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Three positive integers $a,b,c$ have the property that $b > 2a$ and $c > 2b$.
Is it always possible to find a real number $r$, such that each fractional part of the numbers $r\cdot a$, $r\cdot b$ and $r\cdot c$ lies within the interval $(\frac{1}{3}, \frac{2}{3}]$ ?

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    $\begingroup$ This might be an acceptable question on this site, but why not posting it on a more suitable site, such as the mathematics site? $\endgroup$
    – WhatsUp
    Nov 12, 2021 at 14:24
  • $\begingroup$ @WhatsUp, right, it is sometimes hard to decide which site fits best for such questions. I keep it here and see, if somebody wants to give it a try. $\endgroup$
    – ThomasL
    Nov 12, 2021 at 19:09

1 Answer 1

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Yes, it is possible.

Proof:

Let us think of r as a time, the fractional parts of the other variables can then be understood as a circular track of unit length on which a,b,c travel each at their own, constant speed. Two thirds of the track are "neutral" and one third is the target.

Because b travels at more than twice the speed of a it will move more than double the distance in the same time. In particular, while a traverses the target, b will cover more ground than fits in the neutral zone. In other words b will for some of the time be in the target, too.

We can make the same argument w.r.t. b and c, but that does not directly imply that all three will be in the target at the same time. To demonstrate this it will suffice to show that (*) during some of b's cycles a will be in the target for the entire time b is. There are two cases:

Case 1) b is a multiple of a: if b/a>=4 (*) is clearly satisfied because one passage of a through the target translates to at least 4/3 cycles for b which are guaranteed to contain at least one contigouous sweep through the target. But if b/a=3 then we can directly verify that a's very first passage trough the target fully contains one of b's.

Case 2) b is not a multiple of a. Then because their ratio is rational, we will encounter a finite set of phase shifts (i.e. offsets of b's cycles relative to the beginnings of each of a's cycles) which will be of the form 1/n 2/n 3/n etc. Each offset of b contains a shifted period of length >2/3 (at a units) when b is in the target zone. a's target of width 1/3 will be fully inside for a stretch of offsets of width >1/3. If n>=3 these must cover the entire loop. If n=2 the ratio b/a must be 5/2 or more. Now each offset of b contains a target zone of length >=5/6. Therefore a stretch of length >=1/2 of offsets will contain the entire a target. As n=2 this is just enough to cover the full loop.

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  • $\begingroup$ Is there any way to make Case 2 more rigorous? I can see in my mind's eye how the phase shift for b/a >2 will guarantee (*) but I always like to be able to back these things up with an algebraic argument. +1 from me for the formulation of (*) $\endgroup$
    – hexomino
    Nov 14, 2021 at 17:18
  • $\begingroup$ @hexomino I was a bit sloppy with that part of the argument, hope it is better now. $\endgroup$
    – loopy walt
    Nov 14, 2021 at 18:11
  • $\begingroup$ Nice solution using speed travel. $\endgroup$
    – ThomasL
    Nov 15, 2021 at 21:53

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