This is a preprint of: “Trapezoid central configurations”, Montserrat Corbera, Josep Maria Cors, Jaume Llibre, Ernesto Pérez-Chavela, Appl. Math. Comput., vol. 346, 127–142, 2019. DOI: [10.1016/j.amc.2018.10.066] TRAPEZOID CENTRAL CONFIGURATIONS M. CORBERA, J.M. CORS, J. LLIBRE, AND E. PÉREZ-CHAVELA Abstract. We classify all planar four–body central configura- tions 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 po- sition of the bodies, in some cases an specific order of the masses determine the geometry of the configuration, namely acute or ob- tuse trapezoid central configuration. We also prove the existence on non–symmetric trapezoid central configuration with two pairs of equal masses. 1. Introduction Central configurations are particular positions of the masses in the Newtonian n–body problem, where the position and acceleration vec- tors with respect to the center of masses are proportional, with the same constant of proportionality for all masses. They play an impor- tant 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 n X qk − qj mk q̈k = − G mk mj , |qk − qj |3 j=1 j 6= k k = 1, . . . , n, being qk ∈ R2 the position vector of the punctual mass mk in an inertial coordinate system, and G is the gravitational constant 2010 Mathematics Subject Classification. 70F15,70F10,37N05. Key words and phrases. 4-body problem, convex central configurations, trape- zoidal central configurartions. 1 http://www.gsd.uab.cat 2 M. CORBERA, J.M. CORS, J. LLIBRE, AND E. PÉREZ-CHAVELA 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 = {(q1 , . . . , qn ) ∈ R2n : qk 6= qj , for k 6= j}. A configuration of the n bodies (q1 , . . . , qn ) ∈ E is central if there is a positive constant λ such that (1) q̈k = −λ (qk − c) , for k = 1, . . . , n, being c the position Pnvector of the Pn center of mass of the system, which is defined by c = k=1 mk qk / k=1 mk . Two planar central configurations are equivalent if there is a homoth- ecy of R2 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 prob- lem 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 config- urations 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 configura- tions of the four–body problem in some particular cases. For instance, a complete numerical study for the number of classes of central con- figurations 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 . TRAPEZOID CENTRAL CONFIGURATIONS 3 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], Gan- naway [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 symme- try passing through two non–adjacent bodies. The kite non–collinear classes of central configurations having some symmetry for the four– body 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 cen- tral configurations of the four–body problem when the masses satisfy that m1 = m2 6= m3 = m4 was done by Álvarez 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. 4 M. CORBERA, J.M. CORS, J. LLIBRE, AND E. PÉREZ-CHAVELA In [29] Long and Sun shown that any convex central configuration with masses m1 = m2 < m3 = m4 located at the opposite vertices of a quadrilateral and such that the diagonal corresponding to the mass m1 is not shorter than the one corresponding to the mass m3 , 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 para- graph Corbera and Llibre [14] gave a complete description of the fam- ilies 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 re- cently, 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 par- allel 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 mea- sure. • A 3–sides equal trapezoid is an isosceles trapezoid with three sides of the same length. • A parallelogram is an obtuse trapezoid with two pairs of parallel sides. TRAPEZOID CENTRAL CONFIGURATIONS 5 Obtuse trapezoid Right trapezoid Acute trapezoid Parallelogram Rectangle Isosceles trapezoid Rhombus Square Isosceles trapezoid with 3 equal sides Figure 1. Classification of Trapezoids • 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 config- urations. 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 configu- rations. 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. 2. Preliminaries The central configurations (in what follows simply CC by short) can be described in terms of Lagrange multipliers. We denote by q = (q1 , q2 , q3 , q4 ) ∈ (R2 )4 the position of four positive masses m1 , m2 , m3 , m4 on the plane and by rij = ||qi − qj || the mutual distances between the 6 M. CORBERA, J.M. CORS, J. LLIBRE, AND E. PÉREZ-CHAVELA 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) (2) ∇U (q) + λ ∇I(q) = 0, where U is the Newtonian potential Xn mi mj (3) U (q) = , 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 ∆1 + ∆2 + ∆3 + ∆4 = 0. TRAPEZOID CENTRAL CONFIGURATIONS 7 In 1900 Dziobeck [17] proved for planar CC that ∂V 2 = −32 ∆i ∆j . ∂rij From this equality we obtain 2 ∂V ∂V d(rij ) = 2 · = −64rij ∆i ∆j . ∂rij ∂rij drij Fixing the moment of inertia I = I0 and applying Lagrange multi- pliers, we have that the planar non-collinear CC are the critical points of the function (5) U + λM (I − I0 ) − σV. Taking the partial derivatives and using the six mutual distances as variables we obtain −3 −3 m1 m2 (r12 − λ) = 64σ∆1 ∆2 , m3 m4 (r34 − λ) = 64σ∆3 ∆4 , −3 −3 (6) m1 m3 (r13 − λ) = 64σ∆1 ∆3 , m2 m4 (r24 − λ) = 64σ∆2 ∆4 , −3 −3 m1 m4 (r14 − λ) = 64σ∆1 ∆4 , m2 m3 (r23 − λ) = 64σ∆2 ∆3 . 