HYACINTHOS 26709

 [Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C' the pedal triangle of I.

The lines A'A, B'B, C'C intersect again the incircle at A", B", C", resp. (other than A',B',C', resp.)

The circumcircles of IA'A", IB'B", IC'C" are coaxial.

 

 

[Peter Moses]:

Hi Antreas,

X(1323).

Best regards,
Peter Moses.

 

[APH]:

 

I think the generalization is this:


Let ABC be a triangle and A'B'C' the pedal triangle of a point P.

The lines A'A, B'B, C'C intersect again the pedal circle of P, with center O',  at A", B", C", resp. (other than A',B',C', resp.)

If P lies on the Darboux cubic, then the circumcircles of O'A'A", O'B'B", O'C'C" are coaxial.
The axis is the line O'PQ, where Q is the point of concurrence of AA', BB', CC'.

For P = H, O, L [=X20] : Which is the second point of intersection (other than O') of the circles ?

[Peter Moses]:

Hi Antreas,

 

H -> X(403).

 

O -> X(858).

L -> (a^2-b^2-c^2) (4 a^14-7 a^12 b^2-8 a^10 b^4+21 a^8 b^6-4 a^6 b^8-13 a^4 b^10+8 a^2 b^12-b^14-7 a^12 c^2+34 a^10 b^2 c^2-25 a^8 b^4 c^2-52 a^6 b^6 c^2+71 a^4 b^8 c^2-14 a^2 b^10 c^2-7 b^12 c^2-8 a^10 c^4-25 a^8 b^2 c^4+112 a^6 b^4 c^4-58 a^4 b^6 c^4-48 a^2 b^8 c^4+27 b^10 c^4+21 a^8 c^6-52 a^6 b^2 c^6-58 a^4 b^4 c^6+108 a^2 b^6 c^6-19 b^8 c^6-4 a^6 c^8+71 a^4 b^2 c^8-48 a^2 b^4 c^8-19 b^6 c^8-13 a^4 c^10-14 a^2 b^2 c^10+27 b^4 c^10+8 a^2 c^12-7 b^2 c^12-c^14):: 

on line {20, 64}.

 

Best regards,

Peter Moses.