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I found this puzzle posted on a student website I frequent, but no one including the poster nor I could solve it. So I'm posting this puzzle here with hope that some of math whiz in here could illuminate on how to approach it:

There is an alien on the surface of a spherical planet. The alien can run at a top speed of $u$. An alien hunter is hunting the alien in his spaceship which can fly at a top speed of $v$. Once the hunter sees the alien, he fires, and the alien dies. Show that the alien will always die if $v > 10u$.

Note that there's no information given about the planet radius nor the altitude at which the spaceship hovers so I'm suspecting that the solution somehow ignores them or they get canceled out somewhere in the calculation.

p/s: again I repeat that I do not know of the solution myself

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The assumption I would make is that the hunter gets to choose its altitude at will. There is a tradeoff between being high so you see a lot of the surface and being low so you can have a high angular velocity. The answer (I don't have it yet) will be a search strategy, which could vary both altitude and heading, that will find the alien. – Ross Millikan Nov 23 '10 at 4:48
@Ross So I'm assuming the hunter would start off at low attitude in order to get nearer to the alien and then slowly increase the altitude in order to spot the alien faster. The strategy is to find the optimum rate at which the hunter should increase altitude, and then proof that in order to survive that the alien should run at least 10 times faster than the hunter. Am I right? – Lukman Nov 23 '10 at 5:16
Surely if either $u>0$ or $v>0$ and the planet is finite, the alien will always die - the only question is when. The only exceptions would be using carefully constructed movement plans so that the hunter and alien never see each other. That is, assume that neither knows where the other is until they are within sight of each other - at which point the hunter fires and the alien dies. Given enough time, they're bound to blunder into each other. – Steve314 Nov 23 '10 at 7:45
@Steve314: Usually here you have to prove you can catch the alien in finite time (though it can be arbitrarily long) even if it knows your search strategy. For example, if u=v, the alien can just stay at the point opposite the hunter. – Ross Millikan Nov 23 '10 at 14:15
@Ross - understood, but that isn't specified in the question. A strategy that works efficiently even in the worst case (planned or fluke) is valuable - I've suffered enough algorithm theory to understand that. But the question as stated allows a much simpler answer. Besides, in the real world, unpredictability is a normal part of realistic search and evasion strategies for game theory reasons - and as advocates of randomized algorithms will point out, a failure rate in the order of once every thousand universes or so is often more than acceptable. – Steve314 Nov 25 '10 at 0:10
up vote 5 down vote accepted

Let us measure distance in planet radius and time such that the alien can make one radian per unit time on the surface. If I fly along the surface at altitude h, I can see a swath of width $2 \arccos (\frac{1}{1+h})$ and can fly around the equator in $\frac{2 \pi (1+h)}{10}$, so as long as $2 \arccos (\frac{1}{1+h})-\frac{2 \pi (1+h)}{10} \gt 0$ I can make a spiral starting at one pole and push the alien in front of me down to the other pole. At h of about 0.55, this only requires my speed to be about 5.6 instead of 10.

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I'm getting the swath width to be a function of radius $r$ and $h$ as in $r * arccos(\frac{r}{r+h})$ .. am I wrong? Also, I'm expecting to see $\frac{dh}{dt}$ or something ... – Lukman Nov 23 '10 at 5:51
It looks like the radius is taken to be one unit (or 1) in the above answer. – Matt Groff Nov 23 '10 at 5:59
I see. So h is relative to the radius then. But one more thing: how did you decide on $h = 0.55$. Is there any calculation involved to come to that value? – Lukman Nov 23 '10 at 6:38
Yes, I just set r=1. The critical moment comes when I am flying around the equator. I need to prevent the alien from running across my track behind me, so I need to complete the circuit faster than that. To decide on h=0.55, I just varied the speed ratio until $2 \arccos (\frac{1}{1+h})-\frac{2 \pi (1+h)}{v} \gt 0$ was barely satisfied, which I found just by experiment to be h=.55, v=5.6 – Ross Millikan Nov 23 '10 at 13:48
Slight correction, the optimum altitude is closer to 0.535 – Ross Millikan Nov 23 '10 at 14:42

There's a similar puzzle here (with solution).

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The goblin puzzle and the alien puzzle kinda like opposite in many ways so I'm having difficulties to adapt the solution. For example, in the goblin puzzle the destination is clear which is the lakeside, but for this alien puzzle there is no limit. The proof needs to consider if the hunter is very far enough the planet surface (well maybe to the point that he can see the entire surface of one hemisphere). And do I have to break the path to two phases too? – Lukman Nov 23 '10 at 5:30
@~Lukman you could consider flying in from far away, forcing the alien onto one hemisphere, but I did not make use of that. In essence I started at my preferred altitude above the north pole and chased the alien down to the south pole. – Ross Millikan Nov 23 '10 at 13:49

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