A problem V.I. Arnold solved as a primary school student According to a 1995 interview that Vladimir I. Arnold gave to the Notices of the AMS, 
his primary school teacher I.V. Morozkin gave in 1949 (when Arnold was 12 years old) to a Soviet classroom, most likely the 6th grade, the following question

Two women started at sunrise and each walked at a constant velocity. One went from $A$ 
  to $B$ and the other from $B$ to $A$.  They met at noon and, continuing with no stop, 
  arrived respectively at $B$ at 4 p.m. and at $A$ at 9 p.m. At what time was the sunrise 
  that day?

The question is not how to solve this problem, but rather How to solve this problem using what 12 year old kids know (or knew during the Soviet era).
UPDATE. In the same interview, Arnold also said:

I spent a whole day thinking on this
  oldie, and the solution (based on what
  is now called scaling arguments,
  dimensional analysis, or toric variety
  theory, depending on your taste) came
  as a revelation.

It would be of interest to know how, in Arnold's mind, dimensional analysis and toric varieties are related to this problem.
 A: The same without geometry:
Woman $w_{AB}$ did in $4$ hours what woman $w_{BA}$ did in $x$ hours, and woman $w_{BA}$ did in $9$ hours what woman $w_{AB}$ did in $x$ hours. Therefore woman $w_{AB}$ in $36=9\cdot 4$ hours walks as far as woman $w_{BA}$ walks in $9\cdot x$ hours, and for this distance woman $w_{AB}$ needs $x\cdot x$ hours. It follows that $x^2=36$, or $x=6$.
A: Triangle similarity
He could have solved the question by drawing a figure such as the one below. If he knew about similar triangles at the age of $12$, he could easily set up the equation $ \dfrac{x}{4} = \dfrac{9}{x} $. From here, if he knew how to solve equations like this, he could solve $ x^2=36 $. If he did not, he could simply try a few different values of $x$ until he found one that works.

Colored triangles are similar. The problems with this explanation are:


*

*This is essentially drawing a time-position plot. Since this is a $12$ year old who does not know multivariate algebra, it's a bit of a stretch to assume he would know how to use kinematic plots to solve problems. But then again, who knows, maybe he was imagining the women walk and the plot seemed intuitive.

*It requires him to be able to set up and somehow find the solution of the quadratic equation.

*What does this have to do with "scaling arguments, dimensional analysis or toric variety theory"?


Trial and error
He could have simply tried a bunch of possible values until he found a solution.
Let's say he decided to try $9$ am: That means woman $A$ walked the first distance in $3$. So the ratio of distances is $\dfrac{3}{4}$. But if woman $B$ walked the first part in $3$ hours, then the ratio comes out $$\dfrac{3}{9+3}=\dfrac{3}{12}!$$ So $9$ am isn't right.
Luckily the solution is an integer, and somewhere between $4$ and $9$ hours before noon, since the quick woman would have walked more than half the distance in the morning and the slow one would have walked less than half (drawing a $1$D diagram makes this obvious). Even if the solution was not an integer, after exhausting integers he could have tried half hours, then quarter hours, and so on. I'm sure a binary search type of strategy would become obvious if he kept track of how much each sunrise hour was off by. Since he spent "all day" on it, there's plenty of time for numeric solutions.
The problem is that solving it by brute force teaches you absolutely nothing (well, it's arithmetic practice, and you do end up discovering binary search). There's also the question of what this has to do with scaling arguments, dimensional analysis or toric variety theory. It's also the sort of solution that you would expect from perhaps a future clerk, not mathematician.
Incidentally, after trying a bunch of numbers like this, the $\dfrac{x}{4}=\dfrac{9}{x}$ equation does suggest itself.
Sort of trial and error
It's not too great a leap to realize that the distance itself doesn't matter, so let's say he decided to let it be 50 km. Then the speed of woman $A$ is $\dfrac{50}{x+4}$. The speed of woman $B$ is $\dfrac{50}{x+9}$. We know that the ratio of their speed must be $\dfrac{x}{9}$ based on how long it took both women to walk the distance between $A$ and the meeting point. So $\dfrac{\dfrac{50}{x+4}}{\dfrac{50}{x+9}} = \dfrac{9}{x}$ which after some basic manipulation simplifies also to $x^2=36$. When the $50$s cancel, he would see right away that his hunch about distance not mattering was correct.
The problem with this is that requires a $12$ year old to reason about kinematics (ratio of speeds from ratio of times, deriving speed from time and distance) without having the mathematical vocabulary for doing so. It also requires him to not be daunted by the ugly looking equations that come out. Last, it requires him to have a hunch about the distance, otherwise he has to use a variable to represent distance and at that point we are back at multivariate algebra.
This method also does not appear to have anything to do with scaling arguments, dimensional analysis, or toric variety theory, except perhaps in the slightest sense.
Conclusion
None of these options really satisfy me. For instance, I can't really imagine myself using any of these solutions when I was $12$. The only one I would have comprehended would be the straight trial and error, which I would have been too lazy to actually carry out. Though then again, I was never very good at math.
The problem seems to be a simple linear equation system, so I don't see what it has to do with the concepts he refers to. Perhaps he was making a metaphorical point about how this sort of problem is the "tip of the iceberg" of linear algebra that children first get exposed to? Or perhaps he actually figured out linear equations over the course of that day, and that was the revelation?
A: Join the points A and B.
If the two women start back to back at the point AB of the resulting circle, they will meet at the corners of an equilateral triangle if one woman walks twice as fast as the other and stops walking when she meets the other woman again at AB.
But the "twice (as fast)" is in the given problem unknown, say (n-1) (the number of meeting points is then n).
This unknown n is related to 'the' unknown x in two ways :
9 hours / (n-1) =  4 hours (n-1) = x hours
or 
n = 5/2 and x = 6
Composing this equation is actually the hard part, hence the lack of explanation.
A: The distance each woman walks after noon is the distance the other woman has walked from dawn till noon. The two ratios “distance walked before noon : distance walked after noon” are therefore reciprocal.
But since each woman’s speed is constant, the ratios are equal to the ratios “duration of walk before noon : duration of walk after noon”, i.e. x:4 and x:9 respectively.
Hence (x/4)(x/9)=1, whence x=6. The sun rose 6 hours before noon.
Almost no algebra, and nothing a bright 12-year-old wouldn’t know!
A: A trivial solution: system of equations
Call $|AB|=L$, speed of first lady $v_1$, speed of second lady $v_2$ and time passes from dawn to $4pm$ as $t$.
Then we have $4$ equation with $4$ unknown.
$$4v_1+9v_2=L \tag1$$$$ (v_1+v_2)(t-4)=L \tag2$$$$ v_1 t=L\tag 3$$$$ v_2 (t+5)=L\tag 4$$
A: Arnold said his solution was based on what  are “now called scaling arguments” and “came as a revelation.”
Did Arnold draw something like this?straight line with points A, P and P creating "fractions" t over 9 and 4 over t
Since the women (lets call them Ekaterina and Yelena) are walking at constant velocity v_Y=k v_E. Rearranging, v_Y/v_E =k.  Then from the diagram, or otherwise, t/9=4/t  . Hence t^2=36 and t = 6.   Sunrise occurs 6 hours before noon, that is  06.00 or 6 am. (Actually I have omitted one step (AP/t)/(AP/9)=(PB/4)/(PB/t), but this seems trivial.)
