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Seven astronauts land on a small spherical asteroid. They decide to explore it and walk in different directions. All start from the same location. All use the same walking decision procedure:

Walk x kilometers forward, turn 90 degrees left and walk another x kilometers, turn 90 degree left again and walk the last x kilometers.

The seven values of x for the seven astronauts are 30, 40, 50, 60, 70, 80, and 90. Six astronauts finish in the same location, and one astronaut finishes alone.

What is the value of x for the astronaut who finishes alone?
What is the size of the asteroid?

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The part 'walk x kilometers forward' is a bit ambiguous - do the astronauts always follow paths along great circles of the asteroid? – user3490 Mar 15 at 12:29
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@user3490 The only sensible interpretation of movement forward (or in straight lines) on a manifold is along a geodesic. So yes, they should move in great circles. – Lacklub Mar 15 at 12:35
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Did anyone call Bruce Willis or Aerosmith about this puzzle? – Ian MacDonald Mar 15 at 13:57
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@user3490 How can you NOT walk in a great circle along the surface of a sphere? Unless you are turning while you walk, any path "forward" (for any definition of "forward") is along a great circle. – loneboat Mar 15 at 14:54
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That was their decision procedure. What actually happened was that the 90km astronaut shot the others dead and left them piled the starting point. He then calmly walked off, finishing over 90km away. The size of the asteroid part is a red herring. – Hugh Meyers Mar 15 at 15:12

The lonely asteroidnaut might have walked ...

... x = 60 km segments, ...

... especially if the asteroid is [Addendum below has more] ...

... about 1/1,000 the size of Earth, 6.366+ km in radius and 40 km in circumference.

So the six astronauts who reunited had walked segments of ...

... 3/4, 4/4, 5/4, 7/4, 8/4 and 9/4 of the circumference, each of which returns to the starting point when repeated 3 times with left turns. If inconsequential full circles are ignored, these segments reduce to the set: 0, 1/4 and 3/4 times the circumference.

Starting from the north pole, for the sake of narrative convenience,

3 segments of 1/4 (plus 1 and 2) times the circumference has the 50 and 90 km astronauts going to the equator, then quarter-way around it, then back to the north pole;

3 segments of 3/4 (plus 0 and 1) times the circumference has the 30 and 70 km astronauts turning at the the antipodes of where the 50 and 90 km astronauts turned, then likewise returning to the north pole;

3 segments of 1 or 2 full circumferences has the 40 and 80 km astronauts returning to the north pole at the end of every segment;

3 segments of 1/2 (plus 1) times the circumference, however, will take the 60 km astronaut to the south pole, then back to the north pole, and finally again to the south pole.


Addendum from Michael Seifert's comment

The asteroid could actually have an infinite number of sizes that would isolate the same astronaut!

The circumference could be any odd division 1/(2n+1) of 40 km.

This table details the calculation for the 30 and 70 km astronauts.

                    30 & 70 km astronauts      40 & 80 km   50 & 90 km     60 km
                                               astronauts   astronauts     loner
  circum-     total number of
  ference      circumferences --> remainder

 40 km             3/4 & 7/4  -->  r = 3/4        r = 0      r = 1/4      r = 1/2
 40/3              9/4 & 21/4      r = 1/4        r = 0      r = 3/4      r = 1/2
 40/5 (8 km)      15/4 & 35/4      r = 3/4        r = 0      r = 1/4      r = 1/2
 40/7             21/4 & 49/4      r = 1/4        r = 0      r = 3/4      r = 1/2
   :                   :             :              :          :            :
 40/(4m+1)   3(4m+1)/4 & 7(4m+1)/4 r = 3/4        r = 0      r = 1/4      r = 1/2
 40/(4m+3)   3(4m+3)/4 & 7(4m+3)/4 r = 1/4        r = 0      r = 3/4      r = 1/2
   :                   :             :              :          :            :
 
An astronaut will return to the origin if the remainder of their segment after canceling out full circuits (dividing by the circumference) is r = 0, 1/4 or 3/4 of the circumference.

