This is a preprint of: “Tangential trapezoid central configurations”, Pengfei Yuan, Jaume Llibre, Regul. Chaotic Dyn., vol. 25(6), 651–661, 2020. DOI: [10.1134/S156035472006009X] TANGENTIAL TRAPEZOID CENTRAL CONFIGURATIONS JAUME LLIBRE1 AND PENGFEI YUAN2 Abstract. A tangential trapezoid, also called a circumscribed trapezoid, is a trapezoid whose four sides are all tangent to a circle within the trapezoid: the in-circle or inscribed circle. In this paper we classify all planar four-body central configurations, where the four bodies are at the vertices of a tangential trapezoid. 1. Introduction and statement of the results The classical n-body problem concerns the study of the dynamics of n particles interacting among themselves by their mutual attraction according to Newtonian gravity. Let xi ∈ Rd (i = 1, . . . , n) denotes the position vector of the i-body, mi ∈ R+ (i = 1, . . . , n) denotes the mass of the i-body. Rd is the Euclidean space (d = 2 or 3). By Newton’s law of motion and Newton’s gravitational law the equations of the motion of the n-body problem are governed by n X mj (xi − xj ) ẍi = − 3 , 1 ≤ i ≤ n. rij j=1,j6=i Where rij = |xi − xj | is the mutual Euclidean distance between the i-body and the j-body. Here we take the gravitational constant G = 1. The vector x = (x1 , . . . , xn ) ∈ (Rd )n is called the configuration of the system. Define δ(x), the dimension of a configuration x, i.e. the dimension of the smallest affine space of Rd containing all of the points xi . Configurations with δ(x) = 1, 2, 3 are called collinear, planar, spacial, respectively. When n = 2 the n-body problem has been completely solved. However for the n-body problem for n ≥ 3 the complete solution remains open. Let m1 x1 + · · · + mn xn M = m1 + · · · + mn , c= , M be the total mass and the center of masses of the n bodies, respectively. A configuration x is called a central configuration if the acceleration vectors of the n bodies are proportional to their positions with respect to the center of masses 2010 Mathematics Subject Classification. 70F07,70F15. Key words and phrases. Convex central configuration, four-body problem, tangential trapezoid. 1 2 J. LLIBRE AND P. YUAN with the same constant of proportionality, i.e. n X mj (xj − xi ) 3 = λ(xi − c), 1 ≤ i ≤ n, (1) rij j=1,j6=i where λ is the constant of proportionality. A central configuration is invariant under isometries and homotheties with re- spect to the center of masses. So we say that two central configurations x = (x1 , . . . , xn ), x̄ = (x̄1 , . . . , x̄n ) are equivalent if we can pass from one to the other through a dilation or a rotation with respect to the center of the mass. This relation is of equivalence. Therefore when studying central configurations, we only count the classes of central configurations defined by the above equivalence relation. Central configurations play an important role in celestial mechanics. First, the knowledge of central configurations allows us to obtain homographic solutions of the n-body problem (see [35]). We recall that a homographic solution of the n-body problem is a solution such that the configuration remains constant up to rotation and scaling. Second, there is a relation between central configurations and the bifurcations of the hypersurfaces of constant energy h and angular momentum c (see [37, 