we write the system (3) as: for a certain coefficients a i , b i , c i , depending on the other R i j and i jk . Then, the above system is solved for R 14 and R 15 , and has two solutions: Recall that r 4 , r 5 / ∈ l 12 , so the solution (10) is not admissible. Therefore, R 14 and R 15 are given by (11).
Then, we introduce (9) and (11) in equations (8). Using a computer algebra system (concretely the software Mathematica), a generator of the kernel of the matrix is obtained of the form where N i , i = 1, 2, 3 are expressions depending on R i j and i jk and The expressions for N i can be found in the Appendix. Recall that r 4 / ∈ l 23 , so R 24 − R 34 = 0. If K = 0, the system (8) has kernel of dimension one, with the generator given by (12), so there exists non-trivial solutions of Andoyer's equations.
It remains to prove that the same is true when K = 0. We proceed as follows: • First, we isolate one of the R i j from the Eq. (13), introduce the expression obtained into (8) and compute its kernel. • If in the expressions obtained, all the denominators are not null, we are done.
• If there are denominators that can be null, we consider those cases apart, repeating the process. At each step, we introduce the new restrictions into (8) and compute again its kernel.
More concretely, the restriction K = 0 leads to diferent cases: • Case 1. From the equation K = 0, if R 34 = R 12 we isolate R 25 : We introduce the above expression in (8), obtaining the matrix A 1 , which has kernel generated by: where We introduce this expression in A 1 , and the resulting matrix has a kernel generated by and the vector (1, 0, 0, 0, 0). Thus, we have a non-trivial solution for Andoyer's equations. -Case 1b: R 24 = R 12 . In this case, substituting the restriction into (13) we also obtain R 25 = R 12 , and introducing both restrictions into the system (8), we have that all the coefficients corresponding to m 2 are nul. Therefore, (0, 1, 0, 0, 0) is a non-trivial solution.
• Case 2: K = 0 and R 34 = R 12 . Introducing this restriction into (13), we have that where R 12 − R 24 = 0 because r 4 / ∈ l 23 . Therefore, we have R 35 = R 12 , and similarly to the case 1b, introducing the restrictions into into the system (8), we have that all the coefficients corresponding to m 3 are nul. Therefore, (0, 0, 1, 0, 0) is a non-trivial solution.
In all the cases, we have found non-trivial solutions. This concludes the proof of Theorem 1.

Remark 1
In the proof of Theorem 1 we obtain a generator of the kernel of the Andoyer's linear system in all cases. But in the case K = 0, we point out that any solution of the Andoyer's equations have one or more masses (but not all) that vanish. Clearly, those cases cannot be central configurations.
In particular, from the proof of Theorem 1 we have the following equivalence:

Admissible regions
If we restrict to symmetric solutions, there are two possibilities: when the line of symmetry contains only one of the bodies, which must be a vertex of the triangle, or when the line of symmetry contains three bodies, one vertex and the other two bodies.
In the first case, if we take the line of symmetry to be the bisector l 12 (similar results are obtained considering the other bisector lines), the two equations in (3) are equal and can be written as If r 4 , r 5 / ∈ l 12 , then the Eq. (15) has two disconnected components. Taking a system of reference such that the axes correspond to c 12 and l 12 , we can write r 4 = (x, y), and r 5 = (−x, y), Hampton ([15]) shows that for any x such that r 4 is in the interior of the triangle, there exists one value y such that the configuration of the five bodies is central and r 4 , r 5 are located along the curve defined by (15) that connect the vertices r 1 and r 2 , and goes through the barycenter of the triangle (inner symmetric central configurations). Llibre, Mello and Pérez-Chavela ( [17]) show that only for a certain range of values of x such that r 4 is in the exterior of the triangle there exists one value of y for which a central configuration exists and r 4 , r 5 are located along two pieces of a curve outside the triangle that goes through one vertex (outer symmetric central configurations). See Fig. 4.
If r 4 , r 5 ∈ l 12 , then three families of central configurations are obtained, see [18]. In order to study non-symmetric central configurations, using the geometric condition of Theorem 1, we characterize the admissible regions for r 4 , r 5 , when they are not on any bisector. (r 1 , r 2 , . . . , r 5 ) so that r 1 , r 2 , r 3 , form an equilateral triangle. Let r = r 12 = r 13 = r 23 be the side of the triangle, c i j be the line that passes through r i , r j , and l i j be the bisector line of the segment defined by the points r i , r j . We define the following open regions (see Fig. 5): Next result shows that any central configuration can be obtained, by symmetries and rotations, from a central configuration where the bodies r 4 and r 5 are located in the regions given in Definition 1. (r 1 , r 2 , . . . , r 5 ) so that r 1 , r 2 , r 3 , form an equilateral triangle and r 4 , r 5 / ∈ l i j , i, j = 1, 2, 3. Let T i , i = 1, 2, 3 and S j , j = 1, . . . , 4 the regions given in Definition 1.

