Hyacinthos is a Restricted Group with 3 members.
 Hyacinthos

 Restricted Group,
 3 members
Orthologic Triangles  Locus
 1. Let ABC be a triangle, A'B'C' the pedal triangle of point P
and A",B",C" the second intersections of the circles (P,PA'),
(P,PB'), (P,PC') with the pedal circle of P, resp.
Which is the locus of P such that the triangles ABC, A"B"C"
are orthologic?
O is on the locus.
2. Let ABC be a triangle, A'B'C' the pedal triangle of point P,
A*B*C* the circumcevian triangle of P wrt A'B'C' and A",B",C" the
second intersections of the circles (P,PA*), (P,PB*), (P,PC*) with
the pedal circle of P, resp.
Which is the locus of P such that the triangles ABC, A"B"C"
are orthologic?
H is on the locus.
Figures:
http://anthrakitis.blogspot.gr/2013/03/orthologictriangleslocus.html
APH   In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
Dear Antreas
> 1. Let ABC be a triangle, A'B'C' the pedal triangle of point P
> and A",B",C" the second intersections of the circles (P,PA'),
> (P,PB'), (P,PC') with the pedal circle of P, resp.
>
> Which is the locus of P such that the triangles ABC, A"B"C"
> are orthologic?
>
> O is on the locus.
The locus of real point P such that the triangles ABC, A"B"C" are orthologic is Q068 (a circular equilateral quintic) in Higher Degree Related Curves.
http://bernard.gibert.pagespersoorange.fr/relatedcurves.html
It is a result does not appear between the propiedades of Q068 in http://bernard.gibert.pagespersoorange.fr/curves/q068.html
We can express as:
Let ABC be a triangle, A'B'C' the pedal triangle of point P and O' the circumcenter of A'B'C'. Let A", B", C" be the symmetrical points of A, B, C w/r to the line PO'. The triangles ABC, A"B"C" are orthologic if and only if P lies on Q068.
(http://groups.yahoo.com/group/Hyacinthos/files/Hyacinthos21754A.eps)
Angel M.  Writing correct geometric property of the quintic Q068 in the message #21755:
Let ABC be a triangle, A'B'C' the pedal triangle of point P and O' the
circumcenter of A'B'C'. Let A", B", C" be the symmetrical points of A', B', C' w/r to the line PO'. The triangles ABC, A"B"C" are orthologic if and only if P lies on Q068.
A particular situation:
If P=X(3) the orthology centers of ABC y A"B"C" (on NPC) are Q=X(113) and R, ninepointcircleantipode of X(3258).
R= (2*a^8  a^4*(b^44*b^2*c^2+c^4) 2*a^6*(b^2+c^2) + (b^2c^2)^4 )*
(a^6*(b^2 + c^2) + a^4*(3*b^4 + 2*b^2*c^2  3*c^4)+
a^2*(3*b^6  2*b^4*c^2  2*b^2*c^4 + 3*c^6)
(b^2  c^2)^2*(b^4 + 3*b^2*c^2 + c^4)) : .... : ....
Angel M.
 In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@...> wrote:
>
>
>
>  In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
> >
> > 1. Let ABC be a triangle, A'B'C' the pedal triangle of point P
> > and A",B",C" the second intersections of the circles (P,PA'),
> > (P,PB'), (P,PC') with the pedal circle of P, resp.
> >
> > Which is the locus of P such that the triangles ABC, A"B"C"
> > are orthologic?
> >
> > O is on the locus.
>
> Dear Antreas
>
> The locus of real point P such that the triangles ABC, A"B"C" are orthologic is Q068 (a circular equilateral quintic) in Higher Degree Related Curves.
> http://bernard.gibert.pagespersoorange.fr/relatedcurves.html
>
>
> It is a result does not appear between the propiedades of Q068 in http://bernard.gibert.pagespersoorange.fr/curves/q068.html
>
>
> We can express as:
>
> Let ABC be a triangle, A'B'C' the pedal triangle of point P and O' the circumcenter of A'B'C'. Let A", B", C" be the symmetrical points of A, B, C w/r to the line PO'. The triangles ABC, A"B"C" are orthologic if and only if P lies on Q068.
>
>
> (http://groups.yahoo.com/group/Hyacinthos/files/Hyacinthos21754A.eps)
>
>
> Angel M.
>  Dear Angel,
you wrote> Let ABC be a triangle, A'B'C' the pedal triangle of point P
I think you know that more precisely
> and O' the
> circumcenter of A'B'C'. Let A", B", C" be the symmetrical
> points of A', B', C' w/r to the line PO'. The triangles ABC,
> A"B"C" are orthologic if and only if P lies on Q068.