46]). Third, if the n bodies are going to a simultaneous collision or to a total parabolic escape to infinity, then the configuration of n bodies is asymptotic to a central configuration(see [20, 25, 43, 47]). There is an extensive literature on the study of central configurations, see Euler [21], Lagrange [28], Albouy [1, 2], Albouy and Chenciner [3], Albouy and Fu [4], Albouy and Kaloshin [6], Davis et al. [18], Hampton and Moeckel [26], Moeckel [36], Llibre [22, 23, 30], Long [33], Érdi and Czirják [20], Moulton [38], Palmore [39], Schmidt [44], Smale [46], Xia [48, 49], Xie [50], . . . In this paper we are interested in the planar 4- body problem. Simó [45] studied numerically the class of central configurations for the 4-body problem with arbitrary masses . In 2006 the finiteness of central configurations for the 4-body problem has been proved by Hampton and Moeckel [26] with the assistance of a computer. Later on Albouy and Kalsoshin [6] proved this result without using the computer. For m1 = m2 = m3 = m4 Llibre found all the planar central configurations by studying the intersections of two curves and assuming that the central config- urations have a straight line of symmetry, see [30]. Later on Albouy proved the existence of such symmetry and provide a more analytical proof for the central configurations with equal masses. When one of the 4 masses is sufficiently small, Pedersen [40], Barros and Leandro [12, 13] found the classes of central configurations of the 4 body, also see Arenstorf [11], Fernandes et al [23] and Gannaway [24]. The central configurations with three equal masses was studied by Bernat at al. They classified the non-collinear kite central configurations. For more details, see [14], also see [29]. In 2010 Piña and Lonngi [42] found new properties of the symmetric and non- symmetric central configurations for the 4-body problem. 3 Figure 1. A tangential trapezoid. MacMillan and Bartky [34] proved that there is a unique isosceles trapezoid central configuration for the 4-body problem when two pairs of equal masses are located at the adjacent vertices of a trapezoid. Long and Sun [33] studied the con- vex central configurations with three equal masses and they proved that the central configurations must posses a symmetry. Pérez-Chavela and Santoprete [41] gener- alized further and proved that central configurations must posses such symmetry when two equal masses are located at the opposite vertices of a quadrilateral and at most only one of the remaining masses is larger than the equal masses. Later on Albouy [5] obtain the symmetry of convex central configuration with two equal masses at the opposite vertices. For m1 = m2 6= m3 = m4 Álvarez and Llibre [7] characterized the convex and concave central configurations with an axis of symmetry. Using these previous results on the symmetries Corbera and Llibre [15] gave a complete description of the families of central configurations with two pairs of equal masses and two equal masses sufficiently small. Cors and Robert [16] studied the case when 4 masses are located at the vertices of a cyclic quadrilateral, see also [10]. Recently Álvarez and Llibre [8, 9] classified the Hjelmslev and the equilic quadri- lateral central configurations. The trapezoid central configurations have been studied in [16], here we want to see which of these