Ratio between Polygon height to perimeter.

Question

Is there a constant ratio or linear/curvilinear set of ratios of the perimeter of an n-gon and the n-gon maximum height (eg apothem to apothem, circumscribed diameter, etc.) I've only calculated a few regular polygons: The triangle's ratio would be $\sqrt{2}:1$, the square's ratio would be $4:\sqrt{2}$, the hexagon would be $3:1$, and the octagon would be $(2\sqrt{2}+1):1$. Is there an associated sequence following this pattern? If so, please share it with me.

Answer 1

Method 1) The ratio P/h of perimeter (P) to height (h) where we define h as the diameter of the circumcircle (or vertex to opposite vertex of an even numbered regular polygon) the formula is: $$ P/h = N \sin(\pi/N)$$ where N is the number of sides of the polygon.

Method 2) When we define h as the diameter of the incircle (or as twice the apothem or as face to opposite face of an even numbered regular polygon) the formula is: $$ P/h = N \tan(\pi/N)$$ Method 3) When we define h as the orthogonal distance from a face to an opposite vertex of an odd numbered regular polygon (or as apothem plus circumradius) we combine the two above methods and eventually get: $$ P/h = 2 N \tan(\pi/(2 N)) $$

Here is a table of the ratios produced by the 3 methods.

N	Method 1	Method 2	Method 3	-------
3	2.5980	5.1961	3.4641	1.4142
4	2.8284	4.0000	3.3137	2.8284
5	2.9389	3.6727	3.2491	3.0000
6	3.0000	3.4641	3.2154	3.8284
7	3.0371	3.3710	3.1954
8	3.0614	3.3137	3.1825

The last column is the result of converting your ratios into decimal integers.

I've only calculated a few regular polygons: The triangle's ratio would be $\sqrt{2}:1$ , the square's ratio would be $4:\sqrt{2}$ , the hexagon would be $3:1$ , and the octagon would be $(2\sqrt{2}+1):1$

Your examples do not seem to concur with any of the above 3 methods, so maybe I am misunderstanding your question or misinterpreting your definitions or I have made a mistake. Perhaps you could check your figures, to see where we differ. I will try to add to this answer and edit as required in response to any edits you make to clarify the original question.

P.S. I agree with Milton that it would probably be best to treat even and odd polygons as separate cases.

P.P.S All 3 methods converge towards $\pi$ as N gets larger and so does the infinite series $$\lim_{n\to\infty} 2^n \sqrt{2-\sqrt{2+\cdots+ \sqrt 2}} = \pi$$

$$\pi = 2\sqrt 2 \cdot \frac{2}{\sqrt{2+\sqrt2}} \cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt2}}} \cdots.$$ See this old thread. Perhaps this is the connection to the $\sqrt 2$ that you appear to be looking for...