# 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?

• 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. Oct 2 '18 at 11:32

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 $$1$$.
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 $$\sec^2 (\theta)$$. $$1. (i.stack.imgur.com/uA1fI.jpg)$$ $$2. (i.stack.imgur.com/VjIzn.jpg)$$