Construction of Regular Heptadecagon

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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 Euclid's Third Postulate:

construct a circle with center $O$ and radius $OA$.

By Euclid's Second Postulate:

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

By Proposition $11$ of Book $\text{I} $: Construction of Perpendicular Line:

construct $OC$ perpendicular to $OA$.

By Proposition $10$ of Book $\text{I} $: Bisection of Straight Line twice:

construct $OD$ whose length is $\dfrac 1 4$ the length of $OC$.

By Euclid's First Postulate:

join $OA$.

By Proposition $9$ of Book $\text{I} $: Bisection of Angle twice:

construct $\angle ODE$ to be $\dfrac 1 4$ the angle $\angle ODA$.

By Proposition $11$ of Book $\text{I} $: Construction of Perpendicular Line and Proposition $9$ of Book $\text{I} $: Bisection of Angle:

construct $\angle EDF$ to be half a right angle.

Using Proposition $10$ of Book $\text{I} $: Bisection of Straight Line and Euclid's Third Postulate:

construct a semicircle on $AF$ intersecting $OC$ at $G$.

By Euclid's Third Postulate:

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

By Proposition $11$ of Book $\text{I} $: Construction of Perpendicular Line:

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

By Proposition $9$ of Book $\text{I} $: Bisection of Angle:

bisect $\angle LOM$ to obtain angle $\angle NOM$.

By Euclid's First Postulate:

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:

\(\displaystyle \frac {\tan \angle ODE + \tan \angle ODF} {1 - \tan \angle ODE \tan \angle ODF}\) \(=\) \(\displaystyle \tan \angle EDF\) $\quad$ Tangent of Sum $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 1\) $\quad$ Tangent of $\dfrac \pi 4$ $\quad$



Also see


Historical Note

The existence of the construction of the regular heptadecagon was first demonstrated by Carl Friedrich Gauss on $30$th March $1796$, at the age of $19$.

Some sources suggest that it was this discovery that led him to consider mathematics as a career option.

The construction given here is the one given by Herbert William Richmond in $1893$.


Sources