Trapezoid central configurations

We classify all planar four-body central configurations where two pairs of the bodies are on parallel lines. Using Cartesian coordinates, we show that the set of four-body trapezoid central configurations with positive masses forms a two-dimensional surface where two symmetric families, the rhombus and isosceles trapezoid, are on its boundary. We also prove that, for a given position of the bodies, in some cases an specific order of the masses determine the geometry of the configuration, namely acute or obtuse trapezoid central configuration. We also prove the existence on non-symmetric trapezoid central configuration with two pairs of equal masses.


Introduction
Central configurations are particular positions of the masses in the Newtonian n-body problem, where the position and acceleration vectors with respect to the center of masses are proportional, with the same constant of proportionality for all masses. They play an important role in celestial mechanics because, among other properties, they generate the unique known explicit solutions in the n-body problem for n ≥ 3. For general information about central configurations see for instance Albouy and Chenciner [4], Hagihara [22], Moeckel [33], Saari [38,39], Schmidt [41], Smale [43,44] and Wintner [47].
More precisely we consider the planar n-body problem m kqk = − n j = 1 j = k G m k m j q k − q j |q k − q j | 3 , k = 1, . . . , n, being q k ∈ R 2 the position vector of the punctual mass m k in an inertial coordinate system, and G is the gravitational constant that we can take 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 k = q j , for k = j}.
A configuration of the n bodies (q 1 , . . . , q n ) ∈ E is central if there is a positive constant λ such that (1)q k = −λ (q k − c) , for k = 1, . . . , n, being c the position vector of the center of mass of the system, which is defined by c = n k=1 m k q k / n k=1 m k . Two planar central configurations are equivalent if there is a homothecy of R 2 and a rotation of SO(2) with respect to the center of mass which send one into the other. Since this relation is of equivalency, in what follows we shall consider the classes of equivalency of central configurations.
The complete set of planar central configurations of the n-body problem is only known for n = 2, 3. For n = 2 there is only one class of central configurations. For each choice of three positive masses there are five classes of central configurations of the three-body problem, the three collinear central configurations found in 1767 by Euler [18], and the two equilateral triangle central configurations found in 1772 by Lagrange [25].
When n > 3 there are many partial results for the number of classes of central configurations of the n-body problem. In 1910 Moulton [34] showed that there exists exactly n!/2 classes of collinear central configurations for any given set of positive masses, one for each ordering of the masses on the straight line modulo a rotation of π radians. A lower bound of the number of planar non-collinear central configurations was obtained by Palmore in [35].
Although the set of all planar central configurations of the four-body problem is not completely known, we can find in the literature several papers that provide the existence and classification of central configurations of the four-body problem in some particular cases. For instance, a complete numerical study for the number of classes of central configurations for n = 4 and arbitrary masses was done by Simó in [42]. A computer assisted proof of the finiteness of the number of central configurations for n = 4 and any choice of the masses was given by Hampton and Moeckel [23]. Later on Albouy and Kaloshin [7] proved this result analytically, and extend it for for n = 5, except for a zero measure set in the masses space .
Assuming that every central configuration of the four-body problem with equal masses has an axis of symmetry Llibre in [27] obtained all the planar central configurations of the four-body problem with equal masses by studying the intersection points of two planar curves. Later on Albouy in [1,2] gave a complete analytic proof of this result.
When one of the four masses is sufficiently small Pedersen [36], Gannaway [20] and Arenstorf [10] numerically and analytically obtained the number of its classes of central configurations. These studies were completed later on by Barros and Leandro in [11] and [12].
A central configuration is called kite if it has an axis of symmetry passing through two non-adjacent bodies. The kite non-collinear classes of central configurations having some symmetry for the fourbody problem with three equal masses where characterized by Bernat et al. in [13], see also Leandro [26]. The characterization of the convex central configurations with an axis of symmetry and the concave central configurations of the four-body problem when the masses satisfy that m 1 = m 2 = m 3 = m 4 was done byÁlvarez and Llibre in [9].
A planar configuration of the four-body problem can be classified as either convex or concave. A configuration is convex if none of the bodies is located in the interior of the triangle formed by the others. A configuration is concave if one of the bodies is in the interior of the triangle formed by the others.