Proposition 4 Consider a central configuration of five bodies r =
1. There exist a central configuration r that can be obtained from r applying rotations of angle ±2π/3 and reflections through any of the bisector lines l i j , i, j = 1, 2, 3, such that r 1 , r 2 , r 3 still are at the vertices of an equilateral triangle and r 4 ∈ T 1 ∪ T 2 ∪ T 3 . 2. If r 4 ∈ T i , i = 1, 2, 3, then one of the following occurs: Proof Without loss of generality, we can suppose that r 1 , r 2 and r 3 are ordered counterclockwise.
Clearly applying rotations of angle ±2π/3 and reflections through any of the bisector lines l i j , i, j = 1, 2, 3, any central configuration can be transformed to a central configuration such that r 4 ∈ T 1 ∪ T 2 ∪ T 3 . Notice that the bisectors lines are excluded by hypothesis and lines c i j are not allowed by Proposition 2. This concludes statement 1. Fig. 6 The regions defined by F 125 < 0, F 135 > 0 and F 235 > 0 (see proof of Proposition 4) To prove the statement 2 we use the geometric condition established in Theorem 1. For convenience we write equations (7) as following where F klm = (R km − R lm ) klm . Clearly, when r 4 ∈ T i , i = 1, 2, 3, F 124 < 0 and F 234 > 0. Moreover, when r 4 ∈ T 1 ∪ T 2 , F 134 > 0, whereas if r 4 ∈ T 3 , then F 134 < 0. First suppose that r 4 ∈ T 1 ∪ T 2 . If F 125 < 0, then equations (16) have solution only if F 135 > 0 and F 235 > 0. These three inequalities define three regions (see Fig. 6) which share the same double cone defined by the lines c 2 and l 23 as the region T 1 ∪ T 2 . By the Perpendicular Bisector Theorem that is not allowed. Therefore F 125 > 0, and in that case, the equations (16) have solution only if F 135 < 0 and F 235 < 0. That corresponds to the region S 1 ∪ S 2 ∪ S 3 .
When r 4 ∈ T 3 , a similar augment leads to r 5 ∈ S 4 .

On the convexity of Lagrange plus two central configurations
In this section we prove that there no exist convex central configurations of the five body problem containing an equilateral triangle. This result can be proved regardless of the existence of the geometric condition obtained in the previous section. We distinguish between strictly convex configurations, i.e., all the interior angles of the polygon are strictly less than 180 degrees, and convex, but non-strictly convex, configurations. For simplicity, the later will be called non-strictly convex configurations.
In the first case, we will use the Theorem of Cheng and Hsiao, (see Theorem 6.1 in [6]), for five-body strictly convex central configurations:  We notice that from the proof of the theorem, in the case when three interior edges are larger than r 0 , we can obtain that the other two, that must cross each other, do not emanate from the same vertex. That is, the crossing point is not at any vertex r j .
We first suppose that we have a strictly convex configuration with three bodies at the vertices of an equilateral triangle. Without loss of generality we consider that r 1 , . . . , r 5 are disposed counterclockwise. We examine two possibilities, depending whether the three bodies in the equilateral triangle are correlative or not. Without particularization, we consider the cases: (i) The position vectors r 3 , r 4 and r 5 form an equilateral triangle (correlative); (ii) The position vectors r 1 , r 2 and r 4 form an equilateral triangle (non-correlative).
In the case (i) we have that r 34 = r 35 = r 45 . Let S be the open half-cone defined by the half lines starting at r 4 and passing through r 3 and r 5 . As the configuration is strictly convex, it implies that r 1 and r 2 must belong to S outside of the equilateral triangle, see Fig. 7, left. Nevertheless, this situation is not admissible by the Perpendicular Bisector Theorem applied at the vertices of r 4 and r 5 (Fig. 7, right). So, do not exist any central configuration strictly convex of type (i).
In the case (ii) we have that r 12 = r 14 = r 24 , and r 14 , r 24 are interior edges. From Theorem 4, since the two diagonals r 14 and r 24 are equal to one side (r 12 ), the other three interior edges r 13 , r 25 and r 35 must be greater than r 12 . See the forbidden region for r 3 and r 5 in Fig. 8. Nevertheless, this configuration contradicts Theorem 4, since the diagonals r 14 and r 24 do not cross in the interior of the pentagon. So, do not exist any central configuration strictly convex of type (ii).
Next, we suppose that we have a non-strictly convex configuration containing an equilateral triangle. Suppose that the vertices are r 1 , r 2 and r 3 and ordered counterclockwise. Notice that the Perpendicular Bisector Theorem 3 prevents from four aligned bodies in a central configuration. Therefore, if the configuration is non-strictly convex, three of the bodies must lie on the same line. So, there are two possibilities: the three aligned bodies contain two of the vertices or just one. The first case is not possible from Proposition 2. In the second case, from Proposition 3 the configurations is concave. This ends the proof of Theorem 2.

