HYACINTHOS 25555

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[Tran Quang Hung]:

Let ABC be a triangle and F is the first Fermat point.

A'B'C' is cevian triangle of F.

Nbc and Ncb are NPC center of triangle FBC' and FCB', reps.

Line da is perpendicular bisector of segment NbcNcb.

Similarly, we have lines db,dc.

Then da,db,dc are concurrent.

Which is this point ?

Best regards,

Tran Quang Hung.

 

[César Lozada]:

For F=X(13):

Q1 =3*sqrt(3)*(2*a^6-4*(b^2+c^2)* a^4+(b^4-12*b^2*c^2+c^4)*a^2+( b^4-c^4)*(b^2-c^2)) -2*S*(2*a^4+20*(b^2+c^2)*a^2+2 0*b^2*c^2-7*b^4-7*c^4) : :  (barycentrics)

= on lines: {2, 17}, {6671, 8014}

= [ 1.852608295707447, 1.41567166874883, 1.805534113216366 ]

 

For F=X(14)

Q2 =3*sqrt(3)*(2*a^6-4*(b^2+c^2)* a^4+(b^4-12*b^2*c^2+c^4)*a^2+( b^4-c^4)*(b^2-c^2)) + 2*S*(2*a^4+20*(b^2+c^2)*a^2+20 *b^2*c^2-7*b^4-7*c^4) : :  (barycentrics)

= on lines: {2, 18} , {6672, 8015}

= [ -0.947260191490169, 3.99310251131908, 1.313405908605095 ]