## [EMHL] Re: Radical Centers - perspective triangle

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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
Message 1 of 6 , Mar 25, 2013
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]
>
• 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, 2013
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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