Two-sided ruler constructions 3: Rhombuses

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

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

This blog post is all about rhombuses and what they can help us do and a little about what we don’t need them to help us do.

The first thing you can do is just quickly draw a rhombus.

Draw a rhombus anywhere
1. Put the ruler anywhere and draw along both sides.
2. Angle the ruler another way and draw along both sides.
Done! The shape made between both pairs of lines is a rhombus.
Video here

Proof:

It’s definitely a parallelogram, since opposite sides are parallel. This means opposite sides are the same length.

To be a rhombus, adjacent sides need to be the same length too. There are several ways to be sure of this, but I will use the cross-align arcsine lemma.

Draw a diagonal of that parallelogram.

The ends of this line segment have been cross-aligned in both directions, which means the angles where the edges of the parallelogram meets the diagonal are the same. This means that the diagonal cuts the parallelogram into two isosceles triangles, whose outside edges are equal.

End proof!

Seeing the rhombus inside those longer parallel lines immediately makes me want to extend this to a full rhombus tessellation.

Draw a rhombus tessellation
1. Draw a rhombus. While doing so, make the lines that define the rhombus as long as possible.
2. Side-align the lines used to make the rhombus and draw along the other side of the ruler to make parallel lines. Then repeat for the lines you just drew and so on.
Done! You’ve made a rhombus tessellation.
Video here

You can also put a rhombus in a specific spot, rather than just anywhere. You can draw one with a specific diagonal, with a specific angle and with a specific side. These constructions are implicitly part of other constructions in Wernick and Birrell, though not explicitly stated. I think they deserve to have their moment in the limelight.

Draw a rhombus with a specific diagonal
0. Start with a line segment longer than a ruler width.
1. Cross-align the ends of the line segment and draw along both sides of the ruler.
2. Cross-align the ends of the line segment in the other direction and draw along both sides of the ruler.
Done! The shape made between both pairs of lines is a rhombus with the original line segment as the diagonal.
Video here

Draw a rhombus with a specific angle
0. Start with two lines meeting to make an angle.
1. Side-align one arm of the angle and draw along the opposite side of the ruler.
2. Side-align the other arm of the angle and draw along the opposite side of the ruler.
Done! The shape made between both pairs of lines is a rhombus with the original angle as one of its angles.
Video here

Draw a rhombus with a specific side
0. Start with a line segment longer than the ruler width.
1. Cross-align the ends of the line segment and draw along both sides of the ruler.
2. Side-align the original line segment and draw along the opposite side of the ruler.
Done! The shape made between both pairs of lines is a rhombus with the original line segment as one of its sides.
Video here

Note that because of the cross-align arcsine lemma, the shape of each of these rhombuses is completely decided by the specific element you use. For example, you can’t just make any rhombus with a specific angle \(\theta\), but you have to make one where the side length is \(\frac{1}{\sin(\theta)}\). Indeed, there is precisely one rhombus you can draw this way with a specific angle or a specific diagonal. With the side, there are four different rhombuses you can draw, since there are two directions you can cross-align the ends of the line segment and two sides you can side-align on. They’re all congruent though.

Rhombuses have some very cool and useful properties. I’m not going to prove them, just list them. You can find the proofs in any number of places, or prove them yourself.

• All four sides are the same length. (That’s the definition of a rhombus.)
  
• Opposite angles are equal. (This is true for all parallelograms, not just rhombuses.)
  
• The diagonals bisect each other. (This is true for all parallelograms, not just rhombuses.)
  
• The diagonals meet at right angles. (This is not true of all parallelograms, but it is true for kites, and a rhombus is a kite.)
  
• The diagonals bisect the angles. (This is not true of all parallelograms and not true of all kites either.)
  

(You can watch me listing and drawing them in this video.)

The properties of rhombuses above mean the two-sided ruler can do some cool stuff that is traditional fare for ruler and compass constructions.

Draw a right angle anywhere (using a rhombus)
1. Draw a rhombus anywhere as described earlier.
2. Draw the lines joining the opposite corners of the rhombus.
Done! The two diagonals you just drew are at right angles to each other.
Video here

This construction is in both Wernick and Birrell, and it is definitely the easiest way to make a right angle without having to concentrate very hard. It takes six lines: four for the rhombus and two for the diagonals.

I have discovered another way to make a right angle somewhere that also requires drawing six lines. It’s a little harder to describe than the one with the full rhombus, but I still like it.

