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Re: [EMHL] line, circumcenters

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  • 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 1 of 4 , Feb 15, 2013
      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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