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Re: [EMHL] More about X(1000) [Isogonal conjugate of midpoint of X(1),X(57)]

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  • Antreas Hatzipolakis
    You might also want to try these: Let Ja,Jb, Jc be the centers of inversions of BC,CA,AB in the incircle, resp. Are the triangles ABC, JaJbJc perspective? And
    Message 1 of 5 , Mar 9, 2013
      You might also want to try these:

      Let Ja,Jb, Jc be the centers of inversions
      of BC,CA,AB in the incircle, resp.

      Are the triangles ABC, JaJbJc perspective?

      And let Ka, Kb, Kc your centers (ie the centers of the inversions of
      BC,CA,AB in the
      excircles).

      Are JaJbJc and KaKbKc perspective?

      APH


      On Sat, Mar 9, 2013 at 9:54 PM, C�sar Lozada <cesar_e_lozada@...>wrote:

      > **
      >
      >
      > Dear teachers:
      >
      > Let A� the center of the inversive figure of line BC on A-excircle. Build
      > B�
      > and C� ciclycally.
      >
      > Lines (AA�), (BB�) and (CC�) meet in X(1000)
      >
      > Regards.
      >
      > Cesar Lozada
      >
      >


      [Non-text portions of this message have been removed]
    • César Lozada
      ... Yes: At X(3296) ... No. Cesar Lozada
      Message 2 of 5 , Mar 9, 2013
        -----Mensaje original-----
        >> You might also want to try these:

        >>Let Ja,Jb, Jc be the centers of inversions of BC,CA,AB in the incircle,
        >>resp.

        >>Are the triangles ABC, JaJbJc perspective?

        Yes: At X(3296)

        >And let Ka, Kb, Kc your centers (ie the centers of the inversions of
        >BC,CA,AB in the excircles).

        >>Are JaJbJc and KaKbKc perspective?

        No.

        Cesar Lozada
      • Antreas
        Ja, Jb, Jc = the centers of inversion of BC, CA, AB in the Incircle La, Lb, Lc = the Incenters of JaBC, JbCA, JbAB Are the triangles ABC, La,Lb,Lc perspective?
        Message 3 of 5 , Mar 9, 2013
          Ja, Jb, Jc = the centers of inversion of BC, CA, AB in the Incircle

          La, Lb, Lc = the Incenters of JaBC, JbCA, JbAB

          Are the triangles ABC, La,Lb,Lc perspective?

          Excircles version?

          aph
        • Antreas
          [APH] ... Equivalently: Let A B C be the pedal triangle of I (intouch triangle) and Ja,Jb,Jc the midpoints of IA , IB , IC and La, Lb, Lc the incenters of
          Message 4 of 5 , Mar 9, 2013
            [APH]
            > Ja, Jb, Jc = the centers of inversion of BC, CA, AB in the Incircle
            >
            > La, Lb, Lc = the Incenters of JaBC, JbCA, JbAB
            >
            > Are the triangles ABC, LaLbLc perspective?
            >
            > Excircles version?

            Equivalently:

            Let A'B'C' be the pedal triangle of I (intouch triangle)
            and Ja,Jb,Jc the midpoints of IA', IB', IC'
            and La, Lb, Lc the incenters of JaBC, JbCA, JbAB, resp.

            Are the triangles ABC, LaLbLc perspective?

            We have cot(JaBA') = BA'/(r/2), BA' = rcot(B/2) ==>

            cot(JaBA') = 2cot(B/2)
            Similarly cot(JaCA') = 2cot(C/2)

            cotB - 2cot(B/2) / cotC - 2cot(C/2) *

            cotC - 2cot(C/2) / cotA - 2cot(A/2) *

            cotA - 2cot(A/2) / cotB - 2cot(B/2) = 1

            ==> the triangles ABC, JaJbJc are perspective

            Perspector in barycentrics:

            1 / (cotA - 2cot(A/2)) :: =.......

            Now, cot(JaBA') = [cot(LaBA')^2 - 1] / 2cot(LaBA')

            ==> cot(LaBA'): = f(B)

            Similarly we define f(C)

            And we have

            cotB - f(B) / cotC - f(C) *

            cotC - f(C) / cotA - f(A) *

            cotA - f(A) / cotB - f(B) = 1

            ==> the triangles are perspective.

            Perspector in barycentrics:

            1/ (cotA - f(A)) ::

            APH
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