This is a preprint of: “Bicentric quadrilateral central configurations”, Jaume Llibre, Pengfei Yuan,
                         Appl. Math. Comput., vol. 362, 124507: 1–7, 2019.
                         DOI: [10.1016/j.amc.2019.06.021]




                                           BICENTRIC QUADRILATERAL CENTRAL
                                                    CONFIGURATIONS

                                                     JAUME LLIBRE1 AND PENGFEI YUAN2


                                          Abstract. A bicentric quadrilateral is a tangential cyclic quadri-
                                          lateral. In a tangential quadrilateral the four sides are tangents to
                                          an inscribed circle, and in a cyclic quadrilateral the four vertices lie
                                          on a circumscribed circle. In this paper we classify all planar cen-
                                          tral configurations of the 4-body problem, where the four bodies
                                          are at the vertices of a bicentric quadrilateral.




                                          1. Introduction and statement of the results

                                    The well-known Newtonian n-body problem concerns with the mo-
                                 tion of n mass points with positive mass mi moving under their mutual
                                 attraction in Rd in accordance with Newton’s law of gravitation.
                                    The equations of the motion of the n-body problem are :
                                                             Xn
                                                                     mj (xi − xj )
                                                    ẍi = −                        ,      1 ≤ i ≤ n,
                                                            j=1,j6=i
                                                                          rij3

                                 where we have taken the unit of time in such a way that the Newtonian
                                 gravitational constant be one, and xi ∈ Rd (i = 1, . . . , n) denotes the
                                 position vector of the i-body, rij = |xi − xj | is the Euclidean distance
                                 between the i-body and the j-body.
                                    Alternatively the equations of the motion can be written
                                                           mi ẍi = ∇i U(x),       1 ≤ i ≤ n,
                                 where x = (x1 , . . . , xn ), and
                                                                          X       mi mj
                                                              U(x) =                       ,
                                                                        1≤i<j≤n
                                                                                |xi − xj |

                                 is the potential of the system.

                                    2010 Mathematics Subject Classification. 70F07,70F15.
                                    Key words and phrases. Convex central configuration, four-body problem, bi-
                                 centric quadrilateral.
                                                                             1
http://www.gsd.uab.cat
2                          J. LLIBRE AND P. YUAN

  The solutions of the 2-body problem (also called the Kepler problem)
has been completely solved. Unfortunately the solutions for the n-body
(n ≥ 3) is still an open problem.
   For the Newtonian n-body problem the simplest possible motions
are such that the configuration is constant up to rotations and scaling.
Only some special configurations of particles are allowed in such mo-
tions called homographic solutions. Wintner [47] called them central
configurations.
    More precisely, 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 with the same constant of proportionality, i.e.
                Xn
                        mj (xj − xi )
                                      = λ(xi − c),   1 ≤ i ≤ n,        (1)
               j=1,j6=i
                             rij3

where λ is the constant of proportionality.
   Equations (1) are invariant under rotations, dilatations and trans-
lations on the plane. Two central configurations are related if we can
pass from one to the other doing some of the mentioned transforma-
tions. This relation is of equivalence. When we talk about the number
of central configurations we will talk about the number of classes of
equivalence of central configurations.
   Central configurations play an important role in Celestial Mechanics,
for more details see([18, 23, 34, 36, 42, 45, 47].
   There is an extensive literature on the study of central configurations,
see Euler [20], Lagrange [27], Albouy and Chenciner [3], Albouy and Fu
[4], Albouy and Kaloshin [6], Hampton and Moeckel [25], Llibre [30],
Moulton [37] Palmore [38], Schmidt [43], Smale [45], Xia [48, 49], Xie
[50], . . .
  In this paper we are interested in the planar 4-body problem. For the
4-body problem the exact number and classification of central configu-
rations remain open, only some partial results are obtained. There is a
good numerical study on the central configurations of the 4-body pro-
lem, see Simó [44]. The finiteness for the general 4-body problem was
                                                                          3

