HYACINTHOS 26128

 

 

[Le Viet An]: 

 

Let ABC be a triangle and MaMbMc the pedal triangle of O.

Denote: 

 

Pa = AO /\ MbMc, Pb = BO /\ McMa, Pc = CO /\ MaMb

N, Na, Nb, Nc = the NPC centers of MaMbMc, OPbPc, OPcPa, OPaPb, resp.

N, Na, Nb, Nc are concyclic.

Which point is the center of the circle?

[César Lozada]:

 

Z = cos(B-C)*(2*(2*cos(A)-cos(3*A) )*cos(B-C)+cos(2*(B-C))-cos(2* A)-2*cos(4*A)+3/2) : : (trilinears)

= (S^2+SB*SC)*(-5*R^4+(-7*SA+5* SW)*R^2+5*S^2+2*SA^2-4*SB*SC- SW^2) : : (barycentrics)

= (9*R^4-6*SW*R^2+S^2+SW^2)*X( 140)-R^4*X(389)

= On line {140,389}

= [ -2.848704262814094, 5.22984429553289, 1.334789629375803 ]

 

The given circle passes through X(140) (=NPC of medial-triangle) and has radius:

r =( S/16)*sqrt(-2*S^2+9*R^4+2*SW^ 2-8*SW*R^2)*|CyclicProduct(a^ 3/(S^2+SB*SC)) |

 

César Lozada