Let ABC be a triangle, P a point and (a),(Ob),(Oc) the circles

with diameters PA,PB,PC, resp.

The circles (Ob), (Oc) intersect in point A1, other than P, on BC

Similarly B1,C1. The triangle A1B1C1 is the pedal triangle of P.

Let A2B2C2 be the cevian triangle of P.

The circles (Oa),(Ob), (Oc) intersect (O) at A3,B3,C3 [other than A,B,C, resp.]

The lines AA3, BB3, CC3 [= readical axes of

((O),((Oa),((O),((Ob),((O),((Oc), resp.]

bound a triangle A4B4C4.

Loci:

Which is the locus of P such that be perspective the triangles:

1. ABC, A1B1C1 : it is Darboux cubic (since A1B1C1 is the pedal triangle of P)

2. ABC, A4B4C4

3. A1B1C1, A3B3C3

4. A1B1C1, A4B4C4

5. A2B2C2, A3B3C3,

6. A2B2C2, A4B4C4

7. A3B3C3, A4B4C4

I think that in all cases the locus is Darboux cubic + possibly something else.

APH