In [31] MacMillan and Bartky shown that for any assigned order of any four positive masses there is a convex planar central configuration of the four-body problem with that order. Later on, Xia [49] provided a simpler proof of this result. The following convex conjecture stated by Albouy and Fu in [5] (see also [31,37]) is well known between the community working in central configurations: For the planar four-body problem there is a unique convex central configuration of the four-body problem for each ordering of the masses in the boundary of its convex hull.
Already, MacMillan and Bartky in [31] proved that there exists a unique isosceles trapezoid central configuration of the four-body when two pairs of equal masses are located at adjacent vertices. Later on Xie in [50] reproved this result.
The following subconjecture of the convex conjecture is also well known: For the planar four-body problem there exists a unique convex central configuration having two pairs of equal masses located at the adjacent vertices of the configuration and it is an isosceles trapezoid.
In [29] Long and Sun shown that any convex central configuration with masses m 1 = m 2 < m 3 = m 4 located at the opposite vertices of a quadrilateral and such that the diagonal corresponding to the mass m 1 is not shorter than the one corresponding to the mass m 3 , has a symmetry and the quadrilateral is a rhombus. This result was extended by Pérez-Chavela [37] and Santoprete to the case where two of the masses are equal and at most, only one of the remaining mass is larger than the equal masses. Moreover, they proved that there is only one convex central configuration when the opposite masses are equal and it is a rhombus. Later on Albouy et. al. in [8] shown that in the four-body problem a convex central configuration is symmetric with respect to one diagonal if and only if the masses of the two particles on the other diagonal are equal. If these two masses are unequal, then the less massive one is closer to the former diagonal.
Using the results on the symmetries mentioned in the previous paragraph Corbera and Llibre [14] gave a complete description of the families of central configurations of the four-body problem with two pairs of equals masses and two equal masses sufficiently small, proving for these masses the convex conjecture and the subconjecture. More recently, the subconjecture was proved for arbitrary masses by Fernandes et al. in [19].
The co-circular classes of central configurations of the four-body problem, i.e. when the four masses are on a circle have been studied by Cors and Roberts in [15].
A trapezoid is a convex quadrilateral with at least one pair of parallel sides. The parallel sides are called the bases of the trapezoid and the other two sides are called the lateral sides. See Figure 1 for the classification of the nine trapezoids, namely: • An acute trapezoid has two adjacent acute angles on its longer base edge. • An obtuse trapezoid has one acute and one obtuse angle on each base. • A right trapezoid has two adjacent right angles.
• An isosceles trapezoid is an acute trapezoid if its lateral sides have the same length, and the base angles have the same measure. • A 3-sides equal trapezoid is an isosceles trapezoid with three sides of the same length.  • A rhombus is a parallelogram with the four sides with the same length. • A rectangle is a parallelogram with four right angles.
• A square is a rectangle with the four sides with the same length.
In this paper we are interested in studying trapezoid central configurations. See [40] for a really fresh work in the same topic. In section 3 we derive the equations for the trapezoid central configurations in terms of the mutual distances. In section 4 we prove that not all the trapezoid configurations are realizable. In section 5 we characterize, using cartesian coordinates, the set of positions that yield to trapezoid central configurations with positive masses. In section 6 we prove that there exist a one-parameter family of right trapezoid central configurations. Finally, in section 7 we study the set of positive masses which yields to a trapezoid central configurations. We prove, in contrast to the co-circular case, that two pair of equal masses do not imply that the central configuration has some symmetry.

Preliminaries
The central configurations (in what follows simply CC by short) can be described in terms of Lagrange multipliers. We denote by q = (q 1 , q 2 , q 3 , q 4 ) ∈ (R 2 ) 4 the position of four positive masses m 1 , m 2 , m 3 , m 4 on the plane and by r ij = ||q i − q j || the mutual distances between the i-th and the j-th bodies. The vector q is a CC of the 4-body problem if it satisfies the following algebraic equation for some value of λ (the Lagrange multiplier) where U is the Newtonian potential is the moment of inertia, which represents the size of the system, c is the center of mass of the system, and M = m 1 + · · · + m n is the total mass (see [32] for more details).
We observe that generically six mutual distances describe a tetrahedron in R 3 , since in this work we are interested in planar CC, when we write equation (2) in terms of mutual distances, we must add a constraint to maintain the particles on a plane. This constraint arises setting the volume of the tetrahedron equals to zero. Denoting as r = (r 12 , r 13 , r 14 , r 23 , r 24 , r 34 ) ∈ R + 6 the vector of mutual distances, it is well know in the literature (see for instance [39,41]), that the volume V of a tetrahedron is given by the Cayley-Menger determinant .
