I have found a fun problem concerning compass and straight edge constructions, which I am unable to solve completely at the moment.
The problem is the folowing: let $\zeta$ be a primitive elevent root of unity and consider the set $R = \{ z \in \mathbb{C} | z^{11}=1\}$. For every $S \subset R$ we define $z_{S} = \sum_{s\in S}s$. If our starting field is $\mathbb{Q}$, how many subsets $S$ are such that $z_{S}$ is constructible?
For example, if we take $S=\{1, \zeta, \zeta^3 \zeta^4,\zeta^5, \zeta^9 \}$. One can show that this leads to a constructible $z_{S}$. One does this by noting that the extension $\mathbb{Q} \subset \mathbb{Q}(\zeta)$ has a cyclic galois group of order 10 and obviously contains $R$. If a number in this extension is constructible it is necessarily contained in a subfield of degree a power of 2, considering our observation concerning our galois group, this can only be the case if the corresponding galois group is a power of two and because our extension has order ten, if this galois group has order two. This group is generated by the automorphism $\varphi$ which holds $\mathbb{Q}$ fixed and $\varphi(\zeta) = \zeta^9$. When applying some homomorphism properties and modulo $11$ calculations, we conclude that $\varphi(z_{S}) = z_{S}$ Therefore $z_{S}$ is contained in this extension of order two and is therefore constructible (I think, correct me if I'm wrong).
I tried generalizing this for other sets and I think I reduced it to the following combinatorial problem. Let $i \in \{0,1,2, \dots 10\}$. Doing the same calculations as when applying the homomorphism properties in the last example, we find that also $9i, 4i, 3i, 5i$ need to be contained in the sum expression of $z_{S}$. I suppose that choosing $i = 0$ bears no consequences,so we can restrict ourselves to counting the combinations of $i \in \{1, \dots 10 \}$ and multiply these by $2$. My first guess (or at least upper bound I suppose) is that choosing one such $i$ is equivalent to choosing $4$ of them, therefore there are at most $10 \choose 4$ $\cdot 2 = 520$. Any help would be greatly appreciated!