settled by Hampton and Moeckel using an assisted proof by computer
[25], and Albouy and Kaloshin [6] provided an analytical proof.
  For m1 = m2 = m3 = m4 Llibre found all the planar central configu-
rations assuming the central configurations have an axis of symmetry,
see [29]. Later on Albouy proved the existence of such symmetry and
provide a more analytical proof.
  For the case of three equal masses Bernat at al. classified the non-
collinear kite central configurations, see [13], also see [28].
   A quadrilateral is convex if none of the vertices is located in the
interior of the triangle formed by the other three vertices, otherwise it
is concave.
   For m1 = m2 and m3 = m4 , Long and Sun [9] proved some symmetry
of the central configurations. Perez-Chavela and Santoprete [40] proved
that there is a unique convex non-collinear central configuration of
planar 4-body problem when two equal masses are located at opposite
vertices of a quadrilateral and, at most, only one of the remaining
masses is larger than the equal masses.
  When one of the 4 masses is sufficiently small, Pedersen [39], Barros
and Leandro [11, 12] found the classes of central configurations of the
4-body, see also Gannaway [22] and Arenstorf [10].
  For m1 = m2 6= m3 = m4 Álvarez and Llibre [7] characterized the
convex and concave central configurations with an axis of symmetry.
  Corbera and Llibre [14] gave a complete description of the families of
central configurations with two pairs of equal masses when two equal
masses are sufficiently small.
  Recently Álvarez and Llibre [7] classified Hjelmslev quadrilateral cen-
tral configurations. A Hjelmslev quadrilateral is a quadrilateral with
two right angles at opposite vertices.
   A bicentric quadrilateral is a tangential cyclic quadrilateral. In a
tangential quadrilateral the four sides are tangents to an inscribed cir-
cle, and in a cyclic quadrilateral the four vertices lie on a circumscribed
circle, see Figure 1.
  A kite quadrilateral is a quadrilateral that two pairs of adjacent sides
are of equal length.
  In [15] Cors and Robert studied the case when four masses are located
at the vertices of a cyclic quadrilateral, see also [9].
4                              J. LLIBRE AND P. YUAN

                                m4                  m3




                               m1
                                                    m2

                     Figure 1. A bicentric quadrilateral.

  In this paper we want to classify the bicentric quadrilateral cen-
tral configurations, i.e. we will improve the classification of the cyclic
quadrilateral central configurations provided by Cors and Robert show-
ing which of these central configurations are also tangential quadrilat-
erals. We characterize all central configurations of the 4-body problem
with the four bodies at the vertices of a bicentric quadrilateral.
Theorem 1. Without loss of generality we can take positive masses
for the 4-body problem with m1 = 1 and r12 = 1, then

     (a) we have a bicentric quadrilateral central configurations with pos-
         itive massespsatisfying m1 = m3 = p  1, r12 = r23 = 1, √    r14 =
         r34 , r24 = 1 + r14 2
                               , r13 = 2r14 / 1 + r14 2
                                                        , and r14 ∈ ( 33 , 1],
     (b) the shape of a bicentric quadrilateral central configurations is a
         kite quadrilateral with r14 perpendicular to r12 , and r23 perpen-
         dicular to r34 , see Figure 2.

    Theorem 1 is proved in Section 3.
   We note that from statement (b) of Theorem 1 any bicentric quadri-
lateral central configurations is a kite Hjelmslev quadrilateral central
configuration.


                                2. Preliminaries

    Let x = (x1 , x2 , x3 , x4 ) ∈ (R2 )4 .   We associated with x the matrix:
                                                       
                                        1      ··· 1
                                     
                                X = x1         · · · x4  .
                                        0      ··· 0
                                                                               5
                                         m4




                      m1                                   m3




                                    m2
       Figure 2. The shape of a bicentric quadrilateral central
       configuration is also a kite Hjelmslev quadrilateral.

