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The scientists at NASA are investigating an asteroid in space. From particulars transmitted by the Hubble Space Telescope, it is known to the scientists that the asteroid is either a ball or a cube, but they are in the dark about what it actually is. To determine the exact shape, the scientists are sending a spacecraft to the asteroid. Once the spacecraft lands, the rover is programmed to start moving at the landing site and finish at the point symmetric to the landing site with respect to the center of the asteroid. On its way, the rover will keep transmitting its exact location in terms of spatial co-ordinates with respect to the spacecraft, so that the trajectory of the rover movement can be known to the scientists at NASA. Using these data, they plan to determine whether the asteroid is spherical or cubical.

Is their plan sure to work?

Source: Tournament of Towns.

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  • $\begingroup$ I'm sure I've seen this question here before but it looks like it was deleted. $\endgroup$ – 2012rcampion Oct 11 '16 at 15:42
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    $\begingroup$ Aha! Google cached the old post's solution (spoiler alert)! $\endgroup$ – 2012rcampion Oct 11 '16 at 15:48
  • $\begingroup$ @2012rcampion Yes, the previous question has been deleted; that's why I tried to reopen this. $\endgroup$ – Ankoganit Oct 11 '16 at 15:53
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    $\begingroup$ It's an excellent question, I'm glad to see it return =) $\endgroup$ – 2012rcampion Oct 11 '16 at 15:54
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The answer is

NO.

The question is equivalent to analyzing the intersection of a cube and a sphere which share a common center. Thus the question gets reduced to figuring out whether such intersection, which is a curve, can connect two opposite points on the sphere/cube.

Let the edge of the cube has length $1$. Then if you pick the radius of the sphere $1$, the intersection will be a set of $6$ points. If you start gradually increasing it, you can see that when it becomes equal to $\frac{\sqrt{2}}{2}$, the intersection consist of $6$ circles inscribed in the sides of the cube. Then the rover can just move along these circles from one point to its opposite and NASA won't be able to find out the exact shape.

Remark: From the arguments above you can see that $1:\frac{\sqrt{2}}{2}$ is the only edge-radius ratio, for which we can't figure out the shape.

P.S. I see now there is already a link towards a solution posted above in the comments, so I'm making this community Wiki.

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  • $\begingroup$ But why would you program the Rover to behave that way if you wanted to know the topology of the asteroid? $\endgroup$ – Sconibulus Oct 11 '16 at 16:16
  • $\begingroup$ @Sconibulus Well, it is not written in the statement of the problem what the behavior of the rover is, so you must assume it could be anything/non-programmed. Otherwise the question is trivial - just let it move for few seconds and check if there is any curvature. $\endgroup$ – Puzzle Prime Oct 11 '16 at 16:18
  • $\begingroup$ "Once the spacecraft lands, the rover is programmed to start moving at the landing site and finish at the point symmetric to the landing site with respect to the center of the asteroid." I guess I erroneously took this to mean the rover was indeed programmed, and that it would take what it believes to be shortest path to the opposite side. Ehhh... guess not. $\endgroup$ – John Oct 11 '16 at 16:24
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Absolutely would work. If the spacecraft lands on a side of the cube, the rover will pass over at least 2 edges which will be seen spatially as going further away (over the edge) then getting closer again before going down that side. If it's a sphere/ball, there will be a constant shift in the distance between the rover and spacecraft until it reaches the far side.

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I think it should work?
If the asteroid is spherical, the vector describing the rover's movement relative to the craft will be constantly changing, with no variance because spheres are infinity-dimensionally symmetric

If the astroid is a cube, there will certainly be a non-zero interval where the rover's movement is in a straight line, no matter where the craft lands.

EDIT: Proven wrong. Assumed (incorrectly) a straight line path is programmed into the rover.

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  • $\begingroup$ What if the rover rotates on a face of the cube? $\endgroup$ – Ankoganit Oct 11 '16 at 15:51
  • $\begingroup$ @Ankoganit: That's a fault of the rover - Why just don't simply take the shortest path ? $\endgroup$ – user27395 Oct 11 '16 at 16:02

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