Dear Antreas,

> [APH] In general, if Q = (u:v:w) is a fixed point and

> P = (x:y:z) is a variable point, then the locus of P such

> that the trilinear polar of P is perpendicular to QP line,

> is the cubic with trilinear equation (if my calculations

> were correct):

>

> x(y^2(u - wcosB) - z^2(u - vcosC)) + cyclic = 0

>

> If Q = O, then the locus is Darboux cubic, which contains H.

I'm not sure your computation is correct. The barycentric equation I

have is :

[ (SB u + a^2w)y - (SC u + a^2v)z] yz + cyclic = 0

It is a circum-cubic K passing through G and having its asymptotes

parallel to those of the Thomson cubic which is obtained with Q = O.

1. When Q lies at infinity, K decomposes into the line at infinity and

a circum-conic through G.

2. K is a pK iff Q lies on Darboux and then, the pole W lies on

Thomson, the pivot R on Lucas.

We find a family of pK having all their asymptotes parallel to the

Thomson cubic.

question : what are the mappings Q -> W, Q -> R ?

3. K is a nK0 iff Q lies on the circumcircle and then, the pole is Q

and the root is the tripole of the Simson line of Q (a point on the

Simson cubic).

Best regards

Bernard

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