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 −3 −3 −3 −3 −3 −3 (7) (r12 − λ)(r34 − λ) = (r13 − λ)(r24 − λ) = (r14 − λ)(r23 − λ), which must be satisfied for any planar 4-body CC. Solving each of the three pairs of equations for λ we obtain −3 −3 −3 −3 r12 r34 − r13 r24 λ = −3 −3 −3 −3 r12 + r34 − r13 − r24 −3 −3 −3 −3 r r − r14 r23 (8) = −3 13 24−3 −3 −3 r13 + r24 − r14 − r23 −3 −3 −3 −3 r r − r12 r34 = −3 14 23−3 −3 −3 . r14 + r23 − r12 − r34 If we set −3 −3 −3 −3 s1 = r12 + r34 , p1 = r12 r34 , −3 −3 −3 −3 s2 = r13 + r24 , p2 = r13 r24 , −3 −3 −3 −3 s3 = r14 + r23 , p3 = r14 r23 , then equation (8) can be written as p1 − p2 p2 − p3 p3 − p1 (9) λ = = = , s1 − s2 s2 − s3 s3 − s1 8 M. CORBERA, J.M. CORS, J. LLIBRE, AND E. PÉREZ-CHAVELA m2 tH HH HHt H m3 t m4  m1 = 1 t    Figure 2. Central configurations with two parallel lines which means that (s1 , p1 ), (s2 , p2 ), (s3 , p3 ), viewed as points in R+2 , must lie on the same line with slope λ. This in turn, is equivalent to 1 1 1 s1 s2 s3 = 0, p1 p2 p3 a representation that allows to write Dziobeck equation (7) as the nice factorization 3 3 3 3 3 3 3 3 3 3 3 3 D = (r13 − r12 )(r23 − r34 )(r24 − r14 ) − (r12 − r14 )(r24 − r34 )(r13 − r23 )=0 The Dziobeck equation D = 0 must be satisfied for the six mutual distances of every four-body planar central configuration. 3. Equations of trapezoidal central configurations We consider four positive masses m1 , m2 , m3 , m4 located at the ver- tices of a trapezoid, i.e. located by pairs on two parallel lines, which without loss of generality we assume are vertical. Since the central con- figurations 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 m1 = 1 located at the bottom part of the left line, m2 above m1 on the same line and m3 above m4 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 TRAPEZOID CENTRAL CONFIGURATIONS 9 quadrilateral are longer that any of the four sides, that is (10) r13 , r24 > r12 , r14 , r23 , r34 ; 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 r23 is the biggest side and that we exclude the case where all the sides are equal, that is, the square. So, its opposite side, r14 , has to be the smaller one, and r13 , r24 > r23 ≥ r12 , r34 ≥ r14 > 1. Then depending on the relative position of the four masses we have the following four scenarios: p p 2 2 (a) r34 = r12 + pr23 − 1 + pr14 − 1, 2 2 (b) r34 = r12 + pr23 − 1 − pr14 − 1, 2 2 (c) r34 = r12 − pr23 − 1 + pr14 − 1, 2 2 (d) r34 = r12 − r23 − 1 − r14 − 1. Notice that the cases where m3 and m4 are either both above m2 or both below m1 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 r23 is the biggest side or r14 the smaller one. p p 2 2 In the scenario (a), r34 > r12 + r23 − 1 > 1 + r23 − 1 > r23 . For the scenarios (b) and (c) we shall use the following result: Let x, y ≥ 1 be two real numbers, √ p x − x2 − 1 + y 2 − 1 = y if and only if x = y. √ p Moreover if y >px then x − x2 − 1 + y 2 − 1 > y, and if y < x then √ x − x2 − 1 + y 2 − 1 < y. p p 2 2 In (b) p r 34 > p r23 − 1 + r14 − r14 − 1 > r23 , and in (c) r34 < 2 2 r23 − r23 − 1 + r14 − 1 < r14 . p p 2 2 Finally, in (d) r34 < r23 − r23 − 1 + r14 − 1 < r14 from scenario (c). 10 M. CORBERA, J.M. CORS, J. LLIBRE, AND E. PÉREZ-CHAVELA Similar argument works if r14 > 1 is considered the biggest side.  Without loss of generality we label the bodies so that r12 is the longest side. We can also assume that r23 ≥ r14 by an appropriate relabeling. Indeed, equations (6) are invariant if we interchange bodies m1 and m2 and bodies m3 and m4 . The choice r23 ≥ r14 , together with the fact that r12 is the longest side, implies the relation r24 ≥ r13 between the two diagonals. Summarizing, we have proved the following result. Lemma 2. Labeling conveniently the bodies, the mutual distances that can provide trapezoid central configurations can be restricted to the fol- lowing set e = {r ∈ R+ 6 : r24 ≥ r13 > r12 ≥ r23 ≥ r14 ≥ r34 }. Ω Next we give the expression of the masses ratios for the trapezoid e Taking into account the sign of the areas central configurations on Ω. Ai , we have ∆1 = −r34 /2, ∆2 = r34 /2, ∆3 = −r12 /2, ∆4 = r12 /2 (note that we have considered the bodies