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Happy New Year
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 Dear Antreas
Happy New Year to you
and all friends of Hyacinthos.
A locus for the begining of 2013.
If A'B'C' is the cevian triangle of P
and A'BA1, CA'A2 are equilateral triangles
and A* is the mid point of A1, A2 in the
opposite side of BC ralative to A,
and similarly define B*, C* what is the locus
of P such that ABC, A*B*C* are perspective.
Best regards
Nikos Dergiades  Dear Nik and Antreas,
Happy New Year to you and all friends of Hyacinthos.
ND: If A'B'C' is the cevian triangle of P and A'BA1, CA'A2 are equilateral triangles and A* is the mid point of A1, A2 in the
opposite side of BC ralative to A, and similarly define B*, C*
what is the locus of P such that ABC, A*B*C* are perspective?
***
If A_0BC, B_0CA, C_0AB are the equilateral triangles constructed externally on the sides BC, CA, AB respectively, your A* is the midpoint of A'A_0. If P = (u:v:W) in homogeneous barycentric coordinates, then
A* = ((S_B+S_C)(v+w) : (S_C+S/Sqrt[3])v + (S_C+S Sqrt[3])w
: (S_B+S/Sqrt[3])v + (S_B+S Sqrt[3])w),
and similarly for B* and C*.
A*B*C* is perspective with ABC if and only if
Cyclic sum [(S_B+S/Sqrt[3])(S_C+S Sqrt[3]) uv^2
 (S_B+S Sqrt[3])(S_C+S/Sqrt[3]) uw^2]= 0.
Rewriting this equation as
Cyclic sum u/(S_A+S/Sqrt[3])[(S_C+S Sqrt[3])/(S_C+S/Sqrt[3]) v^2
 (S_B+S Sqrt[3])/(S_B+S/Sqrt[3])w^2] = 0 ,
we identify the locus of P as the isocubic K(Q,X(13)), where
Q = ((S_A+S Sqrt[3])/(S_A+S/Sqrt[3]) : ... : ...)
has (6913) search number 0.245203113787
Best regards
Sincerely
Paul  Dear Paul,
we have the same result as for A* . . .
but I am impressed with the factorization
you had in the equation of the cubic.
Happy New Year.
Best regards
Nik
>
> ND: If A'B'C' is the cevian triangle of P and A'BA1, CA'A2
> are equilateral triangles and A* is the mid point of A1, A2
> in the
> opposite side of BC ralative to A, and similarly define B*,
> C*
> what is the locus of P such that ABC, A*B*C* are
> perspective?
>
> ***
>
> If A_0BC, B_0CA, C_0AB are the equilateral triangles
> constructed externally on the sides BC, CA, AB respectively,
> your A* is the midpoint of A'A_0. If P = (u:v:W)
> in homogeneous barycentric coordinates, then
>
> A* = ((S_B+S_C)(v+w) : (S_C+S/Sqrt[3])v + (S_C+S Sqrt[3])w
>
> : (S_B+S/Sqrt[3])v + (S_B+S
> Sqrt[3])w),
> and similarly for B* and C*.
>
> A*B*C* is perspective with ABC if and only if
>
> Cyclic sum [(S_B+S/Sqrt[3])(S_C+S Sqrt[3]) uv^2
>  (S_B+S
> Sqrt[3])(S_C+S/Sqrt[3]) uw^2]= 0.
>
> Rewriting this equation as
>
> Cyclic sum u/(S_A+S/Sqrt[3])[(S_C+S
> Sqrt[3])/(S_C+S/Sqrt[3]) v^2
>  (S_B+S
> Sqrt[3])/(S_B+S/Sqrt[3])w^2] = 0 ,
> we identify the locus of P as the isocubic K(Q,X(13)), where
>
>
> Q = ((S_A+S Sqrt[3])/(S_A+S/Sqrt[3]) : ... : ...)
>
> has (6913) search number 0.245203113787…
>
> Best regards
> Sincerely
> Paul
>
>
>
>
> 
>
> Yahoo! Groups Links
>
>
> Hyacinthosfullfeatured@yahoogroups.com
>
>  Dear Nikos, I also did the calculations but when I was about to post them, I felt gladly overwhelmed by the impressive answer by Paul.
Happy New Year to all Hyacinthos members!
Francisco Javier.
 In Hyacinthos@yahoogroups.com, Nikolaos Dergiades <ndergiades@...> wrote:
>
> Dear Paul,
> we have the same result as for A* . . .
> but I am impressed with the factorization
> you had in the equation of the cubic.
> Happy New Year.
> Best regards
> Nik
>
>
> >
> > ND: If A'B'C' is the cevian triangle of P and A'BA1, CA'A2
> > are equilateral triangles and A* is the mid point of A1, A2
> > in the
> > opposite side of BC ralative to A, and similarly define B*,
> > C*
> > what is the locus of P such that ABC, A*B*C* are
> > perspective?
