Worked solution: Wantzel's algebraic impossibility proof via field extensions (1837)

Step 6 of 7: Conclusion:

3

is not a power of

2

, so

60^\circ

cannot be trisected

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In plain words

Now the two threads meet. Step 3 says any constructible number's degree must be on the list $1, 2, 4, 8, 16,\ldots$ . Step 5 says the number we actually need has degree $3$ . But $3$ is simply not on that list — no amount of doubling ever lands exactly on $3$ .

That clash is the whole proof: it means the specific point needed to trisect $60^\circ$ can never be reached by straightedge and compass, full stop.

[\mathbb{Q}(2\cos20^\circ):\mathbb{Q}]=3 \ne 2^m \implies 60^\circ \text{ cannot be trisected}

Detailed analysis

Step 3 proved that every constructible real number has degree $2^m$ over $\mathbb{Q}$ for some integer $m\ge0$ : that is, its degree must be $1, 2, 4, 8, 16,\ldots$ . Step 5 computed that $[\mathbb{Q}(2\cos20^\circ):\mathbb{Q}] = 3$ , and $3$ never appears in that list — it is odd and greater than $1$ , so it cannot equal any power of $2$ .

Therefore $2\cos 20^\circ$ is not constructible, hence $\cos 20^\circ$ is not constructible, hence the angle $20^\circ$ cannot be produced from a $60^\circ$ angle (and the unit length) by straightedge and compass alone. Since trisecting $60^\circ$ would produce exactly $20^\circ$ , no straightedge-and-compass construction can trisect $60^\circ$ (Wantzel 1837, §II).

This single counterexample refutes any claimed universal method for trisecting angles: had such a method existed, applying it to $60^\circ$ would contradict what was just proved. The classical trisection problem, open since antiquity, was thereby settled — not by finding an ingenious new construction, but by proving decisively that none can exist.