HYACINTHOS 23436

Antreas P. Hatzipolakis

 

[APH]:

I do not remember if I have sent this before....

Let ABC be a triangle and Na,Nb,Nc the NPC centers

of IBC, ICA, IAB, resp.

Let M1,M2,M3 be the midpoints of ANa, BNb, CNc, resp.

The circumcenter of M1M2M3 lies on the Euler line of ABC

(ie it is the intersection of the Euler lines of ABC and M1M2M3).

Point?

APH

[Angel Montesdeoca]:

Dear Antreas,

The circumcenter of M1M2M3  is

(2 a^7-2 a^6 (b+c)+(b-c)^4 (b+c)^3+a^5 (-5 b^2+2 b c-5 c^2)-a (b^2-c^2)^2 (b^2+3 b c+c^2)-2 a^2 (b-c)^2 (2 b^3+3 b^2 c+3 b c^2+2 c^3)+a^4 (5 b^3+b^2 c+b c^2+5 c^3)+a^3 (4 b^4+b^3 c+4 b^2 c^2+b c^3+4 c^4) : ... : ... )

search numbers in ETC
(1.86501930097493, 0.990579410070208, 2.09410059756273)

Best regards
Angel M.