- Mar 1, 2013Dear Antreas,

For the orthic version (P=X(4), case 1.), the circumcircles do not concur, but their radical center is non-ETC -10.266788994809312, which lies on the Euler line, and is in fact the reflection of X(403) in X(4).

For the medial version (P=X(2), case 1.), the circumcircles do not concur. Their radical center is non-ETC 0.675001328037596, for which I can find no special properties.

Randy

--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:

>

> Dear Randy

>

> Thanks.

>

> I guess that the point for the orthic version (ie ABC is orthic triangle of

> reference triangle)

> is not in ETC as well.

>

> Now, as for generalizations (P instead of I):

>

> There are two possible cases since for P = I the reflection of B' in AA' is

> lying on AC etc.:

>

> For A'B'C' = Cevian triangle of P.

>

> Denote:

>

> 1. B'a, C'a = the reflections of B',C' in AA', resp. etc

>

> or

>

> 2. B'a = (perpendicular to AA' from B') /\ AC

> etc

>

> Locus of P such that the circumcircles of the triangles A'aB'aC'a,

> B'bC'bA'b, C'cA'cB'c are concurrent ??.

>

> Quite complicated, I guess.....!

>

> APH

>

>

> On Fri, Mar 1, 2013 at 12:44 AM, rhutson2 <rhutson2@...> wrote:

>

> > **

> >

> >

> > Antreas,

> >

> > The circumcircles are concurrent at non-ETC 0.812149174855220, which is,

> > equivalently:

> >

> > X(1)-Ceva conjugate of X(36)

> > Antigonal conjugate, wrt incentral triangle, of X(1)

> > The point P for which P of the 'orthocentroidal triangle' = X(1).

> >

> > I define the 'orthocentroidal triangle' as:

> > Let A* be the intersection, other than X(4), of the A-altitude and the

> > orthocentroidal circle, and define B*, C* cyclically.

> >

> > A*B*C* is inversely similar to ABC, with similitude center X(6).

> > X(i) of A*B*C* = reflection of X(i) (of ABC) in the centroid of its pedal

> > triangle.

> >

> > As to your second question, the triangles ABC, IaIbIc are not perspective.

> >

> > Best regards,

> > Randy

> >

> >

> > --- In Hyacinthos@yahoogroups.com, "Antreas" wrote:

> > >

> > > Let ABC be a triangle and A'B'C' the cevian triangle of I.

> > >

> > > Denote:

> > >

> > > B'a, C'a = the reflections of B',C' in AA', resp.

> > >

> > > A'a = (perpendicular to BB' from B'a) /\ (perpendicular to CC' from

> > > C'a).

> > >

> > > C'b, A'b = the reflections of C',A' in BB', resp.

> > >

> > > B'b = (perpendicular to CC' from C'b) /\ (perpendicular to AA' from

> > > A'b).

> > >

> > > A'c, B'c = the reflections of A',B' in CC', resp.

> > >

> > > C'c = (perpendicular to AA' from A'c) /\ (perpendicular to BB' from

> > > B'c).

> > >

> > > Are the circumcircles of A'aB'aC'a, B'bC'bA'b, C'cA'cB'c

> > > concurrent?.

> > >

> > > Let Ia,Ib,Ic be the incenters of A'aB'aC'a, B'bC'bA'b, C'cA'cB'c,

> > > resp.

> > >

> > > Are the triangles ABC, IaIbIc perspective?

> > >

> > >

> > > APH

> >

>

>

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