Prove the regular 12-gon is constructible.

Question

Prove, both geometrically and then algebraically, that the regular 12-gon is contructible.

I'm pretty stuck on this one and trying to get my head around constructibility, so far I've seen that proving the 3-gon is contructible could be the first step.

What would be the easiest way to the geometric and the algebraic proofs?

Any help would be greatly appreciated!!

Answer 1

Algebraically, a regular $n$-gon is constructible if and only if $\cos (2\pi/n)$ is a constructible number.

$\cos(\frac{2\pi}{12})=\frac{\sqrt3}{2}$ is clearly constructible.

Answer 2

There is little more to it than just constructing a perpendicular bisector, really.

1. Start with a line segment $AB$.

2. Center the compasses on $A$ and construct the circle through $B$.

3. Center the compasses on $B$ and construct the circle through $A$. Define $C, D$ as the points of I resection of the two circles now constructed.

4. Draw line segments $AC$ and $CD$ thus identifying a $\angle ACD$ which measures 30°.

5. Center the compasses on $C$ and draw a circle through $A$; this circle passes through $E$ on $CD$ so that $\angle ACE$ measures 30° and $\arc AE$ follows suit.

6. Draw chord $AE$ and copy this around the circle centered on $C$.

Answer 3

Inscribe a regular hexagon ( we know this algebraically and geometrically) in a circle and next it is possible to construct angle bisectors of the 6 angles at circle center intersecting circumcircle at new 6 more vertices.