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RE: [EMHL] line, circumcenters

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  • Paul Yiu
    Correction: _________________________________ Dear Antreas, [APH]: Let ABC be a triangle, L a line, A , B , C the orth. projections of A,B,C on L, resp.
    Message 1 of 4 , Feb 15, 2013
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      Correction:
      _________________________________
      Dear Antreas,

      [APH]: Let ABC be a triangle, L a line, A', B', C' the orth. projections
      of A,B,C on L, resp. Denote:

      Ab, Ac = the circumcenters of AB'C, ABC', resp.
      Bc, Ba = the circumcenters of BC'A, BCA', resp
      Ca, Cb = the circumcenters of CA'B, CAB', resp.

      The lines AbAc, BcBa, CaCb are concurrent
      and the radical axes [common chords] of (AB'C, ABC'), (BC'A, BCA'),
      (CA'B, CAB') are concurrent.

      Points for special cases L = OH,OK,OI ??

      *** There are only three circles involved. Your Ba=Ca etc.
      If L = ux+vy+wz = 0, then the radical center is the point

      ((S_A(v-w)^2 + S_B w(w-u) - S_C v(u-v))/(u(b^2S_Bv + c^2S_Cw - S_{BC}u )) : ... : ...).

      For the Euler line, this is ( (a^2(S_{AA}+S_{BC})-4S_{ABC})/(S_{BB}S_{CC}(S_B-S_C)^2) : ... : ...)

      For the Brocard axis: (a^4(b^4+c^4-2^2(b^2+c^2-^2)/(b^2-c^2)^2 : ... : ...)

      *** (a^4(b^4+c^4-2a^2(b^2+c^2-a^2))/(b^2-c^2)^2 : ... : ...)

      For the OI line: (a^2(2a^3-2a^2(b+c)-a(b^2-4bc+c^2)+(b-c)^2(b+c))/((b-c)^2(b+c-a)^2) : ... : ...)

      None of these is in the current edition of ETC.

      Best regards
      Sincerely
      Paul
    • Antreas
      Dear Paul So the the 6 circles are only 3..... now the 12 are only 9..... ABB , AB C AC B, AC C BCC , BC A BA C, BA A CAA , CA B CB A, CB B The 9 circumcenters
      Message 2 of 4 , Feb 15, 2013
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        Dear Paul

        So the the 6 circles are only 3..... now the 12 are only 9.....

        ABB', AB'C
        AC'B, AC'C

        BCC', BC'A
        BA'C, BA'A

        CAA', CA'B
        CB'A, CB'B

        The 9 circumcenters are grouped in three lines passing through O
        and parallel to altitudes.
        Probably they have more properties....

        APH


        --- In Hyacinthos@yahoogroups.com, Paul Yiu <yiu@...> wrote:
        >
        > Correction:
        > _________________________________
        > Dear Antreas,
        >
        > [APH]: Let ABC be a triangle, L a line, A', B', C' the orth. projections
        > of A,B,C on L, resp. Denote:
        >
        > Ab, Ac = the circumcenters of AB'C, ABC', resp.
        > Bc, Ba = the circumcenters of BC'A, BCA', resp
        > Ca, Cb = the circumcenters of CA'B, CAB', resp.
        >
        > The lines AbAc, BcBa, CaCb are concurrent
        > and the radical axes [common chords] of (AB'C, ABC'), (BC'A, BCA'),
        > (CA'B, CAB') are concurrent.
        >
        > Points for special cases L = OH,OK,OI ??
        >
        > *** There are only three circles involved. Your Ba=Ca etc.
        > If L = ux+vy+wz = 0, then the radical center is the point
        >
        > ((S_A(v-w)^2 + S_B w(w-u) - S_C v(u-v))/(u(b^2S_Bv + c^2S_Cw - S_{BC}u )) : ... : ...).
        >
        > For the Euler line, this is ( (a^2(S_{AA}+S_{BC})-4S_{ABC})/(S_{BB}S_{CC}(S_B-S_C)^2) : ... : ...)
        >
        > For the Brocard axis: (a^4(b^4+c^4-2^2(b^2+c^2-^2)/(b^2-c^2)^2 : ... : ...)
        >
        > *** (a^4(b^4+c^4-2a^2(b^2+c^2-a^2))/(b^2-c^2)^2 : ... : ...)
        >
        > For the OI line: (a^2(2a^3-2a^2(b+c)-a(b^2-4bc+c^2)+(b-c)^2(b+c))/((b-c)^2(b+c-a)^2) : ... : ...)
        >
        > None of these is in the current edition of ETC.
        >
        > Best regards
        > Sincerely
        > Paul
        >
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