ordered clockwise). Now from (6) we obtain the following ratios of the masses −3 −3 m2 (λ − r13 ) (λ − r14 ) (11) = −3 = −3 , m1 (r23 − λ) (r24 − λ) −3 −3 m3 (r12 − λ) r12 (r14 − λ) r12 (12) = −3 = −3 , m1 (r23 − λ) r34 (r34 − λ) r34 −3 −3 m4 (r12 − λ) r12 (λ − r13 ) r12 (13) = −3 = −3 . m1 (λ − r24 ) r34 (r34 − λ) r34 We observe that the fact that all masses must be positive places addi- tional constraints on the mutual distances. Using λ = (p2 −p3 )/(s2 −s3 ) and m1 = 1 into the first equation in (11) and after some simplifications we obtain 3 3 3 3 r23 r24 (r13 − r14 ) (14) m2 = 3 3 3 3 . r13 r14 (r24 − r23 ) Doing similar substitutions in (12) and (13) we get 3 2 3 3 r23 r34 (r12 − r14 ) (15) m3 = 2 3 3 3 , r12 r14 (r23 − r34 ) TRAPEZOID CENTRAL CONFIGURATIONS 11 and 3 2 3 3 r24 r34 (r13 − r12 ) (16) m4 = 2 3 3 3 , r12 r13 (r24 − r34 ) respectively. The masses of equations (14), (15) and (16) are positive and well- e except when r12 = r14 and r23 = r34 simultaneously. In defined on Ω, that case, we use λ = (p1 − p2 )/(s1 − s2 ) into equation (12) getting 5 3 3 3 3 r34 (r14 − r24 )(r14 − r13 ) (17) m3 = 5 3 3 3 3 , r14 (r13 − r34 )(r24 − r34 ) e which also is positive and well-defined on Ω. In summary we have proved the next result. Lemma 3. Let e 0 = {r ∈ R+ 6 : r ∈ Ω Ω e and D = 0}. Any point in Ωe 0 defines a four–body trapezoid central configuration with positive masses. Moreover, up to relabeling and rescaling the set Ω e0 contains all trapezoid central configurations. 4. 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 cen- tral 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. Proposition 4. In the planar four–body problem there are no parallelo- gram shape central configurations with positive masses at their vertices, excluding squares and rhombus. Proof. In a parallelogram configuration r12 = r34 and r23 = r14 . From Lemma 3, this parallelogram could be realizable as a central configura- tion if r12 = r23 = r14 = r34 , that is, if it is a rhombus or a square.  The next result is an immediate consequence of Proposition 4. Corollary 5. In the planar four–body problem there are no rectangle shape central configurations with positive masses at their vertices. 12 M. CORBERA, J.M. CORS, J. LLIBRE, AND E. PÉREZ-CHAVELA Proposition 6. In the planar four–body problem there are no 3–sides equal trapezoid shape central configurations with positive masses at their vertices, excluding the square. Proof. A 3–sides equal trapezoid is in particular an isosceles trapezoid, so the length of its diagonals are equal. Assume that r12 is the longest exterior side, that the equal sides are r23 = r14 = r34 = α < r12 and that the diagonals are r24 = r13 = β. Then from de Dziobeck equation D = 0 we get 2 3  β 3 − α3 3 α − r12 = 0. So either r12 = r23 = r14 = r34 = α and r24 = r13 = β which corre- sponds to a square, or β = α which is not possible because it implies r24 = r13 = r12 = r23 = r14 = r34 = α (see Lemma 3). Proceeding in a similar way when the equal sides are r12 = r23 = r14 = α > r34 the Dziobeck equation becomes 2 3  β 3 − α3 3 α − r34 = 0. When r34 = α we get again the square and when β = α we get condi- tion r24 = r13 = r12 = r23 = r14 > r34 . This condition can be satisfied only when the positions of m3 and m4 coincide and the resulting con- figuration is an equilateral triangle. Substituting the above relation into (15) and (16) we get m3 = m4 = 0.  5. The set of realizable trapezoid central configurations In this section we characterize the set of realizable trapezoid central configurations. Proposition 7. The boundaries of Ω e 0 (see Lemma 3) consist of a square, a curve corresponding to the rhombus, a curve containing the isosceles trapezoids and a curve corresponding to degenerate central configurations with m4 = 0. Proof. The possible boundaries of Ω e 0 are the sets where either r24 = r13 , r13 = r12 , r12 = r23 , r23 = r14 , or r14 = r34 . Next we characterize these boundaries. Case A: r24 = r13 . The trapezoids having equal diagonals are the rectangle, the square and the isosceles trapezoid. The rectangle is not a realizable central configuration. TRAPEZOID CENTRAL CONFIGURATIONS 13 Case B: r13 = r12 . After substituting r13 = r12 into equation D = 0 we get the following three subcacases. Note that the configurations coming from this condition are central configurations of the restricted problem; i.e. with one or more masses equal to zero. B.1: r12 = r14 . This implies r12 = r23 = r14 , so the masses m1 , m2 , m3 and m4 are located at the vertices of an equilateral triangle with r34 = 0. B.2: r12 = r23 . In this case r13 = r12 = r23 , this means that the masses m1 , m2 , and m3 are at the vertices of an equilateral triangle. B.3: r24 = r34 . This implies r24 = r13 = r12 = r23 = r14 = r24 which is not possible. Case C: r12 = r23 . After substituting r12 = r23 into equation D = 0 we get the following subcacases. C.1: r13 = r23 . Corresponds to case B.2. C.2: r23 = r24 . This implies r24 = r13 = r12 = r23 , so it corresponds also to case B.2. C.3: r14 = r34 . In this case r12 = r23 and r14 = r34 , so the configura- tion is a kite. Since the configuration must be also a trapezoid it is necessarily a rhombus. Case D: r23 = r14 . The trapezoids having two equal sides are the isosceles trapezoid, the rhombus, the square, the parallelogram and the rectangle. The last two do not correspond to realizable central configurations (see Proposition 6). Case E: r14 = r34 . After substituting r14 = r34 into equation D = 0 we get the following subcacases. E.1: r12 = r23 . Corresponds to case C.3. E.2: r13 = r34 . This implies r13 = r12 = r23 = r14 = r34 , so the configuration is a rhombus. E.3: r24 = r34 . Corresponds to case B.3.  Next we give the shape of the set of realizable trapezoid central configurations. To simplify the computations we parametrize the set of realizable central configurations by using the positions of the masses: q1 = (0, 0), q2 = (0, a), q3 = (1, b), q4 = (1, c), with a ≥ 1 and b > c, and substituting the corresponding mutual distances into the Dziobeck 14 M. CORBERA, J.M. CORS, J. LLIBRE, AND E. PÉREZ-CHAVELA equation D = 0. This equation gives a relation between a, b, c which provides an implicit 2–dimensional surface in R3 . Theorem 8. Let q1 = (0, 0), q2 = (0, a), q3 = (1, b), q4 = (1, c), with a ≥ 1 and b > c, be the positions of the masses m1 , m2 , m3 and m4 respectively, then the set Ω of realizable CC is {(a, b, c) ∈ R3 : a ≥ 1, b > c, D = 0, r24 ≥ r13 > r12 ≥ r23 ≥ r14 ≥ r34 }, and the boundary of Ω is C1 ∪ C2 ∪ C3 ∪ P1 ∪ P2 ∪ P3 where √ √ √ √ C1 = {(2/ 3, 1/ 3, c) : c ∈ (−1/ 3, 1/ 3)}, √ √  √ C2 = { 1 + c2 , c + 1 + c2 , c : c ∈ (−1/ 3, 0)}, √ C3 = {(a, b, c) : a = b + c, f (b, c) = 0, c ∈ (0, 1/ 3)}, √ √ √ P1 = (2/ 3, 1/ 3, 1/ 3), √ √ √ P2 = (2/ 3, 1/ 3, −1/ 3), P3 = (1, 1, 0), and   f (b, c) = (b + c)3 − (c2 + 1)3/2 (b2 + 1)3/2 − (b − c)3 −   (b2 + 1)3/2 − (b + c)3 (c2 + 1)3/2 − (b − c)3 . See Figure 3 for the plot of the set Ω. The points of C1 provide configurations of the restricted problem with m1 , m2 and m3 in an equilateral triangle; the points of C2 and C3 provide rhombus and isosceles trapezoid central configurations, respec- tively; P1 corresponds to an equilateral triangle configuration of the re- stricted problem with a collision of m3 and m4 at one of the vertices; P2 corresponds to a configuration of the restricted problem with the masses at the vertices of a rhombus and such that the positions of m1 , m2 and m3 are the vertices of an equilateral triangle; and P3 corresponds to the square central configuration. p √ Proof. Easily we can compute r = 1 + (a − c)2, r = 1 + b2 , p 24 13 √ r12 = a, r23 = 1 + (a − b)2 , r14 = 1 + c2 , and r34 = b − c. Propo- sition 7 gives the characterization of the central configurations on the boundaries of Ω. We prove the result by using this characterization and the parametrization (a, b, c). On the boundary with the masses m1 , m2 and m3 at the vertices of an equilateral triangle we have r13 = r12√ = r23 . Solving √ the system of equations r13 = r12 = r23 , we get a = 2/ 3 and b = 1/ 3. Substituting TRAPEZOID CENTRAL CONFIGURATIONS 15 this solution into r23 and √ r14 and imposing √ the condition r23 ≥ r14 , we get the condition −1/ √3 ≤ √ c ≤ 1/ 3. It is easy√to check√ that the solution (a, b, c) = (2/ 3, 1/ 3, c) with c ∈ [−1/ 3, 1/ 3] satisfies r24 ≥ r13 ≥ r12 ≥ r23 ≥ r14 ≥ r34 . So the set C1 ∪ P1 ∪ P2 belongs to the boundary of Ω. Moreover it is easy to check that the point P1 correspond to an equilateral triangle configuration with the masses m3 and m4 colliding at the corresponding vertex of the triangle; and the point P2 corresponds to a rhombus configuration such that m1 , m2 and m3 are at the vertices of an equilateral triangle. On the rhombus configurations we have r12 = r23 = r14 = r24 . Solv- ing the system of equations √ r12 =√r23 = r14 = r24 we get the solu- tion s(c) = (a, b) = ( 1 + c2 , c + 1 + c2 ). Imposing that √ this solu- tion satisfies r24 ≥ r13 > r12 we get the condition −1/ 3 < c ≤ 0. So C2 √belongs to √ the √ boundary of Ω. Moreover s(0) = (1, 1) and s(−1/ 3) = (2/ 3, 1/ 3). So the endpoints of C2 are P2 and P3 . On the isosceles trapezoid, configurations (a, b, c) are such that r23 = r14 , r24 = r13 and D = 0. If r23 = r14 and r24 = r13 , then the Dziobeck equation D = 0 becomes 3 3  3 3  3 3  3 3  3 3  r14 − r13 r12 − r14 r13 − r34 − r13 − r12 r14 − r34 = 0. If r14 = r13 , then r24 = r13 = r12 = r23 = r14 . This corresponds to the point P1 (see the proof of Proposition 