This year is always better than last year?(2)

Let $B$ be a vertex and $C$ be a midpoint of one of the sides of a unit $n$ -gon, and let $O$ be the center. Then $B$ , $O$ , and $C$ make a right triangle $\triangle BOC$ , where $BC = \frac{1}{2}$ and $\angle BOC = \frac{\pi}{n}$ , and $BO$ is the radius of the circumcircle and $CO$ is the radius of the incircle.

By trigonometry, $BO = \frac{1}{2} \csc \frac{\pi}{n}$ and $CO = \frac{1}{2} \cot \frac{\pi}{n}$ . Since $BO$ is the radius of the circumcircle and $CO$ is the radius of the incircle, the area between the circles is $f(n) = \pi BO^2 - \pi CO^2 = \frac{\pi}{4} \csc^2 \frac{\pi}{n} - \frac{\pi}{4} \cot^2 \frac{\pi}{n} = \frac{\pi}{4} (\csc^2 \frac{\pi}{n} - \cot^2 \frac{\pi}{n})$ .

However, since $\csc^2 \theta - \cot^2 \theta = 1$ , this area can be simplified to $f(n) = \frac{\pi}{4}$ , which is a constant with respect to the number of sides $n$ of the polygon. Therefore, $f(2017) = f(2018)$ .

Relevant wiki: Pythagorean Theorem

Consider an isosceles triangle formed by one side of the unit regular $n$ -gon with the center of the $n$ -gon. Let the circumradius and inradius of the $n$ -gon be $R$ and $r$ respectively. By Pythagorean theorem, we have $R^2 = r^2 + \left(\frac 12\right)^2 = r^2 + \frac 14$ . Now, we have $f(n) = \pi R^2 - \pi r^2 = \pi (r^2 + \frac 14 - r^2) = \frac \pi 4$ . Implying that $f(n) = \frac \pi 4$ for all $n$ . Therefore, $\boxed{f(2017)=f(2018)}$ .