Four Travelers Problem, Solution

The key to the solution is the concept of uniform motion. When a body travels at a constant speed, its graph is a straight line. In order to draw a graph, one needs a line along which the body travels and a time axis.

Let's denote the four straight lines l1, l2, l3, l4 and identify the Travelers by their numbers - #1, #2, #3, and #4. Draw a line perpendicular to the plane in which the four roads are located and think of it as a time axis. Each of the fellows travels with a constant speed. Therefore, the graphs of their motion are straight lines, say, m1, m2, m3, m4.

The fact that point P = (x, y, t) belongs to mi is equivalent to saying that

1. point Q = (x, y) lies on li.
2. #i passed through point Q at the time t.

From 1. it follows that the projection of mi onto the plane of the roads coincides with li.

Also, since #1 and #2 met, at the time of their encounter they were located at the same planar point. Therefore, by 2., m1 and m2 intersect. It must be remembered that two intersecting lines in space define a unique plane. Since #3 met both #1 and #2, m3 intersects both m1 and m2. Therefore, they all lie in the same plane. But the same argument applies to #4 as well. Hence, all four lines mi, i = 1, 2, 3, 4 lie in the same plane. Finally, lines m3 and m4 could not be parallel because their respective projections on the horizontal plane, l3 and l4, intersect. The fact that the lines m3 and m4 intersect means that #3 and #4 happened to be at the same planar point at some moment in time which means they have indeed met.

As a side bar for the proof, we may observe that at the outset, i.e. at time t = 0, all travellers were located on the same line, the intersection of the base plane and the plane of the graphs. Since selection of the moment t = 0 is arbitrary, the travellers stay collinear at all times. 2D Problems That Benefit from a 3D Outlook 