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Consider these two shapes using squares and circles:

circle and square

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:

circle and triangle

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?

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  • $\begingroup$ For the first pair both ratios are the product of $\frac{Area\ circle}{Area\ inscribed\ square}$ and $\frac{Area\ square}{Area\ inscribed\ circle}$. The same idea also works for combinations of two irregular shapes if their orientations are held fixed. $\endgroup$
    – random
    Oct 2 '18 at 11:32
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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.

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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) $. $$1. (i.stack.imgur.com/uA1fI.jpg)$$ $$2. (i.stack.imgur.com/VjIzn.jpg)$$

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