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Reflection In Orthocenter And Projection From Altitude Foot

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  • Quang Tuan Bui
    Dear All My Friends, Given triangle ABC with orthocenter H, altitudes AHa, BHb, CHc. A , B , C are reflections of A, B, C in H respectively. Ab = orthogonal
    Message 1 of 12 , Feb 28, 2008
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      Dear All My Friends,
      Given triangle ABC with orthocenter H, altitudes AHa, BHb, CHc.
      A', B', C' are reflections of A, B, C in H respectively.
      Ab = orthogonal projection of Ha on AC
      Ac = orthogonal projection of Ha on AB
      A'' = intersection of B'Ab and C'Ac
      Similarly define B'', C''.
      The result:
      Two triangles ABC and A''B''C'' are perspective at perspector P with barycentrics coordinate:

      (a^4 - b^4 - c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :

      Search value: -3.57840086062088, may be not in current ETC?

      Thank you and best regards,
      Bui Quang Tuan



      ---------------------------------
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      [Non-text portions of this message have been removed]
    • Eric Danneels
      Dear Quang, your perspector is X(2883) It is the isogonal conjugate of H wrt the medial triangle of ABC The searchvalue in the ETC is different from you value
      Message 2 of 12 , Feb 28, 2008
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        Dear Quang,

        your perspector is X(2883)
        It is the isogonal conjugate of H wrt the medial triangle of ABC
        The searchvalue in the ETC is different from you value
        I did not check which one is correct

        Kind regards

        Eric


        > Dear All My Friends,
        > Given triangle ABC with orthocenter H, altitudes AHa, BHb, CHc.
        > A', B', C' are reflections of A, B, C in H respectively.
        > Ab = orthogonal projection of Ha on AC
        > Ac = orthogonal projection of Ha on AB
        > A'' = intersection of B'Ab and C'Ac
        > Similarly define B'', C''.
        > The result:
        > Two triangles ABC and A''B''C'' are perspective at perspector P
        with barycentrics coordinate:
        >
        > (a^4 - b^4 - c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
        >
        > Search value: -3.57840086062088, may be not in current ETC?
        >
        > Thank you and best regards,
        > Bui Quang Tuan
      • Quang Tuan Bui
        Dear Peter and Eric, Thank you very much for your messages! After checking I think the perspector could not be X(2883). Please kindly check your construction
        Message 3 of 12 , Feb 28, 2008
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          Dear Peter and Eric,
          Thank you very much for your messages!
          After checking I think the perspector could not be X(2883).
          Please kindly check your construction again, Dear Eric!
          Thank you and best regards,
          Bui Quang Tuan

          Eric Danneels <ERIC.DANNEELS@...> wrote:
          Dear Quang,

          your perspector is X(2883)
          It is the isogonal conjugate of H wrt the medial triangle of ABC
          The searchvalue in the ETC is different from you value
          I did not check which one is correct

          Kind regards

          Eric


          > Dear All My Friends,
          > Given triangle ABC with orthocenter H, altitudes AHa, BHb, CHc.
          > A', B', C' are reflections of A, B, C in H respectively.
          > Ab = orthogonal projection of Ha on AC
          > Ac = orthogonal projection of Ha on AB
          > A'' = intersection of B'Ab and C'Ac
          > Similarly define B'', C''.
          > The result:
          > Two triangles ABC and A''B''C'' are perspective at perspector P
          with barycentrics coordinate:
          >
          > (a^4 - b^4 - c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
          >
          > Search value: -3.57840086062088, may be not in current ETC?
          >
          > Thank you and best regards,
          > Bui Quang Tuan



          ---------------------------------
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          [Non-text portions of this message have been removed]
        • garciacapitan
          Dear Eric and Tuan, I get the same value as Tuan -3.578400860620876159553999905...
          Message 4 of 12 , Feb 29, 2008
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            Dear Eric and Tuan, I get the same value as Tuan

            -3.578400860620876159553999905...



