Is it possible to construct a regular heptagon (a figure with seven sides) with just compass and straightedge? If so, could you please give me directions for how to do this?
[Math] Is it possible to construct a regular heptagon with just compass and straightedge
geometric-constructiongeometrypolygons
Related Solutions
Gleason's article "Angle Trisection, the Heptagon, and the Triskaidecagon" (also available here) mentions a construction due to Plemelj:
Draw the circle with center $O$ passing through $A$ and on it find $M$ so that $AM=OA$. Bisect $OM$ at $N$, and trisect at $P$, and find $T$ on $NP$ so that $\angle NAT=\frac13\angle NAP$. $AT$ is the needed side of the heptagon.
To validate Plemelj's construction, we must prove that $AT=OA\left(2\sin\dfrac{\pi}{7}\right)$. Since $2\cos\dfrac{2\pi}{7}=2-\left(2\sin\dfrac{\pi}{7}\right)^2$, it follows from $x=2\cos\dfrac{2\pi}{7}$ being a root of $x^3+x^2-2x-1=0$ that $2\sin\dfrac{\pi}{7}$ is a root of
$$(2-x^2)^3+(2-x^2)^2-2(2-x^2)-1=0$$
the other roots being $-2\sin\dfrac{\pi}{7}$, $\pm 2\sin\dfrac{2\pi}{7}$, and $\pm 2\sin\dfrac{3\pi}{7}$. The equation can be factored as
$$\left(x^3+\sqrt 7\left(x^2-1\right)\right)\left(x^3-\sqrt 7\left(x^2-1\right)\right)=0$$
The roots corresponding to the first cubic factor are $2\sin\dfrac{\pi}{7}$, $-2\sin\dfrac{2\pi}{7}$, and $-2\sin\dfrac{3\pi}{7}$. Writing the first cubic factor in the form
$$\left(\frac1{x}\right)^3-\frac1{x}=\frac1{\sqrt 7}$$
and making the substitution $\dfrac1{x}=\dfrac2{\sqrt 3}\cos\,\psi$ yields the equation $\cos\,3\psi=\sqrt{\dfrac{27}{28}}$. The desired root corresponds to the choice
$$\psi=\frac13\arccos\sqrt{\frac{27}{28}}=\frac13\arctan\frac1{3\sqrt 3}$$
which yields
$$2\sin\frac{\pi}{7}\cos\,\psi=\frac{\sqrt 3}{2}$$
From the figure, we have $\angle NAP=\arctan\dfrac1{3\sqrt 3}$, so $\angle NAT=\psi$. We thus have $AT\cos\,\psi=AN=\dfrac{\sqrt 3}{2}OA$, and from this we also have $AT=OA\left(2\sin\dfrac{\pi}{7}\right)$.
- Construct $\bigcirc A$ through $O$.
- Let $P_1$ and $P_2$ be the points where $\bigcirc A$ meets $\bigcirc O$.
- Construct $\bigcirc P_1$ through $P_2$.
- Let $C$ be the (other) point where $\bigcirc P_1$ meets $\bigcirc O$.
- Construct $\overleftrightarrow{OP_1}$.
- Let $Q_1$ and $Q_2$ be the points where $\overleftrightarrow{OP_1}$ meets $\bigcirc P_1$.
- Construct $\overleftrightarrow{CQ_1}$.
- Let $B$ be the point where $\overleftrightarrow{CQ_1}$ meets $\bigcirc O$.
- Note that we have constructed $\overleftrightarrow{BC}$.
- Construct $\overleftrightarrow{CQ_2}$.
- Let $D$ be the point where $\overleftrightarrow{CQ_2}$ meets $\bigcirc O$.
- Note that we have constructed $\overleftrightarrow{CD}$.
- Construct $\overleftrightarrow{AB}$.
- Construct $\overleftrightarrow{AD}$.
Square $\square ABCD$, with constructed edge-lines, is inscribed in $\bigcirc O$. (Proof that the quadrilateral is, in fact, a square, is left as an exercise to the reader.)
Edit. Having been asked to elaborate on the square ...
As of Step 2, we know $\triangle P_1 P_2 C$ is equilateral and that $\overline{OA}$ is on the perpendicualr bisector of side $\overline{P_1 P_2}$. Therefore, $\overline{AC}$ is a diameter of $\bigcirc O$, and we have that $\angle ABC$ and $\angle ADC$ (for point $D$ constructed later) are right angles by Thales' Theorem.
As of Step 4, as observed by Jan and Tristan in the comments, $\overline{Q_1 Q_2}$ is a diameter of $\bigcirc{P_1}$, so $\angle Q_1 C Q_2$ is a right angle. Therefore, $\square ABCD$ is at least a rectangle.
Now, define $a := |\overline{OA}|$, so that $|\overline{P_1P_2}| = a\sqrt{3}$ and $|\overline{OQ_1}| = a( 1 + \sqrt{3})$. Since $\angle AOQ_1 = 60^\circ$, if we let $R$ be the foot of the perpendicular from $Q_1$ to $\overleftrightarrow{OA}$, then $|\overline{OR}| = \frac{a}{2}( 1 + \sqrt{3})$ and $$|\overline{Q_1R}| = \frac{a \sqrt{3}}{2}(1+\sqrt{3}) = \frac{a}{2}(3 + \sqrt{3}) = a + |\overline{OR}| = |\overline{CR}|$$ Thus, $\angle Q_1 C R = 45^\circ$ and we may conclude that $\square ABCD$ is a square. $\square$
Best Answer
No, it's not possible; in fact, the regular heptagon is the regular polygon with the least number of sides that is impossible to construct with compass and straightedge alone. It is, however, possible to construct it using a neusis ruler. A related question has been asked (and answered) here.