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Re: [EMHL] Re: Foci of Steiner Ellipse

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  • Steve Sigur
    François, I think Wilson configuration is more accurate. Friendly, Steve ... Steve Sigur Triangle web page:
    Message 1 of 33 , Apr 3, 2006
      François, I think "Wilson configuration" is more accurate.

      Friendly,

      Steve


      On Apr 3, 2006, at 3:01 AM, Francois Rideau wrote:

      > In Steve configuration, foci G1 and G2 were also equicenters but the
      > converse is false of course since in affine geometry there are no
      > foci!

      Steve Sigur

      Triangle web page:
      http://paideiaschool.org/TeacherPages/Steve_Sigur/geometryIndex.htm

      Other math:
      http://paideiaschool.org/TeacherPages/Steve_Sigur/interesting2.htm





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    • Quang Tuan Bui
      Dear Bernard, This is interesting problem. Your circle is nine point circle of GBC, other two circles are nine point circle of GCA, GAB. The nine point circle
      Message 33 of 33 , Apr 5, 2006
        Dear Bernard,
        This is interesting problem. Your circle is nine point circle of GBC, other two circles are nine point circle of GCA, GAB. The nine point circle of ABC also pass through X(115).
        Generally we can take any Cevian point P instead G and there are 17 circles which concur at one point T. (Please refer my message #12524 and please note that I still not have complete proofs for it). Nine point conic also pass through this point.
        Follow is the list of T depend on P for your reference
        P T
        X(1) X(11)
        X(2) X(115)
        X(3) X(125)
        X(4) All 17 circles are the same
        X(5) X(137)
        X(6) X(125)
        X(7) X(11)
        X(8) X(11)
        X(9) X(11)
        X(10) X(115)
        I don't know why that and what about another T when we take another P?.
        I think we should pay enough time to study it.
        Best regards,
        Bui Quang Tuan

        Bernard Gibert <bg42@...> wrote: Dear friends,

        the circle passing through the midpoints of BC, GB, GC contains the
        center X(115) of the Kiepert hyperbola.

        any simple reason for this ?

        obviously, there are two other similar circles.



        Best regards

        Bernard



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