HYACINTHOS 20403

Dear Hyacinthists,

Awhile ago, I found a mapping that has some interesting properties:

Given a point P, let A'B'C' be the pedal triangle of P.
Let O' be the circumcenter of A'B'C', and let A"B"C" be the reflection of A'B'C' in O' (the 'pedal antipodal triangle' of P).
A"B"C" is perspective to ABC at a point I'll call the 'pedal antipodal perspector' of P, or PA(P).
If P lies on the circumcircle, let PA(P) be the isogonal conjugate of P (on the line at infinity).

PA(P) maps P to the rectangular circumhyperbola that is the isogonal conjugate of line OP, where O is the circumcenter of ABC.

The center of this hyperbola lies on the circumcircle of A'B'C'.

The isogonal conjugate of P, P', naturally lies on this same hyperbola, and is the reflection of P in O'.

The isogonal conjugate of PA(P), PA'(P), maps back to the line OP, at a point closer to the circumcircle than P.

If P* is the inverse-in-circumcircle of P, then PA(P*) = PA(P).

PA(X(1)) = PA(X(36)) = X(8)
PA(X(2)) = PA(X(23)) = isogonal conjugate of X(1995)
PA(X(3)) = X(4)
PA(X(4)) = PA(X(186)) = X(68)
PA(X(6)) = PA(X(187)) = isogonal conjugate of X(1384)
PA(X(15)) = PA(X(16)) = X(2)
PA(PU(1)) = PA(PU(2)) = PU(37)

Questions:

Has this mapping been studied before?

If P=p:q:r, what is the formula for PA(P)?

There is a unique point P such that P = PA(P). It is a non-ETC center (search: 2.675235602529272). What are its trilinears?

Best regards,
Randy Hutson