To get at the chord length for an angle of $\frac{1}{2}$º, Ptolemy makes use of a proof from Aristarchus. It starts with this diagram:
This one is a bit different than the other ones we’ve seen as it has an arc in it. But before we get into that, let’s talk about the setup. Ptolemy notes the setup is such that $\angle{ABG}$ is bisected by $\overline{BD}$. In addition, $\overline{D\Theta} \perp \overline{AG}$. Also, $\overline{AB} < \overline{BG}$.
Because $\angle{ABD}$ and $\angle{DBG}$ are equal and on the edge of a circle, the chords they subtend will also be equal. Thus $\overline{AD} = \overline{DG}$.
Ptolemy again turns to Euclid, but with a bit of a twist. The original theorem states that if you have a triangle that has an angle bisected by a line that extends to the base, if you take the ratio of the adjacent sides, that ratio will be the same as the ratio of the two parts of the bisected side.
In our diagram here, the triangle in question is $\triangle{ABG}$. So if we wanted to directly apply Euclid’s theorem, it would be implying:
$$\frac{\overline{AB}}{\overline{BG}} = \frac{\overline{AE}}{\overline{EG}}$$
The twist comes in from remembering that $\overline{AB} < \overline{BG}$. Thus, in that inequality we just stated, the numerator is smaller than the denominator on the left. And since it’s an equals sign, the same must be true for the right. Thus we get that $\overline{AE} < \overline{EG}$.
The next few statements are a bit fuzzy. Ptolemy states that $\overline{AD} > \overline{DE}$ and $\overline{DE} > \overline{DZ}$. There’s no solid proof given for this, but we can still take it as true because the shortest distance between two points is a straight line. Since $\overline{DZ} \perp \overline{AD}$, that’s a straight line. $\overline{DE}$ is at an angle and $\overline{AD}$ a steeper angle still.
This next part’s going to get a bit funny, but it will explain where that odd arc $HE\Theta$ is coming from. What that is, is a segment of another circle with center D as illustrated below.
It is drawn such that the radius is $\overline{DE}$. Then, because of the inequalities we stated above ($\overline{AD} > \overline{DE}$ and $\overline{DE} > \overline{DZ}$) we know that H must fall along $\overline{AD}$ and $\Theta$ must be beyond Z on $\overline{DZ}$.
Now Ptolemy starts turning to areas. Since $\Theta$ is beyond Z, the area of sector $DE\Theta > \triangle{DEZ}$. Similarly, $\triangle{ADE} > DEH$. From this we can state the following:
$$\frac{\triangle{DEZ}}{\triangle{ADE}} < \frac{sector DE\Theta}{sector DEH}$$
That may not seem to follow, but consider it this way: The smaller piece of the first inequality is being divided by the larger piece of the second and the larger piece of the first inequality is being divided by the smaller piece of the second. Thus, we can tell which way the inequality sign goes.
Still not feeling right? Let’s assign some random numbers that fulfill the inequality and check it out. Let’s say $ 5 > 3$ and $7 > 4$.
$$\frac{3}{7} < \frac{5}{4}$$
Even without converting these to decimals, we can clearly see the first one is <1 and the second is >1 making the inequality true. No matter how you put the numbers (so long as the initial inequalities were true) it will work out.
Next Ptolemy compares $\triangle{ADE}$ and $\triangle{DEZ}$ noting that they are related through another theorem of Euclid’s which states that, if two triangles have the same height, the proportion of their areas is equal to the proportion of their bases. Since these two triangles share a common line $\overline{DE}$ which can be used as the height this applies. Stating this mathematically:
$$\frac{\triangle{DEZ}}{\triangle{ADE}} = \frac{\overline{EZ}}{\overline{AE}}$$
Similarly, we can state:
$$\frac{\angle{EDZ}}{\angle{ADE}} = \frac{sector DE\Theta}{sector DEH}$$
No strict proof given for this one, but given that the area of the sectors is a function of the angle, they are proportional and thus the ratios of those proportions equal, so long as they are on the same circle (which they are since they’re both part of that big sketched in circle).
