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).
$\endgroup$ 53 Answers
$\begingroup$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.
$\endgroup$ 6 $\begingroup$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.
$\endgroup$ 10 $\begingroup$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 $$\theta_1^{(2)}=\color{red}{2.6053}4733226364$$ $$\theta_1^{(3)}=\color{red}{2.605325}90502067$$ $$\theta_1^{(4)}=\color{red}{2.60532567}596091$$ $$\theta_1^{(5)}=\color{red}{2.6053256746}1355$$ $$\theta_1^{(6)}=\color{red}{2.60532567460}101$$ $$\cdots$$ $$\theta_1^{(\infty)}=\color{red}{2.60532567460090}$$
$\endgroup$ 5
