# Does there exist a constructible (by unmarked straightedge and compass) angle that cannot be quintsected?

I know that for example an angle of $20^\circ$ cannot be quintsected because an angle of $4^\circ$ cannot be constructed (I'm thinking in terms of (unmarked) straightedge and compass. But an angle of $20^\circ$ cannot be constructed (as above) and I would be interested to see an example of a constructible angle that cannot be quintsected, assuming one exists.

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What about $60^\circ$? – Berci Feb 4 '13 at 21:58
@Berci quintsection of $60^\circ$ is constructible because $12^\circ=72^\circ-60^\circ$ and $72^\circ$ and $60^\circ$ is constructible. – Hanul Jeon Feb 4 '13 at 22:17

The $72$ degree angle is constructible but the $\dfrac{72}{5}$ degree angle cannot be.
This follows from the nice theorem that a regular polygon of $n\ge 3$ sides can be constructed by straight edge and compass if and only if $n$ is of the shape $$2^e p_1p_2\cdots p_k,$$ where the $p_i$ are distinct Fermat primes.
There is no integer $k$ such that the $k$-degree angle is not constructible, while the $5k$ degree angle is. This is basically because $180$ is not divisible by $5^2$. We get a whole degree example for trisection because $3^2$ divides $180$. – André Nicolas Feb 4 '13 at 22:56