I think I have a handle on this problem. Here is the streamlined version omitting all the hair-tearing and going back and forth. First I will restate the question as I understand it.
Four snails, A, B, C and D are moving with constant velocities on an infinite plane. No paths are parallel to each other. There are 6 possible intersections, AB, BC, CD, DA, AC and AD and 6 possible encounters, i.e. 6 different times at which two snails are simultaneously at the intersection of their paths. According to the statement of the question "Five of the (4 × 3)/2 = 6 possible encounters have already occurred. Can we state with certainty that the sixth encounter will also occur?" The short answer to the question the way it's pharesed is "yes and no". The uncertainty that goes with "no" is that because I think that the question should have been phrased "Five
and only five of the (4 × 3)/2 = 6 possible encounters have already occurred. Can we state with certainty that the sixth encounter will also occur?" Then the answer is "yes". I explain why and then provide a method for writing the snail path equations that provide the desired results.
Assumptions
1. We are given a drawing of five points on a plane representing the five encounter points that have already occurred.
2. The sixth intersection point to be found must occur at a later time than any of the five earlier points.
3. We are free to choose the speed and direction of motion of each snail on the path it follows.
The goal is to find the sixth encounter point and demonstrate that the time associated with it is greater than the previous five.
Reasoning
If one pair of the six is removed, the five pairs have two snails (letters) with three encounters and two snails with two encounters. For example, if CD is removed, then A and B appear three times and C and D appear twice.
Conclusion: In the given pattern of five encounters, one can always draw two intersecting lines with three points each. In the example above, one line could be labeled "A" and the other "B" with points (AB, AC, AD) and {BA, BC, BD) intersecting at AB=BA.
With this in mind, there are three possible patterns that can be produced. They resemble the letters X, T and V. These are shown below. In black are the two three-encounter path segments for easy recognition of V, X or T. Red lines indicate how to add the remaining two paths. Numbers indicate the temporal order of encounters and arrows indicate the direction of motion. It should be clear by inspection that no solution can be found for a "T" pattern without situating the sixth point between either arm and the foot of the "T". Thus, five given points in a "T" pattern imply that the sixth intersection point has already occurred. This violates Assumption 2.
Thus, the answer to the question posed by the problem is that
1. If there is no discernible V, X, or T pattern in the given five points, there can be no five encounters.
2. If there is a T pattern, the sixth crossing is one of the given five points.
3. If there is a V or X pattern one can construct a solution and find the sixth point as outlined below.
Construction
We will now see how to construct a path given a possible diagram. Construction means finding four speeds and six crossing times and verify the results. The given quantities are the five points in the plane and, clearly, there is no uniques solution. It is advisable to develop a method that will make the task easy to accomplish. I chose the first diagram on the left. Then
1. I drew a horizontal line below the diagram, called it the x-axis, and extended the four segments until they intersected it.
2. I labeled the intersection, left to right, A, B, C and D. Encounter 1 is AB, encounter 2 is AC and so on.
3. I measured the angles that the segments formed relative to the positive x-axis, ##0 \leq \theta_{\!}i \leq 2\pi## and the ditances ##x_{0,i}## from the leftmost intersection.
4. The guarantee for encounters is that all four ##y##-components be equal. I set the speed ##v_{\!A}=1## so that ##v_{y,i}=v_A\sin\theta_{\!A}##. I calculated the remaining three speeds using ##v_i=\dfrac{v_{\!A}\sin\theta_{\!A}}{\sin\theta_{i}}## and ##x##-components ##v_{x,i}=v_{\!A}\sin\theta_{\!A}\cot\theta_{i}##.
5. The positions as functions of time are $$\mathbf{r_i}=(x_{0,i}+v_{x,i}~t)\mathbf{\hat x}+v_{y,i}~t~\mathbf{\hat y}.$$6. To verify that the temporal sequence is correct, the encounter times are found from ##~~x_{0,i}+v_{x,i}~t_{ij}=x_{0,j}+v_{x,j}~t_{ij}.##
The figure below shows a 3d plot of the result. The third axis is time. I used the typical snail speed of 1 cm/min for ##v_{\!A}##.
