The challenge of two old ladies who hit the road at dawn

The condition of the problem from the famous book by VI Arnold “Tasks for children from 5 to 15 years old” :
At dawn (at the same time), two old women came out to meet each other (along the same road) from A to B and from B to A. They met at noon, but did not stop, but each continued to walk at the same speed, and the first came ( ) at 4 pm and the second (at A) at 9 pm. What time was dawn on that day?

I suggest you listen to (MP3) discussion of this problem on the radio “Moscow speaks” (S.Dorenko, A.Onoshko), and try to solve it, before crawling under the cat, to compare ...

The first option: the system of equations
This is how I solved it, first writing a system of equations like this (s is the distance from A to B, x is the time from dawn to noon):


I did not succeed for a long time to solve either it or a similar system (there were always one more unknowns than equations, and it didn’t work out to get rid of the extra). Finally, writing a system of two equations for the distances of AO and OB (O is the meeting point at noon):
,
managed to get rid of the "extra" unknowns (velocities), multiplying the left and right sides of the two equations. As a result, the product of speeds was reduced, and from the remaining equation t ² = 36, the time sought was easily found:
12-t = 6.
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I must say that I was busy for an hour or two, and then, a few days later, when I returned to the task, I completely got confused and went to the Internet to get a solution, where I found ...

Second option: proportion

Since the task is “childish”, the solution is quite childish. If we make a proportion of the lengths of the segments of the AO / OB path in terms of the first and second old women, we obtain:
,
whence immediately follows the equation t ² = 36. (Even speeds are not needed in proportion, it is enough to understand that the length of the path is proportional to time).

Thus, an elementary school student can also solve the problem: this is a typical olympiad problem, which I hate since childhood (I have to, at least, kill a lot of time to find a solution). Meanwhile, the task is simple, and, if viewed from a practical point of view (getting an answer, rather than finding an elegant solution, which is undoubtedly option 2), today I would use ...

The third option: a quick numerical solution.
I will give it in the form of a Mathcad document:


First, the initial approximations to the unknowns are specified (including the estimated value taken from the ceiling, s = 20 km). Then the standard numerical algorithm searches for a solution to the very system of equations that we wrote out at the very beginning, taking only a couple of minutes. Here s is considered as an unknown variable, resulting in a result equal to the initial approximation of 20 km.

Repeating the calculations for a series of different reasonable s, we find that the solution for s remains equal to the initial approximation, and the desired time x is always 6 hours. This is a very strict solution to the problem (since it is easy to scan the entire interval of reasonable distances AB), which, I confess, I like the most, although it looks a bit dull.

In conclusion, I will give the same calculations for s = 50 km (and using units of measurement in the calculations):