--- In Hyacinthos@yahoogroups.com
, Barry Wolk <wolkbarry@...> wrote:
> --- On Sat, 8/28/10, Francisco Javier <garciacapitan@...> wrote:
> > The perspector is not simple.
> > In its coordinates, shown below, we can see the symbols Ra,
> > Rb, Rc, that stand for the square root of 4(s-b)(s-c),
> > 4(s-c)(s-a), 4(s-a)(s-b), respectively.
> > It is remarkable that these coordinates appear to be non
> > symmetric, as we can think from the problem. I think that
> > the formula can be transformed in some way into a symmetric form.
> > I've checked that this point is on lines FaF'a, FbF'b, FcF'c.
> > Coordinates of the perspector:
> [long formula shipped]
> If a point P is defined geometrically in a symmetric way (usually as a perspector), then there is a general method of getting symmetric coordinates for P. If x,y,z are complicated functions of a,b,c, and a symmetrically defined P has non-symmetric coordinates
> P = (x(a,b,c),y(a,b,c),z(a,b,c) ) = (x1,y1,z1), then also
> P = (z(b,c,a),x(b,c,a),y(b,c,a) ) = (z2,x2,y2), and
> P = (y(c,a,b)),z(c,a,b),x(c,a,b) ) = (y3,z3,x3). Then symmetric coordinates for P are (x1+y3+z2,x2+y1+z3,x3+y2+z1).
> Barry Wolk
> Note: The symmetric coordinates for Q are obtained using the method described by Barry Wolk in the post # 19239 of Hyacinthos.
> Angel Montesdeoca
I now have an update to that post. At the time I didn't know how to handle certain points (such as the circular points at infinity) where the above method gave the meaningless result (0:0:0).
For such points P, apply the symmetrization method to the isotomic conjugate of P. Symmetric coordinates for that conjugate lead immediately to symmetric coordimates for P.
This idea gave the following result: A cyclically symmetric parametrization of the circumcircle is:
P(t) = ( 1/(S_A*S_A - S_B*S_C + t(S_B-S_C)) :
1/(S_B*S_B - S_C*S_A + t(S_C-S_A)) :
1/(S_C*S_C - S_A*S_B + t(S_A-S_B)) )
This just has to be useful for some calculations. The circular points at infinity are at t = +i S and t = -i S.