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.
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4You might want to consider the odd and even cases separately. When $n$ is even you seem to be looking at twice the circumradius (from corner to opposite corner). When $n$ is odd, you seem to be looking at the apothem plus the circumradius (from corner to opposite side; this is not really the maximum diameter of the polygon, e.g. the side of a triangle is longer than the central height). Formulas for both can be found at https://en.wikipedia.org/wiki/Regular_polygon#Circumradius and use trigonometry – Milten Oct 25 '23 at 20:25
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1You might like to check out https://math.stackexchange.com/help/notation – KDP Oct 27 '23 at 00:46
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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...
KDP
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