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Re: [EMHL] a Neuberg conjecture

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  • Bernard Gibert
    Dear friends, ... after rereading that, I feel it is not very clear : I just wanted to say that the 3 conics have 3 points in common (one of them being M) iff
    Message 1 of 16 , Oct 31, 2002
      Dear friends,
      >
      > I conjecture that the three conics form a mesh iff M is a point of the
      > Neuberg cubic. If so, they must have two other (real or not) points N1 & N2
      > in common also on the Neuberg cubic.

      after rereading that, I feel it is not very clear :

      I just wanted to say that the 3 conics have 3 points in common (one of them
      being M) iff M is a point of the Neuberg cubic and then the two other points
      N1 & N2 also lie on it.

      No mesh in it !!!

      Best regards

      Bernard
    • jpehrmfr
      Dear Bernard ... the ... As you wrote in another message, your conjecture is that the three conics have two other common points (real or not) apart M iff M
      Message 2 of 16 , Nov 3, 2002
        Dear Bernard

        > GaGbGc is the antimedial triangle,
        > A',B',C' are the reflections of A / BC, etc.
        >
        > for any point M,
        > Ca is the conic through B, C, A', Ga, M,
        > Cb, Cc defined likewise.
        >
        > I conjecture that the three conics form a mesh iff M is a point of
        the
        > Neuberg cubic. Can somebody help ?

        As you wrote in another message, your conjecture is that the three
        conics have two other common points (real or not) apart M iff M lies
        on the Neuberg cubic.
        This is true but I didn't find an easy proof.
        Here is a summary :
        Suppose that M is barycentric (p : q : r); consider the triangle
        UaUbUc inscribed in the antimedial triangle such as (algebrically)
        GaUc/GbUc = p (2.SA.r + c^2.q)/q/(2.SB.r + c^2.p); Ua,Ub cyclic
        Now, using combination of their equations, we get that the union of
        the lines CM and UaUb is a degenerated conic of the pencil generated
        by Ca and Cb.
        Hence, the three conics will have two other common points (apart M)
        iff Ua, Ub, Uc lie on a same line, ie (Menelaus)
        ((2.SA.r + c^2.q)/(2.SB.r + c^2.p))*cyclic*cyclic = 1, which means
        exactly that M lies on the Neuberg cubic.
        Friendly. Jean-Pierre
      • Barry Wolk
        ... three ... M lies ... triangle ... (algebrically) ... cyclic ... union of ... generated ... (apart M) ... means ... I verified the conjecture by a different
        Message 3 of 16 , Nov 3, 2002
          --- jpehrmfr <jean-pierre.ehrmann@...> wrote:
          > Dear Bernard
          >
          > > GaGbGc is the antimedial triangle,
          > > A',B',C' are the reflections of A / BC, etc.
          > >
          > > for any point M,
          > > Ca is the conic through B, C, A', Ga, M,
          > > Cb, Cc defined likewise.
          > >
          > > I conjecture that the three conics form a mesh iff M is a
          > > point of the Neuberg cubic. Can somebody help ?
          >
          > As you wrote in another message, your conjecture is that the
          three
          > conics have two other common points (real or not) apart M iff
          M lies
          > on the Neuberg cubic.
          > This is true but I didn't find an easy proof.
          > Here is a summary :
          > Suppose that M is barycentric (p : q : r); consider the
          triangle
          > UaUbUc inscribed in the antimedial triangle such as
          (algebrically)
          > GaUc/GbUc = p (2.SA.r + c^2.q)/q/(2.SB.r + c^2.p); Ua,Ub
          cyclic
          > Now, using combination of their equations, we get that the
          union of
          > the lines CM and UaUb is a degenerated conic of the pencil
          generated
          > by Ca and Cb.
          > Hence, the three conics will have two other common points
          (apart M)
          > iff Ua, Ub, Uc lie on a same line, ie (Menelaus)
          > ((2.SA.r + c^2.q)/(2.SB.r + c^2.p))*cyclic*cyclic = 1, which
          means
          > exactly that M lies on the Neuberg cubic.
          > Friendly. Jean-Pierre

          I verified the conjecture by a different method. With M as
          above, and P=(x:y:z), the equation for Ca can be written as
          z/x = (k1 + k2 t)/(k3 + k4 t)
          where t=y/x, for some ki which are polynomials in a,b,c,p,q,r.
          Plugging this into the equation for Cb, we get a cubic
          polynomial Pb(t), with a known root t=q/p. Treating Cc similarly
          gives Pc(t), and we want the two poly's Pb and Pc to have the
          same roots. Comparing their coefficients leads quickly to the
          Neuberg cubic.

          The two additional points lying on the Neuberg cubic will come
          from the other two roots of Pb (or of Pc). They have irrational
          coordinates. but they lie on a line with rational coordinates.
          --
          Barry Wolk


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        • Sergei Markelov
          Dear Jean-Pierre, Nikos, Paul and all Hyacinthos! In the October of 2002 we discussed the problem in subject. ... I have found this Euler s 200 years old paper
          Message 4 of 16 , Jun 5 10:44 AM
            Dear Jean-Pierre, Nikos, Paul and all Hyacinthos!

            In the October of 2002 we discussed the problem in subject.

            In message 5999 Jean-Pierre Ehrmann asked:

            > the former message from Sergei suggests me the following problem :
            > for reasons of continuity, there exist two perpendicular lines D, D'
            > dividing ABC in 4 parts of equal area. How can we construct these
            > lines? (I don't know the answer) I presume that this problem is very
            > old one.


            In message 6000 Paul Yiu answered:

            > Euler has written 2 long papers on this problem:
            >
            > [729] Dilucidationes super problemate geometrico de quadrisectione,
            > M\'emoires de l'acad\'emie des sciences de St. P\'etersbourg,
            > {1803/6}{1809}{26--48} I26, 286--303.
            >
            > [730] Solutio completa problematis de quadrisectione trianguli
            > per duas rectas inter se normales, ibid. {1803/6}{1809}{49--87}
            > I26, 304--333.
            >
            > These papers are in volume 26 of his opera omnia.


            I have found this Euler's 200 years old paper in Russian State Library,
            and scanned it so that those of you who are not happy to have this book
            in local library will still be able to read Euler's original text.
            Article is in Latin (not in Russian).

            http://www.mccme.ru/~markelov/29142-3.pdf

            Size is 4698445 bytes (48 pages). Sorry for slow downloading from
            Russia.

            Interested in this problem should also have a look at the message
            number 6009 by Nikolaos Dergiades:

            > this problem is Leibniz's problem and leads to an equation of eighth
            > degree. F.G-M 1624a. If a point D is on the side AB then we can
            > construct the point E on AC such that 2(ADE) = (ABC) To construct
            > now a line MP that divides the triangle ADE and the quadrilateral
            > DBCE in two equal parts is Huygens' problem (1650) F.G.-M 1674c.
            > The above construction needs the common tangent MP to two
            > hyperbolae but since they have DE as common asymptote this
            > construction leads to a second degree equation and is ruler and
            > compass possible. Now my ancestors would move the point D on AB until
            > the lines DE, MP be perpendicular.


            Another related posting was my message number 5997 with solution to the
            easier problem (constructing line through given point bisecting the
            area of the given triangle).

            Best regards,
            Sergei Markelov
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