From now on we will assume that V = 0. Also in order to avoid collinear configurations we impose that all triples of mutual distances satisfy strictly the triangle inequality (see [15] for more details).
Let A i be the oriented area of the triangle formed by the configuration q where the point q i is deleted, and let ∆ i = (−1) i+1 A i . Since A i > 0 when the vertices are ordered sequentially counterclockwise, for a convex quadrilateral ordered sequentially counterclockwise we obtain ∆ 1 , ∆ 3 > 0 and ∆ 2 , ∆ 4 < 0, satisfying the equation In 1900 Dziobeck [17] proved for planar CC that From this equality we obtain Fixing the moment of inertia I = I 0 and applying Lagrange multipliers, we have that the planar non-collinear CC are the critical points of the function Taking the partial derivatives and using the six mutual distances as variables we obtain Grouping the above equations by row, so that the product of the right-hand side is simply (64σ) 2 ∆ 1 ∆ 2 ∆ 3 ∆ 4 , and since the masses are positive we obtain the well known Dziobeck relation , which must be satisfied for any planar 4-body CC. Solving each of the three pairs of equations for λ we obtain
If we set , then equation (8) can be written as which means that (s 1 , p 1 ), (s 2 , p 2 ), (s 3 , p 3 ), viewed as points in R +2 , must lie on the same line with slope λ. This in turn, is equivalent to a representation that allows to write Dziobeck equation (7) as the nice factorization The Dziobeck equation D = 0 must be satisfied for the six mutual distances of every four-body planar central configuration.

Equations of trapezoidal central configurations
We consider four positive masses m 1 , m 2 , m 3 , m 4 located at the vertices of a trapezoid, i.e. located by pairs on two parallel lines, which without loss of generality we assume are vertical. Since the central configurations are invariant under homotheties we can take the distance between the two parallel lines equal to one, after normalizing the unity of mass we can assume that m 1 = 1 located at the bottom part of the left line, m 2 above m 1 on the same line and m 3 above m 4 on the right line (see Fig. 2). From now on we will use this ordering and normalization of the units of mass and length in this work.
All trapezoidal central configurations are convex, so from the results of McMillan [31], we know first that the diagonals of the respective quadrilateral are longer that any of the four sides, that is (10) r 13 , r 24 > r 12 , r 14 , r 23 , r 34 ; and second that the bigger and the smaller sides of the quadrilateral correspond to opposite sides. We note that in the restricted problem, i.e. when one or more masses are equal to zero, one of the sides of the quadrilateral could be equal to one diagonal.
Lemma 1. The biggest side of the quadrilateral is on the parallel lines.
Proof. Assume that r 23 is the biggest side and that we exclude the case where all the sides are equal, that is, the square. So, its opposite side, r 14 , has to be the smaller one, and r 13 , r 24 > r 23 ≥ r 12 , r 34 ≥ r 14 > 1.
Then depending on the relative position of the four masses we have the following four scenarios: Notice that the cases where m 3 and m 4 are either both above m 2 or both below m 1 are not possible because in these cases one of the diagonals would be smaller than one of the sides.
In all scenarios we will arrive to a contradiction with the fact that r 23 is the biggest side or r 14 the smaller one.
In the scenario (a), r 34 > r 12 + r 2 23 − 1 > 1 + r 2 23 − 1 > r 23 . For the scenarios (b) and (c) we shall use the following result: Let x, y ≥ 1 be two real numbers, Similar argument works if r 14 > 1 is considered the biggest side.
Without loss of generality we label the bodies so that r 12 is the longest side. We can also assume that r 23 ≥ r 14 by an appropriate relabeling. Indeed, equations (6) are invariant if we interchange bodies m 1 and m 2 and bodies m 3 and m 4 . The choice r 23 ≥ r 14 , together with the fact that r 12 is the longest side, implies the relation r 24 ≥ r 13 between the two diagonals.
Summarizing, we have proved the following result.
