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I want to have an analytical way to solve for the number of sides a regular polygon has when given the area and apothem. A regular polygon with n-sides, an area of A, and an apothem of a, has these characteristics related by the formula $na^2\tan (\frac{\pi}{n}) = A$. With some rearranging, you get the equation $\tan (\frac{\pi}{x}) = \frac{c}{x}$ where c is $\frac{A}{a^2}$ I have found that c ranges from $\pi$ to $3\sqrt{3}$. $3\sqrt{3}$ corresponds to a triangle(the smallest number of sides possible in a polygon) and $\pi$ corresponds to a circle(infinite sides from a point of view). So the question now is, of course, what is the method of solving an equation of the form $\tan (\frac{\pi}{x}) = \frac{c}{x}$ where c is a known constant?

KReiser
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John Peter
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2 Answers2

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Notice that you could rewrite this into the form of

$tan(y) = k * y$ for $y=\frac{\pi}{x}$, $k = \frac{c}{\pi}$.

There is no analytical solution to this equation for $k=1$ from what I have found. Therefore, there is no generic way to do this for every $c$. Numerical approximations would be needed.

Here is a post about approximating the value for $k=1$: Solution of tanx = x?

Modify their update function to use $k*y$ instead of just $y$, and once you converge on a value of $y$, solve for $x$.

cmitch
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Let $$f_n = n \tan\left(\frac{\pi}{n}\right), \quad 3 \le n \in \mathbb{N}$$ then $\pi \lt f_n \le 3\sqrt{3}$, and $f_n$ is a monotonically decreasing sequence.

The power series for $f_n$ is $$f_n = \sum_{k=0}^{\infty} \displaystyle \frac{U_{2 k + 1} \pi^{2 k + 1}}{n^{2 k} (2 k + 1)!}$$ where $U_i$ is Euler number or up/down number $i$, forming the OEIS sequence A000111 (of positive monotonically increasing integers).

I do not believe there is an exact algebraic solution to solving $$f_n = c \quad \iff \quad c = \sum_{k=0}^{\infty} \displaystyle \frac{U_{2 k + 1} \pi^{2 k + 1}}{n^{2 k} (2 k + 1)!}$$ for $n$.

If we examine inverse of the relative difference to $\pi$, $$r_n = \frac{\pi}{f_n - \pi} \approx 0.303935 n^2 + 0.000895 n - 1.20688$$ we can use the approximation to solve for $n$ when $c = f_n$ is known, $$n = \left\lceil \sqrt{3.28986823 \frac{\pi}{c - \pi} + 4} \right\rfloor$$ which yields the correct result for $3 \le n \le 257942$.

Perhaps there exists a better invertible approximation $r_n$ that works for all $n$ you are interested in?

None
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  • Thanks, this is pretty useful! Can you please tell me what area of math this is for the sums and approximations. It seems interesting. – John Peter Jul 27 '20 at 21:49
  • @JohnPeter: The power series exist for all trigonometric functions; I grabbed the one for tangent, and examined case $x = \pi/n$. When you have a monotonic series $f_n$, the invertible approximation $r(n)$ only needs to be good enough so that $$\left\lvert \frac{r(n) - f_n}{f_n} \right\rvert \lt \frac{1}{2}$$ for $r(n)$ to yield the correct result when inverted, since $n \in \mathbb{N}$. I don't know which field of maths this is, but I think it falls into numerical methods – even though it works fine for algebraic solutions too. – None Jul 27 '20 at 21:56
  • @JohnPeter: As to how I found $r_n$, I plotted $f_n$, and noticed the $\pi$ limit, then examined the difference to $\pi$, and then the inverted difference, $\pi/(f_n - \pi)$, and noticed that this last one is very, very close to a quadratic function. I suspect a mathematician would say this is the guessing method, but I think of it as systematic exploration of the function, until you find something you can use. – None Jul 27 '20 at 21:59
  • Thank you for the quick reply and solution. – John Peter Jul 27 '20 at 22:03
  • Have a look at https://math.stackexchange.com/questions/3340734/could-these-polynomials-be-identified – Claude Leibovici Jul 28 '20 at 05:34