Loading ...
Sorry, an error occurred while loading the content.

Re: [EMHL] inSteiner foci

Expand Messages
  • Boutte Gilles
    Dear Bernard, Let A be the midpoint between B and C. The locus of a point M such that BC is a bisector of A M and A M (M isogonal conjugate of M) is a
    Message 1 of 16 , Sep 24, 2001
    • 0 Attachment
      Dear Bernard,

      Let A' be the midpoint between B and C. The locus of a point M such
      that BC is a bisector of A'M and A'M' (M' isogonal conjugate of M)
      is a isogonal circular focal cubic, without pivot, with root at infinity on
      BC.

      The foci of the Steiner inellipse are on this cubic K and on the cubic
      reflected
      of K through G.

      Similarly with B' and C'.

      We obtain 6 cubics which are 4 common points : the 4 foci of the
      Steiner inellipse. The 4 foci are on the "second" Brocard cubic.

      Gilles

      Bernard Gibert a *crit :

      > Dear Fred, Jean-Pierre and friends,
      > >
      > > I was just trying a joke : as I knew that Edward, Paul and Bernard
      > > were looking for a nice form of the coordinates of the focii of the
      > > Steiner inellipse, I've sent this little problem because
      > > your "CircumcevianCentroids" are those focii - it's quite funny to
      > > prove that -
      > > I'm sorry for for the time you've lost because of my stupid joke.
      >
      > I don't think Fred lost its time since he confirms the coordinates I had
      > found, although mine are symmetrical but absolutely horrendous.
      > The trilinears are 4.97002 & 0.288722
      >
      > Thanks a lot Fred !
      >
      > (Edward : I've checked those foci are on the "second" Brocard cubic)
      >
      > Bernard
      >
      >
      >
      >
      > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
    • Bernard Gibert
      Dear Gilles and Edward, [GB] ... [cubics denoted by K ,K ] ... That s very good indeed ! Hence the three focals K,K ,K and the second Brocard are in the same
      Message 2 of 16 , Sep 24, 2001
      • 0 Attachment
        Dear Gilles and Edward,

        [GB]
        > Let A' be the midpoint between B and C. The locus of a point M such
        > that BC is a bisector of A'M and A'M' (M' isogonal conjugate of M)
        > is a isogonal circular focal cubic, without pivot, with root at infinity on
        > BC.
        >
        > The foci of the Steiner inellipse are on this cubic K and on the cubic
        > reflected
        > of K through G.
        >
        > Similarly with B' and C'.

        [cubics denoted by K',K"]
        >
        > We obtain 6 cubics which are 4 common points : the 4 foci of the
        > Steiner inellipse. The 4 foci are on the "second" Brocard cubic.
        >
        That's very good indeed !

        Hence the three focals K,K',K" and the second Brocard are in the same pencil
        of focals through A,B,C,the two circular points at infinity and the four
        foci.

        In this pencil there is also a strophoid with node at I, root at X514, focus
        at X106, infinity at X519. It is the locus of foci of inscribed conics whose
        center is on GI.
        The tangents at I are parallel to the asymptotes of the Feuerbach hyperbola
        but I don't think the cubic is its inversive image wrt some circle centered
        at I. It's too late to go on...

        Equation for Edward and §4.3.2 :

        (b-c) x (c^2y^2+b^2z^2) + cyclic + 2(b-c)(c-a)(a-b) xyz = 0

        Similarly, we can define three other strophoids with excenters.

        Best regards

        Bernard
      • jean-pierre.ehrmann@wanadoo.fr
        Dear Hyacinthists, the circumcircle of PBC intersects again the line AB at Ab and the line AC at Ac. A = BAc inter CAb Similarly, define B , C . Find the
        Message 3 of 16 , Sep 26, 2001
        • 0 Attachment
          Dear Hyacinthists,
          the circumcircle of PBC intersects again the line AB at Ab and the
          line AC at Ac.
          A' = BAc inter CAb
          Similarly, define B', C'.
          Find the locus of P such as ABC and A'B'C' are perspective.
          I guess that Bernard will be happy to meet one of his best friends.
          Friendly. Jean-Pierre
        • Bernard Gibert
          Dear Jean-Pierre and Hyacinthists, [JP] ... Good old B2 ! (second Brocard cubic) (b^2 - c^2) x (c^2y^2 + b^2z^2) + cyclic = 0 (barys) It is very nice to denote
          Message 4 of 16 , Sep 27, 2001
          • 0 Attachment
            Dear Jean-Pierre and Hyacinthists,

