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

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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

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