Now remember when above we said,
$$\frac{\triangle{DEZ}}{\triangle{ADE}} < \frac{sector DE\Theta}{sector DEH}$$
We just found some things that were equal to both sides which we can substitute in:
$$\frac{\overline{EZ}}{\overline{AE}} < \frac{\angle{EDZ}}{\angle{ADE}}$$
Next up, Ptolemy uses a cool math trick called “compenendo and dividendo.” I’d never actually encountered it before, but it’s apparently a quick way of dealing with proportions that prevents getting bogged down in a bunch of substitutions if it follows a certain format.
Fortunately, what we’ve just come up with does follow the format $\frac{a}{b} = \frac{c}{d}$ where $a = \overline{EZ}$, $b = \overline{AE}$, $c = \angle{EDZ}$, and $d = \angle{ADE}$. So popping that into the expanded form:
$$\frac{\overline{EZ} + \overline{AE}}{\overline{AE}} < \frac{\angle{EDZ} + \angle{ADE}}{\angle{ADE}}$$
Fortunately, this simplifies a bit as $\overline{EZ} + \overline{AE} = \overline{AZ}$ and $\angle{EDZ} + \angle{ADE} = \angle{ADZ}$. Thus,
$$\frac{\overline{AZ}}{\overline{AE}} < \frac{\angle{ADZ}}{\angle{ADE}}$$
Next up, we’re going to double the numerator on each side. Again, this is totally mathematically legal since we’re doing the same operation to both sides. But instead of just multiplying by 2, we can see a physical meaning in our diagram. Since $\overline{AZ} \perp \overline{DZ}$ this means $\overline{AZ}$ is half of $\overline{AG}$. So $2 (\overline{AZ}) = \overline{AG}$. Similarly $2(\angle{ADZ}) = \angle{ADG}$. Thus, the equation gets rewritten as:
$$\frac{\overline{AG}}{\overline{AE}} < \frac{\angle{ADG}}{\angle{ADE}}$$
And suddenly we’re back in a format of $\frac{a}{b} = \frac{c}{d}$ which means we can apply the dividendo portion of the theorem except now our $a = \overline{AG}$ and $c = \angle{ADG}$. So:
$$\frac{\overline{AG} – \overline{AE}}{\overline{AE}} < \frac{\angle{ADG} – \angle{ADE}}{\angle{ADE}}$$
And again, we need to take a look at what these differences really mean. $\overline{AG} – \overline{AE} = \overline{EG}$ and $\angle{ADG} – \angle{ADE} = \angle{EDG}$. So rewriting our dividendo equation:
$$\frac{\overline{EG}}{\overline{AE}} < \frac{\angle{EDG}}{\angle{ADE}}$$
Before continuing, we’ll make a slight change to what we’re calling the pieces on the right side. In both the numerator, and denominator, we’ve defined those angles with respect to E. But the line that makes up the side of that angle extends past E, so it’s entirely consistent to use B instead. As such we can rewrite the last line as:
$$\frac{\overline{EG}}{\overline{AE}} < \frac{\angle{BDG}}{\angle{ADB}}$$
Next, we’re going to jump a ways back to the first time we looked at Euclid in this post and established $\frac{\overline{AB}}{\overline{BG}} = \frac{\overline{AE}}{\overline{EG}}$. Flipping that over we now have something to substitute in for the left side. But before doing so, we can also get a substitution for the fight side by noting that the ratio of the arcs of angles are directly proportional to their angles. In other words:
$$\frac{\angle{BDG}}{\angle{BDA}} = \frac{arc{BG}}{arc{AB}}$$
This can be substituted in for the right side above. So swapping out both sides we get:
$$\frac{\overline{BG}}{\overline{AB}} < \frac{arc{BG}}{arc{AB}}$$
This is what Ptolemy set out to prove with this exercise. Generalizing it a bit, we can state that the ratio of two chords is always less than the ratio of their respective arcs.
To see why Ptolemy wants this and how an inequality is going to get us to the chord of $\frac{1}{2}$º, let’s apply it:
After all that work up, finally a simple diagram. Shockingly so in some ways. But looking back at the generalized idea we got, all it says is we relate two chords to their respective arcs.
