This blog post is about processes I have figured out for drawing an equilateral triangle and a regular pentagon. I am very proud of them and saved them for last, even though there was enough information in the first three posts to achieve these constructions.
First, I will construct a square, even though I haven’t promised to, because it’s very quick.
Draw a square
1. Draw a rhombus anywhere.
2. Draw the lines joining the opposite corners of the rhombus.
3. Side-align both of the lines just drawn and draw along the opposite sides of the ruler both times.
Done! The four lines drawn at steps 2 and 3 are the sides of a square.
It’s a rhombus, since it was made by overlapping two pairs of parallel lines a ruler-width apart. And it’s a square because it has a right-angle.
End proof!
Next, I’ll do the equilateral triangle. With ruler and compass constructions it takes three lines and two circle arcs to draw an equilateral triangle, and if you follow the same procedure using the two-sided ruler it will take twenty to thirty lines to do the job. But you don’t have to follow the ruler and compass construction. The two-sided ruler can do it more directly. I’ll discuss some properties of the triangle to lead up to it.
The three angles of an equilateral triangle are each \(60^\circ\), and if you bisect one of those angles, you will create two right-angled triangles whose other two angles are \(30^\circ\) and \(60^\circ\). Since the shortest side of those triangles is half the triangle side, that means the shortest side is half the hypotenuse, making \(\sin(30^\circ) = \frac{1}{2}\).
By the cross-align arcsine lemma, this means that if we cross-align a line segment of length \(2\), then the sides of the ruler will meet the line segment in an angle of \(30^\circ\).
As to drawing a line segment of length \(2\), I need a right angle to be sure I’m measuring \(1\) correctly. (Recall that \(1\) was declared to be the ruler width.)
All of this is the explanation for the following construction.
Draw an equilateral triangle
1. Draw a rhombus anywhere. Call the four vertices \(P\), \(Q\), \(R\) and \(S\) in cyclic order.
2. Draw one diagonal of the rhombus, say \(PR\).
3. Side-align the other two vertices \(Q\) and \(S\) and draw along the other side of the ruler. Call by \(Y\) the point where this line meets \(PR\).
4. Side-align \(Q\) and \(S\) on the other side, and draw along the other side of the ruler. You only need enough of this line to see where it meets \(PR\). Call this meeting point \(A\).
5. Cross-align \(A\) and \(Y\) in both directions and draw along the side of the ruler through \(A\) each time. Call by \(B\) and \(C\) the points where these lines meet the line drawn at step 3.
Done! The triangle \(\triangle ABC\) is an equilateral triangle.
The lines \(PQ\) and \(QS\) are perpendicular since they are the diagonals of a rhombus. The distance from \(QS\) to \(Y\) along the perpendicular line \(PQ\) is \(1\) (the ruler width). Similarly the distance from \(QS\) to \(A\) is also \(1\). Hence the length of segment \(AY\) is \(2\).
By the cross-align arcsine lemma, the lines \(AB\) and \(AC\) meet \(AY\) in the angle \(\arcsin\left(\frac{1}{2}\right) = 30^\circ\). Hence \(\angle BAC\) is twice this, which is \(60^\circ\).
The angle \(\angle AYC\) is a right angle since \(BC\) is parallel to \(QS\) which is perpendicular to \(AY\). Hence \(\angle ABY\) is \(60^\circ\). Similarly \(\angle ACY = 60^\circ\).
Therefore all angles of \(\triangle ABC\) are \(60^\circ\) and it’s an equilateral triangle.
End proof!
I think my favourite part of this is that you don’t have to actually draw \(QS\) to do this construction. You can draw \(QS\) if you want, and if you do, there will be a smaller equilateral triangle whose height is \(1\) instead of \(2\).
This smaller triangle makes it possible to fill out the entire page with a triangular grid, also known as an isometric grid.
Draw an isometric grid
1. Draw a rhombus anywhere (though close to the corner of the page is a good place).
2. Draw both diagonals of the rhombus. One diagonal only needs to be just a little longer than a ruler width in each direction. The other diagonal should be as long as possible.
3. Side-align the long diagonal on both sides and draw along the other side of the ruler each time. These lines should be as long as possible. Take note of where these lines meet the short diagonal.
4. Cross-align the two points just mentioned in both directions and draw along both sides of the ruler each time as long as possible. There are now five equilateral triangles between seven long lines.
5. Side-align every line drawn at step 3 and 4 and draw along the other side of the ruler each time to create parallel lines, and continue doing this to the new lines drawn until the grid is as large as you need. (Once there are enough lines on the grid, you can start aligning their intersection points rather than aligning existing lines.)
Done! Unless you want the triangles to be smaller…
6. Choose two adjacent parallel lines, and choose two rhombuses between them, each rhombus made from two adjacent equilateral triangles. One diagonal of each rhombus is drawn already. Draw the other diagonal of each. You only need enough of these lines to see where they cross the other diagonal.
7. Join the two points created at step 6 to make the line halfway between the parallel lines.
8. Side-align this line and continue to side-align the new lines, to draw more halfway lines in that same direction.
9. Connect the intersection points of the new lines and the existing lines to complete the grid.
There is something very meditative about drawing an isometric grid this way, watching the grid appear as you draw more and more lines. I particularly love when there are enough intersection points to start aligning the ruler to them and drawing along both sides.
I am now up to my favourite construction of all: the regular pentagon. Before I do it, I’ll discuss some properties of the regular pentagon.
The total angle sum of any pentagon is \(3 \times 180^\circ = 540^\circ\). In a regular pentagon, all the angles are the same, which means that each angle is \(540^\circ \div 5 = 108^\circ\).
Consider a regular pentagon \(JKLMN\).
Draw two of the diagonals \(JL\) and \(JM\).
Triangles \(\triangle JKL\) and \(\triangle JNM\) are isosceles with one angle \(108^\circ\), and so their other angles are \((180-108)\div 2 = 36^\circ\).
This means that \(\angle LJM = 108 – 2\times 36 = 36^\circ\). I have to say this surprises me every time that those two diagonals perfectly trisect the angle of the pentagon.
The other two angles of \(\triangle LJM\) are \(108-36=72^\circ\).
Draw the diagonal \(KN\), meeting \(JL\) in \(A\) and \(JM\) in \(B\).
The angles \(\angle JAB\) and \(\angle JBA\) are both \(72^\circ\) because \(KN\) is parallel to \(LM\), and \(\angle JNA = 36^\circ\) by the same arguments as earlier. Finally \(\angle AJN = 72^\circ\) since it’s two \(36^\circ\) angles.
This means that \(\triangle JNA\) and \(\triangle AJB\) are similar. Let \(AB\) be length \(a\), \(AJ\) be length \(b\) and \(JN\) be length \(c\). Because \(\triangle AJB\) is isosceles, \(JB = b\), and because \(\triangle JBN\) is isosceles, \(BN = b\) also. Finally because \(\triangle JNA\) is isosceles, \(AN = c \) as well.
Note that \[a+b = c\].
Also by similar triangles, \[\begin{aligned}\frac{b}{c} &= \frac{a}{b}\\ b^2 &= ac\end{aligned}\]
Now draw the perpendicular bisector of \(AB\), meeting \(AB\) in \(C\). Consider the triangle \(\triangle JCB\).
The angle \(\angle CJB\) is \(18^\circ\), since it’s half of \(36^\circ\), and the side opposite this is \(\frac{a}{2}\), and the hypotenuse is \(\frac{a}{2}\left(1+\sqrt{5}\right)\). Therefore \[\sin(18^\circ)=\frac{1}{1+\sqrt{5}}\]
Hence, by the cross-align arcsine lemma, to make an angle of \(18^\circ\), I need to cross-align a length of \(1+\sqrt{5}\). To make a length of \(\sqrt{5}\), I can use the diagonal of a \(1\) by \(2\) rectangle, since that will be \(\sqrt{1^2+2^2}=\sqrt{5}\) by Pythagoras’s Theorem.
All of this is explanation for the following construction.
Draw a regular pentagon
1. Draw a rhombus anywhere. Call the four vertices \(P\), \(Q\), \(R\) and \(S\) in cyclic order.
2. Draw both diagonals the rhombus.
3. Side-align one diagonal, say \(QS\), and draw along the other side of the ruler. You only need enough of this line to see where it meets the other diagonal \(PR\). Call this meeting point \(X\).
4. Side-align \(PR\) and draw along the other side of the ruler. You will need this line to meet \(QS\) and be quite long on the opposite side of \(QS\) to \(X\). Call by \(Y\) the point where this line meets \(QS\).
5. Side-align \(QS\) on the opposite side to before, and draw along the other side of the ruler. You only need enough of this line to see where it meets the line through \(Y\) drawn at step 4. Call this meeting point \(Z\).
6. Side-align \(X\) and \(Z\) so that the ruler is on the other side of \(XZ\) from \(Y\) and draw along the other side of the ruler. You only need enough of this line to see where it meets \(YZ\). Call this meeting point \(J\).
7. Cross-align \(Y\) and \(J\) in both directions and draw along both sides of the ruler each time. This creates a rhombus with \(YJ\) as one diagonal. Call the other two vertices of the rhombus \(L\) and \(M\).
8. Draw the line through \(L\) and \(M\).
9. Side-align \(LM\) so that the ruler is on the same side of \(LM\) as \(J\) and draw along the other side of the ruler. Call by \(K\) where this line meets \(YL\) and by \(N\) where this line meets \(YM\).
If the lines drawn during steps 2 to 5 were extended long enough, they would make two squares with side length \(1\) by the construction of the square above. This means the line segment \(XZ\) is the diagonal of a rectangle with side lengths \(1\) and \(2\) and so has length \(\sqrt{5}\).
At step 6, the two sides of the ruler and the two lines \(PX\) and \(YZ\) form a rhombus (with diagonal \(XJ\)), and so \(JZ\) is the same length as \(XZ\), which is \(\sqrt{5}\). Therefore \(YJ\) has length \(1+\sqrt{5}\).
By the cross-align arcsine lemma, the four lines drawn at step 7 all meet \(YJ\) in the angle \(\arcsin\left(\frac{1}{1+\sqrt{5}}\right)=18^\circ\).
This makes \(\angle MJL = 36^\circ\) and since \(\triangle LJM\) is half a rhombus, it’s isosceles and therefore its other two angles are \(72^\circ\). By alternate angles in parallel lines, \(\angle NMJ = 36^\circ\) and so \(\angle NML = 108^\circ\). Similarly \(\angle KLM = 108^\circ\).
The two lines \(LM\) and \(KN\) are opposite sides of the ruler, as are \(MN\) and \(LJ\). Therefore they enclose a rhombus and so \(LM\) and \(MN\) are the same length. Similarly \(ML\) and \(LK\) are the same length.
Since they meet at \(108^\circ\), the three line segments \(KL\), \(LM\) and \(MN\) are three sides of a regular pentagon.
The triangle \(LJM\) has the correct angles to be the triangle joining one side of a regular pentagon to the opposite vertex, and so \(J\) must be the final vertex of the regular pentagon with the three sides \(KL\), \(LM\) and \(MN\).
End proof!
Every time I do this, it momentarily surprises me once again when the regular pentagon appears seemingly out of nowhere. I just can’t express how cool it is and how proud I am of figuring this out.
I think my favourite part of this construction is actually step 6. There, I do the move necessary to draw the rhombus with angle \(\angle XZJ\) without actually drawing the rhombus or indeed the angle. But you don’t have to draw the rhombus to know that it’s there.
My original version of this construction used the fact that angles \(\angle YLM\) and \(\angle JLM\) are \(72^\circ\). I copied the angles at the point \(L\) to make the five triangles joining the pentagon’s sides to the centre. It wasn’t until quite a bit later I realised I already had three vertices of a rhombus all along.
It’s worth noting that the regular pentagon construction already contains all of the lines needed to draw a big five-pointed star, if you only draw them long enough. I think is cool enough to deserve its own construction title.
Draw a big equilateral five-pointed star
1 to 10. Draw the construction above for a regular pentagon, but make sure the the lines \(YL\) and \(YM\) at step 7, the line \(ML\) at step 8, and the lines \(JN\) and \(JK\) at step 10 are long enough to meet each other.
Done! The five lines just described form an equilateral five-pointed star.
Actually, maybe just one more because you can draw a small five-pointed star with one fewer line than the pentagon.
Draw a small equilateral five-pointed star
1 to 7. Draw the construction above for a regular pentagon up to step 7.
8. Side-align \(L\) and \(M\) so that the ruler is on the same side as \(J\) and draw along the other side of the ruler. Call by \(K\) the point where this line meets \(YL\) and by \(N\) the point where this line meets \(YM\).
9. Draw the lines \(KM\) and \(LN\).
Done! There is now an equilateral five-pointed star with outermost points \(J\), \(K\), \(L\), \(M\) and \(N\).
I do like how there’s a tiny pentagon in the middle of this star, so technically you can draw a regular pentagon with one fewer line than the construction I made earlier. There’s another pentagon you can draw a bit bigger by side-aligning the five lines involved in that star with the ruler outside the star. (That’s how I got the shape of the pentagon for the trigonometry calculations earlier.)
I think that will do. You’ve finally seen the constructions I’m most proud of, and in the previous blog posts (almost) all of my thinking about two-sided ruler constructions from the last couple of months. And I’ve given myself some stars.
