A variable line intersecting $n$ concurrent straight lines

Question

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.

Answer 1

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 $$\left(\frac{1}{k}\sum_{i=1}^{n}\cos\theta_i,\frac 1k\sum_{i=1}^{n}\sin\theta_i\right)$$