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