### Rendezvous

Two solutions (by S. Anderson and R. Fatland) to the 4 Travelers problem use in slightly different ways a fundamental fact pertaining to the uniform motion of two people on two straight roads:

 (1) Assume that two hikers travel with constant speeds on two straight roads towards an intersection. A necessary and sufficient condition for a rendezvous at the crossroads to take place is that, at any moment in time, the distances of the travelers to the intersection are proportional to their speeds.

Let vi = vi(t) and di = di(t), i = 1, 2, be the speeds and the distances to the intersection of the two travelers at time t. Since di/vi is the time needed for traveler #I to reach the intersection, the two will rendezvous iff

 (2) d1/v1 = d2/v2.

Since (2) can be rewritten as

 (3) d1/d2 = v1/v2 (= const),

it is immediately seen to be equivalent to Lemma 1 by S. Anderson which in fact claims that the ratio of the distances to the intersection of the two travellers who are destined for a rendezvous is constant all along the way.

Let's place the velocity vectors at the point of intersection and join their endpoints. (3) says that the triangle thus obtained is similar to the triangle formed by the point of intersection and the locations of the two travelers. From the definition of the vector sum, the line joining the endpoints of the velocity vectors represents their difference v1 - v2 (or v2 - v1, depending on the direction attributed to the line.) This observation is important in understanding R. Fatland's argument.

A stationary observer sees the travelers proceeding along the roads towards (or, as the time passes, away from) the crossroads. However, in the frame of reference that moves along with, say, traveler #2, the latter remains at rest but all the objects in the plane - the roads, their intersections, other travelers - appear to move with the constant velocity -v2. For the objects at rest in the stationary frame, -v2 is the only component of their motion as appears to traveler #1. However, the apparent velocity of traveler #1 will be the sum of two components: -v2 and its native velocity v1. Thus in the view of traveler #2, traveler #1 proceeds with a constant velocity v1 - v2,

i.e. along a line parallel to the line joining the endpoints of the velocity vectors. The two meet iff that line passes through the (stationary) location of traveler #2. Put another way: in the frame of view (the rest frame) of traveler #2, the travelers meet iff traveler #1 appears to proceed along a straight line radial through the location of #2, or iff #1 travels along the line joining the two. As R. Fatland observed, one only needs an angle to describe the position of this line.

The applet below serves to illustrate the foregoing discussion. Two velocity vectors have a common starting point and define the two roads. The travelers are represented by blue draggable circle or, during the animation, dots. The vectors can be also dragged by either of their ends. Their lengths determine the speeds of motion. To make it easier to change the speeds without changing the directions, check the box "Fix lines". The diagram can be also dragged as a whole at any point exclusive of the "special" ones (travelers, vector end points.)

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run? 