HYACINTHOS 26948

[Tran Quang Hung]:

 

Let ABC be a triangle.

 

H is orthocenter

 

(N) is the nine-point circle.

 

Ra is radical axis of (N) and (BHC).

Rb is radical axis of (N) and (CHA).

Rc is radical axis of (N) and (AHB).

 

A*B*C* is triangle bounded by Ra,Rb,Rc.

 

Then the NPC center of triangle A*B*C* lies on Euler line of ABC.

 

Which is this point ?

 

 

[Angel Montesdeoca]:

Dear Tran Quang Hung,

   The NPC of triangle A*B*C* lies on Euler line of ABC with first barycentric coordinate:

(a^2+b^2-c^2) (a^2-b^2+c^2) (2 a^12
            -13 a^10 (b^2+c^2)
            +a^8 (33 b^4+40 b^2 c^2+33 c^4)
            -a^6 (42 b^6+25 b^4 c^2+25 b^2 c^4+42 c^6)
            +4 a^4 (7 b^8-7 b^6 c^2-3 b^4 c^4-7 b^2 c^6+7 c^8)
            -9 a^2 (b^2-c^2)^2 (b^6-2 b^4 c^2-2 b^2 c^4+c^6)
            +(b^2-c^2)^4 (b^4-6 b^2 c^2+c^4))

On the line {2,3}
           
 (6 - 9 - 13) - search numbers    (3.08684042285455, 2.20917733697973, 0.686538438065543).
 
Best regards,
 Angel Montesdeoca