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  • Dear Randy, > [RH] Let ABC be a triangle, and P a point. > Let A'B'C' be the pedal triangle of P. > Let Ba, Ca be the orthogonal projections of A' onto lines CA, AB, resp. > Define Cb, Ab, Ac, Bc cyclically. > > What is the locus of P such that Ba, Ca, Cb, Ab, Ac, Bc lie on a common conic? The locus would include H, for which the conic is the Taylor circle. a quintic with many...
    Bernard Gibert Apr 17, 2013
  • Dear C�sar Lozada, > If P=(p : q : r ) and U=( u : v : w), > > what�s the geometrical meaning of Q = ( p*u : q*v : r*w) You may want to consult �1.2.2 in http://bernard.gibert.pagesperso-orange.fr/files/isocubics.html Best regards Bernard [Non-text portions of this message have been removed]
    Bernard Gibert Apr 15, 2013
  • Dear Randy, > Given two fixed isogonal points, X and X', and two variable isogonal > points, P and P', what is the locus of P such that X, X', P, P' are > concyclic? > > Special cases: X,X' = G,K; O,H; 1st and 2nd Brocard points? I find a bicircular isogonal circum-sextic passing through the in/excenters, X, X', the intersections of (O) and the line XX', their isogonal conjugates...
    Bernard Gibert Apr 1, 2013
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  • Dear Randy, > Are you sure? The Zosma transforms of each of the listed centers on K034 lie on this cubic: You're right and I owe you an apology. I made a stupid mistake using the trilinear macro of Zosma instead of the barycentric macro !!! Nevertheless, the mapping M -> X4 x ctM is still valid. Sorry about that Best regards Bernard [Non-text portions of this message have been...
    Bernard Gibert Mar 22, 2013
  • Dear Randy, > I forgot to include X(1824) in list of centers on the curve. It appears this curve is the Zosma transform of K034. Is this correct? I don't think so. It should be K184 = pK(X76, X76). Best regards Bernard [Non-text portions of this message have been removed]
    Bernard Gibert Mar 22, 2013
  • Dear Randy, > What curve is the inverse-in-polar-circle of the anticomplement of the > Darboux quintic, Q071 > ? It > passes through X(4), X(25), X(51), X(1827), X(1828), X(1843), and the > crosssum of X(3) and X(20). I find no matches on Bernard's site. your curve is the pivotal cubic with pivot X(1843), isopivot X(4) with respect to the orthic triangle. the mapping M -> X(4) x...
    Bernard Gibert Mar 21, 2013
  • Dear Antreas, > I think that, since for P lying on K634 the point of contact of the > antipedal and pedal circles of P > is lying on the circumcircle, the K634 is part of a more general locus: > > Which is the locus of P such that the pedal circle of P, the antipedal > circle of P and the circumcircle, > are concurrent? > > The locus should be K634 (for pedal and antipedal circles...
    Bernard Gibert Mar 16, 2013
  • Dear friends > No, it is "the new cubic" > > S^2 xyz + CyclicSum[ a^2 y z (c^2 y + b^2 z)] = 0 this will be K634, the orthoptic pedal cubic. I will modify the pages K003, K191, K192 and I ask you to carefully check if this corresponds to your findings. many thanks Bernard [Non-text portions of this message have been removed]
    Bernard Gibert Mar 16, 2013
  • Dear friends, > As Bernard Gibert pointed out to us, we were wrong, this is a different cubic, not K191. indeed, although it is a cubic of the pencil generated by K024 and K191, since it is the locus of P whose pedal circle is orthogonal to the orthoptic circle of the Steiner in-ellipse. best regards Bernard [Non-text portions of this message have been removed]
    Bernard Gibert Mar 16, 2013
  • Dear Antreas, Angel, Francisco and all Hyacintists, > Now we have two new geometrical properties of K191 > > http://anthrakitis.blogspot.gr/2013/03/mccay-cubic-circumcevian-triangle.html > http://anthrakitis.blogspot.gr/2013/03/k191-pedal-and-antipedal-circles-tangent.html > > Bernard may include them in the cubic's page: > http://bernard.gibert.pagesperso-orange.fr/Exemples/k191...
    Bernard Gibert Mar 15, 2013