# Neusis – Classical geometry with more than just ruler and compass

A couple of weeks ago, I already wrote about me dealing with Galois theory and about how you can prove the fundamental theorem of algebra using its methods. Since then, I had a closer look at the geometrical implications of Galois’ work. The mathematical folklore is about the classical contructions with ruler and compass, what can and what cannot be done – in particular, every basic text on field theory talks about the impossibility of squaring the circle and doubling the cube and trisection of angles. All very nice and honestly: it’s actually surprising that you can disprove those geometric things with merely algebraic tools.

Just to summarize briefly what’s the core of those folklore arguments: With a ruler and compass you can construct lines and circles (obviously) from certain given points, and by intersection, you get new points. When you consider equations for the circles and for the lines, you will see that you only have to deal with at most quadratic equations. Hence, every new point you construct with ruler and compass will come from a field extension of degree $1$ or $2$. So, if you can construct something with ruler and compass, it must be contained in a field extension of $\mathbb{Q}$ that has degree $2^n$ for some $n$. But then, for squaring the circle, you need to construct $\pi$ which is not algebraic; for doubling the cube, you need to construct $\sqrt[3]{2}$, which has degree $3$ over $\mathbb{Q}$; and for trisection of an arbitrary angle you need numbers of degree $3$ as well. Contradiction. q.e.d.

When the Greek mathematicians tried to solve their classical problems, they didn’t make it with only their classical tools (well… how could they, it’s impossible anyway). Instead, they invented many clever ideas to achieve a solution with slightly extended tools. One of these ideas is the famous quadratrix, which is a curve in the plane that crosses the $x$-axis at distance $\frac{2}{\pi}$ from the origin (modernly speaking). With the usual tools, one can now construct $\sqrt{\pi}$ and so, one has squared the circle. Similar things happen with the sprial of Archimedes.

Something that I have encountered for the first time now, is the so-called neusis. I found it is sometimes also called a construction with compass and marked ruler. The classical Greek ruler is nothing but a stick, it only allows you to draw a straight line. Now that marked ruler has two marks at a certain distance of one another. At first, I wasn’t excited by this: if the distance of the marks is a constructible number, the compass already allows you to draw a straight line of this given length – now big deal about that. But the power of this tool is different: Suppose you have a point and two lines. With the marked ruler, you can draw a line through the given point that crosses the two given lines such that the segment between the two lines has the given length. You can’t do this with the classical unmarked ruler and compass alone, because you can’t know what angle the new line to be drawn needs to have! The marked ruler can be shifted until you have found the correct angle and draw the line then.

Now, we have a new tool that comes quite cheap: Just mark the ruler in some fashion, for instance with the compass set to some unit length. But let’s see how powerful this can get. Let’s say the marks on our ruler have distance $r$, it doesn’t matter if $r$ is constructible or not. We will trisect a given angle $\alpha$. To fix notation, $\alpha$ is set at a point $O$. Let’s draw a circle of radius $r$ around $O$ (see the picture below), such that one of the lines defining $\alpha$ can be considered the diameter of the circle. The other line that defines $\alpha$ will cut the circle at a point $A$. Now, we set our marked ruler on the point $A$ and move it such that one of the two marks is on the circle, the other is on the diameter of the circle (the line through $O$). Thus, we have found a line segment of length $r$ between circle and line, passing the circle at point $C$ and the line at point $B$; and the prolongation of this line also passes through $A$. The angle $\beta$ at $B$ between the new line and the diameter of the circle equals exactly $\frac{\alpha}{3}$.

Trisection of a given angle by a marked ruler construction

Well. Really? Let’s see.

Of course the length r is present all over the place. It’s the distance of $\overline{OA}$, of $\overline{OC}$ (since $A$ and $C$ are on the circle) and of $\overline{BC}$ (by marked-ruler-construction). So the triangle $\triangle OBC$ is isosceles and the angles at $B$ and at $O$ are equal. Similarly the triangle $OAC$ is isosceles with equal angles at $C$ and $A$, let’s call this angle $\gamma$. The third angle of $\triangle OAC$ at $O$ equals $2\pi-2\gamma$. But this angle is also $2\pi-\alpha-\beta$. So, $2\gamma = \alpha+\beta$ and hence $\alpha = 2\gamma-\beta$. Then consider the angles at $C$: one of them is $\gamma$, the other one is $2\pi-2\beta$. But this other one is also $2\pi-\gamma$. So we have $\gamma=2\beta$. Taking everything together, we find $\alpha=4\beta-\beta=3\beta$. q.e.d.

