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Suppose we have $5$ points in plane, each lying on a line for which no three of these lines intersect in one point, and also non of these $5$ points is an intersection point of two lines. At time $t=0$, each of the points starts to move on its own line in an arbitrary direction and with an arbitrary but constant positive speed. Each point keeps going unless it meets another point. When so, the two points reverse their directions and go back the path they've come. Prove that after some finite time $T$, one of the points will never meet anymore points.

This problem popped into my mind some weeks ago. I kept thinking about it for a long while, but I couldn't reach anything. All the answers are welcomed!

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    You want one fixed track (line) for each point. Is that right? The intro might be clearer if you defined the tracks first, then put a moving point on each track.2012-09-11
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    Can each point move at its own distinct constant speed, or are all of the points moving at the same speed?2012-09-11
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    To your first question, yes. Each point has it's own track. Also, each point move at its own distinct constant speed.2012-09-11
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    And I don't quite get the condition "none of these 5 points is an intersection point of two lines". This could make sense to me initially, but after some time has elapsed and two points encounter each other at the same time, then certainly at that moment two tracks are intersecting right there.2012-09-11
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    Note that for an even number of points, this would not be true: you could have the points moving on a $2n$-gon, alternately clockwise and counterclockwise, so that at integer times each point reaches a vertex, collides with its neighbour, and reverses direction.2012-09-11
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    Is this obvious for $3$ points?2012-09-11
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    @mjqxxxx Given the general position of the tracks, if pucks $A$ and $B$ bounce off each other at some point $X$, puck $A$ needs to hit a third puck $C$ after $t$ units of time to avoid going to infinity. With no fourth puck, puck $A$ returns to $X$ at time $2t$, where puck $B$ needs to have simultaneously returned for another bounce. That means puck $B$ needed to change direction at time $t$ as well. And this would require puck $C$ to have been in two places at once.2012-09-11
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    @alex.jordan: I meant that only at the initial position, non of the points is an intersection point of two lines. It's obvious when they start moving, they may meet each other.2012-09-12
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    If we did not forbid the points to lie on intersection lines, then we could have one of them move at speed $0$ and find a configuration, I guess that's why the condition is needed - that made me wonder: Is speed $0$ allowed?2012-09-12
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    @rattle: no, it's not allowed.2012-09-12
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    It's simple to construct a periodic configutarion with $7$ points (or more), so if you can't do it with $5$ points it's a rather special case.2012-09-15
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    @Feanor: would you please show us your construction? Thanks a lot.2012-09-16

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