Suppose you are given a question that goes like this:

Consider three planets that revolve around a star on separate circular orbits sharing the same orbital plane. The first planet, A, takes 385 days to complete its orbit. The second planet, B, takes 814 days, while the third planet, C, takes 1672 days. Suppose, at one moment in time, the three planets and their star are in one straight line among each other (i.e. they are collinear). Calculate how many days the star and planets will be collinear again.

It looks like a simple question. Obvious enough, there are at least two trivial solutions for this kind of question. The first is the linear product of all the figures: 385 x 814 x 1672 = 523988080 days, while the second is their Least Common Multiple (LCM): LCM(385, 814, 1672) = 2165240 days. But surprisingly, it just takes $\frac{2165240}{39}$ days for the planets to be collinear again, albeit not at the same location. The question does not restrict the answer to be an integer, after all.

My "Orbital Collinearity Theorem" :P

By trial and error, I found out that the number of days for the planets to become collinear again can be calculated by the formula:

$\dfrac{LCM(d_1, d_2, ... , d_n)}{M}$

where LCM(…) denotes Least Common Multiple and M is the largest integer for which the following modular congruence holds:

$\frac{d_1}{G} \equiv \frac{d_2}{G} \equiv ... \equiv \frac{d_n}{G} \pmod{M}$

where G is the Greatest Common Divisor of the numbers of days the planets take to complete their respective orbits ($d_1$, $d_2$, ... , $d_n$). In other words, M is the largest integer that can give the same remainder when it divides each of $\frac{d_1}{G}$, $\frac{d_2}{G}$, ... , $\frac{d_n}{G}$. If there is no such M that is larger than 1 (since all integers will give remainder of 0 when divided by 1), the answer will be the LCM, which is very much expected.

For 385, 814 and 1672, the value of M is 39:

$GCD(385, 814, 1672) = 11$

$\frac{385}{11}=35$ and $35 \bmod 39 = 35$

$\frac{814}{11}=74$ and $74 \bmod 39 = 35$

$\frac{1672}{11}=152$ and $152 \bmod 39 = 35$

I don’t really have the proof/working since it is trial and error, but so far after testing with many samples the formula seems to hold for each case. So let’s check if after $\frac{2165240}{39}$ days, the three planets are collinear or not:

  • Planet A: if it takes 385 days to complete its orbit, after $\frac{2165240}{39}$ days, it has traveled $\frac{2165240}{39} \cdot \frac{1}{385}=144 \frac{3080}{15015} = 144 \frac{8}{39}$ cycles

  • Planet B: if it takes 814 days to complete its orbit, after $\frac{2165240}{39}$ days, it has traveled $\frac{2165240}{39} \cdot \frac{1}{814}=68 \frac{6512}{31746} = 68 \frac{8}{39}$ cycles

  • Planet C: if it takes 1672 days to complete its orbit, after $\frac{2165240}{39}$ days, it has traveled $\frac{2165240}{39} \cdot \frac{1}{1672}=33 \frac{13376}{65208} = 33 \frac{8}{39}$ cycles

It should be obvious that the numbers of cycles that the three planets have gone have the same fractional part, which is $\frac{8}{39}$. Meaning that they are all $\frac{8}{39}$ cycle into their current cycle (34th, 69th and 145th). Hence they all form the same angle from their last collinear position, thus they are collinear.

Second sample: minute and hour hands

The minute and hour hands overlap at 12 o’clock. At what time will the hands overlap again?

By considering the hands as planets revolving around the clock center, we can use the formula to solve it. First consider that minute hand takes 60 minutes to complete a cycle, while hour hand takes 720 minutes. LCM(60,720) = 720 and GCD(60, 720) = 60. That will give us M = 11. Thus it will take $\frac{720}{11} = 65.454545454545454545454545454545$ minutes for them to overlap again. And translating that into time gives us 1:05 and 27.27272727272 seconds. Correct. Moreover, this sample is an example of why we must divide the numbers by their GCD to get the M, a step that was not significant with the three planets question; because if we didn’t, M would be 60, giving us 12 minute as the answer.

So, my real question is: Is there official name and proof for this formula/theorem?

p/s: forgive the inappropriate tags since I don't have enough points to add new tags and there are no suitable tags among the existing ones.