So let’s assign some values to them. Since we’re trying to that elusive chord of $\frac{1}{2}$º, let’s work with the smallest chords we’ve yet been able to get at: the chords of $\frac{3}{4}$º and of 1º assigning them to $\overline{AB} and \overline{BG}$ respectively. We don’t have the chord of 1º yet, but this should let us find it.
Applying our chords and arcs to the generalized inequality we can state:
$$\frac{\overline{AG}}{\overline{AB}} < \frac{arc{AG}}{arc{AB}}$$
Before moving on, recall that when I discussed Ptolemy’s math, one of the conventions he used was that he breaks the circumference of a circle into 360 parts which means that the arc will have the same number of parts as the central angle it subtends has degrees1. So keeping that in mind, lets substitute into our equation:
$$\frac{\overline{AG}}{\overline{AB}} < \frac{1}{\frac{3}{4}}$$
$\overline{AG}$ is the chord we’re trying to solve for (since it’s the 1º that we can then divide via the second corollary to get to $\frac{1}{2}$º), so let’s do a bit of algebra to isolate it:
$$\overline{AG} < \frac{4\overline{AB}}{3}$$
When we explored the corollaries, I didn’t go through the math to actually calculate the chord of $\frac{3}{4}$º, but just stated it’s possible to get to. I’ll still not bother showing it just now, but simply state that using the method previously described the chord of $\frac{3}{4}$º is 0;47,8 which is our value for $\overline{AB}$. So we can substitute that into our equation to solve for $\overline{AG}$
$$\overline{AG} < \frac{4(0;47,8)}{3}$$
$$\overline{AG} < 1;2,50$$
Well, that’s a good start. We know $\overline{AG}$ is less than that value, but that still leaves a lot of room. However, if we can come at it from the other side and see what it’s greater than, the hope is we can narrow it down sufficiently that whatever uncertainty remains is below the margin of error for observations and thus not relevant2.
So since we’ve squeezed it on one side with a lower limit, let’s apply the same reasoning to create an upper bound which we can do since we demonstrated we could determine the chord length of $1\frac{1}{2}$º previously as well. However, when we do this, the $1\frac{1}{2}$º chord will be the longer meaning we’ll need to switch our $\overline{AB}$ and $\overline{AG}$ so that $\overline{AB}$ is our 1º chord.
$$\frac{\overline{AG}}{\overline{AB}} < \frac{\frac{3}{2}}{1}$$
$$\frac{\overline{AG}}{\frac{3}{2}} < \frac{\overline{AB}}{1}$$
$$\frac{2\overline{AG}}{3} < \overline{AB}$$
Again, I didn’t show the calculation of the chord of $1\frac{1}{2}$º, but sparing the math it is 1;34,15. Plugging that in:
$$\overline{AB} > \frac{2(1;34,15)}{3}$$
$$\overline{AB} > 1;2,50$$
Compare that to our previous inequality for $\overline{AB}$. To the degree of accuracy we’ve calculated, it is both greater than and less than 1;2,50 which means it’s sufficiently pinned down we can make use of it.
Applying the second corollary we can then determine the chord of $\frac{1}{2}$º is 0;31,25.
We’ve come a long way, but we can now apply the various corollaries to determine every angle between 0º and 180º. Since we’ve approximated the chord of $\frac{1}{2}$º, it’s probably best to use that value as sparingly as possible since repeated addition also increases the uncertainty. As such, we should always start from chords we determined more rigorously and only add and subtract these estimated chords when absolutely necessary.
To ensure that he wouldn’t have to undergo these calculations every time, Ptolemy recorded them in a table for handy reference. These were known as Ptolemy’s Table of Chords.
External References:
- Looking at the diagram, this might seem confusing as we haven’t drawn in a central angle. The chords are existing free of them. This is true, but all chords still have that central angle regardless of whether or not we’ve drawn it in so if it helps you visualize it, go ahead and mentally draw it in yourself. Just don’t let $\angle{BAG}$ trip you up.
- This concept is known as a squeeze theorem. If you haven’t taken calculus, you probably haven’t encountered it before. However, while it is generally taught in calculus, the general idea of it (determining the value of an unknown by squeezing it between two known values that are extremely close on either side), is one that has obviously been around since antiquity as Archimedes used one to approximate $\pi$.