[Math] Can Morley’s theorem be generalized

Question

Morley's theorem states that in any triangle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle.

In a talk some years ago, David Rusin made the provocative claim that Morley's theorem is a rare example of a striking theorem that defies generalization. The first ideas that come to everyone's mind—passing to higher dimensions or hyperbolic geometry for example—don't work.

The proof by Alain Connes yields a mild generalization of sorts, but not a very satisfying one in my opinion. Wikipedia claims that there are "various generalizations" of Morley's theorem, but by this it seems to mean extensions of Morley's theorem, i.e., further equilateral triangles that one can construct. This is not what I would, strictly speaking, call a "generalization."

So is David Rusin correct?

Are there no satisfactory generalizations of Morley's theorem?

Answer 1

Please forgive me if you are aware of this result (as it is linked from the Wikipedia page, albeit in another context), but there is a paper by Richard K. Guy called "The lighthouse theorem, Morley & Malfatti—a budget of paradoxes" in the American Mathematical Monthly. The eponymous theorem could be considered a generalization of Morley's theorem:

Lighthouse Theorem. Two sets of $n$ lines at equal angular distances, one set through each of the points $B$, $C$, intersect in $n^2$ points that are the vertices of $n$ regular $n$-gons.

Naturally, it is not clear how this would qualify as a generalization, but the connecting observation is the following:

The Morley Miracle. The nine edges of the equilateral triangles of the Lighthouse Theorem for $n=3$ are the Morley lines of a triangle.

Properly, the Lighthouse Theorem should be enlarged to include enough observations to make this connection. For example, the $n^2$ lines of the $n$ regular $n$-gons form $n$ families of $\binom{n}{2}$ parallel lines; if $n$ is odd, then the $n$-gons are homothetic. Moreover, there is an angle duplication result that establishes the presence of the trisectors.

From Guy's point of view, the particularly pleasant appearance of Morley's theorem is due to the fact that $\binom{n}{2} = n$ for $n=3$. For comparison, the case $n=2$ is even simpler and may be regarded as the statement that the altitudes of a triangle concur. (The $n$ $n$-gons are an orthocentric system.) The case $n=4$ gives some properties of Malfatti circles. For all of these interpretations, Guy wrestles with the "paradox" that you recover theorems about a triangle even though you don't start with any triangles.

Again, my apologies if you're aware of all of this. I imagine you may be, in which case I justify my answer as simply too long for a comment!