Re: Special Points assoc. with Circular Pivotal Cubics
- 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.
> > Obviously, 3) and 4) below represent the same configuration.[JPE]
> > The relationship with 1) is perhaps more interesting. It seems
> > circumcevian triangle of the pivot point Q is always perspective
> > 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,
> > 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
> > special points are also on the pK. From these points, we can get
> > bunch more special points on the pK.
> > What are the coordinates of these special points?
> In my previous mail, I've noticed that A' = CT inter BT' where T isabove
> 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
> 2) The isogonal conjugate of the common point of the 4 circles
> lies on the cubic too (the common point doesn't lie on the
> 3) I don't think that A'B'C' and the circumcevian triangle of Q are
> Friendly. Jean-Pierre
- 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.
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:
> 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
> 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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