Seven pieces and an equilateral middle

Trisect the angles of any triangle, and the trisectors nearest each side meet in the corners of an equilateral triangle. A sweep over every shape of triangle confirms it to the last digit double-precision arithmetic can see, and a control run on the other trisectors shows the measurement is capable of failing. What a sweep cannot supply is a reason.

There are two classical reasons, and they are opposites. One begins with the triangle and computes; the other begins with the answer and builds. Both are short once the right bookkeeping is in place, and the bookkeeping is the same for both: write the outer angles as $3\alpha$, $3\beta$ and $3\gamma$, so that $\alpha + \beta + \gamma = 60°$.

The trisectors cut the triangle into seven pieces: the equilateral triangle in the middle, three pieces each resting on a side of it with a corner of the outer triangle opposite, and three pieces each resting on a side of the outer triangle with a corner of the equilateral one opposite. Every angle in the picture is one of $\alpha$, $\beta$, $\gamma$, or one of them plus $60°$ or $120°$. That the angles come out this way is the whole of Morley’s theorem in disguise, and the two proofs are two ways of showing that they must.

The bookkeeping of the pieces #

Look at the piece with the outer corner $A$ at its apex, resting on the side $ZY$ of the equilateral triangle. Its angle at $A$ is the middle third of $A$’s angle, which is $\alpha$. Its other two angles are marked $\beta + 60°$ and $\gamma + 60°$, and the three add up correctly: $\alpha + (\beta + 60°) + (\gamma + 60°) = 180°$, since $\alpha + \beta + \gamma = 60°$.

The piece resting on the outer side $AB$, with the Morley point $Z$ at its apex, has angles $\alpha$ at $A$, $\beta$ at $B$ — the thirds of the outer angles nearest that side — and therefore $180° - \alpha - \beta = \gamma + 120°$ at $Z$. So each outer-side piece has one angle that is a third plus $120°$, and each inner-side piece has two that are thirds plus $60°$.

Round the point $Z$, four pieces meet: the equilateral triangle with $60°$, two inner-side pieces with $\alpha + 60°$ and $\beta + 60°$, and one outer-side piece with $\gamma + 120°$. They add to $60 + \alpha + 60 + \beta + 60 + \gamma + 120 = 360$ degrees exactly, as angles round a point must. The labels are consistent everywhere. The figure does not assume them; it measures every angle of every piece off the construction and checks each against its label.

John Conway, who popularised this notation, wrote the three pluses as $0^+$ for $60°$ and $0^{++}$ for $120°$, and the pieces as triples like $(\alpha, \beta^+, \gamma^+)$. The notation makes one thing obvious that the pictures do not: every piece is determined, up to size, by three numbers of the same kind, and the seven triples are symmetric under swapping the roles of $\alpha$, $\beta$ and $\gamma$.

Conway’s proof: build the answer #

The trigonometric proof comes later. Conway’s goes backwards, and it is the one that explains.

Start with an equilateral triangle $XYZ$ of side 1 and a choice of $\alpha$, $\beta$, $\gamma$ adding to $60°$. On the side $ZY$, build outwards the triangle whose base angles are $\beta + 60°$ at $Z$ and $\gamma + 60°$ at $Y$; its apex is a point $A$, with angle $\alpha$ there. Do the same on the other two sides, getting apexes $B$ and $C$. Nothing so far has used any triangle except the equilateral one.

Now join $A$, $B$ and $C$, and ask what the three gaps are — the triangles $ABZ$, $BCX$ and $CAY$ that fill in between. Take $ABZ$. Its angle at $Z$ is what is left of $360°$ after the three pieces already meeting there, which is $\gamma + 120°$. Its sides at $Z$ are known by the law of sines from the two pieces already built: $ZA = \sin(\gamma + 60°)/\sin\alpha$ and $ZB = \sin(\gamma + 60°)/\sin\beta$. Their ratio is $\sin\beta / \sin\alpha$.

A triangle with angles $\alpha$, $\beta$ and $\gamma + 120°$ at $A$, $B$ and $Z$ would have exactly that ratio of sides, by the same law of sines. Two sides and the angle between them determine a triangle, so $ABZ$ is that triangle, and its angles at $A$ and $B$ are $\alpha$ and $\beta$. The same holds for the other two gaps. So the angle at $A$ is split into three parts of $\alpha$ each — one from each of the three pieces meeting there — and the whole angle is $3\alpha$; likewise $3\beta$ at $B$ and $3\gamma$ at $C$.

The figure performs the construction numerically and measures the result: angles of $78°$, $54°$ and $48°$, each cut into three equal parts by the lines to the equilateral triangle’s corners.

Why building proves anything #

The construction produced a triangle with angles $3\alpha$, $3\beta$, $3\gamma$ whose trisectors meet in an equilateral triangle. Any given triangle has angles of that form for some $\alpha$, $\beta$, $\gamma$ adding to $60°$, namely a third of each of its angles. And two triangles with the same angles are the same shape, one a scaled copy of the other.

