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).

  • $\begingroup$ The link you provided has answers which focus on your case, it seems it doesn't have a closed form $\endgroup$ – HBeel Mar 28 '15 at 20:46
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    $\begingroup$ Nah, you dared to put up your stuff for discussion. That is great! $\endgroup$ – mvw Mar 28 '15 at 22:10
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    $\begingroup$ I think not so very many high-school students (or younger) make mathematical scribblings like this for their own amusement. Seems like perfect grist for the math.SE mill. $\endgroup$ – David K Mar 29 '15 at 14:29
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    $\begingroup$ @DavidK: Interestingly, early mathematical scribblings are what sparked my love for the subject. I learned about the golden ratio about the same time that I took my first algebra course in middle school, and I had simply memorized its value to a few digits. One day, on an airplane, my father grabbed some napkins and started drawing some rectangles and, to my astonishment, proved for me that $\varphi=(1+\sqrt{5})/2$ by using the quadratic equation, which was something I had just learned in class. I vividly remember the pure awe and excitement I felt for mathematics on that day... $\endgroup$ – Patrick Shambayati Mar 30 '15 at 1:04

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.

  • $\begingroup$ I would be interested in seeing how numerical methods are used to solve that equation. How would you approach that equation without computer software? $\endgroup$ – Patrick Shambayati Mar 28 '15 at 22:04
  • $\begingroup$ Probably Newton's method. I used Mathematica... $\endgroup$ – rogerl Mar 28 '15 at 22:05
  • $\begingroup$ If it is a segment of a sphere instead of a circle, it results in solution of a cubic, iirc a result from IS&ES Sokolnikoffs ' book. $\endgroup$ – Narasimham Mar 28 '15 at 23:33
  • $\begingroup$ The two answers posted so far are great because they use such different methods of approaching the same value $\endgroup$ – Patrick Shambayati Mar 28 '15 at 23:33
  • $\begingroup$ I'm tempted to post a new question asking for the distance between two disjoint planes that separate a sphere's volume into thirds.... $\endgroup$ – Patrick Shambayati Mar 28 '15 at 23:34

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 $d = 0.5298641692055537$.

Graphic for r = 1


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:

fixed point and root version for r = 1

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.

  • $\begingroup$ Hah! Re-reading my work, I solved $\int \sqrt{x^2-r^2} dx$ instead of $\int \sqrt{r^2-x^2} dx$... Didn't realize that mistake as I was copying. $\endgroup$ – Patrick Shambayati Mar 28 '15 at 21:26
  • $\begingroup$ Regardless of that mistake, can we pursue your work further? How do we apply Newton method? $\endgroup$ – Patrick Shambayati Mar 28 '15 at 21:27
  • $\begingroup$ Something seems wrong with $z=0.789$. The ratio $d/2r$ is the ratio of $d$ to the diameter, and it should be less than $1/3$. $\endgroup$ – Patrick Shambayati Mar 28 '15 at 21:35
  • $\begingroup$ That seems to work now. $\endgroup$ – mvw Mar 28 '15 at 22:00
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    $\begingroup$ Just looking does not give the impression it is a third. Even counting the squares is not too precise. Maybe printing it out and using scissors? :-) $\endgroup$ – mvw Mar 28 '15 at 23:15

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 $$\theta=\frac{7}{4}+\sqrt{3}+\frac{3 \pi }{4}-\frac{1}{12} \sqrt{9 \left(97+40 \sqrt{3}\right)+\pi \left(24 \sqrt{3}+\pi-42 \right)}\approx 2.60533$$


Concerning the precision after one single iteration, let me give the values as a function of the order of the iterative method $$\theta_1^{(2)}=\color{red}{2.6053}4733226364$$ $$\theta_1^{(3)}=\color{red}{2.605325}90502067$$ $$\theta_1^{(4)}=\color{red}{2.60532567}596091$$ $$\theta_1^{(5)}=\color{red}{2.6053256746}1355$$ $$\theta_1^{(6)}=\color{red}{2.60532567460}101$$ $$\cdots$$ $$\theta_1^{(\infty)}=\color{red}{2.60532567460090}$$

  • $\begingroup$ Wow! Such precision after just one iteration. $\endgroup$ – Patrick Shambayati Mar 30 '15 at 1:31
  • $\begingroup$ I wonder which is more difficult to compute: two Newton iterations or one Halley iteration $\endgroup$ – Patrick Shambayati Mar 30 '15 at 10:31
  • $\begingroup$ It is a good point. If you do the calculations by hand, Halley or higher order will be much more simple since we know the exact values of the sine and cosine. If you use a computer, it does not matter. $\endgroup$ – Claude Leibovici Mar 30 '15 at 10:39
  • $\begingroup$ I hope you see that there plenty of ways to solve the problem ! Cheers. $\endgroup$ – Claude Leibovici Mar 30 '15 at 10:42
  • $\begingroup$ You are very welcome. I thank you for the problem since I had a lot of fun playing with it ! Cheers :-) $\endgroup$ – Claude Leibovici Mar 30 '15 at 10:52

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