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15359Re: [EMHL] line/conic intersection

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  • Luís Lopes
    Jul 2, 2007
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      Dear Hyacinthists,

      Dear François,

      It seems the messages are leaving/arriving normally again.

      I see now your solution is to draw a circle through two
      given points and tangent to a line. This is classical.

      One can have two solutions. These are the circumcenters
      of the triangles ABC given <h_a,d_a,m_b>.

      Barukh Ziv some 5 years ago sent me a message showing
      that O belongs to a parabola <P=(F,d)> and to h'_a,
      reflection of h_a on d_a.

      The directrix <d> is parallel to (B,C) and distant m_b^2/4h_a
      from it. The focus <F> is distant \ell/4 from (A,H_a) and
      9h_a/8 from <d>. Here \ell=\sqrt{m_b^2-(h_a/2)^2}.

      Data to have 2 solutions (from Die Konstruktion von
      Dreiecken, Kurt Herterich).

      <h_a=6 cm>, <m_b=5 cm>, and <d_a=6.5 cm>;

      <h_a=3 cm>, <m_b=3.9 cm>, and <d_a=5 cm>;

      Best regards,

      >From: "Francois Rideau" <francois.rideau@...>
      >Reply-To: Hyacinthos@yahoogroups.com
      >To: Hyacinthos@yahoogroups.com
      >Subject: Re: [EMHL] line/conic intersection
      >Date: Mon, 2 Jul 2007 20:21:59 +0200
      >Yes, nice solution and now I remember that I learned it more than fifty
      >years ago.
      >Of course the most important is the discussion which must follows on the
      >number of solutions.
      >About the teaching of the theory of conics, I think the best book was
      >written by the great Henri Lebesgue himself: Les coniques, published in
      >at Gauthier-Villars, just after his death, during the nazi occupation of
      >Poor Henri, nowadays, the only conic still in use in France is the circle
      >roadsigns just before runabouts!
      >On 7/2/07, Luís Lopes <qed_texte@...> wrote:

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