Question

The vertices of a triangle are the points (p, p $\tan \alpha$), (q, q $\tan \beta$), (r, r$\tan \gamma$), where $\alpha +\beta +\gamma =\pi$. If the circumcentre of the triangle is at the origin, then prove that its orthocentre lies on the line $x\left(4\cos \alpha /2\cos \beta /2cos\gamma /2\right)-y(1+4\sin \alpha /2\sin \beta /2\sin \gamma /2)=0$.

Answer 1

Since $OA$=$OB$=$OC$=$R$
$\Rightarrow p^2(1+{\tan }^2\alpha )=R^2$ $\therefore psec\alpha =R$
$\therefore p=R\cos \alpha$ or $R\cos \beta$ or $R\cos \gamma$
$\therefore$ A is $(R\cos \alpha ,R\sin \alpha )$ etc.
If $(\overline{x},\overline{y})$ be the orthocentre, then O, H, G are collinear.
$\therefore$ slope of OH$=$slope of OG
$\frac{\overline{x}}{\overline{y}}=\frac{\cos \alpha +\cos \beta +\cos \gamma }{\sin \alpha +\sin \beta +\sin \gamma }$
Since $\alpha +\beta +\gamma =\pi$
$\therefore \frac{\overline{x}}{\overline{y}}=\frac{1+4\sin \alpha /2\sin \beta /2\sin \gamma /2}{4\cos \alpha /2\cos \beta /2\cos \gamma /2}$
Hence locus of $(\overline{x},\overline{y})$ is as given.