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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.

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+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

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First of all, I'm not sure there is such a thing as a straightedge sufficiently marked to allow for, say, cutting an arbitrary angle into 5 pieces. But even if there is such a thing, I don't see how it would help you to construct, say, a regular 7-sided polygon. To construct that polygon, you need to construct solutions of an irreducible cubic, while the straightedge helps you construct solutions of an irreducible quintic. The fields have nought to do with each other.

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Hi Gerry, I edited my post to reflect the core idea in my question. A marked ruler is known to allow the trisection of an arbitrary angle which is, in some precise sense, equivalent to solving an irreducible cubic (not all, of course). –  user02138 May 19 '12 at 13:08
    
See the last paragraph of Gleason's paper. –  user02138 May 19 '12 at 13:17
    
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
    
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 13:30
    
Ah, yes, I remember that 02138 is your alma mater. Thanks again, Gerry! –  user02138 May 19 '12 at 13:47

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