- 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

> >

> >

>

>

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

>

>

>

> ------------------------------------

>

> Yahoo! Groups Links

>

>

> Hyacinthos-fullfeatured@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 + (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

________________________________________

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 + (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]

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

Yahoo! Groups Links - 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* 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]