            --- In Hyacinthos@yahoogroups.com, "Eric Danneels"
            <ERIC.DANNEELS@...> wrote:
            >
            > Dear Quang,
            >
            > your perspector is X(2883)
            > It is the isogonal conjugate of H wrt the medial triangle of ABC
            > The searchvalue in the ETC is different from you value
            > I did not check which one is correct
            >
            > Kind regards
            >
            > Eric
            >
            >
            > > Dear All My Friends,
            > > Given triangle ABC with orthocenter H, altitudes AHa, BHb, CHc.
            > > A', B', C' are reflections of A, B, C in H respectively.
            > > Ab = orthogonal projection of Ha on AC
            > > Ac = orthogonal projection of Ha on AB
            > > A'' = intersection of B'Ab and C'Ac
            > > Similarly define B'', C''.
            > > The result:
            > > Two triangles ABC and A''B''C'' are perspective at perspector P
            > with barycentrics coordinate:
            > >
            > > (a^4 - b^4 - c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
            > >
            > > Search value: -3.57840086062088, may be not in current ETC?
            > >
            > > Thank you and best regards,
            > > Bui Quang Tuan
            >
          • Eric Danneels
            Dear Quang and Garcia, Quang s searchvalue is correct I made a wrong construction with Ab = orthogonal projection of A (instead of Ha) on AC Ac = orthogonal
            Message 5 of 12 , Feb 29, 2008
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              Dear Quang and Garcia,

              Quang's searchvalue is correct

              I made a wrong construction with
              Ab = orthogonal projection of A' (instead of Ha) on AC
              Ac = orthogonal projection of A' (instead of Ha) on AB....

              Sorry

              Kind regards

              Eric


              --- In Hyacinthos@yahoogroups.com, "garciacapitan"
              <garciacapitan@...> wrote:
              >
              > Dear Eric and Tuan, I get the same value as Tuan
              >
              > -3.578400860620876159553999905...
              >
              >
              >
              > --- In Hyacinthos@yahoogroups.com, "Eric Danneels"
              > <ERIC.DANNEELS@> wrote:
              > >
              > > Dear Quang,
              > >
              > > your perspector is X(2883)
              > > It is the isogonal conjugate of H wrt the medial triangle of ABC
              > > The searchvalue in the ETC is different from you value
              > > I did not check which one is correct
              > >
              > > Kind regards
              > >
              > > Eric
              > >
              > >
              > > > Dear All My Friends,
              > > > Given triangle ABC with orthocenter H, altitudes AHa, BHb,
              CHc.
              > > > A', B', C' are reflections of A, B, C in H respectively.
              > > > Ab = orthogonal projection of Ha on AC
              > > > Ac = orthogonal projection of Ha on AB
              > > > A'' = intersection of B'Ab and C'Ac
              > > > Similarly define B'', C''.
              > > > The result:
              > > > Two triangles ABC and A''B''C'' are perspective at perspector
              P
              > > with barycentrics coordinate:
              > > >
              > > > (a^4 - b^4 - c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
              > > >
              > > > Search value: -3.57840086062088, may be not in current ETC?
              > > >
              > > > Thank you and best regards,
              > > > Bui Quang Tuan
              > >
              >
            • Eric Danneels
              Dear Quang, I would like to present the following generalization: Let PaPbPc be the cevian triangle of P wrt triangle ABC A , B , C are reflections of A, B, C
              Message 6 of 12 , Feb 29, 2008
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                Dear Quang,

                I would like to present the following generalization:

                Let PaPbPc be the cevian triangle of P wrt triangle ABC
                A', B', C' are reflections of A, B, C in P respectively.
                Ab = intersection of AC with parallel to PB through Pa
                Ac = intersection of AB with parallel to PC through Pa
                A'' = intersection of B'Ab and C'Ac
                Similarly define B'', C''.
                Triangles ABC and A''B''C'' are perspective

