HYACINTHOS 26471

[Tran Quang Hung]:

Let ABC be a triangle with NPC center N.

 

A'B'C' is circumcevian triangle of N.

 

N' is isogonal conjugate of N wrt A'B'C'.

 

Na,Nb,Nc are NPC centers of NB'C',NC'A',NA'B'.

 

Prove that NPC center of triangle NaNbNc lies on line NN'.

 

Which is this point ?

 

 

[César Lozada]:

 

 

N’ = (2*cos(2*A)-4*cos(4*A)+12)* cos(B-C)+(8*cos(A)-6*cos(3*A)) *cos(2*(B-C))+(2*cos(2*A)-2* cos(4*A)+3)*cos(3*(B-C))+2* cos(A)*cos(4*(B-C))+cos(7*A)+ 5*cos(A)-3*cos(3*A)-2*cos(5*A) : : (trilinears)

= On lines: {5, 252}, {110, 10203}

= [ 0.329472308293511, -0.70475928795733, 3.976510615896293 ]

 

N* = NPC-of-NaNbNc =

= (205*cos(2*A)+254*cos(4*A)- 103*cos(6*A)-5*cos(8*A)+3*cos( 10*A)-665/2)*cos(B-C)+(-295* cos(A)+323*cos(3*A)-23*cos(5* A)-41*cos(7*A)+7*cos(9*A))* cos(2*(B-C))+(85*cos(2*A)+121* cos(4*A)-53*cos(6*A)+4*cos(8* A)-158)*cos(3*(B-C))+(-88*cos( A)+88*cos(3*A)-10*cos(7*A))* cos(4*(B-C))+(5*cos(2*A)+19* cos(4*A)-7*cos(6*A)-25)*cos(5* (B-C))+(-8*cos(A)+7*cos(3*A)+ cos(5*A))*cos(6*(B-C))-209* cos(A)+233*cos(3*A)+7*cos(9*A) +cos(11*A)-13*cos(5*A)-33*cos( 7*A) : :  (trilinears)

= on line (5, 252)

= [ 0.654426886816252, -0.14233826325378, 3.437163177937188 ]

 

César Lozada