HYACINTHOS 29614

[Kadir Atintas]:

 

Let ABC be a triangle and P be a point. 

 

Denote:

 

Oa, Ob, Oc = the circumcenters of PBC, PCA, PAB, resp. 

Ka, Kb, Kc = the symmedian points of PObOc, POcOa, POaOb, resp. 

Which is the locus of P such that OaObOc, KaKbKc are perspective?

Some perspectors?

 

 

[César Lozada]:

 

 

Locus={Linf} ∪ { uncatalogued cubic pK(X(6), X(524)) : ∑[a^2*((a^2+b^2-2*c^2)*y+(-a^2+2*b^2-c^2)*z)*y*z]=0, through excenters and ETC’s 1, 2, 6, 111, 524, 2930, 5524, 5525, 7312, 7313, 8591, 13574 }

 

ETC-pairs (P,Q(P)=perspector) : (1, 1386), (111, 3)

 

Q( X(2) ) = X(2)X(31744) ∩ X(3)X(5476)

= 3*a^10+2*(b^2+c^2)*a^8-(5*b^4+47*b^2*c^2+5*c^4)*a^6-3*(b^4+b^2*c^2+c^4)*(b^2+c^2)*a^4+2*(b^2+3*b*c+c^2)*(b^2-3*b*c+c^2)*(b^2+c^2)^2*a^2+(b^4-c^4)^2*(b^2+c^2) : :

= lies on these lines: {2, 31744}, {3, 5476}, {14876, 26613}

= [ 2.0109599319241350, 1.5994132279448190, 1.6052430478268270 ]

 

Q( X(13574) ) = X(110)X(6093) ∩ X(5189)X(10748)

= 9*((SB+SC)*(81*R^4*SA+2*SW^3)+3*(SA-5*SW)*SW^2*R^2)*S^4+(81*(SB+SC)*R^2+2*(9*SA-7*SW)*SW)*SA*SW^3*S^2+4*SB*SC*SW^6 : :

= lies on these lines: {110, 6093}, {5189, 10748}

= [ 2.5310765768405630, -4.1700459629412490, 5.35  94301900172930 ]

 

 

César Lozada