Given my blog title, how have I gone this long without discussing triangle geometry? I will rectify this gross negligence in the next few weeks. Let’s begin with a seemingly impossible fact due to Frank Morley.

Start with a triangle *ABC*, with any shape you wish. Now cut its angles into three equal parts, and extend these **angle trisectors** until they meet in pairs as illustrated below, forming triangle *PQR*. Morley’s amazing theorem says that this **Morley triangle**, *PQR*, will always be equilateral!

Why would this be true? Well, Morley’s theorem tells us that this diagram has three nice 60-degree angles in the middle, but we may suspect that, in fact, *all* of the angles are nice! This key insight lets us piece together the following argument, where we build up the diagram *backwards* from its constituent pieces. Draw seven separate “puzzle piece” triangles with angles and side-lengths as shown below, where the original triangle *ABC* has angles \(3\alpha,3\beta,3\gamma\) respectively. (Be sure to check that puzzle pieces with these specifications actually exist! Hint: use the Law of Sines.)

Now, fit the pieces together: all matching edge-lengths are equal (by design), and the angles around vertex *P* add up to \(60+(\gamma+60)+(\alpha+120)+(\beta+60)=360\) and similarly for *Q* and *R*, so the puzzle fits together into a triangle similar to our original triangle *ABC*. But now these pieces must make up the Morley configuration, and since we started with an equilateral in the middle, we’re done with the proof!^{[1]}

But there’s more to the Morley story. Let’s push *A* and *C* to make them swap positions, dragging the trisectors and Morley triangle along for the ride (a process known as **extraversion**). We end up using some *external* angle trisectors instead of only internal ones, but the Morley triangle remains equilateral throughout. This gives us a new equilateral Morley triangle for our original *ABC*!

In fact, each vertex of *ABC* has *six* angle trisectors (three pairs of two), and if you continue applying extraversions you’ll soon uncover that our original triangle *ABC* has 18 equilateral Morley triangles arranged in a stunning configuration! (Click for larger image.)

Here’s a challenge: the diagram above has 27 equilateral triangles, but only 18 of them are Morley triangles (i.e., are made from a pair of trisectors from each vertex). Which ones are they?

### Notes

- This is a slight modification of a proof by Conway and Doyle. [↩]