6). Assume now that r14 6= r13 . Solving √ system r23 = r14 , r24 = √r13 we get a = b + c. So r24 = r13 = b + 1, r12 = b + c, r23 = r14 = c2 + 1, and r34 = b − c and condition 2 D = 0 is equivalent to condition f (b, c) = 0. Cors and Roberts in [15], using a different parametrization, proved the existence of a unique one- parameter 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 m3 = m4 = 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 P1 and P3 . Thus C3 is the last curve in the boundary of Ω and it is a curve joining P1 and P3 , 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 16 M. CORBERA, J.M. CORS, J. LLIBRE, AND E. PÉREZ-CHAVELA Figure 3. The surface Ω of the trapezoid central con- figurations in the abc–space. that there exist a one–parameter family of trapezoid central configu- rations 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. 6. The right trapezoid family We suppose again that we are in the hypotheses of Lemma 2; i.e. that r12 > r23 > r14 > r34 . We assume also that the position of the masses m1 = 1, m2 , m3 and m4 are respectively (0, 0), (0, a), (1, b), (1, 0) with a > 1 andpa > b > 0. Easily we can compute the mutual √ distances a, r23 = 1 + (a − b) , r14 = 1, r34 = b, and r13 = 1 + b2 and r12 = √ 2 r24 = 1 + a2 (the diagonals). First we give the set Ωe (see Lemma 2) on the right trapezoid family parameterized by the positions of the masses (a, b). It is obvious that condition r24 > r12 is always satisfied and that condition r34 6 r14 implies b 6 1. On the other hand, it is easy√to see that conditions r13 > r12 and r12 > r23 imply that a < a1 (b) = 1 + b2 and a > a2 (b) = (b2 + 1)/(2b) respectively. From √ here we get condition a1 (b) > a2 (b) e (see Lemma 2) on which is satisfied for b > 1/ 3. In short, the set Ω the right trapezoid family is   b2 + 1 √ 1 Ωr = (a, b, c) ∈ R :3 6 a < 1 + b2 , √ < b 6 1, c = 0 . 2b 3 TRAPEZOID CENTRAL CONFIGURATIONS 17 (a) (b) Figure 4. (a) The region Ωr and the the curve D = 0 for the right trapezoid family in the a − b plane. (b) The the curve D = 0 for the right trapezoid family on the region Ω. √ We can√ see that √ a 1 is a decreasing function in b ∈ (1/ 3, 1] with a1 (1/ 3) √ = 2/ 3 and a1 (1) √ = 1, whereas √ a2 is an increasing √ function in b ∈ (1/ 3, 1] with √ a2 (1/ 3) = 2/ 3 and a2 (1) = 2 (see Figure 4). Therefore a ∈ (1, 2). Theorem 9. The curve D = 0 is a graph with respect to the variable a in the region Ωr (see Figure 4). In fact, ∂D/∂a evaluated at the curve D = 0 restricted to Ωr is negative. Proof. When b = 1 equation D = 0 becomes  √  2 3/2  3/2  1−2 2 a +1 − 1 a3 − a2 − 2a + 2 = 0, which has a unique real solution with a ≥ 1, the solution a = 1. After straightforward computations we see that substituting√a = a2 (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√= a1 (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 Ωr joining a point of the curve a1 (b) with a point of the curve a2 (b) has at least a point with D = 0. Therefore there exist al least one set in Ωr 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). 18 M. CORBERA, J.M. CORS, J. LLIBRE, AND E. PÉREZ-CHAVELA By simple computations we get ∂D ∂D ∂r12 ∂D ∂r23 ∂D ∂r24 = + + , ∂a ∂r12 ∂a ∂r23 ∂a ∂r24 ∂a where ∂r12 = 1, ∂a ∂r23 a−b r12 − r34 = = , ∂a r23 r23 ∂r24 a r12 = = , ∂a r24 r24 and ∂D 2 3 3  3 3  2 3 3  3 3  = −3r12 r13 − r23 r24 − r34 − 3r12 r24 − r14 r23 − r34 , ∂r12 ∂D 2 3 3  3 3  2 3 3  3 3  = 3r23 r13 − r12 r24 − r14 + 3r23 r12 − r14 r24 − r34 , ∂r23 ∂D 2 3 3  3 3  2 3 3  3 3  = 3r24 r13 − r12 r23 − r34 − 3r24 r12 − r14 r13 − r23 . ∂r24 Rearranging the terms in a convenient way ∂D/∂a can be written in terms of the mutual distances as ∂D = 3(f1 + f2 + f3 + f4 + f5 + f6 ) ∂a where 2 3 3  3 3  f1 = −r12 r24 − r14 r23 − r34 , 2 3 3  3 3  f2 = −r12 r13 − r23 r24 − r34 , 3 3  3 3  f3 = r23 (r12 − r34 ) r13 − r12 r24 − r14 , 3 3  3 3  f4 = r23 (r12 − r34 ) r12 − r14 r24 − r34 , 3 3  3 3  f5 = r12 r24 r13 − r12 r23 − r34 , 3 3  3 3  f6 = −r12 r24 r12 − r14 r13 − r23 . Next, we will see that at the points of Ωr satisfying D = 0 the following conditions hold: f1 + f4 < 0, f2 + f5 < 0 and f3 + f6 6 0. Therefore ∂D/∂a evaluated at the curve D = 0 restricted to Ωr is negative. From D = 0 we get 3 3  3 3  (r12 3 3 − r14 3 ) (r13 3 − r23 3 ) (r24 3 − r34 ) r24 − r14 r23 − r34 = 3 3 . r13 − r12 TRAPEZOID CENTRAL CONFIGURATIONS 19 Then f1 + f4 can be written as 3 3 3 3 g1 (r12 − r14 ) (r24 − r34 ) f1 + f4 = 3 3 , r13 − r12 where 3 3  3  g1 = −r34 r23 r13 − r12 − r12 (r12 − r23 ) r12 r23 (r12 + r23 ) + r13 . Since r13 > r12 > r23 on Ωr and g1 < 0 on Ωr , it follows that f1 +f4 < 0 evaluated at the curve D = 0 restricted to Ωr . Proceeding as above, using that D = 0 expression f2 + f5 can be written as 3 3 3 3 g2 (r13 − r12 ) (r23 − r34 ) f2 + f5 = 3 3 , r12 − r14 where 3 2 2 g2 = −r12 (r24 − r12 )(r14 + r12 r24 + r12 r24 ). Since r24 > r12 on Ωr , g2 < 0 on Ωr , we obtain that f2 + f5 < 0 evaluated at the curve D = 0 restricted to Ωr . Finally, using that D = 0 expression f3 + f6 can be written as 3 3 3 3 g3 (r13 − r12 ) (r24 − r14 ) f3 + f6 = 3 3 , r24 − r34 where 3 3 3 3 g3 = −r24 r12 (r23 − r34 ) + r23 (r12 − r34 )(r24 − r34 ). There exists a curve in Ωr such that g3 = 0, so the previous arguments are not valid to prove that f3 + f6 evaluated at the curve D = 0 restricted to Ωr 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 Ωr for which D and g3 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. 