HYACINTHOS 25504

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and A'B'C' the pedal triangle of O.

1. Denote:

Na, Nb, Nc = the NPC centers of NBC, NCA, NAB, resp.
N1, N2, N3 = the NPC centers of N'B'C', N'C'A', N'A'B', resp, where N ' = the NPC of A'B'C'

(ie N1, N2, N3 are the complements of Na, Nb, Nc, resp.)

 

Ma, Mb, Mc = the midpoints of NaN1, NbN2, NcN3, resp.

The circumcenter of MaMbMc lies on the Euler line of ABC.

2. Denote:

Oa, Ob, Oc = the circumcenters of NBC, NCA, NAB, resp.
O1, O2, O3 = the circumcenters of N'B'C', N'C'A', N'A'B', resp, where N ' = the NPC of A'B'C'

(ie O1, O2, O3 are the complements of Oa, Ob, Oc, resp.)

M1, M2, M3 = the midpoints of OaO1, ObO2, OcO3, resp.

The NPC of M1M2M3 lies on the Euler line of ABC.

 

[César Lozada]:

1)      Z1 = 9*(cos(2*A)+cos(4*A)-11/6)* cos(B-C)-10*(cos(A)-cos(3*A))* cos(2*(B-C))+(cos(2*A)-6)*cos( 3*(B-C))-cos(5*A)-11*cos(A)+ 11*cos(3*A) : : (trilinears)

= on line {2,3}

= [ 1.175103165240290, 0.30248329052179, 2.888897666050696 ]

 

2)      Z2 = 11*(cos(2*A)+cos(4*A)-3/2)* cos(B-C)-2*(5*cos(A)-6*cos(3* A))*cos(2*(B-C))-(cos(2*A)+5)* cos(3*(B-C))+cos(5*A)-9*cos(A) +11*cos(3*A) :: (trilinears)

= on line {2,3}

= midpoint of X(140) and X(10289)

= [ 3.824072288380420, 2.94446436622948, -0.162767289350071 ]

 

César Lozada