> >
> > ***
> >
> > If A_0BC, B_0CA, C_0AB are the equilateral triangles
> > constructed externally on the sides BC, CA, AB respectively,
> > your A* is the midpoint of A'A_0.Â IfÂ P = (u:v:W)
> > in homogeneous barycentric coordinates, then
> >
> > A* = ((S_B+S_C)(v+w) : (S_C+S/Sqrt[3])v + (S_C+S Sqrt[3])w
> >
> > Â Â Â Â : (S_B+S/Sqrt[3])v + (S_B+S
> > Sqrt[3])w),
> > andÂ similarly for B* and C*.
> >
> > A*B*C* is perspective with ABC if and only if
> >
> > Cyclic sum [(S_B+S/Sqrt[3])(S_C+S Sqrt[3]) uv^2
> > Â Â Â Â Â Â Â  (S_B+S
> > Sqrt[3])(S_C+S/Sqrt[3]) uw^2]= 0.
> >
> > Rewriting this equation as
> >
> > Cyclic sumÂ u/(S_A+S/Sqrt[3])[(S_C+S
> > Sqrt[3])/(S_C+S/Sqrt[3]) v^2
> > Â Â Â Â Â  (S_B+S
> > Sqrt[3])/(S_B+S/Sqrt[3])w^2] = 0 ,
> > we identify the locus of P as the isocubic K(Q,X(13)), where
> >
> >
> > Q = ((S_A+S Sqrt[3])/(S_A+S/Sqrt[3])Â : ... : ...)
> >
> > has (6913) search number 0.245203113787â¦
> >
> > Best regards
> > Sincerely
> > Paul
> >
> >
> >
> >
> > 
> >
> > Yahoo! Groups Links
> >
> >
> > Â Â Hyacinthosfullfeatured@yahoogroups.com
> >
> >
>  Dear Nik and Francisco Javier,
ND: If A'B'C' is the cevian triangle of P and A'BA1, CA'A2
are equilateral triangles and A* is the mid point of A1, A2
in the opposite side of BC ralative to A, and similarly define B*,
C*, what is the locus of P such that ABC, A*B*C* are perspective?
PY: If A_0BC, B_0CA, C_0AB are the equilateral triangles
constructed externally on the sides BC, CA, AB respectively,
... A*B*C* is perspective with ABC if and only if P [lies on]
the isocubic K(Q,X(13)), where
Q = ((S_A+S Sqrt[3])/(S_A+S/Sqrt[3]) : ... : ...)
has (6913) search number 0.245203113787â¦
***
I think we can go a little bit further and get a Euclidean constructible example. The point $Q$ can be regarded as the isogonal conjugate of the barycentric product of $X(15)$ (isodynamic point) and $X(17)$ (Napoleon point):
$$
Q = (X(15)*X(17))^*.
$$
It is an interior point of triangle $ABC$ if all angles are less than $120^\circ$.
In this case, the barycentric square root $\sqrt Q$ is a point on the locus. With $P = \sqrt Q$, the perspector is the barycentric square root of
$$
(1/((S_A+S \sqrt 3)(S_A+S/\sqrt 3)) : ... : ...)
= (1/(S_A+S \sqrt 3)^2 : ... : ...)*
((S_A+S Sqrt[3])/(S_A+S/Sqrt[3]) : ... : ...)
= (1/(S_A+S \sqrt 3)^2 : ... : ...)*Q,
$$
and is the barycentric product $X(17)*P$.
Best regards
Sincerely
Paul  Dear Nikos, Paul
Happy New Year!!!
1. We can take A'B'C' as the pedal triangle of P and ask for the locus.
2. We can take A*,B*,C* as triangle centers (Circumcenters, Orthocenters
etc) of
the triangles A'A1A2, B'B1B2, C'C1C2 and ask of the loci of P such taht
ABC,A*B*C* be perspective.
I think that interesting is the case of the circumcenters.
Season's Greetings
Antreas
On Tue, Jan 1, 2013 at 2:15 AM, Nikolaos Dergiades <ndergiades@...>wrote:
> **
>
>
> Dear Antreas
> Happy New Year to you
> and all friends of Hyacinthos.
>
> A locus for the begining of 2013.
> If A'B'C' is the cevian triangle of P
> and A'BA1, CA'A2 are equilateral triangles
> and A* is the mid point of A1, A2 in the
> opposite side of BC ralative to A,
> and similarly define B*, C* what is the locus
> of P such that ABC, A*B*C* are perspective.