P lies on Linf + Circumcircle + Q068 + sextic
Qx = b^2 c^4 x^4 y^2  a^2 c^4 x^3 y^3  b^2 c^4 x^3 y^3 + c^6 x^3 y^3 
a^2 c^4 x^2 y^4 + a^2 b^2 c^2 x^4 y z  b^4 c^2 x^4 y z 
b^2 c^4 x^4 y z + a^2 b^2 c^2 x^3 y^2 z  b^4 c^2 x^3 y^2 z +
b^2 c^4 x^3 y^2 z  a^4 c^2 x^2 y^3 z + a^2 b^2 c^2 x^2 y^3 z +
a^2 c^4 x^2 y^3 z  a^4 c^2 x y^4 z + a^2 b^2 c^2 x y^4 z 
a^2 c^4 x y^4 z  b^4 c^2 x^4 z^2 + a^2 b^2 c^2 x^3 y z^2 +
b^4 c^2 x^3 y z^2  b^2 c^4 x^3 y z^2 + 6 a^2 b^2 c^2 x^2 y^2 z^2 +
a^4 c^2 x y^3 z^2 + a^2 b^2 c^2 x y^3 z^2  a^2 c^4 x y^3 z^2 
a^4 c^2 y^4 z^2  a^2 b^4 x^3 z^3 + b^6 x^3 z^3  b^4 c^2 x^3 z^3 
a^4 b^2 x^2 y z^3 + a^2 b^4 x^2 y z^3 + a^2 b^2 c^2 x^2 y z^3 +
a^4 b^2 x y^2 z^3  a^2 b^4 x y^2 z^3 + a^2 b^2 c^2 x y^2 z^3 +
a^6 y^3 z^3  a^4 b^2 y^3 z^3  a^4 c^2 y^3 z^3  a^2 b^4 x^2 z^4 
a^4 b^2 x y z^4  a^2 b^4 x y z^4 + a^2 b^2 c^2 x y z^4 
a^4 b^2 y^2 z^4 = 0
But if P lies on Circumcircle then the points A', B', C'
are collinear and
if P lies on Qx then the points A", B", C" are collinear.
Best regards
Nikos Dergiades  Dear Nikos
If P is lying on your Qx, what kind of circle is the pedal circle of
P, on which
should be lying three collinear points A",B",C"?
APH
On Sat, Mar 16, 2013 at 10:15 AM, Nikolaos Dergiades
<ndergiades@...> wrote:> Dear Angel,
> you wrote
>> Let ABC be a triangle, A'B'C' the pedal triangle of point P
>> and O' the
>> circumcenter of A'B'C'. Let A", B", C" be the symmetrical
>> points of A', B', C' w/r to the line PO'. The triangles ABC,
>> A"B"C" are orthologic if and only if P lies on Q068.
>
> I think you know that more precisely
> P lies on Linf + Circumcircle + Q068 + sextic
> Qx = b^2 c^4 x^4 y^2  a^2 c^4 x^3 y^3  b^2 c^4 x^3 y^3 + c^6 x^3 y^3 
> a^2 c^4 x^2 y^4 + a^2 b^2 c^2 x^4 y z  b^4 c^2 x^4 y z 
> b^2 c^4 x^4 y z + a^2 b^2 c^2 x^3 y^2 z  b^4 c^2 x^3 y^2 z +
> b^2 c^4 x^3 y^2 z  a^4 c^2 x^2 y^3 z + a^2 b^2 c^2 x^2 y^3 z +
> a^2 c^4 x^2 y^3 z  a^4 c^2 x y^4 z + a^2 b^2 c^2 x y^4 z 
> a^2 c^4 x y^4 z  b^4 c^2 x^4 z^2 + a^2 b^2 c^2 x^3 y z^2 +
> b^4 c^2 x^3 y z^2  b^2 c^4 x^3 y z^2 + 6 a^2 b^2 c^2 x^2 y^2 z^2 +
> a^4 c^2 x y^3 z^2 + a^2 b^2 c^2 x y^3 z^2  a^2 c^4 x y^3 z^2 
> a^4 c^2 y^4 z^2  a^2 b^4 x^3 z^3 + b^6 x^3 z^3  b^4 c^2 x^3 z^3 
> a^4 b^2 x^2 y z^3 + a^2 b^4 x^2 y z^3 + a^2 b^2 c^2 x^2 y z^3 +
> a^4 b^2 x y^2 z^3  a^2 b^4 x y^2 z^3 + a^2 b^2 c^2 x y^2 z^3 +
> a^6 y^3 z^3  a^4 b^2 y^3 z^3  a^4 c^2 y^3 z^3  a^2 b^4 x^2 z^4 
> a^4 b^2 x y z^4  a^2 b^4 x y z^4 + a^2 b^2 c^2 x y z^4 
> a^4 b^2 y^2 z^4 = 0
>
> But if P lies on Circumcircle then the points A', B', C'
> are collinear and
> if P lies on Qx then the points A", B", C" are collinear.
