Source

I got this problem from Rustan Leino, who was sent it by Howard Lederer.

I solved it and wrote up my solution.

Problem

You're on a government ship, looking for a pirate ship. You know that the pirate ship travels at a constant speed, and you know what that speed is. Your ship can travel twice as fast as the pirate ship. Moreover, you know that the pirate ship travels along a straight line, but you don't know what that line is. It's very foggy, so foggy that you see nothing. But then! All of a sudden, and for just an instant, the fog clears enough to let you determine the exact position of the pirate ship. Then, the fog closes in again and you remain (forever) in the thick fog. Although you were able to determine the position of the pirate ship during that fog-free moment, you were not able to determine its direction. How can you navigate your government ship so that you eventually capture the pirate ship?

Solution
Reveal

Let $L$ be the location of the pirate ship at the fog-free moment. Go at top speed toward $L$ until you're 2/3 of the way to $L$. Then, staying at top speed, continually turn so that your heading is always 120 degrees clockwise from a heading that would direct you straight toward $L$.

Let $s$ be the pirate ship's speed; your speed is $2s$. Let $D$ be one third of the distance between your ships at the fog-free moment.

In the time it takes you to get 2/3 of the way to $L$, the pirate ship goes half of that distance. So, when you get 2/3 of the way to $L$, both ships are the same distance $D$ away from $L$. Call this moment time $0.$

By maintaining a heading 120 degrees clockwise from $L$ starting at time $0,$ the speed at which you move away from $L$ is $(2s) \cos (60^\circ) = s.$ So, starting at time $0,$ your distance from $L$ always remains the same as the pirate ship's distance from $L$.

Thus, your course starting from time $0$ is a counterclockwise spiral centered at $L$, always maintaining the same distance from $L$ as the pirate ship. As soon as your angle relative to $L$ matches the constant heading the pirate ship is using, you'll come upon the pirate ship.

Let $r(t)$ be the distance between you and $L$ at time $t.$ Because you move at constant sped $s$ away from $L$, we have $r(t) = D + st$ for $t \geq 0.$

Let $\theta(t)$ be your angle relative to $L$ at time $t.$ The component of your speed that's counterclockwise around $L$ is $(2s)\cos(30^\circ) = s\sqrt{3}$. To convert this to an angular speed one must divide by the distance from $L.$ So, your angular speed $d\theta(t)/dt$ is $s\sqrt{3}/r(t)$ radians per unit time, for $t \geq 0.$

So, in the first $T$ units of time during the spiral, your angle changes by \begin{eqnarray*} \int_0^T (d\theta(t)/dt)dt & = & \int_0^T \frac{s\sqrt{3}}{r(t)} \;dt \\ & = & \int_0^T \frac{s\sqrt{3}}{D + st} \; dt \\ & = & \left. \ln(D + st)\sqrt{3} \right|_0^{T} \\ & = & \sqrt{3}[\ln(D + sT) - \ln D] \\ & = & \sqrt{3}\left[\ln\left(1 + \frac{sT}{D}\right)\right]. \end{eqnarray*}

Given this, you can set $T = (D/s)(e^{2\pi/\sqrt{3}} - 1)$ to get $\int_0^T (d\theta(t)/dt)dt = 2\pi.$ This value of $T$ is always non-negative and finite; therefore, you cover all angles within finite time $T.$ Thus, you eventually encounter the pirate ship.