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:

> http://f1.grp.yahoofs.com/v1/wBlNUXTk_prMpAQuRSkfBeorkY6PtiaKnCxn-V4o6o9X1W2FkiOwaDALdn3PHq-ILiA4gk809bSkFT7ISn2Ot_Ts4e_Z1Y7w/Hyacinthos21808A.eps

> )

>

>

> 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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