The sides of a triangle ABC, inscribed in a hyperbola $xy = c^2$, makes angles $\alpha, \beta, \gamma $ with an asymptote.
Prove that the normals at A, B, C will meet in a point if $cot2\alpha + cot2\beta + cot 2\gamma = 0$
My approach: Assuming $A \left(ct_1,\frac{c}{t_1}\right), B \left(ct_2,\frac{c}{t_2}\right) and C \left(ct_3,\frac{c}{t_3}\right)$, we can work out $t_1t_2=-cot\alpha$ and so on
Equation of a normal to rectangular hyperbola is $t^3x-ty=ct^4-c$ which is fourth degree equation which doesn't help the cause. Using determinants for concurrency doesn't yield any fruitful results.
I worked out the answer myself and since I couldn't get the solution elsewhere online, I have decided to post it
We have already worked out $t_1t_2=-cot\alpha, t_2t_3=-cot\beta,$ and $t_3t_1=-cot\gamma.$
Also, equation of normal to hyperbola is $ct^4-xt^3+ty-c=0$
From the equation we get $S_4=t_1t_2t_3t_4=-1.$ We can write $t_4=\frac{-1}{t_1t_2t_3}$
Also, $S_2=t_1t_2+t_2t_3+t_3t_1+t_4(t_1+t_2+t_3)=0$
We can write, $S_2=t_1t_2+t_2t_3+t_3t_1+\frac{-1}{t_1t_2t_3}(t_1+t_2+t_3)=0$ which is $\sum \left( t_1t_2-\frac{1}{t_1t_2} \right)=0$
$\sum \left( cot\alpha-tan\alpha \right)=0 \Rightarrow \sum cot2\alpha=0$