So the given triangle is a scaled copy of the assembled one, the scaling carries trisectors to trisectors, and the given triangle’s Morley triangle is a scaled copy of the equilateral triangle the construction began with. The theorem is proved by showing that the configuration it describes can be built for every choice of angles, and a triangle, being determined by its angles up to size, has no room to be anything else.

That style of argument — construct the answer, then show it matches the question — is a standard remedy when a direct proof arrives at an identity nobody can see coming. Its weakness is that it hides where the equilateral triangle came from: it was put there at the start. Its strength is that every step is a triangle with known angles, and the only fact about triangles used is the law of sines once per gap.

The trigonometric proof: compute the sides #

The direct proof goes forwards, from the triangle to the lengths, and needs one identity.

The identity is $\sin 3\theta = 4 \sin\theta \sin(60° + \theta) \sin(60° - \theta)$, which follows from the triple-angle formula $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ and a product-to-sum rearrangement. It is exactly what turns a sine of a whole angle into a product of sines of its thirds, and Morley’s theorem is about thirds.

Scale the triangle so that its circumradius is $R$. By the law of sines, the side $BC$ is $2R\sin 3\alpha$. In the small triangle $BXC$, with angles $\beta$ and $\gamma$ at the base, the side $BX$ is $BC \cdot \sin\gamma / \sin(\beta + \gamma)$. Replace $\sin 3\alpha$ by the product and $\sin(\beta + \gamma)$ by $\sin(60° - \alpha)$, and a factor cancels: $BX = 8R \sin\alpha\sin\gamma\sin(60° + \alpha)$. The same computation gives $BZ = 8R\sin\gamma\sin\alpha\sin(60° + \gamma)$ from the other side.

Now in the triangle $BXZ$ the two sides at $B$ are known and so is the angle between them, $\beta$. Rather than grind through the law of cosines, recognise the triangle. The sides $BX$ and $BZ$ are in the ratio $\sin(60° + \alpha) : \sin(60° + \gamma)$, and the angles $\beta$, $60° + \alpha$ and $60° + \gamma$ add to $180°$ — so there is a triangle with those three angles, and by the law of sines its sides are in the ratio $\sin\beta : \sin(60° + \alpha) : \sin(60° + \gamma)$. Put its angle $\beta$ at $B$ and its two neighbouring sides match $BX$ and $BZ$ in ratio. Two sides and the included angle fix a triangle, so $BXZ$ is that triangle scaled, and its third side, opposite $\beta$, is in the same proportion:

$$XZ = 8R \sin\alpha\sin\beta\sin\gamma.$$

The expression is symmetric in $\alpha$, $\beta$ and $\gamma$, so the other two sides of Morley’s triangle, computed the same way, have the same length. The triangle is equilateral.

How large the middle is #

The formula is checked here on every shape of a grid.

Measured in circumradii, Morley’s side is a function of the shape alone, and the grid of 91 shapes puts every one on the diagonal. The largest, at the equilateral outer triangle, is $8\sin^3 20° \approx 0.3201$ circumradii, against the outer triangle’s own side of $\sqrt 3 \approx 1.732$ — so the inner triangle is never more than about eighteen and a half per cent of the outer one’s size. The formula makes the degenerate end visible too: as a triangle flattens, one third-angle goes to nothing, its sine with it, and the Morley triangle shrinks to a point.

The formula also records what cannot be done with it. The points $X$, $Y$, $Z$ are placed from computed angles, since an angle cannot in general be trisected with straightedge and compass. The side $8R\sin\alpha\sin\beta\sin\gamma$ is therefore typically not a constructible length, even when the outer triangle’s sides are whole numbers. An equilateral triangle of a length Euclid could not draw, sitting inside a triangle he could, is part of why the theorem waited until 1899.

The two proofs meet in this number. Conway’s construction starts from an equilateral triangle of side 1 and produces an outer triangle; the trigonometric proof starts from an outer triangle of circumradius $R$ and produces a side $8R\sin\alpha\sin\beta\sin\gamma$. Run Conway’s construction and measure the outer triangle’s circumradius, and it is $1/(8\sin\alpha\sin\beta\sin\gamma)$: the two computations are inverse to each other, as they must be.

A third route, through turning #

There is a proof in which the trisection itself becomes an algebraic operation. Place the triangle in the complex plane. A rotation about a point is multiplication by a number of length one, and trisecting an angle at a vertex means taking a cube root of the rotation through that angle. The three Morley points can then be written as fixed points of products of such rotations, and the condition for three points $X$, $Y$, $Z$ to form an equilateral triangle, taken in order, is the single equation $X + \omega Y + \omega^2 Z = 0$, where $\omega$ is a cube root of unity.