            [JP]
            > the circumcircle of PBC intersects again the line AB at Ab and the
            > line AC at Ac.
            > A' = BAc inter CAb
            > Similarly, define B', C'.
            > Find the locus of P such as ABC and A'B'C' are perspective.
            > I guess that Bernard will be happy to meet one of his best friends.

            Good old B2 ! (second Brocard cubic)
            (b^2 - c^2) x (c^2y^2 + b^2z^2) + cyclic = 0 (barys)

            It is very nice to denote that the perspector Q is on B2 as well and is what
            I call Psi(P) : if F1 & F2 are the foci of the inSteiner ellipse, Psi is the
            product of the symmetry about the line F1F2 and the inversion wrt the circle
            with diameter F1F2.
            This is a really nice involution with many properties.

            JP : I'm happy to find a geometrical description of B2 and to see that our
            work on orthopivotal cubics reappears.

            Thank you

            Bernard
          • jean-pierre.ehrmann@wanadoo.fr
            Dear Hyacinthists, P* is the isogonal conjugate of P The lines PB and PC intersects again the Kiepert hyperbola respectively at B and C . Find the locus of P
            Message 5 of 16 , Sep 27, 2001
            • 0 Attachment
              Dear Hyacinthists,

              P* is the isogonal conjugate of P
              The lines PB and PC intersects again the Kiepert hyperbola
              respectively at B' and C'.
              Find the locus of P for which the lines AP*, B'C' intersect on the
              Brocard axis OK.
              Friendly. Jean-Pierre

              PS : I apologize if my message is duplicated, but i never received
              the first one
            • Bernard Gibert
              Dear Jean-Pierre and Hyacinthists, ... [JP] ... It s the day of B2 ! Thanks JP. in the style of the same nobody : let A = B C / AP*, B & C likewise what
              Message 6 of 16 , Sep 27, 2001
              • 0 Attachment
                Dear Jean-Pierre and Hyacinthists,
                >
                [JP]
                > P* is the isogonal conjugate of P
                > The lines PB and PC intersects again the Kiepert hyperbola
                > respectively at B' and C'.
                > Find the locus of P for which the lines AP*, B'C' intersect on the
                > Brocard axis OK.

                It's the day of B2 ! Thanks JP.

                in the style of the same nobody :

                let A" = B'C' /\ AP*, B" & C" likewise

                what is the locus of P such that A",B",C" are collinear ?
                ---------------------------------------------------------
                generalization in the style of Antreas :

                replace OK with any line through O and Kiepert with the isogonal conjugate
                of the line.

                what are the 2 loci above ?
                ---------------------------------------------------------

                generalization of the generalization in the style of Antreas :

                replace OK with any line d and Kiepert with the isoconjugate of the line.

                what are the 2 loci above ?

                Best regards

                Bernard
              • Boutte Gilles
                Dear Bernard The 2 loci are the same : an isogonal cubic without pivot, root at the isogonal of the trilinear pole of the lline d. Friendly Gilles
                Message 7 of 16 , Sep 27, 2001
                • 0 Attachment
                  Dear Bernard

                  The 2 loci are the same : an isogonal cubic without pivot, root at the
                  isogonal
                  of the trilinear pole of the lline d.

