HYACINTHOS 25762

[Tran Quang Hung]:

 

Let ABC be a triangle. 

 

A'B'C' be pedal triangle of NPC center N.

 

Euler line of triangle AB'C' cuts B'C' at A''.

 

Similarly, we have B'',C''.

 

Then A'A'',B'B'',C'C'' are concurrent at point P and isogonal conjugate of P wrt A'B'C' lies on Euler line of triangle ABC.

 

[César Lozada]:

P = X(12006) = midpoint  {3,143}

 

P*’ = isogonal(P) wrt A’B’C’ =

= (S^2+SB*SC)*(2*SA^2+4*(2*R^2- SW)*SA-R^2*(5*R^2+2*SW)+3*S^2+ SW^2) : :  (barycentrics)

= On lines: {2,3}, {137,10095}

= [ 1.658034190727441, 0.78414033176986, 2.332551548807966 ]

 

César Lozada