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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

SADBOYS
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    I think there is a mistake in your problem. The side should not appear in the sine function. Suppose that it's true for a particular triangle. If I increase the size of the triangle by a factor of $2$, the $OA, OB, OC$ increase by a factor of two as well. Which is fine in the first equation. But the argument of sines wil double, and I assume it might change the equation. The only valid solution would be $O=A=B=C$ – Andrei Dec 01 '18 at 15:22
  • The mistake was that the angles were written in lowercase and I thought those were the sides – SADBOYS Dec 01 '18 at 15:26

1 Answers1

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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 $O$ $$0=\frac{\sin(2A)\vec{OA}+\sin(2B)\vec{OB}+\sin(2C)\vec{OC}}{\sin(2A)+\sin(2B)+\sin(2C)}.$$

Jack D'Aurizio
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