Beads on a frictionless frame of wire I was doing a physics problem and though I have the solution, I have derived a general result which I would like to get verified. I will state the problem here :

Two beads q1, q2 of equal mass m can slide on a fixed frictionless massless rod, bent at right angle. Initially, the beads are held at rest at distances d and 2d from the corner as shown and then released simultaneously. When one bead reaches the corner, where will the other bead be ? Treat the beads as particles. The setup is isolated from any external interference.


I solved this problem by considering small intervals of time dt and I found that after each dt, the ratio of their distances from the corner remains 1:2 , so the line joining these beads is parallel to the initial line joining the beads at all moments in time. And if we were to take any ratio of initial distances from the corner (not just 1:2) , the result would be the same i.e. the beads meet at the corner. Now, if these conclusions are correct, I think there is a beautiful something (symmetry?) going on here which could be explained by using less Mathematics and more Physics. Is there some intuitive way to arrive at this result ? I did think that the wires(frame) does zero work, but got stuck at a later point.
I was hoping I could get some help.
 A: It is not clear whether it is gravitational attraction that causes the beads to move, or some other force, but in fact this is irrelevant. The force could be due to gravitational attraction or electrostatic attraction or an invisible spring or anything else.
Newton's Third Law tells us that whatever force $q_1$ exerts on $q_2$, then $q_2$ exerts an equal and opposite force on $q_1$. If we consider $q_1$ and $q_2$ together, then this "internal" force between them cancels out and the remaining forces on the system are a normal force $N_1$ on $q_1$ and a normal force $N_2$ on $q_2$.
The ratio $N_1 : N_2$ is $d : 2d$ and the net force therefore acts along a line from $C$ to $O$ where $C$ is mid-way between $q_1$ and $q_2$ and $O$ is the corner. But since $q_1$ and $q_2$ have equal masses, their centre of mass is at $C$, so the centre of mass of the system moves in a straight line towards $O$. So the beads reach $O$ at the same time.
Your intuition that this result does not depend on the initial distances is correct.