Lemma 2. Labelling conveniently the bodies, the mutual distances that can provide trapezoid central configurations can be restricted to the following set Next we give the expression of the masses ratios for the trapezoid central configurations on Ω. Taking into account the sign of the areas A i , we have ∆ 1 = −r 34 /2, ∆ 2 = r 34 /2, ∆ 3 = −r 12 /2, ∆ 4 = r 12 /2 (note that we have considered the bodies ordered clockwise). Now from (6) we obtain the following ratios of the masses We observe that the fact that all masses must be positive places additional constraints on the mutual distances. Using λ = (p 2 −p 3 )/(s 2 −s 3 ) and m 1 = 1 into the first equation in (11) and after some simplifications we obtain .
In summary we have proved the next result.

Lemma 3.
Let Any point in Ω defines a four-body trapezoid central configuration with positive masses. Moreover, up to relabelling and rescaling the set Ω contains all trapezoid central configurations.

The trapezoids which are not realizable as central configuration
In this section we prove that the vertices of the parallelogram, the rectangle and the 3-sides equal trapezoid are not realizable as central configurations of the four-body problem with the exception of the square and the rhombus.
Assume that the bodies are ordered sequentially as in Figure 2.   Proof. The possible boundaries of Ω are the sets where either r 24 = r 13 , r 13 = r 12 , r 12 = r 23 , r 23 = r 14 , or r 14 = r 34 . Next we characterize these boundaries.
Case A: r 24 = r 13 . The trapezoids having equal diagonals are the rectangle, the square and the isosceles trapezoid. The rectangle is not a realizable central configuration. , See Figure 3 for the plot of the set Ω.
The points of C 1 provide configurations of the restricted problem with m 1 , m 2 and m 3 in an equilateral triangle; the points of C 2 and C 3 provide rhombus and isosceles trapezoid central configurations, respectively; P 1 corresponds to an equilateral triangle configuration of the restricted problem with a collision of m 3 and m 4 at one of the vertices; P 2 corresponds to a configuration of the restricted problem with the masses at the vertices of a rhombus and such that the positions of m 1 , m 2 and m 3 are the vertices of an equilateral triangle; and P 3 corresponds to the square central configuration.
On the boundary with the masses m 1 , m 2 and m 3 at the vertices of an equilateral triangle we have r 13 = r 12 = r 23 . Solving the system of equations r 13 = r 12 = r 23 , we get a = 2/ √ 3 and b = 1/ √ 3. Substituting this solution into r 23 and r 14 and imposing the condition r 23 ≥ r 14 , we get the condition −1/ √ satisfies r 24 ≥ r 13 ≥ r 12 ≥ r 23 ≥ r 14 ≥ r 34 . So the set C 1 ∪ P 1 ∪ P 2 belongs to the boundary of Ω. Moreover it is easy to check that the point P 1 correspond to an equilateral triangle configuration with the masses m 3 and m 4 colliding at the corresponding vertex of the triangle; and the point P 2 corresponds to a rhombus configuration such that m 1 , m 2 and m 3 are at the vertices of an equilateral triangle.
On the isosceles trapezoid, configurations (a, b, c) are such that r 23 = r 14 , r 24 = r 13 and D = 0. If r 23 = r 14 and r 24 = r 13 , then the Dziobeck equation D = 0 becomes  [15], using a different parametrization, proved the existence of a unique oneparameter family of isosceles trapezoid central configurations which is characterized by a differentiable function of one of the parameters in the parameter space. Moreover they prove that the endpoints of this family are the square configuration and the configuration consisting of an equilateral triangle with the masses m 3 = m 4 = 0 at one of the vertices. In our parametrization the differentiable function of the parameter is the function b(c) defined implicitly by f (b, c) = 0 and the endpoints of the curve are the points P 1 and P 3 . Thus C 3 is the last curve in the boundary of Ω and it is a curve joining P 1 and P 3 , see Figure 3.
Unfortunately we are not able to prove that Ω is a differentiable function over the two positions of the masses, as was stablish in the co-circular case, see [15]. Nevertheless, in the next section we prove that there exist a one-parameter family of trapezoid central configurations that divides Ω in two disjoint regions, namely the region that contains the trapezoid central configurations and the one that contains the obtuse trapezoid central configurations.
