Dear Antreas, Fred and Darij

[FL]

> Let ABC be a triangle, P a point, and (Oa, Ob, Oc)

> the Circumcenters of the triangles PBC, PCA, PAB, resp.

>

> The locus of P such that the perpendiculars

> from A,B,C to POa, POb, POc are concurrent is:

> infinity line*circumcircle*KjP cubic.

Consider three lines La, Lb, Lc going respectively through A, B, C

and such as (La,POa) = (Lb,POb) = (Lc,POc) = phi where (L1,L2) is

the measure modulo Pi of the oriented angle of lines L1, L2.

Then La, Lb, Lc concur if and only if

(AP,BC)+(BP,CA)+(CP,AB) = Pi/2-phi

Thus, we get the MacCay cubic for phi = 0, the Kjp cubic for phi =

Pi/2 and, for any theta, a member of the pencil generated by Kjp and

MacCay.

Friendly. Jean-Pierre