A Puzzling Fortnight - Day 9

You know that at time t after the lightning, the ship has to be on a circle radius v*t, where v is the ship's speed.

So you go in a spiral path with distance from the ship exactly equal to vt, expressed in polar co-ordinates (r,theta)=(v*t, cos Mt+alpha) where M and alpha are arbitrarily chosen so that your ship can actually be on such a curve at some t, and M low enough that your ship is fast enough to traverse the curve in one 360-degree rotation. (I'm too lazy/busy to do the calculus right now)

OK, my first thought is at any given moment you can identify a circle on whose circumference the pirates are (center = the point where you glimpsed them; radius = time elapsed * rate of pirates movement). The radius of the circle expands with time. So all you have to do is go around that "expanding circle" until you catch the pirates; since it's expanding, your path will be a spiral. Now I have to prove that you have enough time to catch them even though their path is straight and yours is a spiral.

I don't like calculus so I'm going to try and just prove it intuitively:

Just to make the proof a little bit simpler, park your ship for exactly long enough that if the pirates were going towards you they would have hit you by now. OK, from then on:

At any given moment, you are on some point of the circle where the pirates are, at that moment. One component of your motion (one vector, if you like) has to be outward (i.e. perpendicular to the circle's tangent at the point you are on), in order to stay on the circle as it expands. This component has to match the pirates' speed. The second component of your motion is devoted to going around the circle (i.e. this vector is parallel to the tangent line at the point you are on; clockwise or counterclockwise doesn't matter as long as you are consistent). The magnitude of this vector = whatever speed you have left over--i.e. your-speed - pirates'-speed. As long as this is more than 0, you are steadily covering the circle and will eventually go all the way around its circumference, at which point you must have found the pirates.

So in short, as long as your speed is even a tiny bit greater than the pirates', you will catch them eventually.
edit: On second thought, realized that this is not quite true: if the rate at which you traverse the circumference decreases rapidly enough (as a result of the circle's rate of increase), you will never traverse it entirely, despite being constantly in motion along it. So we have to compare the rate of increase of the circle's circumference with the rate of your travel along it (ugh, this is edging closer to calculus):

At any given moment, the "momentary" difference in radius is the limit of t*v where v is your "outward" speed, or the length of the vector corresponding to your outward motion. "t" is time traveled, and t approaches 0. The momentary increase in circumference is 2*pi*t*v (the distance to the pirates is fractionally less, since you only have to cover some fraction of the circumference). Therefore, the "momentary distance" traveled along the circumference must be at least 2*pi*t*v. for you to make headway.

The momentary difference you travel along the circumference is determined by the other vector of your motion: the vector parallel to the tangent. Call its length xv. Your total speed is not the sum of the lengths of these vectors, as I hastily said before, but the length of the vector that corresponds to the distance between your origin and the diagonal of the parallelogram formed by these two vectors (i.e. the vector that corresponds to your actual direction of travel).

So, basically, we want to make sure that x>=2*pi, so that 2*pi*t*x*v >= 2*pi*t*v, so that you are constantly making headway along the circumference as it expands.

Now, the "outward" and "sidelong" vectors are perpendicular by definition, so that makes finding their diagonal easier. Making "v" (the pirates' speed) our unit, for convenience, we have a right triangle here: one side has length 1, one side has length x; hypotenuse (corresponding to our speed, in terms of the pirates' speed) has length sqrt(x^2 + 1). Given that x must be at least 2*pi, the hypotenuse must be at least sqrt(4*pi^2 +1) According to Google, this is upwards of 6.36, so unless I am making some new mistake (which I very possibly am), this method of catching them would not work unless you went over 6.36 times the pirates' speed.

One way you could think about this is as follows. Put the possible pairs for (A, X) in some order, such as:

(0, 0),
(1, -1), (1, 0), (1, 1),
(-1, -1), (-1, 0), (-1, 1),
(2, -2), (2, -1), (2, 0), (2, 1), (2, 2),
(-2, -2), (-2, -1), (-2, 0), (-2, 1), (-2, 2),
(3, -3), ... etc.

The actual ordering doesn't matter. I just want a way to refer to the "first pair", "second pair", and so on, in such a way that every pairing gets a number, starting at 1.

Then, at my first shot, I look at the first pair -- (0,0) in my list -- and pretend those were the starting (A, X) values, and shoot wherever the frog would be -- in this case, (0,0). For my second shot, I look at the second pair -- in this case (1, -1) -- and pretend those were the starting values. Since the frog started at 1 and has moved once to the left, I ought to shoot at zero again. For the third shot, I have the pair (1, 0), so I'd assume the frog has moved twice (0 spaces) starting at 1, and I'd shoot 1. For the fourth shot (starting at (1,1)), you can see I'd want to shoot at 4.

Since the frog's initial (A,X) started in the list, I'll eventually hit it (on the turn corresponding to where its initial (A,X) was in the list, if not before).

that's the thing, there's nowhere in the problem that states X<=A or anything along those lines
so the pairs would really be
(0,0) (0,1) (0,2) (0,3) ... (0,-1) (0,-2) (0,-3)...
(1,0) (1,1) (1,2) (1,3) ... (1,-1) (1,-2) (1,-3)...
(-1,0) (-1,1) (-1,2) (-1,3) ... (-1,-1) (-1,-2) (-1,-3)...

the original problem fixes A=0 (and this has been posted on TL before), and i had no problem with that one, if it's 0,0 shoot at 0, then assume 0,1, shoot at 1, then 0,-1 so shoot at -2, etc.
but with multiple starting locations arbitrary but fixed starting location i don't see how this works

I want to claim that as long as you have the ability to travel faster than the pirate ship, that you can always catch them (using the method proposed by qrs), and that there is in fact no worry about the circle "expanding faster than you can keep up with" (though I don't think this should be a priori obvious).

My argument is as follows. I'm going to let a be the speed of the pirate ship, and v be our speed; I am assuming v > a. The pirate ship travels radially outward from its starting position (which is in the center of our polar coordinate system, coordinates r and T -- here I'll use T to stand for theta). For simplicity, I'll wait until the pirate ship reaches my current radius, and then begin traveling outwards with a radial velocity a (so that I am always on the same "expanding circle" as the pirate ship) and around with an angular velocity W. (Here W is literally dT/dt, where t is time.) I will travel so as to make W as large as possible given my maximum speed of v. Evidently, as r increases, W must decrease. We have the fundamental relationship that if our ship is at a distance r from center, then, given that our total speed is v and our radial speed is a,

v^2 = a^2 + r^2 W^2,

from decomposing my velocity into its "radial" and "angular" components (as shown in qrs's picture). In addition, we may solve for r in terms of a; at a given time t, since I am always traveling radially outward with speed a, we must have

r = at + R,

where R is my initial distance from the pirate ship (I'm choosing time so that I start moving at time t=0. Thus, for instance, the flash of lightning occurs at time t = -R/a).

Now, the total angle through which I travel is the integral from time t=0 to time t=(infinity) of W (with respect to time). I would like this to be at least 2pi. The above two equations let me solve for W in terms of t. I find

W = (v^2 - a^2)^(1/2) * (integral from t = 0 to t = infinity) 1/(at+R) dt.

This integral diverges (an antiderivative of the integrand for t > 0 is (1/a)log(at + R)). It follows that I can go as far as I want around the "expanding circle", while still keeping on this circle (so I am at the same r-coordinate as the pirates), and will therefore catch them.

You could even use the above integral to compute, for given v, a, how long it will take at most to catch the pirates. In JeeJee's case of v = 2a, I am finding that it ought take no more than (R/a)*e^(2pi) from the flash of light.