HYACINTHOS 10039

Dear Paul,

> [APH]: Let ABC be a triangle and P a point.
> The line AP intersects the circumcircle of triangle
> PBC at A' [other than P].
> Let A" be the orthogonal projection of A' on BC.
> Similarly define B" and C".
>
> [FvL]: Triangle A"B"C" is always perspective to ABC, and the mapping
> of P to this perspector is called by Antreas the REHHAGEL mapping.
> What if we take the desmic mate A1B1C1 of A'B'C' and the orthogonal
> projections of A1B1C1 to ABC as A*B*C*? Is A*B*C* perspective to ABC?
>
> [PY]: If P = (u:v:w), A*B*C* is perspective with ABC at
>
> Q = (1/(-a^2S_A vw + b^2S_B wu + c^2S_c uv + b^2c^2 u^2) : ... : ...).
>
> Here are some examples:
>
> P Q
> -----------------------------
> I, X(36) X(8)
> X(4), X(186) X(68)
>
> *** Note that X(36) is the inverse of I in the circumcircle, and X
> (186) is that of X(4). Also, for P = X(15), X(16) [isodynamic
> points], Q = G, the centroid.
>
>
> This point Q(P) has appeared before. Nik [Hyacinthos 6325] has found
> that the reflection of the pedal triangle of P in its own circumcenter
> is perspective with ABC at Q(P). It is indeed true that inverses in
> the circumcircle have the same Q.

Thank you very much!

Let us call A'B'C' the CircleCevian triangle of P. Because A'B'C' is a
Jacobi-triangle as Darij pointed out, A1B1C1 is just the CircleCevian
triangle of P*.
This means that the REHHAGEL mapping is isogonal conjugacy followed by Nik's
mapping (which I was prepared to call MICHELS mapping - MICHELS was coach of
Dutch soccer teams winning the Euro Champs in 1988 and being runner up of
the World Champs of 1974).

Note that A'A1 // B'B1 // C'C1 // PP*

Kind regards,
Sincerely,
Floor.