Parallel lines divide a circle's area into thirds 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).
 A: If the angle at the origin formed by the segments to the two ends of the chord on the right is $\theta$, then the area of the right-hand segment is
$$\frac{R^2}{2}(\theta-\sin\theta)$$
(see Wikipedia, for example). Setting this equal to $\frac{1}{3}\pi R^2$ and solving gives $\theta\approx 2.6$. This means that the angle formed by the upper segment and the $x$-axis is $\frac{\theta}{2}\approx 1.3$, so that $\frac{d}{2} = R\cos 1.3\approx 0.2675R$. From the equations, it seems likely that there is no closed form in elementary functions.
A: The area $A_1$ of the first quadrant of the inner third is
\begin{align}
A_1 = \frac{1}{12} \pi r^2
&= \int\limits_0^{d/2} y(x) \, dx \\
&= \int\limits_0^{d/2} \sqrt{r^2 - x^2} \, dx \quad (*) \\
&= r \int\limits_0^{d/2} \sqrt{1 - (x/r)^2} \, dx \\
&= r^2 \int\limits_0^{d/(2r)} \sqrt{1- u^2} \, du \\
&= \frac{r^2}{2} \left[u \sqrt{1-u^2} + \arcsin u \right]_{u=0}^{u=d/(2r)} \\
&= 
\frac{r^2}{2}\left(
\frac{d}{2r}\sqrt{1-\left(\frac{d}{2r}\right)^2} + 
\arcsin \left(\frac{d}{2r} \right)
\right) \\
\end{align}
Note: My solution deviates from yours at $(*)$. Because $x^2 + y^2 = r^2 \Rightarrow y = \sqrt{r^2 - x^2}$.
This gives the equation
$$
\sin\left(
\frac{\pi}{6} - \frac{d}{2r} \sqrt{1 - \left(\frac{d}{2r}\right)^2}
\right) = \frac{d}{2r}
$$
IMHO this equation in the unknown $d$ is not solvable using elementary functions.
Numerical Solution (Root finding):
With $z = d/(2r)$ we get
$$
F(z) := 
\sin\left(
\frac{\pi}{6} - z \sqrt{1-z^2}
\right) - z = 0
$$
and can apply the Newton method or some other solver to find a root.
Maxima gives:
$$
z \approx 0.2649320846027768
$$
For $r = 1$ this means $d = 0.5298641692055537$.

Testing:
Numerical integration of $(*)$ for $r = 1$  gives
$$
A_1 = 0.2617993877991494
$$
on the other hand
$$
\frac{\pi}{12} = 0.2617993877991494
$$
Numerical Solution (Fixed Point):
Another way to pose the problem is as a fixed point equation:
$$
f(z) := 
\sin\left(
\frac{\pi}{6} - z \sqrt{1-z^2}
\right) = z
$$
This you can solve already roughly with a function plot program like Gnuplot py proper panning and zooming:

The image shows where $y = x$ and $f(x)$ cross, the $x$ coordinate is the fixed point. I added the root version $F(x)$ as which has the root there.
Otherwise one can fixed point iteration (depending if the fixed point is attractive or not one might need to do this with a transformed $g$).
Newton iteration convergers much faster, meaning needs less iterations to gets digits of the result. 