                Kind regards

                Eric


                --- In Hyacinthos@yahoogroups.com, "garciacapitan"
                <garciacapitan@...> wrote:
                >
                > Dear Eric and Tuan, I get the same value as Tuan
                >
                > -3.578400860620876159553999905...
                >
                >
                >
                > --- In Hyacinthos@yahoogroups.com, "Eric Danneels"
                > <ERIC.DANNEELS@> wrote:
                > >
                > > Dear Quang,
                > >
                > > your perspector is X(2883)
                > > It is the isogonal conjugate of H wrt the medial triangle of ABC
                > > The searchvalue in the ETC is different from you value
                > > I did not check which one is correct
                > >
                > > Kind regards
                > >
                > > Eric
                > >
                > >
                > > > Dear All My Friends,
                > > > Given triangle ABC with orthocenter H, altitudes AHa, BHb,
                CHc.
                > > > A', B', C' are reflections of A, B, C in H respectively.
                > > > Ab = orthogonal projection of Ha on AC
                > > > Ac = orthogonal projection of Ha on AB
                > > > A'' = intersection of B'Ab and C'Ac
                > > > Similarly define B'', C''.
                > > > The result:
                > > > Two triangles ABC and A''B''C'' are perspective at perspector
                P
                > > with barycentrics coordinate:
                > > >
                > > > (a^4 - b^4 - c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
                > > >
                > > > Search value: -3.57840086062088, may be not in current ETC?
                > > >
                > > > Thank you and best regards,
                > > > Bui Quang Tuan
                > >
                >
              • Quang Tuan Bui
                Dear Francisco and Eric, Thank you very much for your messages! I am very happy to receive your messages. I would like to thank Peter J.C. Moses for remark
                Message 7 of 12 , Feb 29, 2008
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                  Dear Francisco and Eric,
                  Thank you very much for your messages! I am very happy to receive your messages.
                  I would like to thank Peter J.C. Moses for remark that the barycentrics can be more simple as:

                  (S^2 + 2*SA^2)*(2*SB*SC - S^2)/SA^2 : :

                  and this perspector is intersection of lines {{2,1105},{4,6}}

                  One small remark:
                  We construct A* = intersection of B'Ac and C'Ab similary construct B*, C* then two triangles ABC and A*B*C* are perspective at isogonal conjugate of X(1593)

                  Thank you Dear All again!
                  Have you very nice weekend!
                  Bui Quang Tuan


                  Eric Danneels <ERIC.DANNEELS@...> wrote: Dear Quang and Garcia,

                  Quang's searchvalue is correct

                  I made a wrong construction with
                  Ab = orthogonal projection of A' (instead of Ha) on AC
                  Ac = orthogonal projection of A' (instead of Ha) on AB....

                  Sorry

                  Kind regards

                  Eric


                  --- In Hyacinthos@yahoogroups.com, "garciacapitan"
                  wrote:
                  >
                  > Dear Eric and Tuan, I get the same value as Tuan
                  >
                  > -3.578400860620876159553999905...
                  >
                  >
                  >
                  > --- In Hyacinthos@yahoogroups.com, "Eric Danneels"
                  > wrote:
                  > >
                  > > Dear Quang,
                  > >
                  > > your perspector is X(2883)
                  > > It is the isogonal conjugate of H wrt the medial triangle of ABC
                  > > The searchvalue in the ETC is different from you value
                  > > I did not check which one is correct
                  > >
                  > > Kind regards
                  > >
                  > > Eric
                  > >
                  > >
                  > > > Dear All My Friends,
                  > > > Given triangle ABC with orthocenter H, altitudes AHa, BHb,
                  CHc.
                  > > > A', B', C' are reflections of A, B, C in H respectively.
                  > > > Ab = orthogonal projection of Ha on AC
                  > > > Ac = orthogonal projection of Ha on AB
                  > > > A'' = intersection of B'Ab and C'Ac
                  > > > Similarly define B'', C''.
                  > > > The result:
                  > > > Two triangles ABC and A''B''C'' are perspective at perspector
                  P
                  > > with barycentrics coordinate:
                  > > >
                  > > > (a^4 - b^4 - c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
                  > > >
                  > > > Search value: -3.57840086062088, may be not in current ETC?
                  > > >
                  > > > Thank you and best regards,
                  > > > Bui Quang Tuan
                  > >
                  >