>
> Best regards
> Nikos Dergiades  Dear Antreas,
this Qx comes from the calculations
but I said nonsense because I thik it is
obvious that this sextic is valid only for
P = I.
This point I lies on Q068 but it must be
excluded because the line PO' is not defined.
There is only one line passing through I
such that the reflections A",B",C"
give a triangle orthologic with ABC.
Thank you Antreas.
Regards
Nikos
 Στις Σάβ., 16/03/13, ο/η Antreas Hatzipolakis <anopolis72@...> έγραψε:
> Από: Antreas Hatzipolakis <anopolis72@...>
> Θέμα: Re: [EMHL] Re: Orthologic Triangles  Locus
> Προς: Hyacinthos@yahoogroups.com
> Ημερομηνία: Σάββατο, 16 Μάρτιος 2013, 10:33
> Dear Nikos
>
> If P is lying on your Qx, what kind of circle is the pedal
> circle of
> P, on which
> should be lying three collinear points A",B",C"?
>
> APH
>
> On Sat, Mar 16, 2013 at 10:15 AM, Nikolaos Dergiades
> <ndergiades@...>
> wrote:
> > Dear Angel,
> > you wrote
> >> Let ABC be a triangle, A'B'C' the pedal triangle of
> point P
> >> and O' the
> >> circumcenter of A'B'C'. Let A", B", C" be the
> symmetrical
> >> points of A', B', C' w/r to the line PO'. The
> triangles ABC,
> >> A"B"C" are orthologic if and only if P lies
> on Q068.
> >
> > I think you know that more precisely
> > P lies on Linf + Circumcircle + Q068 + sextic
> > Qx = b^2 c^4 x^4 y^2  a^2 c^4 x^3 y^3  b^2 c^4 x^3
> y^3 + c^6 x^3 y^3 
> > a^2 c^4 x^2 y^4 + a^2 b^2 c^2 x^4 y z  b^4 c^2
> x^4 y z 
> > b^2 c^4 x^4 y z + a^2 b^2 c^2 x^3 y^2 z  b^4 c^2
> x^3 y^2 z +
> > b^2 c^4 x^3 y^2 z  a^4 c^2 x^2 y^3 z + a^2 b^2
> c^2 x^2 y^3 z +
> > a^2 c^4 x^2 y^3 z  a^4 c^2 x y^4 z + a^2 b^2 c^2
> x y^4 z 
> > a^2 c^4 x y^4 z  b^4 c^2 x^4 z^2 + a^2 b^2 c^2
> x^3 y z^2 +
> > b^4 c^2 x^3 y z^2  b^2 c^4 x^3 y z^2 + 6 a^2 b^2
> c^2 x^2 y^2 z^2 +
> > a^4 c^2 x y^3 z^2 + a^2 b^2 c^2 x y^3 z^2  a^2
> c^4 x y^3 z^2 
> > a^4 c^2 y^4 z^2  a^2 b^4 x^3 z^3 + b^6 x^3 z^3 
> b^4 c^2 x^3 z^3 
> > a^4 b^2 x^2 y z^3 + a^2 b^4 x^2 y z^3 + a^2 b^2
> c^2 x^2 y z^3 +
> > a^4 b^2 x y^2 z^3  a^2 b^4 x y^2 z^3 + a^2 b^2
> c^2 x y^2 z^3 +
> > a^6 y^3 z^3  a^4 b^2 y^3 z^3  a^4 c^2 y^3 z^3 
> a^2 b^4 x^2 z^4 
> > a^4 b^2 x y z^4  a^2 b^4 x y z^4 + a^2 b^2 c^2 x
> y z^4 
> > a^4 b^2 y^2 z^4 = 0
> >
> > But if P lies on Circumcircle then the points A', B',
> C'
> > are collinear and
> > if P lies on Qx then the points A", B", C" are
> collinear.
> >
> > Best regards
> > Nikos Dergiades
>
>
> 
>
> Yahoo! Groups Links
>
>
> Hyacinthosfullfeatured@yahoogroups.com
>
>  Sorry there are two lines.
ND
> Dear Antreas,
> this Qx comes from the calculations
> but I said nonsense because I thik it is
> obvious that this sextic is valid only for
> P = I.
> This point I lies on Q068 but it must be
> excluded because the line PO' is not defined.
