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Re: Special Points assoc. with Circular Pivotal Cubics

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  • Jeff Brooks
    Dear Jean-Pierre, Thank you very much for your results in message 11474. I never noticed the construction you gave for triangle A B C or the complex
    Message 1 of 31 , Aug 25, 2005
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      Dear Jean-Pierre,

      Thank you very much for your results in message 11474. I never
      noticed the construction you gave for triangle A'B'C' or the complex
      associations used in your proof. I am eager to explore this further.

      I think I created confusion using the word "special" to refer to all
      the various concurrences. Only in the case 3) below did I mean for
      the concurrence to lie on the circumcircle and here I should have
      said the circumcevian triangle of Q* or the isogonal conjugate of Q.
      Of course, P can be any point in this case. Interestingly, the
      circles A'B'C, B'C'A, C'A'B and ABC concur at the same circumcircle
      point. I would like to establish a construction like the one you
      gave in message 11474 using these facts eventhough they are not
      specific to the circular isocubic case.

      Sincerely,
      Jeff Brooks



      [JB]
      > > Obviously, 3) and 4) below represent the same configuration.
      > >
      > > The relationship with 1) is perhaps more interesting. It seems
      > the
      > > circumcevian triangle of the pivot point Q is always perspective
      > with
      > > triangle A'B'C' at a point on the circumcircle for all points P.
      > > Call this circumcircle point a special point.
      > >
      > > Assuming P lies on the non-isogonal circular pK with pivot Q,
      > then
      > > A'B'C' is perspective with the reference triangle ABC at another
      > > special point and circles ABC', BCA', CAB' concur at yet another
      > > special point. It seems the ISOGONAL CONJUGATES of these
      > particular
      > > special points are also on the pK. From these points, we can get
      > a
      > > bunch more special points on the pK.
      > >
      > > What are the coordinates of these special points?

      [JPE]
      > In my previous mail, I've noticed that A' = CT inter BT' where T is
      > second intersection of the line QB with the circle CPQ and T' is
      > second intersection of the line QC with the circle BPQ.
      > For any points P and Q, the four circles ABC', BCA', CAB', A'B'C'
      > have a common point.
      > Now, if P lies on the circular isocubic with pivot Q :
      > 1) A'B'C' and ABC are perspective and the isogonal conjugate of the
      > perspector lies on the cubic (the perspector doesn't lie on the
      > circumcircle : his locus is obviously the isogonal conjugate of the
      > cubic)
      > 2) The isogonal conjugate of the common point of the 4 circles
      above
      > lies on the cubic too (the common point doesn't lie on the
      > circumcircle)
      > 3) I don't think that A'B'C' and the circumcevian triangle of Q are
      > perspective
      > Friendly. Jean-Pierre
    • Francois Rideau
      Dear Jeff You have a circle of center O and a point A on it. You have a line L on A and you want the other point of intersection B of line L with
      Message 31 of 31 , Oct 26, 2007
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        Dear Jeff
        You have a circle <Gamma> of center O and a point A on it.
        You have a line L on A and you want the other point of intersection B of
        line L with circle <Gamma>.
        To avoid to be stuck, follow this way and make a macro with it:
        Draw through the center O the line D perpendicular to L.
        Then B is the symmetric of A wrt line D.
        Friendly
        Francois
        PS
        You can also be stuck with 2 circles with a known point of intersection when
        you want to draw the other.
        I let you find the solutionand the macro

        On 10/26/07, Jeff <cu1101@...> wrote:
        >
        > Dear Friends,
        >
        > In Hyacinthos message 11474, Jean-Pierre gave the following
        > construction and proof:
        >
        > -----
        > [JPE]
        > Construction of A' (cyclically for B', C')
        > T = second intersection of the line BQ with the circle CPQ
        > T'= second intersection of the line CQ with the circle BPQ
        > then A' = CT inter BT'
        >
        > Proof : If a,b,c,p,q,a' are the complex affixes of A,B,C,P,Q,A', we
        > have (p-a')/(p-c) = (p-b)/(p-q) hence the direct similitude with
        > center P mapping Q to B maps C to A' and the direct similtude with
        > center P mapping Q to C will map B to A'. Hence the construction
        > above.
        > -----
        >
        > I'm having problems making a GSP construction here. Given the four
        > points B,C,P,Q I construct the circles CPQ and BPQ and the lines BQ
        > and CQ respectively with no problem. But, I notice that points T and
        > T' will get stuck at point Q as the point P is dragged about.
        >
        > I've seen this same situation occur in a different setting but don't
        > know how to solve the problem. Can anyone help?
        >
        > Sincerely, Jeff
        >
        >
        >


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