HYACINTHOS 24797

[Antreas P. Hatzipolakis]:

 

Inspired from Seiichi's posting Hyacinthos 24793

Let ABC be a triangle and Ja, Jb, Jc the ex-Schiffler points (*)

Denote:

Ma, Mb, Mc = the midpoints of AJa, BJb, CJc, resp.

The centroid of the MaMbMc lies on the line JaJbJc =  Euler line of ABC.

(*)

Ja is the point of concurrence of the Euler lines of IaBC, IaCA, IaAB and similarly Jb,Jc, where IaIbIc is the excentral triangle.

[Peter Moses]:

Hi Antreas,

 

4 a^12-21 a^10 b^2+45 a^8 b^4-50 a^6 b^6+30 a^4 b^8-9 a^2 b^10+b^12+10 a^10 b c+10 a^9 b^2 c-34 a^8 b^3 c-34 a^7 b^4 c+42 a^6 b^5 c+42 a^5 b^6 c-22 a^4 b^7 c-22 a^3 b^8 c+4 a^2 b^9 c+4 a b^10 c-21 a^10 c^2+10 a^9 b c^2+30 a^8 b^2 c^2+14 a^7 b^3 c^2-15 a^6 b^4 c^2-36 a^5 b^5 c^2+8 a^3 b^7 c^2+12 a^2 b^8 c^2+4 a b^9 c^2-6 b^10 c^2-34 a^8 b c^3+14 a^7 b^2 c^3+22 a^6 b^3 c^3+4 a^5 b^4 c^3-2 a^4 b^5 c^3-2 a^3 b^6 c^3+14 a^2 b^7 c^3-16 a b^8 c^3+45 a^8 c^4-34 a^7 b c^4-15 a^6 b^2 c^4+4 a^5 b^3 c^4-12 a^4 b^4 c^4+16 a^3 b^5 c^4-3 a^2 b^6 c^4-16 a b^7 c^4+15 b^8 c^4+42 a^6 b c^5-36 a^5 b^2 c^5-2 a^4 b^3 c^5+16 a^3 b^4 c^5-36 a^2 b^5 c^5+24 a b^6 c^5-50 a^6 c^6+42 a^5 b c^6-2 a^3 b^3 c^6-3 a^2 b^4 c^6+24 a b^5 c^6-20 b^6 c^6-22 a^4 b c^7+8 a^3 b^2 c^7+14 a^2 b^3 c^7-16 a b^4 c^7+30 a^4 c^8-22 a^3 b c^8+12 a^2 b^2 c^8-16 a b^3 c^8+15 b^4 c^8+4 a^2 b c^9+4 a b^2 c^9-9 a^2 c^10+4 a b c^10-6 b^2 c^10+c^12::
on Euler line.

 

Best regards,

Peter Moses.

 

[APH]:

Thanks, Peter !

 

GENERALIZATION:

Let ABC be a triangle and Ja, Jb, Jc the ex-Schiffler points.

Denote:

A*, B*, C* = Points of AJa, BJb, CJc, resp. such that:

A*Ja / A*A = B*Jb / B*B = C*Jc / C*C = t.

The locus of the centroid of A*B*C* as t varies is the line JaJbJc = Euler line.

APH