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According to his interview to the Notices of the AMS, when Vladimir I. Arnold was 12 years old (in 1949) his teacher I.V. Morozkin, gave to his classroom (apparently 6th grade of a soviet primary school) the following question (see

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?

My 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).

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closed as off-topic by Andy Putman, Douglas Zare, Anton Petrunin, Michael Renardy, Stefan Kohl Dec 30 '13 at 22:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Andy Putman, Douglas Zare, Anton Petrunin, Michael Renardy, Stefan Kohl
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This is off-topic for MathOverflow. It would probably get a good reception at – Andy Putman Dec 30 '13 at 21:47
I think Soviet 12 year olds might have been taught about harmonic means, proportions, and work related problems, which is appropriate for the problem. Also, trial and error and common sense converge to a solution quickly. Gerhard "Probably Could Solve At Ten" Paseman, 2013.12.30 – Gerhard Paseman Dec 30 '13 at 22:07
This isn't that hard. Take a look at the MathCounts Target Round problems , aimed for 11-14 year olds working at an average of 3 minutes a problem. They're at a comparable level of difficulty. Obviously, most 12 year olds don't know how to set up a word problem leading to the quadratic formula and solve it, but it's well with in range of a smart 12 year old, and Russian schools were extensively tracked, so Arnold's class was probably mathy kids. – David Speyer Dec 30 '13 at 23:00
For the record, I do this in about 3-4 minutes as follows: Drew a picture. Let $x$ be the distance from A to the meeting point, let $y$ be the distance from the meeting point to $x$. The sunrise (in hours before noon) is at time $4x/y$ and also $9y/x$, so $4x^2=9y^2$, we get $x:y = 3:2$, so sunrise is at 6 before noon. I was probably twice as fast when I was training for math competitions. – David Speyer Dec 30 '13 at 23:05
How about, "V. I. Arnold's 6th grade problem?" High school usually doesn't include 6th grade. – Douglas Zare Dec 30 '13 at 23:54

Isn't it conceivable that 12 year olds could solve simple quadratic equations with integer coefficients by inspection as my generation learned, especially when the numbers are arranged, as here, to give a simple whole number solution?

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To the best of my knowledge, they didn't know how to solve such equations in 6th grade. – smyrlis Dec 30 '13 at 21:55
But here it reduces to factoring 108 as a product of two integers whose sum is 24, surely well within reach of a 12 year old Arnold. – 7891user Dec 30 '13 at 22:00
Can you be more specific? – smyrlis Dec 30 '13 at 22:03
The first part of the problem uses merhods that we learned in primary school (translating text into formula and manipulating the results with simple algebra) to get the equation $s^2-24s+108=0$. – 7891user Dec 30 '13 at 22:10
If you define $x = 12-s$, so $x$ is the number of hours between sunrise and noon, then the equation is simpler: $x/4= 9/x$. – Douglas Zare Dec 30 '13 at 23:36

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