> There is only one line passing through I
> such that the reflections A",B",C"
> give a triangle orthologic with ABC.
>
> Thank you Antreas.
> Regards
> Nikos
>
>
>  Στις Σάβ., 16/03/13, ο/η Antreas Hatzipolakis
> <anopolis72@...>
> έγραψε:
>
> > Από: Antreas Hatzipolakis <anopolis72@...>
> > Θέμα: Re: [EMHL] Re: Orthologic Triangles  Locus
> > Προς: Hyacinthos@yahoogroups.com
> > Ημερομηνία: Σάββατο, 16 Μάρτιος
> 2013, 10:33
> > Dear Nikos
> >
> > If P is lying on your Qx, what kind of circle is the
> pedal
> > circle of
> > P, on which
> > should be lying three collinear points A",B",C"?
> >
> > APH
> >
> > On Sat, Mar 16, 2013 at 10:15 AM, Nikolaos Dergiades
> > <ndergiades@...>
> > wrote:
> > > Dear Angel,
> > > you wrote
> > >> Let ABC be a triangle, A'B'C' the pedal
> triangle of
> > point P
> > >> and O' the
> > >> circumcenter of A'B'C'. Let A", B", C" be the
> > symmetrical
> > >> points of A', B', C' w/r to the line PO'. The
> > triangles ABC,
> > >> A"B"C" are orthologic if and only if P lies
> > on Q068.
> > >
> > > I think you know that more precisely
> > > P lies on Linf + Circumcircle + Q068 + sextic
> > > Qx = b^2 c^4 x^4 y^2  a^2 c^4 x^3 y^3  b^2 c^4
> x^3
> > y^3 + c^6 x^3 y^3 
> > > a^2 c^4 x^2 y^4 + a^2 b^2 c^2 x^4 y z  b^4 c^2
> > x^4 y z 
> > > b^2 c^4 x^4 y z + a^2 b^2 c^2 x^3 y^2 z  b^4
> c^2
> > x^3 y^2 z +
> > > b^2 c^4 x^3 y^2 z  a^4 c^2 x^2 y^3 z + a^2 b^2
> > c^2 x^2 y^3 z +
> > > a^2 c^4 x^2 y^3 z  a^4 c^2 x y^4 z + a^2 b^2
> c^2
> > x y^4 z 
> > > a^2 c^4 x y^4 z  b^4 c^2 x^4 z^2 + a^2 b^2 c^2
> > x^3 y z^2 +
> > > b^4 c^2 x^3 y z^2  b^2 c^4 x^3 y z^2 + 6 a^2
> b^2
> > c^2 x^2 y^2 z^2 +
> > > a^4 c^2 x y^3 z^2 + a^2 b^2 c^2 x y^3 z^2  a^2
> > c^4 x y^3 z^2 
> > > a^4 c^2 y^4 z^2  a^2 b^4 x^3 z^3 + b^6 x^3 z^3
> 
> > b^4 c^2 x^3 z^3 
> > > a^4 b^2 x^2 y z^3 + a^2 b^4 x^2 y z^3 + a^2 b^2
> > c^2 x^2 y z^3 +
> > > a^4 b^2 x y^2 z^3  a^2 b^4 x y^2 z^3 + a^2 b^2
> > c^2 x y^2 z^3 +
> > > a^6 y^3 z^3  a^4 b^2 y^3 z^3  a^4 c^2 y^3 z^3
> 
> > a^2 b^4 x^2 z^4 
> > > a^4 b^2 x y z^4  a^2 b^4 x y z^4 + a^2 b^2 c^2
> x
> > y z^4 
> > > a^4 b^2 y^2 z^4 = 0
> > >
> > > But if P lies on Circumcircle then the points A',
> B',
> > C'
> > > are collinear and
> > > if P lies on Qx then the points A", B", C" are
> > collinear.
> > >
> > > Best regards
> > > Nikos Dergiades
> >
> >
> > 
> >
> > Yahoo! Groups Links
> >
> >
> > Hyacinthosfullfeatured@yahoogroups.com
> >
> >
>
>
> 
>
> Yahoo! Groups Links
>
>
> Hyacinthosfullfeatured@yahoogroups.com
>
>   In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@...> wrote:
>
Made the correction:
>
> Writing correct geometric property of the quintic Q068 in the message #21755:
>
> Let ABC be a triangle, A'B'C' the pedal triangle of point P and O' the
> circumcenter of A'B'C'. Let A", B", C" be the symmetrical points of A', B', C' w/r to the line PO'. The triangles ABC, A"B"C" are orthologic if and only if P lies on Q068.
>
>
> A particular situation:
>
> If P=X(3) the orthology centers of ABC y A"B"C" (on NPC) are Q=X(113) and R, ninepointcircleantipode of X(3258).
