TITLE: Parallel lines divide a circle's area into thirds
QUESTION [9 upvotes]: When I was young I came up with a geometry problem and drew it in a notebook:
Suppose we have a circle with radius $r$ and area $A$. Let two parallel lines be equidistant from the center of the circle and divide the circle's area into thirds. What is the distance $d$ between these two lines?

(Note: the following work contains the laughable mistake of solving $\int \sqrt{x^2-r^2} \; dx$ instead of $\int \sqrt{r^2-x^2} \; dx$. I'll go ahead and leave my work anyway:)
Later on, in high school, I found the notebook again and used my new calculus  tools to approach the problem, which I recorded on the next few pages of that notebook. I recognized that
$$\frac{1}{12}A=\frac{1}{12}(\pi r^2)=\int_0^{d/2} \sqrt{x^2-r^2}\; dx$$
$$=\frac{x}{2}\sqrt{x^2-r^2}-\frac{r^2}{2}\ln \left|x+\sqrt{x^2-r^2}\right|\biggl|_0^{d/2}$$
$$= \frac{d}{4}\sqrt{\frac{1}{4}d^2-r^2} -\frac{r^2}{2}\left[ \ln\left|{\frac{d}{2}+\sqrt{\frac{1}{4}d^2-r^2}} \right| -\ln\left|\sqrt{-r^2} \right| \right] $$
But we know that $$\frac{1}{4}d^2-r^2<0$$The final equation I wrote down was
$$r^2= \frac{3d}{\pi}i\sqrt{r^2-\frac{1}{4}d^2} -\frac{6r^2}{\pi}\left[ \ln\left|{\frac{d}{2}+i\sqrt{r^2-\frac{1}{4}d^2}} \right| -\ln\left|ir\right| \right] $$
and this is where I probably slammed the notebook shut in frustration.

I'm more mathematically mature now (college student) and want to finally get an answer to this problem. There are probably a few different ways to approach this problem. I found a version of this problem on MSE here, but mine is the particular case $n=2$. Can anybody help me put this decade-old problem to rest? I mainly posted this because I want to know if there is some elegant solution out there to this seemingly simple problem (e.g. solution without numerical methods).

REPLY [4 votes]: Setting the problem as rogerl did, you have to solve for $\theta$ $$f(\theta)=\theta-\sin(\theta)-\frac{2\pi}3=0$$ By inspection $f(\frac{2\pi}3)=-\frac{\sqrt{3}}{2}<0$ and $f(\pi)=\frac{\pi }{3}>0$. So, use $\theta_0=\frac 12 \Big(\frac{2\pi}3+\pi \Big)=\frac{5\pi}6$ and perform one iteration of Newton method. This will give $$\theta_1=2-\sqrt{3}+\left(\frac{1}{6}+\frac{1}{\sqrt{3}}\right) \pi\approx 2.60535$$ while the solution is $\approx 2.60533$.
If instead of Newton, you use Halley method, the first iterate would be $$\theta_1=\frac{9 \left(13+8 \sqrt{3}\right)+\left(354+201 \sqrt{3}-\pi \right) \pi }{18
   \left(2+\sqrt{3}\right)^3}\approx 2.60533$$
Another approach could be to expand as a Taylor series the function around $\frac{5\pi}6$. This will give $$f(\theta)=\frac{1}{6} (\pi -3)+\left(1+\frac{\sqrt{3}}{2}\right) \left(\theta-\frac{5 \pi
   }{6}\right)+\frac{1}{4} \left(\theta-\frac{5 \pi }{6}\right)^2+O\left(\left(\theta-\frac{5
   \pi }{6}\right)^3\right)$$ and solving the quadratic $$\theta=\frac{1}{6} \left(-12-6 \sqrt{3}+2 \sqrt{3 \left(9+12 \sqrt{3}+\frac{31104-3456 \pi
   }{1728}\right)}+5 \pi \right)\approx 2.60533$$
Another approach could be to build the simplest Pade approximant at $\theta=\frac{5\pi}6$. This would give $$sin(\theta)=\frac{\frac{1}{2}-\frac{7 \left(x-\frac{5 \pi }{6}\right)}{4
   \sqrt{3}}}{1-\frac{(x-\frac{5 \pi }{6})}{2 \sqrt{3}}}$$ and then the problem reduces to a quadratic equation, the solution of which being $$\theta=\frac{7}{4}+\sqrt{3}+\frac{3 \pi }{4}-\frac{1}{12} \sqrt{9 \left(97+40
   \sqrt{3}\right)+\pi  \left(24 \sqrt{3}+\pi-42 \right)}\approx 2.60533$$
Edit
Concerning the precision after one single iteration, let me give the values as a function of the order of the iterative method $$\theta_1^{(2)}=\color{red}{2.6053}4733226364$$ $$\theta_1^{(3)}=\color{red}{2.605325}90502067$$ $$\theta_1^{(4)}=\color{red}{2.60532567}596091$$  $$\theta_1^{(5)}=\color{red}{2.6053256746}1355$$ $$\theta_1^{(6)}=\color{red}{2.60532567460}101$$ $$\cdots$$ $$\theta_1^{(\infty)}=\color{red}{2.60532567460090}$$