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                • Moses, Peter J. C.
                  Dear Eric & Tuan, A = {p - q - r, 2 q, 2 r} Ab = {p q, 0, r (p + q + r)} Ac = {p r, q (p + q + r), 0} A = {p q r (3 p + q + r), q (-p + q - r) (p q + q^2 + q
                  Message 8 of 12 , Mar 1, 2008
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                    Dear Eric & Tuan,

                    A' = {p - q - r, 2 q, 2 r}
                    Ab = {p q, 0, r (p + q + r)}
                    Ac = {p r, q (p + q + r), 0}
                    A" = {p q r (3 p + q + r), q (-p + q - r) (p q + q^2 + q r + 2 r p), r (-p - q + r) (2 p q + q r + r p + r^2)}

                    ABC & A"B"C" are perspective at p (p - q - r) (p^2 + p q + 2 q r + r p) ::
                    When P = I, the perspector is X(958),
                    When P = G, the perspector is X(2),
                    When P = O, the perspector is X(578),
                    When P = H, the perspector is the point Tuan found, (S^2 + 2 SA^2) (2 SB SC - S^2) / SA^2 ::

                    Ab, Ac ... live on an conic centered on p (q + r) (p^2 + q^2 + r^2 + 2 p q + q r + 2 r p) ::
                    When P = G, the conic is an ellipse (Sqrt[7] / 4 size of Steiner Circum) centered on G.
                    When P = H, the conic is an ellipse centered on X(389)
                    When P = X(20), the conic is centered on X(2883).

                    Best regards,
                    Peter.


                    ----- Original Message -----
                    From: Eric Danneels
                    To: Hyacinthos@yahoogroups.com
                    Sent: Friday, February 29, 2008 9:07 PM
                    Subject: [EMHL] Re: Reflection In Orthocenter And Projection From Altitude Foot


                    Dear Quang,

                    I would like to present the following generalization:

                    Let PaPbPc be the cevian triangle of P wrt triangle ABC
                    A', B', C' are reflections of A, B, C in P respectively.
                    Ab = intersection of AC with parallel to PB through Pa
                    Ac = intersection of AB with parallel to PC through Pa
                    A'' = intersection of B'Ab and C'Ac
                    Similarly define B'', C''.
                    Triangles ABC and A''B''C'' are perspective

                    Kind regards

                    Eric

                    --- In Hyacinthos@yahoogroups.com, "garciacapitan"
                    <garciacapitan@...> wrote:
                    >
                    > Dear Eric and Tuan, I get the same value as Tuan
                    >
                    > -3.578400860620876159553999905...
                    >
                    >
                    >
                    > --- In Hyacinthos@yahoogroups.com, "Eric Danneels"
                    > <ERIC.DANNEELS@> wrote:
                    > >
                    > > Dear Quang,
                    > >
                    > > your perspector is X(2883)
                    > > It is the isogonal conjugate of H wrt the medial triangle of ABC
                    > > The searchvalue in the ETC is different from you value
                    > > I did not check which one is correct
                    > >
                    > > Kind regards
                    > >
                    > > Eric
                    > >
                    > >
                    > > > Dear All My Friends,
                    > > > Given triangle ABC with orthocenter H, altitudes AHa, BHb,
                    CHc.
                    > > > A', B', C' are reflections of A, B, C in H respectively.
                    > > > Ab = orthogonal projection of Ha on AC
                    > > > Ac = orthogonal projection of Ha on AB
                    > > > A'' = intersection of B'Ab and C'Ac
                    > > > Similarly define B'', C''.
                    > > > The result:
                    > > > Two triangles ABC and A''B''C'' are perspective at perspector
                    P
                    > > with barycentrics coordinate:
                    > > >
                    > > > (a^4 - b^4 - c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
                    > > >
                    > > > Search value: -3.57840086062088, may be not in current ETC?
                    > > >
                    > > > Thank you and best regards,
                    > > > Bui Quang Tuan
                    > >
                    >





