For which $\alpha$ will take the cake ever be again with chocolate on the bottom and cream on the top Question: A bored kid left alone at home decides to take a chocolate cream cake (chocolate on the bottom, cream on top) and his protractor and spend the day as follows: He cuts a slice of angle $\alpha$ put it back up-side-down (i.e. cream on the bottom) rotate the cake clockwise by $\alpha$ and repeat the same procedure again and again. For which $\alpha$ will take the cake ever be again with chocolate on the bottom and cream on the top?
My approach: Well, intuitively $\alpha$ could be 180 degrees, and 360 degrees, my approach is that $\alpha$ could be anything that divides 360 degrees without a remainder. Not quite sure how to show/prove that mathematically, maybe there could be some trick here, any help is appreciated.
 A: The answer is any angle,
this article covers it all,
https://mathstrek.blog/2013/02/08/thoughts-on-a-problem-iii/
A: EDIT: this answer assumes model bad model of piece inverting (that putting piece up-side-down is equal to reversing colors). In real pie we not only inverse colors but also mirror them. See @user2323232's link for correct answer.
You need $\alpha$ to be rational (I assume slice is half-open segment, otherwise we will never get the entire cake right).
Lets reformulate a bit: we take a red line, start at point $0$, then color segment $[0, \alpha)$ in green, then color segment $[\alpha, 2\alpha)$ and so on. Point at $x$ degrees is cream-top after $n$ steps iff even number of points $x, x + 360, x + 720, \ldots$ is green (it means that we made an even number of flips of this point)
If $\alpha = \frac{n}{m}$ is rational, then after $720m$ steps, we will have segment $[0, 720n)$ green and the rest is red. Then for any $x$ there will be exactly $2n$ green points: $x, x + 360, \ldots, x + 360 \cdot (2n - 1)$. So after $720m$ steps our cake will be cream-top again.
Lets assume that after $k$ steps our cake is cream-top. We will have green segment $[0, k\alpha)$. If $k\alpha = 720n$ for some integer $n$, then $\alpha$ is rational. Otherwise choose $n$ such that $720n < k\alpha < 720(n + 1)$. Take some $x^\prime \in (720n, k\alpha)$ such that $x^\prime + 360 > k\alpha$ and let $x = x^\prime - 720n$. Then point at $x$ degrees is cream-bottom, as $x, x + 360, \ldots, x + 720n$ are green, but $x + 720n + 360$ isn't. So $k$ steps didn't left us with cream-top cake.
