## Re: [EMHL] Re: Concurrent circles

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• 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
Message 1 of 44 , Feb 28, 2013
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]
• [Tran Quang Hung]:. Let ABC be a triangle with orthocenter H and centroid G. A ,B ,C are on HA, HB, HC such that GA _|_ GA, GB _|_ GB, GC _|_ GC. Then the
Message 44 of 44 , Jul 26
[Tran Quang Hung]:.

Let ABC be a triangle with orthocenter H and centroid G.

A',B',C' are on HA, HB, HC such that GA' _|_ GA, GB' _|_ GB, GC' _|_ GC.

Then the circles (A',A'A), (B',B'B), (C',C'C) are concurrent at a point.

1. Which is this point ?

2. I see that A',B',C' and G are concyclic. Is this new circle ?