Pierpont proved that a regular $n$-gon is constructible by (singly) marked straightedge and compass if and only if $n = k \, p_1 \cdots p_{s}$, where $k = 2^{a_1} 3^{a_2}$ for $a_i \geq 0$ and $p_i = 2^{b_1} 3^{b_2} + 1 > 3$ is prime with $b_i \geq 0$.
It has been known since the time of Archimedes that a marked straightedge allows for angle trisection. Let a $q$-sector be an object which allows for angle $q$-section.
Does this result generalize to the following?
Let $q$ be a prime. A regular $n$-gon is constructible by $q$-sector, straightedge and compass if and only if $n = k \, p_1 \cdots p_{s}$, where $k = 2^{a_1} 3^{a_2} \cdots q^{a_m}$ for $a_i \geq 0$ and $p_i = 2^{b_1} 3^{b_2} \cdots q^{b_m} + 1 > q$ is prime with $b_i \geq 0$.
Update: Gleason's paper provides the complete answer for constructible $n$-gons. Here, it is shown that a regular $n$-gon is constructible by straightedge, compass and $p$-sector for each prime $p$ dividing $\varphi(n)$, the Euler totient of $n$.
Thus, I must modify my conjecture to the following:
Let $q$ be a prime. A regular $n$-gon is constructible by $\{ 3, 5, \dots, q \}$-sectors, straightedge and compass if and only if $n = k \, p_1 \cdots p_{s}$, where $k = 2^{a_1} 3^{a_2} \cdots q^{a_m}$ for $a_i \geq 0$ and $p_i = 2^{b_1} 3^{b_2} \cdots q^{b_m} + 1 > q$ is prime with $b_i \geq 0$.
One direction is certainly true by using the multiplicativity of the Euler totient function. The question is now whether the other direction also holds.