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
- point Q = (x, y) lies on li.
- #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,
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
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