This is a tricky construction, but it’s pretty much just moving around triangles. I can see how you can invent this trick. But those Greeks have found many more deep constructions using the marked ruler, just just look like magic to me. I can prove them, after I have thought about them long enough – but getting the idea to look at just that, this seems amazing. Besides, my proofs heavily rely on algebraic notation and on theorems that can’t possibly have been clear to the ancient Greeks. Maybe they just believed in the truth of those theorems and accepted them – they wanted to construct points after all, so they wanted to draw things; not invent any deep theory of algebra. But that doesn’t diminish their efforts and insights. I am very interested in the way they proved their results, sadly I couldn’t find any readable account of this.

Let’s try to find cubic roots, $\sqrt[3]{k}$ for instance, with $0. We start with an isosceles triangle $\triangle ABC$, where $\overline{AC} = \overline{BC} = 1$ and $\overline{AB} = k/4$. Prolonging the line $\overline{AC}$ beyond $A$, we find the point $D$ with $\overline{AD}=1$. Then, we draw a line through $D$ and $B$, and we prolong the line $\overline{AB}$ beyond $B$. Then, we can use our marked ruler through the point $C$ such that we’ll find a segment between the lines through $\overline{BD}$ and through $\overline{AB}$ with length $1$. This segment cuts $\overline{BD}$ at the point $Q$ and $\overline{AB}$ at the point $R$, $\overline{QR}=1$. Then, $\overline{BR}=\sqrt[3]{k}$, amazingly.

Construction of a cube root of k via a marked-ruler construction

Once you have accepted this, you can find cube roots of any $k>0$ by scaling with a factor $2^n$ (and by constructing $\sqrt[3]{2}$ first). The tricky part is to prove the correctness of this construction in the first place. I don’t see how you could get this idea anyway… but it works just fine as you will see.

The proof of how to find a cube root of k with a marked ruler

We’ll need two new lines for our proof: A line at a right angle to $\overline{AB}$ through $C$, meeting $\overline{AB}$ in $M$; and a parallel line to $\overline{AB}$ through $C$, meeting the line $\overline{DB}$ in $E$. Then the triangles $\triangle ABD$ and $\triangle CED$ are similar, so we have $\frac{\overline{CE}}{2}=\frac{\overline{AB}}{1}$ and thus $\overline{CE}=2\overline{AB}=\frac{k}{2}$. The triangles $\triangle ECQ$ and $\triangle BQR$ are similar as well (by the parallel lines and the respective angles at $Q$ being equal). So, $\frac{\overline{CE}}{\overline{CQ}}=\frac{\overline{BR}}{1}$ and hence $\overline{CQ}=\frac{\overline{CE}}{\overline{BR}}=\frac{k}{2\overline{BR}}$. Now, we can use the Pythagorean theorem twice in a row: $\overline{CR}^2=\overline{CM}^2+\overline{MR}^2 = \overline{CA}^2-\overline{AM}^2+\overline{MR}^2$. This yields $(1+\frac{k}{2\overline{BR}})^2=1-(\frac k8)^2+(\overline{BR}+\frac k8)^2$. By multiplying out and sorting all the terms, we find $4\overline{BR}^4+k \overline{BR}^3-k^2-4k \overline{BR}=0$, and looking at this long enough, this is $(4\overline{BR}+k)(\overline{BR}^3-k)=0$. Now, the first term gives $\overline{BR}=-\frac k4$, but this is negative and $\overline{BR}$, being a line segment, needs to have positive length. So, the length of $\overline{BR}$ needs to be a positive solution to the second term, and so $\overline{BR}=\sqrt[3]{2}$, as claimed (the other two solutions are not real). q.e.d.

Isn’t that a pretty proof?

Algebraically, by the way, the marked ruler allows you a little more than the classical ruler: It gives you points in field extensions of degree $3$ and $4$. It still doesn’t allow you to construct any real number you want. The quadratrix on the other hand only gives you a field extension with the transcendental number $\pi$, no cubic roots.

Finally, I need to give credit to the book that showed me these things. The book by Cox on Galois theory, that I mentioned a couple of weeks ago already, has these things in an optional section and mostly as exercises. I have spent many an evening recently on those exercises – this was a lot of fun in geometry of all things (I have never liked geometry… at all.). The images that I have built in here, rely heavily on Cox’ notation and on the images to his exercises.