>
>
> R= (2*a^8  a^4*(b^44*b^2*c^2+c^4) 2*a^6*(b^2+c^2) + (b^2c^2)^4 )*
> (a^6*(b^2 + c^2) + a^4*(3*b^4 + 2*b^2*c^2  3*c^4)+
> a^2*(3*b^6  2*b^4*c^2  2*b^2*c^4 + 3*c^6)
> (b^2  c^2)^2*(b^4 + 3*b^2*c^2 + c^4)) : .... : ....
>
>
>
> Angel M.
>
Let A", B", C" be the symmetrical points of A', B', C'
instead of:
Let A", B", C" be the symmetrical points of A, B, C
This is the corrected EPS:
http://groups.yahoo.com/group/Hyacinthos/files/Hyacinthos21754Acorrected.eps
Angel M.  Dear Antreas
2. Let ABC be a triangle, A'B'C' the pedal triangle of point P,
A*B*C* the circumcevian triangle of P wrt A'B'C' and A",B",C" the
second intersections of the circles (P,PA*), (P,PB*), (P,PC*) with
the pedal circle of P, resp.
The locus of P such that the triangles ABC, A"B"C" are orthologic is the septic Qnnn:
a^2 c^4 x^4 y^3 + b^2 c^4 x^4 y^3 + c^6 x^4 y^3 
a^2 c^4 x^3 y^4 + b^2 c^4 x^3 y^4  c^6 x^3 y^4 +
2 b^2 c^4 x^4 y^2 z  2 a^2 c^4 x^2 y^4 z 
2 b^4 c^2 x^4 y z^2  2 b^4 c^2 x^3 y^2 z^2 +
2 b^2 c^4 x^3 y^2 z^2 + 2 a^4 c^2 x^2 y^3 z^2 
2 a^2 c^4 x^2 y^3 z^2 + 2 a^4 c^2 x y^4 z^2 +
a^2 b^4 x^4 z^3  b^6 x^4 z^3  b^4 c^2 x^4 z^3  2 a^4 b^2 x^2
y^2 z^3 + 2 a^2 b^4 x^2 y^2 z^3 + a^6 y^4 z^3  a^4 b^2 y^4 z^3 +
a^4 c^2 y^4 z^3 + a^2 b^4 x^3 z^4 + b^6 x^3 z^4 
b^4 c^2 x^3 z^4 + 2 a^2 b^4 x^2 y z^4  2 a^4 b^2 x y^2 z^4 
a^6 y^3 z^4  a^4 b^2 y^3 z^4 + a^4 c^2 y^3 z^4=0,
(+ line at infinity (A'B'C' undefined triangle) + circumcircle (A', B' and C' points aligned) + a sextic ).
Points of the septic Qnnn:
A, B, C (triple points)
X(1) double , X(4)
excenters (which are double)
feet of the altitudes.
The sextic (without real points ???)is:
b^2 c^4 x^4 y^2  a^2 c^4 x^3 y^3  b^2 c^4 x^3 y^3 +
c^6 x^3 y^3  a^2 c^4 x^2 y^4 + a^2 b^2 c^2 x^4 y z 
b^4 c^2 x^4 y z  b^2 c^4 x^4 y z + a^2 b^2 c^2 x^3 y^2 z 
b^4 c^2 x^3 y^2 z + b^2 c^4 x^3 y^2 z  a^4 c^2 x^2 y^3 z +
a^2 b^2 c^2 x^2 y^3 z + a^2 c^4 x^2 y^3 z  a^4 c^2 x y^4 z +
a^2 b^2 c^2 x y^4 z  a^2 c^4 x y^4 z  b^4 c^2 x^4 z^2 +
a^2 b^2 c^2 x^3 y z^2 + b^4 c^2 x^3 y z^2  b^2 c^4 x^3 y z^2 +
6 a^2 b^2 c^2 x^2 y^2 z^2 + a^4 c^2 x y^3 z^2 +
a^2 b^2 c^2 x y^3 z^2  a^2 c^4 x y^3 z^2  a^4 c^2 y^4 z^2 
a^2 b^4 x^3 z^3 + b^6 x^3 z^3  b^4 c^2 x^3 z^3 
a^4 b^2 x^2 y z^3 + a^2 b^4 x^2 y z^3 + a^2 b^2 c^2 x^2 y z^3 +
a^4 b^2 x y^2 z^3  a^2 b^4 x y^2 z^3 + a^2 b^2 c^2 x y^2 z^3 +
a^6 y^3 z^3  a^4 b^2 y^3 z^3  a^4 c^2 y^3 z^3  a^2 b^4 x^2 z^4  a^4 b^2 x y z^4  a^2 b^4 x y z^4 + a^2 b^2 c^2 x y z^4 
a^4 b^2 y^2 z^4=0.
