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It's well-known that you can't trisect an arbitrary angle with a classical compass and straightedge construction. However, it is possible (and pretty easy) to trisect a right angle (construct two adjacent 60$^{\circ}$ angles, bisect both of them, and three of the resulting 30$^{\circ}$ angles give you your right angle). I also know it is possible to 5-sect a right angle, although it's a bit more involved; the construction I know involves part of the construction of a regular pentagon. However, not all $n$-sections of a right angle are possible. For example, constructing a 90-section of a right angle would be equivalent to a construction of a 1$^{\circ}$ angle, which is definitely impossible.

Question: What $n$-sections of right angles are possible to construct?

I have a suspicion that the answer will involve the constructible polygons in some way, since I'm seeing that same pattern here; you can trisect and 5-sect a right angle, plus repeated bisections gives you 3, 4, 5, 6, 8, 10, 12, 16, ...etc -sections pretty easily. But I don't know about some of the others, e.g. 17-section, 15-section (combination of 5-section and trisection, one of which will need to be done on a non-right angle), 65537-section, ...etc.

realityChemist
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$n$-secting a right angle is equivalent to constructing an angle of $\dfrac {\pi}{2n}$ radians. And that is equivalent to being able to construct a segment of length $\sin\left(\dfrac {\pi}{2n}\right)$, given a segment of length $1$. This turns out to be possible only if $n = 2^mp_1p_2\ldots$ where $m \in \Bbb N$ and the $p_1, p_2, \ldots$ are $1$ or distinct Fermat primes (primes of the form $2^{2^k}+1$ - of which only $k = 0,1,2,3,4$ are known to give primes, with $k=5,\ldots,32$ and some higher $k$ known to be composite).

Paul Sinclair
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