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I would like to calculate x number of points around a circle using the circle's radius/diameter.

For example, if I wanted 24 equidistant points around a circle with radius 30, how could I go about that?

The circle's origin is (0, 0), radius 30 and I need to calculate 24 (X, Y) equidistant points around it.

Example with 17 points: http://i.imgur.com/Jibwr.png

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    Are you familiar with complex numbers?2012-11-02

2 Answers 2

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If you start from the right wing of the $x$ axis, your points of the regular $n$-gon are $P_k=(r\cos\left(\frac{k\cdot 2\pi}{n}\right), r\sin\left(\frac{k\cdot 2\pi}{n}\right))$ where $k=0,..,n-1$.

If the first angle, measured from the right wing of $x$ axis, counterclockwise, is not $0$ but some $\alpha_0$, then $P_k=(r\cos\left(\frac{k\cdot 2\pi}{n}+\alpha_0\right), r\sin\left(\frac{k\cdot 2\pi}{n}+\alpha_0\right))$

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    Thanks @FrenzYDT.! That clears a lot up.2012-11-02
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First of all, observe that multiplication of coordinates preserves equality of distances: if points $a$ and $b$ are equidistant, then so are $\alpha a$ and $\alpha b$ for any number $\alpha$. Thus, we can reduce the problem to the case when the radius of the circle is equal to one.

Now, consider a function $f(\cos \varphi, \sin \varphi) = (\cos(n\varphi), \sin(n\varphi))$. As you can see, it wraps the circle around itself $n$ times, and $\left|\frac{\mathrm{d}}{\mathrm{d}\varphi}f\right| = \left|(-n\sin(n\varphi), n\cos(n\varphi))\right| = \sqrt{n^2\sin(n\varphi)^2 + n^2\cos(n\varphi)^2} = n \sqrt{\sin(n\varphi)^2 + \cos(n\varphi)^2} = n,$ so as $f$ wraps a circle around itself, it does so with constant velocity, so each 'lap' has the same length. Thus, for example, all points that $f$ maps to $(\cos \varphi_0, \sin \varphi_0)$ must be equidistant, and there are exactly $n$ of them.

So now all that's left for you to do is to solve the equation $f(\cos \varphi, \sin \varphi) = (\cos \varphi_0, \sin \varphi_0)$, and you'll arrive to Berci's answer.

One last thing, if you're familiar with complex numbers, $f(z) = z^n$, so any set of $n$ equidistant points on a unit circle is a set of $n$-th roots of some number of length one.

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    No problem :) ${}$2012-11-02