Monday 2 November 2015

mathematics - Star-Lord and the Space Police


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.



Answer



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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