> Best regards
> Nikos Dergiades
>
>
[Nontext portions of this message have been removed]  Dear Paul,
A little generalization:
If A'B'C' is the cevian triangle of a point P
and A"BC, B"CA, C"AB are similar isosceles triangles
outside of ABC with base agnle w then the triangles
ABC, A*B*C* where A* is the mid point of A'A" . . .
are perspective if and only if the locus of P
is the isocubic with pivot S = (1/(S_A+S_w):. . :. .)
and pole Q = ((S_A+3S_w)/(S_A+S_w):..:..).
The point S lies on the Kiepert hyperbola
and corresponds to A", B", C".
The point S'= ((1/(S_A+3S_w):..:..)) lies also
on the Kiepert hyperbola and corresponds to the
Centroids of triangles A"BC, B"CA, C"AB.
The point Q is the barycentric product of S
and the isotomic of S'.
Best regards
Nik
> I think we can go a little bit further and get a Euclidean
> constructible example. The point $Q$ can be regarded as the
> isogonal conjugate of the barycentric product of $X(15)$
> (isodynamic point) and $X(17)$ (Napoleon point):
> $$
> Q = (X(15)*X(17))^*.
> $$
> It is an interior point of triangle $ABC$ if all angles are
> less than $120^\circ$.
>
> In this case, the barycentric square root $\sqrt Q$ is a
> point on the locus. With $P = \sqrt Q$, the perspector is
> the barycentric square root of
> $$
> (1/((S_A+S \sqrt 3)(S_A+S/\sqrt 3)) : ... : ...)
> = (1/(S_A+S \sqrt 3)^2 : ... : ...)*
> ((S_A+S Sqrt[3])/(S_A+S/Sqrt[3]) : ... : ...)
> = (1/(S_A+S \sqrt 3)^2 : ... : ...)*Q,
> $$
> and is the barycentric product $X(17)*P$.
>
> Best regards
> Sincerely
> Paul
>
>
>
>
> 
>
> Yahoo! Groups Links
>
>
> Hyacinthosfullfeatured@yahoogroups.com
>
>  Dear Antreas,
if A'B'C' is the pedal triangle and not
the cevian then the locus is a cubic
but I think complicated. The same holds
if we take circumenters.
I think that the most interesting case is:
If A'B'C' is the cevian triangle of a point P,
A1A'B, A2CA'are similar isosceles triangles
outside of ABC with base angle w,
(the bases are opposite of A1 and A2)
A* is the centroid of triangle A'A1A2
and similarly define B*, C*
then the triangles ABC, A*B*C* are perspective
if and only if the locus of P
is the isocubic with pivot S = (1/(S_A+S_w):. . :. .)
and pole Q = ((S_A+5S_w)/(S_A+S_w):..:..).
Best regards
Nikos
> Dear Nikos, Paul
>
> Happy New Year!!!
>
> 1. We can take A'B'C' as the pedal triangle of P and ask for
> the locus.
>
> 2. We can take A*,B*,C* as triangle centers (Circumcenters,
> Orthocenters
> etc) of
> the triangles A'A1A2, B'B1B2, C'C1C2 and ask of the loci of
> P such taht
> ABC,A*B*C* be perspective.
>
> I think that interesting is the case of the circumcenters.
>
> Season's Greetings
>
> Antreas
>
> On Tue, Jan 1, 2013 at 2:15 AM, Nikolaos Dergiades <ndergiades@...>wrote:
>
> > **
> >
> >
> > Dear Antreas
> > Happy New Year to you
> > and all friends of Hyacinthos.
> >
> > A locus for the begining of 2013.
> > If A'B'C' is the cevian triangle of P
> > and A'BA1, CA'A2 are equilateral triangles
> > and A* is the mid point of A1, A2 in the
> > opposite side of BC ralative to A,
> > and similarly define B*, C* what is the locus
> > of P such that ABC, A*B*C* are perspective.
> > Best regards
> > Nikos Dergiades
> >
> >
>
>
> [Nontext portions of this message have been removed]
>
>
>
> 
>
> Yahoo! Groups Links
>
>
> Hyacinthosfullfeatured@yahoogroups.com
>
>  Dear Nikos, Paul and friends,
> [ND] A little generalization:
If A* is the homothetic of A" under h(A', k), B* and C* likewise, then the locus of P such that ABC and A*B*C* are perspective (at X) is pK(Ω, S) with
> If A'B'C' is the cevian triangle of a point P
> and A"BC, B"CA, C"AB are similar isosceles triangles
> outside of ABC with base agnle w then the triangles
> ABC, A*B*C* where A* is the mid point of A'A" . . .
> are perspective if and only if the locus of P
> is the isocubic with pivot S = (1/(S_A+S_w):. . :. .)
> and pole Q = ((S_A+3S_w)/(S_A+S_w):..:..).