Draw a right angle anywhere (using a triangle)
1. Draw a line anywhere.
2. Angle the ruler in a new direction and draw along both sides to produce two parallel lines that both meet the existing line. For ease of later description, I’ll call by \(A\) and \(B\) the points where the two parallel lines meet the first line.
3. The points \(A\) and \(B\) were cross-aligned at step 2. Cross-align them in the other direction and draw along the side of the ruler through \(B\).
4. Side-align the line you just drew so that the other side of the ruler is on the far side of \(B\) relative to \(A\), and draw along the far side of the ruler. Call by \(C\) the point where this line meets the original line and call by \(X\) the point where it meets the other line through \(A\).
5. Draw the line through \(X\) and \(B\).
Done! The angle \(\angle ABX\) is a right angle.
Video here

Obviously you don’t have to actually label the points with letters. They are just there to make it easier to describe in text form what line to draw (and to describe triangles and lengths in the proof later).

Proof:

At step 3, you followed the procedure to double the segment \(AB\), so that \(AB = BC\). Hence, when those segments are cross-aligned, the sides of the ruler meet them at the same angle. That means \(\angle XBA\) is the same as \(\angle XCB\), making \(\triangle AXC\) an isosceles triangle.

The line joining the vertex of an isosceles triangle to the midpoint of its base always meets the base at a right angle.

Note I could have proved this directly by noting that triangles \(\triangle AXB\) and \(\triangle CXB\) are congruent, so that \(\angle ABX = \angle CBX\).

End proof!

I really like how this method is based not on a rhombus but on an isosceles triangle. Though I do admit there is a rhombus there if I drew my lines long enough: the rhombus with diagonal \(BX\). But the right angle is at one of the vertices of the rhombus rather than at the centre. Alternatively I could see the isosceles triangle as half a rhombus: if I had drawn the other side of the ruler at step 3, and cross-aligned \(BC\) in the other direction too, then I could have drawn a big rhombus whose diagonals would meet in the right angle I made. But still, the bit I liked was that I directly used the idea that cross-aligning the same length always produces the same angle, which is actually more fundamental than making a rhombus.

I do want to note that I could have done steps 3 and 4 the other way around. That is, I could have drawn a third parallel line to find point \(C\) and then cross-aligned \(BC\) to draw the other side of the isosceles triangle. They take the same amount of lines and require the same amount of cross-aligning. I just happen to like the symmetry of the final diagram that this version produces.

For all that I love it, I still find the one based on making a rhombus first easier to do, and it does use the same number of lines, and tends to take up less space, so in later constructions, the rhombus is how you’ll see me making right angles. I just wanted to have both here for comparison. Also it clearly makes the point that there is more than one way to do a thing, which is something that is important to me about all mathematics and so is a point I will make more as I go. Two-sided ruler constructions seem to be able to make that point rather well.

Bisect a line segment (longer than ruler width)
0. Start with a line segment longer than the ruler width.
1. Draw a rhombus with the line segment as its diagonal.
2. Draw the other diagonal of the rhombus.
Done! The second diagonal bisects the line segment.
Video here

Note this is actually better than simply bisecting the line segment, because this process actually draws a perpendicular bisector, a point which Wernick makes a big deal of when he presents this process.

Bisect an angle
0. Start with two lines meeting to make an angle.
1. Draw a rhombus with the angle as one of its angles.
2. Draw the diagonal of the rhombus that passes through the vertex of the angle.
Done! This diagonal bisects the angle.
Video here

Again, this is presented in Wernick and it makes a heap of sense given the propensity of the two-sided ruler to make rhombuses. However I’ve found an alternative that draws lines on the outside of the angle instead, which has a certain charm for me.

Bisect an angle (from the outside)
0. Start with two lines meeting to make an angle. Call the vertex of the angle \(A\).
1. Side-align one arm of the angle so that the ruler is on the outside of the angle and draw along the opposite side of the ruler.
2. Side-align the other arm of the angle so that the ruler is on the outside of the angle and draw along the opposite side of the ruler. You only need to draw enough of this line to find where it meets the first line you drew. Call this point \(B\)
3. Draw the line through \(B\) and \(A\).
Done! The line you just drew bisects the angle.
Video here

Proof:

Call the original angle’s vertex \(A\). The line segment \(AB\) has already been cross-aligned in both directions. Thus the two other lines through \(B\) meet \(AB\) in the same angle.

Since the arms of the original angle are parallel to these lines, they also meet \(AB\) in the same angle.

End proof!

Here we have another case of being able to prove something from the cross-align arcsine lemma, rather than referring to a rhombus. Thinking on it, I could have proven the other way to bisect an angle in the same way, noting that there the diagonal of the rhombus has been cross-algined in both directions too.

So, somehow, for all that I said rhombuses are very cool, the moral of this section seems to be that they are not nearly as cool to me as the cross-align arcsine lemma.

At this stage, we have all the tools necessary to draw a square, an equilateral triangle, and a regular pentagon. However, I think these are so cool and I am so proud of them that I am saving them for the last blog post.

The next blog post is about copying line segments and cutting them into parts, something that until now we’ve only been able to do with line segments if they’re long enough.