## 9179Re: [EMHL] Orthologic Triangles

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• Feb 2, 2004
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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