HMMT February

Solution:

Let $\alpha_{k}$ be a random variable that represents the turn made after step $k$, choosing $\alpha_{k}$ uniformly at random on the complex plane among the arc of the unit circle containing $1$ from $-i$ to $i$. It is well known that $\mathbb{E}\left[\alpha_{k}\right]=\frac{2}{\pi}$. We have that
$$a_{n}=\sum_{i=1}^{n} \sum_{j=1}^{n} \mathbb{E}\left[\prod_{k=1}^{i-1} \alpha_{k} \prod_{k=1}^{j-1} \alpha_{k}^{-1}\right]$$
Separating the sum based on $|i-j|$,
$$a_{n}=n+2 \sum_{t=1}^{n-1}(n-t) \mathbb{E}\left[\alpha_{1} \cdots \alpha_{t}\right]=n+2 \sum_{t=1}^{n-1}(n-t)\left(\frac{2}{\pi}\right)^{t}$$
Since terms with large $t$ get very small, we can write
$$\lim _{n \rightarrow \infty} \frac{a_{n}}{n}=\lim _{n \rightarrow \infty} 1+2 \sum_{t=1}^{\infty}\left(1-\frac{t}{n}\right)\left(\frac{2}{\pi}\right)^{t}=1+2 \sum_{t=1}^{\infty} \lim _{n \rightarrow \infty}\left(1-\frac{t}{n}\right)\left(\frac{2}{\pi}\right)^{t}=1+2 \sum_{t=1}^{\infty}\left(\frac{2}{\pi}\right)^{t}$$
This gives that
$$x=\frac{\pi+2}{\pi-2} .$$
To estimate this quantity we use the approximation $\pi \approx 22 / 7$, which gives us
$$x \approx \frac{22+14}{22-14}=4.5$$