# On Constructions by Marked Straightedge and Compass

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.

• +1 for making me look up angle trisection with a marked ruler. For some reason I didn't know about this until today. – Jyrki Lahtonen May 19 '12 at 14:04

• Ha. I took classes from Prof Gleason when I lived in 02138. But it seems to me that that last paragraph confirms my suggestion that a quinquisector wouldn't help you construct a regular 7-gon, thus firmly answering your question in the negative. Indeed, it seems to me to give a complete answer to the question of which $n$-gons you can construct with a $q$-dissector. – Gerry Myerson May 19 '12 at 13:29