Minimum number of circular segments.

Question

Let K be any natural number. Consider the unit square, and the circle of diameter 1 inside of the square. We then consider circular segments of area $\frac{1}{2K}$ and claim that there exists a constant $c$ such that $\#$(mutually disjoint circular segments of area $\frac{1}{2K}$)$\geq cK^{\frac{1}{3}}$.
Why is this correct? I tried explaining to myself by geometry but not with the best of results. Any hint would be appreciated.

Answer 1

Forget about the unit square, it's irrelevant.

Let K be a positive real. The maximum area of disjoint segments arises if the segments bound a polygon inscribed in the circle, and since the areas of a set of segments for a given K are equal that polygon is regular.

Let $n$ be the number of segments. Let $\theta$ be the angle subtended at the centre of the circle by each segment, so $\theta = 2\pi/n$.

The area of a segment is $\pi r^2/n - \frac{1}{2}r^2 \sin\theta$; $r = \frac{1}{2}$, so

$\frac{1}{2K} = \pi /4n - \sin\theta / 8$
$\frac{4}{K} = 2\pi /n - \sin\theta$
$\frac{4}{K} = \theta - \sin\theta$

Now $\sin \theta = \theta - \theta^3/3! + \theta^5/5! - ...$

For small $\theta$ (large $n$) we can ignore the 3rd & subsequent terms. So

$$\begin{align} \frac{4}{K} & \approx \theta^3/6\\ \frac{24}{K} & \approx \theta^3\\ \left(\frac{24}{K}\right)^\frac{1}{3} & \approx \theta = 2\pi/n\\ \left(\frac{K}{24}\right)^\frac{1}{3} & \approx n / 2\pi\\ n & \approx 2\pi\left(\frac{K}{24}\right)^\frac{1}{3}\\ n & \approx \left(\frac{2\pi}{24^\frac{1}{3}}\right) K^\frac{1}{3} \end{align}$$

So $n$ is proportional to $K^\frac{1}{3}$ for sufficiently large $n$. I'll let you clean up the fine details relating to bounds. :)