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If I want to cut a circle into $n$ equal parts,

For example: enter image description here

In this case ($n=3$), $x_1≈-0.264$ and $x_2≈0.264$, where $x_1$ and $x_2$ are values that sactisfy $\int_{-1}^{x_1}\sqrt{1-x^2}\mathrm{d}x=\int_{x_1}^{x_2}\sqrt{1-x^2}\mathrm{d}x=\int_{x_2}^{1}\sqrt{1-x^2}\mathrm{d}x$

And my problem is to find a generalised formula for $x_1,x_2,x_3...x_{n-1}$ when I cut the circle into $n$ equal parts.

The furthest I have gone is that $x_1$ is a value that sactisfy $\arcsin{(x_1)}+x_1\sqrt{1-x_1^2}=\frac{\pi}{n}-\frac{\pi}{2}$ which I couldn't solve for $x_1$

KReiser
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YesSpoon3
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    Do you have a reason to think a formula for the answer exists? – aschepler Aug 08 '23 at 00:22
  • If you have a formula for $x_1,$ then you have determined the area $\pi/n$ and the angle $\alpha$ at the center of the segment. This is a function of the angle $\alpha$ and sine of the angle $\sin(\alpha)$. I do not see how you can solve this in a closed algebraic expression. It is probably out of reach for compass and straightedge, and likely also for more general algebraic expressions. – Marius S.L. Aug 08 '23 at 00:45
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    There is a calculator for doing this at https://planetcalc.com/8943/ and they also give the equations (but they note that the equations can't be solved algebraically, only numerically). E.g., for a circle of radius $100$ cut into seven equal pieces, the widths of the pieces, to two decimal places, are $40.09,25.56,23.10,22.49,23.10,25.56,40.09$. – Gerry Myerson Aug 08 '23 at 02:25
  • Your equation is a Kepler equation as if $x=\sin(w)$, then the equation to solve is $w+\sin(w)\cos(w)=\frac\pi n+\frac \pi 2$. There is a Mathematica closed form and linked integral/series representations in this answer – Тyma Gaidash Aug 13 '23 at 01:19
  • Are you talking about the circle (so equal arclengths) or the disk (so equal areas)? It is not clear without reading on. This is a typical confusion with the ambiguity of the word circle. – Ted Shifrin Aug 13 '23 at 01:33

1 Answers1

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For $n$ equal parts, the values of positive $x_k$'s are the solutions of the equation you wrote

$$\arcsin{(x_k)}+x_k\sqrt{1-x_k^2}=\frac{\pi}{2}-k\frac{\pi}{n}$$

I shall use $x=x_k$ and $A=\frac{\pi}{2}$$-k\frac{\pi}{n}$

A quite good approximation of the lhs is given by almost the simplest Padé approximant of it built around $x=0$, that is to say

$$\arcsin{(x)}+x\sqrt{1-x^2} \sim \frac{2 x \left(19 x^2-60\right)}{9 x^2-60}$$ whose error is $\frac{29 }{2800}x^7$.

To give an idea, consider the norm

$$\Phi=\int_{-1}^{+1} \Bigg(\arcsin{(x)}+x\sqrt{1-x^2}- \frac{2 x \left(19 x^2-60\right)}{9 x^2-60} \Bigg)^2\,dx=1.10061\times 10^{-4}$$

So, an approximation is given by the solution of the cubic equation $$38 x^3-9 A x^2-120 x+60A=0$$ which has three real roots.

Using the trigonometric solution, the solution to be retained (positive and smaller than $1$) is given by

$$x=\frac{1}{38} \left(3 A-2 \sqrt{9 A^2+1520} \cos \left(\frac{1}{3} \left(\cos ^{-1}\left(\frac{3 A \left(9 A^2-12160\right)}{\left(9 A^2+1520\right)^{3/2}}\right)+\pi \right)\right)\right) $$

For example, for $n=7$, the results would be

$$\left( \begin{array}{ccc} \text{approximation} & \text{solution} \\ \pm 0.112437 & \pm 0.112437 \\ \pm 0.343474 & \pm 0.343478 \\ \pm 0.598821 & \pm 0.599061 \\ \end{array} \right)$$

Another solution would be possible : expand the lhs as a series

$$\arcsin{(x)}+x\sqrt{1-x^2} =\sum_{m=0}^\infty a_m\, x^{2m+1}$$

Truncate to some order and use power series reversion to obtain $$x=\sum_{m=0}^\infty b_m\, A^{2m+1}$$ where the first coefficients are

$$\left\{\frac{1}{2},\frac{1}{48},\frac{13}{3840},\frac{493}{645120 },\frac{37369}{185794560},\frac{4732249}{81749606400},\frac{90 1188997}{51011754393600}\right\}$$

Claude Leibovici
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