Browse Groups

• 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
View Source
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
> > >
> >
> >
>

[Non-text portions of this message have been removed]
Your message has been successfully submitted and would be delivered to recipients shortly.
• Changes have not been saved
Press OK to abandon changes or Cancel to continue editing
• Your browser is not supported
Kindly note that Groups does not support 7.0 or earlier versions of Internet Explorer. We recommend upgrading to the latest Internet Explorer, Google Chrome, or Firefox. If you are using IE 9 or later, make sure you turn off Compatibility View.