Four boats chasing

Simple solution

Due to symmetry, the positions of the four boats at any time $t$ will be the vertices of a square. Let $s$ be the side of that square. Since each boat is always moving exactly parallel to its side of the square, we have $\frac{ds}{dt} = -\frac{dx}{dt} = -v.$ The boats meet when $s = s_0 - vt = 0$ . Therefore $t = \frac{s_0}v = \frac{\SI{1}{km}}{\SI{10}{km/h}} = \SI{0.1}{hr} = \SI{6}{min}.$

---

Alternative solution

Consider the center of the lake the origin of a coordinate system $(0,0)$ . Use polar coordinates: the position of boat 1 is $(r\cos \theta, r\sin\theta)$ , where $r$ and $\theta$ are functions of the time $t$ . The position of boat 2 is $(r\cos(\theta - 90^\circ), r\sin(\theta - 90^\circ)) = (r\sin\theta, -r\cos\theta)$ .

The vector from boat 1 to boat 2 is therefore $\vec d = r(\sin\theta - \cos\theta, -\sin\theta - \cos\theta).$ and as boat 1 is traveling in that direction at speed $v$ , its velocity is $\vec v = \frac v d \vec d = \frac v{\sqrt 2}(\sin\theta - \cos\theta, -\sin\theta - \cos\theta).$ Splitting this in a radial and tangential component, we get $\vec v = \vec v_{rad} + \vec v_{tan} = -\frac v{\sqrt 2}(\cos\theta,\sin\theta) - \frac v{\sqrt 2}(-\sin\theta, \cos\theta).$ The radial component gives the change in $r$ , and the tangential component the change in $\theta$ : $\frac{dr}{dt} = -\frac v{\sqrt 2},\ \ \ \ \frac{d\theta}{dt} = -\frac v{\sqrt 2 r}.$ Solving the first of these equations with $r_0 = 1/\sqrt 2$ and $v = 10$ we get $r_t = \frac 1{\sqrt 2} (1 - 10 t);$ we see that $r_t = 0$ when $t = 1/10$ hours, or 6 minutes.

The second equation becomes $\frac{d\theta}{dt} = -\frac{10}{1 - 10t},\ \ \ \ \theta_t = \theta_0 + \ln (1 - 10t).$ This value tends to $-\infty$ as $t \to 1/10$ : the boats circle each other infinitely many times before actually meeting! Each boat will have completed a full revolution when $\ln (1 - 10t) = -2\pi\ \ \ \ \ t = \frac{1 - e^{-2\pi}}{10} \approx 0.09981\ \text{hours},$ that is, 99.8% of the total way.

Imagine that you are in one of the boats. Let's look at the positions of your boat and your neighbor's boat (the one that you are chasing) at time $t$ and $t+\Delta t$ . In the time $\Delta t$ your neighbor moved in the direction perpendicular to the straight line connecting the two of you, and accordingly, to the first order in $\Delta t$ , the distance between the two of you did not change. In other words your neighbor is neither getting away from your nor is he/her getting closer to you. On the other hand, you gain a distance of $v \Delta t$ , (where $v$ is your velocity), because you are moving directly in his direction. (You also have to adjust the direction of your motion, so that at the time $t+\Delta t$ you are still aiming at your neighbor. However, adjusting the direction does not change the distance between the two of you.)

Since you are always gaining distance at the rate of your full velocity, and you neighbor's motion does not influence the distance between the two of you, you will have to travel exactly 1km to meet your neighbor. That will take 6 minutes.

Note: There is no need to determine the orbit. A similar argument (but a bit more complex geometry) works for any number of boats on the corners of a regular polygon.

It couldn't seem to work up a great intuition for this problem, so here's how I did it. Consider the initial conditions:

$(x_1,y_1) = (0,0) \hspace{1cm} (x_2,y_2) = (D,0) \\ (v_{x1},v_{y1}) = (v,0) \hspace{1cm} (v_{x2},v_{y2}) = (0,v)$

After the first increment:

$(x_1,y_1) = (v \Delta t,0) \hspace{1cm} (x_2,y_2) = (D,v \Delta t)$

The original distance was $D$ . The new distance is (neglecting square infinitesimals):

${D'}^2 = (D - v \Delta t)^2 + (v \Delta t)^2 \approx D^2 - 2 v D \Delta t \\ D' = \sqrt{D^2 - 2 v D \Delta t }$

So we have a relationship between the distances at the present iteration and the previous iteration. Running this on a computer (with the original distance set to 1) and incrementing the time gives a final time of very nearly 6 minutes. Also, here's a plot of the spiral trajectory (made with an alternate program which calculates the 2D velocities at each time step using the 2D geometry).

Here is what it looks like when you start with a unit-hexagon (6 boats).

I thought of the boats as points on the complex plane, where each boat was moving towards a 90 degree rotation of it's current position, which gave the differential equation $\begin{aligned} \frac{dx}{dt} = ix - x && x(0) = 0.5 + 0.5i \end{aligned}$ where $x(0)$ was given for the top right boat, with solution $x(t) = (0.5 + 0.5i)e^{(-1+i)t}$ I then evaluated the path length to obtain the answer $\int_{0}^{\infty} |-e^{(-1+i)t}| dt = \int_{0}^{\infty} e^{-t} dt = 1 \ \mathrm{km}$ $\frac{1 \ \mathrm{km}}{10 \ \mathrm{km/h}} = 6 \ \mathrm{minutes}$

Consider any boat. It starts off at a distance of 1 km from its counterclockwise neighbour and travels directly towards it at a constant speed of 10 km/h and so takes 1/10th of an hour or 6 minutes to reach it.

Suppose we have a regular n-gon of side A met. with a boat placed at each vertex and each boat moving exactly as stated in the problem. We find that with each move made simultaneously by the boats, the shape of the n-gon is maintained but its size goes on shrinking.

The distance to be covered = A and the relative velocity of any two boats near each other = $v - v \times \cos \frac{360^\circ}{n}$ by compounding velocities. Hence the time taken to converge = $\dfrac{A}{v (1- \cos \frac{360^\circ}{n})}$ in the general case. In our case, n=4 so $t=\dfrac{A}{v (1- \cos \frac{360^\circ}{4})}$ = 1/(10) hr. = 6 min.

The velocity vector towards the centre is always constant at $\frac {1}{\sqrt{2}}$ km/hr. Distance b/w starting and the centre point is $\frac {10}{\sqrt{2}}$ and hence the solution. Apply Time equals Displacement/velocity.

I calculated the length of the spiral that the boat makes towards the center. I did that through geometrical series. In step dt the boat travels distance "a" along 1km edge of the initial square.

In the second step it travels distance a(1-a) of the rotated nested square.

In the next step dt it travels $a(1-a)^2$ and so on. It forms a geometrical series with a1=a and q= 1-a with the infinite sum for dt->0 of S=1km. Having v=10km/h and s=1km the time is 6 minutes.