Best regards
Angel Montesdeoca
 In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> 1. Let ABC be a triangle, A'B'C' the pedal triangle of point P
> and A",B",C" the second intersections of the circles (P,PA'),
> (P,PB'), (P,PC') with the pedal circle of P, resp.
>
> Which is the locus of P such that the triangles ABC, A"B"C"
> are orthologic?
>
> O is on the locus.
>
> 2. Let ABC be a triangle, A'B'C' the pedal triangle of point P,
> A*B*C* the circumcevian triangle of P wrt A'B'C' and A",B",C" the
> second intersections of the circles (P,PA*), (P,PB*), (P,PC*) with
> the pedal circle of P, resp.
>
> Which is the locus of P such that the triangles ABC, A"B"C"
> are orthologic?
>
> H is on the locus.
>
> Figures:
> http://anthrakitis.blogspot.gr/2013/03/orthologictriangleslocus.html
>
> APH
>  They are orthologic (the orth. center (ABC, (B3C2, C1A3, A2B1)) is (X(74))For P = HWhich is the locus of P such that ABC and triangle bounded bySimilarly C1, A3 and A2, B1C2 = the reflection of Cb in PC'Cb = the orthogonal projection of C' on PB'B3 = the reflection of Bc in PB'Bc = the orthogonal projection of B' on PC'Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P.Denote:
(B3C2, C1A3, A2B1) are 1. orthologic 2. parallelogic?and parallelogic (the parallelogic center (ABC, (B3C2, C1A3, A2B1)) is X(110)APH  APHWhich is the other than O orthologic center?ABC, OaObOc are orthologic.Oa, Ob, Oc = the reflections of O in La, Lb, Lc resp.La, Lb, Lc = the reflections of the Euler line in BC,CA,AB, resp.Denote:Starting point:Let ABC be a triangle.(ie the orthologic center (OaObOc, ABC))Generalization:
Let ABC be a triangle and P a point.
Denote:
La, Lb, Lc = the reflections of the Euler line in BC,CA,AB, resp.
Pa, Pb, Pc = the reflections of P in La, Lb, Lc resp.
Which is the locus of P such that ABC, PaPbPc are orthologic ?  APHThe orthologic center (A1B1C1, A2B2C2) is N of A'B'C' [= O of ABC]A1B1C1, A2B2C2 are orthologic.Let A2B2C2 be the pedall triangle of O of A'B'C'Let ABC be a triangle and A'B'C' the antipedal triangle of I.That is:tiangle in order to get a point on the Euler line of ABC.This is interesting because we can take as reference triangle the orthicWhich point is this wrt tr. ABC and which wrt orthic tr. of ABC?Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C'are orthologic?Which is the locus of P such that A'B'C', A"B"C"A"B"C" = the pedal triangle of O'.O' = the circumcenter of A'B'C' ( = the center of the pedal circle of P)A'B'C' = the pedal triangle of PDenote:Let ABC be a triangle and P a point.On the locus:1. P = H (O' = N)
Orthologic center (A'B'C', A"B"C") = ?
2. P = Ο (O' = N)
Orthologic center (A"B"C", A'B'C') = the O of A'B'C' = the NPC center of ABC
Orthologic center (A'B'C', A"B"C") = ? on the Euler line of the orthic triangle(ABC is the pedal triangle of H of A'B'C'').Now:Let A1B1C1 be the pedal triangle of N of A'B'C' [ = O of ABC]
The orthologic center (A2B2C2, A1B1C1) lies on the Euler line
of ABC. Point ?  [Antreas]:APHThe orthologic center (A1B1C1, A2B2C2) is N of A'B'C' [= O of ABC]A1B1C1, A2B2C2 are orthologic.Let A2B2C2 be the pedall triangle of O of A'B'C'Let ABC be a triangle and A'B'C' the antipedal triangle of I.That is:tiangle in order to get a point on the Euler line of ABC.This is interesting because we can take as reference triangle the orthicWhich point is this wrt tr. ABC and which wrt orthic tr. of ABC?Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C'are orthologic?Which is the locus of P such that A'B'C', A"B"C"A"B"C" = the pedal triangle of O'.O' = the circumcenter of A'B'C' ( = the center of the pedal circle of P)A'B'C' = the pedal triangle of PDenote:Let ABC be a triangle and P a point.On the locus:1. P = H (O' = N)
Orthologic center (A'B'C', A"B"C") = ?