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.
  Dk is twice the signed area of the triangle whose vertices contain all
bodies except the k-th body, and from the quadrilateral of Figure 1 we
have
       D1 , D3 > 0,    D2 , D4 < 0,             D1 + D2 + D3 + D4 = 0.
The equations for the central configurations (1) of the 4-body problem
were written by Dziobek [16] (see also equations (8) and (16) of Moeckel
[34] or [22]) as the following 12 equations with 12 unknowns.
                             1            Di Dj
                             3
                                = c1 + c2       ,                            (2)
                            rij           mi mj
                                     ti − tj = 0,
for 1 ≤ i < j ≤ 4, where
                                     4
                                     X
                             ti =              Dj rij2 .
                                    j=1,j6=i

The unknowns of equations (3) are the mutual distances rij , the vari-
ables Di , and the constants ck (k = 1, 2).
  The first six Dziobek’s equation (2) are
              −3                                  −3
      m1 m2 (r12 − c 1 ) = c 2 D1 D2 ,    m3 m4 (r34 − c 1 ) = c 2 D3 D4 ,
              −3                                  −3
      m1 m3 (r13 − c 1 ) = c 2 D1 D3 ,    m2 m4 (r24 − c 1 ) = c 2 D2 D4 ,   (3)
              −3                                  −3
      m1 m4 (r14 − c 1 ) = c 2 D1 D4 ,    m2 m3 (r23 − c 1 ) = c 2 D2 D3 .
6                             J. LLIBRE AND P. YUAN

Multiplying the two equations which appear in each of the three rows
of equations (3) we obtain the same expression c22 D1 D2 D3 D4 , which
provide the Dziobek relation:
      −3         −3            −3         −3            −3         −3
    (r12 − c1 )(r34 − c1 ) = (r13 − c1 )(r24 − c1 ) = (r14 − c1 )(r23 − c1 ). (4)
We can solve c1 from the Dziobek relation (4) and we have
                                    −3 −3      −3 −3
                                  r12  r34 − r13  r24
                          c1 =  −3       −3    −3     −3
                               r12 + r34 − r13 − r24
                                  r −3 r −3 − r14
                                               −3 −3
                                                  r23
                             = −3 13 24  −3    −3     −3                     (5)
                               r13 + r24 − r14 − r23
                                 r −3 r −3 − r12
                                              −3 −3
                                                 r34
                             = −3 14 23 −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 ,
equation (5) become
                       p1 − p2    p2 − p3         p3 − p1
                      c1 =      =             =            ,                (6)
                        s1 − s2   s2 − s3         s3 − s1
which imply that the points (s1 , p1 ), (s2 , p2 ), (s3 , p3 ) of the plane are
on the same straight line with slope c1 , i.e.
                                 1 1 1
                                 s1 s2 s3        = 0.
                                 p1 p2 p3
So we can write Dziobek equation as
      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.
                                                                          (7)
Lemma 2. If the four masses are at the vertices of a bicentric quadri-
lateral as in Figure 1, then
                             r23 = 1 − r14 + r34 ,
                                      2
                                     r14 + r13 r24 − r14
                             r34 =                       .                   (8)
                                          1 + r14
Proof. For a bicentric quadrilateral as it is shown in Figure 1 the four
sides are tangents to an inscribed circle, and the four vertices lie on a
circumscribed circle.
                                                                                    7

  According to the Pitot theorem [26], we have
                             r12 + r34 = r23 + r14 ,                              (9)
and by the Ptolemy’s theorem [17], we have
                            r13 r24 = r12 r34 + r23 r14 .                       (10)
From (9) we can get
                             r23 = r12 + r34 − r14 ,                            (11)
then substitute (11) into (10), we obtain
                                    2
                                   r14 + r13 r24 − r14
                           r34 =                       .                        (12)
                                        1 + r14
                                                                                   