trapezoid central configurations are tangential. A tangential trapezoid, also called a circumscribed trapezoid, is a trapezoid whose four sides are all tangent to a circle within the trapezoid: the in-circle or inscribed circle. Without loss of generality we take m1 = 1 and assume that the positions of four bodies at the vertices of a trapezoid are x1 = (0, 0), x2 = (1, 0), x3 = (c, a), x4 = (b, a), (2) 4 J. LLIBRE AND P. YUAN Σ3 Σ1 Σ2 Figure 2. Tangential trapezoid central configurations Σ1 , Σ2 , Σ3 . where a > 0. Lemma 1. If the configuration of 4 masses is a tangential trapezoid with the ver- tices x1 , x2 , x3 and x4 , as it is shown in Figure 1 , then √ (1 − b)(a2 + 2(b + a2 + b2 )) c= . (3) 4 − a2 − 4b Lemma 1 is proved in section 3. We characterize all planar 4-body problem central configurations, where the four bodies are at the vertices of a tangential trapezoid. Theorem 2. We take positive masses for the 4-body problem with m1 = 1. Then we have a tangential trapezoid central configuration given by (2) for each value of (a, b) in the arc Σ1 , Σ2 , Σ3 , see Figure 2. (a) The arc Σ1 and Σ2 is symmetric with √ respect to the a-axis.√ The arc Σ1 goes from the point (1, 0) to the point ( 2 , 2 ), it is open at ( 23 , 12 ) and closed 3 1 at (1, 0). On Σ1 we have m1 = m3 = 1, m2 = m4 > 1, and m2 = m4 increases √ from 1 to ∞. The√ arc Σ2 goes from the point (1, 0) to the point ( 23 , − 12 ), it is open at ( 23 , − 12 ) and closed at (1, 0). On Σ2 we have m1 = m3 = 1, m2 = m4 < 1, and m2 = m4 decreases from 1 to 0. (b) The value (a, b) = (1, 0) corresponds to the tangential trapezoid given by the square with four equal masses at its vertices. √ (c) The arc Σ3 goes from the point (1, 1) to the point ( 23 , 11 10 ), it is closed at √ 3 11 (1, 1) and open at ( 2 , 10 ). On Σ3 we have m2 > 1 > m3 > m4 , and m2 increases from 1 to 1.24789.., m3 decreases from 1 to 0.29184.., m4 decreases from 1 to 0. (d) The value (a, b) = (1, 1) ∈ Σ3 corresponds to the tangential trapezoid given by the square with four equal masses at its vertices. 5 Theorem 2 is proved in section 3. 2. Dziobek’s equations In this section we present the equations of the central configurations provided by Dziobek for the 4-body problem. Let x = (x1 , x2 , x3 , x4 ) ∈ (R2 )4 . We associated with x the matrix:   1 1 1 1 X = x1 x2 x3 x4  . 0 0 0 0 Xk denotes the matrix obtained deleting from the matrix X its k-th column and its last row. Then let Dk = (−1)k+1 det (Xk ) for k = 1, . . . , 4. The equations for the central configurations (1) of the 4-body problem were written by Dziobek [17] (see also equations (8) and (16) of Moeckel [35] or [24]) as the following 12 equations with 12 unknowns. Di Dj 1 (4) ei,j = c1 + c2 − 3 = 0, mi mj rij ti − tj = 0, for 1 ≤ i < j ≤ 4, with 4 X 2 ti = Dj rij . j=1,j6=i The unknowns of equation (4) are the mutual distances rij , the variables Di , and the constants ck (k = 1, 2). The first six Dziobek’s equation (4) are −3 −3 m1 m2 (r12 − c1 ) = c2 D1 D2 , m3 m4 (r34 − c1 ) = c2 D3 D4 , −3 −3 (5) m1 m3 (r13 − c1 ) = c2 D1 D3 , m2 m4 (r24 − c1 ) = c2 D2 D4 , −3 −3 m1 m4 (r14 − c1 ) = c2 D1 D4 , m2 m3 (r23 − c1 ) = c2 D2 D3 , where 1 c b b c 0 D1 = 0 a a = a(c − b), D2 = a a 0 = a(b − c), 1 1 1 1 1 1 0 1 b c 1 0 D3 = 0 0 a = a, D4 = a 0 0 = −a. 1 1 1 1 1 1 Multiplying equations (5) in order that each one at the right has the expres- sion c22 D1 D2 D3 D4 , and since the masses must be positive we obtain the Dziobek relation: −3 −3 −3 −3 −3 −3 (r12 − c1 )(r34 − c1 ) = (r13 − c1 )(r24 − c1 ) = (r14 − c1 )(r23 − c1 ). (6) The relation holds for every planar central configuration of the 4-body problem. 