# Diameter of circle with n points where adjacent points are m distance apart

How do I calculate the diameter of a circle that has n evenly-spaced points on its circumference where adjacent points are m distance apart?

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It seems like this should be trivial, but I must not be using the right terms when I google for it. –  antony.trupe Nov 1 '12 at 2:06

If you have $n$ points uniformly distributed on a circle then the angle between the points is $\frac{2\pi}{n}$. If the distance between adjacent points is $m$ then the chord of the sector with angle $\frac{2\pi}{n}$ has length $m$.

Trigonometry then yields $$r = \frac{m\sin\left(\frac{(n-2)\pi}{2n}\right)}{\sin\left(\frac{2\pi}{n}\right)}$$

Alternatively, this formula will probably be simpler $$r = \frac{m}{2\sin\left(\frac{\pi}{n}\right)}$$

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beautiful. I probably could have gotten there in a few days, and maybe gotten a little smarter in the process, but I would much rather be writing logic, so thanks! –  antony.trupe Nov 1 '12 at 2:15

The circle is divided into $n$ wedges with angle $$\theta = \frac{2 \pi}{n}.$$ Consider the triangle formed by the center of the circle (point $A$), one of the two adjacent points (point $B$), and the midpoint of the line joining the two adjacent points (point $C$). Now, $\angle BAC = \theta / 2 = \pi / n$, $AB = d/2$, and $BC = m / 2$, where $d$ is the diameter of the circle, so $$AB \sin \left(\angle BAC\right) = BC \Rightarrow \frac{d}{2} \sin\left(\frac{\pi}{n}\right) = \frac{m}{2} \Rightarrow d = \frac{m}{\sin \left(\pi / n\right)}.$$

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ooohhh, pretty. –  antony.trupe Nov 1 '12 at 3:52