A stealth missile $M$ is launched from space station. You, at another space station far away, are trusted with the mission of intercepting $M$ using a single cruise missile $C$ at your disposal . 


You know the target missile is traveling in straight line at constant speed $v\_m$. You also know the precise location and time at which it was launched. $M$, built by state-of-the-art stealth technology however, is invisible (to you or your $C$). So you have no idea in which direction it is going. Your $C$ has a maximum speed $v\_c>v\_m$.


Can you control trajectory of $C$ so that it is guaranteed to intercept $M$ in finite time? Is this possible? 




---


I can think of 3 apparent possibilities:


1. Precise interception is possible. (It is possible in two dimensions, by calibrating your missile's trajectory to the parameters of certain logarithm spiral).
2. Precise interception is impossible, but for any $\epsilon\gt 0$, paths can be designed so that $C$ can get as close to $M$ as $\epsilon$ in finite time.
3. There's no hope, and your chance of intercepting or getting close to $M$ diminishes as time goes by.




---


This question is inspired by a similar problem in two dimensions by Louis A. Graham in his book *Ingenious Mathematical Problems and Methods*. 



Intercept the missile
There is no hope. Timothy Budd already explained why staying on the sphere $S\_t$ of possible locations of $M$ will not work; so this is to explain why leaving it will not help. What is enough to show is the following **claim**: at large times, if $t\_1<t\_2$ and $C\_{t\_{1,2}}$ are at $\epsilon$-neighborhoods of $S\_{t\_{1,2}}$ respectively, then the radial projections of $C\_{t\_{1,2}}$ onto the unit sphere are at distance at most $c\cdot \log {t\_2/t\_1}$ from each other. With the **claim**, essentially the same argument will work.


(Details: think about the task in terms of gradually ruling out possible directions in which $M$ can go. One can only exclude a direction if at some time $t$, $C$ is at the $\epsilon$-neighborhood of the corresponding point of $S\_t$. Suppose Alice has a radar that she can direct to any point of $S\_t$ and rule out the $2\epsilon$-neighborhood of it. Let Bob follow any strategy, and let Alice do this: whenever Bob's $C$ is in the closed $\epsilon$-neighborhood of $S\_t$, she directs the radar to the projection of Bob's $C$ on $S\_t$; in between those times, she turns her radar at constant speed. Clearly Alice will know more than Bob, yet by the **claim**, the angular speed of the radar will be bounded by $c/t$, and by Budd's argument, Alice will not be able to rule out everything.)


But the **claim** is fairly obvious: between $t\_1$ and $t\_2$, $C\_t$ cannot get more than $(v\_c-v\_m)(t\_2-t\_1)+2\epsilon$ behind $S\_t$, otherwise it wouldn't be able to catch up. If $t\_2-t\_1<t\_1\cdot \frac{v\_c-v\_m}{100v\_m}$, then this means that $C\_t$ is always at distance at least $\frac12v\_m t$ from the origin, and then the speed of the projection is bounded by $\mathrm{const}/t$ as in Budd's answer. If $t\_2/t\_1\geq 1+\frac{v\_c-v\_m}{100v\_m}$, then we can just bound the distance between projections of $C\_{t\_{1,2}}$ by $2$, and take the constant $c$ to be $2\cdot \log\left(1+\frac{v\_c-v\_m}{100v\_m}\right)$


