“Neusis” constructions (1)

As you know, the geometry of the ancient Greeks, as exemplified by Euclid, restricted constructions to the use of unmarked straight-edge, and compass.  So if you start with two arbitrary points on the plane, and designate them (0,0) and (1,0), the constructible points (p,q) are those for which p and q are in the quadratic closure of \mathbb{Q}; that is, the smallest superfield of \mathbb{Q} which is closed under the taking of square roots.  This is because ruler and compass constructions allow for addition, subtraction, multiplication and division, as well as square roots.

And there are of course limits to what can be constructed.  For example, it can be shown that although any quadratic equation can be solved by ruler and compass, no irreducible cubic equation can be solved, as no finite sequence of arithmetic and square roots will produce a cube root.  This can be made precise by field theory, as was done first by Pierre Wantzel in 1837.  For a modern treatment, Charles Hadlock’s “Field Theory And Its Classical Problems” can hardly be bettered.

This means that two of the classical problems: duplication of the cube (in other words, constructing \sqrt[3]{2}), and trisecting a general angle, can’t be done using just ruler and compass.  The first because x^3-2 is irreducible, the second because x=2\cos(\pi/9) (a length from the trisection of \pi/3) satisfies the irreducible equation x^3-3x-1=0.

Enter neusis!

The word “neusis” means “verging”, and employs a straight-edge with two marks on it.  A neusis construction starts with a point and two lines, and fits the straight-edge on the point so that the two marks lie on the lines.  For example, let A=(-2,\sqrt{3}), and we have a straight-edge with marks a distance 2 apart.  Lie the straight-edge on A so that the marks line up with the positive x and y axes.  It looks like this:

Rendered by QuickLaTeX.com

What is the x-intercept?

Suppose the line through A cuts the x-axis at (s,0).  The equation of the line between this point and A will be

    \[ y=\frac{-\sqrt{3}}{s+2}(x-s) \]

and so the y-intercept will be \sqrt{3}s/(s+2).  We thus have

    \[ \left(\frac{\sqrt{3}s}{s+2}\right)^{\!2}+s^2=2^2=4 \]

and so

    \[ \frac{3s^2}{(s+2)^2}+s^2-4=0. \]

Multiplying both sides and simplifying results in the equation

    \[ s^4+4s^3+3s^2-16s-16=(s+1)(s^3+3s^2-16)=0. \]

As you see then, the x-intercept satisfies a cubic equation, and in general a neusis construction involves the roots of a quartic equation.  It can be shown that the field of neusis-constructible points is the superfield of \mathbb{Q} which is closed under both square and cube roots.

In a while I’ll follow with some further posts of neusis constructions.

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