Re: [EMHL] Cubics
>>2). 3 circles, one through B & C and tangent at B to OB, at C to OC[APH]
>>the other two defined similarly.
>Now, let Ka, Kb, Kc be the centers of the three circles.The circumradius of this triangle is R' = R/4cosAcosBcosC.
>The triangle KaKbKc in general is not in perspective with ABC
Probably some of its notable points (O,H, etc) is a notable
point for the reference triangle (ABC) too.
>Let P be a point on the plane of a triangle ABC, and Ab, Ac the centers^^^ ^^^
>of the circles passing through B, C and tangent to PB,PC at B, C resp.
>Let A' = BAb /\ CAc. Similarly we define B', C'.
>Which is the locus of the points P such that A'B'C' be in perspective
>with ABC? (locus of perspectors?)
>More simply (since AAb is _|_ to PB, and CCc is _|_ to PC):
- 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...