In the last post, we used several theorems we’d developed to arrive at Menelaus’ theorem. However, at the very end Ptolemy simply mentions another version of the theorem, but does not derive it. I simply took his word that it worked, but as that alternative form is used first thing in the next chapter, I want to make sure at the very use, we know how to use it, even if we don’t go through how it’s derived.
First, let’s set up a generic Menelaus configuration on a sphere which is the intersection of the arcs of four great circles:
The first thing we should understand is that, of the four arcs, two are interior to the other two. To help visualize that, let’s label things:
Here we can more clearly see that we have the four arcs, m, n, r, and s where r and s are bound by m and n. In addition, because of how they interact, each is broken into two parts.
If we want to state the form of Menelaus’ theorem that we derived in the last post in terms of these variables, it would be:
$$\frac{Crd \; arc \; 2 m_2}{Crd \; arc \; 2 m_1} = \frac{Crd \; arc \; 2 r_2}{Crd \; arc \; 2 r_1} \cdot \frac{Crd \; arc \; 2 n_2}{Crd \; arc \; (n_2 + n_1)}$$
I’ll note that the translation of the Almagest I’m using makes frequent reference to another work, A History of Ancient Mathematical Astronomy by O. Neugebauer. In that work, Neugebauer has numbered the two variations on Menelaus’ theorems and because he derived them in reverse order, the above is often referred to as Menelaus’ Theorem II (MTII). For the sake of consistency and clarity, I have also adopted this notation.
Here, I’ve written out $arc \; n$ as $n_2 + n_1$ because I wanted to drive the point home that it’s the full arc, and I didn’t just forget a subscript.
In this version of Menelaus’ theorem, what we’ve done is related four outer parts ($m_2$, $m_1$, $n_2$, and $n$) to two inner parts ($r_1$ and $r_2$). And if we look carefully, we didn’t even use any of the $s$ segments.
Next, we can also notice that we can easily do some algebra and flip this equation around. We can easily move one side to another simply by taking it across the equals sign and flipping it over1. So we could easily rewrite the equation in this form:
$$\frac{Crd \; arc \; 2 r_2}{Crd \; arc \; 2 r_1} = \frac{Crd \; arc \; 2 m_2}{Crd \; arc \; 2 m_1} \cdot \frac{Crd \; arc \; 2 n}{Crd \; arc \; 2n_2}$$
Or, if we felt like it, this one:
$$\frac{Crd \; arc \; 2 m_1}{Crd \; arc \; 2 m_2} \cdot \frac{Crd \; arc \; 2 r_2}{Crd \; arc \; 2 r_1} \cdot \frac{Crd \; arc \; 2 n_2}{Crd \; arc \; 2n} = 1$$
We’ve got options.
Now, let’s look at the other equation Ptolemy tossed in at the very end of the chapter. Written using the notation here we get:
$$\frac{Crd \; arc \; 2m}{Crd \; arc \; 2m_1} = \frac{Crd \; arc \; 2r}{Crd \; arc \; 2r_1} \cdot \frac{Crd \; arc \; 2s_2}{Crd \; arc \; 2s}$$
In this one, we’ve related four inner parts ($r$, $r_1$, $s$, $s_2$) to to outer ones ($m$ and $m_1$).
Again, we could play algebra games with this and change the form. I won’t write it out here.
Notice in this form of the equation, $n$ doesn’t come into play at all. In both forms, out of the four great circle arcs, we only end up using three. Thus, if we start seeing three great circle arcs intersecting, we should probably start thinking Menelaus.
I’ll again note that Neugebauer has numbered this version of the theorem as Menelaus Theorem I (MTI).