                  Friendly

                  Gilles


                  Bernard Gibert a *crit :

                  > Dear Jean-Pierre and Hyacinthists,
                  > >
                  > [JP]
                  > > P* is the isogonal conjugate of P
                  > > The lines PB and PC intersects again the Kiepert hyperbola
                  > > respectively at B' and C'.
                  > > Find the locus of P for which the lines AP*, B'C' intersect on the
                  > > Brocard axis OK.
                  >
                  > It's the day of B2 ! Thanks JP.
                  >
                  > in the style of the same nobody :
                  >
                  > let A" = B'C' /\ AP*, B" & C" likewise
                  >
                  > what is the locus of P such that A",B",C" are collinear ?
                  > ---------------------------------------------------------
                  > generalization in the style of Antreas :
                  >
                  > replace OK with any line through O and Kiepert with the isogonal conjugate
                  > of the line.
                  >
                  > what are the 2 loci above ?
                  > ---------------------------------------------------------
                  >
                  > generalization of the generalization in the style of Antreas :
                  >
                  > replace OK with any line d and Kiepert with the isoconjugate of the line.
                  >
                  > what are the 2 loci above ?
                  >
                  > Best regards
                  >
                  > Bernard
                  >
                  >
                  >
                  >
                  > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
                • Bernard Gibert
                  Dear Gilles, ... Good ! I was just attempting to symmetrize JP s idea. Best regards Bernard
                  Message 8 of 16 , Sep 27, 2001
                  • 0 Attachment
                    Dear Gilles,
                    >
                    > The 2 loci are the same : an isogonal cubic without pivot, root at the
                    > isogonal
                    > of the trilinear pole of the lline d.
                    >
                    Good ! I was just attempting to "symmetrize" JP's idea.

                    Best regards

                    Bernard

                    >
                    > Bernard Gibert a *crit :
                    >
                    >> Dear Jean-Pierre and Hyacinthists,
                    >>>
                    >> [JP]
                    >>> P* is the isogonal conjugate of P
                    >>> The lines PB and PC intersects again the Kiepert hyperbola
                    >>> respectively at B' and C'.
                    >>> Find the locus of P for which the lines AP*, B'C' intersect on the
                    >>> Brocard axis OK.
                    >>
                    >> It's the day of B2 ! Thanks JP.
                    >>
                    >> in the style of the same nobody :
                    >>
                    >> let A" = B'C' /\ AP*, B" & C" likewise
                    >>
                    >> what is the locus of P such that A",B",C" are collinear ?
                    >> ---------------------------------------------------------
                    >> generalization in the style of Antreas :
                    >>
                    >> replace OK with any line through O and Kiepert with the isogonal conjugate
                    >> of the line.
                    >>
                    >> what are the 2 loci above ?
                    >> ---------------------------------------------------------
                    >>
                    >> generalization of the generalization in the style of Antreas :
                    >>
                    >> replace OK with any line d and Kiepert with the isoconjugate of the line.
                    >>
                    >> what are the 2 loci above ?
                    >>
                    >> Best regards
                    >>
                    >> Bernard
                    >>
                    >>
                    >>
                    >>
                    >> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
                    >
                    >
                    >
                    >
                    >
                    > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
                    >
                    >
                    >
                  • jean-pierre.ehrmann@wanadoo.fr
                    Dear Bernard, Gilles and other Hyacinthists, As you ve noticed, if we consider a line D, an isoconjugation i and the second common points Pa, Pb, Pc of AP, BP,
                    Message 9 of 16 , Sep 27, 2001
                    • 0 Attachment
                      Dear Bernard, Gilles and other Hyacinthists,

                      As you've noticed, if we consider a line D, an isoconjugation i and
                      the second common points Pa, Pb, Pc of AP, BP, CP with the
                      circumconic i(D), the following properties are equivalent, for P not
                      on D and not on i(D) :
                      1) i(P)A, PbPc, D concur
                      2) i(P)B, PcPa, D concur
                      3) i(P)C, PaPb, D concur
                      4) UPa, VPb, WPc concur(U,V,W = common points of D with the sides of
                      ABC)
                      5) P lies on the i(D)-invariant circumcubic with root the tripole of
                      D and parameter 0 (which means without xyz term or such as the
                      tangent at U, V, W concur)
                      6).... (using duality or anything else)

                      It becomes quite nice with Kjp ( D = line at infinity, i(D) =
                      circumcircle ) : P lies on Kjp if AP* and B'C' are parallel, ...
                      A'B'C' = circumcevian triangle of P.(May be, Bernard, you can write
                      that in the isocubics paper)

                      Starting from this remark, I've immediately generalized by
                      projectivity and then I've applied that to B2 to propose you this
                      locus in nobody's style.