Checking the equation is algebra, and the reason it holds is a fact about rotations. Turning about $A$ through twice its angle is the same as reflecting in one side through $A$ and then in the other, and likewise at $B$ and at $C$; composing the three full turns, the reflections cancel in pairs, so the three turns together do nothing at all. The trisecting turns are cube roots of those full turns, and when three cube roots compose to a cube root of doing nothing, the cube roots of unity appear, and with them the equation that says $X$, $Y$, $Z$ are the corners of an equilateral triangle. This is the proof Alain Connes gave in 1998, phrased with affine maps of a line over any field where cube roots exist, and it shows that the theorem is not really about Euclidean geometry at all: it is a fact about how rotations compose, and it holds in settings where angles and lengths have no meaning.

What each proof explains #

The trigonometric proof explains why the three sides are equal: each is given by the same symmetric expression, and symmetry does the rest. It explains nothing about why that expression appears. The identity for $\sin 3\theta$ has to be known in advance, and the cancellation that produces the symmetric product is a pleasant surprise rather than a reason.

Conway’s proof explains why the angles work: every piece is determined by a triple of the same kind, the triples fit round each point with nothing left over because $\alpha + \beta + \gamma = 60°$, and the only thing to check is one law-of-sines ratio per gap. It explains nothing about why anyone would think to begin with an equilateral triangle, except that the theorem said to.

There is also a synthetic proof, due to M. T. Naraniengar in 1909, that constructs the equilateral triangle directly inside the given one using reflections and inscribed angles, without trigonometry and without running backwards. It is longer than both and uses the angle an arc subtends several times. None of the three proofs is obviously the reason; each is a reason, and the theorem has more of them than most.

What the figures cannot show #

The piece and assembly figures draw two shapes each, and the proofs are about every shape. What the figures establish is that the bookkeeping of angles is right on those shapes — every label matches its measured angle, and the assembly closes up with the right outer angles — which is the evidence that the general argument has been transcribed correctly, not the argument itself.

The identity figure plots both sides on a grid of six hundred angles; that they agree everywhere follows from the triple-angle formula, not from the plot. And the size figure checks the formula on 91 shapes. A formula checked on 91 shapes and proved for all is what the section above supplies; the figure is the part that could have caught an error in transcribing it.

Nor do the figures show why the nearest trisectors are the right ones. Both proofs use that choice at every step — the pieces’ angles are thirds because the lines are the nearest trisectors — and the other choices lead to different pieces with different angles. What happens with those other choices is the subject of the essay on the eighteen equilateral triangles, where most of them turn out to give equilateral triangles too.

Still open: a proof as simple as the statement #

Morley’s theorem has been proved dozens of ways since 1899: by trigonometry, by Conway’s assembly, by synthetic constructions, by complex numbers — where the trisection becomes taking cube roots and the equilateral triangle appears as a relation among cube roots of unity — and by the algebra of reflections. Alain Connes gave one in 1998 using the group of affine transformations of a line, in which the theorem becomes a statement that a certain product of rotations has order three.

What nobody has found is a proof as short as the statement that also explains it. Each known proof either computes an identity whose symmetry appears at the end, or builds the answer and checks that it fits. The question of why trisecting the angles produces an equilateral triangle — why sixty degrees, why the nearest trisectors — is answered by each proof in its own terms and by none of them in a sentence. That is a question about exposition rather than mathematics, and it is open in the sense that the subject keeps producing new attempts. A collection of proofs maintained by Alexander Bogomolny lists more than two dozen, and new ones still appear. The statement fails on a sphere and in the hyperbolic plane, where the angles of a triangle do not add to a half turn, so any explanation has to use flatness somewhere — and every known proof uses it only through the one equation $\alpha + \beta + \gamma = 60°$. A proof that made that dependence feel inevitable would be the one worth having.

Two directions, one triangle #

The trisectors cut every triangle into seven pieces whose angles are thirds of the outer angles plus multiples of sixty degrees, and both proofs rest on that description. The trigonometric one computes forwards and finds each side of the middle piece to be $8R\sin\alpha\sin\beta\sin\gamma$, the same for all three by symmetry. Conway’s builds backwards from an equilateral middle, checks that the gaps close by one law-of-sines ratio each, and concludes that every triangle is the one it built.

The first is an identity; the second is a jigsaw. Between them they account for the theorem completely, and neither accounts for why the jigsaw’s middle piece had to be equilateral — which, in the end, is simply that the thirds add to sixty degrees, and sixty degrees is the angle of an equilateral triangle.

That sentence is not a proof, but every proof uses it. In the trigonometric one it appears as $\sin(\beta + \gamma) = \sin(60° - \alpha)$, the cancellation that produced the symmetric product; in Conway’s it is what makes the four angles round each Morley point add to a full turn; in the complex one it is why the cube roots of the three rotations compose to a cube root of the identity. Three proofs, three languages, and one number doing the work in each.