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Re: [EMHL] Points [correction]

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  • xpolakis@otenet.gr
    [APH] ... 2a^x + aby + acz = bax + 2b^y + bcz = cax + cby + 2c^z == x y z ... == (x:y:z) = ([3bc - a(b + c)]/a ::) in
    Message 1 of 1 , Nov 2, 2001
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      [APH]
      >> Let ABC be a triangle, P a point, and A'B'C', A"B"C" the cevian,
      >> pedal triangles of P, resp.
      >>
      >> Find the P's inside the triangle such that:
      >>
      >> PB" PC" PC" PA" PA" PB"
      >> 1. ---- + --- = --- + --- = --- + ---
      >> AC' AB' BA' BC' CB' CA'
      >>
      >>
      >> PB" PC" 1 PC" PA" 1 PA" PB" 1
      >> 2. (---- + ---) * --- = (---- + ----) * --- = (---- + ----) * ---
      >> AC' AB' BC BA' BC' CA CB' CA' AB
      >>
      >> where the line segments are not signed.

      [Correction of the solution of (1)]:
      >Let P = (PA": PB": PC") = (x:y:z) in actual normals
      >[ x, y, z > 0, for P inside ABC]
      >
      >We have:
      >
      >BA' = zh_b / (ysinB + zsinC), A'C = yh_c / (ysinB + zsinC), etc
      >
      >where h_a, h_b, h_c are the altitudes of ABC.
      >
      >Therefore the (1) is equivalent to:
      >
      >a(2ax + by + cz) = b(ax + 2by + cz) = c(ax + by + 2cz)
      >
      >==>

      2a^x + aby + acz =

      bax + 2b^y + bcz =

      cax + cby + 2c^z

      ==>

      x y z
      ----------------- = ----------------- = ----------------- ==>
      | 1 ab ac | | 2a^2 1 ac | | 2a^2 ab 1 |
      | | | | | |
      | 1 2b^2 bc | | ba 1 bc | | ba 2b^2 1 |
      | | | | | |
      | 1 cb 2c^2 | | ca 1 2c^2 | | ca cb 1 |


      ==> (x:y:z) = ([3bc - a(b + c)]/a ::) in normals


      Sorry!

      Antreas
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