The picture probably explains my question best.I need to find a way to divide a circle into 3 parts of equal area with only 2 lines that intersect each other on the outline of the circle.Also I need to check, if whatever diameter is between those lines, also splits circles with a different diameter into equal parts.And lastly, and probably the most difficult question: How do I have to calculate the angle between x lines that all intersect in one point, so that the circle is split into x+1 parts with area = 1/(x+1) of the circle?I tried my best, but couldn"t even find a single answer or the right strategy to tackle the question...


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edited Mar 18 at 20:50

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Given the angle $\theta$, split by the diameter containing $B$, consider the following diagram:


$\overline{BO}$ is the line through the center and $\overline{BA}$ is the chord cutting off the lune whose area we wish to compute.

The area of the circular wedge subtended by $\angle BOA$ is$$\frac{\pi-\theta}2r^2\tag1$$The area of $\triangle BOA$ is$$\frac12\cdot\overbrace{r\sin\left(\frac\theta2\right)}^\text{altitude}\cdot\overbrace{2r\cos\left(\frac\theta2\right)}^\text{base}=\frac{\sin(\theta)}2r^2\tag2$$Therefore, the area of the lune is $(1)$ minus $(2)$:$$\frac{\pi-\theta-\sin(\theta)}2r^2\tag3$$To get the area divided into thirds, we want$$\frac{\pi-\theta-\sin(\theta)}2r^2=\frac\pi3r^2\tag4$$which means we want to solve$$\theta+\sin(\theta)=\frac\pi3\tag5$$whose solution can be achieved numerically (e.g. use $M=\frac\pi3$ and $\varepsilon=-1$ in this answer)$$\theta=0.5362669789888906\tag6$$Giving us


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