Neusis constructions (3): the regular heptagon

The regular heptagon (seven sided polygon) is the smallest polygon which can't be constructed used Euclidean compass and straight-edge.  However, it can be constructed by using a marked ruler, and the first such construction was given by the the great Renaissance French mathematician François Viète (1540 – 1603). 

Before we show how it can be done, we need a little algebra.

Constructing a heptagon is equivalent to constructing the point z=e^{2\pi i/7} in the complex plane.  This value z satisfies z^7-1=0 and this equation can be factorized as

    \[ (z-1)(z^6+z^5+z^4+z^3+z^2+z+1)=0. \]

Since z\ne 1, it must satisfy the second term, and since z\ne 0 we can divide through by z^3 to obtain

    \[ z^3+z^2+z+1+z^{-1}+z^{-2}+z^{-3}=0. \]

A bit of fiddling shows that this last equation can be written as

    \[ \left(z+\frac{1}{z}\right)^3+\left(z+\frac{1}{z}\right)^2-2\left(z+\frac{1}{z}\right)-1=0. \]

So if

    \[ w=z+\frac{1}{z} \]

we have w^3+w^2-2w-1=0.  However, with z=e^{2\pi i/7}, then

    \[ z+z^{-1}=2\cos\left(\frac{2\pi}{7}\right). \]

\Thus if we can solve the cubic equation w^3+w^2-2w-1=0, we will have w=2\cos(2\pi/7) and from that we can construct the heptagon.

We have shown how an angle can be trisected using neusis constructions, and we also know that 4\cos^3\theta-3\cos \theta=\cos(3\theta). Multiplying by 2 means that this equation can be written as

    \[ (2\cos \theta)^3-3(2\cos \theta)-2\cos(3\theta)=0. \]

This means we can solve any equation

    \[ z^3-3z-A=0 \]

by putting A=2\cos(3\theta), and then z=2\cos \theta.

We now have to coerce the equation w^3+w^2-2w-1=0 into the form z^3-3z-A=0.  This is actually standard algebra, started by putting w=y-1/3.  Then

    \begin{align*} w^3+w^2-2w-1&=\left(y-\frac{1}{3}\right)^3+\left(y-\frac{1}{3}\right)^2-2\left(y-\frac{1}{3}\right)-1\\ \&=y^3-\frac{7}{3}y-\frac{7}{27}. \end{align*}

Now write y=rx so that

    \[ r^3x^3-\frac{7}{3}rx-\frac{7}{27}=0. \]

or that

    \[ x^3-\frac{7}{3r^2}x-\frac{7}{27r^3}=0. \]

We require

    \[ \frac{7}{3r^2}=3 \]

and so

    \[ r=\frac{\sqrt{7}}{3} \]

and the equation to be solved is

    \[ x^3-3x-\frac{1}{\sqrt{7}}=0. \]

What this means is that if we can construct an angle \theta for which 2\cos\theta=\frac{1}{\sqrt{7}}, then the value x will be 2\cos(\theta/3).  Then y=2\sqrt{7}\cos(\theta/3)/3 and so finally

    \[ w = \frac{2\sqrt{7}}{3}\cos\left(\frac{\theta}{3}\right)-\frac{1}{3}. \]

This means that w/2=\cos(2\pi/7).

So to start we need to construct

    \[ \cos\theta=\frac{1}{2\sqrt{7}}=\frac{1}{\sqrt{28}}. \]

\We will need a triangle like this:

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Scaling by \sqrt{28}/6 produces

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And this second triangle is easily constructed from an equilateral triangle with unit sides:

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So the construction starts as follows: let AOB be the diameter of a unit circle (centred at (0,0)), and place an equilateral triangle OBC, with C being on the circle.  Let D=(1/3,0), and join DC.  Using Archimedes' neusis construction, trisect the angle CDB.  This will involve a circle centred at D with radius DC, and a line which goes through C, crosses this new circle at F and the diameter AOB at G, and for which FG=DC.  It looks like this so far:

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Now let M be the mid point of OG and draw the perpendicular to AOB through M which crosses the original circle at P and Q. Then AP and AQ are two sides of a regular heptagon.

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To see why this works, note that angle FGA=\cos(\theta/3) and that DC=FG=\sqrt{7}/3.  Thus GD = 2(\sqrt{7}/3\cos(\theta/3)) and so OG=2(\sqrt{7}/3\cos(\theta/3))-1/3.   But this is the value of w from above which satisfies w^3+w^2-2w-1=0.  Then OG/2=w/2 which is the cosine of 2\pi/7.  And so the angle we need is obtained by the perpendicular.

Viète's construction is slightly different in details to this; what is remarkable is how he managed it without the tools of modern algebra (even though one of Viète's great innovations was to use letters for variables).  You can read about his construction here.

One problem with the construction given above is that it requires the use a straight-edge with marks a distance \sqrt{7}/3 apart.  This is not insurmountable: \sqrt{7}/3 is a constructible value, and so such marks can be generated by Euclidean methods.  But it is more elegant, in some ways, to use a straight-edge on which the marks are a distance 1 apart.  And in fact the above construction can easily be modified. 

Start with the circle, the equilateral triangle, and the line DC.  But instead of trisecting the angle CDB, we first create a radius OE parallel to DC, and use Archimedes' trisection on the original circle to trisect angle EOB: a line through E which crosses the circle and the line AB at points X and Y and with XY=1.  But since angles EOB and CDB are equal, we can construct G with a line through C which is parallel to the new line.  Like this:

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This is pretty much what Viète did, which I think is a remarkable tour-de-force of geometric reasoning, especially given the mathematical tools he had available to him.

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