20 M. CORBERA, J.M. CORS, J. LLIBRE, AND E. PÉREZ-CHAVELA Let e1 = e1 (a, b, r13 , r23 , r24 ) = 0, (18) e2 = e2 (a, b, r23 , r24 ) = 0, be the system defined by the two equations D = 0 and g3 = 0 after performing the substitutions r12 = a, r14 = 1 and r34 = b. Here we think that the mutual distances r13 , r23 and r24 are the positive solutions of system 2 e3 = e3 (a, b, r13 ) = r13 − (b2 + 1) = 0, 2 e4 = e4 (a, b, r23 ) = r23 − ((a − b)2 + 1) = 0, 2 e5 = e5 (a, b, r24 ) = r24 − (a2 + 1) = 0. Using resultants we eliminate the variables r13 , r23 and r24 from the equations (18) in the following way. We eliminate the variable r13 from e1 by doing the resultant R1 = Res[e1 , e3 , r13 ]. Then we eliminate the variable r23 from R1 and e2 by doing the resul- tants S1 = Res[R1 , e4 , r23 ], S2 = Res[e2 , e4 , r23 ], and the variable r24 from S1 and S2 by doing the resultants T1 = Res[S1 , e5 , r24 ], T2 = Res[S2 , e5 , r24 ]. Here T1 = 16a2 b2 Te1 and T2 = b4 Te2 , were Te1 and Te2 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 Te1 = 0, Te2 = 0 contains all solutions with a, b 6= 0 of system (18), or equivalently all the solutions with a, b 6= 0 of system D = 0, g3 = 0 (thinking D and g3 as a function of a, b via the mutual distances rij ). Now we solve system Te1 = 0, Te2 = 0 by using resultants again. We compute Res[Te1 , Te2 , a] and we get the polynomial 20 4 16 W (b) = (b − 1)16 b96 (b2 + 1) (b2 − b + 1) (b2 + b + 1) 2 4 2 (19) (3b2 − 4) (21b4 + 12b2 + 16) (21b4 + 24b2 + 16) W1 (b) W2 (b) W3 (b) W4 (b), where W1 , W2 , W3 and W4 are polynomials of degrees 162, 202, 210, and 214 respectively. From properties of resultants the set of zeroes of W contains the component b of all the solutions of system Te1 = 0, Te2 = 0. Recall that we are only interested √ in solutions belonging to Ωr , so we only consider zeroes with 1/ 3 < b 6 1. We compute TRAPEZOID CENTRAL CONFIGURATIONS 21 analytically the zeroes of the first eight factors of W and numerically the zeroes of the √ remaining four factors of W and we get the following solutions with 1/ 3 < b 6 1 b = 0.61283303 . . . , b = 0.69216326 . . . , b = 0.71614387 . . . , b = 0.76874157 . . . , b = 0.79099409 . . . , b = 0.82966657 . . . , b = 0.91953907 . . . , b = 1. Next we compute Res[Te1 , Te2 , b] and we get the polynomial 40 4 8 w(a) = (a − 1)24 a64 (a2 + 1) (a2 − a + 1) (a2 + a + 1) 2 4 2 (20) (3a2 − 1) (21a4 + 6a2 + 1) (21a4 + 54a2 + 49) w1 (a) w2 (a) w3 (a) w4 (a), where w1 , w2 , w3 and w4 are polynomials of degrees 162, 202, 210, and 214 respectively. The set of zeroes of w contains the component a of all the solutions of system Te1 = 0, Te2 = 0. Since we are only interested √ in solutions belonging to Ωr , we only consider zeroes with 1 6 a 6 2. As above we compute analytically the zeroes of the first eight factors of w and numerically the zeroes of the remaining√ four factors of w and we get the following solutions with 1 6 a 6 2 a = 1, a = 1.04304633 . . . , a = 1.07124596 . . . , a = 1.08484650 . . . , a = 1.09217286 . . . , a = 1.10559255 . . . , a = 1.16459040 . . . . We consider D and g3 as a function of (a, b) by substituting the ex- pressions of the mutual distances rij . The possible solutions (a, b) of system D = 0, g3 = 0 √ are the pairs (a, b) = (a∗ , b∗ ) formed by √ a zero a∗ of w with a∗ ∈ [1, 2], and a zero b∗ of W with b∗ ∈ (1/ 3, 1]. Substituting all possible pairs (a, b) = (a∗ , b∗ ) into D = 0, g3 = 0 we see that (a, b) = (1, 1) is the unique pair that provides a solution of the system. This point belongs to the boundary of Ωr . Therefore the function g3 does not change its sign on the solutions of D = 0 that belong to Ωr . √ √ It is easy to the check that the point (a, b) = (2/ 3, 1/ 3), which belongs to the boundary of√ Ωr , satisfies √ D = 0. Moreover the function g3 evaluated at (a, b) = (2/ 3, 1/ 3) is negative. Hence g3 6 0 in Ωr . This end the proof of the theorem.  22 M. CORBERA, J.M. CORS, J. LLIBRE, AND E. 