## 1710Re: [EMHL] Cubics

Expand Messages
• Oct 31, 2000
• 0 Attachment
Dear 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
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
>
• Show all 42 messages in this topic