HYACINTHOS 25925

[Antreas P. Hatzipolakis]:

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

Denote:

Ja, Jb, Jc = the incenters of AB'C', BC'A', CA'B', resp.

The circumcircles of IA'Ja, IB'Jb, IC'Jc are coaxial.

 

2nd point of intersection?

 

 

[Peter Moses]:

 

Hi Antreas,

 

 a ((b-c) (a+b-c) (a-b+c) (a^2-b^2-4 b c-c^2)+2 b (b-c) c (5 a^2-2 a b-3 b^2-2 a c-10 b c-3 c^2) Sin[A/2]+2 c (a-b+c) (2 a^3+6 a^2 b+3 a b^2+b^3-9 a b c-2 a c^2-b c^2) Sin[B/2]-2 b (a+b-c) (2 a^3-2 a b^2+6 a^2 c-9 a b c-b^2 c+3 a c^2+c^3) Sin[C/2])::

 

 on lines {{1,167},{517,12814},{3616,705 7},{5571,10503},{6728,10492},{ 11033,11192}}.

{X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (1,2089,177),(1,8091,8422),(1, 8241,11234),(1,11044,11191).

X(10215)-Ceva conjugate of X(177).

inverse in the incircle of X(177).

inverse in the Conway circle of  X(12554).

Searches {-1.49100712766482258377242751 861,-1.25302028366845282120011 539583,5.196297198753995325690 32538512}.

 

Best regards,

Peter Moses.