                      In fact, it appears that, among the non-pivotal isocubics,the ones
                      such as the tangents at U,V,W concur have a lot of special
                      properties, which are similar or dual of the properties of pivotal
                      isocubics.

                      So, I propose to use a special term - if it doesn't still exist - for
                      this kind of isocubics : something like "rooted" or root-pivotal
                      isocubics.

                      What do you think of that ?
                      Friendly. Jean-Pierre

                      PS : As my mail is long enough, I didn't repeat the previous ones.
                    • jean-pierre.ehrmann@wanadoo.fr
                      Dear Hyacinthists, I wrote ... not ... of ... of ... The root is not the tripole of D but his isoconjugate (the perspector of the conic) and the concurrent
                      Message 10 of 16 , Sep 27, 2001
                      • 0 Attachment
                        Dear Hyacinthists,
                        I wrote

                        > As you've noticed, if we consider a line D, an isoconjugation i and
                        > the second common points Pa, Pb, Pc of AP, BP, CP with the
                        > circumconic i(D), the following properties are equivalent, for P
                        not
                        > on D and not on i(D) :
                        > 1) i(P)A, PbPc, D concur
                        > 2) i(P)B, PcPa, D concur
                        > 3) i(P)C, PaPb, D concur
                        > 4) UPa, VPb, WPc concur(U,V,W = common points of D with the sides
                        of
                        > ABC)
                        > 5) P lies on the i(D)-invariant circumcubic with root the tripole
                        of
                        > D and parameter 0 (which means without xyz term or such as the
                        > tangent at U, V, W concur)

                        The root is not the tripole of D but his isoconjugate (the perspector
                        of the conic) and the concurrent tangents are the tangent at the
                        common points of the trilinear polar of the root with the sidelines
                        of ABC.

                        Friendly. JP
                      • Bernard Gibert
                        Dear Jean-Pierre and other Hyacinthists, this is very interesting indeed ! I will only add several minor remarks... ... *** The point they concur at is the
                        Message 11 of 16 , Sep 28, 2001
                        • 0 Attachment
                          Dear Jean-Pierre and other Hyacinthists,

                          this is very interesting indeed !
                          I will only add several minor remarks...
                          >
                          > As you've noticed, if we consider a line D, an isoconjugation i and
                          > the second common points Pa, Pb, Pc of AP, BP, CP with the
                          > circumconic i(D), the following properties are equivalent, for P not
                          > on D and not on i(D) :
                          > 1) i(P)A, PbPc, D concur
                          > 2) i(P)B, PcPa, D concur
                          > 3) i(P)C, PaPb, D concur
                          > 4) UPa, VPb, WPc concur(U,V,W = common points of D with the sides of
                          > ABC)

                          *** The point they concur at is the pole of the line PP* in the circumconic
                          i(D).

                          > 5) P lies on the i(D)-invariant circumcubic with root the tripole of
                          > D and parameter 0 (which means without xyz term or such as the
                          > tangent at U, V, W concur)
                          > 6).... (using duality or anything else)
                          >
                          > It becomes quite nice with Kjp ( D = line at infinity, i(D) =
                          > circumcircle ) : P lies on Kjp if AP* and B'C' are parallel, ...
                          > A'B'C' = circumcevian triangle of P.(May be, Bernard, you can write
                          > that in the isocubics paper)

                          *** I will, dear JP ! That's another characterization of your favourite
                          cubic.
                          >
                          > Starting from this remark, I've immediately generalized by
                          > projectivity and then I've applied that to B2 to propose you this
                          > locus in nobody's style.

                          *** I've tried with other known lines/conics/cubics :
                          isogonal examples,
                          D=Euler line, i(D)=Jerabek, the cubic is B3 (third Brocard cubic)
                          D=OI, i(D)=Feuerbach, the cubic is the "Pelletier" cubic.

                          In fact, any line through O corresponds to a rectangular hyperbola and to a
                          circular focal cubic.

                          I'll try to investigate and try to relate our "new" cubics to lines and/or
                          conics.

