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• 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
Mar 15 1 of 9
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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, nine-point-circle-antipode of X(3258).

R= (2*a^8 - a^4*(b^4-4*b^2*c^2+c^4)- 2*a^6*(b^2+c^2) + (b^2-c^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.pagesperso-orange.fr/relatedcurves.html
>
>
> It is a result does not appear between the propiedades of Q068 in http://bernard.gibert.pagesperso-orange.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 ... 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
Mar 16 1 of 9
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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 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
Mar 16 1 of 9
View Source
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
Mar 16 1 of 9
View Source
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
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
• Sorry there are two lines. ND
Mar 16 1 of 9
View Source
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
> >
> >
> >     Hyacinthos-fullfeatured@yahoogroups.com
> >
> >
>
>
> ------------------------------------
>
> Yahoo! Groups Links
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
• ... Made the correction: 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
Mar 16 1 of 9
View Source
--- In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@...> wrote:
>
>
> 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, nine-point-circle-antipode of X(3258).
>
>
> R= (2*a^8 - a^4*(b^4-4*b^2*c^2+c^4)- 2*a^6*(b^2+c^2) + (b^2-c^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.
>

Made the correction:

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
Mar 16 1 of 9
View Source
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/orthologic-triangles-locus.html
>
> APH
>
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