This is a question from the Cambridge Mathematical Tripos in 1871, Scanned copy added at the end of the post.
A ship $A$ sees another ship $B$ whose course is not known. Given that they have the same speed, prove that chance of them coming within a distance $d$ of each other is always $\frac{2sin^{-1}(\frac{d}{a})}{\pi}$ no matter the course of $A$, provided that it's inclination to $AB$ is not greater than $cos^{-1}(\frac{d}{a})$where $AB=a$
This is a summary of my method so far:
First I constructed a vector triangle $ABC$ where $C$ was the intersection of the courses of $A$ and $B$. In this triangle I denoted the angle between $A$ and $AB$ as $\theta$ and $B$ and $AB$ as $\phi$.
I then used trigonometry to get an expression for $d$ which I subsequently differentiated with respect to time and set as a minimum so I could get $t=\frac{b+c}{2v}$ where $v$ is the velocity of the ships.
Using this I got $d=acos(\frac{\theta + \phi}{2})$ which implies $\phi=2cos^{-1}(\frac{d}{a})-\theta$
Could someone help me proceed and post a solution themselves if they have a better method?
It would also be helpful if you could point out if this is possible to complete having only high school maths knowledge.
I've just learned about double integrals and their application in probability so I think I can finish off the problem now.
Continuing from where I got to, the chance would be:
$$\frac{\int^{cos^{-1}(\frac da)}_{-cos^{-1}(\frac da)} \int^{2\pi - \phi}_{\phi} \; d\theta d\phi}{\int^{cos^{-1}(\frac da)}_{-cos^{-1}(\frac da)} \int^{2\pi}_{0} \; d\theta d\phi} $$
Which is equal to:
$$\frac{2\int^{cos^{-1}(\frac da)}_{-cos^{-1}(\frac da)}(\pi - \phi) \; d\theta}{4\pi cos^{-1}(\frac da)}$$
This works out to:
$$\frac{\pi - 2cos^{-1}(\frac da)}{\pi}=\frac{2sin^{-1}(\frac da)}{\pi}$$
I think this is right, would anyone mind double checking or even better suggesting a different method if one exists?