ADGEOM 3939

 

 

[Tran Quang Hung]:

Dear geometers,

 Let ABC be a triangle with orthocenter H.

 Fermat axis of triangle HBC,HCA,HAB meets BC,CA,AB at A',B',C'.

 Then AA',BB',CC' are concurrent. Which is this point ?

 

Best regards,

Tran Quang Hung.

 

[César Lozada]:

 

Q = polar conjugate of X(252)

= (S^2+SB*SC)*(SA^2-3*S^2)*SB*SC : : (barys)

= On lines: {2, 10979}, {4, 569}, {53, 311}, {94, 2052}, {275, 11538}, {472, 8836}, {473, 8838}, {1568, 6750}, {1625, 1993}, {3078, 10003}

= polar conjugate of X(252)

= {X(53), X(467)}-Harmonic conjugate of X(324)

= [ 0.103478799694122, 0.06249012912866, 3.549642638805712 ]

 

César Lozada