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1  
The same logic follows if 10 km is equal to 3/4, or 5/4, or 7/4, etc. of the asteroid's circumference. For example, if the asteroid's circumference is 40/3 km, the asteroidnauts are then walking 9/4, 12/4, 15/4, 18/4, 21/4, 24/4, and 27/4 of the asteroid's circumference; and so for each segment, most of the astronauts (except for the one who walks 60 km) are still going 0, 1/4, or 3/4 of the way around. So the asteroid's circumference could also be 40/3 km, 40/5 = 8 km, 40/7 km, 40/9 km, etc. – Michael Seifert Mar 16 at 15:02
    
Wish I'd thought of that, @MichaelSeifert! Convenient how the residues 1/4 and 3/4 alternate their roles. Never too late, this is now part of the answer and duly credited. – humn Mar 16 at 22:44

Small note: the original question does not state that the astronauts must walk in a straight line/forward on the second and third leg of the journey. I will assume that is what the questioner meant.

Designate the starting point as a pole.

In order to return to the starting-point — on the pole — after making 90 degree turns and walking the same distance on all legs of the journey, while moving in a straight line, the astronauts must...

...walk on the equator or a longitude for the second leg.

The reason is that all the other latitudes do not follow a straight line. There is a fallacy in assuming that just because you turned to the east/west then you will keep moving in an east/westerly direction when you walk in a straight line. This is not true.

The easy way to prove this is to get to the Earth's south pole. Walk 1 km to the north (all directions are north there), and then turn left. This happens to be west. Walk 1 km and turn left again. This time you are not ending up facing south, but actually south-west. Walk 1 km and stop. You are now ~1 km from your starting point.

Unless your second leg was in a quarter-circle, then you will miss your starting-point. There is only one "place" on Earth where you can walk in a straight line and in an east-westerly direction at the same time and that is...

...on the equator.

From this we can conclude that each leg of a returning astronaut's journey must be...

...quarters of the circumference, so that they always end up on a pole or the equator at the end of a leg.

At the same time, any one such leg of the journey must not...

...end up on the opposing pole from where they started. Because that means they will walk to the opposing pole, back to the starting pole, and then to the opposing pole again when they walk the third leg. So the astronaut with a leg distance that is 2/4 (or 6/4, or 10/4 and so on) of the circumference is the lone astronaut.

Circling the asteroid is allowed because that results in...

the same distance being traveled from the pole as if they had not circled it. Walking 5/4 of a lap is the same as walking 1/4.

This then means that the following distances — measured in quarters of the circumference — are "allowed" in order to return to the starting point:

1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15...

...or more clearly written...

0, 1, x, 3, 4, 5, x, 7, 8, 9, x, 11, 12, 13, x, 15...

And now the answer becomes trivial. One quarter of the asteroid circumference is...

10 km, leading to a total circumference of 40 km

...and the astronaut that ended up alone is the one that walked...

60 km

...on each leg of the journey.

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The asteroid could also have a circumference of

180km

Then, six astronauts starting at an arbitrary point which we will call the "north pole" walk

south, then east, then north again, returning to the north pole (the starting point).

But the seventh walks

90km south, ending on the south pole. Whichever way he turns, he will walk north again, then turning again he can only walk south. Thus he ends up on the south pole.

This relies on a different projection than the other answers,

mercator

such that on the second leg of the route "straight" is defined as

"constant latitude from the north pole, or constant radius from the starting position" rather than a Great Circle route.

The question does not specify which assumptions are correct/allowed, despite the comments below it.

EDIT : this answer does not uniquely determine the circumference of the asteroid.

If it were 100km, 120km, 140 or 160km, it would also leave one astronaut walking pole to pole, while those walking shorter distances would describe the classic "triangular" path. While the 90km walker would pass the south pole and start heading North again. If he turns left he'll walk West (probably thinking he's heading East), then when he turns left a second time he'll head south again, cross the S pole and return to the N pole. For a 100km circumference planet, the 60,70 and 80km walkers will join him but the 50km walker will not.

