Constant area ratio of circles and regular polygons Consider these two shapes using squares and circles:

On the left: there are two squares, one inscribed and one circumscribed in the same circle. The ratio of the two square areas is $2$:
$$
\cfrac{Area_{big-square}}{Area_{small-square}} = \cfrac{ (2R)^2}{ (R \sqrt{2})^2} = 2
$$
On the right: there are two circles, one inside and the other outside the square (both circles touch the square). The ratio of the two circle areas is also $2$:
$$
\cfrac{Area_{big-circle}}{Area_{small-circle}} = \cfrac{\pi R^2}{\pi r^2} = \cfrac{\pi (r \sqrt{2})^2}{\pi r^2} = 2
$$
So the ratio is the same and equals $2$ in both cases.
Interestingly, this also happens when we have triangles:

On the left: the radius r is $1/3$ of the height of the big triangle, and $2/3$ of the height of the small triangle (see apothem). So we have:
$$
\cfrac{Area_{big-triangle}}{Area_{small-triangle}} = \cfrac{\frac{(3r)^2}{\sqrt{3}}}{\frac{(3r/2)^2}{\sqrt{3}}} = \cfrac{9r^2}{\frac{9r^2}{4}} = 4
$$
On the right: The small radius r is $1/3$ of the triangle height, and the big radius R is $2/3$ of the triangle height. The ratio is:
$$
\cfrac{Area_{big-circle}}{Area_{small-circle}} = \cfrac{\pi (2r)^2}{ \pi r^2} = 4
$$
Again, the ratio is the same in both cases and equal $4$.
Question: is this relationship between the constant area ratios documented somewhere? Does this apply to other regular polygons as well?
 A: It is relatively simple to prove such a relationship. Consider a regular polygon with $n$ sides. The distance from the center to the vertices is $R$, the distance to the middle of the sides is $r$. The angle between the lines from the center to two consecutive vertices is $\frac{2\pi}{n}$, while the angle between the line from the center to one vertex and the line to an adjacent middle is $\frac{2\pi}{2n}=\frac{\pi}{n}$. Now, we can write $$\cos\frac{\pi}{n}=\frac{r}{R}$$ Notice that area is proportional to $r^2$ and $R^2$, so the ratio $$\frac{Area_{large}}{Area_{small}}=\frac{R^2}{r^2}=\frac{1}{\cos^2\frac{\pi}{n}}$$
For $n=3$ the ratio is $4$, for $n=4$ the ratio is $2$. As $n\to\infty$, the ratio goes to $1$. 
You can now also write explicitly areas of the inscribed and circumscribed polygons. 
A: Googling didn't give me any source where it may have been documented. 
I can prove it applies to other polygons too.  We start by noting that areas of all polygons are $A= k*r^2$ where $k$ is fixed for a polygon and $r$ is radius of circumscribing  circle.  (You can derive general formula for area of a polygon as exercise.) 
In the figure whose image link is given at bottom,  you can see that the angle $\theta = \frac{\pi}{n}$ . (Construct $n$ triangles to get the angle) 
Also, $$ r = r' \cos(\theta) $$
So,  ratio of area of the circles would be $ \sec^2 (\theta) $. 
Similarly,  in the figure 2 here,  again the ratio of radii is fixed for a polygon ;$ \sec(\theta) $. And as the area is proportional to $r^2$. So,  the ratio of areas of the polygons will also be $ \sec^2 (\theta) $. 
$$1. (i.stack.imgur.com/uA1fI.jpg)$$
$$2. (i.stack.imgur.com/VjIzn.jpg)$$
