For P = I the concurrence point is the same point L that in Hyacinthos message
#17947


--- In Hyacinthos@yahoogroups.com, "xpolakis" <anopolis72@...> wrote:
>
> Let ABC be a triangle and P = (x:y:z) a point.
>
> Denote:
>
> A* :=(Perpendicular from B to CP) /\ (Perpendicular from C in BP)
>
> B* :=(Perpendicular from C to AP) /\ (Perpendicular from A to CP)
>
> C* :=(Perpendicular from A to BP) /\ (Perpendicular from B to AP)
>
> [The triangles A*BC, B*CA, C*AB share the same Orthocenter P]
>
> Oa := The Circumcenter of A*BC
> Ob := The Circumcenter of B*CA
> Oc := The Circumcenter of C*AB
>
> P* := The Point of Concurrence of the Circumcircles
> of A*BC, B*CA, C*AB
> [We have seen recently this point in Hyacinthos]
>
> La := The Reflection of P*Oa in AI
>
> Lb := The Reflection of P*Ob in BI
>
> Lc := The Reflection of P*Oc in CI
>
> I think that for P = I the lines La, Lb, Lc are
> concurrent. Point?
>
> In general, which is the locus of P such that the La,Lb,Lc
> are concurrent?
>
>
> Antreas
>