- Dear All My Friends,

We use following notations:

Y*Z = barycentric product of Y and Z

Crd(Y,Z) = Crossdifference of Y and Z

1/Z = isotomic conjugate of Z

Y/Z = Y*(1/Z)

Given two fixed points with barycentrics P = (p : q : r) and U = (u : v : w). X is variable point with barycentrics X = (x : y : z). We construct point Y as:

Y = P*Crd(P,X)/Crd(U,X) and denote the transform as Y = fPU(X)

Barycentrics of Y = fPU(X) as following:

p*(q*z - r*y)/(v*z - w*y) : q*(r*x - p*z)/(w*x - u*z) : r*(p*y - q*x)/(u*y - v*x)

The transform Y = fPU(X) is conjugation: fPU(fPU(X)) = X

Are there somethings interesting from this conjugation?

Best regards,

Bui Quang Tuan - Dear Tuan,

> We use following notations:

This is similar to the Cundy-Parry transformations in

> Y*Z = barycentric product of Y and Z

> Crd(Y,Z) = Crossdifference of Y and Z

> 1/Z = isotomic conjugate of Z

> Y/Z = Y*(1/Z)

>

> Given two fixed points with barycentrics P = (p : q : r) and U = (u : v : w). X is variable point with barycentrics X = (x : y : z). We construct point Y as:

> Y = P*Crd(P,X)/Crd(U,X) and denote the transform as Y = fPU(X)

>

> Barycentrics of Y = fPU(X) as following:

>

> p*(q*z - r*y)/(v*z - w*y) : q*(r*x - p*z)/(w*x - u*z) : r*(p*y - q*x)/(u*y - v*x)

>

> The transform Y = fPU(X) is conjugation: fPU(fPU(X)) = X

>

> Are there somethings interesting from this conjugation?

http://pagesperso-orange.fr/bernard.gibert/Classes/cl037.html

your Y is the intersection of the lines UX and PX* where X* is the homologue of X in the isoconjugation that swaps P and U.

Best regards

Bernard

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

Thank you very much! You are right!

I see following fact but not sure if is it true:

If pK is any pK cubic and U is any point of pK then we can always find unique P on pK such that fPU(pK)=pK

Best regard,

Bui Quang Tuan

--- On Wed, 7/14/10, Bernard Gibert <bg42@...> wrote:

> From: Bernard Gibert <bg42@...>

> Subject: Re: [EMHL] A Conjugation Based On Crossdifference

> To: Hyacinthos@yahoogroups.com

> Date: Wednesday, July 14, 2010, 6:49 PM

> Dear Tuan,

>

> > We use following notations:

> > Y*Z = barycentric product of Y and Z

> > Crd(Y,Z) = Crossdifference of Y and Z

> > 1/Z = isotomic conjugate of Z

> > Y/Z = Y*(1/Z)

> >

> > Given two fixed points with barycentrics P = (p : q :

> r) and U = (u : v : w). X is variable point with

> barycentrics X = (x : y : z). We construct point Y as:

> > Y = P*Crd(P,X)/Crd(U,X) and denote the transform as Y

> = fPU(X)

> >

> > Barycentrics of Y = fPU(X) as following:

> >

> > p*(q*z - r*y)/(v*z - w*y) : q*(r*x - p*z)/(w*x - u*z)

> : r*(p*y - q*x)/(u*y - v*x)

> >

> > The transform Y = fPU(X) is conjugation: fPU(fPU(X)) =

> X

> >

> > Are there somethings interesting from this

> conjugation?

>

> This is similar to the Cundy-Parry transformations in

>

> http://pagesperso-orange.fr/bernard.gibert/Classes/cl037.html

>

> your Y is the intersection of the lines UX and PX*

> where X* is the homologue of X in the isoconjugation that

> swaps P and U.

>

> Best regards

>

> Bernard

> - Dear Tuan,

> If pK is any pK cubic and U is any point of pK then we can always find unique P on pK such that fPU(pK)=pK

fPU(pK) is in general a quartic unless pK also contains P.

fPU globally fixes pK(P x U, P) and pK(P x U, U) just like the cubics K003 and K004 in CL037.

Best regards

Bernard

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