Πέμπτη 29 Δεκεμβρίου 2011

McCay Cubic


Let ABC be a triangle, P a point and A1B1C1 the circumcevian triangle of P.


The perpendiculars from A1,B1,C1 to the sidelines of ABC, BC,CA,AB resp. intersect again the circumcircle at A2,B2,C2, resp. (other than A1,B1,C1).

Denote:

Ab := the orthogonal projection of A2 on B1B2
Ac := the orthogonal projection of A2 on C1C2

(O1) := the circumcircle of A2AbAc.

Similarly (O2), (O3).

Which is the locus of P such that (O1),(O2),(O3) are concurrent?

APH, 29 December 2011

-------------------------------------------------------------

The locus is Line at infinity + Circumcircle + McCay cubic.

If ABC is right angled
A = pi/2 then there is no locus.
For every point P the circles are concurrent at the intersection of lines B1B2, C1C2.

Nikos Dergiades, Hyacinthos #20610

Δεν υπάρχουν σχόλια:

Δημοσίευση σχολίου

Another relationship between Napoleon cubic and Neuberg cubic

Another relationship between Napoleon cubic K005 and Neuberg cubic K001 The world of Triangle Geometry is very intrincate. There are many...