I would like to know two series of rational polygons (in the sense of polygons with rational vertices) that converge to the unit circle. More precisely one series P_n of rational polygons all contained in the circle and one series Q_n all containing the circle, such that the area of the n-th complement Q_n-P_n goes to 0.
For example: give a sequence of ever finer (finite) partitions of the circumference by rational points (using any parametrization of Pythagorean triples). Each partition gives an inscribed polygon, and scaling that polygon up by a rational factor (which can get closer and closer to 1 as the number of sides rises) gives a polygon containing the circle. Since the rational scalars approach 1, the areas of the inner and outer polygons converge.
But I do not want to work out some specific sequence of partitions for myself, and corresponding scalars, since I suspect simpler neater solutions are already known.
Is some simple way known?
Eventually i would also like to know higher dimensional analogues -- rational polytopes converging to the $n$-dimensional spheres. But I will accept an answer for the 2-D case.
Here is an easily described solution, though it could obviously be improved for practical calculations. Consider the first quadrant of the circle. For each natural number $n$ partition the arc at $n$ points, namely the intersections of the circle with the lines from the point $\langle 0,-1\rangle$ to the points $\langle i/n, 0\rangle$ as $i$ ranges from $0$ to $n$. Then the inner Riemann sum (i.e. using rectangles, each with its upper right corner at a partition point) gives a polygon contained in the circle, while the outer one gives a polygon containing the circle. A simple calculation shows these inner and outer polygons come as close to each other as desired when $n$ goes to infinity.
Namely for any $d>0$ take $n$ with $1/n<d/2$. The difference between the inner and outer sums for this $n$ is less than $d$.
For verification notice for fixed $n$, the $x$ coordinates of the partition points are $2in/(n^2+i^2)$. These intervals have monotone decreasing length as you go from $0$ to $1$. (One proof: $2x/(1+x^2)$ has negative second derivative all over $[0,1]$.) So the left Riemann sum for this partition exceeds the right Riemann sum by less than the area of the first summand in the left Riemann sum, and that is $2/n$.