## Re: [EMHL] Happy New Year

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• 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
Message 1 of 14 , Jan 2, 2013
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
> >
> >
>
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
• Dear Nikos, Paul and friends, ... 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
Message 2 of 14 , Jan 2, 2013
Dear Nikos, Paul and friends,

> [ND] 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'.

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

pole Ω = (SA k + (2-k)Sw) / (SA + Sw) : : ,

pivot S = 1 / (SA + Sw) : : , on the Kiepert hyperbola,

isopivot : SA k + (2-k)Sw : : , on the line X2-X69.

When k varies, the locus of Ω is a line through X2,

When w varies, the locus of Ω is a nodal circum-cubic with node X2, the nodal tangents being X2-X6 and Euler line.

This cubic is a member of the pencil generated by :
the three medians (k = 1)
Kiepert + Euler (k = 0)
line X2-X6 + isotomic of Euler (k = 2).

----------------------------------------------------------

The locus of X is another pK(Ω', S') with

pole Ω' = 1 / [(SA + Sw)(SA k + (2-k)Sw))] : : ,

pivot S' = (SA k + Sw) / [(SA + Sw)(SA k + (2-k)Sw))] : : ,

isopivot : 1 / (SA k + Sw) : : , on the Kiepert hyperbola.

Best regards and Happy New Year to you all.

Bernard

[Non-text 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
Message 3 of 14 , Jan 2, 2013
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'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'.

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

pole Ω = (SA k + (2-k)Sw) / (SA + Sw) : : ,

pivot S = 1 / (SA + Sw) : : , on the Kiepert hyperbola,

isopivot : SA k + (2-k)Sw : : , on the line X2-X69.

When k varies, the locus of Ω is a line through X2,

When w varies, the locus of Ω is a nodal circum-cubic with node X2, the nodal tangents being X2-X6 and Euler line.

This cubic is a member of the pencil generated by :
the three medians (k = 1)
Kiepert + Euler (k = 0)
line X2-X6 + isotomic of Euler (k = 2).

----------------------------------------------------------

The locus of X is another pK(Ω', S') with

pole Ω' = 1 / [(SA + Sw)(SA k + (2-k)Sw))] : : ,

pivot S' = (SA k + Sw) / [(SA + Sw)(SA k + (2-k)Sw))] : : ,

isopivot : 1 / (SA k + Sw) : : , on the Kiepert hyperbola.

Best regards and Happy New Year to you all.

Bernard

[Non-text portions of this message have been removed]

------------------------------------

• Happy New year 2013 . From Sevilla  Ricardo http://personal.us.es/rbarroso/trianguloscabri/ [Non-text portions of this message have been removed]
Message 4 of 14 , Jan 2, 2013
Happy New year 2013 .

From Sevilla

Ricardo

http://personal.us.es/rbarroso/trianguloscabri/

[Non-text 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*
Message 5 of 14 , Jan 2, 2013
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)k-4 SA Sw : : , on the line X4-X69.

If k = 0, T = X69 and this is the Lucas cubic.

Best regards

Bernard

[Non-text portions of this message have been removed]
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