2. P = Ο (O' = N)
Orthologic center (A"B"C", A'B'C') = the O of A'B'C' = the NPC center of ABC
Orthologic center (A'B'C', A"B"C") = ? on the Euler line of the orthic triangle(ABC is the pedal triangle of H of A'B'C'').Now:Let A1B1C1 be the pedal triangle of N of A'B'C' [ = O of ABC]
The orthologic center (A2B2C2, A1B1C1) lies on the Euler line
of ABC. Point ?[Peter Moses]:>Which is the locus of P such that A'B'C', A"B"C" are orthologic?
Circumcircle, infinity, K003. McCay cubic1)Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C' = X(143) of ABC
Orthologic center (A'B'C', A"B"C") = on ABCs lines {{4,93},{6,24},{25,195},...} = X(79) of Orthic?2)Orthologic center (A'B'C', A"B"C") = X(1209) of ABC, X(3651) of Orthic. (probably!)Best regardsPeter  [Antreas]:Which point is this wrt tr. ABC and which wrt orthic tr. of ABC?Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C'are orthologic?Which is the locus of P such that A'B'C', A"B"C"A"B"C" = the pedal triangle of O'.O' = the circumcenter of A'B'C' ( = the center of the pedal circle of P)A'B'C' = the pedal triangle of PDenote:Let ABC be a triangle and P a point.On the locus:1. P = H (O' = N)
Orthologic center (A'B'C', A"B"C") = ?
2. P = Ο (O' = N)
Orthologic center (A"B"C", A'B'C') = the O of A'B'C' = the NPC center of ABC
Orthologic center (A'B'C', A"B"C") = ? on the Euler line of the orthic triangle[Peter Moses]:>Which is the locus of P such that A'B'C', A"B"C" are orthologic?
Circumcircle, infinity, K003. McCay cubic1)Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C' = X(143) of ABC
Orthologic center (A'B'C', A"B"C") = on ABCs lines {{4,93},{6,24},{25,195},...} = X(79) of Orthic?2)Orthologic center (A'B'C', A"B"C") = X(1209) of ABC, X(3651) of Orthic. (probably!)Best regardsPeter
**************************Thanks, Peter,What a surprise! A new point in such a simple construction !!!!I wouldn't bet a cent for it as a new point :)Description of the point:Let ABC be a triangle and A'B'C', A"B"C" the pedal triangles of H,N, resp.A'B'C', A"B"C" are orthologic with orthologic center (A'B'C', A"B"C")on {4,93},{6,24},{25,195},.. lines of ABC.If you compute other "things" of the point (coordinates etc) please let me know(to inform Clark)Season's GreetingsAPH
 The midpoint of the line segment joining the orthologic centers (AhBhCh, AnBnCn) and (AoBoCo, AnBnCn) is the orthocenter of AnBnCn.of H,O,N respLet ABC be a triangle and AhBhCh, AoBoCo, AnBnCn the pedal trianglesBut let's first rename the pedal triangles.Another observation in the same configuration.***************************************[Antreas]:Which point is this wrt tr. ABC and which wrt orthic tr. of ABC?Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C'are orthologic?Which is the locus of P such that A'B'C', A"B"C"A"B"C" = the pedal triangle of O'.O' = the circumcenter of A'B'C' ( = the center of the pedal circle of P)A'B'C' = the pedal triangle of PDenote:Let ABC be a triangle and P a point.On the locus:1. P = H (O' = N)
Orthologic center (A'B'C', A"B"C") = ?
2. P = Ο (O' = N)
Orthologic center (A"B"C", A'B'C') = the O of A'B'C' = the NPC center of ABC
Orthologic center (A'B'C', A"B"C") = ? on the Euler line of the orthic triangle[Peter Moses]:>Which is the locus of P such that A'B'C', A"B"C" are orthologic?
Circumcircle, infinity, K003. McCay cubic1)Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C' = X(143) of ABC
Orthologic center (A'B'C', A"B"C") = on ABCs lines {{4,93},{6,24},{25,195},...} = X(79) of Orthic?2)Orthologic center (A'B'C', A"B"C") = X(1209) of ABC, X(3651) of Orthic. (probably!)Best regardsPeterWhich is this point (the orthocenter of the pedal triangle of N)?And is it a general property of all (pairs of isogonal conj.) points on the locus (the McCay cubic)?ie that, that midpoint is the orthocenter of the pedal triangle of the center of the common pedal circle of the isog. conj. points?(Or at least is it lying on the Euler line of the triangle)?Hmmm....... neither I understand at first glance what I wrote ("the orthocenter
of the pedal triangle of the center of the common pedal circle of the isog. conj.
