9174Re: [EMHL] Orthologic Triangles
- Feb 1, 2004Dear Jean-Pierre and Fred,
> [JP] Consider three lines La, Lb, Lc going respectively through A, B, CAll the cubics of the pencil are K60+ with three asymptotes concurring
> > 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.
There is only one K60++ in the pencil, the one passing through G.
For which angle phi is it obtained ?
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