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Updated List of Multitude of Latest Autodesk Free Product & Cracking Codes.Q:

Arithmetic and circle/curve question

Assume that $x_1,x_2,\dots,x_{n-1}$ are fixed points of a simple $n$-fold rotation $\rho$. Prove that $\rho$ is a rotation if and only if $\sum\limits_{i=1}^{n-1} (x_i-x_{i-1})=0$.

So in the case of a circle, $x_i=r\cos(2\pi i/n)$ and $r$ is a fixed radius. In this case, $\rho^n=1$, so $\rho$ is a rotation. I’m having a little trouble getting started.
Any pointers would be appreciated.

A:

Let’s assume that $\rho$ is a rotation, so $\rho^n=1$. Note that $\rho$ is affine, hence $\rho(x_i)-x_i=a(x_i-x_{i-1})$ for some number $a$ such that $a eq 0$. Therefore,
$$\sum_{i=1}^{n-1} (\rho(x_i)-x_i)=a\sum_{i=1}^{n-1} (x_i-x_{i-1}).$$
Since we know that $\rho^n=1$, we know that $a=1$ and hence $\sum_{i=1}^{n-1} (\rho(x_i)-x_i)=0$. On the other hand, if $\sum_{i=1}^{n-1} (\rho(x_i)-x_i)=0$, then $a=0$ and $\rho$ is a translation.

Q: