Menelaus’ Theorem and Variants

Let ABC be a triangle in the Euclidean plane. Let $\ell$ be a line, meeting AB at P, BC at Q and AC at R. Then Menelaus’ theorem states that

$\frac{AP \cdot BQ \cdot CR}{PB \cdot QC \cdot RA}=-1.$

Here XY denotes the distance from point X to point Y and we are using “signed distances” which means that, for each line in the diagram, we pick an orientation of that line and measure distances to be positive or negative according to whether or not they are aligned with that orientation.

Menelaus’ theorem, according to Wikipedia, actually dates back all the way to the first century of the common era. Nowadays, I think it is safe to guess that most math majors don’t learn this result. Instead, they learn the following: Let E be a smooth plane cubic. Let $\ell$ be a line, meeting E at P, Q and R. Then $\mathcal{O}(P+Q+R) \cong \mathcal{O}(1)$. I’ll call this the “modern Menelaus’ theorem” or MMT.

But a triangle is just a singular cubic. And, as I’ll explain below the fold, Menelaus’ theorem is just the degeneration of the MMT to this singular case.

It is a little hard to state the MMT in a way that isn’t circular. If you ask for a definition of the line bundle $\mathcal{O}(1)$, you’ll usually be told that it is the restriction, from $\mathbb{P}^2$ to E, of the antitautological line bundle. After unpacking this definition for a bit, you’ll realize that what this means is that it is the line bundle $\mathcal{O}(P_0+Q_0+R_0)$, where $P_0$, $Q_0$ and $R_0$ are three colinear points on E. However, this doesn’t mean that the MMT is vacuous. There is a group of line bundles on E, called the Picard group of E. The Picard group of E has countably many components, each of which is isomorphic to E. This isomorphism is canonical for the component which corresponds to degree 1 line bundles and noncanonical for the others. As a toplogical group, the connected component of the identity is $S^1 \times S^1$ and the quotient of $Pic(E)$ by the connected component of the identity is $\mathbb{Z}$. We write $Pic^d(E)$ for the component which goes to d under the latter map. There is a point, called $\mathcal{O}(1)$, in $Pic^3(E)$ such three points P, Q and R on E are colinear if and only if $P+Q+R=\mathcal{O}(1)$, where we have used the isomorphism between $Pic^1$ and E on the left hand side.

Now, it turns out that $Pic$ is a functor, so we can take a family of cubics in which E degenerates to a triangle and get a corresponding family of groups. In general, understanding the what Picard does to singular curves is an interesting research topic — if you want to look into this, the key terms are “limit linear series” and “Neron models”. We’ll just describe what happens in the triangle case. I’ll call the triangle T and its Picard group $Pic(T)$.

The group of connected components of $Pic(T)$ is $\mathbb{Z}^3$; we’ll denote the components as $Pic^{i,j,k}(T)$. The three components $Pic^{0,0,1}(T)$, $Pic^{0,1,0}(T)$ and $Pic^{0,0,1}(T)$ are canonically identified with the interiors of the three sides of the triangle T. That is to say, each of these components is a projective line with two points deleted, these two points being the intersection of one side of T with the other two sides. As a group and as an algebraic variety, the connected component of the identity is $\mathbb{C}^*$. Let’s state Menelaus’ Theorem again:

MMT for the triangle: There is a point, which we will again call $\mathcal{O}(1)$, in $Pic^{1,1,1}(T)$ such that, if $P$, $Q$ and $R$ lie in $Pic^{0,0,1}(T)$, $Pic^{0,1,0}(T)$ and $Pic^{0,0,1}(T)$ then $P$, $Q$ and $R$ correspond to colinear points if and only if $P+Q+R=\mathcal{O}(1)$.

Pretty nice. Just a few more details to clean up to get to Menelaus’ original statement. First of all, where do the ratios of distances come from? Well, if A and B are two points on the line $\ell$, then the function $AP/PB$ gives an isomorphism between $\ell$ and $\mathbb{P}^1$ which takes A and B to 0 and $\infty$. So we see that there is some constant $\alpha$ so that P, Q and R are colinear if and only if

$\frac{AP \cdot BQ \cdot CR}{PB \cdot QC \cdot RA}=\alpha.$

How do we see that $\alpha=-1$? Consider the three colinear points obtained by intersecting $AB$, $BC$ and $AC$ with the line at infinity. QED

A few concluding notes:

If you can visualize how $Pic^1(E)$ degenerates to $Pic^{1,0,0}(T) \cup Pic^{0,1,0}(T) \cup Pic^{0,0,1}(T)$, you will have a pretty good understanding of the theory of Tate curves. If you can visualize what happens to the univeral covers of these spaces, both topologically and holomorphically, you will have an excellent understanding.

We said that the formula $AP/PB$ gives a holomorphic map from a planar line to $\mathbb{P}^1$. The same formula also works when A and B are two points of a circle and P ranges over the rest of the circle (which is a genus zero curve). (Exercise: Prove this. Hint: naively, you’d think that the square root in the definition of distance would give $AP/PB$ a branch singularity. Start by figuring out why this naive intuition is wrong.) Then deduce a variant of Menelaus’ theorem which, given a point $P$ on a line and two points $Q$ and $R$ on a circle, gives a formula for when P, Q and R are colinear.

One thought on “Menelaus’ Theorem and Variants”

1. Allen Knutson says:

It is a little hard to state the MMT in a way that isn’t circular.

Making it seem to be about conics, rather than cubics?