HYACINTHOS 20404

Dear Randy, I'm not able to answer the second question.

If P=(x:y:z), your PA(P) is

{(-a^2 c^2 x y - b^2 c^2 x y + c^4 x y - 2 a^2 c^2 y^2 + a^2 b^2 x z -
b^4 x z + b^2 c^2 x z + a^4 y z - a^2 b^2 y z -
a^2 c^2 y z) (a^2 c^2 x y + b^2 c^2 x y - c^4 x y - a^2 b^2 x z +
b^4 x z - b^2 c^2 x z + a^4 y z - a^2 b^2 y z - a^2 c^2 y z -
2 a^2 b^2 z^2), -(2 b^2 c^2 x^2 + a^2 c^2 x y + b^2 c^2 x y -
c^4 x y + a^2 b^2 x z - b^4 x z + b^2 c^2 x z + a^4 y z -
a^2 b^2 y z - a^2 c^2 y z) (a^2 c^2 x y + b^2 c^2 x y - c^4 x y -
a^2 b^2 x z + b^4 x z - b^2 c^2 x z + a^4 y z - a^2 b^2 y z -
a^2 c^2 y z -
2 a^2 b^2 z^2), -(2 b^2 c^2 x^2 + a^2 c^2 x y + b^2 c^2 x y -
c^4 x y + a^2 b^2 x z - b^4 x z + b^2 c^2 x z + a^4 y z -
a^2 b^2 y z - a^2 c^2 y z) (-a^2 c^2 x y - b^2 c^2 x y +
c^4 x y - 2 a^2 c^2 y^2 + a^2 b^2 x z - b^4 x z + b^2 c^2 x z +
a^4 y z - a^2 b^2 y z - a^2 c^2 y z)}

which is the isotomic conjugate of

{-2 b^2 c^2 x^2 - a^2 c^2 x y - b^2 c^2 x y + c^4 x y - a^2 b^2 x z +
b^4 x z - b^2 c^2 x z - a^4 y z + a^2 b^2 y z +
a^2 c^2 y z, -a^2 c^2 x y - b^2 c^2 x y + c^4 x y - 2 a^2 c^2 y^2 +
a^2 b^2 x z - b^4 x z + b^2 c^2 x z + a^4 y z - a^2 b^2 y z -
a^2 c^2 y z,
a^2 c^2 x y + b^2 c^2 x y - c^4 x y - a^2 b^2 x z + b^4 x z -
b^2 c^2 x z + a^4 y z - a^2 b^2 y z - a^2 c^2 y z - 2 a^2 b^2 z^2}

Best regards,

Francisco Javier García Capitán

 

--- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@...> wrote:
>
> 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
>