HYACINTHOS 20405

Thank you, Francisco.

Another property I forgot to mention:

If P* is the inverse-in-circumcircle of P, then PA(P) is the isogonal conjugate of the {P,P*}-Harmonic conjugate of X(3).

Best regards,
Randy Hutson

>________________________________
> From: Francisco Javier <garciacapitan@gmail.com>
>
>
> 
>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.
>
>--- 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
>>
>