HYACINTHOS 29381

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C' the pedal triangle of O,

Denote:

Na, Nb, Nc = the NPC centers of NBC, NCA, NAB, resp.

 

A'Na intersects again the NPC (N) at A*
B'Nb intersects again the NPC (N) at B*

C'Nc intersects again the NPC (N) at C*

A*B*C*, NaNbNc are orthologic,
The orthologic center (A*B*C*, NaNbNc) lies on the (N)


[Peter Moses]:


Hi Antreas,

>The orthologic center (A*B*C*, NaNbNc) lies on the (N)

MIDPOINT OF X(4) AND X(32749) =

= (b - c)^2*(b + c)^2*(-a^12 + 2*a^10*b^2 + 3*a^8*b^4 - 12*a^6*b^6 + 13*a^4*b^8 - 6*a^2*b^10 + b^12 + 2*a^10*c^2 + 4*a^8*b^2*c^2 - 10*a^6*b^4*c^2 - 4*a^4*b^6*c^2 + 12*a^2*b^8*c^2 - 4*b^10*c^2 + 3*a^8*c^4 - 10*a^6*b^2*c^4 - 3*a^4*b^4*c^4 - 6*a^2*b^6*c^4 + 7*b^8*c^4 - 12*a^6*c^6 - 4*a^4*b^2*c^6 - 6*a^2*b^4*c^6 - 8*b^6*c^6 + 13*a^4*c^8 + 12*a^2*b^2*c^8 + 7*b^4*c^8 - 6*a^2*c^10 - 4*b^2*c^10 + c^12)*(-a^14 + 5*a^12*b^2 - 11*a^10*b^4 + 15*a^8*b^6 - 15*a^6*b^8 + 11*a^4*b^10 - 5*a^2*b^12 + b^14 + 5*a^12*c^2 - 16*a^10*b^2*c^2 + 21*a^8*b^4*c^2 - 12*a^6*b^6*c^2 - 5*a^4*b^8*c^2 + 12*a^2*b^10*c^2 - 5*b^12*c^2 - 11*a^10*c^4 + 21*a^8*b^2*c^4 - 11*a^6*b^4*c^4 - 3*a^4*b^6*c^4 - 5*a^2*b^8*c^4 + 9*b^10*c^4 + 15*a^8*c^6 - 12*a^6*b^2*c^6 - 3*a^4*b^4*c^6 - 4*a^2*b^6*c^6 - 5*b^8*c^6 - 15*a^6*c^8 - 5*a^4*b^2*c^8 - 5*a^2*b^4*c^8 - 5*b^6*c^8 + 11*a^4*c^10 + 12*a^2*b^2*c^10 + 9*b^4*c^10 - 5*a^2*c^12 - 5*b^2*c^12 + c^14) : :

= lies on the nine-point circle and this line:  {4, 32749}

 

= midpoint of X(4) and X(32749).


Best regards,
Peter Moses.