Question

Let $\tan \alpha ,\tan \beta ,\tan \gamma ;\alpha ,\beta ,\gamma \ne \frac{(2n-1)\pi }{2}$,$n\in N$ be the slopes of three line segment $OA,$ $OB$ and $OC,$ respectively, where $O$ is origin. If the circumcentre of $△ABC$ coincides with origin and its orthocentre lies on $y-$ axis, then the value of ${\left(\frac{\cos 3\alpha +\cos 3\beta +\cos 3\gamma }{\cos \alpha \cos \beta \cos \gamma }\right)}^2$ is equal to

Answer 1

As origin is circumcentre, so
$OA=OB=OC=R$
The coordinates of the vertices of the triangle are
$A=(R\sin \alpha ,R\cos \alpha )B=(R\sin \beta ,R\cos \beta )C=(R\sin \gamma ,R\cos \gamma )$

Since orthocentre and circumcentre both lies on $y-$axis, so centroid also lies on $y-$axis
$R[\cos \alpha +\cos \beta +\cos \gamma ]=0\Rightarrow \cos \alpha +\cos \beta +\cos \gamma =0\Rightarrow {\cos }^3\alpha +{\cos }^3\beta +{\cos }^3\gamma =3\cos \alpha \cos \beta \cos \gamma$
Now,
$\frac{\cos 3\alpha +\cos 3\beta +\cos 3\gamma }{\cos \alpha \cos \beta \cos \gamma }=\frac{4({\cos }^3\alpha +{\cos }^3\beta +{\cos }^3\gamma )-3(\cos \alpha +\cos \beta +\cos \gamma )}{\cos \alpha \cos \beta \cos \gamma }=12\therefore {\left(\frac{\cos 3\alpha +\cos 3\beta +\cos 3\gamma }{\cos \alpha \cos \beta \cos \gamma }\right)}^2=144$