Lets say you have a point, draw a ray of length 1 from it, let the angle of that ray be $\theta$. Now, draw another ray of length $1$ from the end of the previous ray with the same angle as before, $\theta$. Repeat this process.
Example:
Let $\theta = \pi/2$. Draw a point, and then draw a ray of length $1$ from it, at an angle of $\pi/2$ (straight to the left.) Then, draw another ray of length $1$ and at an angle of $\pi/2$ from the end of the previous ray, repeat this and eventually you will end up with a square.
My question is, is there any way to find and recognize values for $\theta$ which will produce non-trivial closed shapes? (Non-trivial as in not a regular polygon wich never intersects itself.) (Intersection with previous rays is allowed.)
Best Answer
Once you have three points, that is, once you have actually constructed two line segments so that the angle between line segments is demonstrated, then these three points uniquely determine a circle that passes through all three points.
Assuming you turn $\theta$ radians in the same direction every time, the third line segment has the same geometric relation to the second segment as the second segment had to the first, and the three endpoints of these two segments will determine the exact same circle.
Continue like this and every endpoint of every line segment will land on the same circle again.
Now imagine an observer sitting at the center of this circle, watching the tip of your pencil as it traces the line segments you are drawing and measuring the angle through which you traveled from their point of view. Traveling $2\pi$ radians would take you exactly once around the circle.
In order for you to return back to the starting point in order to make a closed shape, the observer at the center of the circle will have to see you travel an exact multiple of $2\pi$ radians in whatever number of steps you took. Any other angle will not bring you back to the starting point. Suppose you drew $n$ line segments and the observer saw you travel a total of $2m\pi$ radians. Then on each line segment you traveled $2m\pi/n$ radians, that is, each segment is the base of an isosceles triangle whose apex is at the center of the circle, the apex angle is $2m\pi/n$, the two base angles are each $(\pi/2) - (m\pi/n)$, the interior angle between two line segments (the bases of two adjacent isosceles triangles) is $\pi - 2m\pi/n$, and the exterior angle between two line segments (how much you have to change your direction of travel if you travel to the end of one segment and then start traveling along the next one) is $2m\pi/n$.
It was not clear to me whether $\theta$ was the interior angle between two line segments or the exterior angle between the two line segments, but each of these angles is a rational multiple of $\pi$. That is, you get a closed figure if and only if $\theta$ is a rational multiple of $\pi$.
So to tell whether your angle will produce a closed figure, divide it by $\pi$. You have a closed figure if and only if $\theta/\pi$ is a rational number.
You get a regular polygon in all cases where the exterior angle between segments is $\pi/k$ for some integer $k \geq 3$. In all other cases where you get a closed figure, the figure is a star. There are no other closed figures.