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The plot of the masses m2 (upper line), m3 (medium line) and m4 (bottom line) along the right trapezoid family. In short, the set of realizable right trapezoid central configurations is Ω0r = {(a, b, c) ∈ Ωr : D = 0}, 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/ 3, 1/ 3) corresponds to a right trapezoid central configurations with masses m1 = 1, m4 = 0 and √  √  7 8 3 − 9 49 + 8 7 m2 = = 0.94993335 . . . , (21) 2511 2  √  m3 = 8 3 − 9 = 0.15417163 . . . , 63 such that√ the masses m1 , m2 , m3 form an equilateral triangle with edge length 2/ 3. 7. 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) m4 ≤ m3 ≤ m1 = 1 m4 ≤ m2 . TRAPEZOID CENTRAL CONFIGURATIONS 23 (1,1,1) m3 1.0 Rhombus 0.5 - 0.0  Isosceles trapezoid 1.0 (0,1,0) P2 - m4 0.5 Equilateral triangle - 0.0 0.0 0.5 1.0 m2 (1,0,0) Figure 6. The image of Ω in m2 m3 m4 –space under equations (14), (15) and (16) with m1 = 1. Moreover, also from Santoprete [40], we know that if m3 = m1 = 1 or m2 = m4 , then the central configuration is a rhombus and the remain- ing two masses have to be equal. And if m3 = m4 , 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 con- figuration with only a pair of equal masses remains unknown, namely, m2 = m3 and m2 = m1 = 1. In the next two subsections we are going to show the existence of these two classes of trapezoid central config- uration. 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 equilateral triangle family. By substituting the points of C1 into (14), (15) and (16) we get m4 = 0 and (23) √  √ 2 √  8 3 − 9k13 k23 2 3 − 3c 8 3 − 9k13 m2 = µ2 (c) = , m3 = µ3 (c) = √  , 9k13 (k23 − 8) 27 3c + 1 k13 k22 √ p √ where k1 = c2 + 1 and√k2 = √ 3c2 − 4 3c√+ 7. The function µ2 is defined for all√ c ∈ [−1/ 3, 1/ 3), µ2 (−1/ 3) = 0 and √ µ2 (c) → 1 when c → 1/ 3; moreover it is increasing for c ∈ [−1/ 3, c0 ), it has a maximum at c0 = 0.27448350 . . . with µ2 (c0 ) = 1.0912476 . . . and 24 M. CORBERA, J.M. CORS, J. LLIBRE, AND E. PÉREZ-CHAVELA √ it is decreasing √ √for c ∈ (c√ 0 , 1/ 3). The function µ3 is defined√for all c ∈ (−1/ 3, 1/ 3], µ3 (1/ 3)√= 0, √ µ3 (c) → 1/2 when c → −1/ 3 and it is decreasing for c ∈ (−1/ 3, 1/ 3]. The plot of the masses on C1 is given in Figure 7 (a). The rhombus family. By substituting the points of C2 into (14), (16) and (17) we get m1 = m3 = 1 and   3/2 3/2 ((c − k)2 + 1) ((c + k)2 + 1) − k 3 m2 = m4 = µr (c) = −   , k 3 − ((c − k)2 + 1)3/2 ((c + k)2 + 1)3/2 √ where k = 1 + c2 . 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 µr is√an increasing function defined for all c ∈ [−1/ 3, 0], such that µr (−1/ 3) = 0 and µr (0) = 1. The plot of the masses on C2 is given in Figure 7 (b). The isosceles trapezoid family. On the isosceles trapezoid family we know that m1 = m2 = 1 and m3 = m4 = 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 C3 is given in Figure 7 (c). Note that if we approach to P2 over the set C2 then m3 → 1, whereas if we approach to P2 over the set C1 then m3 → 1/2. Thus the limit of m3 as we approach to P2 depends on the path you take and m3 has a non removable discontinuity at P2 . 7.1. m2 = m3 . Take f = m2 − m3 . We are interested in the solutions of f = 0. On C2 (corresponding to the rhombus family) we have r12 = r23 = r34 = r14 and m1 = m3 = 1. We know from equation (14) that on C2 3 3 3 3 3 3 3 r23 r24 (r13 − r14 ) r24 (r13 − r12 ) (24) m2 = 3 3 3 3 = 3 3 3 . r13 r14 (r24 − r23 ) r13 (r24 − r12 ) Therefore 3 3 3 3 3 3 m2 < 1 ⇐⇒ r24 (r13 − r12 ) < r13 (r24 − r12 ) 3 3 3 3 3 3 ⇐⇒ −r24 r12 < −r13 r12 ⇐⇒ r24 > r13 . TRAPEZOID CENTRAL CONFIGURATIONS 25 1 1 æ æ æ æ 1 æ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ 0.8 æ ææ æ æ æ 0.8 æ æ æ ææ æ æ ææ æ 0.8 æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ 0.6 æ æ ææ æ æ 0.6 ææ æ æ ææ æ æ ææ 0.6 æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ 0.4 æ ææ æ æ 0.4 ææ æ æ ææ æ æ æ 0.4 æ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ 0.2 æ ææ 0.2 æ æ ææ æ æ ææ 0.2 æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ c æ ææ æ c æ ææ æ c æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ ææ æ æ 1 1 1 1 - -0.4 -0.2 0.2 0.4 - -0.4 -0.2 0 0.2 0.4 3 3 3 3 (a) (b) (c) Figure 7. (a) The plot of the masses m2 (continuous line) and m3 (dashed line) on the boundary C1 . (b) The plot of the masses m2 = m4 (continuous line) and m1 = m3 = 1 (dashed line) along the rhombus family. (c) The plot of the masses m1 = m2 = 1 (dashed line) and m3 = m4 (continuous line) along the isosceles trapezoid family. The last inequality follows from the fact that r24 > r13 > 1. Therefore on C2 we have f = m2 − m3 < 0. Now on C3 (corresponding to the isosceles trapezoid family) we have r23 = r14 , r24 = r13 and m1 = m2 = 1. We also know that on this family m3 = m4 . 