                          > In fact, it appears that, among the non-pivotal isocubics,the ones
                          > such as the tangents at U,V,W concur have a lot of special
                          > properties, which are similar or dual of the properties of pivotal
                          > isocubics.
                          >
                          > So, I propose to use a special term - if it doesn't still exist - for
                          > this kind of isocubics : something like "rooted" or root-pivotal
                          > isocubics.

                          *** I must admit I'm not thrilled neither with "rooted" (since any
                          nK=non-pivotal isocubic has a root) nor with "root-pivotal" (since it might
                          create confusion with pK= pivotal isocubic). For the paper I propose to
                          denote them by nK_0 and let's try to find something better for the name.
                          >
                          > What do you think of that ?

                          THAT'S A GREAT IDEA ! BRAVO, JP !

                          best regards

                          Bernard
                        • jean-pierre.ehrmann@wanadoo.fr
                          Dear Bernard and other Hyacinthists, we ve seen that if i is an isoconjugation D a line intersecting the sides of ABC at U, V, W i(D) the circumconic
                          Message 12 of 16 , Sep 28, 2001
                          • 0 Attachment
                            Dear Bernard and other Hyacinthists,

                            we've seen that if
                            i is an isoconjugation
                            D a line intersecting the sides of ABC at U, V, W
                            i(D) the circumconic isoconjugate of D
                            if the line AP intersects again i(D) at A',..
                            the following properties are equivalent
                            1)Ai(P), B'C', D concur
                            2)Bi(P), C'A', D concur
                            3)Ci(P), A'B', D concur
                            4)UA', VB', WC' concur
                            5)P lies on the nK_0 i-invriant cubic with root the perspector of i
                            (D) (= isoconjugate of the tripole of D)

                            Now if I, J is the pair of isoconjugate points lying on D, if we use
                            the involution on D with fixed points I, J, we can associate to the
                            nK_0 a pivotal isocubic.

                            In fact the following properties are equivalent :
                            1)(Ai(P) inter D) and (B'C' inter D) are harmonic conjugate wrt I,J
                            2)(Bi(P) ...)
                            3)(Ci(P)...)
                            4)U'A', V'B', W'B' concur where U' is the harmonic conjugate of U wrt
                            I,J
                            5)P lies on the pivotal isocubic invariant by i with pivot the pole
                            of D wrt i(D)

                            I propose to call this pivotal cubic the pivotal sister of the nK_0
                            If the nK_0 is ux(ry^2+qz^2)+.. = 0, then the pivotal sister is
                            (-u^2/p+v^2/q+w^2/r)ux(ry^2-qz^2) + ... = 0
                            For instance, the pivotal sister of Kjp is the Mac Cay cubic, the
                            pivotal sister of B2 is the Napoleon cubic ...
                            Of course, it is a strange family, because a pivotal isocubic is the
                            sister of 3 nK_0.
                            I'm sure that Fred will find a very nice isomorphism between those
                            cubics.

                            Friendly. Jean-Pierre
                          • Boutte Gilles
                            Dear Jean-Pierre and other Hyacinthists, The tripole of D is the common point of the tangents at U,V,W. Regards Gilles
                            Message 13 of 16 , Sep 28, 2001
                            • 0 Attachment
                              Dear Jean-Pierre and other Hyacinthists,

                              The tripole of D is the common point of the tangents at U,V,W.

                              Regards

                              Gilles

                              jean-pierre.ehrmann@... a *crit :

