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Star-Lord is on a deserted planet along with two space policemen. If he runs into either of them, he will get immediately arrested. Fortunately for Star-Lord, somewhere on two opposite ends of the planet he has hidden a jetpack and a battery for it, which are visible only to him.

Assuming all characters have the same speed, have full information about others' locations and can take decisions in real time, can the policemen prevent Star-Lord from getting his jetpack and its battery and flying away from the planet? What if there is only one policeman?

Remark: The policemen are aware that Star-Lord has hidden his jetpack and its battery in two antipodal positions. However, their exact locations are unknown to them. The policemen can not see the items even if SL picks one them or they stand on top of the other.

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  • $\begingroup$ do you mean there are two jetpacks in antipodal position exactly 90º away from him? $\endgroup$
    – hjhjhj57
    Jul 17, 2015 at 3:24
  • $\begingroup$ There is just one jetpack and one battery for it, which are in antipodal positions. However, the position of Star-Lord with respect to them is known only to him. $\endgroup$
    – Puzzle Prime
    Jul 17, 2015 at 3:26
  • $\begingroup$ and where are the two policemen relative to the meridian SL, the jetpack, and the battery form? $\endgroup$
    – hjhjhj57
    Jul 17, 2015 at 3:27
  • $\begingroup$ They are in arbitrary positions. Basically the question is can Star-Lord escape 1) always 2) sometimes 3) never. $\endgroup$
    – Puzzle Prime
    Jul 17, 2015 at 3:28
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    $\begingroup$ No torture at all; I'm just trying to make the question as clear as possible for future readers! Thanks for posting! $\endgroup$
    – GentlePurpleRain
    Jul 17, 2015 at 3:52

1 Answer 1

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Here's my answer:

To guarantee SL will never reach one of his items, the policemen need to guarantee he'll stay in one hemisphere. For if he doesn't it's easy to give initial conditions with which he can get both of his items and escape.

Once this is settled, we can easily say the problem can be solved

always.

The strategy being:

While one of the policemen chases him, as in the famous dog-rabbit problem, the other one must mirror his movements relative to the great circle perpendicular his and SL's initial positions. This way, you guarantee he'll never leave one hemisphere.

Bonus:

With this, we see you need only one policemen to keep him at the planet.

Second Bonus FTW:

In fact, the second policeman can catch SL. Once SL is trapped in an hemisphere, send the second policeman to the pole of the hemisphere (the furthest point from the great circle). After this, since their speeds are all equal, this policeman can always be in the same meridian as SL (meridians in this case are the great circles orthogonal to the equator) and, additionally, move towards him, since he moves through parallels faster than SL.

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  • $\begingroup$ This sounds interesting... How do you keep SL in one hemisphere though? I'm pretty sure that if you have one policeman only, you can not do such thing. $\endgroup$
    – Puzzle Prime
    Jul 17, 2015 at 4:23
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    $\begingroup$ Spoiler: Imagine you're SL and there's a policeman. Draw the great circle from which each point is equidistant from you and the policeman. If he mirrors your movements you'll never be able to cross said circle. $\endgroup$
    – hjhjhj57
    Jul 17, 2015 at 4:25
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    $\begingroup$ I thought of a couple more complicated case-based strategies before arriving to this solution :) $\endgroup$
    – hjhjhj57
    Jul 17, 2015 at 4:28
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    $\begingroup$ Yes, if we assume that SL and the policemen are physical entities (not dots), this strategy works for catching SL. If we assume they are dots, then their projector can end up an infinite spiral around the equator. Good job! $\endgroup$
    – Puzzle Prime
    Jul 17, 2015 at 4:58
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    $\begingroup$ @qwertylpc, he can't really touch the jetpack and the battery if they are on the great circle - in the moment he ends up on it, the space policeman would be also there and catch him. Basically he is restricted in an open hemisphere. $\endgroup$
    – Puzzle Prime
    Jul 17, 2015 at 16:28

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