Shan-Yu and Mulan are playing a game. Let $\theta$ be an angle with $0^\circ<\theta<180^\circ$ known to both players. Initially, Shan-Yu makes a paper triangle $\mathcal{T}$ with measurements of his choice. Then, they repeatedly perform the following steps:
If $\mathcal{T}$ has at least one angle measuring exactly $\theta$, then the game stops and Mulan wins.
Otherwise, Mulan chooses a point $P$ on the perimeter of $\mathcal{T}$, different from its three vertices. She then makes a straight cut from $P$ to the opposite vertex of $\mathcal{T}$, splitting it into two triangles.
Shan-Yu discards one of the two triangles. The remaining triangle becomes the new $\mathcal{T}$.
For which real values of $\theta$ can Mulan guarantee her victory in finitely many steps, no matter how Shan-Yu plays?
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The answer is $\boxed{\theta=\frac{\pi}{n},n\in\mathbb{N}\setminus\{1\}}$.
We say that an angle is good if it is an integer multiple of $\theta$. Clearly, Mulan can win if $\pi$ is good. Otherwise, Shan-Yu sets up $\mathcal{T}$ without any good angle. Consider the first time Mulan marks $P$ such that both triangle $ABP$ and $ACP$ has a good angle. By definition, none of $\angle B$ and $\angle C$ is a good.
- If both $\angle APB$ and $\angle APC$ are good, then so is $\pi$, a contradiction.
- If both $\angle APB$ and $PAC$ are good, then so is $\angle C$, a contradiction.
- If both $\angle APC$ and $PAB$ are good, then so is $\angle B$, a contradiction.
- If both $\angle PAB$ and $\angle PAC$ are good, then so is $\angle BAC$, a contradiction.
Hence, Shan-Yu can prevent Mulan from winning by always picking a remaining triangle without any good angle.
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