Non-symmetric Lagrange plus two central configurations
In this section we show numerically the existence of non-symmetric central configurations of the five body problem in which three of the masses are located at the vertices of an equilateral triangle. As we mention before, it is known that there exist two families of symmetric central configurations of this problem: a family inside the equilateral triangle (inner central configurations, Hampton [15]), and a family outside the triangle (outer central configurations, Llibre et al. [17]). See Fig. 4.
We can consider that r 1 , r 2 and r 3 are the vertices of the equilateral triangle ordered counterclockwise. By Theorem 1, the admissible configurations for central configurations are those satisfying (3) such that the solution of the Andoyer's equations give all m i > 0. By Proposition 4, it is enough to study their location for r 4 and r 5 in the regions T i , i = 1, 2, 3 and S j , j = 1, . . . , 4 respectively. See Fig. 5. The inner symmetric central configurations are such that r 4 ∈ T 1 and r 5 ∈ S 1 , whereas the outer ones are such that r 4 ∈ T 2 and r 5 ∈ S 2 . We assume a system of reference of axes c 12 and l 12 in which r 1 = (−1, 0), r 2 = (1, 0), r 3 = (0, √ 3), r 4 = (x, y) and r 5 = (u, v). For each one of the regions T i , i = 1, 2, 3, we take a grid of values (x, y) (by taking a step of 5 × 10 −4 and 10 −3 for variables x and y respectively) within a ball (we will see that for values of (x, y) far away we did not find any solution of the equations). Then, for any fixed (x, y) we solve the system (3) for the variables (u, v) by an iterative scheme. For each solution, we substitute the values (u, v, x, y) into the system of Andoyer's equations (2), where the equations corresponding to f 13 and f 23 are removed. The linear system for the masses is solved by using singular value decomposition (SVD) to determine the kernel of the matrix (with a precision up to 10 −7 ). We keep the values for which a generator of the kernel has all the components of the same sign.
By the explorations performed, we have obtained that there exist central configurations only when (r 4 , r 5 ) ∈ T 1 × S 1 and (r 4 , r 5 ) ∈ T 2 × S 2 . That is, there only exist non-symmetric central configurations in the regions surrounding the curves of symmetric central configurations. In Fig. 9 we show the locations r 4 , r 5 for which there exist non-symmetric central configurations. We notice that in the case of outer central configurations (with r 4 , r 5 outside the equilateral triangle), we only find central configurations confined inside a ball of center r 3 and radius the length of the side of the triangle. Therefore we conjecture that Proposition 4 can be improved by deleting T 3 , S 3 and S 4 from the statement, and restricting the regions T 2 and S 2 by r 3 j ≤ r 12 .

Discussion and conclusions
We study central configurations in the five-body problem when a proper subset of the configuration form an equilateral triangle and so is also a central configuration of the three-body problem: Lagrange plus two configuration. In that sense the study can be categorized inside the stacked central configuration class. As far as we know, all the five-body stacked central configurations found till now have two main characteristics: are symmetric and concave, with the exception of one symmetric family that is convex, but non-strictly convex, since three of the bodies are aligned.
Under the previous considerations we give a geometric characterization, similar to Dziobek's condition for four bodies: any Lagrange plus two central configuration must satisfy a system of two equations that depends only on the location of the bodies. Using this geometric condition, we are able to limit the regions where the two bodies out of the Lagrange configuration can be placed.
Furthermore, we prove that is impossible to have convex central configurations in the five-body problem with three bodies at the vertices of an equilateral triangle. In particular, there are no (5,2)-stacked central configurations strictly convex.
Finally, using the geometric characterization of the configuration, we carry out an exhaustive numerical study to explore the existence of non-symmetric Lagrange plus two central configurations. As far as we now these are the first non-symmetric central configurations exhibited in the five-body problem.
Among other things, it is still an open question the existence of non-symmetric five-body central configurations with three aligned bodies.

Data Availability
The data that support the findings of this study are available from the corresponding author upon reasonable request.