Construction of Regular Heptadecagon

Theorem
It is possible to construct a regular hepadecagon (that is, a regular polygon with $17$ sides) using a compass and straightedge construction.

Construction

 * HeptadecagonConstruction.png

The construction will inscribe a regular hepadecagon inside any arbitrary circle.

By :
 * construct a circle with center $O$ and radius $OA$.

By :
 * produce $OA$ to $B$, hence making $AB$ a diameter of this circle.

By :
 * construct $OC$ perpendicular to $OA$.

By twice:
 * construct $OD$ whose length is $\dfrac 1 4$ the length of $OC$.

By :
 * join $OD$.

By twice:
 * construct $\angle ODE$ to be $\dfrac 1 4$ the angle $\angle ODA$.

By and :
 * construct $\angle EDF$ to be half a right angle.

Using and :
 * construct a semicircle on $AF$ intersecting $OC$ at $G$.

By :
 * construct a semicircle with center $E$ and radius $EG$, intersecting $AB$ at $H$ and $K$.

By :
 * construct $HL$ and $KM$ perpendicular to $OA$, intersecting the circle $ACB$ at $L$ and $M$.

By :
 * bisect $\angle LOM$ to obtain angle $\angle NOM$.

By :
 * join $NM$.

$NM$ is one of the sides of a regular hepadecagon which has been inscribed inside circle $ACB$.

Proof
It remains to be demonstrated that the line segment $NM$ is the side of a regular hepadecagon inscribed in circle $ACB$.

This will be done by demonstrating that $\angle LOM$ is equal to $\dfrac {2 \pi} {17}$ radians, that is, $\dfrac 1 {17}$ of the full circle $ACB$.

For convenience, let the radius $OA$ be equal to $4 a$.

By Pythagoras's Theorem, $AD = a \sqrt {17}$.

By definition of tangent, $OE = a \arctan \left({\dfrac {\angle ODA} 4}\right)$.

By construction, $\angle EDF = \dfrac \pi 4$ radians.

Thus:

Also see

 * Construction of Regular Prime $p$-Gon Exists iff $p$ is Fermat Prime