How to equally divide a circle with parallel lines? How can I "draw" $n$ parallel lines in such a way that they will divide a circle (disc) in $n+1$ equal areas ?
 A: Even the case $n=2$ is hard, according to this link. 
A: Building on Phira's answer: you have to solve the equation
$$\varphi - \sin\varphi = \frac{2\pi k}{n+1}$$
This has to be solved numerically. You can use Newton's method for this. Let $a = \frac{2\pi k}{n+1}$. Then you want to solve $f(x) = 0$, where $f(x) = x - \sin x - a$. So choose an initial approximation $x_0$ (for instance, $x_0 = a$), and then iterate the equation
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n - \sin x_n - a}{1 - \cos x_n}$$
as many times as you need to get the desired accuracy.  
A: The area of the segment is $\frac{r^2}2 (\varphi - \sin \varphi)$ where $\varphi$ is the angle under which one views the chord of the segment from the center.
The distance from the center is $r \cos \frac{\varphi} 2$.
So, you have to solve the equations $\frac{1}2 (\varphi - \sin \varphi)=\frac {\pi k} {n+1}$ and then plug the values of $\varphi$ into the above equation.
A: Take a circle of radius $r$ centered at $(2r;0)$ with its equation being $(x-r)^2+y²=r^2$.
Try to find the cartesian function of the top half circle: $y=\sqrt{r^2-(x-r)^2}$ or $y=\sqrt{-x^2 + 2 \cdot x \cdot r}$
Finding the n-th of a circle or of a half-circle is basically the same.
The area of the half-circle is $1/2 \cdot \pi \cdot r^2$. You want to know where to cut it to get areas of $\frac{1/2 \cdot \pi \cdot r^2}{n}$
That's quite simple calculus:
$$\int_{0}^{v_i} \sqrt{-x^2 + 2 \cdot x \cdot r} \mathrm{d} x = 1/2 \cdot \pi \cdot r^2 \cdot \frac{i}{n} \text{with } 0 < i \leq n $$
find $v_i$, the vertical cuts.
A: For any circle surely the distance between the lines drawn is a constant fraction of the diameter of the circle, depending on the divisions required. Is this too simple?
