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9174Re: [EMHL] Orthologic Triangles

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  • Bernard Gibert
    Feb 1, 2004
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      Dear Jean-Pierre and Fred,

      > [JP] Consider three lines La, Lb, Lc going respectively through A, B, C
      > > and such as (La,POa) = (Lb,POb) = (Lc,POc) = phi where (L1,L2) is
      > > the measure modulo Pi of the oriented angle of lines L1, L2.
      > > Then La, Lb, Lc concur if and only if
      > > (AP,BC)+(BP,CA)+(CP,AB) = Pi/2-phi
      > > Thus, we get the MacCay cubic for phi = 0, the Kjp cubic for phi =
      > > Pi/2 and, for any theta, a member of the pencil generated by Kjp and
      > > MacCay.

      All the cubics of the pencil are K60+ with three asymptotes concurring
      at G.

      There is only one K60++ in the pencil, the one passing through G.

      For which angle phi is it obtained ?

      Best regards

      Bernard

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