                              > Dear Hyacinthists,
                              > I wrote
                              >
                              > > As you've noticed, if we consider a line D, an isoconjugation i and
                              > > the second common points Pa, Pb, Pc of AP, BP, CP with the
                              > > circumconic i(D), the following properties are equivalent, for P
                              > not
                              > > on D and not on i(D) :
                              > > 1) i(P)A, PbPc, D concur
                              > > 2) i(P)B, PcPa, D concur
                              > > 3) i(P)C, PaPb, D concur
                              > > 4) UPa, VPb, WPc concur(U,V,W = common points of D with the sides
                              > of
                              > > ABC)
                              > > 5) P lies on the i(D)-invariant circumcubic with root the tripole
                              > of
                              > > D and parameter 0 (which means without xyz term or such as the
                              > > tangent at U, V, W concur)
                              >
                              > The root is not the tripole of D but his isoconjugate (the perspector
                              > of the conic) and the concurrent tangents are the tangent at the
                              > common points of the trilinear polar of the root with the sidelines
                              > of ABC.
                              >
                              > Friendly. JP
                              >
                              >
                              >
                              >
                              > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
                            • fred.lang@eivd.ch
                              Dear Jean-Pierre, Bernard, Gilles and all Hyacinthists, The subject of JP is a special case of jacobian: General: If C(1), C(2) and C(3) are three independant
                              Message 14 of 16 , Sep 29, 2001
                              • 0 Attachment
                                Dear Jean-Pierre, Bernard, Gilles and all Hyacinthists,

                                The subject of JP is a special case of jacobian:

                                General:
                                If C(1), C(2) and C(3) are three independant conics (i.e. C3 not in
                                the pencil generated by C1 and C2), they form a net of conics.
                                Let p(i) be the polar of P relative to C(i).
                                The locus of P such that the p(i) concur is a cubic J, known as the
                                jacobian of the net of conics.

                                J is the locus of the double points of the degenerated conics of the
                                net.

                                Special:
                                Take for C(1), C(2) two generators of the pencil of conics which
                                defines the isoconjugation i and put C(3) = i(D), then J is the nK_0
                                i-invariant cubic of JP.

                                Example:
                                C(1) = (I Ia)(Ib Ic), C(2) = (I Ib)(Ia Ic), C(3) = (O), i = isogonal
                                transformation and J = KjP.

                                In the article, I sended to Paul for FG, you'll see that:
                                1) Every non-pivotal isocubic is the jacobian of a net.
                                2) If C(3) is circumscribed to the autopolar triangle of C(1), C(2),
                                then you receive a nK_0 cubic, and in this case the circumcircle of
                                ABC is in the net.
                                Later I plan:
                                3)properties of the special case nK_0
                                4) group structure (difficult, but I have the elliptic invariant of
                                the cubics)

                                Indeed, in Durège you have 100 pages of properties relative to the
                                jacobian...

                                I think it could be a good idea to write something together for FG
                                (JPE + BG + FL +...)

                                Regards.

                                Fred.




                                --- In Hyacinthos@y..., jean-pierre.ehrmann@w... wrote:

                                > we've seen that if
                                > i is an isoconjugation
                                > D a line intersecting the sides of ABC at U, V, W
                                > i(D) the circumconic isoconjugate of D
                                > if the line AP intersects again i(D) at A',..
                                > the following properties are equivalent
                                > 1)Ai(P), B'C', D concur
                                > 2)Bi(P), C'A', D concur
                                > 3)Ci(P), A'B', D concur
                                > 4)UA', VB', WC' concur
                                > 5)P lies on the nK_0 i-invriant cubic with root the perspector of i
                                > (D) (= isoconjugate of the tripole of D)
                                >
                                > Now if I, J is the pair of isoconjugate points lying on D, if we use
                                > the involution on D with fixed points I, J, we can associate to the
                                > nK_0 a pivotal isocubic.
                                >
                                > In fact the following properties are equivalent :
                                > 1)(Ai(P) inter D) and (B'C' inter D) are harmonic conjugate wrt I,J
                                > 2)(Bi(P) ...)
                                > 3)(Ci(P)...)
                                > 4)U'A', V'B', W'B' concur where U' is the harmonic conjugate of U
                                wrt
                                > I,J
                                > 5)P lies on the pivotal isocubic invariant by i with pivot the pole
                                > of D wrt i(D)
                                >
                                > I propose to call this pivotal cubic the pivotal sister of the nK_0
                                > If the nK_0 is ux(ry^2+qz^2)+.. = 0, then the pivotal sister is
                                > (-u^2/p+v^2/q+w^2/r)ux(ry^2-qz^2) + ... = 0
                                > For instance, the pivotal sister of Kjp is the Mac Cay cubic, the
                                > pivotal sister of B2 is the Napoleon cubic ...
                                > Of course, it is a strange family, because a pivotal isocubic is the
                                > sister of 3 nK_0.
                                > I'm sure that Fred will find a very nice isomorphism between those
                                > cubics.
                                >
                                > Friendly. Jean-Pierre
                              • jean-pierre.ehrmann@wanadoo.fr
                                Dear Fred, thank you for your very interesting remarks. They should lead to alot ofnice results. ... the ... nK_0 ... The tangent to i(D) intersect BC, CA, AB
                                Message 15 of 16 , Sep 29, 2001
                                • 0 Attachment
                                  Dear Fred,
                                  thank you for your very interesting remarks.
                                  They should lead to alot ofnice results.

