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Let points $A,B,C$ are represented by $(a\cos\theta_1,a\sin\theta_1),(a\cos\theta_2,a\sin\theta_2),(a\cos\theta_3,a\sin\theta_3)$ respectively and $\cos(\theta_1-\theta_2)+\cos(\theta_2-\theta_3)+\cos(\theta_3-\theta_1)=\frac{-3}{2}$.Then prove that $(i)$ triangle $ABC$ is a equilateral triangle and (ii)orthocenter of triangle $ABC$ is at the origin.


For the triangle to be equilateral,$AB=BC=CA$.I found
$AB=2a\sin\frac{(\theta_1-\theta_2)}{2},BC=2a\sin\frac{(\theta_2-\theta_3)}{2},CA=2a\sin\frac{(\theta_3-\theta_1)}{2}$
For $AB=BC=CA$,we need to have $\theta_1-\theta_2=\theta_2-\theta_3=\theta_3-\theta_1$

But i am stuck here,i dont know how to prove it.And i could not prove the second part,Please help me.

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1 Answer 1

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Let $a=R$ (it is the circumradius of $ABC$). By the cosine theorem, $$ \cos(\theta_1-\theta_2) = \frac{OA^2+OB^2-AB^2}{2 OA\cdot OB} = 1-\frac{1}{2}\left(\frac{AB}{R}\right)^2 $$ hence our triangle fulfills: $$ AB^2+AC^2+BC^2 = 9R^2 $$ so the centroid of $ABC$ and the circumcenter of $ABC$ are the same point.

That gives that $ABC$ is equilateral, so $O\equiv G\equiv H$.

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