Re: [EMHL] Cubics
- 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:
I have read the interesting paper of Pinkernell, Journal of Geometry, vol
(1996) pp 141-161 and another about general quadratic transformation and I
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
Euler-pencil (locus of the pivot = Euler line), Pinkernell give a geometric
He defines, for each d-pedal cubic, a corresponding d-cevian cubic, (the
the pivot is jerabek'hyperbola )
and a 2-1 correspondence, the d-pedal and the (-d)-pedal corresponds to the
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 ?
You have a high production of problems.
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.
> 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]:
> 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!
- Here is an old unanswered Hyacinthos message:
Let P be a point on the plane of triangle ABC.
The parallels to AB, AC through P intersect BC at Ab, Ac resp.
Let A1,A2 be the traces of the circumcenter,incenter of PAbAc on AbAc.
Similarly we define the points B1,B2; C1,C2.
Which are the loci of P such that (1) A1B1C1 (2) A2B2C2 be in perspective with ABC?
I found that the loci are the following cubics:
(1) K044 - Euler Central Cubic
(2) K033 - Spieker Central Cubic
Changing circumcenter, incenter by an arbitrary point Q=(u:v:w), the locus is the cubic
(cyclic sum) (u(v+w)y z((u+v-w)y -(u-v+w)z))) = 0
Central cubic with center the complement of Q.
--- In Hyacinthos@yahoogroups.com, xpolakis@... wrote:
> Let P be a point on the plane of triangle ABC.
> The parallels to AB, AC through P intersect BC at Ab, Ac resp.
> Let A1,A2 be the traces of the circumcenter,incenter of PAbAc on AbAc.
> Similarly we define the points B1,B2; C1,C2.
> Which are the loci of P such that (1) A1B1C1 (2) A2B2C2 be in perspective
> with ABC?
> I found that the loci are these cubics:
> xcos(B+u) + zcosu ycos(C+v) + xcosv zcos(A+w) + ycosw
> ----------------- * ------------------ * ------------------ = 1
> xcos(C-u) + ycosu ycos(A-v) + zcosv zcos(B-w) + xcosw
> where u := B-C, v := C-A, w := A-B for the (1), and
> 2u := B-C, 2v := C-A, 2w := A-B for the (2).
> But maybe these are Darboux cubics of some triangles of the reference
> triangle ABC...