<Tri-2025-2218>, hint

Let H(A) lie on the circumcircle of ABC and symmetric to H about BC.

Since A'HH(A) is isosceles OA'+A'H =OA'+A'H(A)=R. A' is the point on BC,

intersectin of BC with OH(A)

Vector AA'=OA'-OA, then AA'+BB'+CC'=(OA'+OB'+OC')-(OA+OB+OC)=0

which means OA+'OB'+OC'=0. because OA+OB+OC=0.  IF H=O , equilateral,

OA'+OB'+OC=1/2(OB+OC+OC+OA+OA+OB)=0