                    [Non-text portions of this message have been removed]
                  • Quang Tuan Bui
                    Dear Eric, Thank you for interesting generalization. If I am not wrong, if barycentrics of P = (p : q : r) then results are: Two triangles ABC and A B C
                    Message 9 of 12 , Mar 1, 2008
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                      Dear Eric,

                      Thank you for interesting generalization. If I am not wrong, if barycentrics of P = (p : q : r) then results are:

                      Two triangles ABC and A''B''C'' are perspective at:
                      p*(p - q - r)*(p*(p + q + r) + 2*q*r) : :

                      We construct A* = intersection of B'Ac and C'Ab similary construct B*, C* then two triangles ABC and A*B*C* are perspective at:
                      1/(p*(p + q + r) + 2*q*r) : :

                      Barycentric product of these two point is a point:
                      p*(p - q - r) : :

                      Best regards,
                      Bui Quang Tuan


                      Eric Danneels <ERIC.DANNEELS@...> wrote: Dear Quang,

                      I would like to present the following generalization:

                      Let PaPbPc be the cevian triangle of P wrt triangle ABC
                      A', B', C' are reflections of A, B, C in P respectively.
                      Ab = intersection of AC with parallel to PB through Pa
                      Ac = intersection of AB with parallel to PC through Pa
                      A'' = intersection of B'Ab and C'Ac
                      Similarly define B'', C''.
                      Triangles ABC and A''B''C'' are perspective

                      Kind regards

                      Eric




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                      [Non-text portions of this message have been removed]
                    • Quang Tuan Bui
                      Dear Eric and All My Friends, We can also take one more point and get one cubic locus. Detail as following: Barycentrics of P = (p : q : r), X = (x : y : z).
                      Message 10 of 12 , Mar 1, 2008
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                        Dear Eric and All My Friends,
                        We can also take one more point and get one cubic locus. Detail as following:
                        Barycentrics of P = (p : q : r), X = (x : y : z). XaXbXc is cevian triangle of X wrt ABC
                        A', B', C' are reflections of A, B, C in P respectively
                        Ab = intersection of AC with parallel to XB through Xa
                        Ac = intersection of AB with parallel to XC through Xa
                        A'' = intersection of B'Ab and C'Ac
                        A* = intersection of B'Ac and C'Ab

                        The locus of X such that ABC and A''B''C'' are perspective is union of some lines and one cubic C(P).
                        The locus of X such that ABC and A*B*C* are perspective is union of some lines and the same cubic C(P).
                        Barycentric equation of C(P):

                        CyclicSum[y*z*p*((r + p - 3*q)*y - (q + p - 3*r)*z)] = 0

                        This cubic always passes A, B, C, X(2), P

                        - If P = X(3), C(P) passes:
                        X(2), X(3), X(64), X(69), X(253), X(1073), X(3146)
                        - If P = X(5), C(P) passes:
                        X(2), X(5), X(20), X(69), X(264), X(343)
                        - If P = X(10) , C(P) passes:
                        X(2), X(7), X(10), X(65), X(75), X(145), X(226), X(3175)
                        These cubics are not listed in ETC and in Bernard cubics sites.

                        Any good idea about these cubics?

                        Thank you and best regards,
                        Bui Quang Tuan


                        Eric Danneels <ERIC.DANNEELS@...> wrote: Dear Quang,

                        I would like to present the following generalization:

                        Let PaPbPc be the cevian triangle of P wrt triangle ABC
                        A', B', C' are reflections of A, B, C in P respectively.
                        Ab = intersection of AC with parallel to PB through Pa
                        Ac = intersection of AB with parallel to PC through Pa
                        A'' = intersection of B'Ab and C'Ac
                        Similarly define B'', C''.
                        Triangles ABC and A''B''C'' are perspective

