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## 9173Re: [EMHL] Orthologic Triangles

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• Feb 1, 2004
Dear Fred, Darij and Antreas
> Dear Jean-Pierre, Antreas and Darij
> > [FL]
> > > Let ABC be a triangle, P a point, and (Oa, Ob, Oc)
> > > the Circumcenters of the triangles PBC, PCA, PAB, resp.
> > >
> > > The locus of P such that the perpendiculars
> > > from A,B,C to POa, POb, POc are concurrent is:
> > > infinity line*circumcircle*KjP cubic.
> > [JPE]
> > 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.
> > Friendly. Jean-Pierre
> [FL]
> That's fine.
> A link between nK0 and pK.
>
> I have the pencil MacCay -4s tan(phi) Kjp.
>
> This cubic is an isocubic iff phi = 0 or phi = Pi/2
>
> Remark: MacCay and Kjp are isogonal cubic but this cubic is not
> (if phi non 0 or Pi/2).
>
> Question (probably difficult) to JPE: Locus of perspector?

This is the same cubic with -phi instead of phi - ie the isogonal
conjugate of the locus of P -.
Of course, in the case of Mac Cay or Kjp, the locus of the common
point of La, Lb, Lc and the locus of P are the same ones.
Friendly. Jean-Pierre
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