Strategy optimisation

Question

This is a question from the Singapore Invitational Mathematics Challenge 2016. The question paper can be found here. (Part C:Question 2) http://www.nushigh.edu.sg/qql/slot/u90/file/simc/2016QuestionPaper.pdf

It is said that there is a boat in distress(say boat $A$) within the radius of $R$ of an island. The rescue boat(say boat $B$) from the island and boat $A$ can both travel a maximum of $10$ km. Boat $A$ is rescued when the trajectories of the two boat meet. Boat $B$ knows the exact location of the boat $A$, but boat $A$ knows neither where is the island nor the location of boat $B$.

The question is to find an optimal strategy of boat $A$ to move so that its chance to be rescued is maximised when

(i) $R=40$

(ii)$R=15$

For more details, you can refer to the question paper.

I believe it is not the best idea for boat $A$ to stay still but I cannot find any better strategy, as boat $A$ do not have any information on location.

I hope some kind, brilliant soul can help me out.

Answer 1

This is a partial answer. The rest of it is a bit too much for me, but i hope it'll encourage another kind soul after me.

Some disclaimers first:

Let $P(S)$ be the probability of A being saved.

Also, It can easily be proven that the probability of the actual distance $D$ between A and the island is less than or equal to any $d$ is:

$$ \mathbb{P}(D<d) = 1 * \frac{\int_{D<d}p.dp}{\int_{D<R}p.dp} = > \frac{\pi d^2}{\pi R^2} = \left(\frac{d}{R}\right)^2 $$ I'm assuming $D$ is always greater than zero for the rest of the answer, as that case is plain silly: A gets rescued no matter what.

Since the link specifies that boat A can still send, but not receive, radio signals to boat B, and that boat A does not know where he is but has a compass so he can accurately plan a trajectory, that means that boat B will know A's entire trajectory. Thus, boat B's path will be a straight line to A's resting place, and we can assume the rescue will be successful if boat A is there by sunset. Even if they cross paths before, that doesn't matter at all because they'll surely meet again.

Thus, A's trajectory is mostly irrelevant: All that matters is it's resting place.

And for A, that means he can only choose a distance $X \le 10 $ and row on a random direction for that amount.

¿So for how long does A have to row?

1. Let's say $X = 0$, or that A decides to just wait. We need the probability $P(S|_{X = 0})$ of A Surviving if it decides to stay still. That will happen if and only if the actual distance $D$ to the island was not over $10$ (Meaning $P(S|_{X=0 \cap D<10}) = 1$), so

$$ P(S|_{X = 0}) = $P(S|_{X=0 \cap D<10}) * \mathbb{P}(D<10) = 1 * \frac{\int_{D<10}p.dp}{\int_{D<R}p.dp} = \frac{\pi 10^2}{\pi R^2} = \frac{100}{R^2} $$

That is, $\frac{1}{16}$ in case (i), or $\frac{4}{9}$ in (ii)

2. Now let's say A decides to row a certain $X \le 10$ distance in a straight line. $P(S|_{X = x})$ will be a little more complicated to obtain.

Now, there is a chance to be rescued as long as $ D \le 10 + X $ but unfortunately now $P(S|_{X = x \cap D <10 + x})$ is not constant. Since it's a bit too complicated to explain in words, i'll let this drawing do the talk (or most of it) for me:

It can be deduced from there that A will be rescued as long as the point he chooses gets inside B's range, meaning:

$$P(S|_{X = x \cap D = d}) = 1_{\{d < 10-x\}} + \frac{\theta(x, d)}{2\pi} 1_{\{10-x < d < 10 + x\}}$$

Where $\theta(x, d)$ is the angle depicted in the diagram above.

We can use the Law of cosines :

$$c^2 = a^2 + b^2 - 2ab\cos\left(\alpha\right)$$

To obtain an expression of $\theta$ as a function of $d$ and $x$, when $10-x < d < 10+x:

$$10^2 = x^2 + d^2 - 2xd\cos\left(\frac{\theta}{2}\right)$$ $$\theta = 2 \cos^{-1}\left(\frac{d^2 + x^2 - 100}{2xd}\right)$$

Which means:

$$P(S|_{X = x \cap D = d}) = 1_{\{d < 10-x\}} + \frac{2 \cos^{-1}\left(\frac{d^2 + x^2 - 100}{2xd}\right)}{2\pi} 1_{\{10-x < d < 10 + x\}}$$