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concurrence

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  • Alexey.A.Zaslavsky
    Dear colleagues! Let given the triangle ABC. A1B1C1 is the circumcevian triangle of its centroid, A2, B2, C2 --- the second common points of its circumcircle
    Message 1 of 4 , Mar 1 3:30 AM
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      Dear colleagues!
      Let given the triangle ABC. A1B1C1 is the circumcevian triangle of its
      centroid, A2, B2, C2 --- the second common points of its circumcircle and
      three lines passing through its vertex and parallel to opposite sidelines
      (i.e. A2B2C2 is the orthotriangle of antimedial triangle). Then A1A2, B1B2
      and C1C2 concur. Its common point X is in the Euler line of ABC and the
      polar of X wrt the circumcircle is the trypolar of the point isotomic
      conjugated to the Lemoine point of ABC.
      What is the number of X in ETC?

      Sincerely Alexey




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    • Wilson Stothers
      ... its ... circumcircle and ... sidelines ... A1A2, B1B2 ... the ... isotomic ... Dear Alexey, The answer is 22 (the tripolar of the isotomic conjugate of
      Message 2 of 4 , Mar 1 4:20 AM
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        --- In Hyacinthos@yahoogroups.com, "Alexey.A.Zaslavsky" <zasl@...>
        wrote:
        >
        > Dear colleagues!
        > Let given the triangle ABC. A1B1C1 is the circumcevian triangle of
        its
        > centroid, A2, B2, C2 --- the second common points of its
        circumcircle and
        > three lines passing through its vertex and parallel to opposite
        sidelines
        > (i.e. A2B2C2 is the orthotriangle of antimedial triangle). Then
        A1A2, B1B2
        > and C1C2 concur. Its common point X is in the Euler line of ABC and
        the
        > polar of X wrt the circumcircle is the trypolar of the point
        isotomic
        > conjugated to the Lemoine point of ABC.
        > What is the number of X in ETC?
        >

        Dear Alexey,

        The answer is 22

        (the tripolar of the isotomic conjugate of Lemoine Point is the
        de Longchamps Line. This meets the Euler Line at X(858). The
        inverse in the Circumcircle of this is the point you want, and
        it is X(22) from ETC)

        regards

        Wilson
      • Jeff Brooks
        ... of ... and ... And if this is true we should be able to write: p=N[x22]; q=N[x4]; g[u_,v_,w_]:= (p-u)/Conjugate[p-u] (q-u)/Conjugate[q-u] T[u_,v_,w_]:=
        Message 3 of 4 , Mar 1 4:39 AM
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          --- In Hyacinthos@yahoogroups.com, "Wilson Stothers" <wws@...> wrote:
          >
          > --- In Hyacinthos@yahoogroups.com, "Alexey.A.Zaslavsky" <zasl@>
          > wrote:
          > >
          > > Dear colleagues!
          > > Let given the triangle ABC. A1B1C1 is the circumcevian triangle
          of
          > its
          > > centroid, A2, B2, C2 --- the second common points of its
          > circumcircle and
          > > three lines passing through its vertex and parallel to opposite
          > sidelines
          > > (i.e. A2B2C2 is the orthotriangle of antimedial triangle). Then
          > A1A2, B1B2
          > > and C1C2 concur. Its common point X is in the Euler line of ABC
          and
          > the
          > > polar of X wrt the circumcircle is the trypolar of the point
          > isotomic
          > > conjugated to the Lemoine point of ABC.
          > > What is the number of X in ETC?
          > >
          >
          > Dear Alexey,
          >
          > The answer is 22
          >
          > (the tripolar of the isotomic conjugate of Lemoine Point is the
          > de Longchamps Line. This meets the Euler Line at X(858). The
          > inverse in the Circumcircle of this is the point you want, and
          > it is X(22) from ETC)
          >
          > regards
          >
          > Wilson
          >

          And if this is true we should be able to write:

          p=N[x22];
          q=N[x4];

          g[u_,v_,w_]:= (p-u)/Conjugate[p-u] (q-u)/Conjugate[q-u]

          T[u_,v_,w_]:= (g[v,w,u] - g[w,u,v])

          k=1;
          M[u_,v_,w_]:= (v-w)^k

          f[n_,u_,v_,w_]:= (M[w,u,v] T[v,w,u] w + n M[v,u,w] T[w,u,v] v)/
          (M[w,u,v] T[v,w,u] + n M[v,u,w] T[w,u,v])

          d1[u_,v_,w_]:=1;

          pp1=Chop[N[f[d1[z1,z2,z3],z1,z2,z3],12],10^-6]
          pp2=Chop[N[f[d1[z2,z3,z1],z2,z3,z1],12],10^-6]
          pp3=Chop[N[f[d1[z3,z1,z2],z3,z1,z2],12],10^-6]

          and get the same answer.

          Sincerely, Jeff
        • Jeff Brooks
          Please disregard, this does nothing but give the inverse in the circumcircle ______________________________________________ And if this is true we should be
          Message 4 of 4 , Mar 1 5:35 AM
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            Please disregard,

            this does nothing but give the inverse in the circumcircle

            ______________________________________________

            And if this is true we should be able to write:

            p=N[x22];
            q=N[x4];

            g[u_,v_,w_]:= (p-u)/Conjugate[p-u] (q-u)/Conjugate[q-u]

            T[u_,v_,w_]:= (g[v,w,u] - g[w,u,v])

            k=1;
            M[u_,v_,w_]:= (v-w)^k

            f[n_,u_,v_,w_]:= (M[w,u,v] T[v,w,u] w + n M[v,u,w] T[w,u,v] v)/
            (M[w,u,v] T[v,w,u] + n M[v,u,w] T[w,u,v])

            d1[u_,v_,w_]:=1;

            pp1=Chop[N[f[d1[z1,z2,z3],z1,z2,z3],12],10^-6]
            pp2=Chop[N[f[d1[z2,z3,z1],z2,z3,z1],12],10^-6]
            pp3=Chop[N[f[d1[z3,z1,z2],z3,z1,z2],12],10^-6]

            and get the same answer.

            Sincerely, Jeff
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