Question

The vertices of a triangle are $A(p,p\tan \alpha ),B(q,q\tan \beta ),C(r,r\tan \gamma )$. If circumcentre $O$ of triangle $ABC$ is at the origin and $H(\overline{x},\overline{y})$ be its orthocentre, then show that
$\frac{\overline{x}}{\overline{y}}=\frac{\cos \alpha +\cos \beta +\cos \gamma }{\sin \alpha +\sin \beta +\sin \gamma }$

Answer 1

Let $x^2+y^2=R^2$ be the equation of the circle circumscribing the triangle $ABC$ as its centre $O$ is $(0,0)$. The vertices lie on it

$\therefore p^2+p^2{\tan }^2\alpha =R^{2}orp\sec \alpha =R$

$\therefore p=R\cos \alpha$

Hence the point $A(p,p\tan \alpha )$ is $A(R\cos \alpha ,R\sin \alpha )$

Similarly $B$ and $C$ are
$(R\cos \beta ,R\sin \beta )$ and $(R\cos \gamma ,R\sin \gamma )$

If $G$ be the centroid of triangle $ABC$, then
$G=(\frac{R\sum \cos \alpha }{3},\frac{R\sum \sin \alpha }{3}),O=(0,0)$

and $H(\overline{x},\overline{y})$

From geometry we know that $O,G,H$ are collinear
Hence slope of $OG=$ slope of $OH$

$\therefore \frac{\overline{y}-0}{\overline{x}-0}=\frac{\frac{R\sum \sin \alpha }{3}-0}{\frac{R\sum \cos \alpha }{3}-0}\therefore \frac{\overline{x}}{\overline{y}}=\frac{\cos \alpha +\cos \beta +\cos \gamma }{\sin \alpha +\sin \beta +\sin \gamma }$