I apologize in advance for the long post. You can freely skip to the last paragraph.

I was motivated by this question given in 5th grade mathematics competition that I was solving with my advanced students:

Three satellites are orbiting the Earth. The first one circles around the Earth in $75$ minutes, the second one in $105$ minutes and the third one in $135$ minutes. How much time is needed, counting from the moment they simultaneously passed over South Pole, to meet again at the same spot.

Now, obviously, the answer is $\operatorname{lcm}(75,105,135) = 4725$.

Let us now assume the orbits are coplanar and we are trying to determine when the satellites are going to first align. Like in previous example let us assume they start at the same point and assume uniform circular motion. It seems natural to work inside $\mathbb R/\mathbb Z$, and convert time to frequency. So, we need to find the smallest $T\in\mathbb R_{>0}$ such that

$$f_1T +\mathbb Z = f_2T +\mathbb Z =\ldots =f_nT + \mathbb Z\tag{1}$$

This is equivalent to finding the smallest $T\in\mathbb R_{>0}$ such that $$(f_k - f_1)T\in\mathbb Z,\quad\forall k=2,3,\ldots,n$$ or even more conveniently, equivalent to finding integers $m_k\neq 0$ such that $$\frac{m_2}{f_2 - f_1} = \frac{m_3}{f_3 - f_1} = \ldots = \frac{m_n}{f_n - f_1}\tag{2}$$

This is pretty nice as it will lead to computing $T$. But, before I continue, let me also note that now we have that $T$ is the generator of $\displaystyle\bigcap_{k=2}^n \frac 1{f_k-f_1}\mathbb Z.$ This is not surprising at all, actually, as satellites $1$ and $k$ will meet (align) exactly at times $$T = \frac m{f_k-f_1},\, m\in\mathbb Z$$ so above intersection will exactly correspond to the times all satellites meet simultaneously.

Returning to $(2)$, we can conclude that $$\frac{f_k-f_1}{f_2-f_1}=\frac{m_k}{m_2}\in\mathbb Q,\quad\forall k = 2,\ldots,n$$ Thus, $(2)$ can be solved if and only if $$\frac{f_k-f_1}{f_2-f_1}\in\mathbb Q,\quad\forall k = 2,\ldots,n\tag{3}$$ This actually gives us computational method, as I mentioned earlier:

1) Calculate $\frac{f_k-f_1}{f_2-f_1} = \frac {a_k}{b_k}$ for all $k$, where $a_k$ and $b_k$ are relatively prime.

2) Set $T = \frac{l}{f_2 - f_1}$, where $l = \operatorname{lcm}(b_3,\ldots,b_n)$

This works because we have $$\frac{1}{f_2 - f_1} = \frac{\frac{a_3}{b_3}}{f_3 - f_1} = \ldots = \frac{\frac{a_n}{b_n}}{f_n - f_1}\implies \frac{l}{f_2 - f_1} = \frac{m_3}{f_3 - f_1} = \ldots = \frac{m_n}{f_n - f_1}$$ where $m_k = l\frac{a_k}{b_k},\, k>2$, and such $T$ is really the smallest because $m_3,\ldots,m_n$ are relatively prime.

Also, note that we have nice necessary condition to have a solution in the first place:

$$\frac{f_k-f_1}{f_2-f_1} = q_k \implies f_k = (1-q_k)f_1 + q_kf_2\implies f_k\in\mathbb Q(f_1,f_2),\quad k >2$$ so we can immediately conclude that, for example, satellites with frequencies $$f_1 = 1,\ f_2=\sqrt 2,\ f_3=\sqrt 3$$ will never simultaneously meet after starting from the same point.

I was positively surprised that such a problem (in physics?) can be completely characterized (and computed) algebraically.

I tried to generalize this to the problem when satellites do not necessarily start at the same point, but had no luck. So, after the long introduction, my question is:

For fixed positive constants $f_1,\ldots,f_n$ (such that $f_i\neq f_j$ for $i\neq j$) and $\alpha_1,\ldots,\alpha_n$, can we find $T\in\mathbb R_{>0}$ such that $$f_1T + \alpha_1 + \mathbb Z = f_2T + \alpha_2 + \mathbb Z = \ldots = f_nT + \alpha_n + \mathbb Z$$ in general or at least find necessary conditions for existence?


1) For $n=2$ and $\alpha_1 = \alpha_2 = 0$ $$fT + \mathbb Z = gT+\mathbb Z$$ has solutions $T = \frac m{f-g}$ for any $m\in\mathbb Z$.

2) For $\alpha_1 = \alpha_2 = \alpha_3 = 0$ and $f_1 = 1$, $f_2 = \sqrt 2$, $f_3 = \sqrt 3$ there are no solutions to $$T+\mathbb Z = T\sqrt 2+\mathbb Z = T\sqrt 3+\mathbb Z$$ as previously noted.

3) Another example might be $$T + \mathbb Z = T\sqrt 2 +\sqrt 3+\mathbb Z = T\sqrt 3 +\frac 1 2+\mathbb Z$$ This kind of system interests me most, as (I think) I have solved the problem for trivial $\alpha$'s.

P.S. If you read my long introduction, feel free to correct any mistakes, or suggest better method.

  • $\begingroup$ from the stuff in the middle it seems the $f_j$ are also constants. So the question becomes, what are, if possible to figure out, necessary and sufficient conditions on two $n$-tuples, $f_j$ and $\alpha_j,$ such that there exists a real $T$ that solves the system above. $\endgroup$ – Will Jagy Nov 2 '15 at 23:15
  • $\begingroup$ @WillJagy, I've edited the last paragraph to make it more clear. It is as you interpreted it. $\endgroup$ – Ennar Nov 2 '15 at 23:32
  • $\begingroup$ You seem to have done the analysis already. $f_1T + \alpha_1 + k_1 = f_2T + \alpha_2 + k_2$ , $T_{1,2} = \frac{\alpha_1-\alpha_2 + k_1-k_2}{f_2-f_1}$ , assuming $f_1 < f_2$ , $T_{1,2}$ is the time at which satellites 1 and 2 align. $\Delta T_{1,2} = \frac{k_1-k_2}{f_2-f_1}$ If satellites 1,2 and 3 ever align then the next alignment is when $\Delta T_{1,2} = \Delta T_{1,3}$ or $\frac{k_1-k_2}{f_2-f_1} = \frac{k_1-k_3}{f_3-f_1}$ , $\frac{f_3-f_1}{f_2-f_1} = \frac{k_1-k_3}{k_1-k_2}$. If the ratio of frequencies is irrational then they will not align again. $\endgroup$ – arthur Nov 7 '15 at 1:22
  • $\begingroup$ @arthur, thank you for your comment, but I'm interested to know if any solution exists at all. If I have at least one alignment, then I can use my analysis to get all the alignments, as you say, but how to get a particular solution? $\endgroup$ – Ennar Nov 8 '15 at 10:04

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