This blog post is about the final two constructions that prove two-sided ruler constructions are capable of making any point you can make with ruler and compass constructions. Obviously you can’t draw the curve of a circle without a compass, but you can find any specific point on a circle anyone asks for. In particular, if someone tells you where the centre of a circle is and one point on the circumference so you know its radius, you can find where that circle meets a specific line, and where it meets another circle similarly defined.
Wernick and Birrell both present the constructions in their writing, calling them “intersection criteria” in the sense that these criteria must be satisfied in order to be sure the two-sided ruler is sufficient to do all that the ruler and compass can do. They are complicated constructions and I have to be honest and say I don’t really understand them very well, especially the one for the intersection of two circles. In particular, several of the arguments in the proofs are very obscure to me.
What I’ve done instead is find my own ways of doing the constructions and proving them that do make sense to me. I reckon this is what Birrell did with hers too. Her circle-meets-line construction is different to Wernick’s, and while her circle-meets-circle construction is identical, the proof is very different. I think she made sense of them in her own way, which is only right.
Unfortunatly they still didn’t make sense to me, so what follows are my own constructions proved in my own way.
First up, the places where a line meets a circle.
Consider a circle with centre \(C\) and radius \(r\), and a line a distance of \(d\) away from the centre of the circle. You’ll definitely need \(d\) to be shorter than \(r\) or the line won’t meet the circle at all. But if it is short enough, then you’ll be able to make two right-angled triangles with hypotenuse \(r\) and short side \(d\).
Let \(\theta\) be the angle between the radius and the line. Then \(\sin(\theta) = \frac{d}{r}\). By the cross-align arcsine lemma, in order to produce such an angle, you need to make a length of \(\frac{r}{d}\) and cross-align its ends.
If you draw a line parallel to that original line but through \(C\), the hypotenuse-radiuses meet this line at the same angle as the original line.
If we can make a distance of \(\frac{r}{d}\) along that line starting at \(C\) and cross-align it, then we’ll be drawing the exact angle we need in the exact place we need it.
But that means we somehow need to do division of specific lengths. We’ve done it with whole numbers, but we want to divide by a possibly irrational length \(d\). Is that possible? Yes, using parallel lines.
Consider a triangle \(\triangle APB\) with another line parallel to \(AB\) meeting the two sides \(PA\) and \(PB\) at \(Q\) and \(S\) respectively.
Then \(\triangle QPB\) is similar to \(\triangle APB\) and so matching sides are in propotion:
\[\frac{QP}{AP}=\frac{SP}{BP}\]
If \(AP=1\) (that is, the ruler width), then we have
\[QP =\frac{SP}{BP}\]
Lo and behold, we have managed to do division!
Note you can rearrange that proportional relationship like this:
\[\frac{QP}{SP}=\frac{AP}{BP}\].
Now if \(SP=1\) then
\[QP =\frac{AP}{BP}\]
Again, we have managed to do division.
Note that in both cases, the result of the division had to be on the opposite arm of the angle from the denominator of the fraction.
Also note that in order to measure a distance of exactly one ruler width, you need to have a right angle, since to measure the width of the ruler you need to travel along a line that is at a right angle to the side of the ruler.
Therefore, to effectively do the division we want with the two-sided ruler, we can do one of the following constructions.
Mark the length r/d given the lengths r and d measured along the same arm of a right angle
0. Start with a right angle with the lengths \(r\) and \(d\) measured from the vertex along the same arm of the angle. Call the vertex of the angle \(C\) and the ends of those lengths \(R\) and \(D\) respectively.
1. Side-align the arm of the angle with \(R\) with the ruler inside the angle and draw along the other side of the ruler. Call by \(U\) the point where this line meets the arm of the right angle not through \(R\).
2. Draw the line through \(D\) and \(U\).
3. Draw the line through \(R\) that is parallel to \(DU\) so that it meets the other arm of the angle. Call this meeting point \(Q\).
Done! The segment \(CQ\) is of length \(\frac{r}{d}\).
Mark the length r/d given the lengths r and d measured along different arms of a right angle
0. Start with a right angle with the lengths \(r\) and \(d\) measured from the vertex along the different arms of the angle. Call the vertex of the angle \(C\) and the ends of those lengths \(R\) and \(D\) respectively.
1. Side-align the arm of the angle with \(R\) with the ruler inside the angle and draw along the other side of the ruler. Call by \(U\) the point where this line meets the arm of the right angle not through \(R\).
2. Draw the line through \(D\) and \(R\).
3. Draw the line through \(U\) that is parallel to \(DR\) so that it meets the other arm of the angle. Call this meeting point \(Q\).
Done! The segment \(CQ\) is of length \(\frac{r}{d}\).
You may notice in both of these constructions, that parallel line and the lines used to construct it were very close together, which can make it hard to do this accurately with physical tools. Only while doing the videos for later constructions in this post did I realise you can arrange everything to be more spread out by putting the point \(U\) on the other side of the angle vertex. I’ll only show you for the case where \(r\) and \(d\) are on different arms of the angle, because that’s the situation when I use it later.
Mark the length r/d given the lengths r and d measured along different perpendicular lines
0. Start with two perpendicular lines with the lengths \(r\) and \(d\) measured from their meeting along point different lines. Call the vertex of the angle \(C\) and the ends of those lengths \(R\) and \(D\) respectively.
1. Side-align the line with \(R\) with the ruler on the opposite side from \(D\) and draw along the other side of the ruler. Call by \(U\) the point where this line meets the original line through \(D\).
2. Draw the line through \(D\) and \(R\).
3. Draw the line through \(U\) that is parallel to \(DR\) so that it meets the other original line. Call this meeting point \(Q\).
Done! The segment \(CQ\) is of length \(\frac{r}{d}\).
In the specific case of using this to find where a circle meets a line, we already have perpendicular lines to \(CD\) through both \(C\) and \(D\), and in that case, there’s a neat shortcut.
One of the ways to make a line parallel to an existing line from the previous blog post is to use three equally spaced points on the line and draw lines from a central point through all three of them. In the first construction above, we already have lines from \(C\) through \(D\) and \(U\). All we need on top of this is to bisect \(DU\) with a line through \(C\).
But if we already have a line through \(D\) perpendicular to \(CD\), then when we find \(U\) we automatically make a rectangle which has \(DU\) as a diagonal. And the diagonals of a rectangle bisect each other! So we just have to draw the other diagonal to make the bisecting line we want.
Mark the length r/d given the lengths r and d measured along the same arm of a right angle (using a rectangle)
0. Start with a right angle with the lengths \(r\) and \(d\) measured from the vertex along the same arm of the angle. Call the vertex of the angle \(C\) and the ends of those lengths \(R\) and \(D\) respectively. Also suppose there is a line through \(D\) perpendicular to \(CD\).
1. Side-align the arm of the angle with \(R\) and draw along the other side of the ruler. Call by \(U\) the point where this line meets the arm of the angle not through \(R\), and call by \(Y\) the point where it meets the line through \(D\) that is pependicular to \(CD\).
2. Draw the line through \(C\) and \(Y\).
3. Draw the line through \(R\) and \(U\). You only need enough of this line to find where it meets the line \(CY\). Call this meeting point \(X\).
4. Draw the line through \(D\) and \(X\). You only need enough of this line to find where it meets the line \(CR\). Call this meeting point \(Q\).
Done! The segment \(CQ\) is of length \(\frac{r}{d}\).
Now we have everything we need to find where a circle meets a line. If the centre of the circle is already on the line, then I just need to copy the length to the other side of some angles, and if I know one point where the line meets the circle, then I can make a perpendicular and copy a length or an angle angle to find the other one. The trickiest situation is when the centre of the circle and the point defining the radius of the circle are both off the line.
I am not going to describe every single line you need to draw to do this construction. Instead I am just going to string together earlier constructions. If you do the whole procedure by hand, then it’s possible to reuse lines from earlier parts of the procedure, but I am not going to explicitly tell when to do that. You can see me doing it in the video though.
Find the points where a line meets a circle defined by its centre and one point on its circumfernce
0. Start with a line, and two points not on the line. Choose one point to be the centre of a circle and the other to be a point on its circumference. Call the centre of the circle \(C\) and the other point \(S\).
1. Draw the line through \(C\) and \(S\). Call by \(r\) the distance \(CS\).
2. Draw lines through \(C\) parallel and perpendicular to the existing line. Call by \(D\) where the perpendicular line meets the existing line and call by \(d\) the length \(CD\).
3. Copy the length \(CS\) from one arm of the angle \(\angle SCD\) to the other. Call by \(R\) the resulting point on \(CD\). Note that if \(R\) is between \(C\) and \(D\), this means the radius \(r\) is less than the distance \(d\) to the line and so the circle does not meet the line and you can stop the procedure.
4. Mark the length \(\frac{r}{d}\) along the line just drawn at step 4 starting at \(C\). Call the end of that length \(Q\).
5. Cross-align \(C\) and \(Q\) in both directions and draw along the side of the ruler through \(C\) both times. Call the places where those lines meet the original line \(A\) and \(B\)
Done! The points \(A\) and \(B\) are where the circle with centre \(C\) and radius \(CS\) meets the original line.
Note that instead of copying the radius onto the line \(CD\) you can copy it onto the line through \(C\) parallel to the original line and use the procedure for finding \(\frac{r}{d}\) when \(r\) and \(d\) are on different arms of a right angle. However, I like putting \(r\) and \(d\) on the same line so that I can tell if the circle meets the line at all.
If you tally up all the lines I drew in the video above, you will get a total of seventeen lines, which I have to say is pretty good, considering just doing the perpendicular and parallel line through \(C\) uses nine.
Now it’s time to tackle finding the places where two circles meet.
Consider two circles with centres \(A\) and \(B\) that are \(c\) apart, and with radiuses \(a\) and \(b\) respectively. Suppose that the two circles meet in two points \(G\) and \(H\).
The quadrilateral \(AGBH\) is a kite since the sides \(AG\) and \(AH\) are both the radius \(a\) and the sides \(BG\) and \(BH\) are both the radius \(b\). That means its diagonals \(AB\) and \(GH\) meet at right angles. Let’s call that meeting point \(X\).
If we could locate the point \(X\) and then draw a perpendicular to \(AB\) through it, then the problem of finding \(G\) and \(H\) becomes the problem of finding where the circle with centre \(A\) and radius \(a\) (or centre \(B\) and radius \(b\)) meets that line.
Note the kite \(AGBH\) may actually be a dart with a reflex angle at \(A\) or \(B\) if the centre of the smaller radius circle is inside the other circle.
(It may even be actually a triangle if \(B\) happens to be exactly in the right spot for \(GH\) to pass through it.)
Either way, the angle at the centre of the larger radius circle will always be less than 180°. We’ll call the larger radius \(a\) and the smaller radius \(b\), with matching centres \(A\) and \(B\) from now on.
Consider the triangle \(\triangle AGB\). If \(AB\) is the base, then the height of this triangle is \(GX\). Call the angle \(\angle GAB\) by \(\beta\). Also call the distance \(AX\) by \(x\).
So the position of \(X\) is half the distance between the two circles \(\frac{c}{2}\), plus the distance \(\frac{(a+b)(a-b)}{2c}\).
I worked on and off for several weeks, trying to find a way to create that distance and I finally came up with a way which I have to say I am extremely proud of.
Consider this diagram:
The vertical line segment \(AB\) on the side is \(c\) long and has midpoint \(C\).
There’s a horizontal line segment \(AP\) that is \(b\) long and a horizontal line segment \(BQ\) that is \(a\) long.
There’s a vertical (dotted) line from \(A\) meeting the line \(BQ\) in \(M\).
There’s a horizontal (dotted) line at \(C\) meeting the line \(PQ\) in the point \(Y\).
There is a line through \(Y\) perpendicular to \(PQ\) meeting \(AB\) in the point \(X\).
The point \(X\) is a distance of \(\frac{a^2-b^2}{2c}\) from \(C\).
Proof:
Since \(CY\) is positioned halfway from \(A\) to \(B\), the length of \(CY\) is exactly halfway between the lengths of \(AP\) and \(BQ\) and so its length is \(\frac{a+b}{2}\).
The length of \(QM\) is \(a-b\), and so the slope of the line \(PQ\) is \(\frac{c}{a-b}\). That is, for every \((a-b)\) it goes across to the right, it goes up \(c\).
The line \(YX\) is perpendicular to \(PQ\), so its slope is \(-\frac{a-b}{c}\). That is, for every \(c\) across to the right, it goes down \((a-b)\).
From \(Y\) to \(X\) the line \(YX\) travels across to the right \(\frac{a+b}{2}\), which means it must go down that much times the slope, which is xh ti is \(\frac{a+b}{2}\times \frac{a-b}{c}= \frac{(a-b)(a+b)}{2c}\).
End proof!
Isn’t that cool?! I couldn’t believe it when it happened. There were two key moments in making it happen. First, I realised that the diagram with that line \(PQ\) had the elements \(\frac{a+b}{2}\) and \(\frac{c}{a-b}\) in it already. The second was realising I could flip that slope by making a right angle. Even then I was amazed it all pulled together.