6 J. LLIBRE AND P. YUAN We can solve c1 from the Dziobek relation and we have −3 −3 −3 −3 r12 r34 − r13 r24 c1 = −3 −3 −3 −3 r12 + r34 − r13 − r24 −3 −3 −3 −3 r13 r24 − r14 r23 (7) = −3 −3 −3 −3 r13 + r24 − r14 − r23 −3 −3 −3 −3 r14 r23 − r12 r34 = −3 −3 −3 −3 . r14 + r23 − r12 − r34 Defining −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 (7) becomes p1 − p2 p2 − p3 p3 − p1 c1 = = = , (8) s1 − s2 s2 − s3 s3 − s1 which imply that the point (s1 , p1 ), (s2 , p2 ), (s3 , p3 ) are on the same line with slope c1 , i.e. 1 1 1 s1 s2 s3 = 0. p1 p2 p3 So we can write Dziobek equation as the following 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. (9) The equation D = 0 must be satisfied for every planar central configuration of the 4-body problem. 3. Proofs of Lemma 1 and Theorem 2 For proving Lemma 1 we shall use Pitot’s Theorem which states that in a tan- gential quadrilateral the two sums of lengths of opposite sides are the same, for a proof see for instance [27]. Proof of Lemma 1. By Pitot’s and Pythagoras Theorems and Figure 1 we get that p p 1+c−b= a2 + b2 + (1 − c)2 + a2 . Isolating c from this equality we obtain the conclusion of the lemma.  7 Figure 3. The graphic of Dziobek= 0 for a = 1. Figure 4. The graphic of Dziobek= 0 for a = 0.95. Proof of Theorem 2. From (2) we have √ p 1, r14 = a2 + b2 , r24 = a2 + (1 − b)2 , r12 = s √ (1 − b)2 (a2 + 2(b + a2 + b2 ))2 r13 = a2 + , (a2 + 4b − 4)2 s √ (1) 2 (a2 (b − 2) + 2(b − 1)(b − 2 + a2 + b2 ))2 r23 = a + , (a2 + 4b − 4)2 s √ (a2 − (b − 1)( a2 + b2 − b))2 r34 = . (a2 + 4b − 4)2 Substituting these expressions together with the values of Dk for k = 1, 2, 3, 4 into the last six equations of (4), we found that they are identically zero for a tangential trapezoid configuration. 8 J. LLIBRE AND P. YUAN Figure 5. The graphic of Dziobek= 0 for a = 0.9. Dividing conveniently two different equations of (5) we obtain −3 m2 (r23 − c1 ) D2 −3 = , m1 (r13 − c1 ) D1 −3 m3 (r23 − c1 ) D3 (11) −3 = , m1 (r12 − c1 ) D1 −3 m4 (r24 − c1 ) D4 −3 = . m1 (r12 − c1 ) D1 Since m1 = 1, using (7) and (11), we have 3 3 3 3 D2 r23 r24 (r13 − r14 ) m2 = 3 r 3 (r 3 − r 3 ) , D1 r13 14 23 24 3 3 3 3 D3 r23 r34 (r12 − r14 ) (12) m3 = 3 3 3 3 , D1 r12 r14 (r23 − r34 ) 3 3 3 3 D4 r24 r34 (r12 − r13 ) m4 = 3 3 3 3 . D1 r12 r13 (r24 − r34 ) Substituting these masses into the first six Dziobek equations (4), and taking only the numerators of these six equations because the denominators do not vanish, we have e1,2 =D2 (c2 D13 r12 3 3 2 3 r13 r14 (r24 3 − r23 3 3 ) − r23 3 r24 (c1 r12 3 − 1)(r13 3 − r14 )), e1,3 =D3 (c2 D12 r12 3 3 3 3 r13 r14 (r34 3 − r23 3 3 ) − r23 3 r34 (c1 r13 3 − 1)(r12 3 − r14 )), 3 3 3 3 3 3 3 3 e2,3 =D2 D3 (r23 r24 r34 (c1 r23 − 1)(r12 r14 )(r14 − r13 )− c2 D12 r12 3 3 6 3 r13 r14 (r23 3 − r24 3 )(r23 3 − r34 )), (13) e1,4 =D4 (c2 D12 r12 3 3 3 3 r13 r14 (r34 3 ) − r24 3 3 − r24 3 r34 (c1 r14 3 − 1)(r12 3 − r13 )), 3 3 3 3 3 3 3 3 e2,4 =D2 D4 (−r23 r24 r34 (c1 r24 − 1)(r12 − r13 )(r13 − r14 )− c2 D12 r12 3 6 3 3 r13 r14 (r23 3 − r24 3 )(r24 3 − r34 )), e3,4 =D3 D4 (c2 D12 r12 6 3 3 3 r13 r14 (r23 3 − r34 3 )(r34 3 − r24 )− 3 3 3 3 3 3 3 3 r23 r24 r34 (c1 r34 − 1)(r12 − r13 )(r12 − r14 )). 