                         3. Proof of Theorem 1

   Since we study classes of central configurations, without loss of gen-
erality, we can assume m1 = 1, r12 = 1, and that r12 is the longest side
of the quadrilateral.
   From (3) we can obtain the masses expressed in terms of the distances
rij and the areas Dk , i.e.
                                    3 3       3     3
                                D2 r23 r24 (r13 − r14  )
                          m2 =      3 3       3     3
                                                          ,
                                D1 r13 r14 (r23 − r24 )
                                     3 3           3
                                D3 r23  r34 (1 − r14  )
                           m3 =       3     3     3
                                                        ,                       (13)
                                 D1 r14 (r23 − r34 )
                                     3 3           3
                                D4 r24  r34 (1 − r13  )
                           m4 =       3     3     3
                                                        .
                                 D1 r13 (r24 − r34 )

  Substituting these masses into the first six Dziobek equations (3),
and taking only the numerators of these six equations because the
denominators do not vanish, we have
   e1 = 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 )),
                  2 3 3 3        3      3         3 3        3         3     3
   e2 = D3 (c2 D1 r12 r13 r14 (r34 − r23 ) − r23 r34 (c1 r13 − 1)(r12 − r14 )),
                  3 3 3          3            3 3     3        3
   e3 = D2 D3 (r23  r24 r34 (c1 r23  − 1)(r12  r14 )(r14  − r13  )−
            2 3 3 6         3       3     3       3
        c2 D1 r12 r13 r14 (r23 − r24 )(r23 − r34 )),
   e4 = 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 )),
                                                                                  (14)
8                              J. LLIBRE AND P. YUAN



                           3 3 3           3           3     3      3    3
         e5 = D2 D4 (−r23     r24 r34 (c1 r24  − 1)(r12   − r13 )(r13 − r14 )−
                   2 3 6 3         3       3     3       3
              c2 D1 r12 r13 r14 (r23 − r24 )(r24 − r34 )),
         e6 = D3 D4 (c2 D12 r126 3 3         3
                                  r13 r14 (r23  − r343     3
                                                       )(r34 − r243
                                                                    )−
               3 3 3           3           3       3     3      3
              r23 r24 r34 (c1 r34 − 1)(r12 − r13 )(r12 − r14 )).
We remark that the last six equations of (2) are identically zero when
we substitute the coordinates of the four bodies of Figure 1.
   Notice that Di (i = 1, 2, 3, 4) is non-zero, so we can eliminate the
Di′s which appear as a factor in equation (14). First, we solve the first
two equations with respect to c1 and c2 , and then we substitute c1 and
c2 in the last four equations of (14). We obtain
                       D 6 3 3 3            3     3     3
                  e3 = r23  r24 r34 (r14 − r12 )(r14 − r13 ),
                        d
                  e4 = 0,
                       D 3 6 3 3            3     3     3
                  e5 = r23  r24 r34 (r13 − r12 )(r13 − r14 ),       (15)
                        d
                       D 3 3 6 3            3     3     3
                  e6 = r23  r24 r34 (r13 − r12 )(r12 − r14 ),
                        d
where D = 0 is the Dziobek equation (7), 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 denominators of c1 and c2 .
  From MacMillan and Bartky [33] we know that for every convex
central configuration we have the following inequalities
                            r13 , r24 > r12 , r23 , r34 , r14 .                   (16)

   Since D1 , D3 > 0, D2 , D4 < 0, and r13 > r12 , r13 > r14 , the solutions
of system (15) are satisfied if and only if D = 0.
   Solving the Dziobeck equation (8), after substituting r12 = 1, r23 , r34
from Lemma 1, we get the solution
                                      2r14
                                r13 =      .                           (17)
                                       r24
Then from (8) and (17), for a bicentric quadrilateral central configura-
tion, we have
                                                    2r14
                 r12 = r23 = 1, r14 = r34 , r13 =        .             (18)
                                                     r24
So the bicentric quadrilateral central configurations are also kite quadri-
laterals and from (13) we obtain m1 = m3 .
                                                                                        9
                   r24

                   2

                                                         √
                                                      (1, 2)
                              ( √13 , √23 )

                   1




                                     1
                                                                           r14
                                     2
                                                  1

                                     Figure 3.

