Probably you have heard of the impossibility of trisecting an angle. (Hold on for a quick rant about the importance of understanding your hypotheses…) What's actually true is that you can't trisect a generic angle if you accept the restriction of using the old-fashioned tools of Euclidean geometry: the compass and straight-edge. There are a lot of constructions that can't be done using just a straight-edge and compass — angle trisection, duplication of a cube¹Duplicating the cube is also known as the Delian problem — the problem comes from a pronouncement by the oracle of Apollo at Delos that a plague afflicting the Athenians would be lifted if they built an altar to Apollo that was twice as big as the existing altar. The existing altar was a cube, one meter on a side, so they carefully built a two meter cube — but the plague raged on. Apparently what Apollo wanted was a cube that had double the volume of the present altar — it's side length would have to be \(\sqrt[3]{2} \approx 1.25992\) and since this was Greece and it was around 430 B.C. and there were no electronic calculators, they were basically just screwed., squaring a circle, constructing a regular heptagon, et cetera.

If you allow yourself to use a ruler — i.e. a straight-edge with marks on it (indeed you really only need two marks a unit distance apart) then angle trisection can be done via what is known as a neusis construction. Nevertheless, because of the central place of Euclid's Elements in mathematical training throughout the centuries, and thereby, a very strong predilection towards that which is possible via compass and straight-edge alone, it is perhaps not surprising that a perfectly beautiful result that involved trisecting angles went undiscovered until 1899, when Frank Morley stated his Trisector Theorem. There is much more to this result than we will state here — so much more that the name “Morley's Miracle” that has been given to the Trisector theorem is truly justified — but even the simple, initial part of this beautiful theory is arguably miraculous! To learn more about Morley's theorem and its extension see  [8].

So, let's state the theorem!

Start with an arbitrary triangle \({\triangle}ABC\). Trisect each of its angles to obtain a diagram something like that in <<Unresolved xref, reference "fig_morley_setup"; check spelling or use "provisional" attribute>>.

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The six angle trisectors that we've just drawn intersect one another in quite a few points.

Among the points of intersection of the angle trisectors there are three that we will single out — the intersections of adjacent trisectors. In <<Unresolved xref, reference "fig_morley_1st_triangle"; check spelling or use "provisional" attribute>> the intersection of adjacent trisectors are indicated, additionally, we have connected them together to form a small triangle in the center of our original triangle.

\({\triangle}ABC\)

Are you ready for the miraculous part? Okay, here goes!

In other words, that little blue triangle in <<Unresolved xref, reference "fig_morley_1st_triangle"; check spelling or use "provisional" attribute>> that kind of looks like it might be equilateral actually does have all three sides equal to one another. Furthermore, it doesn't matter what triangle we start with, if we do the construction above we'll get a perfect \(60^\circ - 60^\circ - 60^\circ\) triangle in the middle!

Sources differ, but it is not clear whether Morley ever proved his theorem. The first valid proof (according to R. K. Guy in  [8] was published in 1909 by M. Satyanarayana  [15]. There are now many other proofs known, for instance the cut-the-knot website (http://www.cut-the-knot.org/) exposits no less than nine different proofs. The proof by Satyanarayana used trigonometry. The proof we'll look at here is arguably the shortest ever produced and it is due to John Conway. It is definitely a “Book proof”!

Let us suppose that an arbitrary triangle \({\triangle}ABC\) is given. We want to show that the triangle whose vertices are the intersections of the adjacent trisectors is equilateral — this triangle will be referred to as the Morley triangle. Let's also denote by \(A\), \(B\) and \(C\) the measures of the angles of \({\triangle}ABC\). (This is what is generally known as an “abuse of notation” — we are intentionally confounding the vertices (\(A\), \(B\) and \(C\)) of the triangle with the measure of the angles at those vertices.) It turns out that it is fairly hard to reason from our knowledge of what the angles \(A\), \(B\) and \(C\) are to deduce that the Morley triangle is equilateral. How does the following plan sound: suppose we construct a triangle, that definitely does have an equilateral Morley triangle, whose angles also happen to be \(A\), \(B\) and \(C\). Such a triangle would be similar²In Geometry, two objects are said to be similar if one can be made to exactly coincide with the other after a series of rigid translations, rotations and scalings. In other words, they have the same shape if you allow for differences in scale and are allowed to slide them around and spin them about as needed. to the original triangle \({\triangle}ABC\) — if we follow the similarity transform from the constructed triangle back to \({\triangle}ABC\) we will see that their Morley triangles must coincide; thus if one is equilateral so is the other!

One of the features of Conway's proof that leads to its great succinctness and beauty is his introduction of some very nice notation. Since we are dealing with angle trisectors, let \(a\), \(b\) and \(c\) be angles such that \(3a=A\), \(3b=B\) and \(3c=C\). Furthermore, let a superscript star denote the angle that is \(\pi/3\) (or \(60^\circ\) if you prefer) greater than a given angle. So, for example, \begin{equation*} a^\star = a + \pi/3 \end{equation*} and \begin{equation*} a^{\star\star} = a + 2\pi/3. \end{equation*}

Now, notice that the sum \(a+b+c\) must be \(\pi/3\). This is an immediate consequence of \(A+B+C=\pi\) which is true for any triangle in the plane. It follows that by distributing two stars amongst the three numbers \(a\), \(b\) and \(c\) we will come up with three quantities which sum to \(\pi\). In other words, there are Euclidean triangles having the following triples as their vertex angles:

In a nutshell, Conway's proof consists of starting with an equilateral triangle of unit side length, adding appropriately scaled versions of the six triangles above and ending up with a figure (having an equilateral Morley triangle) similar to \({\triangle}ABC\). The generic picture is given in <<Unresolved xref, reference "fig_morley_conway_puzzle"; check spelling or use "provisional" attribute>>. Before we can really count this argument as a proof, we need to say a bit more about what the phrase “appropriately scaled” means. In order to appropriately scale the triangles (the small acute ones) that appear green in <<Unresolved xref, reference "fig_morley_conway_puzzle"; check spelling or use "provisional" attribute>> we have a relatively easy job — just scale them so that the side opposite the trisected angle has length one; that way they will join perfectly with the central equilateral triangle.

\({\triangle}ABC\)

The triangles (these are the larger obtuse ones) that appear purple in <<Unresolved xref, reference "fig_morley_conway_puzzle"; check spelling or use "provisional" attribute>> are a bit more puzzling. Ostensibly, we have two different jobs to accomplish — we must scale them so that both of the edges that they will share with green triangles have the correct lengths. How do we know that this won't require two different scaling factors? Conway also developed an elegant argument that handles this question as well. Consider the purple triangle at the bottom of the diagram in <<Unresolved xref, reference "fig_morley_conway_puzzle"; check spelling or use "provisional" attribute>> — it has vertex angles \((a,b,c^{\star\star})\). It is possible to construct triangles similar (via reflections) to the adjacent green triangles \((a, b^\star, c^\star)\) and \((a^\star, b, c^\star)\) inside of triangle \((a,b,c^{\star\star})\). To do this just construct two lines that go through the top vertex (where the angle \(c^{\star\star}\) is) that cut the opposite edge at the angle \(c^\star\) in the two possible senses — these two lines will coincide if it should happen that \(c^\star\) is precisely \(\pi/2\) but generally there will be two and it is evident that the two line segments formed have the same length. We scale the purple triangle so that this common length will be 1. See <<Unresolved xref, reference "fig_morley_conway_puzzle_scaling"; check spelling or use "provisional" attribute>>.

Of course the other two obtuse triangles can be handled in a similar way.