HYACINTHOS 23955

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Antreas P. Hatzipolakis

 

[APH]:

 

Let ABC be a triangle and A'B'C' the cevian triangle of a point P.

Denote:

Nab, Nac = the NPC centers of ABA', ACA', resp.

Ma = the midpoint of NabNac

Similarly Mb, Mc

Which is the locus of P such that the centroid of MaMbMc lies on the Euler line ?

The Euler line?

 
[César Lozada]:

Which is the locus of P such that the centroid of MaMbMc lies on the Euler line ? The Euler line?

No. The locus is the cubic (SA^2+8*SB*SC-3*S^2)*x*(y^2-z^ 2) + … = 0 (barycentrics), passing through vertices of the anticomplementary triangle, G and no other ETC center.

 

For P=G the centroid of MaMbMc is

Gm = 9*SB*SC+13*S^2 : : (barycentrics)

= 13*X(3)+11*X(4)

= midpoint of X(i),X(j) for these {i,j}: {2,5066}, {5,547}, {140,381}, {376,3853}, {546,549}, {3828,9955}                   

= reflection of X(i) in X(j) for these (i,j): (3628,547), (3860,5066), (3861,381)

= On lines:  (2,3), (1327,8252), (1328,8253), (1587,6495), (1588,6494), (3582,3614), (3584,7173), (3655,7989), (3656,7988), (3828,9955), (4669,5844), (4677,8227), (4745,9956), (5355,7603), (5663,6688)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k):  (2,5,5066), (2,381,8703), (2,3545,3830), (2,3830,549), (2,8703,140),…

= [ 1.071996004348358, 0.19964812892372, 3.007679159876411 ]

 

César Lozada