Loading ...
Sorry, an error occurred while loading the content.

ex-extra perspectors: query

Expand Messages
  • Dick Tahta
    A query - with apologies for cutting across current concerns - about ex-extra perspectors which were discussed at length by Barry and Steve some months ago:
    Message 1 of 42 , Nov 6, 2000
    • 0 Attachment
      A query - with apologies for cutting across current concerns - about
      ex-extra perspectors which were discussed at length by Barry and Steve some
      months ago:

      First a brief recap. The ex-extra perspector PP0 of a point P0 is defined
      as the perspector - if it exists - of the triangle of ex-points (harmonics)
      of P0 and the triangle P1,P2,P3 of extra-points (extraversions) of P0.
      When the quadrangle Pi (i=0-3) has a desmic mate then there is indeed an
      ex-extra perspector PP0 and this in turn also has an ex-extra perspector
      PPP0. Moreover, this (after some gruesome algebra) turns out to be P0
      itself. (Thus Nag-Nag-Nag = Nag - if, that is, I have the notation
      right.)

      Meanwhile there are certainly some non-desmic Pi for which the above is
      also true. My query is whether when the ex-extra perspectors exist PPP0
      = P0 for all P0?

      Dick Tahta
    • Francisco Javier
      Here is an old unanswered Hyacinthos message: APH: Let P be a point on the plane of triangle ABC. The parallels to AB, AC through P intersect BC at Ab, Ac
      Message 42 of 42 , Jan 16, 2011
      • 0 Attachment
        Here is an old unanswered Hyacinthos message:

        APH:

        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...
        >
        > Antreas
        >
      Your message has been successfully submitted and would be delivered to recipients shortly.