Constructible angles of integral degrees.

For which positive integer `n` is `n^o` constructible (classically with straightedge and compass)? There is a satisfactory answer:

Let `n` be a positive integer. Then the angle `n^o` is constructible if and only if `n` is a multiple of `3`.

The proof is straightforward as well, provided we believe some other facts. First note the classic fact that

A regular pentagon is constructible, hence `72^o` is constructible.

And that

If `A` and `B` are any two constructible angles, then so is any integer combination `xA + yB`, where `x, y in ZZ`.

So, since we can also construct an equilateral triangle, we can construct `60^o`. Hence we can construct `72^o - 60^o = 12^o`. Now, using the fact that

If `A` is a constructible angle, then so is its bisection `A/2`.

This shows we can construct `6^o` and `3^o`. Now using the fact of integer combinations of constructible angles, we have the first direction:

For any integer `x`, the angle `3x^o` is constructible.

Now we show if integer `n` is not divisible by `3`, then `n^o` is not constructible.

 

We show this by contradiction. Suppose to the contrary that some integer `n` is not divisible by `3` but is constructible. Then by repeated subtraction of `3^o`, we can construct the angles `1^o` or `2^o`, as `n mod 3` is `1` or `2`. And since we can bisect, constructing `2^o` means we can construct `1^o`. So far, this says if such integer `n` that is not a multiple of `3` but `n^o` is constructible, then we can construct `1^o`. But wait, this is crazy! Because of the venerated fact that

The angle `20^o` is not constructible.

We should not be able to construct `1^o`, otherwise by integer combination of constructible angles, we could construct any integral angles!