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Re: [EMHL] Ismail Erciyes' question

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  • Zak Seidov
    My formulas are rather lengthy (and i d not liked to give them, but as comparison and check with Barry Wolk s formula they may be of some sense); let: x[B,C]:=
    Message 1 of 7 , Aug 7, 2003
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      My formulas are rather lengthy
      (and i'd not liked to give them,
      but as comparison and check
      with Barry Wolk's formula
      they may be of some sense);

      let:
      x[B,C]:=
      R ((1/2)Sqrt[Sin[2B]*Sin[2C]]-
      Cos[B]*Cos[C]);
      R is circumradius;
      then:
      areaDEF=
      x[A,B]*x[A,C]*Sin[A]+
      x[B,A]*x[B,C]*Sin[B]+
      x[C,B]*x[C,A]*Sin[C];
      %%%%%%%%%%
      two numerical cases:
      {a, b, c} = {4, 5, 6} :
      areaDEF/areaABC =
      3 (-45 + 30 Sqrt[7] - 22 Sqrt[15] +
      7 Sqrt[105])/320 Sqrt[7] = 0.0740417
      %%%%%%%%
      {a, b, c} = {13, 14, 15} :
      areaDEF/areaABC =
      (3/236600)* (154 (120 - 7 Sqrt[5]) -
      23 Sqrt[154](5 + 7 Sqrt[5])) = 0.129013.

      Barry's formula gives (happily)
      the same numerical values
      Zak

      --- In Hyacinthos@yahoogroups.com,
      Barry Wolk <wolkbarry@y...> wrote:
      > --- Darij Grinberg <darij_grinberg@w...> wrote:
      > > Given an acute-angled triangle ABC, the A-altitude intersects
      > > the circle with diameter BC at the point D. (In fact, there
      > > are two intersections, but we take the one which lies on the
      > > same half-plane of the line BC as A.) Analogously define the
      > > points E and F.
      > >
      > > What is area(ABC) : area(DEF) ?
      >
      > Let x = (1-cot B)(1-cot C) / sqrt((cot B)(cot C)),
      > with y and z defined cyclically. Then
      > area(DEF) /area(ABC) = (cos A)(cos B)(cos C)(2-x-y-z)
      >
      > Routine barycentrics gave an answer, but reducing that answer to
      > a short formula was tricky.
      >
      > --
      > Barry Wolk
      >
      >
      > __________________________________
      > Do you Yahoo!?
      > Yahoo! SiteBuilder - Free, easy-to-use web site design software
      > http://sitebuilder.yahoo.com
    • Steve Sigur
      Bernard, This ellipse is mentioned in Kapetis, but lightly, showing a nice method by which the axes of an inconic may be constructed. I have always liked this
      Message 2 of 7 , Aug 10, 2003
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        Bernard,

        This ellipse is mentioned in Kapetis, but lightly, showing a nice
        method by which the axes of an inconic may be constructed. I have
        always liked this conic and did not know that the axes are parallel to
        Jerabek. How did you decide this by the way?

        Back after a long absence, friendly from USA,

        Steve


        >
        > Dear friends,
        >
        > the in-conic with perspector X69 (isotomic of H) is an ellipse centered
        > at O whose axes are parallel to the asymptotes of the Jerabek
        > hyperbola. It contains X125 (center of the Jerabek hyperbola) and
        > obviously its reflection E in O which is not mentionned in ETC.
        >
        > its 1st bary : S_A [ (b^2 - c^2)^2 + a^2 (b^2 + c^2 - 2a^2) ]^2
        >
        > the normals dropped from X20 to this ellipse pass through the cevians
        > of X69 and E.
        >
        > two isogonal conjugate points have their polar lines in this ellipse
        > which are parallel if and only if they lie on the McCay cubic.
        >
        > does someone have any reference about this ellipse ?
        >
        > Best regards
        >
        > Bernard
      • Bernard Gibert
        Dear Steve, ... I expect it s a known result : for any point M on the Lucas cubic, the axes of the inconic with perspector M are parallel to the asymptotes of
        Message 3 of 7 , Aug 11, 2003
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          Dear Steve,

          > This ellipse is mentioned in Kapetis, but lightly, showing a nice
          > method by which the axes of an inconic may be constructed. I have
          > always liked this conic and did not know that the axes are parallel to
          > Jerabek. How did you decide this by the way?

          I expect it's a known result :

          for any point M on the Lucas cubic, the axes of the inconic with
          perspector M are parallel to the asymptotes of the rectangular
          circum-hyperbola through M.

          Best regards

          Bernard

          PS : Antreas kindly sent me the Kapetis book. Unfortunately, I can't
          read Greek.
          Where do you find the information about this ellipse ?
          Antreas, please provide a translation. Thanks.

          [Non-text portions of this message have been removed]
        • Antreas P. Hatzipolakis
          ... Dear Bernard Right now I don t see it in my chaotic library! Only the 2nd volume I bought recently. Antreas PS: When more copies arrive from Thessaloniki
          Message 4 of 7 , Aug 12, 2003
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            Bernard Post-Scripted:

            >PS : Antreas kindly sent me the Kapetis book. Unfortunately, I can't
            >read Greek.
            >Where do you find the information about this ellipse ?
            >Antreas, please provide a translation. Thanks.

            Dear Bernard

            Right now I don't see it in my chaotic library!
            Only the 2nd volume I bought recently.

            Antreas

            PS: When more copies arrive from Thessaloniki
            (the town where the book was published) in the bookstore,
            then I will send you that volume too
            APH

            --
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