Question

If orthocentre of triangle $ABC$ is $H_1$, orthocentre of triangle $B{H}_{1}C$ is $H_2$, orthocentre of triangle $B{H}_{2}C$ is $H_3$, and so on, where $A\equiv (1,1),B\equiv (1,2)$ and $C\equiv (2,3)$.

If coordinates of $H_{10}$ are $(\alpha ,\beta )$ and $V_1=\alpha +\beta .$
If $L$ is/are the lines inclined at an angle of ${60}^0$ with line $L_1\equiv x+\sqrt{3}y+1=0$ and $V_2=$ reciprocal of sum of slope of all possible lines $L$, then $V_1+V_2=$

Answer 1

Orthocentre of $\Delta ABC$ is $H_1$

Orthocentre of $\Delta B{H}_{1}C$ is $A$ which is equivalent to $H_2$
Similarly,

$H_{10}=A(1,1)$
So $V_1=2$
$M_1=-\frac{1}{\sqrt{3}}\Rightarrow \theta =-{30}^{\circ }$
So possible values of slope of L are $=\left(tan\right({30}^{\circ }),tan(-{90}^{\circ }\left)\right)$
$=\frac{1}{\sqrt{3}},-\infty$
$V_2=\frac{1}{\frac{1}{\sqrt{3}}+(-\infty )}=0$
i.e $V_1+V_2=2$