- Oct 31, 2000Dear Andreas,

your generalization is exactly the one of Guido M. Pinkernell in Journal of

Geometry Vol 55 (1996) " Cubics Curves in the Triangle Plane".

Your cubic is denoted as the t-pedal cubic, he defines a (2,1)

correspondance from the t-pedal to the t-cevian cubic:

See my message of the 8 oct: "Cubic Kn is an identified object", I wrote:

*******BEGIN******

I have read the interesting paper of Pinkernell, Journal of Geometry, vol

55

(1996) pp 141-161 and another about general quadratic transformation and I

arrive

to the follswing:

The cubic Kn is the 2-cevian cubic of Pinkernell.

It's definition could be:

"The set of points P such that F, P, P' are collinear points",

where F = X265 and P->P' is the quadratic transformation given by

(x,y,z) -> ( 1/(x(1-4cosA^2)) , 1/(y(1-4cosB^2)) , 1/(z(1-4cosC^2)) )

In Pinkernell's paper, you can see the definition of d-pedal cubic, they

form the

Euler-pencil (locus of the pivot = Euler line), Pinkernell give a geometric

construction.

He defines, for each d-pedal cubic, a corresponding d-cevian cubic, (the

locus of

the pivot is jerabek'hyperbola )

and a 2-1 correspondence, the d-pedal and the (-d)-pedal corresponds to the

same

d-cevian.

Example:

1-pedal is Darboux.

2-pedal is Neuberg.

0-pedal is MacKay.

(-1)-pedal is Thomson (= 17-point)

infinite-pedal is ortho-cubic.

0-cevian = 0-pedal.

1-cevian = Lucas.

2-cevian = Kn.

infinite-cevian = threee altitudes of ABC.

Is this not funny ?

******END*******

You have a high production of problems.

Regards Fred.

xpolakis@... wrote :

> I wrote:

>

> >Darboux Cubic:

> >Let ABC be a triangle, and A'B'C' the pedal triangle of a point P.

> >The locus of the points P such that A'B'C' is in perspective with ABC

> >(or in other words, A'B'C' is the cevian triangle of some point Q)

> >is a cubic: Darboux cubic.

> >

> >Generalization:

> [...]

>

> Another one:

> Let ABC be a triangle, and A'B'C' the pedal triangle of a point P.

> Let Pa, Pb, Pc be three points on the lines PA', PB', PC' respectively

> such that PPa, PPb, PPc be proportional to line segments PA', PB', PC',

> that is PPa / PA' = PPb / PB' = PPc / PC' := t.

> For t fixed, the locus of P such that PaPbPc be in perspective with

> ABC is an isogonal cubic with pivot (cosA - tcosBcosC ::), a point

> lying on the Euler line.

> [or to be precise: that's what I found by calculations made by hand]:

>

> Equation:

> x(y^2 - z^2)(cosA - tcosBcosC) + [cyclically] = 0

>

> For t = 1 (that is Pa = A', Pb = B', Pc = C') we have the Darboux cubic.

> [the pivot is the de Longchamps point].

>

> For t = -2 we have the Napoleon cubic

> [the pivot is the 9PC center N = (cos(B-C) ::)]

>

> and so on!

>

> Antreas

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