# 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 l_{1}, l_{2}, l_{3}, l_{4} 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, m_{1}, m_{2}, m_{3}, m_{4}.

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., m_{1} and m_{2} intersect. It must be remembered that two intersecting lines in space define a unique plane. Since #3 met both #1 and #2, m_{3} intersects both m_{1} and m_{2}. Therefore, they all lie in the same plane. But the same argument applies to #4 as well. Hence, all four lines m_{i}, _{3} and m_{4} could not be parallel because their respective projections on the horizontal plane, l_{3} and l_{4}, intersect. The fact that the lines m_{3} and m_{4} 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

### 2D Problems That Benefit from a 3D Outlook

- Four Travellers, Solution
- Desargues' Theorem
- Soddy Circles and Eppstein's Points
- Symmetries in a Triangle
- Three Circles and Common Chords
- Three Circles and Common Tangents
- Three Equal Circles
- Menelaus from 3D
- Stereographic Projection and Inversion
- Stereographic Projection and Radical Axes
- Sum of Squares in Equilateral Triangle

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

71625269