So, to find the shared chord of two circles, I need to create this diagram, (or at least something with enough of the same elements that puts the correct distance in the correct spot) and then the perpendicular line to \(AB\) through \(X\) is the shared chord I want. There are many ways to do this, some easier than others, and some that are easier if certain lines are long enough, or you reuse certain lines that had been previously drawn in other constructions. I actually think it’s really cool that there are many ways to do it, and that you might choose a different one depending on the conditons. I also think it’s very nice that I don’t have to remember a long tangled sequence of moves, but a principle.
To draw the three parallel lines all perpendicular to \(AB\) involved in the diagram, you can use the constructions at the end of the previous blog post. However, you don’t actually have to draw the middle parallel line through the midpoint of \(AB\) to do the construction of the place where circles meet, since the midpoint of \(PQ\) can be found on its own.
The fiddliest part of the diagram isn’t drawing those perpendicular lines, but it’s the fact that the radiuses are measured along these perpendicular lines starting at the wrong centre! I need the radius that goes with centre \(A\) to be at \(B\) and the radius that goes with centre \(B\) to be at \(A\). If I keep them at their own centres, then the line \(PQ\) faces the wrong way and the point \(X\) ends up at the wrong end of \(AB\). I have three ways of dealing with that which I like, and I’ll show you all of them.
The first strategy is to use the original matching radiuses where they are, which produces a point at the wrong end of the line joining the centres, and then copy the length to the other end.
Given the radiuses of two circles measured at right angles to the line joining their centres, mark the place where the shared chord meets that line (by reflecting a length)
0. Start with the circle centres \(A\) and \(B\) with lines perpendicular to \(AB\) at both \(A\) and \(B\). Also start with point \(P\) on the perpendicular through \(A\) and point \(Q\) on the perpendicular through \(B\), with both \(P\) and \(Q\) on the same side of \(AB\). The points \(P\) and \(Q\) will define the radiuses of the circles with centres \(A\) and \(B\) respectively.
1. Draw the line through \(P\) and \(Q\).
2. Draw the perpendicular bisector of \(PQ\) so that it meets \(AB\). Call this meeting point \(W\).
3. Copy the length \(BW\) so that one end of the new length is at \(A\) and the other end is on \(AB\) on the same side of \(A\) as \(B\). Call this second endpoint \(X\).
Done! The point \(X\) is the correct distance from \(A\).
Note there are many choices you can make for how to copy that length at step 3. My favourite is to put your two parallel lines on the opposite side of the original line \(AB\) from \(P\) and \(Q\) and to reuse the line \(AP\) during the process. You can see me doing it this way in the video.
This strategy is the one I’m going to use in the final construction because I love it so much and I find it the easiest to use, but I still want to share the other strategies because I’m still proud of coming up with them.
The second strategy is to use the fact that the line in the wrong direction is on the opposite side of the perpendicular bisector of \(AB\) from the line in the right direction.
Given the radiuses of two circles measured at right angles to the line joining their centres, mark the place where the shared chord meets that line (by reflecting an angle)
0. Start with the circle centres \(A\) and \(B\) with lines perpendicular to \(AB\) at both \(A\) and \(B\). Also start with point \(P\) on the perpendicular through \(A\) and point \(Q\) on the perpendicular through \(B\), with both \(P\) and \(Q\) on the same side of \(AB\). The points \(P\) and \(Q\) will define the radiuses of the circles with centres \(A\) and \(B\) respectively.
1. Draw the line through \(P\) and \(Q\).
2. Draw the perpendicular bisector of \(AB\). Call by \(Y\) the place where the bisector meets \(PQ\).
3. Draw a perpendicular line to \(PQ\) through \(Y\).
4. Copy the angle between the perpendicular to \(PQ\) and the perpendicular bisector of \(AB\) to the other side of the perpendicular bisector of \(AB\). Call by \(X\) the point where this line meets \(AB\).
Done! The point \(X\) is the correct distance from \(A\).
Note that you don’t actually have to draw the perpendicular line to \(PQ\) through \(Y\), just set up one point you know is on the line. This is because the process of copying the angle requires you to side-align that perpendicular line and draw on the other side of the ruler. And you can just as easily side-align two points as a whole line.
This saving is quite nice, but if the two radiuses are close together, then the angle we’re copying is very narrow, so you need a lot of space and it’s hard to get it right since small variations in your aligning produce big variations in where the line meets another line far away.
The third strategy is to copy the radiuses to the opposite side of \(AB\).
Given the radiuses of two circles measured at right angles to the line joining their centres, mark the place where the shared chord meets that line (by reflecting the radiuses)
0. Start with the circle centres \(A\) and \(B\) with lines perpendicular to \(AB\) at both \(A\) and \(B\). Also start with point \(P\) on the perpendicular through \(A\) and point \(Q\) on the perpendicular through \(B\), with both \(P\) and \(Q\) on the same side of \(AB\). The points \(P\) and \(Q\) will define the radiuses of the circles with centres \(A\) and \(B\) respectively.
1. Draw the perpendicular bisector of \(AB\). Call by \(Z\) the midpoint of \(AB\)
2. Draw the line \(PZ\) so that it meets \(BQ\). Call this meeting point \(L\). Draw the line \(QZ\) so that it meets \(AP\). Call this meeting point \(M\).
3. Draw the line \(LM\). Call by \(Y\) the point where this line meets the perpendicular bisector of \(AB\).
4. Draw the line through \(Y\) perpendicular to \(LM\) so that it meets \(AB\). Call this meeting point \(X\).
Done! The point \(X\) is the correct distance from \(A\).
Triangles \(\triangle APZ\) and \(\triangle BLZ\) are similar because of the parallel lines \(AP\) and \(BL\) and congruent because of the equal lengths \(AZ\) and \(BZ\). Therefore the edges \(AP\) and \(BL\) are equal.
Similarly \(BQ\) and \(AM\) are equal.
The perpendicular bisector of \(AB\) must also bisect \(LM\) becasuse of the equally spaced parallel lines.
Therefore the earlier diagram has been created with the circle’s radiuses measured at the opposite centre, so the perpendicular bisector of \(LM\) will meet \(AB\) in the correct point.
End proof!
Note that at step 1 you could you could have found the midpoint of \(AB\) without a perpendicular line and then later at step 5 found a perpendicular bisector of \(LM\). However, if \(LM\) is shorter than the ruler width this second part would take a lot more lines than the way we have for finding the perpendicular bisector of \(AB\) and so make it a much longer construction.
Also note that while the second method allows you to save a line in the construction, this third method is much easier to do when the two radiuses are close together. On the other hand, if the radiuses are quite long, then a small amount of error aligning \(P\) and \(Q\) with the midpoint of \(AB\) will result in a large amount of error in the positions of \(L\) and \(M\) and so give a final result that is not accurate. So keep that in mind!
However you choose to do it, we now have all the tools needed to find where two circles meet.
I’ve presented two versions of the construction below, depending on how far apart the centres of the circles are. The steps are not described in detail, but just tell you do use constructions I’ve described earlier, and there are various choices for how you do those constructions. Indeed, the choices you make for some parts will impact how you do other parts because you may want to reuse lines. (Indeed, I could have saved a line in the centres-further-apart-than-ruler-width version by drawing a perpendicular bisector of one segment to help with the perpendicular bisector of another, but didn’t see it at the time. And I’m not redoing that video any more times!)
Find the points where two circles meet, each circle defined by its centre and one point on its circumference (centres closer than ruler width)
0. Start with the centres of the two circles and one point on the circumference of each circle. Call the centres of the two circles \(A\) and \(B\) and the points on their circumferences respectively \(R\) and \(S\).
1. Draw the line through \(A\) and \(B\), the line through \(A\) and \(R\), and the line through \(B\) and \(S\).
2. Draw lines through \(A\) and \(B\) perpendicular to \(AB\).
3. Copy the length \(AR\) onto the line through \(A\) drawn at step 2, and the length \(BS\) onto the line through \(B\) drawn at step 2, so that both lengths are on the same side of \(AB\). Call the end of the lengths at \(A\) and \(B\) by \(P\) and \(Q\) respectively.
4. Mark the place where the line joining the intersections of the two circles meets \(AB\). Call this point \(X\).
5. Draw a line through \(X\) perpendicular to \(AB\) (or parallel to \(AP\), since it’s the same thing).
6. Find the points where the line drawn at step 5 meets the circle with centre \(A\) and with \(R\) on its circumference (or where it meets the circle with centre \(B\) and with \(S\) on its circumference). Call these points \(G\) and \(H\).
Done! The points \(G\) and \(H\) are where the circle with centre \(A\) and radius \(AR\) and the circle with centre \(B\) and radius \(BS\) meet.
Find the points where two circles meet, each circle defined by its centre and one point on its circumference (centres further apart than ruler width)
0. Start with the centres of the two circles and one point on the circumference of each circle. Call the centres of the two circles \(A\) and \(B\) and the points on their circumferences respectively \(R\) and \(S\).
1. Draw the line through \(A\) and \(B\), the line through \(A\) and \(R\), and the line through \(B\) and \(S\).
2. Draw lines through \(A\) and \(B\) perpendicular to \(AB\).
3. Copy the length \(AR\) onto the line through \(A\) drawn at step 2, and the length \(BS\) onto the line through \(B\) drawn at step 2, so that both lengths are on the same side of \(AB\). Call the end of the lengths at \(A\) and \(B\) by \(P\) and \(Q\) respectively.
4. Mark the place where the line joining the intersections of the two circles meets \(AB\). Call this point \(X\).
5. Draw a line through \(X\) perpendicular to \(AB\) (or parallel to \(AP\), since it’s the same thing).
6. Find the points where the line drawn at step 5 meets the circle with centre \(A\) and with \(R\) on its circumference (or where it meets the circle with centre \(B\) and with \(S\) on its circumference). Call these points \(G\) and \(H\).
Done! The points \(G\) and \(H\) are where the circle with centre \(A\) and radius \(AR\) and the circle with centre \(B\) and radius \(BS\) meet.
One thing that fascinates me about all this is how when you string multiple processes together, the description can seem very short and clean, but actually when you do it all properly, it’s viciously complicated. Look at all the lines I drew to do this construction, especially when the centres of the circles were close together! I wonder how many processes and concepts we teach our students are like that, and feel just like that tangle of lines to the students.
Interestingly, it was me trying to cut down on the number of lines involved that inspired many of the constructions in earlier blog posts. For example, the constructions for drawing a line halfway between two parallel lines were inspired by the constructions for finding where circles meet. And I wasn’t planning to properly present the copy-a-length-to-an-arbitrary-place construction except I found it was super useful for the circles meeting.
I will say again that no matter how you choose to do these circle constructions, it’s very easy to introduce errors along the way when doing it by hand. I have certainly wasted many hours trying to get things to meet where I know they should. You may possibly content yourself with knowing it’s theoretically possible and not actually do it yourself with your own hands.
Anyway, I am now happy that I understand how I would do these circle constructions without circles. I didn’t doubt that the two-sided ruler was able to do what the ruler and compass could do, but I really deeply believe it now. Also I’m also proud of all of the problem-solving I did to come to that understanding.
Speaking of problem-solving I’m proud of…
There is one blog post left, and it contains my constructions for making an equilateral triangle and a regular pentagon, which I am deeply proud of. That’s why I saved them for last.
This blog post is about drawing lines parallel to other lines, but through specific points. One of the fundamental things the two-sided ruler can do is draw lines parallel to other lines that are exactly one ruler width away. But what if you want your parallel line to be more or less than that distance away? Well it can be done.
Near the end of the last blog post, I noted that the construction for drawing a line perpendicular to an existing line through a point off the line had a couple of parallel lines in its diagram, and I wondered if I could put that diagram in the correct spot to draw the parallel lines where I want them.
And it is possible. Here’s the construction I came up with that’s in the spirit of the perpendicular line construction.
Draw a line parallel to an existing line through a point not on that line (using a perpendicular line)
0. Start with a line and a point not on that line. Call the point \(Q\).
1. Align the point \(Q\) on one side of the ruler so that the ruler meets the original line, and draw along both sides of the ruler. Call by \(A\) the point where the line through \(Q\) meets the original line and by \(Y\) the point where the other parallel line meets the original line. Note it’s important to make sure \(Y\) and \(A\) are on opposite sides of the position of \(Q\) if you projected it down to the line.
2. Side-align the line you just drew through \(Y\) and draw along the other side of the ruler to make a third parallel line. You only need enough of this line to find where it meets the original line. Call this meeting point \(B\).
3. The points \(Y\) and \(B\) were already cross-aligned in step 2. Cross-align them in the other direction and draw along the side of the ruler through \(B\). Call by \(P\) the point where this line meets the line \(QA\).
4. Draw the line through \(P\) and \(Y\).
5. Draw the line through \(Q\) and \(B\) and find where it meets the line \(PY\). Call this point \(X\).
6. Draw the line through \(A\) and \(X\) and find where it meets the line \(BP\). Call this point \(S\).
7. Draw the line through \(Q\) and \(S\).
Done! The line \(QS\) passes through \(Q\) and is parallel to the original line \(AB\).
I have carefully made the labels of the points match up with the earlier construction, but the points and lines were drawn almost in the reverse order, so I think I need a new proof.
Proof:
Triangle \(\triangle APB\) was constructed by the same method for drawing a line perpendicular to a specific point on a line, and so, just as in that proof, it’s an isosceles triangle. In particular the sides \(AP\) and \(BP\) are equal, the angles \(\angle PAB\) and \(\angle PBA\) are equal, and the line \(PY\) meets \(AB\) in a right angle and bisects the angle \(APB\).
