- 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.

[JPE]

> >

> > 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?

> In my previous mail, I've noticed that A' = CT inter BT' where T is

above

> 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

> 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 - 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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