Theorem 9. The curve D = 0 is a graph with respect to the variable a in the region Ω r (see Figure 4). In fact, ∂D/∂a evaluated at the curve D = 0 restricted to Ω r is negative.
which has a unique real solution with a ≥ 1, the solution a = 1. After straightforward computations we see that substituting a = a 2 (b) into D we get a function of b that is zero at b = 1 and b = 1/ √ 3 and positive for b ∈ (1/ √ 3, 1). In a similar way substituting a = a 1 (b) into D we get a function of b that is zero at b = 0 and b = 1/ √ 3 and negative for b ∈ (1/ √ 3, 1]. Therefore each curve in Ω r joining a point of the curve a 1 (b) with a point of the curve a 2 (b) has at least a point with D = 0. Therefore there exist al least one set in Ω r satisfying D = 0. Next we see that this set is a graph in the variable a that joins the points (1, 1) and (2/ √ 3, 1/ √ 3), see Figure 4(a). Rearranging the terms in a convenient way ∂D/∂a can be written in terms of the mutual distances as ∂D ∂a Next, we will see that at the points of Ω r satisfying D = 0 the following conditions hold: f 1 + f 4 < 0, f 2 + f 5 < 0 and f 3 + f 6 0. Therefore ∂D/∂a evaluated at the curve D = 0 restricted to Ω r is negative. Then f 1 + f 4 can be written as where g 1 = −r 34 r 23 r 3 13 − r 3 12 − r 12 (r 12 − r 23 ) r 12 r 23 (r 12 + r 23 ) + r 3 13 . Since r 13 > r 12 r 23 on Ω r and g 1 < 0 on Ω r , it follows that f 1 +f 4 < 0 evaluated at the curve D = 0 restricted to Ω r .
Finally, using that D = 0 expression f 3 + f 6 can be written as where g 3 = −r 24 r 12 (r 3 23 − r 3 34 ) + r 23 (r 12 − r 34 )(r 3 24 − r 3 34 ). There exists a curve in Ω r such that g 3 = 0, so the previous arguments are not valid to prove that f 3 + f 6 evaluated at the curve D = 0 restricted to Ω r is negative.
In order to avoid the last obstacle, by using resultants we will prove that there are no values (a, b) in the interior of Ω r for which D and g 3 are zero simultaneously.
Let Res[P, Q, x] denote the resultant of the polynomials P (x, y) and Q(x, y) with respect to x. The resultant Res[P, Q, x] is a polynomial in the variable y satisfying the following property: if (x, y) = (x * , y * ) is a solution of system P (x, y) = 0, Q(x, y) = 0 then y = y * is a zero of Res[P, Q, x]. In other words, the set of zeroes of Res[P, Q, x] contains the components y of all solutions of the system P (x, y) = 0 and Q(x, y) = 0. We observe that it may contain additional solutions that are not related with the solutions of the system P (x, y) = 0 and Q(x, y) = 0. Here T 1 = 16a 2 b 2 T 1 and T 2 = b 4 T 2 , were T 1 and T 2 are polynomials of total degree 64 and 16, respectively, in the variables a and b. Note that by the properties of resultants the set of solutions of the new system of equations T 1 = 0, T 2 = 0 contains all solutions with a, b = 0 of system (18), or equivalently all the solutions with a, b = 0 of system D = 0, g 3 = 0 (thinking D and g 3 as a function of a, b via the mutual distances r ij ). Now we solve system T 1 = 0, T 2 = 0 by using resultants again. We compute Res[ T 1 , T 2 , a] and we get the polynomial Next we compute Res[ T 1 , T 2 , b] and we get the polynomial (20) w(a) = (a − 1) 24 a 64 (a 2 + 1) 40 (a 2 − a + 1) 4 (a 2 + a + 1)  In short, the set of realizable right trapezoid central configurations is and it is plotted in Figure 4. In Figure 5 we plot the masses along the right trapezoid family parameterized by the parameter a. We note that the limit case (a, b) = (1, 1) correspond to the square with equal masses, and the limit case (a, b) = (2/ such that the masses m 1 , m 2 , m 3 form an equilateral triangle with edge length 2/ √ 3.

Trapezoid CC with a couple of equal masses
In this section we will study the trapezoid CC with a pair of equal masses. In [15] Cors and Roberts shown that for a given order of the mutual distances in any co-circular central configuration the set of masses is completely ordered. A similar result for the trapezoid central configurations has been obtained recently by Santoprete [40]. Although in that case the masses are not totally ordered. With our particular choice of labeling, from [40] any trapezoid central configuration satisfies (22) m 4 ≤ m 3 ≤ m 1 = 1 m 4 ≤ m 2 .  (14), (15) and (16) with m 1 = 1.