Consider the triangles \(\triangle APX\) and \(\triangle BPX\).
The sides \(AP\) and \(BP\) are equal, and the angles \(\angle APX\) and \(\angle BPX\) are equal, and the side \(PX\) is shared. Therefore \(\triangle APX\) and \(\triangle BPX\) are congruent.
Hence, \(\angle PAX\) and \(\angle PBX\) are equal.
Consider triangles \(\triangle APS\) and \(\triangle BPQ\).
The sides \(AP\) and \(BP\) are equal, the angles \(\angle PAX\) and \(\angle PBX\) are equal and the angles \(\angle APS\) and \(\angle BPQ\) are equal because they are shared. Therefore \(\triangle APS\) and \(\triangle BPQ\) are congruent.
Hence \(PS\) and \(PQ\) are the same length, making triangle \(QPS\) an isosceles triangle. The line \(PY\) bisects the angle \(\angle QPS\) and so it must meet the opposite side \(QS\) in a right angle.
Now \(QS\) and \(AB\) are both perpendicular to \(PY\) and are therefore parallel to each other.
End of proof!
Note I could have argued that last part differently by noticing that the ratio \(PQ:PA\) is the same as the ratio \(PS:PB\) and going via similar triangles. I didn’t, because I wanted to match the spirit of the previous construction.
However, there is a way to do almost this same construction without making a right angle anywhere, and indeed without making any isosceles triangles or bisecting any angles. You just need the base \(AB\) to be bisected and it will work. Indeed, when Wernick and Birrell describe this construction, they just say to arrange three equally spaced points on the line using the process for doubling a line longer than a ruler width, but no particular instructions of where to put them.
Their construction is indeed particularly useful if you do already happen to have three equally spaced points on that original line. My construction requires you to make the points yourself or the later lines won’t go through the right points, so theirs is very nice if you don’t want to make them yourself.
Draw a line parallel to an existing line through a point not on that line (using three existing equally spaced points on the line)
0. Start with a line and a point not on that line. Also start with three equally-spaced points on the line. Call the point not on the line \(Q\). Call the three points on the line \(A\), \(Y\) and \(B\) in order.
1. Draw the line through \(Q\) and \(A\).
2. Draw any line through \(B\) that meets \(QA\) in a point other than \(Q\) and \(A\). Call this point \(P\).
3. Draw the line through \(P\) and \(Y\).
4. Draw the line through \(Q\) and \(B\) and find where it meets the line \(PY\). Call this point \(X\).
5. Draw the line through \(A\) and \(X\) and find where it meets the line \(PB\). Call this point \(S\).
6. Draw the line through \(Q\) and \(S\).
Done! The line \(QS\) passes through \(Q\) and is parallel to the original line \(AB\).
It’s truly remarkable to me that this construction works. My favourite part is that it will work no matter which direction \(A\), \(Y\) and \(B\) are labelled and no matter where the point \(P\) is on the line \(QA\) . The diagrams produced will look similar from a distance, but will be labelled very differently. (Wernick and Birrell don’t draw all the other diagrams, but I wanted to be sure it really worked.)
The point \(A\) is closest to \(Q\). The point \(P\) is on the far side of \(Q\) from \(A\).
The point \(A\) is furthest from \(Q\). The point \(P\) is on the far side of \(Q\) from \(A\).
The point \(A\) is closest to \(Q\). The point \(P\) between \(Q\) and \(A\).
The point \(A\) is furthest from \(Q\). The point \(P\) is between \(Q\) and \(A\).
The point \(A\) is closest to \(Q\). The point \(P\) is on the far side of \(A\) from \(Q\).
The point \(A\) is futhest from \(Q\). The point \(P\) is on the far side of \(A\) from \(Q\).
Wenrick asserts that this construction works by citing the “harmonic properties of a complete quadrilateral” and Birrell properly proves it works using Ceva’s theorem, a very cool theorem that I learned because she used it. However, I’m pretty sure Ceva’s theorem won’t apply if the points \(P\) and \(X\) are on opposite sides of the original line, so I’ve just proved it directly. I did take inspiration from the proof of Ceva’s theorem, which uses areas.
Proof:
Consider triangles \(\triangle AYX\) and \(\triangle BYX\). If \(AY\) and \(BY\) are considered their bases, then note these bases are on the same line and the triangles also share the opposite vertex \(X\), so they have the same height. Since the bases and heights are equal, they must have the same area. I’ll use the modulus straight line brackets to mean the area of a triangle so I can write \(|\triangle AYX|=|\triangle BYX|\).
Now consider triangles \(\triangle AYP\) and \(\triangle BYP\). They also have aligned equal bases \(AY\) and \(BY\) and shared vertex \(P\), so \(|\triangle AYP|=|\triangle BYP|\).
Now consider triangles \(\triangle AXP\) and \(\triangle BXP\). If \(X\) and \(P\) are on opposite sides of \(Y\), then \(\triangle AXP\) is made by joining \(\triangle AYX\) and \(\triangle AYP\) and \(\triangle BXP\) is made by joining \(\triangle BYX\) and \(\triangle BYP\). If \(X\) and \(P\) are on the same side of \(Y\), then \(\triangle AXP\) is made by removing \(\triangle AYX\) from \(\triangle AYP\) and \(\triangle BXP\) is made by removing \(\triangle BYX\) from \(\triangle BYP\). Either way, since the triangles involved in each calculation have the same areas as the matching triangles in the other calculation, the resulting triangles have the same area too. That is, \(|\triangle AXP|=|\triangle BXP|\).
Now consider the two line segments \(QA\) and \(QP\) on the same line. Two triangles can be created with each of these segments as bases by choosing the opposite vertex to be \(X\) or \(B\).
The triangles \(\triangle QAX\) and \(\triangle QPX\) have aligned bases and shared vertex \(X\) so they have the same height. This means their areas are in the same proportion as \(QA\) and \(QP\). That is, \[\frac{QA}{QP}=\frac{|\triangle QAX|}{|\triangle QPX|}\].
Similarly, the triangles \(\triangle QAB\) and \(\triangle QPB\) have aligned bases and shared vertex \(B\) so they have the same height and their areas are in the same proportion as \(QA\) and \(QP\). That is, \[\frac{QA}{QP}=\frac{|\triangle QAB|}{|\triangle QPB|}\].
Now, if two fractions are equal, you can subtract (or add) the numerators and denominators and produce another equal fraction.
(A quick proof if you don’t believe it, because it does feel surprising… Suppose \(\frac{a}{b} = \frac{c}{d}\). Begin by multiplying both parts of these equations by both \(b\) and \(d\): \[\begin{aligned}\frac{a}{b} &= \frac{c}{d} \\ ad &=bc \\ ad-ab &=bc -ab\\ a(d-b) &= (c-a)b \\ \frac{a}{b} &= \frac{c-a}{d-b} \end{aligned}\] So there you go.)
Notice that removing \(\triangle QAX\) from \(\triangle QAB\) leaves \(\triangle BAX\) and removing \(\triangle QPX\) from \(\triangle QPB\) leaves triangle \(BPX\). Therefore \[\frac{QA}{QP}=\frac{|\triangle BAX|}{|\triangle BPX|}\].
Similar reasoning on the other side of the diagram concludes that \[\frac{SB}{SP}=\frac{|\triangle ABX|}{|\triangle APX|}\].
Note that \(\triangle BAX\) is the same as \(\triangle ABX\) and \(|\triangle APX| = |\triangle BPX|\), so \[\frac{SB}{SP}=\frac{|\triangle BAX|}{|\triangle BPX|}=\frac{QA}{QP}\]
Consider triangles \(\triangle APB\) and \(\triangle QPS\). They share the angle at \(P\) and their matching sides on either side of that angle are in the same proportion. Therefore the triangles are similar.
Hence \(\angle PAB = \angle PQS\), which means lines \(AB\) and \(QS\) are parallel.
End proof!
This is an epic proof with a lot of ninja algebra, and ninja triangle-finding so I’m proud of figuring it out.
A very interesting thing about this construction is that after you’ve found the three equally-spaced points, you don’t use both of the sides of the ruler at once, so you could do it with a one-sided ruler. That’s pretty cool in its own way. Still, for some reason I find the construction fiddly. I can’t put my finger on why, but there it is.
I have discovered an alternative way to use three equally spaced points on the line to draw a parallel line. It’s a modification of Wenrick’s method of copying a line segment, and I think it’s rather cool.
Draw a line parallel to an existing line through a point not on that line (using three existing equally spaced points on the line, and an intermediate parallel line)
0. Start with a line and a point not on that line. Also start with three equally-spaced points on the line. Call the point not on the line \(Q\). Call the three points on the line \(W\), \(X\) and \(Y\) in order.
1. Side-align the original line and draw along the other side of the ruler to make a parallel line. It works best if this line is on the opposite side of the original line from \(Q\).
2. Draw the line through \(Q\) and \(W\) and find where it meets the parallel line you drew at step 1. Call this meeting point \(S\).
3. Draw the line through \(Q\) and \(X\) and ifind where it meets the parallel line you drew at step 1. Call this meeting point \(B\).
4. Draw the line through \(S\) and \(X\) .
5. Draw the line through \(B\) and \(Y\) and find where it meets the line \(SX\). Call this point \(A\).
6. Draw the line through \(Q\) and \(A\).
Done! The line \(QA\) passes through \(Q\) and is parallel to the original line \(WY\).
Since \(WX\) is parallel to \(SB\) the triangles \(\triangle SQB\) and \(\triangle WQX\) are similar. Therefore the matching sides are in the same proportion. In particular \[\frac{QX}{QB}=\frac{WX}{SB}\]
Also the triangles \(\triangle SAB\) and \(\triangle XAY\) are similar, so that \[\frac{AY}{AB} =\frac{XY}{SB}\].
But the lengths of \(WX\) and \(XY\) are the same, so
Now triangles \(\triangle QBA\) and \(\triangle XBY\) have a shared angle at \(B\) and the enclosing sides are in the same proportion, to they are similar triangles.
Therefore \(\angle BQA = \angle BXY\) and so \(QA\) must be parallel to \(XY\).
End proof!
I did say earlier that it works best if that intermediate parallel line is on the opposite side of the original line from \(Q\) but it does work even if it’s on the same side as \(Q\). I just find it a bit harder to draw. Here’s what it looks like in all three situations:
The intermediate parallel line is on the far side of the original line from \(Q\).
The intermediate parallel line is between the original line and \(Q\).
The intermediate parallel line is on the far side of \(Q\) from the original line.
The proof above is designed for the case where that first parallel line is indeed on the opposite side of the original line from \(Q\), and the algebra in the middle needs to be modified to make it match the other cases.
I’m honestly surprised Wenrick didn’t include this method in his paper, since it is very similar to his method of copying a length. Me, I like it very much. There’s something cool to me about drawing an intermediate parallel line and somehow copying the parallelness to another line. Plus it uses the same number of lines as the previous construction (six, on top of any you used to make the three equally-spaced points).
Just like for Wenrick’s copy-a-length construction, it’s worth noting that this will work just fine with an intermediate parallel line that’s some other distance than one ruler-width away from the original line. So, if you happen to have such a line already, then this method is more efficient than the previous one. It’s only more efficient by one line, but that’s still better!
I do have one more way I’ve designed to construct a parallel line through a given point and it’s my favourite of all of them. I made it when I first looked at Wernick and Birrell’s construction, because it occurred to me that I have to draw three parallel lines to make those equally-spaced points and surely there must be a way to use those lines directly to construct the actual line I want. There is a way, and here it is.
Draw a line parallel to an existing line through a point not on that line (using three parallel lines)
0. Start with a line and a point not on that line. Call the point \(Q\).
1. Align the point \(Q\) on one side of the ruler so that the ruler meets the original line, and draw along both sides of the ruler. Call by \(A\) the point where the line through \(Q\) meets the original line and by \(Y\) the point where the other parallel line meets the original line.
2. Side-align the line you just drew through \(Y\) and draw along the other side of the ruler to make a third parallel line. Call by \(B\) the point where this line meets the original line.
3. Draw the line through \(Q\) and \(B\) and find where it meets the middle parallel line through \(Y\). Call this point \(X\).
4. Draw the line through \(A\) and \(X\) and find where it meets the third parallel line through \(B\). Call this point \(S\).
5. Draw the line through \(Q\) and \(S\).
Done! The line \(QS\) passes through \(Q\) and is parallel to the original line \(AB\).
The three parallel lines through \(A\), \(Y\) and \(C\) all meet the original line at the same angle, so the lengths \(AY\) and \(YB\) are equal by the cross-align arcsine lemma.
Triangles \(\triangle AQB\) and \(\triangle YBX\) are similar because of the shared angle and the parallel lines. Since \(YB\) is half of \(AB\), this means \(YX\) is half as long as \(AQ\).
Triangles \(\triangle ABS\) and \(\triangle AYX\) are also similar, and \(AY\) is half as long as \(AB\), which means \(YX\) is also half as long as \(BS\).
Therefore \(AQ\) and \(BS\) are the same length.
