A student asked what angles are constructible, considering
only angles that measure a whole number of degrees (which I'll call integral
angles). The answer is very simple: The only constructible integral angles are
those which measure 3n degrees, where n is any natural number.

To see this, first note that angles of measure 60 and 72
(vertex angles in a regular pentagon) are constructible. Since angles may be
bisected repeatedly by construction, this means that angles of measure 15 =
60/4 and 9 = 72/8 degrees may be constructed. Since 3 = 9*2 – 15, an angle of 3
degrees may be constructed by constructing two adjacent 9 degree angles, and
then a 15 degree angle inside the resulting 18 degree angle. By adding n 3-degree
angles together, a 3n degree angle can be constructed.

The proof that only these integral angles can be constructed
seems to require the more advanced result that there is at least one integral
angle that is not constructible. For example, an angle of 20 degrees is
non-constructible. (This is the standard proof, using Galois theory, that
demonstrates the impossibility of angle trisection.) If there were an integral
angle of measure n that is not a multiple of 3 that is constructible, then
since (n,3) = 1, there is a linear combination with integral coefficients of n
and 3 that gives 1, and so an angle of 1 degree would be constructible, and
hence an angle of 20 degrees would be constructible, which it is not..

## 1 comment:

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