Two congruent circles of area 18 intersect; the region of their intersection has area 5. What is the angle $\theta$ of the central angle between the centers of the circles and the points of intersection of the circles?

The radius of the circles is $\sqrt{18/\pi}$. $O$ is the center of one of the circles, and $P$ and $Q$ are the points of intersection. $\mathrm{m}\angle\mathit{POQ} = \theta$. According to the Law of Cosines, \begin{equation*} \left\vert \overline{\mathit{PQ}} \right\vert^{2} = 2\left(\frac{18}{\pi}\right)\Bigl(1 - \cos\theta\Bigr) . \end{equation*} The altitude of $\triangle\mathit{OPQ}$ from $O$ is \begin{equation*} \sqrt{\frac{18}{\pi}} \sin\left(\frac{\pi}{2} - \frac{\theta}{2}\right) = \sqrt{\frac{18}{\pi}} \cos\left(\frac{\theta}{2}\right) . \end{equation*} So, \begin{equation*} \frac{\theta}{2\pi}\Bigl(18\Bigr) - \frac{1}{2} \left(2\left(\frac{18}{\pi}\right)\Bigl(1 - \cos\theta\Bigr)\right) \left(\sqrt{\frac{18}{\pi}} \cos\left( \frac{\theta}{2}\right)\right) = \frac{5}{2} . \end{equation*} What is the solution - or an approximation of a solution - to this trigonometric equation?


It easier to do it this way: if radius of circle is $r$, the area of sector is $0.5r^2 \theta$.

$A_{\triangle{OPQ}}=0.5r^2 \sin \theta$.

Area of intersection $A=r^2 \theta-r^2 \sin \theta=5 \rightarrow \theta-\sin \theta=\frac{5}{r^2}=\frac{5\pi}{18}$. The solution $\theta \approx 1.837$

  • $\begingroup$ Sorry but the area of a circular sector is $r^2 \theta/2$. $\endgroup$ – Jean Marie May 9 '19 at 17:36
  • $\begingroup$ @JeanMarie: But of course, thanks for correcting me! $\endgroup$ – Vasya May 9 '19 at 17:41
  • $\begingroup$ Thus we agree finally... $\endgroup$ – Jean Marie May 9 '19 at 17:42

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Fig. 1 : Squares have unit area. We are looking for the value of $\theta$ = angle(QOP).

The area in question is the sum of the areas of 2 so-called "lunules", (one of them being delimited by points $P,I,Q$) whose formula


is classical (https://web.calstatela.edu/faculty/hmendel/Ancient%20Mathematics/HippocratesOfChios/Lunules/IntroductionToLunules.htm).

Thus, as $R^2=\frac{18}{\pi}$, the equation to be solved is :


or $$\theta-\sin\theta=\dfrac{5\pi}{18}$$

or $$\theta=\underbrace{\sin\theta+\dfrac{5\pi}{18}}_{f(\theta)}$$

Applying fixed point iterative method ($\theta_{n+1}=f(\theta_n)$ with $\theta_0=\pi/2$), one obtains :

$$\theta_1=1.82744..., \theta_2=1.83991..., \theta_3=1.83667...,$$

converging to $\Theta= 1.837349240792971...$ (radians), i.e. approximately 105°.

Some comments about the method (see the theory in https://en.wikipedia.org/wiki/Fixed-point_iteration) :

The convergence factor is the absolute value of $f'(\Theta)=\cos(\Theta)\approx -0.26$ ; it means that the absolute value of the error $e_n=\Theta-\theta_n$ is divided by approximately four at each step.

Furthermore, as $f'(\Theta)<0$, the convergence is alternate.

  • $\begingroup$ I appreciate the reference to the solution via fixed points. $\endgroup$ – A gal named Desire May 9 '19 at 17:38
  • 1
    $\begingroup$ I have added how I have obtained the initial equation (lunules have been studied at least for 2500 years...). $\endgroup$ – Jean Marie May 9 '19 at 18:01

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