                                  > The subject of JP is a special case of jacobian:
                                  >
                                  > General:
                                  > If C(1), C(2) and C(3) are three independant conics (i.e. C3 not in
                                  > the pencil generated by C1 and C2), they form a net of conics.
                                  > Let p(i) be the polar of P relative to C(i).
                                  > The locus of P such that the p(i) concur is a cubic J, known as the
                                  > jacobian of the net of conics.
                                  >
                                  > J is the locus of the double points of the degenerated conics of
                                  the
                                  > net.
                                  >
                                  > Special:
                                  > Take for C(1), C(2) two generators of the pencil of conics which
                                  > defines the isoconjugation i and put C(3) = i(D), then J is the
                                  nK_0
                                  > i-invariant cubic of JP.

                                  The tangent to i(D) intersect BC, CA, AB respectively at U, V, W
                                  lying on a same line
                                  Thus a conic is member of your net iff the polar of A, B, C wrt the
                                  conic go respectively through U,V,W.
                                  >
                                  > Example:
                                  > C(1) = (I Ia)(Ib Ic), C(2) = (I Ib)(Ia Ic), C(3) = (O), i =
                                  isogonal
                                  > transformation and J = KjP.
                                  >
                                  > In the article, I sended to Paul for FG, you'll see that:
                                  > 1) Every non-pivotal isocubic is the jacobian of a net.
                                  > 2) If C(3) is circumscribed to the autopolar triangle of C(1), C
                                  (2),
                                  > then you receive a nK_0 cubic, and in this case the circumcircle of
                                  > ABC is in the net.

                                  I don't understand that; I think that the only circumcubic in the net
                                  is the circumconic with perspector the root of the cubic - but, may
                                  be, you were talking of the special case of Kjp -

                                  > Later I plan:
                                  > 3)properties of the special case nK_0
                                  > 4) group structure (difficult, but I have the elliptic invariant of
                                  > the cubics)
                                  >
                                  > Indeed, in Durège you have 100 pages of properties relative to the
                                  > jacobian...
                                  >
                                  > I think it could be a good idea to write something together for FG
                                  > (JPE + BG + FL +...)

                                  Why not!!
                                  >
                                  > Regards.
                                  >
                                  > Fred.
                                • fred.lang@eivd.ch
                                  Dear JP, ... Yes. ... of ... net ... That s right, It s circumconic and not circumcircle. Sorry. Bye Fred
                                  Message 16 of 16 , Oct 1, 2001
                                  • 0 Attachment
                                    Dear JP,
                                    --- In Hyacinthos@y..., jean-pierre.ehrmann@w... wrote:

                                    > The tangent to i(D) intersect BC, CA, AB respectively at U, V, W
                                    > lying on a same line
                                    > Thus a conic is member of your net iff the polar of A, B, C wrt the
                                    > conic go respectively through U,V,W.

                                    Yes.

                                    > > 2) If C(3) is circumscribed to the autopolar triangle of C(1), C
                                    > (2),
                                    > > then you receive a nK_0 cubic, and in this case the circumcircle
                                    of
                                    > > ABC is in the net.
                                    >
                                    > I don't understand that; I think that the only circumconic in the
                                    net
                                    > is the circumconic with perspector the root of the cubic - but, may
                                    > be, you were talking of the special case of Kjp -

                                    That's right, It's circumconic and not circumcircle.
                                    Sorry.
                                    Bye
                                    Fred
                                  Your message has been successfully submitted and would be delivered to recipients shortly.