OK, I've worked out the proof half-way but I still can't connect it to the last piece. I'm appending the half-proof here with hope that any of you guys can help me complete it:

Let $L = LCM(d_1, d_2, ... , d_n)$. Then it's a given that L is the least number of days for the planets to be collinear again at the same position (i.e. all of them have revolves $c_1, c_2, ... , c_n$ integral number of cycles around the star). Proof by contradiction: suppose there exists another number $K < L$ for which the planets are collinear at the same position, then L must be a common multiple of $d_1, d_2, ... , d_n$, but $L$ is the least of their common multiples, hence contradiction.

Yet, we still have a possibility that L could be the M-th time the planets are collinear regardless of position. If such $M > 1$ exists, then the number of cycles that the planets have undergone after L days, $c_1, c_2, ... , c_n$, should have the same remainder if divided by M. Therefore, the following modular congruence must hold:

$\frac{L}{d_1} \equiv \frac{L}{d_2} \equiv ... \equiv \frac{L}{d_n} \pmod{M}$

That's where I am now. I'm trying to connect the above congruence to:

$\frac{d_1}{G} \equiv \frac{d_2}{G} \equiv ... \equiv \frac{d_n}{G} \pmod{M}$

It's trivial to do so with $d_1$ and $d_2$ only because $LCM(d_1,d_2)GCD(d_1,d_2) = d_1d_2$ but the relationship between LCM and GCD becomes less simplistic for more than two numbers. The simplest I could find is $\displaystyle a_1 ... a_n = GCD(a_1, a_2, ... a_n) LCM \left( \frac{a_1 ... a_n}{a_1}, \frac{a_1 ... a_n}{a_2}, ... \frac{a_1 ... a_n}{a_n} \right)$ (from another question in Math SE) but it cannot be used directly in the formula because the GCD and LCM terms are different.

Can anyone help?

  • $\begingroup$ How about rotation periods of $1$, $\pi$ and $\dfrac{3 \pi}{4-\pi}$? $\endgroup$ – Henry Feb 13 '14 at 15:36

Here's how you'd start working on it. Suppose $t$ is a time in which planets $i$ and $j$ are collinear. Then $t/d_i - t/d_j$ is some integer $n$. Thus $$t = \frac{nd_id_j}{d_i-d_j}.$$ Therefore you're looking for the smallest $t$ in $$\bigcap_{i < j} \frac{d_id_j}{d_i-d_j} \mathbb{Z}_+.$$ We can take the common denominator and find the answer (up to sign) $$\frac{LCM(\{d_id_j\prod_{(l,k)\neq (i,j)} (d_k-d_l) : i<j\})}{\prod_{k < l} (d_k-d_l)}.$$ Of course, that form is not as explicit as yours.

EDIT: Here is the explicit result (up to sign) for $n=3$, where the orbits are denoted $x,y,z$: $$\frac{LCM(xy(z-x)(z-y),xz(y-x)(y-z),yz(x-y)(x-z))}{(x-y)(x-z)(y-z)}.$$

  • $\begingroup$ Sorry, can you please extend your formulas to cover at least 3 planets? Do I find $t = \frac{nd_hd_id_j}{d_h-d_i-d_j}$ instead? And how will the 3rd formula looks like with three $d_n$ parameters? $\endgroup$ – Lukman Nov 22 '10 at 5:51
  • $\begingroup$ Also, I'm looking for an explanation of why the formula for finding next collinearity has something to do with LCM, GCD and modular congruence, all three being discrete arithmetic concepts. $\endgroup$ – Lukman Nov 22 '10 at 5:59
  • $\begingroup$ Re your last question, this is not very surprising. If you require the time to be integral then you get exactly the LCM, so why should your variant be different. $\endgroup$ – Yuval Filmus Nov 22 '10 at 8:39
  • $\begingroup$ Yes, I understand about having LCM in the equation. But in the formula in the question also involve GCD and modular congruence, so I'm just being curious how can the three concepts mix and match together to give the solution. Your formula works too, but on the other hand it seems like a series of multiplications of the x, y, z and their pairwise differences and suddenly voila, you get the answer. But I'm looking for explanation and reason why the formula is like that. Sorry I'm no math major so my knowledge on theorems is a bit limited. But thanks nevertheless for an alternate formula. $\endgroup$ – Lukman Nov 22 '10 at 9:00

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