Today I was trying to solve a problem about triangulation and I came with this
For which $n≥3$ is it possible to draw a regular $n$-gon in the plane ($\mathbb{R}^2$) such that all vertices have irrational coordinates?
My approach was assume WLOG irrational vertices $(x_0,y_0)$ and $(x,y)$. Then, if $(x,y)$ is a rotation of $(x_0, y_0)$ in $\frac{2\pi}{n}$ about origin, we have $$(x,y)=\left(x_0\cos\frac{2\pi}{n}-y_0\sin\frac{2\pi}{n}, x_0\sin\frac{2\pi}{n}+y_0\cos\frac{2\pi}{n}\right)$$ And my idea was analize that expression, but I do not know how. Is there a best approach? Thanks in advance for your help
The idea is correct. Here is how to complete. Only countable number of rotations will give polygons which have rational vertices (do you know how to show that?). So almost all rotations will give polygons where all vertices are irrational (the same argument works if "irrational" is replaced by "transcendental").