# Intercept the missile

If $C$ stays on the growing sphere of possible locations of $M$ then there seems to be no hope. Suppose $v_m = 1$ and $v_c = \sqrt{2}$, $t$ is the time since launch and we are trying to get to within distance $\epsilon$ from $M$.

As soon as $C$ reaches the $t$-multiple of the unit sphere, say at time $t=t_0$, it will follow a trajectory $t\cdot x(t)$ with $\|x(t)\|=1$ on the unit sphere with velocity $\|\dot{x}(t)\|=1/t$ (by Pythagoras). The area of the positions of $M$ that $C$ can exclude by moving on the sphere is roughly $$ \int_{t_0}^t \frac{\mathrm{d}t}{t} \frac{\epsilon}{t} = \frac{\epsilon}{t_0} - \frac{\epsilon}{t} < \frac{\epsilon}{t_0}$$ So to exclude an area of order $4\pi$ we are in trouble if $\epsilon < 4\pi t_0$.

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 $2\cdot \log\left(1+\frac{v_c-v_m}{100v_m}\right)$