Consider triangles \(\triangle AQB\) and \(\triangle SBQ\). Side \(BQ\) is shared, sides \(AQ\) and \(SB\) are the same length, and angles \(\angle AQB\) and \(\angle QBS\) are equal since they are alternate angles around a transversal to parallel lines.
Therefore \(\triangle AQB\) and \(\triangle SBQ\) are congruent, which means \(\angle ABQ = \angle BQS\). These are alternate angles around a transversal to \(AB\) and \(QS\).
Hence \(AB\) and \(QS\) are parallel.
End proof!
If this proof reminded you of the proof for Wernick’s method of copying a length, it’s not a coincidence, since the diagram is pretty much the same, only sideways, as it were. Indeed this is why I labelled the points the way I did for the intermediate parallel line way to make a parallel line, because that diagram too is pretty much this one but sideways.
(I actually didn’t read Wernick’s paper until long after I made this construction, and I noticed the connection when I came to write it all up.)
There’s another connection this construction and the other ones we’ve seen so far. It’s not just that I took the opportunity of the three parallel lines from Wernick and Birrell’s construction. If you take that point \(P\) from their construction and move it away from \(Q\) as far as you can, then the three lines through \(P\) stretch out further and further and become closer and closer to being parallel. This construction is the same construction but with the point \(P\) at infinity! This warms my finite geometer’s heart. I’m sure there’s some sort of projective geometry theorem that makes it all work (maybe those harmonic properties of the complete quadrilateral that Wernick mentioned), though I am very happy with my similar triangles and areas.
So, I’ve got four ways to draw a parallel line through a point not on the line.
The next thing I want to do is simultaneously draw a perpendicular line and a parallel line through a point not on the original line. This is a thing that’s useful in some later constructions I want to describe.
You could naively just draw a parallel line through the point and separately draw a perpendicular line through the point. The best versions of those constructions use six lines each, so I can definitely do it in twelve lines.
But actually if you already have one of the parallel or perpendicular lines, the other line can be drawn by going perpendicular to the line you just drew. That is, if you have a line parallel to another, then a perpendicular to that line is perpendicular to the original line; and if you have a line perpendicular to another, then a perpendicular to that line is parallel to the original.
And the construction for a perpendicular line through a point takes four lines. So the estimate is now six plus four (which is ten) lines. And it turns out either way I can reuse one of the lines from the first part when doing the second part, which gets me down to nine lines.
I can’t seem to make it happen in any less than nine, and honestly nine is pretty good considering that Wernick’s original method for drawing a parallel line through a point uses eight lines if you include the lines to make the equally spaced points. So here’s two constructions for drawing a perpendicular and parallel line at the same time, though I have to say I like the first one better.
Draw lines parallel then perpendicular to an existing line through a point not on that line
0. Start with a line and a point not on that line. Call the point \(Q\).
1. Align the point \(Q\) on one side of the ruler so that the ruler meets the original line, and draw along both sides of the ruler. You will need the line on the other side of the ruler from \(Q\) to be long enough to meet the perpendicular through \(Q\) once it’s drawn. Call by \(A\) the point where the line through \(Q\) meets the original line and by \(Y\) the point where the other parallel line meets the original line.
2. Side-align the line you just drew through \(Y\) and draw along the other side of the ruler to make a third parallel line. Call by \(B\) the point where this line meets the original line.
3. Draw the line through \(Q\) and \(B\) and find where it meets the middle parallel line through \(Y\). Call this point \(X\).
4. Draw the line through \(A\) and \(X\) and find where it meets the third parallel line through \(B\). Call this point \(S\).
5. Draw the line through \(Q\) and \(S\).
6. The line \(QA\) was side-aligned at step 2 to draw the line \(YX\). Side-align it on the other side and draw along the opposite side of the ruler. You only need enough of this line to see where it meets \(QS\). Call this meeting point \(L\).
7. The points \(L\) and \(Q\) were cross-aligned at step 6. Cross-align them in the other direction and draw along the side of the ruler through \(L\). You only need enough of this line to see where it meets \(XY\). Call this meeting point \(M\).
8. Draw the line through \(M\) and \(Q\) long enough to meet the original line.
Done! The line \(QS\) passes through \(Q\) and is parallel to the original line \(AB\). The line \(MQ\) passes through \(Q\) and is perpendicular to the original line.
Draw lines perpendicular then parallel to an existing line through a point not on that line
0. Start with a line and a point not on that line. Call the point \(A\).
1. Align the point \(A\) on one side of the ruler so that the ruler meets the original line, and draw along both sides of the ruler. You will need the line on the other side of the ruler from \(A\) to be long enough to reach both the perpendicular and parallel lines through \(A\) once they are drawn. Call by \(P\) the point where the line through \(A\) meets the original line and by \(R\) the point where the other parallel line meets the original line.
2. The points \(P\) and \(R\) were already cross-aligned at step 1. Cross-align them the other way and draw along the side of the ruler through \(P\).
3. Side-align \(A\) and \(R\) and draw along the opposite side of the ruler from \(A\). You only need enough of this line to see where it meets the original line \(PR\). Call this meeting point \(X\).
4. The points \(X\) and \(R\) were already cross-aligned at step 3. Cross-align them in the other direction and draw along the side of the ruler through \(R\). Call by \(B\) the point where this line meets the line drawn through \(P\) at step 2.
5. Draw the line \(AB\). Make sure it’s long enough to meet the line through \(R\) you drew at step 1. Call the meeting point \(L\)
6. The points \(A\) and \(L\) were cross-aligned at step 1. Cross-align them in the other direction and draw along the side of the ruler through \(A\).
7. Side-align the line you just drew so that the ruler is on the opposite side from \(L\), and draw along the other side of the ruler. You only need enough of this line to see where it meets \(LR\). Call this meeting point \(M\).
8. Draw the line through \(M\) and \(A\).
Done! The line \(AB\) passes through \(A\) and is perpendicular to the original line \(PR\). The line \(AM\) passes through \(A\) and is parallel to the original line.
The point of this is that if you are going to string different constructions together to achieve a particular goal, you can choose your constructions carefully and also sometimes reuse lines from earlier parts in later parts to make the whole process a little more efficient and less full of overlapping lines everywhere.
In a long complicated construction, I would choose one or the other of these two orders for making a parallel and perpendicular line, or possibly something different entirely, depending on what existing lines and points I already had at my disposal.
This will come up in the next blog post, which is about the final two constructions that prove the two-sided ruler can do anything a ruler and compass can do.
Before I do that, though, I want to share three more constructions involving parallel lines, which I found when trying to make the constructions in the next blog post easier to do.
Draw a parallel line exactly halfway between two existing parallel lines
0. Start with two parallel lines.
1. Angle the ruler so it meets both parallel lines and draw along both sides of the ruler.
2. Side-align one of these new lines on the other side and draw along the other side of the ruler to make a third line parallel to the two drawn at step 2.
3. Two identical parallelograms have been created. Draw both diagonals of both parallelograms, creating one point in each where the diagonals meet.
4. Draw the line joining these two points.
Done! The line drawn at step 4 is parallel to both original lines and halfway between them.
The diagonals of a parallelogram bisect each other, so the two points where the diagonals of the two parallelograms meet are each halfway between the two original parallel lines.
End proof!
You don’t even need the two parallelograms to be next to each other or the same as each other and it will still work. It’s just that you can save a line by putting the parallelograms next to each other.
This construction is particularly nice if you already have a line segment meeting both parallel lines in a right angle, because then the middle parallel line must be the perpendicular bisector of the line segment.
If your line segment is longer than the ruler width, then you can very quickly draw all three of the parallel lines from the previous picture. This is the same as picking a specific distance apart for your parallel lines and also drawing one halfway between.
Draw two parallel lines a specific distance (longer than ruler width) apart and the parallel line halfway between them
0. Start with a line segment longer than the ruler width. Call the endpoints \(A\) and \(B\)
1. Cross-align the line segment and draw along the side of the ruler through \(A\).
2. Cross-align the line segment in the other direction and draw along the side of the ruler through \(B\). Call by \(X\) the point where these two lines cross.
3. Side-align \(AX\) and draw along the other side of the ruler. Call by \(P\) the point where this line meets \(BX\).
4. Side-align \(BX\) and draw along the other side of the ruler. Call by \(Y\) the point where this line meets the line drawn at step 3. Call by \(Q\) the point where this line meets \(AX\).
5. Draw \(AP\), \(BQ\) and \(XY\).
Done! The three lines \(AP\), \(BQ\) and \(XY\) are parallel, equally spaced, and perpendicular to \(AB\).
If you imagine drawing in the other sides of the ruler at step 1 and 2, then you will create four identical rhombuses joined together into one big rhombus. The opposite-direction diagonals of these rhombuses are all at right angles to each other.
End proof!
Isn’t that lovely? Even if you don’t want the perpendicular bisector, i’s the quickest way I can think of to draw the perpendiculars at both ends of a line segment. And if you only want the perpendicular bisector, it uses exactly the same number of lines as the construction all the way in the first blog post for bisecting a line segment longer than the ruler width.
It’s worth noting that the the line joining the places where \(PY\) and \(QY\) meet \(AP\) and \(BQ\), the line \(PQ\) and the line \(AB\) are also three equally spaced parallel lines, which I think is nice. Not least because it now looks like the construction above for finding the line halfway between two parallel lines.
Of course this only works if \(AB\) is long enough to cross-align. If it’s too short, then you just have to construct the perpendiculars separately. Though if you pay attention, you can reuse quite a few of those lines…
Draw two parallel lines a specific distance (shorter than ruler width) apart and the parallel line halfway between them
0. Start with a line segment, which can be shorter than the ruler width. Call the endpoints \(A\) and \(B\)
1. Side-align the line segment and draw along that side of the ruler to extend the line segment in both directions.
2. Align \(A\) on one side of the ruler so that the opposite side meets the original line, and draw along the opposite side. Call by \(P\) the point where that line meets the extended original line segment.
3. The points \(A\) and \(P\) were cross-aligned at step 2. Cross-align them in the other direction and draw along the side of the ruler through \(A\).
4. Side-align the line you just drew so that the ruler is on the opposite side of \(A\) from \(P\) and draw along the other side of the ruler. Call by \(X\) the point where it meets the line through \(P\) drawn at step 2.
5. Draw the line through \(X\) and \(A\).
6. Align \(B\) on one side of the ruler so that the opposite side meets the original line, and draw along the opposite side. Call by \(Q\) the point where that line meets the extended original line segment.
7. The points \(B\) and \(P\) were cross-aligned at step 5. Cross-align them in the other direction and draw along the side of the ruler through \(B\). Call by \(K\) the point where this line meets \(AX\).
8. Side-align the line you just drew so that the ruler is on the opposite side of \(B\) from \(Q\) and draw along the other side of the ruler. Call by \(Y\) the point where it meets the line through \(Q\) drawn at step 5, and call by \(L\) where it meets \(AX\).
9. Draw the line through \(Y\) and \(B\). Call by \(M\) the point where it meets the line through \(A\) drawn at step 3, and call by \(N\) the point where it meets the line through \(X\) drawn at step 4.
10. Draw \(LB\) and \(YK\). Call by \(S\) the point where they meet. Also draw \(AN\) and \(MX\). Call by \(T\) the point where they meet.
11. Draw the line through \(S\) and \(T\).
Done! The three lines \(AX\), \(BY\) and \(ST\) are parallel, equally spaced, and perpendicular to \(AB\).
If you count the lines drawn here, it’s fourteen. Considering the processes I made for drawing the perpendicular bisector of a line shorter than a ruler width both took eleven lines, I think this is pretty good! It’s only got three extra lines and two of those were part of the goal!
Speaking of only using a few lines, an idea occurred to me while checking all the blog posts for typos, and since the place where I was recording was still set up, and I didn’t want my thought to disappear, I thought I’d come back and add it.
The first construction for dawing the parallel line halfway between two others uses eight lines, but I can do it in one fewer line by drawing four parallel lines instead of three. It’s a little harder to describe in text, so I have to label points this time.
Draw a parallel line exactly halfway between two existing parallel lines (using intermediate parallel lines)
0. Start with two parallel lines.
1. Angle the ruler so it meets both parallel lines and draw along both sides of the ruler. Call the four vertices of the parallelogram so created \(A\), \(B\), \(K\) and \(L\) in cyclic order.
2. Side-align both of the lines drawn at step 1 and draw along the other side of the ruler to make new lines parallel to them. Call by \(C\) the point where the new line on the far side of \(B\) from \(A\) meets \(AB\) and by \(M\) the point where the new line on the far side of \(L\) from \(K\) meets \(LK\). You only need enough of these new lines to see where \(C\) and \(M\) are.
3. Draw the line through \(C\) and \(L\). Call the point where it meets \(BK\) by \(X\). Draw the line through \(B\) and \(M\). Call the point where it meets \(AL\) by \(Y\).
5. Draw the line through \(X\) and \(Y\).
Done! The line \(XY\) is parallel to both original lines and halfway between them.
Consider triangles \(\triangle LMY\) and \(\triangle ABY\).
Because \(AL\) and \(BM\) are transversals to the original parallel lines, it must be that \(\angle YML = \angle YBA\) and \(\angle YLM =\angle YAB\).
Also, \(AB\) and \(LM\) were cross-aligned at the same angle (which is the angle \(AL\) meets the original lines), so they are equal in length by the cross-align arcsine lemma.