> The point S lies on the Kiepert hyperbola
> and corresponds to A", B", C".
> The point S'= ((1/(S_A+3S_w):..:..)) lies also
> on the Kiepert hyperbola and corresponds to the
> Centroids of triangles A"BC, B"CA, C"AB.
> The point Q is the barycentric product of S
> and the isotomic of S'.
pole Ω = (SA k + (2k)Sw) / (SA + Sw) : : ,
pivot S = 1 / (SA + Sw) : : , on the Kiepert hyperbola,
isopivot : SA k + (2k)Sw : : , on the line X2X69.
When k varies, the locus of Ω is a line through X2,
When w varies, the locus of Ω is a nodal circumcubic with node X2, the nodal tangents being X2X6 and Euler line.
This cubic is a member of the pencil generated by :
the three medians (k = 1)
Kiepert + Euler (k = 0)
line X2X6 + isotomic of Euler (k = 2).

The locus of X is another pK(Ω', S') with
pole Ω' = 1 / [(SA + Sw)(SA k + (2k)Sw))] : : ,
pivot S' = (SA k + Sw) / [(SA + Sw)(SA k + (2k)Sw))] : : ,
isopivot : 1 / (SA k + Sw) : : , on the Kiepert hyperbola.
Best regards and Happy New Year to you all.
Bernard
[Nontext portions of this message have been removed]  Dear Nik, Bernard, and Francisco Javier,
Thank you for these generalizations. FJ, I did not realize that the locus contains G!
Best regards
Sincerely
Paul
________________________________________
From: Hyacinthos@yahoogroups.com [Hyacinthos@yahoogroups.com] on behalf of Bernard Gibert [bg42@...]
Sent: Wednesday, January 02, 2013 8:00 AM
To: Hyacinthos@yahoogroups.com
Subject: Re: [EMHL] Re: Happy New Year and Nik's locus for 2013
Dear Nikos, Paul and friends,
> [ND] A little generalization:
If A* is the homothetic of A" under h(A', k), B* and C* likewise, then the locus of P such that ABC and A*B*C* are perspective (at X) is pK(Ω, S) with
> If A'B'C' is the cevian triangle of a point P
> and A"BC, B"CA, C"AB are similar isosceles triangles
> outside of ABC with base agnle w then the triangles
> ABC, A*B*C* where A* is the mid point of A'A" . . .
> are perspective if and only if the locus of P
> is the isocubic with pivot S = (1/(S_A+S_w):. . :. .)
> and pole Q = ((S_A+3S_w)/(S_A+S_w):..:..).
> The point S lies on the Kiepert hyperbola
> and corresponds to A", B", C".
> The point S'= ((1/(S_A+3S_w):..:..)) lies also
> on the Kiepert hyperbola and corresponds to the
> Centroids of triangles A"BC, B"CA, C"AB.
> The point Q is the barycentric product of S
> and the isotomic of S'.
pole Ω = (SA k + (2k)Sw) / (SA + Sw) : : ,
pivot S = 1 / (SA + Sw) : : , on the Kiepert hyperbola,
isopivot : SA k + (2k)Sw : : , on the line X2X69.
When k varies, the locus of Ω is a line through X2,
When w varies, the locus of Ω is a nodal circumcubic with node X2, the nodal tangents being X2X6 and Euler line.
This cubic is a member of the pencil generated by :
the three medians (k = 1)
Kiepert + Euler (k = 0)
line X2X6 + isotomic of Euler (k = 2).

The locus of X is another pK(Ω', S') with
pole Ω' = 1 / [(SA + Sw)(SA k + (2k)Sw))] : : ,
pivot S' = (SA k + Sw) / [(SA + Sw)(SA k + (2k)Sw))] : : ,
isopivot : 1 / (SA k + Sw) : : , on the Kiepert hyperbola.
Best regards and Happy New Year to you all.
Bernard
[Nontext portions of this message have been removed]

Yahoo! Groups Links  Happy New year 2013 .
From Sevilla
Ricardo
http://personal.us.es/rbarroso/trianguloscabri/
[Nontext portions of this message have been removed]  Dear all,
If A'B'C' is the pedal triangle of P (instead of cevian triangle of P), and A", B", C" as before,
If A* is the homothetic of A" under h(A', k), B* and C* likewise, then the locus of P such that ABC and A*B*C* are perspective (at X) is a central cubic passing through X3 with asymptotes parallel to the altitudes of ABC and concurring on the Brocard axis, very similar to the Darboux cubic which is obtained when k = 0.
The locus of the perspector X is pK(X2, T) where
T = (a^4+b^4+c^4)k4 SA Sw : : , on the line X4X69.
If k = 0, T = X69 and this is the Lucas cubic.
Best regards
Bernard
[Nontext portions of this message have been removed]
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