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Construction problem

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  • xpolakis@otenet.gr
    Let ABC be a triangle. Construct a line intersecting CA, CB at Ca, Cb, resp, and another line intersecting BA, BC at Ba, Bc, resp., such that: (ACa = BCb = ABa
    Message 1 of 4 , Sep 4, 2001
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      Let ABC be a triangle. Construct a line intersecting CA, CB at Ca, Cb, resp,
      and another line intersecting BA, BC at Ba, Bc, resp., such that:

      (ACa = BCb = ABa = CBc) AND (CaCb = BaBc)


      A
      /\
      / \
      / \
      / \
      Ba Ca
      / \
      / \
      / \
      / \
      / \
      / \
      B-------Cb----Bc-------C


      A trig. solution leads to a quadratic equation:

      Let ACa = BCb = ABa = CBc = x

      Triangle (CaCbC): (CaCb)^2 = (b-x)^2 + (a-x)^2 - 2(b-x)(a-x)cosC (1)

      Triangle (BaBcB): (BaBc)^2 = (c-x)^2 + (a-x)^2 - 2(c-x)(a-x)cosB (2)

      CaCb = BaBc, and (1) and (2) ==>

      (b-x)^2 - (c-x)^2 = 2(b-x)(a-x)cosC - 2(c-x)(a-x)cosB

      and from this we get the x.

      Synthetic solution??


      Antreas

      PS: Probably the point of intersection A' of CaCb, BaBc,
      (if there is only one such real point !), and the similarly defined
      ones B', C', are of some interest.
      (for example: Are the lines AA', BB', CC' concurrent?)
    • martin lukarevski
      Dear Hyacinthians, let me give you the following construction problem. Construct the triangle ABC if given in position, the line on which the base BC is
      Message 2 of 4 , Oct 5, 2001
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        Dear Hyacinthians,


        let me give you the following construction problem.

        Construct the triangle ABC if given in position, the
        line on which the base BC is located,and the feet of
        the Cevians of O to the two lateral sides.

        Best regards,
        Martin


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      • Atul Dixit
        Dear all Given the first Brocard triangle of triangle ABC,how to construct the triangle ABC ? My solution has become unnecessarily long.Will anybody tell me a
        Message 3 of 4 , Aug 31 9:41 AM
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          Dear all
          Given the first Brocard triangle of triangle ABC,how to construct
          the triangle ABC ?

          My solution has become unnecessarily long.Will anybody tell me a precise
          solution of it?

          Yours faithfully
          Atul.A.Dixit



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        • Dick Klingens
          Johnson, page 280: === The solution depends onthe fact that any triangle is inversively similar to its first Brocard triangle. We locate the Brocard points of
          Message 4 of 4 , Aug 31 10:00 AM
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            Johnson, page 280:

            ===
            The solution depends onthe fact that any triangle is inversively similar to
            its first Brocard triangle.
            We locate the Brocard points of the given triangle, also its first Brocard
            triangle; then by similarity we locate on the circumcircle of the given
            triangle the Brocard points of the required triangle.
            The vertices of the latter can then be found at once.
            ===

            Is this what you did?

            Regards,
            Dick Klingens

            ::-----Original Message-----
            ::From: Atul Dixit [mailto:atul_dixie@...]
            ::Sent: Saturday, August 31, 2002 6:42 PM
            ::To: Hyacinthos@yahoogroups.com
            ::Subject: [EMHL] Construction problem
            ::
            ::
            ::Dear all
            :: Given the first Brocard triangle of triangle ABC,how to construct
            ::the triangle ABC ?
            ::
            ::My solution has become unnecessarily long.Will anybody tell me a precise
            ::solution of it?
            ::
            ::Yours faithfully
            ::Atul.A.Dixit
            ::
            ::
            ::
            ::_________________________________________________________________
            ::Send and receive Hotmail on your mobile device: http://mobile.msn.com
            ::
            ::
            ::
            ::
            ::
            ::Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
            ::
            ::
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