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Re: [EMHL] Concurrent tripolars

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  • Floor van Lamoen
    Dear all, ... This particular example has been discussed in the thread Don Wallace Locus . Kind regards, Floor.
    Message 1 of 5 , Apr 1, 2001
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      Dear all,

      xpolakis@... wrote:
      >
      > Let P be a point and P', P" its isogonal, isotomic conjugates, resp.
      > We have six lines: PP', PP", P'P", tripolars of P,P',P", which can form
      > C(6,3) = 20 triads of lines.
      >
      > For each one of the triads we can ask for the locus of P such that
      > the members of the triad concur.
      >
      > For example:
      > Which is the locus of P such that the tripolars of P,P',P" concur?
      > I found that the locus is a conic (hyperbola) passing through the four
      > Ix and G. [equation: a^2(b^2-c^2)x^2 + cycl. = 0]

      This particular example has been discussed in the thread "Don Wallace
      Locus".

      Kind regards,
      Floor.
    • Steve Sigur
      ... Floor is not mentioning that, before Hyacinthos, he also posted (and solved) this in the geometry forum. Steve
      Message 2 of 5 , Apr 2, 2001
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        >> For example:
        >> Which is the locus of P such that the tripolars of P,P',P" concur?
        >> I found that the locus is a conic (hyperbola) passing through the four
        >> Ix and G. [equation: a^2(b^2-c^2)x^2 + cycl. = 0]
        >
        > This particular example has been discussed in the thread "Don Wallace
        > Locus".
        >
        > Kind regards,
        > Floor.


        Floor is not mentioning that, before Hyacinthos, he also posted (and
        solved) this in the geometry forum.

        Steve
      • xpolakis@otenet.gr
        ... Yes, the Don Wallace locus, The locus of P such that P, gP [= isog. conj. of P], tP [= isog. conj. of P] are collinear (AMM 43(1936) 500), is the
        Message 3 of 5 , May 1 10:13 AM
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          [APH]:
          >> Let P be a point and P', P" its isogonal, isotomic conjugates, resp.
          >> We have six lines: PP', PP", P'P", tripolars of P,P',P", which can form
          >> C(6,3) = 20 triads of lines.
          >>
          >> For each one of the triads we can ask for the locus of P such that
          >> the members of the triad concur.
          >>
          >> For example:
          >> Which is the locus of P such that the tripolars of P,P',P" concur?
          >> I found that the locus is a conic (hyperbola) passing through the four
          >> Ix and G. [equation: a^2(b^2-c^2)x^2 + cycl. = 0]

          [FvL]:
          >This particular example has been discussed in the thread "Don Wallace
          >Locus".

          Yes, the Don Wallace locus, "The locus of P such that P, gP [= isog. conj.
          of P], tP [= isog. conj. of P] are collinear" (AMM 43(1936) 500), is the
          hyperbola a^2(b^2 - c^2)x^2 + b^2(c^2 - a^2)y^2 + c^2(a^2 - b^2)z^2 = 0

          But is the locus of P, such that the tripolars of P, gP, tP are concurrent,
          the Wallace hyperbola, as I wrote ?

          According to my new calculations:

          The locus of P such that the tripolars of :

          - P, gP, tP concur, is the quartic:

          (b^2 - c^2)/x^2 + (c^2 - a^2)/y^2 + (a^2 - b^2)/z^2 = 0,

          - P, gP gtP concur, is the conic (Wallace hyperbola):

          a^2(b^2 - c^2)x^2 + b^2(c^2 - a^2)y^2 + c^2(a^2 - b^2)z^2 = 0

          The reader may find the loci of P such that the tripolars of

          - P, tP, tgP

          - P, tgP, gtP

          - P, gtgP, tgtP

          etc, are concurrent.


          Antreas
        • Steve Sigur
          ... This equation is in trilinears, I assume. When I do this in barycentrics, I get (b^2 - c^2)x^2 + (c^2 - a^2)y^2 + (a^2 - b^2)z^2 = 0 which is better named
          Message 4 of 5 , May 1 7:28 PM
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            on 5/1/01 1:13 PM, xpolakis@... wrote:

            > Yes, the Don Wallace locus, "The locus of P such that P, gP [= isog. conj.
            > of P], tP [= isog. conj. of P] are collinear" (AMM 43(1936) 500), is the
            > hyperbola a^2(b^2 - c^2)x^2 + b^2(c^2 - a^2)y^2 + c^2(a^2 - b^2)z^2 = 0


            This equation is in trilinears, I assume. When I do this in barycentrics, I
            get

            (b^2 - c^2)x^2 + (c^2 - a^2)y^2 + (a^2 - b^2)z^2 = 0

            which is better named as the Kiepert hyperbola of the medial triangle.

            ABC is self polar with respect to it. It has many point on it which may be
            found by applying (: x-y+z:) to any point (:y:) on the Kipert hyperbola. In
            particular the 4 incenters lie on it, as well as H and L.

            Apologies if this has already been pointed out.

            The Triangle Book nears completion, when I look forward to being a day to
            day Hyacinthian again.

            Friendly from Princeton,

            Steve
          • xpolakis@otenet.gr
            ... isot. ... Yes in trilinears, or better in normals. ... ^^^^^^ antimedial ... ^ G L lies on it, since L(ABC) = H(antimedial), and every rect. hyperbola
            Message 5 of 5 , May 2 5:16 AM
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              [APH]:
              >> Yes, the Don Wallace locus, "The locus of P such that P, gP [= isog. conj.
              >> of P], tP [= isog. conj. of P] are collinear" (AMM 43(1936) 500), is the
              isot.
              >> hyperbola a^2(b^2 - c^2)x^2 + b^2(c^2 - a^2)y^2 + c^2(a^2 - b^2)z^2 = 0

              [SS]:
              >This equation is in trilinears, I assume. When I do this in barycentrics, I

              Yes in trilinears, or better in normals.

              >get
              >
              >(b^2 - c^2)x^2 + (c^2 - a^2)y^2 + (a^2 - b^2)z^2 = 0
              >
              >which is better named as the Kiepert hyperbola of the medial triangle.
              ^^^^^^
              antimedial


              >ABC is self polar with respect to it. It has many point on it which may be
              >found by applying (: x-y+z:) to any point (:y:) on the Kipert hyperbola. In
              >particular the 4 incenters lie on it, as well as H and L.
              ^
              G

              L lies on it, since L(ABC) = H(antimedial), and every rect. hyperbola
              passes through the orthocenter of the triangle.

              >Apologies if this has already been pointed out.
              >
              >The Triangle Book nears completion, when I look forward to being a day to
              >day Hyacinthian again.

              APH
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