9 Notice that Di (i = 1, 2, 3, 4) is non-zero, so we can eliminate Di from equations (13). First, we solve the first two equations with respect to c1 and c2 , and then we substituted c1 and c2 in the last four equations of (13). We obtain D 6 3 3 3 3 3 3 e2,3 = r r r (r − r12 )(r14 − r13 ), d 23 24 34 14 e1,4 = 0, D 3 6 3 3 3 3 3 (14) e2,4 = r23 r24 r34 (r13 − r12 )(r13 − r14 ), d D 3 3 6 3 3 3 3 e3,4 = r23 r24 r34 (r13 − r12 )(r12 − r14 ), d where D = 0 is the Dziobek equation (9), and 3 3 3 3 3 3 3 3 3 3 3 3 3 3 d = r12 (r13 r23 (r24 − r34 ) + r14 r24 (r34 − r23 )) + r13 r14 r34 (r23 − r24 ), is the denominator which comes from the denominator of c1 and c2 . In conclusion the tangential trapezoid central configurations must satisfy (e2,3 , e2,4 , e3,4 ) = (0, 0, 0). Substituting (10) into e2,3 = 0, e2,4 = 0, e3,4 = 0, we get that these three equations are satisfied if and only if the following equations E2,3 =DW1 W3 = 0, (15) E2,4 =DW2 W3 = 0, E3,4 =DW1 W2 = 0, have solutions respectively, where p W1 =1 − a2 + b2 , s √ 2 (b − 1)2 (a2 + 2(b + a2 + b2 ))2 W2 =1 − a + , (a2 + 4b − 4)2 s √ p (b − 1)2 (a2 + 2(b + a2 + b2 ))2 2 2 W3 = a + b − a + 2 , (a2 + 4b − 4)2 and D = 0 is the Dziobek equation (9). The solutions of {E2,3 , E2,4 , E3,4 } = 0 and of Dziobek =0 having positive masses are the tangential trapezoid central configurations of the four-body problem. First we solve equations {E2,3 , E2,4 , E3,4 } = 0. Using Mathematica, we obtain that √ 3 1 a= , b = − , m2 = 1.5124659.., m3 = 1, m4 = 1.5124659... 2 2 Next we find solutions of Dziobek =0. In order to solve Dziobek =0, we use Math- ematica. We plot the graphic of Dziobek =0 when a = 1, 0.95, 0.9, see Figures 3, 4, and 5. When a = 1 we solve Dziobek = 0 and get b = 0, and from (12) we obtain m1 = m2 = m3 = m4 = 1. 10 J. LLIBRE AND P. YUAN When a = 0.95 we solve Dziobek = 0 and get the three solutions b = −0.3122499.., b = 0.3122499.., b = 1.02234... Then from (12) we have m2 =0.49701.., m3 =1, m4 =0.49701.., for(a = 0.95, b = −0.31224..), m2 =2.01201.., m3 =1, m4 =2.01201.., for (a = 0.95, b = 0.31224..), m2 =1.09839.., m3 =0.7557.., m4 =0.63245.., for(a = 0.95, b = 1.02234..). When a = 0.9 we solve Dziobek = 0 and get the three solutions b = −0.4358899.., b = 0.4358899.., b = 1.05179... Therefore from (12) we have m2 =0.20834.., m3 =1, m4 =0.20834.., for(a = 0.9, b = −0.43588..), m2 =4.79976.., m3 =1, m4 =4.79976.., for(a = 0.9, b = 0.43588..), m2 =1.14928.., m3 =0.51298.., m4 =0.27025.., for(a = 0.9, b = 1.05179..). √ When a goes from 1 to 23 , the solutions for Dziobek = 0 has two symmetric solutions in the interval (− 12 , 12 ) and one solution in the interval (1, 11 10 ) with positive masses, and we have 1 m1 = m3 = 1, m2 = m4 > 1 for 0≤b< , 2 1 m1 = m3 = 1, m2 = m4 < 1 for − < b ≤ 0, 2 11 m2 > m1 = 1 > m3 > m4 for 1≤b< . 10 √ 3 1 We find that m2 = m4 as (a, b) → ( 2 , 2) , and m2 = m4 → 0 as (a, b) → √ ( 23 , − 12 ). √ 3 In summary studying the graphic of D = 0 with a varying from 1 to 2 , we obtain the statements of Theorem 2.  Acknowledgements We thank to the reviewer his/her detailed report which help us to improve the presentation of our results. 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