Prove that $\sin(2A)\overrightarrow{OA}+\sin(2B)\overrightarrow{OB}+\sin(2C)\overrightarrow{OC} =0 $
Where $O$ is the circumradius of the $ABC$ triangle
How to approach this type of problem?
How do we demonstrate that for P ( a point in the triangle ) we get :
$\overrightarrow{PO}=\frac{\sin(2A)\overrightarrow{PA}+\sin(2B)\overrightarrow{PB}+\sin(2C)\overrightarrow{PC}}{\sin(2A)+\sin(2B)+\sin(2C)} =0 $ only if O = P
The exact trilinear coordinates of the circumcenter $O$ are $[R\cos A;R\cos B;R\cos C]$, hence the trilinear coordinates are $[\cos A;\cos B;\cos C]$ and the barycentric coordinates are $[a\cos A; b\cos B; c\cos C]$ or $[\sin(2A);\sin(2B);\sin(2C)]$. It follows that $$\vec{O}=\frac{\sin(2A)\vec{A}+\sin(2B)\vec{B}+\sin(2C)\vec{C}}{\sin(2A)+\sin(2B)+\sin(2C)}$$ and if the reference system is centered at $O$ $$0=\frac{\sin(2A)\vec{OA}+\sin(2B)\vec{OB}+\sin(2C)\vec{OC}}{\sin(2A)+\sin(2B)+\sin(2C)}.$$