HYACINTHOS 26435

[Tran Quang Hung]:

 

Let ABC be a triangle with symmedian point L.

 La,Lb,Lc are the symmedian points of triangle LBC,LCA,LAB, resp.

 LbLc cuts LA at A'. Similarly, we have B',C'.

 Then triangle LaLbLc and A'B'C' are perspective and the perspector lies on the Brocard axis of ABC.

 

 Which is this point ?

 

[César Lozada]:

Perspector = (7*a^2+4*(b^2+c^2))*a^2 : :  (barycentrics)

= 3*S^2*X(3)-11*SW^2*X(6)

= On lines: {3,6}, {111,7954}, {1992,7820}, {3619,7826}, {3620,7889}, {3630,7822}, {3856,5305}, {5304,7753}, {5306,10109}, {5354,8585}, {7794,11008}, {7798,12150}, {7916,10583}

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (6, 187, 7772), (6, 5008, 574), (6, 5024, 5041), (39, 5210, 574), (574, 5008, 32)

= [ 0.814623682275958, 1.35369374678525, 2.327511726928765 ]

 

César Lozada