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9179Re: [EMHL] Orthologic Triangles

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  • fredlangch
    Feb 2 11:55 AM
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      Dear Jean-Pierre and Bernard,
      --- In Hyacinthos@yahoogroups.com, "jpehrmfr"
      <jean-pierre.ehrmann@w...> wrote:
      > Dear Bernard
      > > > [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.
      > [BG]
      > > 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 ?
      > If I didn't mistake, if U = (AG,BC)+(BG,CA)+(CG,AB), then
      > SU = -(a^2-b^2)(b^2-c^2)(c^2-a^2)/(a^2b^2+b^2c^2+c^2a^2)
      > I don't know if it is possible to get a nice expression for U, may
      > be using cot(AG,BC) = 1/2(cot C-cot B).
      > Friendly. Jean-Pierre

      I arrive at the same expression for phi:

      phi = -(a^2-b^2)(b^2-c^2)(c^2-a^2)/(a^2b^2+b^2c^2+c^2a^2)/s

      Good evening.
      Fred
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