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You are asleep on your boat on open sea. When you wake up you discover you have been robbed. A quick inspection of security cam footage reveals that the pirates who robbed you left your ship exactly an hour ago. The sea is flat, extends indefinitely, and is fully covered in a thick persistent fog. You have no idea in which direction the pirates fled. But you do know that these pirates always continue in a straight line at full speed away from their victim. Their maximum speed on open water is 20 nautical miles per hour. Your boat can reach 21 nautical miles per hour.

How do you catch the pirates?

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  • $\begingroup$ Can we assume that they know what is their smartest move? $\endgroup$ – dmg Jan 19 '15 at 17:06
  • $\begingroup$ @dmg - not sure what you mean by "smartest move" (the only choice the pirates have is the choice of escape direction). But your strategy should cover the worst-case scenario. (In other words: your strategy should guarantee you to catch the pirates.) $\endgroup$ – Johannes Jan 19 '15 at 17:17
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    $\begingroup$ By using radar. $\endgroup$ – A E Jan 19 '15 at 21:54
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    $\begingroup$ Catching is not defined. $\endgroup$ – alecail Jan 20 '15 at 15:40
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If we assume the ocean is flat and extends indefinitely in all directions, there is a strategy that guarantees we can catch the pirates in at most 800,000 years.

Put our current location as the origin of a coordinate system. We will describe our position in polar coordinates, as a function of time: $(r(t),\theta(t))$ (where we have arbitrarily chosen a direction to be $\theta=0$, and $t=0$ is when we realized we had been robbed).

We begin by traveling in the $\theta=0$ direction for 20 hours, putting our position at $(420, 0)$. We are then the same distance from the origin as the pirates. Next, we will travel in a spiral, in a manner so that $r'(t)=20$ at all times. This guarantees we will always be the same distance from the origin as the pirates. For $t\geq 20$, we will have $r(t)=420+20(t-20)=20t+20$.

Our speed is $$ \sqrt{(r')^2+r^2(\theta')^2} = 21\text{ mph}, $$ and $r'(t)=20$ for $t>20$, so $$ \theta'(t)=\sqrt{\frac{41}{r^2}}=\frac{\sqrt{41}}{20+20t}. $$

If there is a $t\geq 20$ for which $\theta(t)$ is the angle in which the pirates fled, we will catch them. This means we will certainly catch the pirates by the time $\theta$ has increased from $0$ to $2\pi$. If $t_0$ is the time this happens, we have $$ 2\pi=\int_{20}^{t_0}\theta'(t)\,dt=\int_{20}^{t_0}\frac{\sqrt{41}}{20+20t}dt. $$ Solving for $t_0$ gives $$ t_0=21\mathrm{exp}\left(\frac{40\pi}{\sqrt{41}}\right)-1\approx 7,005,043,026. $$ This means we can catch the pirates in at most 7,005,043,026 hours, or about 800 millennia. Better later than never!

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    $\begingroup$ Dang; same answer beat me to it moments $\endgroup$ – kaine Jan 19 '15 at 17:18
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    $\begingroup$ so, YOU can never catch the pirates. maybe one of your descendants catch theirs! $\endgroup$ – Rafe Jan 19 '15 at 17:19
  • $\begingroup$ Damn, so quick! $\endgroup$ – dmg Jan 19 '15 at 17:20
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    $\begingroup$ How large would your range of vision have to be to achieve it in a reasonable time? (Know this is not fully specified, but hope there is an interesting conclusion) $\endgroup$ – Dennis Jaheruddin Jan 20 '15 at 13:55
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    $\begingroup$ The problem is that given 800 millennia you would never catch the pirates (though you might catch some particles). At open sea they (and their boat) would have already disintegrated. This is not a correct solution unless mortality and parts of thermodynamics are put on hold. $\endgroup$ – Xeoncross Jan 20 '15 at 22:19
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Stay put for about 45 days, after which the pirates would have circumnavigated the globe and returned to your current position.