Therefore \(\triangle LMY\) is congruent to \(\triangle ABY\) and therefore \(AY\) is the same length as \(LY\). That is, \(Y\) is halfway along \(AL\).
Similar reasoning concludes that \(X\) is halfway along \(BK\). Hence the line joining \(XY\) is halfway between the two original lines.
End proof!
I really like how symmetrical and simple this calculation is, and I’m sad that I only thought of it after having finished the circles constructions because it could have saved me a few lines. But I’m not redoing those videos now! I am content to just put it in here.
That concludes my discussion of parallel (and perpendicular) lines. The next blog post is about circles.
This blog post is all about drawing perpendicular lines exactly in the place you want them. It feels surprising that you could do these with just the two-sided ruler, but it is very much possible.
I’ve already made perpendicular lines a few times. In the post about rhombuses, I had two constructions of a right angle just anywhere, and also the perpendicular bisector of a line segment longer than a ruler width. And in the post about copying and cutting, I noticed that the construction for copying a length from one arm of an angle to another had some perpendicular lines in it. In this post I’ll be taking inspiration from all of these to put perpendiculars in very specific places.
The first construction is to draw a line perpendicular to an existing line through a point on that line. This is my construction and actually you’ve seen it before when I made a right angle just anywhere, it’s just that this time I have one of the lines already made.
Draw a line perpendicular to an existing line through a point on that line
0. Start with a line and a point on that line. Call the point \(B\).
1. Align \(B\) on one side of the ruler so that the opposite side meets the original line, and draw along the opposite side. Call by \(A\) the point where that line meets the original line.
2. The points \(A\) and \(B\) were cross-aligned at step 1. Cross-align them in the other direction and draw along the side of the ruler through \(B\).
3. Side-align the line you just drew so that the ruler is on the opposite side of the line from \(A\) and draw along the other 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 line through \(A\) drawn at step 1.
The diagram here is the exact same one for the way to make a right angle using a triangle, so the proof that the angle is a right angle is exactly the same too.
I deeply love this construction. It takes exactly four lines including the perpendicular line you were hoping to draw. You don’t even need to make that third line long enough to meet the original line at point \(C\), just make a little mark where the ruler meets the line through \(A\). I only needed to talk about point \(C\) to make my proof easier to describe.
Wernick and Birrell don’t do this construction. Instead they make two points \(A\) and \(C\) on either side of \(B\) so that \(B\) is their midpoint, and then draw a rhombus with diagonal \(AC\) and draw in the other diagonal. But I noticed that the very lines you need to draw to create those points on either side of \(B\) already show you where the perpendicular should be. You don’t need to draw any more!
If you happen to already have the middle point from three equally spaced points, and the outside two are further than a ruler width apart, then I concede you can very quickly find the perpendicular by drawing a rhombus. But you only need half the rhombus.
Draw a line perpendicular to an existing line through the middle point of three equally-spaced points on the line
0. Start with a line and three equally spaced points on the line so that the outside two are further than a ruler width apart. Call the three points in order \(A\), \(B\) and \(C\).
1. Cross-align \(A\) and \(C\) and draw along the side of the ruler through \(A\).
2. Cross-align \(A\) and \(C\) in the other direction and draw along the side of the ruler through \(C\). Call by \(X\) the point where the two lines drawn at step 1 and 2 meet.
The sides \(AX\) and \(CX\) of triangle \(\triangle AXC\) were created by cross-aligning \(AC\) so they both meet \(AC\) in the same angle by the cross-align arcsine lemma. So, \(\triangle AXC\) is an isosceles triangle and the line joining \(X\) and the midpoint of \(AC\) must meet \(AC\) in a right angle.
End proof!
Again, if I had drawn in both sides of the ruler when I cross-aligned, I would have drawn a rhombus and could have used the fact that its diagonals bisect each other at right angles. But I don’t need to draw the whole rhombus! I can save two lines and do this in just three lines, which makes me happier than it probably has a right to.
At this point, I think I should address the fact that we have a method for finding the perpendicular bisector of a line segment longer than the ruler width, but not one for shorter line segments.
Since we do have a way to bisect short segments and a way to draw a perpendicular through a specific point, it can definitely be done. It takes six lines to bisect a segment and four lines to draw a perpendicular to a line through a point on the line, so that’s ten lines in total. Except you’ll have to extend the line segment to draw that perpendicular, so that’s eleven lines total. Try as I might, I just cannot find a quicker way than that, so I’m just going to leave it at that.
Draw the perpendicular bisector to a line segment
0. Start with a line segment. Call the endpoints \(A\) and \(C\).
1. Bisect the line segment by the process shown before. That is: – Draw a line parallel to the segment – Mark three equally spaced points on the line and connect the outside ones to \(A\) and \(C\) – Connect the middle point to the intersection point just created and find where that line meets the segment. Call the midpoint \(B\).
2. Side-align the segment \(AB\) and draw along that side of the ruler to extend the segment. You only need to extend it in one direction (even though in the video I extended it in both.)
3. Draw a perpendicular line to the segment through \(B\) the midpoint using the process shown before. That is – Cross-align \(B\) and some other point \(P\) on the extended segment and draw the along the side of the ruler through \(P\). – Cross-align in the other direction and draw along the side of the ruler through \(B\). – Side-align that line and draw along the other side of the ruler making a line meeting the line just drawn through \(P\) at \(X\). – Draw \(XB\).
Done! The line \(XB\) is the perpendicular bisector of \(AC\).
Actually I think I will share one of the other ways I found to draw the perpendicular bisector of a line segment using eleven lines that I find rather charming.
Draw the perpendicular bisector to a line segment (by copying a length first)
0. Start with a line segment of any length. Call its endpoints \(A\) and \(B\)
1. Side-align the segment \(AB\) and draw along that side of the ruler to extend the segment in both directions.
2. Align \(A\) on one side of the ruler in such a way that the other side of the ruler meets the extended line segment beyond \(B\), and draw along both sides. Call by \(C\) the point where the line not through \(A\) meets the extended line segment.
3. Side-align the line segment and draw along the opposite side of the ruler to make a parallel line. Repeat on the other side of the line segment. Call by \(P\) and \(Q\) the points where the line through \(A\) drawn at step 2 meets these parallel lines. It works best if \(P\) is the point on the opposite side of \(A\) from \(B\).
4. Draw the line through \(P\) and \(C\) so that it meets the parallel line through \(Q\). Call this meeting point \(X\).
5. Draw the line through \(X\) and \(B\) so that it meets the parallel line through \(P\). Call this meeting point \(Y\).
6. Draw the line through \(Q\) and \(Y\) so that it meets the extended line segment \(AB\). Call this meeting point \(Z\).
7. Cross-align the points \(Z\) and \(B\) so that the sides of the ruler are not parallel to \(PQ\) and draw along both sides of the ruler. Call by \(L\) the point where the line just drawn through \(B\) meets \(PQ\), and by \(M\) the point where the line just drawn through \(Z\) meets the line drawn through \(C\) at step 2.
8. Draw the line through \(L\) and \(M\).
Done! The line \(LM\) is the perpendicular bisector of \(AB\).
Steps 2 to 6 are the construction for copying line segment \(AC\) to \(ZB\). Removing \(AB\) from both line segments must produce the same lengths, so \(ZA\) is the same length as \(CB\).
Since \(ZB\) and \(AC\) are the same length, cross-aligning \(ZB\) will produce lines that meet the extended line segment \(ZC\) in the same angles as when cross-aligning \(AC\).
This means \(ZM\) and \(CM\) meet \(ZC\) in the same angle and so \(ZMC\) is an isosceles triangle. This means the perpendicular bisector of \(ZC\) will pass through \(M\).
Also \(AL\) and \(BL\) meet \(AB\) in the same angle and so \(ALB\) is an isosceles triangle. This means the perpendicular bisector of \(AB\) will pass through \(L\).
However, \(AB\) and \(ZC\) share a centre, since \(ZA\) is the same as \(CB\), and so they have the same perpendicular bisector.
Both \(L\) and \(M\) are on that perpendicular bisector, so \(LM\) is the perpendicular bisector.
End proof!
One reason I like this construction is that the process for longer line segments also begins with cross-aligning some points on the line segment (the endpoints), so it’s as if we’ve tried to do that process and then when it didn’t work we pivoted to something new. And yet by the end there was a rhombus anyway and the perpendicular bisector was its diagonal after all.
Next, we want to draw a line perpendicular to an existing line through a point not on the line. In both Wernick and Birrell, they do this by first drawing a perpendicular through a point on the line, and then drawing a line parallel to that one through the point in question. (Drawing parallel lines is in the next blog post, but Wenick and Birrell do parallel lines before perpendicular lines.) When reading that, I was sure there had to be a quicker way. And there is.
Draw a line perpendicular to an existing line through a point not on that line
0. Start with a line and a point not on that line. Call the point \(A\).
1. Align the point \(A\) on one side of the ruler so that the ruler meets the original line, and draw along both sides of the line. Call by \(P\) the point where the line through \(A\) meets the original line and by \(R\) the point where the other parallel line meets the original line.
2. The points \(P\) and \(R\) were already cross-aligned at step 1. Cross-align them in the other direction and draw along both sides of the ruler. This creates a rhombus with \(PR\) as a diagonal. Call by \(Q\) the vertex of the rhombus on \(PA\) and by \(S\) the final vertex of the rhombus.
3. Draw the line \(AS\) and find where it meets the rhombus diagonal \(PR\). Call this point \(X\).
4. Draw the line \(QX\) and find where it meets the line \(PS\). Call this point \(B\).
5. Draw the line \(AB\).
Done! The line \(AB\) passes through \(A\) and meets the original line \(PR\) in a right angle.
The diagram here is the exact same diagram as when we copied a length from one arm of an angle to the other in the previous blog post. Therefore \(PA\) is the same length as \(PB\) and \(\triangle APB\) is an isosceles triangle.
Hence, the angle bisector of the angle at \(P\) must meet the opposite side \(AB\) in a right angle.
End proof!
Note there’s a bit of ambiguity in my description as to which side of \(A\) you put the ruler and therefore which side of the line \(AP\) the point \(R\) is. The reason I didn’t address it is that the construction works perfectly well either way. This is what it looks like if the ruler is on the other side of \(A\).
The reason I didn’t present it this way is that if you’re not careful, the line \(AS\) might take a very long distance to actually meet the original line, so it’s safest to do it the other way.
As I noted in the proof, this is pretty much exactly the method of of copying a length from one side of an angle to the other based on Tisdell’s method. However, I actually created this method before seeing Tisdell’s method of copying a line segment and then noticed how similar they were afterwards. When I made this, I was trying to find a way to do what Wernick and Birrell told me to do and make a perpendicular line anywhere and then a line parallel to that line through \(A\), while also trying to reuse as many lines as possible rather than draw new ones at every step, and also avoid drawing lines that weren’t necessary. Though I didn’t draw it, the first perpendicular is \(QS\) when that rhombus was made. And you can see that \(AB\) is indeed parallel to \(QS\).
Seeing it now, I wonder if you could make a line parallel to an existing line through a specific point by putting this diagram in the right location. You can, and it makes a neat connection between perpendiculars and parallels that helped me decide to tell the story in the order I have, rather than do parallels first like Wernick and Birrell did.
Before I do that, I have to show you another way to find a line perpendicular to another through a point not on the line. I came up with it while writing a part of the next blog post and I’ve come back here to put it in because I like it so much. It only draws six lines total, as opposed to seven.
Draw a line perpendicular to an existing line through a point not on that line (without drawing a rhombus)
0. Start with a line and a point not on that line. Call the point \(A\).
1. Align the point \(A\) on one side of the ruler so that the ruler meets the original line, and draw along both sides of the ruler. Call by \(P\) the point where the line through \(A\) meets the original line and by \(R\) the point where the other parallel line meets the original line.
2. The points \(P\) and \(R\) were already cross-aligned at step 1. Cross-align them the other way and draw along the side of the ruler through \(P\).
3. Side-align \(A\) and \(R\) (it doesn’t matter which side), and draw along the opposite side of the ruler from \(A\). You only need enough of this line to see where it meets the original line \(PR\). Call this meeting point \(X\).
4. The points \(X\) and \(R\) were already cross-aligned at step 3. Cross-align them in the other direction and draw along the side of the ruler through \(R\). Call by \(B\) the point where this line meets the line drawn through \(P\) at step 2.
5. Draw the line \(AB\).
Done! The line \(AB\) passes through \(A\) and meets the original line \(PR\) in a right angle.
Consider the triangles \(\triangle APR\) and \(\triangle BPR\).
The lines \(AR\) and \(BR\) meet \(PR\) in the same angle since they were produced by cross-aligning \(PR\), so \(\angle ARP = \angle BRP\).
The lines \(AP\) and \(BP\) meet \(PR\) in the same angle since they were produced (or at least imagined) while cross-aligning \(XR\), so \(\angle APR = \angle BPR\).
Finally the side \(PR\) is shared. Therefore \(\triangle APR\) and \(\triangle BPR\) are congruent.
Hence \(AP = BP\) and the triangle \(\triangle APB\) is isosceles. Since the angles between \(PR\) and both of \(AP\) and \(BP\) are the same, \(PR\) is the angle bisector of the angle at \(P\) in isosceles triangle \(\triangle APB\), therefore it must meet the base \(AB\) in a right angle.