  Since the four bodies are co-circular, and a convex quadrilateral is
cyclic if and only if its opposite angles are supplementary [46], we can
deduce r14 and r12 are perpendicular, r23 and r34 are perpendicular.
  Then we have
                                                                       q
                                                 2r14                           2
  r12 = r23 = 1,       r34 = r14 ,       r13 = p      2
                                                               r24 =       1 + r14 . (19)
                                                1 + r14
 In order to find the bicentric quadrilateral central configurations with
positive masses, from (13), (16), (19) and since r12 = 1 is the longest
side of the quadrilateral the following conditions hold
                            1 = r23 < r24 ,
                            r34 , r14 < r13 , r24 ,
                                                      2r14
                            r14 ≤ 1, 1 < r13 =             .
                                                      r24
After some simplification, we obtain
                           r14 ≤ 1 < r24 < 2r14 < 2.                                (20)
Denote Λ = {(r14 , r24 ) | r14 ≤ 1 < r24 < 2r14 < 2}.
   By simple
      p       computation,
                        √     the
                                √ intersection point of r24 = 2r14 with
              2
r24 = 1 + r14   pis (1/   3, 2/  3),
                                 √ and the intersection point of√r14 =
                        2
1 with r24 =      1 + r14√is (1, 2), then we obtain r14 ∈ [1/ 3, 1].
Then for each r14 ∈ (1/ 3, 1) we have a bicentric quadrilateral central
configurations with positive masses satisfying (19).
10                         J. LLIBRE AND P. YUAN
                           √1             m4      √1
                             3                      3



                                                        m3
                      m1


                                 1                1


                                         m2
                                     Figure 4.

  Now we shall study if we have
                            √ or not a bicentric quadrilateral central
configuration when r14 = 1/ 3 and r14 = 1.
                   √
  Case for r14 = 1/ 3. From (19), we obtain
                                 √                        √
    r12 = r23 = 1, r14 = r34 = 1/ 3, r13 = 1, r24 = 2/ 3          (21)

  Next we check the masses for this central configuration. From (13)
and (21), we have that m2 > 0, m3 = 1, m4 = 0, and the central
configuration in this case is the kite of Figure 4 but with m4 = 0.
     Case for r14 = 1. From (19), we obtain
                                                               √
               r12 = r23 = r34 = r14 = 1,        r13 = r24 =       2,   (22)
so the configuration is a square. Thus the bicentric quadrilateral central
configuration satisfying (22) corresponds to the square central config-
uration with m1 = m2 = m3 = m4 = 1.
  In summary we have bicentric quadrilateral central configurations
with positive masses for rij satisfying
                                      q
                                                        2r14
  r12 = r23 = 1, r34 = r14 , r24 = 1 + r+214 , r13 = p           ,
                                                       1 + r+214
                 √
where r14 ∈ (1/ 3 < r14 ≤ 1]. This proves the statement (a) of
Theorem 1.


                           Acknowledgements

  This work is supported by the Ministerio de Economa, Industria
y Competitividad, Agencia Estatal de Investigacin grant MTM2016-
77278-P (FEDER), the Agència de Gestió d’Ajuts Universitaris i de
                                                                                  11

Recerca grant 2017 SGR 1617, and the European project Dynamics-
H2020-MSCA-RISE-2017-777911.
  The second author is partially supported by Fundamental Research
Funds for the Central Universities (NO.XDJK2015C139), China Schol-
arship Council(N0.201708505030).

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  1
   Departament de Matemàtiques, Universitat Autònoma de Barcelona,
08193 Bellaterra, Barcelona, Catalonia, Spain
  E-mail address: jllibre@mat.uab.cat

  2
    School of Mathematics and Statistics, Southwest University, 400715,
Chongqing, China
  E-mail address: yuanpengfei@swu.edu.cn