Two-sided ruler constructions 2: Fundamentals

This is the second in a series of blog posts about two-sided ruler constructions. Here are all the blog posts in the series:

1. Introduction
2. Fundamentals (you are here)
3. Rhombuses
4. Copying and cutting
5. Perpendicular lines
6. Parallel lines
7. Circles without circles
8. Equilateral triangle and regular pentagon

This blog post is about the fundamentals of what a two-sided ruler can do, leading up to some basic constructions.

So, what can a two-sided ruler do?

It can draw lines.

Draw a line anywhere
1. Put the ruler down and draw along one side.

Draw a line through a specific point
0. Start with an existing point.
1. Align the point on one side of the ruler and draw along that side.

Draw a line through two specific points
0. Start with two existing points.
1. Align both points on one side of the ruler and draw along that side.

A one-sided ruler can do all of this of course. What a two-sided ruler can do that a one-sided ruler can’t do is draw two parallel lines a ruler width apart.

Draw two parallel lines anywhere (a ruler width apart)
1. Put the ruler down and draw along both sides.

Draw a line parallel to an existing line (and one ruler width away)
0. Start with an existing line.
1. Align one side of the ruler along the existing line, and draw along the other side.
Done! The second line is parallel and one ruler width away from the first line.
Video here

You don’t have to actually draw the first line either. If you have two points, you can align them both on one side of the ruler and then draw along the opposite edge. This draws a line that is a ruler width away from the line that joins the two points without actually drawing the line that joins the two points – you don’t have to draw the line to know where it is.

Draw a line a ruler width away from both of two points
0. Start with two existing points.
1. Align both points on the same side of the ruler, and draw along the other side.
Done! The line just drawn is parallel to the line joining the two points.
Video here

There is a second way to use two existing points to draw lines using the two-sided ruler, and that’s to align the two points on opposite sides of the ruler.

(Of course, this can only be possible if the points are at least a ruler width apart or you won’t be able to fit the ruler between them. For points aligned on the same side of the ruler, it doesn’t matter how far apart they are.)

This move of aligning one point on one side of the ruler and the other point on the other side of the ruler is so common in the later constructions, it needs its own name. I will call it cross-aligning, as in, I’ll write an instruction like, “Cross-align these two points.” I’ll also talk about cross-aligning a line segment, by which I mean cross-aligning the endpoints of the line segment.

While I’m at it, I’ll call the action of aligning two points on the same side of the ruler, or aligning a line along one side of the ruler, side-aligning.

I’m going to need that saving in word count because there are two ways to both side-align and cross-align and there will be times I’ll have to use up some words to specify which I want.

You might not believe there’s only two ways to cross-align two points, thinking surely you can fit the ruler in there in lots of ways. I urge you to give it a try and you will soon believe it, just like I did. On top of that, I can prove it.

Proof:

Let the width of the ruler be \(1\), and consider two points \(A\) and \(B\) a distance of \(d\) apart with \(d > 1\). Cross-align \(A\) and \(B\) in one direction and draw along both sides of the ruler. Imagine a line from \(B\) perpendicular to both sides of the ruler, meeting the opposite side in the point \(C\). Let the angle \(\angle CAB\) be \(\theta\).

The triangle \(\triangle ABC\) is a right-angled triangle, and the side opposite the angle marked \(\theta\) is \(1\) since it’s the ruler width, while its hypotenuse is \(d\). Therefore \(\sin(\theta)=\frac{1}{d}\).

End proof!

At the most basic level, this means that given a specific distance between two points, cross-aligning them will always produce a specific angle between the edge of the ruler and the line joining the points. And it works the other way too: given a specific angle between the edge of the ruler and another line, the ruler-edges will always cut off the same length segment of line between them.

More precisely what this means is that the two-sided ruler can calculate sine and arcsine. (Technically it’s cosecant and arccosecant, but that’s much too hard to say.)

Anyway, that’s my Lemma.

The cross-align arcsine lemma.

Let the width of the ruler be \(1\).

When two points that are \(d > 1\) apart are cross-aligned, the acute angle that the ruler edges make with the line joining the points is \[\arcsin\left(\frac{1}{d}\right)\]

When the sides of the ruler meet a line in an acute angle \(\theta\), then the distance between the points where the line meets the sides of the ruler is \[\frac{1}{\sin(\theta)}\].

There’s video explaining the lemma and its proof here.

I’ll be using the full trigonometricality of it in the last two blog posts, but for now the most important thing to remember is that cross-aligning a specific length always produces a specific angle, and vice versa.

Right now, I can do the first traditional construction, which is to double a length, at least when the length is longer than the ruler width.

Double a line segment (longer than ruler width)
0. Start with a line segment longer than the ruler width.
1. Cross-align the opposite ends of the line segment, and draw along one side of the ruler.
2. Side-align the line you just drew, and draw along the opposite side of the ruler to produce a parallel line.
3. Finally, side-align the original line segment and draw along that side of the ruler to extend the segment to where it meets the parallel line you just drew.
Done! This extended segment is twice as long as the original.
Video here

This procedure works because of the cross-align arcsine lemma, since I know a certain angle will always cut off a certain length. It’s in Wernick, but I thought of it before reading that paper, and he doesn’t use my lemma to prove it.

You can repeat the procedure as many times as you like to multiply the length of a line segment by any natural number, though perhaps you’ll want to extend the line segment first rather than second. It will also produce as many points as you want on an existing line all an equal distance apart by just putting your ruler anywhere to make the first segment.

I can also do another traditional construction, which is to double an (acute) angle.

Double an angle
0. Start with two lines meeting to make an acute angle.
1. Side-align one arm of the angle and draw along the opposite side of the ruler. You only need enough of this line to see where it meets the other arm of the angle.
2. This point just made and the vertex of the angle have already been cross-aligned. Cross-align them in the other direction and draw along the side of the ruler through the vertex.
Done! This line just drawn and the nearest arm make the same angle as the original.
Video here

This works because whenever you cross-align any line segment you always get the same angle, due to the cross-align arcsine lemma.

I find it very interesting that neither Wernick nor Birrell contain this construction. Instead, Wernick has a much more complicated construction built from making right angles and doubling lengths, both of which take quite a few lines to draw. I think it’s because they were excited about rhombuses, which on reflection is not that surprising, because rhombuses are pretty cool.

The next blog post is all about rhombuses.