Moreover, also from Santoprete [40], we know that if m 3 = m 1 = 1 or m 2 = m 4 , then the central configuration is a rhombus and the remaining two masses have to be equal. And if m 3 = m 4 , then the central configuration is an isosceles trapezoid and again the two remaining masses are necessarily equal. Figure 6 shows the full set of masses for which a trapezoid central configuration exist.
From the previous results only two cases of trapezoid central configuration with only a pair of equal masses remains unknown, namely, m 2 = m 3 and m 2 = m 1 = 1. In the next two subsections we are going to show the existence of these two classes of trapezoid central configuration. Something remarkable is that we will proved analytically the existence of non-symmetric trapezoid central configurations with two equal masses. As far as we know this result was known numerically, but we think that is the first time that this result is proved analytically in the four-body problem. Now we study the value of masses along the boundary of Ω.
The rhombus family. By substituting the points of C 2 into (14), (16) and (17) we get m 1 = m 3 = 1 and Note that on the rhombus family expression (15) is not well defined and we should take expression (17) instead of it. We can see that µ r is an increasing function defined for all c ∈ [−1/ √ 3, 0], such that µ r (−1/ √ 3) = 0 and µ r (0) = 1. The plot of the masses on C 2 is given in Figure 7 (b).
The isosceles trapezoid family. On the isosceles trapezoid family we know that m 1 = m 2 = 1 and m 3 = m 4 = µ (see for instance [15]), but since we do not have an explicit expression of the solutions of f (b, c) = 0, we cannot give the explicit expression of µ as a function of the parameter c. Studying numerically the function µ we see that it is a decreasing function in c ∈ (0, 1/ √ 3) such that µ → 1 when c → 0 and µ → 0 when c → 1/ √ 3. The plot of the masses on C 3 is given in Figure 7 (c).
Note that if we approach to P 2 over the set C 2 then m 3 → 1, whereas if we approach to P 2 over the set C 1 then m 3 → 1/2. Thus the limit of m 3 as we approach to P 2 depends on the path you take and m 3 has a non removable discontinuity at P 2 .
7.1. m 2 = m 3 . Take f = m 2 − m 3 . We are interested in the solutions of f = 0. On C 2 (corresponding to the rhombus family) we have r 12 = r 23 = r 34 = r 14 and m 1 = m 3 = 1.
Numerically we show that for any fixed path connecting C 2 and C 3 in Ω the solution of f = 0 is unique. Curve representing the zeros of f in Ω goes from P 3 to (2/ √ 3, 1/ √ 3, −0.351839354 . . . ) ∈ C 1 (see Figure  8). To verify the last statement, we observe that the function f on C 1 becomes f =  We apply Sturm Theorem to conclude that f = 0, as a polynomial of degree 24, has a unique real solution in c ∈ (−1/ √ 3, 1/ √ 3), namely c = −0.351839354 . . . 7.2. m 2 = m 1 = 1. It is clear that m 2 = m 1 = 1 along the isosceles trapezoid central configuration family, that is, on the boundary C 3 . Moreover m 1 = m 2 = m 3 = m 4 = 1 on P 3 , that is, on the square configuration, and m 2 − m 1 < 0 on C 2 ∪ P 2 .

Conclusions
Using the positions of the masses we have classified the set of trapezoid central configurations. This set is a two-dimensional surface whose boundaries are known families consisting in a rhombus, an isosceles trapezoid and an equilateral triangle with a zero mass off the triangle. Although a specific ordering of the masses has not hold for any trapezoid central configuration, we can split the two-dimensional surface in three disjoint regions where the set of masses is totally ordered. Somewhat we must remark that we have proved analytically the existence of non-symmetric trapezoid central configurations with a pair of equal masses.
There exist a one-parameter family of right trapezoid central configurations that also splits the two-dimensional surface in two disjoint regions, namely the acute and the obtuse regions. Along such a nonsymmetric family the masses are completely ordered, that is, the family belong to one of the previous three regions, concretely the middle one, where the set of masses is totally ordered. Moreover, when the pair of equal masses belong to biggest parallel side, only acute trapezoid central configurations are allow. On the other hand, when the two equal masses belongs to the non-parallel side, only obtuse trapezoid central configurations are allow.