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 m_i is equivalent to saying that

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

From 1. it follows that the projection of m_i onto the plane of the roads coincides with l_i.

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 m_i, 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

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