<s>[INST] Sum of the areas of all n-sided polygons inscribed within each other Imagine you take an $n$-sided regular polygon with side length $x$, and connect the midpoints of each of its sides to construct a smaller but identical $n$-sided regular polygon within it.
Now perform the same process with the smaller polygon, and then with the next one, and the next and so on.
If you continue this, endlessly inscribing smaller and smaller regular $n$-sided polygons, what is the sum of all of their individual areas in terms of $x$ and $n$?
Below is an image describing this question when n = 6:

I know there must be a common ratio between each of the areas, so i'm thinking about turning this into a geometric series of the form $\frac{a}{1-r}$ (since $r$ < 1), where $a$ = the area of the first polygon, however I do not know how to find the specific values for $a$ and $r$ in terms of $x$ and $n$.
Any help with this question would be greatly appreciated!
 [/INST] Let us first find $a$.

Let $O$ be the center of the regular polygon. Then, a circle centered at $O$ circumscribes the polygon.  Each side is a chord of the circle and hence the perpendicular from the center bisects the side. Also, $\triangle AOM\cong\triangle BOM$ so $\angle AOM=\angle BOM=\dfrac{\pi}{n}$. Now, in $\triangle OAM $,$$\frac{AM}{MO}=\frac{x/2}{r}=\tan\frac{\pi}{n}$$ so $$r=\frac{x}{2\tan\frac{\pi}{n}}$$ so the area of triangle $OAB$ is $\dfrac12OM\times AB=\dfrac{x^2}{4\tan\frac{\pi}{n}}$. There are exactly $n$ such triangles so the total area is, $$\fbox{$a= \dfrac{nx^2}{4\tan\frac{\pi}{n}}$}.$$
Now,

to find the side length of the smaller second polygon, join BC. By the midpoint theorem on $\triangle ABC$, the side length of the second polygon is $\dfrac{BC}{2}$. Now, to find $BC$, note that $\angle CAB=\dfrac{(n-2)\pi}{n}$. By the cosine rule in $\triangle ABC$,$$\frac{x^2+x^2-BC^2}{2x^2}=\cos\angle CAB$$$$\implies 1-\frac{BC^2}{2x^2}=\cos\frac{(n-2)\pi}{n}=\cos\left(\pi-\frac{2\pi}{n}\right)$$$$=-\cos\frac{2\pi}{n}$$ so that $$\frac{BC^2}{2x^2}=1+\cos\frac{2\pi}{n}=2\cos^2\frac{\pi}{n}$$ which implies that $$BC=2x\cos\frac{\pi}{n}$$ or that the side of the second polygon is $\dfrac{BC}{2}= x\cos\dfracπn.$ Thus, its area will be (by similarity) $a\cos^2\dfrac{\pi}{n}$, which implies that the common ratio for the geometric series is $\cos^2\dfrac{\pi}{n}.$
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