I used to be able to do these

@venda
To be unambiguous: Can the “straight line” be planet-planet-sun OR planet-sun-planet?.

When will it next occur? Ah ok assuming same planet planet sun....


Ok I think I found a way that will be labor intensive for 6,5 and 2. In 1 year each planet will travel 5/30, 6/30 and 15/30 of its orbit. This is like ring theory where every complete time around the ring starts all over again. So you just keep adding a fraction for each year throwing put any whole number that results. Of course this would only find a whole year answer easily.

@iamatiger said
@venda
To be unambiguous: Can the “straight line” be planet-planet-sun OR planet-sun-planet?.

@venda said
At one time I could do this type of thing easily.
Now apparently not because recently I came across this and I can't work out how to get to the answer so I must be going wrong somewhere.The given answer doesn't agree with what I get it to.
I'm sure someone here can solve the puzzle and show me the working out please:-
A man walks from one town to another.
On the first day h ...[text shortened]... urth day half of the remaining distance.He now has 14 miles left.
How far has he travelled?

@deepthought said
Just a point on this type of problem. They are deliberately constructed to be confusing. They tend to require fluid intelligence, which is more of a young persons thing, as one ages crystallized intelligence (roughly speaking experience) becomes more important. In so far as they are useful it is because they force one to be careful in analysing data, which in real world problems often isn't that neat and tidy.

OK, I think I have a solution.

I found a formula for planetary alignment they way I was assuming (all planets on a single radius away from the sun):

For orbits of period p1, p2, p3, where p1 < p2 < p3:
alignment of p1 with p2:
t = k / (1/p1 - 1/p2), where k is any integer

alignment of p1 with p3:
t = j / (1/p1 - 1/p3), where j is any integer

The three planets are aligned with both equations are satified:
(1/p1 - 1/p3) / (1/p1 - 1/p2) = j / k

Substituting our values for p1, p2, and p3
(1/2 - 1/6) / (1/2 - 1/5) = j/k
j/k = 10 / 9

back into one of the original equations, let's use j
t = 10 / (1/2 - 1/6)
t = 30

That's the number I also got and showed earlier. This problem states that alignment doesn't need to be in a radius, however.

I think we can solve this by halving the orbital periods. If the planets are orbiting twice as fast, then instead of being across from each other they would be aligned instead. Solving again:

j/k is still 10/9

t = 10 / (1/1 - 1/3) = 15

I think this means that the solution is always half of the radial alignment solution? It doesn't seem quite right to me, but I can't see why not.

@forkedknight said
OK, I think I have a solution.

I found a formula for planetary alignment they way I was assuming (all planets on a single radius away from the sun):

For orbits of period p1, p2, p3, where p1 < p2 < p3:
alignment of p1 with p2:
t = k / (1/p1 - 1/p2), where k is any integer

alignment of p1 with p3:
t = j / (1/p1 - 1/p3), where j is any integer

The three plan ...[text shortened]... half of the radial alignment solution? It doesn't seem quite right to me, but I can't see why not.

To simplify:
@venda I think your equation for two planets is correct, you just have to solve for j/k instead of applying the equation again for a third planet.

t = j * p1 * p3 / 2 (p3 - p1)
j / k = (p2*p3 - p2*p1) / (p2*p3 - p1*p3)

The blog/page I found with the planetary equations also has a formula for syzygy, which is planetary alignment, not necessarily including the sun / center of orbit:
https://www.mathpages.com/home/kmath161/kmath161.htm

I thought it would be interesting to plot your example problem:
https://imgur.com/tgF8lLj

The planets are aligned when the plot crosses 0

Not sure how well these links will work on this forum, but I thought it was interesting.

@forkedknight said
That's not quite true.

@wolfgang59 said
I was talking about the same alignment occuring after X years,
where X is the lowest common multiple of the two orbital times.

Two planets that are lined up PPSun will generally line up PSunP in half the time it takes them to get back to PPSun.

If the orbital period of the slow planet is S and the fast planet period is F then the next PSunP will happen when the fast planet has done exactly half an orbit more than the slow planet.

T/F = T/S + 1/2
1/F-1/S = 1/(2T)
T = FS/(2S-2F)

So this should be soluble with the same techniques

@iamatiger said
Two planets that are lined up PPSun will generally line up PSunP in half the time it takes them to get back to PPSun.

If the orbital period of the slow planet is S and the fast planet period is F then the next PSunP will happen when the fast planet has done exactly half an orbit more than the slow planet.

T/F = T/S + 1/2
1/F-1/S = 1/(2T)


So this should be soluble with the same techniques

@iamatiger said
T = FS/(2S-2F)

What this is saying is that the planets will be in line with the sun whenever
T=JFS/(2S-2F)
Where J is an integer

Using ForkedKnight’s cunning approach we can add a third planet M: M and F will line up when
T = FS/(2S-2F)

What this is saying is that the planets will be in line with the sun whenever
T=KFM/(2M-2F)

and therefore all 3 planets are ...[text shortened]... 1: line up is SMFX
At T9, F=9, M=3, S=3/2: line up is MFXS
At T12, F=12, M=4, S=2: line up is SMFX