## Re: [EMHL] Re: The nonpivotal isocubics nK(W,G*W)

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• Dear Angel Nice results! This configuration is rich in locus problems, but I am not sure if we get such nice results for all! Now, we can replace the parallels
Message 1 of 4 , Mar 23, 2013
Dear Angel

Nice results!

This configuration is rich in locus problems, but I am not sure if we get
such nice results for all!

Now, we can replace the parallels with perpendiculars.

For the case of W = H, we have to draw perpendiculars through the
feet of H (not of P, since we get the ABC)

That is:

Suppose we are given a triangle ABC and two other points W=(p:q:r) and
P=(x:y:z), not on its sidelines,
and WaWbWc, PaPbPc the cevian triangles of W,P, resp.

A* = (Perpendicular through Wb to PaPb) /\ (Perpendicular through Wc to
PaPc),
B* = (Perpendicular through Wc to PbPc) /\ (Perpendicular through Wa to
PbPa),
C* = (Perpendicular through Wa to PcPa) /\ (Perpendicular through Wb to
PcPb).

When W=H, which is the locus of P such that:

1. ABC, A*'B*C*

2. WaWbWc [= HaHbHc], A*B*C*

are perspective?

and also 3. A*B*C*, PaPbPc.

But I see more..... and will come back later, when I have free time.....

APH

On Sat, Mar 23, 2013 at 5:03 AM, Angel <amontes1949@...> wrote:

> **
>
>
> Dear Antreas
>
>
> Suppose we are given a triangle ABC and two other points W=(p:q:r) and
> P=(x:y:z), not on its sidelines, and WaWbWc, PaPbPc the cevian triangles of
> W,P, resp.
>
> A' = (Parallel through Pb to WaWb) /\ (Parallel through Pc to WaWc),
> B' = (Parallel through Pc to WbWc) /\ (Parallel through Pa to WbWa),
> C' = (Parallel through Pa to WcWa) /\ (Parallel through Pb to WcWb).
>
> When W=H, which is the locus of P such that:
>
>
> 1. ABC, A'B'C'
>
> 2. HaHbHc, A'B'C'
>
> are perspective?
>
> 1. The triangles ABC and A'B'C' are perspective if and only if P lies on
> the circumconic of perspector the crosspoint of P and G , p(q+r)yz+
> q(r+p)zx+ r(p+q)xy=0, or lies on the quartic p^2(q^2- r^2)y^2z^2+ q^2(r^2-
> p^2)z^2x^2+ r^2(p^2- q^2)x^2y^2=0.
>
> Points of the quartic: W, the vertices de ABC (are double points), and the
> vertices of the antimedial triangle, Ga, Gb and Gc.
> The tangents to quartic at Ga,Gb, Gc determine a triangle perspective with
> ABC of the perspector
> 1/( p^2(q^2- r^2)):1/( q^2(r^2- p^2)): 1/( r^2(p^2- q^2)).
>
> In particular, if W=H
> The triangles ABC and A'B'C' are perspective if and only if P lies on the
> circumcircle or lies on the quartic Q066 ( Stammler quartic,
> http://bernard.gibert.pagesperso-orange.fr/curves/q066.html)
>
> If P lies on circumcircle the perspector Q of the triangles ABC and A'B'C'
> is the isogonal conjugate of P.
> If P lies on the Stammler quartic the perspector Q of the triangles ABC
> and A'B'C' lies on the JERABEK HYPERBOLA
>
> ---------------------------------------------------------
> Another geometry property of the quartic Q066:
> (EPS file:
> )
>
>
> Let P be a point and PaPbPc its cevian triangle, HaHbHc
> the orthic triangle (cevian triangle of H).
> The parallel through the point Pb to the sideline HaHb and
> the parallel through the point Pc to the sideline HaHc
> intersect at A'. The points B' and C' defined likewise.
>
> The triangles ABC and A'B'C' are perspective if and
> only if P lies on the Stammler quartic (together with the circumcircle).
> ----------------------------------------------------------
>
> 2. The triangles WaWbWc and A'B'C' are perspective if and only if P lies
> on the cevians of W.
>
> Best regards
> Angel Montesdeoca
>
> --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> >
> >
> >
> > --- In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@> wrote:
> > >
> > > -------------------------------------------------
> > > Another geometry property of K211:
> > >
> > > Let P be a point and PaPbPc its cevian triangle, HaHbHc
> > > the orthic triangle (cevian triangle of H).
> > > The parallel through the point Pb to the sideline HaHb and
> > > the parallel through the point Pc to the sideline HaHc
> > > intersect at A'. The points B' and C' defined likewise.
> > >
> > > The triangles PaPbPc and A'B'C' are perspective if and
> > > only if P lies on on K211 (together with the three
> > > circm-rectangular hyperbolas passing through Ga, Gb, Gc).
> >
> > Naturally one may ask which are the rest two loci we get by combinations
> > of the triangles in the configuration, ie
> >
> > Which is the locus of P such that:
> >
> > 1. ABC, A'B'C'
> >
> > 2. HaHbHc, A'B'C'
> >
> > are perspective?
> >
> > Furthermore, we can define more points and ask for loci.
> >
> > For example:
> >
> > Denote A* = PbHc /\ PcHb and similarly B*,C*.
> >
> > Which is the locus of P such that:
> >
> > A*B*C* is perspective with:
> >
> > 1. ABC 2. HaHbHc. 3. PaPbPc. 4. A'B'C' ?
> >
> >
> > APH
> >
>
> _
>

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