End proof!
It’s worth showing what this looks like when you place the ruler on the other side of the point \(A\).
Whichever one you choose, they both tend to require a lot of space if the point \(A\) is close to the original line, so fair warning.
I was inspired to create this one by thinking about the isosceles triangles that made some of the other constructions work, and wondering if perhaps I could get the perpendicular I wanted by constructing two different isosceles triangles, or if you like, constructing a kite.
I do love all the cross-aligning that’s happening here. The saving of an extra line might not feel much of a saving compared to the care with which you have to enact all that precise ruler work. Also the extra space required when \(A\) is close to the line might prohibit using it too. It’s up to you to decide what works best for you.
And before anyone says it, I have indeed drawn three of the four sides of a rhombus with diagonal \(PR\), but I didn’t draw the fourth one so I still think it counts as not drawing a rhombus.
That completes the ways to make lines at right angles to each other. The next blog post is about how to make lines parallel to each other.
This 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. To be able to do these constructions, we’re going to need some help from similar and congruent triangles, which were the standard way to prove almost everything in classical ancient Greek geometry.
The first construction is for copying a line segment along the line it is part of.
Copy a line segment next to itself
0. Start with a line segment of any length that is part of a longer line. Call the endpoints of the line segment \(A\) and \(B\)
1. Side-align the line segment and draw along the opposite side of the ruler to make a parallel line. Repeat on the other side of the line segment.
2. Draw a line through \(A\) so that it meets both parallel lines on either side. Call these points \(P\) and \(Q\).
3. Draw the line \(PB\) and find where it meets the parallel line through \(Q\). Call this point \(X\).
4. Draw the line \(QB\) and find where it meets the parallel line through \(P\). Call this point \(Y\).
5. Draw the line \(XY\) and find where it meets the extended line segment. Call this point \(C\).
This procedure is in Wernick, and I laughed out loud when I saw it because it is so clever. Here’s Wenrick’s proof in my own words.
Proof:
Triangles \(\triangle APB\) and \(\triangle QPX\) are similar because they share the angle at \(P\) and the other edges are parallel. Therefore their matching edges are in the same proportion.
Since the three parallel lines are the same distance apart, that means \(PA\) is half the length of \(PQ\). Hence \(AB\) is half the length of \(QX\).
By similar reasoning, triangles \(\triangle BYC\) and \(\triangle QYX\) are similar and \(BC\) is half as long as \(QX\).
Therefore \(BC\) is the same length as \(AB\), since they’re both half of \(QX\).
End proof!
Isn’t that clever? I love so much how you make a length twice as long and then make one half as long as that. Also the diagram is so neat and spacious. And you can just repeat it as many times as you want to multiply the line segment by any natural number.
It’s worth noting that the process will work the same even if the three parallel lines aren’t equal distances apart. The proportion between the segments on the middle line and the segments on the outside lines will be the same as the proportion between the distances between the lines, so the two line segments will still work out the same as each other. So, if you happen to already have parallel lines on either side of the original line segment, you don’t have to draw new ones a ruler width away.
It even works when the two extra parallel lines are on the same side of the original line, rather than on opposite sides, though the picture isn’t nearly so symmetrical.
It’s also worth noting that you can actually start the new equal-length segment at any point on the extended line \(AB\), by using that point in the place of \(B\) at step 4. That’s why Wernick called this one copy a line segment instead of double.
Copy a line segment along its line to an arbitrary point
0. Start with a line segment of any length that is part of a longer line, and another point. Call the endpoints of the line segment \(A\) and \(B\), and the other point \(C\).
1. Side-align the line segment and draw along the opposite side of the ruler to make a parallel line. Repeat on the other side of the line segment.
2. Draw a line through \(A\) so that it meets both parallel lines on either side. Call these points \(P\) and \(Q\).
3. Draw the line \(PB\) and find where it meets the parallel line through \(Q\). Call this point \(X\).
4. Draw the line \(QC\) and find where it meets the parallel line through \(P\). Call this point \(Y\).
5. Draw the line \(XY\) and find where it meets the extended line segment. Call this point \(D\).
There is another process for doubling a line segment mentioned in Wernick and Birrel. I’m describing it a little differently here.
Double a line segment
0. Start with a line segment of any length.
1. Side-align the line segment and draw along that side of the ruler to extend the segment in the direction you want to double it. Call the enpoint of the line segment on the extended side \(B\) and the other endpoint \(A\).
2. Side-align the extended line segment and draw along the other side of the ruler to make a parallel line.
3. Draw a line through \(A\) that meets the parallel line you just drew. Call the point where it meets that line \(P\).
4. Align the point \(P\) on the side of the ruler so that the opposite side of the ruler is on the same side as \(B\). Draw along the opposite side of the ruler to make a parallel line. You need enough of the line to be able to side-align it later and to find where it meets the line drawn at step 2. Call this meeting point \(Q\). Note it will work best if you make sure \(PQ\) is longer than \(AB\).
5. Side-align the line drawn in step 4 on the other side, and draw along the opposite side of the ruler to make a parallel line. You only need enough of this line to find where it meets the line drawn at step 2. Call this meeting point \(R\).
6. Draw the line \(QB\) and find where it meets the line \(PA\). Call this meeting point \(X\).
7. Draw the line \(XR\) and find where it meets the extended original line segment. Call this meeting point \(C\).
Triangles \(\triangle AXC\) and \(\triangle PXR\) are similar because they share the angle at \(X\) and the opposite edges are parallel. Therefore the proportions \(AX:PX\) and \(CX:RX\) are the same.
Triangles \(\triangle AXB\) and \(\triangle PXQ\) are also similar for the same reason as before. Therefore the proportions \(AX:PX\) and \(AB:PQ\) are the same.
Triangles \(\triangle BXC\) and \(\triangle QXR\) are also similar for the same reason as before. Therefore the proportions \(CX:RX\) and \(BC:QR\) are the same.
Therefore the proportions \(AB:PQ\) and \(BC:QR\) must be the same. But the lengths \(PQ\) and \(QR\) are the same which means that \(AB\) and \(BC\) must be the same length too, to keep the same proportions.
End proof!
I love how we solved the problem of doubling a segment that’s too short by making a different segment that’s longer than the ruler width, doubling that, and relating the two together. Indeed, in Wernick and Birrell, they actually describe the process by saying to just create three equally spaced points on that first parallel line you drew using the process for lines longer than the ruler width. I wanted to include it all in one process.
If you want it to turn out the most nicely, you do have to decide on how widely spaced those three points are based on how long the original segment is. You’ll probably only use this process if the segment is shorter than the ruler width, in which case it won’t really matter that much, but you never know, so I’m just telling you just in case.
It’s worth noting again that this will work fine if your parallel line isn’t one ruler width away, so if you happen to already have a parallel line, you can use it just fine. It will also work if you just happen to already have three equally spaced points on that parallel line. Note, though, that if the spacing is shorter than the original segment, then the point \(X\) will be on the other side of the parallel line.
Finally, just like last time, you can modify this process to make as many copies of \(AB\) as you like by first making more copies of \(PQ\).
This process can also be modified to divide a segment into equal parts, instead of multiplying it.
Bisect a line segment
0. Start with a line segment of any length. Call the endpoints \(A\) and \(C\).
1. Side-align the line segment and draw along the opposite side to make a parallel line.
3. Draw a line through \(A\) that meets the parallel line you just drew. Call the meeting point \(P\).
4. Align the point \(P\) on the side of the ruler so that the opposite side of the ruler is on the same side as \(C\). Draw along the opposite side of the ruler to make a parallel line. You need enough of the line to be able to side-align it later and to find where it meets the line drawn at step 2. Call this meeting point \(Q\). Note it will work best if you make sure \(PQ\) is longer than \(AB\).
5. Side-align the line drawn in step 4 on the other side, and draw along the opposite side of the ruler to make a parallel line. You only need enough of this line to find where it meets the line drawn at step 2. Call this meeting point \(R\).
6. Draw the line \(RC\) and find where it meets the line \(PA\). Call this meeting point \(X\).
7. Draw the line \(XQ\) and find where it meets the extended original line segment. Call this meeting point \(B\).
The proof is identical to the proof of the previous construction, because except for the lengths of the segments involved, the diagram is identical to the diagram in the previous construction, even though we drew the lines in a slightly different order.
Again if you happen to have an existing line parallel to the segment, you can use that and it will still work. And if you happen to have equally spaced points on that parallel line, you can use them instead of making your own (though the point \(X\) may be on the other side of the parallel line if the points are too close together.) And just like before you can modify it to make as many copies of \(PQ\) as you want before joining \(C\) to the outermost point, and so divide the segment into as many pieces as you desire.
If you count the number of lines drawn to do this construction, including making the parallel line and the equally spaced points, you’ll get a total of six lines. This is one more than the number of lines it took to perpendiculalry bisect a line segment longer than the ruler width by drawing a rhombus around it. However, if you already have a parallel line or a parallel line and some points marked on it, this process will be much quicker because you won’t have to draw those lines yourself.
I find this “line saving” calculation fascinating. You can make arguments about a procedure being equally or less efficient than another, but in the right context, it might be more efficient because of being able to reuse already-drawn lines. It feels so satisfying to me to find these moments in the more complicated constructions.
There’s only one more construction I want to do in this post, which is to copy a length from one arm of an angle to the other. Both Wernick and Birrell have the same construction for this, which involves drawing a perpendicular line (a construction I’ve put later). But Chris Tisdell has a different construction, which is more direct and I like it very much. That’s the one I present here, except I don’t draw all the lines he draws.
Copy a length from one arm of an angle to the other
0. Start with two lines forming an angle, and a line segment along one of the arms starting at the vertex. Call the vertex of the angle \(P\), and the other end of the line segment \(A\).
1. Draw a rhombus with the angle as one of its angles and bisect that angle, as described earlier. Call by \(Q\) the vertex of the rhombus on same arm as \(A\), call by \(R\) the vertex of the rhombus between the two angle arms, and call by \(S\) the vertex of the rhombus on the other arm.
2. Draw the line \(AS\) and find where it meets the angle bisector \(PR\). Call this point \(X\).
3. Draw the line \(QX\) and find where it meets the second arm of the angle \(PS\). Call this point \(B\).
Consider triangles \(\triangle PSX\) and \(\triangle PQX\).
Side \(PS\) is the same length as side \(PQ\), since they’re two sides of a rhombus.
Angle \(\angle SPX\) is the same as \(\angle QPX\) since we set up to bisect the original angle.
Side \(PX\) is shared.
Therefore \(\triangle PSX\) and \(\triangle PQX\) are congruent.
Therefore \(\angle PXS\) is the same as \(\angle PXQ\).
Also note that \(\angle SXB\) is the same as \(\angle QXA\) since they’re vertically opposite.
Adding these angles together, \(\angle PXB\) is the same as \(\angle PXA\).
Now consider triangles \(\triangle PXB\) and \(\triangle PXA\).
We just showed \(\angle PXB\) is the same as \(\angle PXA\).
We already said \(\angle BPX\) is the same as \(\angle APX\) since we set up to bisect the original angle.
Side \(PX\) is still shared.
Therefore \(\triangle PXB\) and \(\triangle PXA\) are congruent.
Hence length \(PB\) is the same as length \(PA\).
End proof!
I had a lot of fun making this proof, since I hadn’t done triangle congruence proofs for a while. I’m sure there are other ways to do it, but this is mine.
I was a bit worried how it might look when the length of \(PA\) was shorter than the side length of the rhombus. Here’s the diagram in that case, with all the point labels done as described in the construction.
If you follow the proof with this diagram, all the arguments work exactly the same, except for one moment. In the previous proof, you know \(\angle PXB\) is the same as \(\angle PXA\) because they were made by joining other angles together. In this case, you know they’re the same because you subtract angles from each other. But it still works.
There’s something very interesting about these diagrams too. The triangle \(\triangle APB\) is isosceles, which means the bisector of the angle at \(P\) will meet the opposite side \(AB\) in a right angle. So we haven’t just copied the length \(PB\) to the other side of the angle, we’ve also dropped a perpendicular from \(A\) to the line \(PR\). This is going to be very important in the next blog post, which is about perpendicular lines.
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.
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.
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.
First, drawing a right angle.
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.
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 \(A\).
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.
The second traditional construction is bisecting a line segment, but we can only do it for lengths longer than the ruler width at the moment.
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.
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.
The final traditional construction we can now do is bisecting an angle.
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.
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 \(B\)
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-aligned 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.
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.
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.
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.
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.
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.
This blog post series is about geometric constructions using a tool that is two long parallel straight edges, basically exactly like a physical ruler, except without any marks to measure lengths with (or at least, you don’t use the marks).
This tool is variously known as a parallel straightedge, a fixed-width straightedge, a double-edged straightedge, or a double straightedge. I like to call it a two-sided ruler, or just a ruler if you already know I’m talking about a ruler with two sides. (Though, I do have to say the phrase “double-edged” does have a certain appeal because it makes it sound like a sword.)
The reason I have to specify that the ruler has two sides is that in classical ancient Greek geometry, the tools were a compass (for drawing circles) and a one-sided ruler (for drawing lines).
