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Re: Area problem

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  • shokoshu2
    Better I add a pic: http://pic-hoster.net/code/af0e139a00fff39403ed36b97bcabf36.gif So, we can immediately ask for several natural possible loci: 1. Green
    Message 1 of 3 , Jun 1, 2010
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      Better I add a pic:
      http://pic-hoster.net/code/af0e139a00fff39403ed36b97bcabf36.gif

      So, we can immediately ask for several natural possible loci:

      1. Green areas equal (fiendish)
      2. Yellow areas equal (trivial, incenter)
      3. Arcs DE=FG=HI (middle difficulty, I give it a try)
      4. Arcs EF=GH=ID (same as 2)
      5. Triangle aress ADE=BFG=CHI (straight lines DE,FG,HI
      have been omitted for clarity - middle difficulty)

      Obviously, these can be used to combine two loci into
      a point, but only 3+5 makes good sense. One solution
      surely is the circumcenter, others might exist.

      (Pic approximates case 3)

      Hauke
    • shokoshu2
      [superseding reply - why does Yahoo have no edit mode, as any lousy BBS can do?] ... Algebraically I get the Brocard axis as solution. Note that when
      Message 2 of 3 , Jun 2, 2010
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        [superseding reply - why does Yahoo have no
        edit mode, as any lousy BBS can do?]

        > 3. Arcs DE=FG=HI (middle difficulty, I give it a try)

        Algebraically I get the Brocard axis as solution.
        Note that when DE=FG=HI=0, the circumcircle arises
        naturally (trivial). Somehow I took the wrong surd
        at Albuquerque, as I get quarter-circle arcs for
        the cosine circle (at least P=symmedian and
        r=abc/(a^2+b^2+c^2) results) which is of course nonsense:
        http://mathworld.wolfram.com/KenmotuPoint.html
        THIS are the quarter-circle arcs. <shrug>


        Hauke
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