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Re: [EMHL] Fermat Points

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  • Wilson Stothers
    Dear Bernard, I know nothing about angular coordinates, so hesitate to say much Your proof below suggests that, if F = X13, then it is also X13 of its cevian
    Message 1 of 10 , Dec 8, 2005
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      Dear Bernard,

      I know nothing about angular coordinates, so hesitate to say much
      Your proof below suggests that, if F = X13, then it is also X13 of
      its cevian and anticevian triangle. This isn't always so, it can
      be X14 for one or other.


      > > We have the following fact :
      > >
      > > If F is a Fermat Point of triangle ABC, then F is also a
      > > Fermat Point of its Cevian, and of its Anticevian Triangle.
      >
      > this is direct consequence of the fact that the angular
      coordinates
      > of a point wrt ABC, its cevian and anticevian triangles are the
      same.

      Is this true?
      I'm guessing at the definition, but I think there may be
      supplements involved. Probably I just guessed wrongly.

      > for X13 : 120°, 120°, 120°
      >
      > [snip]
      >
      >

      Yours

      Wilson
    • Steve Sigur
      Ah, that is what I wanted to know. But it seems to me that their angular coordinates would not change, so both points should be the same. It puzzles me that
      Message 2 of 10 , Dec 8, 2005
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        Ah, that is what I wanted to know. But it seems to me that their
        angular coordinates would not change, so both points should be the
        same. It puzzles me that the second points are different. Intriguing
        if they are.

        Oh I know, it must be true that if we get the pedal and antipedal of
        Fs, then the three Fs's are the same. This makes 12 points from these
        six triangles, which I bet have an interesting relationship, Can't
        wait until I have some time with Sketchpad.

        Hmm, algebra at work.

        Steve


        On Dec 8, 2005, at 9:55 AM, Wilson Stothers wrote:

        > the three triangles will have different second
        > Fermat Points, so the lines FnFs are not parallel.



        [Non-text portions of this message have been removed]
      • Steve Sigur
        Bernard, The two Fermat points each have a cevian and anticevian triangle. The 4 Euler lines of these triangles seem to be parallel to that of ABC (observed on
        Message 3 of 10 , Dec 9, 2005
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          Bernard,

          The two Fermat points each have a cevian and anticevian triangle. The
          4 Euler lines of these triangles seem to be parallel to that of ABC
          (observed on Sketchpad, not proved). Is there a curve, all of whose
          points have Euler line parallel to that of ABC?

          Steve
        • Bernard Gibert
          Dear Steve, ... It is true. ... this was answered by Wilson yesterday under the thread [EMHL] The Cubic K278! Best regards Bernard [Non-text portions of this
          Message 4 of 10 , Dec 9, 2005
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            Dear Steve,

            > The two Fermat points each have a cevian and anticevian triangle. The
            > 4 Euler lines of these triangles seem to be parallel to that of ABC
            > (observed on Sketchpad, not proved).

            It is true.


            > Is there a curve, all of whose
            > points have Euler line [of their cevian and anticevian triangles, I
            > suppose] parallel to that of ABC?

            this was answered by Wilson yesterday under the thread [EMHL] The
            Cubic K278!


            Best regards

            Bernard



            [Non-text portions of this message have been removed]
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