# A curious limit: area between a regular n-gon with side length 1 and an inscribed curve with the same perimeter

A curve $$S$$ with polar equation $$r=a\cos{(n\theta)}+b$$ is inscribed in a regular n-gon with side length $$1$$. $$S$$ touches the midpoint of each side of the n-gon. $$S$$ has the same perimeter as the n-gon.

Example with $$n=6$$:

I want to evaluate $$\lim_{n\to\infty}(\text{Area of n-gon}-\text{Area enclosed by }S)$$.

Context:
I know that if a circle is inscribed in a regular n-gon with side length $$1$$, the difference in areas approaches $$\dfrac{\pi}{12}$$ as $$n\to\infty$$. Then I wondered, if we replace the circle with a particular curve that has the same perimeter as the n-gon, what would the difference in areas approach?

My attempt:
$$r_\text{max}=a+b=\frac{1}{2}\cot{\left(\frac{\pi}{n}\right)}\implies b=\frac{1}{2}\cot{\left(\frac{\pi}{n}\right)}-a$$

Setting the perimeters equal, we have $$\int_{0}^{2\pi}\sqrt{\left(a\cos{(n\theta)}+\frac{1}{2}\cot{\left(\frac{\pi}{n}\right)}-a\right)^2+\left(-an\sin{(n\theta)}\right)^2}\,d\theta=n.$$

Then I tried unsuccessfully to express $$a$$ in terms of $$n$$, so that I could use that to evaluate the limit:

$$\lim_{n\to\infty}(\text{Area of n-gon}-\text{Area enclosed by }S)$$ $$=\lim_{n\to\infty}\left(\frac{n}{4}\cot{\left(\frac{\pi}{n}\right)}-\frac{1}{2}\int_{0}^{2\pi}\left(a\cos{(n\theta)}+\frac{1}{2}\cot{\left(\frac{\pi}{n}\right)}-a\right)^2\text{d}\theta\right)=\text{ ?}$$

Playing with desmos, it seems that $$a\to\dfrac{0.977...}{n}$$ as $$n\to\infty,$$ and the limit is approximately $$1.24$$.

• How can it be? > I know that if a circle is inscribed in a regular n-gon with side length $1$, the difference in areas approaches $\dfrac{\pi}{12}$ as $n\to\infty$. May 31, 2022 at 13:27
• @IvanKaznacheyeu It can be, because the area of the $n$-gon tends to $\infty$. May 31, 2022 at 13:48
• OK, I've lost the point in constant side length May 31, 2022 at 13:49

The perimeter equation can be rewritten (letting $$t=n\theta$$) as: $$\int_{0}^{2\pi n}\sqrt{\left(a\cos{t}+\frac{1}{2}\cot\frac{\pi}{n}-a\right)^2+(an)^2\sin^2{t}}\,{dt\over n}=n.$$
We can expand the integrand as a power series in $$n$$, retaining for $$a$$ only the leading term in the expansion: $$a= k/n$$, where $$k$$ is a positive constant. The result is: $$\sqrt{\left(a\cos{t}+\frac{1}{2}\cot\frac{\pi}{n}-a\right)^2+(an)^2\sin^2{t}}= \frac{n}{2 \pi }+\frac{\pi k^2 \sin ^2(t)+k \cos (t)-k-\frac{\pi }{6}}{n}+O\left(\frac{1}{n^3}\right).$$ We can then insert this into the equation, dropping the last term. We get: $$n^2+\pi^2 k^2 -2\pi\left(k+\frac{\pi}{6}\right)=n^2.$$ The $$n^2$$ terms cancel out and we can solve the equation for $$k$$: $$k={1+\sqrt{1+\pi^2/3}\over\pi}\approx 0.977593.$$ Finally, we can compute the last limit setting $$a=k/n$$: $$\lim_{n\to\infty}(\text{Area of n-gon}-\text{Area enclosed by }S)=k+{\pi\over12}.$$
• Nice, but how can we justify that $a\sim \frac{k}{n}$?
• @Dan That is just the first term in the expansion of $a$ in powers of $1/n$. You could go on and find the other terms as well, if needed. May 29, 2022 at 6:31
• Thanks, I get it now. By the way, if we change the n-gon to a circle with circumference n, then $\lim_{n\to\infty}(\text{Area of circle}-\text{Area enclosed by }S)=\dfrac{2}{\pi}$. I proved this using the method in your answer.