## Re: Cofactor Triangle

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• Dear friends Now I think my affine construction of the dual line is false, so forget it but all the rest is still true. PQR and its cofactor triangle beeing
Message 1 of 4 , Sep 1, 2006
Dear friends
Now I think my "affine construction " of the dual line is false, so forget
it but all the rest is still true. PQR and its cofactor triangle beeing
perspective is the outcome of a general result about 2 triangles PQR and
P'Q'R' dual wrt some conic {Gamma}, here the equalizer. They are
perspective and if the perspector is O then the axis of perspective is the
polar of O wrt {Gamma}.
In case {Gamma} is a circle, it is clear that PQR and P'Q'R' are orthologic,
the 2 centers of orthology are the same point to say the center of the
circle {Gamma}, so fixed by the affine map f sending PQR to P'Q'R'.
It rest to look at the general case when {Gamma} is not a circle and to
understand why the center of {Gamma} is the fixed point of f and the fixed
lines of f are real and their directions conjugate wrt {Gamma}.
I notice that the perspector O of PQR and P'Q'R' is named "eigencenter" in
MathWorld, why?
Is there a connection with eigenvalues of some linear map?
Friendly
François

On 9/1/06, Francois Rideau <francois.rideau@...> wrote:
>
> Dear friends
> Given the affine reference triangle ABC and an arbitrary point D not on
> the sidelines of ABC, we look at the projective frame {A,B,C;D} of which D
> is the unit point.
> In this frame, every point M of the (extended projective) plane have
> homogeneous coordinates (x:y:z) and we have:
> A(1:0:0), B(0:1:0), C(0:0:1), D(1,1,1)
> In this frame, the dual line (L) of M(x:y:z) has equation:
> x.X + y.Y + z.Z = 0
> (L) is simply the polar line of M wrt the equalizer conic of equation: x.x+
> y.y + z.z = 0
> But as it is not very easy to construct a polar line wrt an imaginary
> conic,I give you (without proof!),a neat construction of (L) so you could
> check all what I said!
> Let {Gamma} the circumconic shaped with the equalizer, i.e sharing with
> the equalizer the same points at infinity.
> An equation of {Gamma} in the {A,B,C;D} frame is : y.z + z.x + x.y = 0
> Then {Gamma} is just the circumconic with center D, so very easy to
> construct.
> Let (L') be the polar line of M wrt {Gamma}, also easy to construct, then
> (L) is the image of (L') in the dilation of center D and ratio -1/2, also
> easy to construct.
> I am eager to have a more projective way to construct (L) from A, B, C, D,
> M.
> Of course with Cabri, I have a macro giving the fixed point and the
> invariant lines of any affine map f.
> For example, if you choose D as the circumcenter, then PQR and its
> cofactor triangle P'Q'R' are orthologic and perspective!
> Check it!
>
> Friendly
> François
>
>
>
> On 9/1/06, Francois Rideau <francois.rideau@...> wrote:
> >
> > Dear friends
> > We have already noticed the surprising fact that with the barycentric
> > definition of the cofactor triangle P'Q'R' of PQR wrt ABC that the fixed
> > point of the affine map f sending PQR to P'Q'R' is always the centroid G of
> > ABC.I also notice that the invariant lines of the map f are always real
> > (through G) and the directions of these 2 lines are conjugate wrt the
> > equalizer conic of barycentric equation: x.x + y.y + z.z =0 and so also
> > conjugate wrt every shaped conic with the equalizer, for instance the
> > Steiner ellipses.
> > Hence if ABC is equilateral, then triangles PQR and P'Q'R' are always
> > orthologic! Check that with your dynamic geometry software!
> > So in this special case, we could apply the Sondat theorem for triangles
> > PQR and P'Q'R' for they are also perspective.
> > By the way, I remember to have seen a recent and (russian?)
> > semi-synthetic proof of the Sondat theorem, different than the Sollertinski
> > projective synthetic complicated proof.
> > But I have lost it! Please, could you give me the reference!
> > I have also said the definition of cofactor triangle and most of their
> > properties are still valid in the plane with any homogeneous coordinates wrt
> > projective frame {A,B,C,D} where ABC is the affine reference triangle and D
> > an arbitrary unit point.
> > What can be said in this general case and look at the cases where D is
> > the incenter or the circumcenter.
> > Friendly
> > François
> >
> >
> > On 9/1/06, Francois Rideau < francois.rideau@...> wrote:
> > >
> > >
> > >
> > > Dear friends
> > > I just saw the definition of the unary cofactor triangle P'Q'R' of a
> > > triangle PQR wrt a triangle ABC in MathWorld.
> > > This definition uses trilinears and so the euclidian structure of the
> > > plane.
> > > What can be said if we replace trilinears by barycentrics, using only
> > > the affine structure of the plane?
> > > Is there a name for such triangle P'Q'R'? affine cofactor triangle?
> > > Some properties of the unary cofactor triangle are preserved. For
> > > example, PQR and P'Q'R' are still perspective. What's the name for the
> > > perpector O?
> > > The axis of perpective is the dual of O wrt ABC
> > > In a sense PQR and P'Q'R' are duals wrt ABC, for the dual of P wrt ABC
> > > is Q'R', the dual of P' wrt ABC is QR and so one...
> > > Besides, the centroid G of ABC is the fixed point of the affine map
> > > sending P to P', Q to Q', R to R' and so one...
> > > I think all these properties are already known and I need some
> > > references.
> > > Thanks in advance
> > > Friendly
> > > PS
> > > Of course, all these definitions and properties are still valid in a
> > > projective plane, using homogeneous coordinates wrt some projective frame
> > > {A,B,C,D} where D is an arbitrary unit point.
> > > François
> > >
> >
> >
>

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