In the rest frame of traveller 1: Travellers T2, T3, and T4 all approach along straight lines we can define in terms of angles (say a2, a3, and a4). Of course once they meet T1 they depart along the same lines with the new headings (a2 + 180 deg) etcetera.

We can determine a2 by observing T2 at just about any time (since a2 is constant) so let's choose the moment that T2 meets T3. This gives us T3's angle a3 as well, the same: a2 = a3. Similarly since T2 meets T4 we have a2 = a4.

Thus from T1's perspective, T2, T3, and T4 are all travelling along the same line. Given adequate constraints, T2, T3, and T4 must all meet one another. "Adequate contraints" are provided by the conditions "T2 meets T3 and T4" and the condition that no two roads (in the absolute rest frame) are parallel.

Demonstrating the constraints are adequate: Not done; just wanted to point out the T1-frame idea.

best regards
Rob

Your solution is very nice. I think you can complete it easily without worring about "adequate constraints" (see below) to get a very good and complete solution.

>In the rest frame of traveller 1: Travellers T2, T3, and T4
>all approach along straight lines we can define in terms of
>angles (say a2, a3, and a4). Of course once they meet T1
>they depart along the same lines with the new headings (a2 +
>180 deg) etcetera.
>
I work with physics all the time, so of course as soon as I saw this idea, I got the feeling that I should have thought of it myself -- but I didn't.

>We can determine a2 by observing T2 at just about any time
>(since a2 is constant) so let's choose the moment that T2
>meets T3. This gives us T3's angle a3 as well, the same: a2
>= a3. Similarly since T2 meets T4 we have a2 = a4.
>
You have just shown that all 4 travelers are collinear in T1's frame (since T1 is also of course on the line containing T2, T3, and T4). The angles aren't really necessary for this, just the observation that the apparent motions are radial to T1, and that T2 meets T3, and T2 meets T4 and points OTHER THAN the position of T1.

The travelers remain collinear in the rest frame as well, of course, since their positions are merely offset by the current coordinates of T1. Therefore, all 4 travellers are collinear in the rest frame.

What is very interesting about this is that in T1's frame, the paths of the travelers are no longer along the roads (which appear now to be moving), nevertheless, the properties of collinearity and constant velocity are preserved. Thus there exists a point of view from which the problem is trivial, because the paths in T1's frame are forced to collapse onto a single line, even though the roads themselves do not.

>Thus from T1's perspective, T2, T3, and T4 are all
>travelling along the same line. Given adequate constraints,
>T2, T3, and T4 must all meet one another. "Adequate
>contraints" are provided by the conditions "T2 meets T3 and
>T4" and the condition that no two roads (in the absolute
>rest frame) are parallel.

Here is where you don't need any further constraints: All 4 travelers proceed forward on a line moving forward across the plane. Since the line obviously does not rotate in T1's frame, it does not rotate in the rest frame either, and so the line sweeps across every point in the plane, including the points of intersection of all the roads. Since no two are parallel, they all meet each other eventually.

Thanks for a very interesting perspective on this problem.

--Stuart Anderson

I agree with Stuart: the argument shows that in one frame the four of them are collinear. They remain collinear in any other frame, the original in particular.

Many thanks,

Alex

In a sense, all solutions are equivalent; perhaps some more so than the others. If that idea of equivalence is meaningful then your solution is rather related to Stuart's:

https://www.cut-the-knot.org/4travelers/FourTravelers.shtml

His Lemma 1 is equivalent to saying that in the rest frame of #1, #2 moves along a direction towards #1.

His Lemma 2 is equivalent to saying that either a2=a3=a4, or #1 has a position where he meets all the rest without moving.

Thank you,
Alex