HYACINTHOS 26497

[Tran Quang Hung]:

 

 

Let ABC be a triangle. 

 

A',B',C' are midpoints of BC,CA,AB.

 

N is NPC center of ABC.

 

A'',B'',C'' are isogonal conjugate of N wrt AB'C',BC'A',CA'B', respectively.

 

Then orthocenter of A''B''C'' lies on Euler line of ABC. 

 

Which is this point ?

 

 

[César Lozada]:

 

H” =  complement of X(13150)

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

= On lines: {2, 3}, {6750, 12300}

= complement of X(13150)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (1884, 7546, 444), (1884, 13852, 4238), (2675, 7557, 6622), (3147, 13362, 7401), (6906, 7557, 13383), (8368, 11099, 14035)

= [ 7.715803135082416, 6.82592872869296, -4.646118392610322 ]

 

César Lozada