Two-sided ruler constructions 7: Circles without circles

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

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

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 this meeting point \(U\).
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 \(\frac{r}{d}\).
Video here

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 this meeting point \(U\).
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 \(\frac{r}{d}\).
Video here

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 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 this meeting point \(U\).
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 \(\frac{r}{d}\).
Video here

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 \(\frac{r}{d}\).
Video here

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.
Video here

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 \(x\).

By the cosine rule in triangle \(\triangle AGB\),

\[\begin{aligned} b^2 &= a^2 + c^2 – 2ac \cos(\beta) \\ 2ac \cos(\beta) &= a^2+c^2-b^2 \\ \cos(\beta) &=\frac{a^2-b^2+c^2}{2ac}\end{aligned}\]

Using the right-angled triangle \(\triangle AGX\),

\[\cos(\beta) = \frac{x}{a}\]

Therefore

\[\begin{aligned}\frac{x}{a} &= \frac{a^2-b^2+c^2}{2ac}\\ x &= \frac{a^2-b^2+c^2}{2c}\\ \phantom{x} &= \frac{a^2-b^2}{2c}+\frac{c^2}{2c} \\ \phantom{x} &= \frac{(a+b)(a-b)}{2c}+\frac{c}{2}\end{aligned}\]

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 \(\frac{a-b}{c}\).

The line \(YX\) is perpendicular to \(PQ\), so its slope is \(-\frac{c}{a-b}\). That is, for every \(c\) downwards, it goes across \((a-b)\).

From \(Y\) to \(X\) the line \(YX\) travels downwards \(\frac{a+b}{2}\), which is \(\frac{a+b}{2}\div c\) relative to \(c\). Therefore it goes across \((a-b)\times\frac{a+b}{2}\div c = \frac{(a-b)(a+b)}{2c}\).

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 \(\frac{a-b}{c}\) 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 \(B\) 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 just use the original matching radiuses, 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 \(A\).
Video here

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 \(A\).
Video here

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 \(A\).
Video here

Proof:

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.
Video here

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.
Video here

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 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.