= Solution
A <straightedge and compass construction> uses finitely many steps drawing lines through known points and circles with known centres and constructible radii, and adjoining their intersection points. The coordinates obtained by line–line intersections require only <field> operations; line–circle and circle–circle intersections require at most a <quadratic extension> after eliminating a linear equation. Thus all coordinates lie in a tower of <quadratic extensions> of the initial coordinate <field>. The degree of any finite intermediate coordinate <field> containing that initial <field> divides a <power of two> and is therefore $\boxed{2^r}$ for some $r\ge0$.
For the source's notation this assumes $K\supseteq K_0$, as is customary when previously constructed points are retained. If only the coordinates of a final subset are called $K$, replace it by the compositum $K_0K$; otherwise the expression $[K:K_0]$ need not be defined. The power-of-two condition is necessary and is not being asserted as sufficient for an arbitrary algebraic number.
Starting from a unit cube, <impossibility of doubling the cube by straightedge and compass>[doubling its volume] would construct a length $\sqrt[3]2$. Its <polynomial> $X^3-2$ is Eisenstein at two, so its degree over $\mathbb Q$ is three, which cannot divide a <power of two>. Thus \b[doubling the cube is impossible by ruler and compass].
Set $c=\cos\theta$, $s=\sin\theta$. The real part of $(c+is)^5$ is $c^5-10c^3s^2+5cs^4$. Replacing $s^2$ by $1-c^2$ gives
$$
\boxed{\cos5\theta=16c^5-20c^3+5c.}
$$
Choose the given angle to have cosine $5/6$. It is constructible, since its <unit circle> point has coordinates $(5/6,\sqrt{11}/6)$. If a fifth of that angle were constructible and had cosine $c$, the identity would imply
$$
96c^5-120c^3+30c-5=0.
$$
This <polynomial> is Eisenstein at five: all nonleading coefficients are divisible by five and the constant term is not divisible by 25, while 96 is not divisible by five. Hence $[\mathbb Q(c):\mathbb Q]=5$. But the initial angle <field> is $\mathbb Q(\sqrt{11})$, itself of degree two, so adjoining any finite collection of constructible coordinates to it would still give total degree over $\mathbb Q$ a <power of two>. It could not contain a degree-five element. Thus \b[<general angle quintisection is impossible with straightedge and compass>].
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