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The astronauts are not walking in a projection. They are walking on an actual asteroid. – Taemyr Mar 15 at 14:42
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While this is true, how are they determining where to walk? If they are following their APS displays, or magnetic compasses work on this asteroid, then this answer works. Note also that the question does not specify how direction is determined. – Brian Drummond Mar 15 at 14:47
    
Direction is determined by walking forwards and seeing where you end up, which is along a great circle path. I don't see why the question should have to specify that the astronauts didn't take the time to define an arbitrary pole and set up a co-ordinate system, because them doing so seems like a completely unwarranted assumption. – Zandar Mar 15 at 19:42
    
This works if the astronauts follow curved lines of latitude, but only if they start at a pole. It doesn't work if they "Walk x kilometers forward". – Octopus Mar 15 at 21:53

Call the starting point the North Pole. If the astronaut is standing straight (normal to the tangent plane), then that normal passes through the centre of the sphere. The straight-ahead trajectory of the astronaut then must be a great circle.

If the first leg stops at the South Pole, then the second leg takes the astronaut back to the North Pole, and the final leg goes back to the South Pole again.

If the first leg stops at the equator, turning 90 degrees has the astronaut oriented along the equator. The second leg sits along a single latitude, and the last 90 degree turn orients the astronaut facing longitudinally again, this time on a return path (possibly along a different longitude from the first leg) ending back at the North Pole.

If the asteroid has circumference 2x20km = 40km, the astronaut who walked 60km per leg ends up at the South Pole and the rest end up at the North Pole.

Thanks @MichaelKarnerfors and @Etoplay for your thoughts about the non-equatorial latitudes.

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Actually, if the assumptions in my answer hold, then for any circumference that is not $2x$ for any $x$ corresponding to the astronauts that walk along great circles, then if 6 astronauts walk in great circles and one doesn't, the one that doesn't could end up in a different spot. – Lawrence Mar 15 at 14:53
    
Perhaps the proper way to solve this is to transform the problem to polar coordinates with centre at the centre of the asteroid. Start at $(r,0,0)$, move to $(r,\theta_1,\phi_1)$ on the first leg, calculate final position $(r,\theta_2,\phi_2)$ for the second leg and then $(r,\theta_3,\phi_3)$ for the third. Normalise the angles to 0 to 360 degrees (or equivalently $0$ to $2\pi$) for the final destination, then for each $x$, make the angles for other 6 correspond and determine the conditions for which the 'chosen $x$' doesn't correspond. – Lawrence Mar 15 at 14:59
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The latitudes are not straight lines. If you follow a latitude and are not at the equator you have to move an arc. That means Turning 90 degree doesn't let you walk along an latitude. – Etoplay Mar 15 at 15:51
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@Lawrence Your answer is almost correct. There is one latitude that does follow a straight line: the equator. All the other latitudes do not. The simple way to prove this is to place a polar grid on an arbitrary point on a globe, and then try to align any of the this new grid's longitudes with the old grid's latitudes. It cannot be done... with the one exception being the equator. With this limitation in mind, your solution works, but the answer is not 180 km :D – Michael Karnerfors Mar 16 at 11:09
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@MichaelKarnerfors You're right - using 20km as half the circumference (i.e. walking from pole to pole), the x=60km astronaut ends up at the South pole while the others all end up at the North pole. I'll edit my answer to suit. Thanks! – Lawrence Mar 16 at 12:40

Answer:

The astronaut with x=60km ends on a different point than the other astronauts and the circumference of the asteroid is 40km.

Why?

We need to notice following things:
Let c be the circumference of the sphere.
If we follow the path with x=1/4c,3/4c or c (of course 5/4c,7/4c,...) we will end where we started.
If x=1/2 (x=3/2c,...) we will end on the opposite side.

The astronaut with...
x=30km ends at the starting point (3/4c)
x=40km ends at the starting point (1c)
x=50km ends at the starting point (5/4c => 1/4c)
x=60km ends at the opposite point (3/2c => 1/2c)
x=70km ends at the starting point (7/4c => 3/4c)
x=80km ends at the starting point (2c)
x=90km ends at the starting point (9/4c => 1/4c)

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