If I have a circle with a given radius, how calculate I find a regular polygon's vertices, if I know they are all on the circles edge (like in the images below)?
[Math] Convert a circle to a polygon
circlespolygons
Related Solutions
A proof depends by the definition of a circle that we use.
If the definition is:
the locus of points equidistant from a given point called center $C$
consider a regular polygon with $C$ as center of symmetry. For a point on the polygon the distance $d$ from $C$ is such that: $$ r\le d\le r\cos \dfrac{\theta}{2} $$ where: $r$ is the distance of a vertex form $C$ and $\theta$ is the angle of vertex $C$ subtended by a side.
If the number of sides $n \rightarrow \infty$ than $\theta \rightarrow 0$ and : $$ \lim_{\theta \to 0}r\cos \dfrac{\theta}{2}=r $$ so, at the limit, all points of the polygon have the same distance $r$ from $C$.
if you want use the internal angle $\alpha=\dfrac{\pi}{n}(n-2)$, note that $\theta = \pi -\alpha$ and: $$ n \rightarrow 0 \iff \alpha \rightarrow \pi \iff \theta \rightarrow 0 $$
or use:
$$ r\le d\le r \sin \dfrac {\alpha}{2} $$
Polar integration seems to be the best method for creating a general formula for this.
Re-expressing both circles as polar yields this: $r1 = 1$ and $r2 = cos\theta+\sqrt(a^2-sin^2\theta)$ based off of the general formula for a polar circle, centered at $(1, 0)$ with radius $a$. Really all we need here are the points of intersection which can easily be solved by setting each equation equal and following algebraic and trigonometric rules
$$cos\theta+\sqrt(a^2-sin^2\theta) = 1$$
$$a^2-sin^2\theta=1+cos^2\theta-2cos\theta$$
Replacing $sin^2$ and $cos^2$ with 1 and solving for theta yields:
$$\theta=\pm arcos(a^2-2/-2)$$
From here it should be noted that we must add $\pi$ to the resulting angle because our result accounts for the part of the circle outside of the central unit circle. So our angles from which we base our integral are
$$\theta = \pi \pm arcos(a^2-2/-2)$$
At this point I believe you can evaluate this regularly
$$.5\int_{\theta_1}^{\theta_2} 1^2-[cos\theta+\sqrt({a^2-sin^2\theta})]^2 \,d\theta$$
There may be some errors I missed in my answer however this general method should work
Best Answer
These points on the unit circle given by $x^2+y^2=1$ will form a regular $n$-gon:
$$(x,y)=(\cos(\frac{2k\pi}{n}),\sin(\frac{2k\pi}{n}))$$
for $k=0, 1,\cdots, n-1$.
And they're in fact the roots of unity which means the complex roots of the equation $z^n=1$. You need to know complex numbers to fully understand this answer.
But you can also use trigonometry to achieve this. Note that you want to divide a circle into $n$ sections each of $\frac{360}{n}$ degrees. Now use trigonometry to see why that's true. When $k$ ranges from $0$ to $n-1$, you are moving $\frac{360}{n}$ degrees forward on the circle. Does it make sense?