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    $\begingroup$ +1. Would be perfect if you included the calculation also. $\endgroup$ – justhalf Jan 20 '15 at 0:43
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    $\begingroup$ A creative way to force me to sharpen the problem statement. Thx! $\endgroup$ – Johannes Jan 20 '15 at 13:31
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    $\begingroup$ "What!? We weren't expecting the pirates for another 6 days! Man the guns!!!" $\endgroup$ – Xeoncross Jan 20 '15 at 22:21
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    $\begingroup$ @Brendan Long - Jabe probably used miles per hour instead of nautical mile per hour. $\endgroup$ – Aura Jan 20 '15 at 22:21
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    $\begingroup$ @CoreyOgburn the question specified very particularly that the sea has no boundary, which makes this a very reasonable answer. That is, the special case was built into the question, unlike the transparent die. $\endgroup$ – msh210 Jan 23 '15 at 0:03
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Drive 20 hours in a direction we will denote with as having $\theta=0$. Drive in a spiral pattern such that you are $20t+20$ nautical miles away from you starting position. $t$ is time in hours from right this second.

After 20 hours, your distance from the starting position will be $r=20t+20$ while your angle will be... more difficult to find. If we assume the spiral is of the form $r = ae^{b\theta}$, $a=420$ and $b\theta = \ln\frac{t+1}{21}$. The arc length for such a curve is $s = a\sqrt{1+b^2}e^{b\theta}/b=r\sqrt{1+b^2}/b$. The derivative of this with respect to t is $21=v=20\sqrt{1+b^2}/b$. This means $b = \frac{20}{\sqrt{41}}$.

$$\theta = \frac{\sqrt{41}}{20}\ln\frac{t+1}{21}$$

For $\theta =2\pi$ then $t=7005043026$ hours

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    $\begingroup$ Upvote as consolation prize! $\endgroup$ – Johannes Jan 19 '15 at 17:25
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    $\begingroup$ "you starting position" -> your starting position $\endgroup$ – David Conrad Jan 20 '15 at 16:47
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Assuming there is no wind, your boat was completely still on the water before it was boarded by pirates, the pirates stepped off your boat at its center of gravity, and the pirates used motorized boats, the escape vector of the pirates will be the exact opposite of the direction your boat will be drifting in once they "step off" your boat after robbing you.

Like when you step off a boat in real life, the boat is pushed in the opposite direction that you leave it.

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  • $\begingroup$ Haha great idea! +1 $\endgroup$ – avalancha Jan 21 '15 at 15:37
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    $\begingroup$ I don't believe that this necessarily hold true. It seems that if the pirate's ship was along side of your vessel, their exiting steps, providing opposite momentum to your boat, would be toward their own ship, but their direction of departure could still be toward the front or back of your boat, or anywhere in between (somewhere around 180 degrees.) $\endgroup$ – MrWonderful Jan 22 '15 at 18:19
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Wait 22.5 days, one minute and twenty seconds. By this time, the pirates are on the opposite point on the globe (assuming it's Earth) and they have stopped, because there is no way they could go away from you. Then choose a random direction. As the pirates always go away from you, they choose the same direction as you have. After 454 days, you will meet and the booty is yours!

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Since you were initially at rest, apply the law of conservation of momentum:

Initially, (boat + money) have velocity 0.

After being robbed, money has velocity 21 mph and boat has velocity v in the exact opposite direction.

Since the mass of the boat is a lot higher than the mass of the money,v is very small, however is it non-zero. Throw the anchor in the water, it will eventually come at rest at the bottom of the sea. Then, wait for a couple of hours and see in which direction does your boat stray from the anchor. Then take the anchor back and go in the opposite direction. You will catch the pirates in 20 * (1 + wait_time_in_hours) total time.

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    $\begingroup$ This is identical to an answer from over a year ago, and is invalid for the same reason as that one, which comments pointed out. $\endgroup$ – Nij Sep 1 '16 at 20:25

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