Problem Statement:- A variable line cuts $n$ given concurrent straight lines at $A_1,A_2,A_3,\ldots A_n$, such that $\displaystyle\sum_{i=1}^{n}{\frac{1}{{OA}_{i}}}$ is a constant. Show that it always passes through a fixed point, $O$ being the point of intersection of the lines.
Attempt at a solution:-
Let the point of intersection of the $n$ concurrent lines be the origin. Then the general equation of the n lines will be of the form $y=m_i x$
Now let the equation of the variable line be $y=mx+c$.
Then the point of intersection of the variable line $y=mx+c$ with the $n$ concurrent lines $y=m_ix$, where $i\in\{1,2,3,\ldots,n\}$ is $$(x,y)=\left(\frac{c}{m_i-m}, \frac{c\cdot m_i}{m_i-m}\right)$$
Now, $$O{A}_i=\sqrt{\left(\frac{c}{m_i-m}\right)^2+\left(\frac{c\cdot m_i}{m_i-m}\right)^2}=\left|\frac{c}{m_i-m}\right|\sqrt{1+{m_i}^2}$$
$$\therefore \frac{1}{OA_i}=\left|\frac{m_i-m}{c}\right|\cdot\frac{1}{\sqrt{1+{m_i}^2}}$$
As per the problem statement
$$\sum_{i=1}^{n}{\frac{1}{{OA}_{i}}}=\sum_{i=1}^{n}{\left|\frac{m_i-m}{c}\right|\cdot\frac{1}{\sqrt{1+{m_i}^2}}}=k\text{ (say})$$
My gut feeling says that the modulus must open as follows
$$\pm\sum_{i=1}^{n}{\frac{m_i-m}{c}\cdot\frac{1}{\sqrt{1+{m_i}^2}}}=k$$
which will happen only when all the $(m_i-m)$ have the same sign which can be either positive or negative.
So we would get either $$\left(\sum_{i=1}^{n}{\frac{1}{\sqrt{1+{m_i}^2}}},\sum_{i=1}^{n}{\frac{m_i}{\sqrt{1+{m_i}^2}}}\right)$$ or $$\left(-\sum_{i=1}^{n}{\frac{1}{\sqrt{1+{m_i}^2}}},-\sum_{i=1}^{n}{\frac{m_i}{\sqrt{1+{m_i}^2}}}\right)$$ as the fixed points.
But I dont know how to deal with the modulus appropriately here, your help is appreciated.
And as always do post other fantastic solution that you can think of.
Considering in polar coordinates seems a key.
Let $O$ be the origin, and let $\theta=\theta_1,\theta=\theta_2,\cdots,\theta=\theta_n$ be the given fixed $n$ lines.
Let the variable line be $\frac 1r=a\cos\theta+b\sin\theta$.
We may suppose that the line cuts the given fixed lines at $(r_1,\theta_1),(r_2,\theta_2),\cdots, (r_n,\theta_n)$.
Now, we have $$\sum_{i=1}^{n}\frac{1}{r_i}=\sum_{i=1}^{n}(a\cos\theta_i+b\sin\theta_i)=a\sum_{i=1}^{n}\cos\theta_i+b\sum_{i=1}^{n}\sin\theta_i=k\ (\text{say})$$
Hence, we get that the line always passes through $$\left(\frac{1}{k}\sum_{i=1}^{n}\cos\theta_i,\frac 1k\sum_{i=1}^{n}\sin\theta_i\right)$$