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

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  • Antreas Hatzipolakis
    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 2:22 AM
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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 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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