There are any amount of things you can read about classical ruler and compass constructions, and you can find out about a dizzying array of cool art and mathematics that can be made using them. However, you can construct all the same things with a two-sided ruler that you can with a ruler and compass (except the actual curve of a circle). Also it seems to me that rulers are much easier to come by and easier to manipulate with your hands than compasses, so I have come to love the idea of two-sided ruler constructions. Indeed, anyone who has been near me recently will likely tell you that I’ve been obsessed with them.
According to Florian Cajori [3], two-sided ruler constructions were comprehensively described by August Adler in 1890 [1], though that work is hard to come by and is in German anyway, so I haven’t read it. However, Sandra Kay Birrell did a great job of describing them in her 1983 Masters thesis [2] and I have read that. (The part on double-edged straightedge constructions begins on page 50.) She says this chapter was inspired by a paper by William Wernick [5], which actually has a few nice extra constructions that Birrell didn’t include. Chris Tisdell has done some videos on parallel straightedge constructions too [4], though I watched them long after beginning my own investigations.
In the blog posts that follow are my own idiosyncratic way of making sense of two-sided ruler constructions. It’s in a different order than those before me have done. Indeed, it’s actually been quite tough to put it all in an order that makes sense, and I have to say I fully believe now that Euclid’s most spectacular achievement when he wrote his Elements was deciding what order to put his propositions in.
Anyway, this presentation contains many constructions of my own design, some constructions I’ve modified from other people’s designs, and some of my own proofs of constructions that other people designed. You can assume anything where I haven’t explicitly said where it came from is mine.
If you get nothing else from it, I hope you can see how much I enjoyed thinking about and playing with this, because I really did. Maybe you’ll catch some of that joy too. Or at least seek it in your own places.
This YouTube playlist has all the videos of all the constructions in order.
Index of constructions and other results
This is a more comprehensive list of the constructions, definitions and other results mentioned in the blog posts, in case you (or me) are looking for something specific.
[1] Adler, August. “Ueber die zur Ausfithrung geometrischer Constructionsaufgaben zweiten Grades not- wendigen Hilfsmittel.” Wiener Sitzungsberichte d. Akademie d. Wiss., Math.-Naturw. Classe, 99 (1890): 846-859.
[2] Birrell, Sandra Kay. “Euclidean constructions: alternate tools to the traditional compass and straightedge.” Master’s thesis, California State University, Northridge, 1983. https://scholarworks.calstate.edu/downloads/0c483n19f
[3] Cajori, Florian. “A Forerunner of Mascheroni.” The American Mathematical Monthly 36, no. 7 (1929): 364–65. https://doi.org/10.2307/2298942
In the previous post, I factorised a quadratic equation with a fraction coefficient by thinking of numbers that add to 1/6 and multiply to -2. This is how I did it:
2/6+(-1/6)=1/6, 2/6×(-1/6)=1/3×(-1/6)=-1/18, which is not low enough. 3/6+(-2/6)=1/6, 3/6×(-2/6)=1/2×(-1/3)=-1/6, which is lower but not low enough, and I’ve got quite a long way to go. 7/6+(-6/6)=1/6, 7/6×(-6/6)=7/6×(-1)=-7/6, so much closer. 8/6+(-7/6)=1/6, 8/6×(-7/7)=… yeah that won’t work out right. 9/6+(-8/6)=1/6, 9/6×(-8/6)=3/2×(-4/3)=-2 yay!
But how could I be sure that counting in sixths would eventually get me to the right place? What if it was some other denominaror I needed?
Well, it turns out that if you want two numbers with specific rational sum and product, they are guaranteed to be able to be written with a specific denominator, so you will be able to try only numbers with that denominator.
The theorem
Theorem: Let \(a\), \(b\), \(c\) and \(d\) be integers with \(b\) and \(d\) not zero. Suppose there are rational numbers \(x\) and \(y\) such that \(x+y=\frac{a}{b}\) and \(xy=\frac{c}{d}\). Then it is possible to write both \(x\) and \(y\) with denominator \(bd\).
I would like to show my proof for this theorem. There might be an easier way than I did it, but I definitely enjoyed my way, so I want to share it. It’s actually two proofs, but the first one makes me feel a little uncomfortable. I’ll do that one first.
Proof 1
Since \(x\) and \(y\) are rational, let \(x=\frac{p}{q}\) and \(y=\frac{r}{s}\) for integers \(p\), \(q\), \(r\), and \(s\) with \(q\) and \(s\) not zero.
Thus both the sum and the product can be written with denominator \(qs\). The numbers \(x\) and \(y\) themselves can also be written with that denominator, as
In other words, the numbers \(x\) and \(y\) can be written with the same denominator as the common denominator of the sum and the product. The sum \(\frac{a}{b}\) and the product \(\frac{c}{d}\) have \(bd\) as a common denominator, so that means \(x\) and \(y\) can be written with this denominator.
Interlude
I am certain this proof is watertight, except for maybe tidying up the idea that if a two fractions can be written with a common denominator, then they can be written with any of the common denominators.
But still it feels a bit uncomfortable somehow. It gives a whiff of circular reasoning, maybe, or at least I don’t directly talk about the original denominators \(b\) and \(d\) anywhere until I reveal them at the last moment, which feels sneaky. I’d much prefer a proof that begins with the equations and ends with solutions with the correct denominators. And so I have this proof too.
Using completing the square or the quadratic formula,
\[x = \frac{ad\pm\sqrt{(ad)^2-4b^2cd}}{2bd}\]
Now \((ad)^2-4b^2cd\) is an integer, so \(\sqrt{(ad)^2-4b^2cd}\) is either unreal, irrational or an integer. Since \(x\) is rational, that means it must be an integer. Let it be \(m\), so that
\[x = \frac{ad\pm m}{2bd}\]
Suppose \(ad\) is odd. Then \((ad)^2\) is odd, and so is \((ad)^2-4b^2cd\), and therefore so is \(\sqrt{(ad)^2-4b^2cd}=m\). But now \(ad\pm m\) is even.
Alternatively suppose \(ad\) is even. Then \((ad)^2\) is a multple of 4, and so is \((ad)^2-4b^2cd\), which means \(\sqrt{(ad)^2-4b^2cd}=m\) is even. But now again \(ad\pm m\) is even.
Hence \(x\) can be written with an integer numerator and denominator as
Because of the symmetry in the original equations, this is also the two solutions for \(y\). Thus both \(x\) and \(y\) can be written with denominator \(bd\).
Conclusion
I particularly enjoyed that second proof. I was convinced when I did it the first time that the denominator had to be \(2bd\) but then I realised that the numerator had to be even and so it came down to \(bd\) after all, which was very satisfying.
And I am very happy that this means I can now factorise monic quadratics with rational coefficients by directly working through numbers with one specific denominator that add to the x-coefficient. It just feels like such a bold move to me, and now I know it just seems bold, because it provably will definitely work.
This blog post is about a way to define addition and multiplication on a number line using the geometry of the plane that surrounds the line.
Recently I talked about how numbers have multiple purposes and one of those purposes is locating. When you draw a number line, that’s pretty much what you’re doing – saying that the numbers are exactly locations on that line. There is an issue with this idea, which is this: how do you add or multiply locations?
The usual way of dealing with this is to think of numbers as not just locations but journeys too, so that when we see something like 10+3, the first number is a location and the second one is a journey and the answer 13 is the location we arrive at by beginning at 10 and travelling onwards 3. This is fine and I like it very much, actually, but there is still the huge problem of how to multiply locations or indeed how to multiply journeys. You can say that 3×10 is three lots of a journey of 10, which makes sense, but now the 3 is neither a journey nor a location but a literal amount and we have three different things so far that the numbers mean. Is there a way that preserves the location-ness of everything?
Yes, there is.
There is indeed a way to define addition and multiplication of locations on a number line that preserves their fundamental location-ness, by looking outwards to the other lines of the plane your line is part of. The system doesn’t directly use lengths or angles, but only uses the most fundamental geometrical actions of drawing lines through two points, finding where two lines meet, and drawing lines parallel to other lines. This is my favourite thing about it, that it’s fully based on the relationships between the points and the lines as objects. As a pure mathematician and a finite geometer specifically, geometry isn’t really about measurements at all but is all about relationships, so something that focuses on the relationships deeply appeals to me.
I created this method in October 2020, heavily based on the method invented by Marshall Hall Jr in the late 1950s. Hall’s method works by adding coordinates to all the points in the plane including the ones on your number line, making equations for the lines, and referring to the coordinates of specific points on specific lines to define the arithmetic. I studied his method in 2001 while doing the honours year in my undergraduate maths degree, but it wasn’t until 19 years later that I realised there was a way to do it without referring to coordinates, by focusing my attention on the number line itself rather than the coordinate axes. My method for multiplication also has striking similarities to René Descartes’ original definition for the multiplication of lengths published in the 1630s, even though I didn’t mean it to and only found this out later. An important difference between them is that his method has the two factors on different sides of a triangle instead of along the same line. I find it very interesting that Hall’s book is called “The Theory of Groups” and Descartes’ book is called “Geometry”, highlighting the deep connection between geometry and algebra which all three methods point to.
Anyway, enough historical notes. Let’s get to it.
How to do geometric arithmetic
You can watch me doing both processes live in a video here, or you can read a text description and see screenshots from the video below.
Setting up
To do parallel line arithmetic, you need to set up a few things.
First, choose a line to be your number line. The points on the number line are your numbers.
Next, you’ll need to choose two different points on the line to call 0 and 1. The point 0 will be important for defining addition and both 0 and 1 will be important for defining multiplication.
Finally, you’ll need to choose two lines other than the number line itself that are not parallel to the number line and not parallel to each other. It doesn’t really matter where they are because the specific lines themselves aren’t important, only their directions, since during the constructions for addition and multiplication, you won’t be making them intersect with any lines. Instead you will draw several lines parallel to each of these lines. I call one direction “home” and the other “away”. (The reason why I chose these words specifically rather than “in” and “out” for example is because I am Australian and “Home and Away” means something to me.) To make it easier to focus when I’m drawing diagrams for the arithmetic, I usually put my home and away lines a bit away from the part of my number line I drew.
Addition
To add two numbers geometrically, you follow this process.
First, you need to have your two points a and b, and the point 0 on your number line. You don’t need a and b to be different from each other or different from 0, but I’ve drawn them as different to make it easier to show how the process works.
Create a journey from 0 to a first following the away direction and then following the home direction. That is, draw a line through 0 parallel to the away line.
Then draw a line parallel to the home line that passes through a.
Find the point where these two lines meet. If you follow the journey from 0 to a first along the away direction and then along the home direction, this point is where the journey turns from going away to going home.
Now draw a line parallel to the number line, through this turning point. This is the turning line for all additions that go some number plus a.
Now you are ready to do b+a.
Draw a line through b parallel to the away line and find where it meets the turning line. This is where the journey from b will turn and return home to the number line.
Now draw a line through this turning point parallel to the home line, and find where it meets the number line.
This point is the point b+a.
Multiplication
To multiply two numbers geometrically, you follow this process, which is similar to the process for addition, but with two very important differences.
First, you need to have your two points a and b, and the two points 0 and 1 on your number line. Just like before, you don’t need a and b to be different from each other or different from 0 or 1, but I have chosen them that way to make it easier to show how the process works.
Create a journey from 1 to a first following the away direction and then following the home direction. (This is the first point of difference between multiplication and addition, that the journey begins at 1 and not 0.)
That is, draw a line through 1 parallel to the away line.
Then draw a line parallel to the home line that passes through a.
Find the point where these two lines meet. If you follow the journey from 1 to a first along the away direction and then along the home direction, this point is where the journey turns from going away to going home.
Now draw a line that passes through both this turning point and 0. This is the turning line for all multiplications that go some number times a. (This is the second point of difference between multiplication and addition, that the turning line passes through 0 rather than being parallel to the number line.)
Now you are ready to do b×a.
Draw a line through b parallel to the away line and find where it meets the turning line. This is where the journey from b will turn and return home to the number line.
Now draw a line through this turning point parallel to the home line, and find where it meets the number line.
This point is the point b×a.
Thoughts
So that’s David Butler’s methods of geometric addition and multiplication. I love it so much, especially the multiplication. Addition is a little more complicated than you’d expect if you are used to adding numbers by joining journeys head to tail, but multiplication somehow feels so much simpler than it has a right to be.
My favourite thing to do is to convince myself that various algebraic properties of numbers have to be true using these two definitions of addition and multiplication. For example, this diagram shows that a+b = b+a (at least for positive numbers).
And this diagram shows that b+b = b×2 (at least for numbers more than 1).
And this diagram shows that when you multiply two negative numbers, you get a positive number. (I know those numbers are negative because they’re on the opposite side of 0 from 1.)
Yeah none of them are formal proofs, but I still love them.
The truly remarkable thing is that it doesn’t matter how you choose your home and away directions, it will stillwork! For example, here are three diagrams showing 1+1=2 and 2×2=4.
The geometry of the real plane is so neatly structured that it works every time. If our number line was in a plane with a different structure, this might not work the same every time. Indeed, you may end up with entirely different rules for how addition or multiplication work, such as multiplication not being associative.(That is, (a×b)×c not being the same as a×(b×c), which is very annoying!)
I may do future blog posts about some of that other stuff, but for now, I’m enjoying just revelling in the coolness that is the existence of a method for multiplying locations, and the even higher coolness of watching geometry cause the algebraic properties of numbers. I hope you enjoyed it too.
