The forest

@Satvik Golechha – 😐 👍. I'll just say yes. Hahaha.

@Trevor Arashiro – One good problem leads to another. :)

@Trevor Arashiro – Yes, that's right. First, without loss of generality, fix a starting point on the circumference. Draw a diameter through this point, and let the central angle $\theta$ measured relative to the starting point vary uniformly from $0$ to $\pi$ . (By symmetry we only have to consider "exit" points on one half of the circle). This will cover all the "exit" points on one half of the circle. Now, using the Cosine rule, the length $L(\theta)$ of a chord for a given value of $\theta$ is given by

$L^{2} = 2r^{2}(1 - \cos(\theta)) = 2r^{2}(1 - (2\cos^{2}(\frac{\theta}{2}) - 1)) =$

$4r^{2}(1 - \cos^{2}(\frac{\theta}{2})) = 4r^{2}\sin^{2}(\frac{\theta}{2})$

$\Longrightarrow L = 2r\sin(\frac{\theta}{2})$ for $0 \le \theta \le \pi.$

Thus, using the MVT, the expected length of a path out of the forest will be

$\displaystyle\dfrac{1}{\pi}\int_{0}^{\pi} 2r\sin(\frac{\theta}{2}) d\theta = \dfrac{4r}{\pi}\int_{0}^{\frac{\pi}{2}} \sin(u) du = \dfrac{4r}{\pi}(-\cos(\frac{\pi}{2}) - (-\cos(0))) = \dfrac{4r}{\pi}.$

Note that the substitution $u = \frac{\theta}{2}$ was made, giving us $2 du = d\theta$ with $u$ going from $0$ to $\frac{\pi}{2}$ as $\theta$ went from $0$ to $\pi$ .

@Brian Charlesworth – Yoda, I am. Speak backwards, I do.