                        Kind regards

                        Eric




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                      • michgarl
                        Dear Eric In barycentrics coordinates P = [m:n:p] the perspector is [2np(n+p) +m(n²+p²-m²):2mp(m+p)+n(p²+m²-n²):2mn(m+n)+p(m²+n²-p²)] King regards
                        Message 11 of 12 , Mar 3, 2008
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                          Dear Eric
                          In barycentrics coordinates P = [m:n:p]
                          the perspector is
                          [2np(n+p) +m(n²+p²-m²):2mp(m+p)+n(p²+m²-n²):2mn(m+n)+p(m²+n²-p²)]

                          King regards

                          Michel

                          --- In Hyacinthos@yahoogroups.com, "Eric Danneels" <ERIC.DANNEELS@...>
                          wrote:
                          >
                          > Dear Quang,
                          >
                          > I would like to present the following generalization:
                          >
                          > Let PaPbPc be the cevian triangle of P wrt triangle ABC
                          > A', B', C' are reflections of A, B, C in P respectively.
                          > Ab = intersection of AC with parallel to PB through Pa
                          > Ac = intersection of AB with parallel to PC through Pa
                          > A'' = intersection of B'Ab and C'Ac
                          > Similarly define B'', C''.
                          > Triangles ABC and A''B''C'' are perspective
                          >
                          > Kind regards
                          >
                          > Eric
                          >
                          >
                          > --- In Hyacinthos@yahoogroups.com, "garciacapitan"
                          > <garciacapitan@> wrote:
                          > >
                          > > Dear Eric and Tuan, I get the same value as Tuan
                          > >
                          > > -3.578400860620876159553999905...
                          > >
                          > >
                          > >
                          > > --- In Hyacinthos@yahoogroups.com, "Eric Danneels"
                          > > <ERIC.DANNEELS@> wrote:
                          > > >
                          > > > Dear Quang,
                          > > >
                          > > > your perspector is X(2883)
                          > > > It is the isogonal conjugate of H wrt the medial triangle of ABC
                          > > > The searchvalue in the ETC is different from you value
                          > > > I did not check which one is correct
                          > > >
                          > > > Kind regards
                          > > >
                          > > > Eric
                          > > >
                          > > >
                          > > > > Dear All My Friends,
                          > > > > Given triangle ABC with orthocenter H, altitudes AHa, BHb,
                          > CHc.
                          > > > > A', B', C' are reflections of A, B, C in H respectively.
                          > > > > Ab = orthogonal projection of Ha on AC
                          > > > > Ac = orthogonal projection of Ha on AB
                          > > > > A'' = intersection of B'Ab and C'Ac
                          > > > > Similarly define B'', C''.
                          > > > > The result:
                          > > > > Two triangles ABC and A''B''C'' are perspective at perspector
                          > P
                          > > > with barycentrics coordinate:
                          > > > >
                          > > > > (a^4 - b^4 - c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
                          > > > >
                          > > > > Search value: -3.57840086062088, may be not in current ETC?
                          > > > >
                          > > > > Thank you and best regards,
                          > > > > Bui Quang Tuan
                          > > >
                          > >
                          >
                        • Bernard Gibert
                          Dear Tuan, [snip] ... this is the pivotal cubic with pivot taaP and isopivot P (tX = isotomic conj. of X, aX = anticomplement of X) thus, C(X3) = pK(X1073,
                          Message 12 of 12 , Mar 4, 2008
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                            Dear Tuan,

                            [snip]

                            > [QTB] Barycentric equation of C(P):
                            >
                            > CyclicSum[y*z*p*((r + p - 3*q)*y - (q + p - 3*r)*z)] = 0
                            >
                            > This cubic always passes A, B, C, X(2), P

                            this is the pivotal cubic with pivot taaP and isopivot P
                            (tX = isotomic conj. of X, aX = anticomplement of X)

                            thus,
                            C(X3) = pK(X1073, X253)
                            C(X5) = pK(X343, X69)
                            C(X10) = pK(X226, X7)



                            Best regards

                            Bernard



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