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• Re: [EMHL] A Conjugation Based On Crossdifference

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• Dear Tuan, ... This is similar to the Cundy-Parry transformations in http://pagesperso-orange.fr/bernard.gibert/Classes/cl037.html your Y is the intersection
Jul 14, 2010 1 of 4
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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

[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
Jul 15, 2010 1 of 4
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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, ... 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
Jul 15, 2010 1 of 4
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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

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