How can I "draw" $n$ parallel lines in such a way that they will divide a circle (disc) in $n+1$ equal areas ?
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What is the problem with doing it? Do you know about Cartesian coordinates $Oxy$ and trigonometric functions? – SBF Oct 28 '11 at 09:02
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Building on Phira's answer: you have to solve the equation $$\varphi - \sin\varphi = \frac{2\pi k}{n+1}$$ This has to be solved numerically. You can use Newton's method for this. Let $a = \frac{2\pi k}{n+1}$. Then you want to solve $f(x) = 0$, where $f(x) = x - \sin x - a$. So choose an initial approximation $x_0$ (for instance, $x_0 = a$), and then iterate the equation $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n - \sin x_n - a}{1 - \cos x_n}$$ as many times as you need to get the desired accuracy.
TonyK
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Looks awfully like the Kepler equation, that one. A few methods for numerically solving it were discussed here. – J. M. ain't a mathematician Oct 28 '11 at 10:50
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The area of the segment is $\frac{r^2}2 (\varphi - \sin \varphi)$ where $\varphi$ is the angle under which one views the chord of the segment from the center.
The distance from the center is $r \cos \frac{\varphi} 2$.
So, you have to solve the equations $\frac{1}2 (\varphi - \sin \varphi)=\frac {\pi k} {n+1}$ and then plug the values of $\varphi$ into the above equation.
Phira
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4@pedja: It is common usage to mean radians when one just writes $\varphi$ for an angle. – Phira Oct 28 '11 at 10:25
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1also I think that you should write $\varphi_k$ instead of $\varphi$ in final formulae – Pedja Oct 28 '11 at 10:29
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Take a circle of radius $r$ centered at $(2r;0)$ with its equation being $(x-r)^2+y²=r^2$.
Try to find the cartesian function of the top half circle: $y=\sqrt{r^2-(x-r)^2}$ or $y=\sqrt{-x^2 + 2 \cdot x \cdot r}$
Finding the n-th of a circle or of a half-circle is basically the same.
The area of the half-circle is $1/2 \cdot \pi \cdot r^2$. You want to know where to cut it to get areas of $\frac{1/2 \cdot \pi \cdot r^2}{n}$
That's quite simple calculus:
$$\int_{0}^{v_i} \sqrt{-x^2 + 2 \cdot x \cdot r} \mathrm{d} x = 1/2 \cdot \pi \cdot r^2 \cdot \frac{i}{n} \text{with } 0 < i \leq n $$
find $v_i$, the vertical cuts.
SteeveDroz
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1Umm...yes, OK (maybe) -- but how do you find $v_i$? You did say it was "quite simple". – TonyK Oct 28 '11 at 16:23
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After integrating: $r^2 \cdot \arctan(\frac{\sqrt{v_i}}{\sqrt{2 \cdot r - v_i}}) - \frac{\sqrt{v_i} \cdot (r - v_i) \cdot \sqrt{2 \cdot r - v_i}}{2} = 1/2 \cdot \pi \cdot r^2 \cdot \frac{i}{n}$. (Of course, it may be easier to read with proper values for $r$, $i$ and $n$). Then, isolating $v_i$ is algebra I won't do it here. – SteeveDroz Oct 31 '11 at 11:37
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It's not algebra. The equation has to be solved numerically. But I'm glad to see you deleted "quite simple"! – TonyK Nov 02 '11 at 13:06
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Hi @SteeveDroz, I refer to your solution to this problem awhile back. Instead of taking the integral from 0 to v(i), shouldn't the integral be taken from r to v(i) because your circle is of radius r centered at (2r, 0)? – Leockl Feb 29 '20 at 07:16
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For any circle surely the distance between the lines drawn is a constant fraction of the diameter of the circle, depending on the divisions required. Is this too simple?
C W
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1Why would it be a constant fraction? If you have evenly spaced lines then the area of the circle sectors are pretty clearly not equal – MT_ Apr 27 '16 at 06:24