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[EMHL] Re: Radical Centers - perspective triangle

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  • rhutson2
    Dear Antreas, They are indeed perspective (ETC search -17.765195072303639), though I could find no relationship to existing ETC centers, except that the line
    Message 1 of 6 , Mar 25 1:32 PM
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      Dear Antreas,

      They are indeed perspective (ETC search -17.765195072303639), though I could find no relationship to existing ETC centers, except that the line though the perspector and the radical center of (Ja),(Jb),(Jc) (ETC search 5.603197179118705) passes through X(1790).

      Randy

      --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
      >
      > Dear Randy
      >
      > The triangles OBC,OCA,OAB are special triangles (isosceles)
      > and I think we can get perspectivities for other triads of circles
      > respective to BC,CA,AB.
      >
      > For example:
      >
      > Denote
      >
      > Ja = the excenter of the excircle respective to BC of the triangle OBC
      > Jb = the excenter of the excircle respective to CA of the triangle OCA
      > Jc = the excenter of the excircle respective to AB of the triangle OAB
      >
      > Ra = the radical center of (O), (Jb), (Jc)
      > Rb = the radical center of (O), (Jc), (Ja)
      > Rc = the radical center of (O), (Ja), (Jb)
      >
      > The triangles ABC, RaRbRc [I would bet that they] are perspective.
      >
      > Antreas
      >
      >
      > On Mon, Mar 25, 2013 at 6:17 AM, rhutson2 <rhutson2@...> wrote:
      >
      > > **
      > >
      > >
      > > Dear Antreas,
      > >
      > > The perspector is the isogonal conjugate of X(340) = X(92)-isoconjugate of
      > > X(323) = intersection of lines X(13)X*(14) and X(14)X*(13), where X* =
      > > inverse-in-circumcircle.
      > >
      > > Best regards,
      > > Randy
      > >
      > >
      > > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@>
      > > wrote:
      > > >
      > > > Let ABC be a triangle and (Na),(Nb),(Nc) the NPCs of OBC, OCA,OAB, resp.
      > > >
      > > > Denote:
      > > >
      > > > Ra = the radical center of (O), (Nb), (Nc)
      > > >
      > > > Rb = the radical center of (O), (Nc), (Na)
      > > >
      > > > Rc = the radical center of (O), (Na), (Nb)
      > > >
      > > > The triangles ABC, RaRbRc are perspective.
      > > >
      > > > Perspector?
      > > >
      > > > APH
      > > >
      > >
      > >
      > >
      >
      >
      > [Non-text portions of this message have been removed]
      >
    • rhutson2
      And if we substitute incircles for excircles, then RaRbRc is homothetic to ABC at (ETC search -0.387264139784864). Interestingly, the line through this
      Message 2 of 6 , Mar 25 2:41 PM
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        And if we substitute incircles for excircles, then RaRbRc is homothetic to ABC at (ETC search -0.387264139784864). Interestingly, the line through this perspector and the radical center of (Ia),(Ib),(Ic) (ETC search 3.463390186710557) also passes through X(1790). Also, the two perspectors both lie on line X(25)X(41), and are {X(i),X(j)}-harmonic conjugates for (i,j) in {(25,1973), (41,2187), (42,205)}.

        Randy

        --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@...> wrote:
        >
        > Dear Antreas,
        >
        > They are indeed perspective (ETC search -17.765195072303639), though I could find no relationship to existing ETC centers, except that the line though the perspector and the radical center of (Ja),(Jb),(Jc) (ETC search 5.603197179118705) passes through X(1790).
        >
        > Randy
        >
        > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@> wrote:
        > >
        > > Dear Randy
        > >
        > > The triangles OBC,OCA,OAB are special triangles (isosceles)
        > > and I think we can get perspectivities for other triads of circles
        > > respective to BC,CA,AB.
        > >
        > > For example:
        > >
        > > Denote
        > >
        > > Ja = the excenter of the excircle respective to BC of the triangle OBC
        > > Jb = the excenter of the excircle respective to CA of the triangle OCA
        > > Jc = the excenter of the excircle respective to AB of the triangle OAB
        > >
        > > Ra = the radical center of (O), (Jb), (Jc)
        > > Rb = the radical center of (O), (Jc), (Ja)
        > > Rc = the radical center of (O), (Ja), (Jb)
        > >
        > > The triangles ABC, RaRbRc [I would bet that they] are perspective.
        > >
        > > Antreas
        > >
        > >
        > > On Mon, Mar 25, 2013 at 6:17 AM, rhutson2 <rhutson2@> wrote:
        > >
        > > > **
        > > >
        > > >
        > > > Dear Antreas,
        > > >
        > > > The perspector is the isogonal conjugate of X(340) = X(92)-isoconjugate of
        > > > X(323) = intersection of lines X(13)X*(14) and X(14)X*(13), where X* =
        > > > inverse-in-circumcircle.
        > > >
        > > > Best regards,
        > > > Randy
        > > >
        > > >
        > > > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@>
        > > > wrote:
        > > > >
        > > > > Let ABC be a triangle and (Na),(Nb),(Nc) the NPCs of OBC, OCA,OAB, resp.
        > > > >
        > > > > Denote:
        > > > >
        > > > > Ra = the radical center of (O), (Nb), (Nc)
        > > > >
        > > > > Rb = the radical center of (O), (Nc), (Na)
        > > > >
        > > > > Rc = the radical center of (O), (Na), (Nb)
        > > > >
        > > > > The triangles ABC, RaRbRc are perspective.
        > > > >
        > > > > Perspector?
        > > > >
        > > > > APH
        > > > >
        > > >
        > > >
        > > >
        > >
        > >
        > > [Non-text portions of this message have been removed]
        > >
        >
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