- 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

>

[Non-text portions of this message have been removed] - The circumcircles of AaAbAc, BaBbBc, CaCbCc are concurrent on the NPC circle of A'B'C'.AaAbAc, BaBbBc, CaCbCc = the pedal triangles of A", B", C", wrt triangle A'B'C' resp.Let ABC be a triangle, P a point andÂ A'B'C' the pedal triangle of P.Denote:L = the Euler line of A'B'C'A", B", C" = orthogonal projections of A, B, C, on L, resp.The point of concurrence is the reflection point of L wrt the medial triangle of A'B'C'.Problem:Let ABC be a triangle and L a line.For which points P's the L is the Euler line of the pedal triangle of P ?APH