points") ! :). So let me rewrit itLet ABC be a triangle and P,P* two isogonal conjugate points on the McCay
cubic and let Q be the center of the common pedal circle of P,P* (= midpoint
of PP*)Denote:T1, T2, T3 = the pedal triangles of P, P*, Q, resp.Let Mp be the midpoint of the line segment joining the orthologcic centers (T1, T3) and (T2, T3).Is Mp the orthocenter of T3 for all P's (on the cubic)?Or in general, is it lying on the Euler line of T3 (orthocenter or not) ?Which is its locus as P moves on the McCay cubic?APH  From: Randy Hutson
Dear Antreas and Peter,
This point appears in César E. Lozada's PerspectiveOrthologicParallelogic.pdf as the orthologic center of orthic and reflection triangles, also reflection of X(54) in X(973), with trilinears a*((S^2SA^2)*(SWR^2)+2*S^2*SA)/SA : : and search value 12.481825032289.
This point is also the orthic isogonal conjugate of X(1594), and X(79) of orthic IF ABC is acute.
Best regards,
Randy Dear Antreas and Peter,
This point appears in César E. Lozada's PerspectiveOrthologicParallelogic.pdf as the orthologic center of orthic and reflection triangles, also reflection of X(54) in X(973), with trilinears a*((S^2SA^2)*(SWR^2)+2*S^2*SA)/SA : : and search value 12.481825032289.
This point is also the orthic isogonal conjugate of X(1594), and X(79) of orthic IF ABC is acute.
Best regards,
Randy********************
Thanks, Randy!!Yes!!!! since the pedal trangle of N and the reflection triangles are homotheticCesar's table published in Dec 2013
http://www.marveloustriangles.blogspot.gr/2014/12/sometables.htmland later also appeared the point in question in the paper
Jesus Torres, The triangle of reflections,Forum Geometricorum, 14 (2014) 265294 [in Theorem 4.3].(published in 7 Oct. 2014)APH [Antreas]:Which point is this wrt tr. ABC and which wrt orthic tr. of ABC?Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C'are orthologic?Which is the locus of P such that A'B'C', A"B"C"A"B"C" = the pedal triangle of O'.O' = the circumcenter of A'B'C' ( = the center of the pedal circle of P)A'B'C' = the pedal triangle of PDenote:Let ABC be a triangle and P a point.On the locus:1. P = H (O' = N)
Orthologic center (A'B'C', A"B"C") = ?
2. P = Ο (O' = N)
Orthologic center (A"B"C", A'B'C') = the O of A'B'C' = the NPC center of ABC
Orthologic center (A'B'C', A"B"C") = ? on the Euler line of the orthic triangle[Peter Moses]:>Which is the locus of P such that A'B'C', A"B"C" are orthologic?
Circumcircle, infinity, K003. McCay cubic1)Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C' = X(143) of ABC
Orthologic center (A'B'C', A"B"C") = on ABCs lines {{4,93},{6,24},{25,195},...} = X(79) of Orthic?2)Orthologic center (A'B'C', A"B"C") = X(1209) of ABC, X(3651) of Orthic. (probably!)Best regardsPeter
**************************Thanks, Peter,What a surprise! A new point in such a simple construction !!!!I wouldn't bet a cent for it as a new point :)Description of the point:Let ABC be a triangle and A'B'C', A"B"C" the pedal triangles of H,N, resp.A'B'C', A"B"C" are orthologic with orthologic center (A'B'C', A"B"C")on {4,93},{6,24},{25,195},.. lines of ABC.If you compute other "things" of the point (coordinates etc) please let me know(to inform Clark)Season's GreetingsAPH
[Peter Moses]:
A few extra properties of the point:Barys: a^2 (a^2+b^2c^2) (a^2b^2+c^2) (a^42 a^2 b^2+b^42 a^2 c^2b^2 c^2+c^4) (a^4 b^22 a^2 b^4+b^6+a^4 c^22 a^2 b^2 c^2b^4 c^22 a^2 c^4b^2 c^4+c^6)::On lines {{4,93},{6,24},{25,195},{49,143},{52,539},{70,6145},{113,5446},{155,3060},{403,3574},{648,1179},{1112,2914},{1209,1216},{1843,5965},{1986,3575}}.Reflection of X(i) in X(j) for these {i,j}: {54,973},{1493,143},{2914,1112}.3 X[1209]  2 X[1216].3 X[54]  5 X[3567].6 X[973]  5 X[3567].X(4)ceva conjugate of X(1594).X(4)crosspoint of X(3518).X(3)crosssum of X(3519).X(2216)isoconjugate of X(3519).Trilinear product X(1594) X(2964).Barycentric product X(1594) X(1994).X(79) Orthic triangle.{{1,21},{7,79},...} of Tangential triangle.Best regards,Peter.