Harvard-MIT Math Tournament

Solution:
The radius of the inscribed circle is $\frac{1}{2} \cot \frac{\pi}{n}$, the radius of the circumscribed circle is $\frac{1}{2} \csc \frac{\pi}{n}$, and the area of the $n$-gon is $\frac{n}{4} \cot \frac{\pi}{n}$. The diagram below should help you verify that these are correct.

Then $A_{n} = \pi \left( \frac{1}{2} \csc \frac{\pi}{n} \right)^{2} - \frac{n}{4} \cot \frac{\pi}{n}$, and $B_{n} = \frac{n}{4} \cot \frac{\pi}{n} - \pi \left( \frac{1}{2} \cot \frac{\pi}{n} \right)^{2}$, so
$$\frac{A_{n}}{B_{n}} = \frac{\pi \left( \csc \frac{\pi}{n} \right)^{2} - n \cot \frac{\pi}{n}}{n \cot \frac{\pi}{n} - \pi \left( \cot \frac{\pi}{n} \right)^{2}}$$
Let $s$ denote $\sin \frac{\pi}{n}$ and $c$ denote $\cos \frac{\pi}{n}$. Multiply numerator and denominator by $s^{2}$ to get
$$\frac{A_{n}}{B_{n}} \sim \frac{\pi - n c s}{n c s - \pi c^{2}}$$
Now use Taylor series to replace $s$ by $\frac{\pi}{n} - \frac{\left( \frac{\pi}{n} \right)^{3}}{6} + \ldots$ and $c$ by $1 - \frac{\left( \frac{\pi}{n} \right)^{2}}{2} + \ldots$. By l'Hôpital's rule it will suffice to take just enough terms so that the highest power of $n$ in the numerator and denominator is determined, and that turns out to be $n^{-2}$ in each case. In particular, we get the limit
$$\frac{A_{n}}{B_{n}} = \frac{\pi - n \frac{\pi}{n} + n \frac{2}{3} \left( \frac{\pi}{n} \right)^{3} + \ldots}{n \frac{\pi}{n} - n \frac{2}{3} \left( \frac{\pi}{n} \right)^{3} - \pi + \pi \left( \frac{\pi}{n} \right)^{2} + \ldots} = \frac{\frac{2}{3} \frac{\pi^{3}}{n^{2}} + \ldots}{\frac{1}{3} \frac{\pi^{3}}{n^{2}} + \ldots} \rightarrow \mathbf{2}$$