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!
 A: Each interior angle of a regular n-gon is given by $(2n - 4) \times 90^0 /n$.
The radius connecting the center to a vertex will divide that angle in halves.
The ray from center perpendicular to a side will cut that side into halves.
The above info will be enough to determine r.
s = The area of that triangle, can therefore be found in terms of x and n.
a = 2ns.
A: 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 $\cos^2\dfrac{\pi}{n}.$
A: Let the side length of the outermost polygon and side length of the first constructed polygon be $x_0$ and $x_1$ respectively. The external angles of the polygons will each be $\frac{2\pi}{n}$, so the internal angles will each be $\pi - \frac{2\pi}{n}$. Consider the triangle shown in the diagram below:

Applying the cosine law, we get
$$\begin{align}
(x_1)^2 &= \left( \frac{x_0}{2} \right)^2 + \left( \frac{x_0}{2} \right)^2 - 2\left( \frac{x_0}{2} \right)\left( \frac{x_0}{2} \right) \cos \left( \pi - \frac{2\pi}{n} \right) \\
\\
&= \frac{(x_0)^2}{2} + \frac{(x_0)^2}{2} \cos \left( \frac{2\pi}{n} \right) \\
\\
&= \frac{(x_0)^2}{2} \left[ 1 + \cos \left( \frac{2\pi}{n} \right) \right] \\
\\
&= \frac{(x_0)^2}{2} \left[ 2 \cos^2 \left( \frac{\pi}{n} \right) \right] \\
\\
&= (x_0)^2 \cos^2 \left( \frac{\pi}{n} \right) \\
\\
x_1 &= x_0 \cos\left( \frac{\pi}{n} \right)
\end{align}$$
Thus the common ratio of the sides is $\cos \left( \frac{\pi}{n} \right)$, so the common ratio of the areas will be $\boxed{\cos^2 \left( \frac{\pi}{n} \right)}$
