Have I discovered something or have I not been paying attention to circular or near circular geometry?

Question

It appears I have discovered something for myself:

Take a polygon with an odd number of sides, e.g. a nonagon (9 sides for clarification.) Draw it so a horizontal side is facing down and a point is at the top. Then, draw a polygon with twice the number of sides so that the bottom side is in the same space as the bottom side of the smaller polygon. then draw a circle the size of the larger polygon (circumscribed) and its center should land right on the topmost point of the smaller polygon. When this is done with circles, where one is half the size of the other, and is tangent to the inner surface of the larger one, one end touches the circle, and the other passes through the radius.

But polygons aren't exactly circular, and their heights aren't proportional in the same way circles are. Another thing is that the triangle and hexagon are pretty obvious, as one stacks to form an analog of the other, but a pentagon doesn't stack to make a decagon in the same way.

If anyone else has figured this out, please explain, though I feel the explanation is rather simple. I am a stickler for knowing what is known and what is unknown, so not knowing the possibility of someone else knowing this bugs me.

The phenomenon I've seen:

edit Another thing I've noticed is that when I draw a line along any of the smaller polygon's sides, each line passes through at least 2 points of the larger, or at least comes very close to the latter.

Answer 1

We can easily prove the following:

Draw a regular $(2n+1)$-gon ($n=1$ is trivial, so let $n>1$). Select a point (call it $A$), and with this point as the centre, draw a circle passing through the 2 points (say $B$ and $C$) whose midpoint is opposite to $A$. [This is possible since $AB = AC$ by symmetry.] Then, a regular $(4n+2)$-gon can be inscribed in the said circle, with $BC$ as one side.

Proof:
Join $AB$ and $AC$. Note that our $(2n+1)$-gon is divided into two $(n+1)$-gons and a triangle. Let $\angle ABC = \angle ACB = x$ and $\angle BAC = y$. We have that $2x+y = 180^\circ$. Also, any angle in the $(2n+1)$-gon is $(2n-1)180^\circ/(2n+1)$. Now, the sum of angles in the $(n+1)$-gon is $$(n-1)(2n-1)180^\circ/(2n+1) + 2((2n-1)180^\circ/(2n+1) - x) \\= (n+1)(2n-1)180^\circ/(2n+1)-2x$$

[I have skipped a lot of steps here, so I encourage you to verify this]. This should be equal to $(n-1)180^\circ$. Equating, we get $x = 180^\circ n/(2n+1)$. So, $\angle CAB = y = 180^\circ - 2x = 180^\circ/(2n+1) = \color{green}{360^\circ/(4n+2)}$.

This proves the claim, hence the observed properties.

Answer 2

The apothem of a $2n$-sided polygon, of side length a, is given by

$ p = \dfrac{a}{ 2 \tan ( \dfrac{\pi}{ 2n} )}$

The height of a $n$-side polygon, of side length a, is given by

$ h = \dfrac{a}{ 2 \tan( \dfrac{\pi}{n} ) } + \dfrac{a}{ 2 \sin ( \dfrac{\pi}{n}) } $

So we only have to show that $p = h $.

For that, note that

$ h = \dfrac{a}{2} \cdot \dfrac{ \sin( \dfrac{ \pi}{n}) + \tan( \dfrac{\pi}{n} ) }{ \tan( \dfrac{ \pi}{n} ) \sin ( \dfrac{ \pi}{n} ) } = \dfrac{a}{2} \cdot \dfrac{\cos(\dfrac{\pi}{n}) + 1}{\sin ( \dfrac{\pi}{n})} = \dfrac{a}{2} \cdot \dfrac{ \cos^2(\dfrac{\pi}{2 n}) }{ \sin(\dfrac{\pi}{2n})\cos(\dfrac{\pi}{2n}) } = p$

Answer 3

The reason is the angle at $C$ is one half of the central angle at $O$ (see figure below). Hence $C$ is the center of a polygon with the same side $AB$ but doubled number of sides.