2 3 3 r34 (r13 − r12 ) m3 = m4 < 1 ⇐⇒ 2 3 3 r12 (r13 − r34 ) 2 3 3 2 3 3 ⇐⇒ r34 (r13 − r12 ) < r12 (r13 − r34 ) 3 2 2 2 2 ⇐⇒ −r13 (r12 − r34 ) < r12 r34 (r12 − r34 ) 3 2 2 ⇐⇒ −r13 (r12 − r34 )(r12 + r34 ) < r12 r34 (r12 − r34 ) 3 2 2 ⇐⇒ −r13 (r12 + r34 ) < r12 r34 . That is, on C3 we have f = m2 − m3 > 0. Therefore for any given path connecting C2 and C3 in Ω, there exist (a, b, c) ∈ Ω such that f = 0 or equivalently m2 = m3 . Numerically we show that for any fixed path connecting C2 and C3 in Ω the solution of f = √ 0 is unique. √ Curve representing the zeros of f in Ω goes from P3 to (2/ 3, 1/ 3, −0.351839354 . . . ) ∈ C1 (see Figure 8). To verify the last statement, we observe that the function f on C1 becomes  2 !   √1 −c √ 2 −4 3c+7 3/2 8 3/2 6 (3c ) √ 3 3 − (c2 + 1) − 3 √ 3/2 − √ 3/2 8−(3c2 −2 3c+1) 8−(3c2 −4 3c+7) f= . (c2 + 1)3/2 26 M. CORBERA, J.M. CORS, J. LLIBRE, AND E. PÉREZ-CHAVELA 1.0 m1 0.8 0.6 m2 = m3 0.4 0.2 m4 1 2 a 3 Figure 8. (a) In black, the curve m2 = m3 on the re- gion Ω. (b) The curves m1 = 1, m2 = m3 , and m4 parameterized by a on the curve in (a). 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. m2 = m1 = 1. It is clear that m2 = m1 = 1 along the isosceles trapezoid central configuration family, that is, on the boundary C3 . Moreover m1 = m2 = m3 = m4 = 1 on P3 , that is, on the square configuration, and m2 − m1 < 0 on C2 ∪ P2 . On the boundary C1 ∪ {P1 , P2 } we have m4 = 0 and m1 , m2 and m3 at the vertices of an equilateral triangle. Let m2 = µ2 (c) be the value of the mass √ m2 on C1 , see (23). We have seen that µ√ 2 (c) → 1 when c → 1/ 3, and that µ2 is decreasing for c ∈ (c0 , 1/ 3) with m2 − m1 > 0 near P1 . On the other hand at c0 = 0.27448350 . .√. , so √ the point Q = (2/ 3, 1/ 3, 0) corresponding to the right trapezoid we have m2 = 0.15417163 · · · < 1, see (21) which implies m2 − m1 < 0. Therefore there exists a point T on C1 such that m2 = m1 . Again applying Sturm Theorem to a polynomial of degree 22 √ in c, we √ can conclude that T is unique, and its coordinates are (2/ 3, 1/ 3, c1 ) where c1 = 0.0517595932 . . . e1 = {(a, b, c) ∈ C1 : c < c1 } and C Let C e0 = {(a, b, c) ∈ C1 : c > c1 }. 1 Therefore, for any given path connecting C e1 and P3 ∪ C3 ∪ P1 ∪ C e0 in 1 Ω, there exist (a, b, c) ∈ Ω such that m2 = m1 = 1. TRAPEZOID CENTRAL CONFIGURATIONS 27 0.20 m3 0.15 0.10 0.05 m4 1.131 2 a 3 (a) (b) Figure 9. (a) In black, the curve m2 = 1 on the region Ω. (b) The curves of m3 and m4 , parameterized by a on the curve in (a). Numerically we show that for any fixed path connecting C e1 and e0 in Ω the solution of m2 − 1 = 0 is unique. Curve repre- P3 ∪ C3 ∪ P1 ∪ C 1 senting the zeros of m2 − 1 joins the boundaries C3 and C1 . This curve goes√ from√(1.13102016 . . . , 0.896392974 . . . , 0.234627188 . . . ) ∈ C3 to (2/ 3, 1/ 3, c1 ) ∈ C1 (see Figure 9 (a)). The values of the masses m3 and m4 along this curve are plotted in Figure 9 (b). 8. Conclusions Using the positions of the masses we have classified the set of trape- zoid 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 trape- zoid central configuration, we can split the two–dimensional surface in three disjoint regions where the set of masses is totally ordered. Some- what 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 con- figurations that also splits the two–dimensional surface in two disjoint regions, namely the acute and the obtuse regions. Along such a non– symmetric family the masses are completely ordered, that is, the family belong to one of the previous three regions, concretely the middle one, 28 M. CORBERA, J.M. CORS, J. LLIBRE, AND E. PÉREZ-CHAVELA where the set of masses is totally ordered. Moreover, when the pair of equal masses belong to biggest parallel side, only acute trapezoid cen- tral 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. Acknowledgements The first three authors are partially supported by a FEDER-MINECO grant MTM2016-77278-P and a MINECO grant MTM2013-40998-P. The second and third authors are also supported by an AGAUR grant number 2014SGR-568. The fourth author is supported by Fondo Mex- icano de Cultura A.C. References [1] Albouy, A., Symétrie des configurations centrales de quatre corps, C. R. Acad. Sci. Paris, 320 (1995), 217–220. 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A, Math. 142 (2012), 665–672. TRAPEZOID CENTRAL CONFIGURATIONS 31 M. Corbera: Departamento de Matemáticas, Universidad de Vic, Spain E-mail address: montserrat.corbera@uvic.cat J.M. Cors: Departamento de Matemáticas, Universidad Politécnica de Catalunya, Spain E-mail address: cors@epsem.upc.edu J. Llibre: Departamento de Matemáticas, Universidad Autónoma de Barcelona, Spain E-mail address: jllibre@mat.uab.cat E. Pérez-Chavela: Departamento de Matemáticas, ITAM, Rı́o Hondo 1, Col. Progreso Tizapán, Ciudad de México 01080, México E-mail address: ernesto.perez@itam.mx