# Cutting a circle

Two ways to cut a circle into equal parts : sector cuts and parallel cuts.

Below you can find two calculators which calculate how to cut a circle into equal parts - traditional and non-traditional way. By the traditional way, I assume cutting a circle into sectors, just like you usually cut a pie or pizza. And by the non-traditional way, I assume cutting a circle into equal vertical slices with parallel lines or with parallel chords, if you like. Both calculators present a drawing that illustrates the result. And you can find all formulas and math in the article below the calculators.

#### Cutting a Circle into equal sectors

Digits after the decimal point: 2
Angle of a Sector

Length of an Arc of a Sector

Length of a Chord of a Sector

#### Cutting a Circle into equal parts with Parallel Cuts

Digits after the decimal point: 2
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### Cutting a Circle into Sectors

Ok, you need to cut a circle into several sectors (even non-even numbers). To do this, you need to find the parameters of a sector. It is a straightforward task:

1. Find the angle of a sector in radians by dividing 2π (representing 360 degrees in radians) by a number of sectors.

$\alpha=\frac{2\pi}{N}$

1. Find the length of an arc of a sector by multiplying a radius by an angle of a sector in radians.

$a=\alpha R$

1. Find the length of a chord of a sector by using Law of cosines (a chord is the base of the isosceles triangle, with two radiuses as legs and sector angle as apex angle).

$c=R^2+R^2-2RR \cos \alpha$

This completely defines all N equal sectors.

### Cutting a Circle with Parallel Cuts

This way is more interesting. For simplicity, I will consider half of a circle since it is symmetrical.

Let's slice it with vertical slices. In this case, we need to find the x-coordinates of parallel chords, which should split our circle into equal-area parts. (see points x1 and x2 on the picture above). Let's derive the general formula for an area of a left slice.

Our half-circle can be thought of as a function y=f(x), where x - is the coordinate along the abscissa axis, and y is the function equal to the value of the corresponding half-circle point.

Using the Pythagorean theorem, the y function is

$y=\sqrt{R^2 - x^2}}$

To find an area of a left slice, you need to integrate this function from -R to x. The antiderivative of our function is :

$F(x)=\frac{1}{2} (x\sqrt{R^2-x^2} + R^2 \arctan(\frac{x}{\sqrt{R^2-x^2}}))+C$

We need to find the value of constant. Obviously, at the point where x equals -R area should be zero. If we plug -R instead of x into the formula above, we get

$F(-R)=-\frac{\pi R^2}{4}+C=0$, hence

$C=\frac{\pi R^2}{4}$

Our final integral is

$F(x)=\frac{1}{2} (x\sqrt{R^2-x^2} + R^2 \arctan(\frac{x}{\sqrt{R^2-x^2}}))+\frac{\pi R^2}{4}$

Now how do we find x of the first cut? We know we area we should get - Nth part of the total area (note the half-circle)

$S=\frac{\frac{\pi R^2}{2}}{N}=\frac{\pi R^2}{2N}$

Thus we can equate

$\frac{1}{2} (x\sqrt{R^2-x^2} + R^2 \arctan(\frac{x}{\sqrt{R^2-x^2}}))+\frac{\pi R^2}{4}=\frac{\pi R^2}{2N}$

Which gives us

$\frac{1}{2} (x\sqrt{R^2-x^2} + R^2 \arctan(\frac{x}{\sqrt{R^2-x^2}}))+\frac{\pi R^2}{4}-\frac{\pi R^2}{2N}=0$

This is a transcendental equation, and we need to use numerical methods to solve it, for example, Bisection method or Newton's method. Here I used Newton's method.

The next points of cut can be found with the same approach. We need to cut two times more for second point $S_2=2\frac{\pi R^2}{2N}$, three times more for third point $S_3=3\frac{\pi R^2}{2N}$ and so on.

Then we can find all other parameters, like chord length, using the point coordinates.

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