Morley's Trisector Theorem: Adjacent Angle Trisectors of Any Triangle Meet in an Equilateral Triangle

What Morley's theorem actually claims

Take any triangle ABC in the Euclidean plane, with interior angles A, B and C adding to 180°. Each corner has two interior trisectors — the two rays that cut it into three equal wedges — so the figure has six trisectors in all. Morley's theorem picks out three particular crossings among them.

Look at side BC. One trisector from B hugs that side (the one at B⁄3 from it) and one trisector from C hugs it too (at C⁄3 from it). Those two meet at a point; call it P. Repeat for side CA to get Q, and for side AB to get R. Morley's theorem is the claim that PQR is equilateral, for every triangle ABC. PQR is called the first Morley triangle.

The word adjacent is not decoration; it is the whole content of the statement. In the triangle the animation opens on — angles 95°, 52° and 33°, circumradius 2.319606 — the adjacent pairing gives three sides that agree to fourteen significant figures at 0.553813. Pair the same three sides with the far trisector from each end instead, and the three lengths come out 0.629472, 0.774449 and 0.687307. That is not a near-miss; it is simply a different, unremarkable triangle. Two cevians from different corners always cross inside the triangle, so the six trisectors meet in twelve interior points, and the theorem lives entirely in the choice of which three of those twelve you take.

The hypotheses are light. ABC may be scalene, isosceles, right-angled or obtuse; it may be enormous or microscopic. The one thing it may not be is degenerate: unlike Napoleon's theorem, which survives collapsing all three vertices onto a line, Morley's construction needs three genuine interior angles to trisect. The formula above shows what happens as you approach that limit — the Morley triangle shrinks toward a point while staying exactly equilateral the whole way down.

Building the figure, one rotation at a time

The construction is short enough to write as arithmetic. List the vertices counterclockwise. At A, the interior angle is the positive rotation that carries the ray A→B onto the ray A→C; call it A. Then the two trisectors leaving A point along

Rot(A⁄3)(B − A)   and   Rot(2A⁄3)(B − A),

the first hugging side AB, the second hugging side AC. The same two lines appear at B starting from B→C, and at C starting from C→A. The three Morley points are then three line-line intersections:

• P = (the B⁄3 ray from B) ∩ (the 2C⁄3 ray from C) — the two hugging BC;
• Q = (the C⁄3 ray from C) ∩ (the 2A⁄3 ray from A) — the two hugging CA;
• R = (the A⁄3 ray from A) ∩ (the 2B⁄3 ray from B) — the two hugging AB.

Two practical consequences follow, and both are visible in the animation. First, every trisector is a cevian: extended, it crosses the opposite side, so the entire figure — six segments, three points, one small green triangle — lives inside ABC. The bounding box of the whole picture is just the bounding box of the triangle. Second, because each trisector direction is an explicit rotation of a vertex difference rather than the output of a case analysis, the construction is a continuous function of the three vertices. That is why the outer triangle can be dragged through wildly different shapes without the figure jumping, flipping or losing a point.

Written this way the construction takes a few lines of arithmetic. Drawing it with the classical Greek instruments, as the next section but one explains, is impossible.

How big is the Morley triangle?

There is a clean closed form. With R the circumradius of ABC,

L = 8R · sin(A⁄3) · sin(B⁄3) · sin(C⁄3).

The derivation is worth following because it is also the shortest proof of the theorem itself. Work in triangle BPC. By construction ∠PBC = B⁄3 and ∠PCB = C⁄3, so the third angle is ∠BPC = 180° − (B+C)⁄3 = 120° + A⁄3. The law of sines in BPC then gives BP = a·sin(C⁄3) ⁄ sin(120° + A⁄3), and since sin(120° + x) = sin(60° − x), the denominator is sin(60° − A⁄3).

Now feed in the triple-angle identity in its product form, sin3θ = 4·sinθ·sin(60° − θ)·sin(60° + θ). Applying it to a = 2R·sinA = 2R·sin(3·A⁄3) turns the side length into a = 8R·sin(A⁄3)·sin(60° − A⁄3)·sin(60° + A⁄3). The awkward factor cancels and what is left is remarkably tidy:

BP = 8R · sin(A⁄3) · sin(60° + A⁄3) · sin(C⁄3).

For the animation's opening triangle (A = 95°, C = 33°, R = 2.319606) that predicts BP = 1.8581, and the drawn figure measures 1.8581 — agreement to four decimals, which is as far as the comparison can be pushed because the animation stores its vertices to four decimals too. The symmetric formula for BR follows by swapping roles, and the law of cosines in triangle BPR, whose included angle at B is exactly B⁄3, collapses to PR = 8R·sin(A⁄3)·sin(B⁄3)·sin(C⁄3). That expression is symmetric in A, B and C, so the other two sides give the same number, and the triangle is equilateral. The symmetry is where the theorem actually happens.

The formula also settles how large the Morley triangle can get. The product sin(A⁄3)sin(B⁄3)sin(C⁄3) is maximised, subject to A+B+C = 180°, when all three are equal, so the extreme case is the equilateral triangle with L⁄R = 8sin³20° = 6sin20° − √3 ≈ 0.320070. Since an equilateral triangle of circumradius 1 has side √3 ≈ 1.732051, the Morley triangle there is 18.48% of the original's side — and that is the best it ever does. For the 170°, 6°, 4° sliver in the table it is down to 0.54% of the circumradius.

The twist: the figure cannot be drawn with compass and straightedge

Every classical triangle theorem the Greeks might have found involves objects they could construct. Morley's does not, and that is why it waited until 1899.

Pierre Wantzel proved in 1837 that a general angle cannot be trisected with compass and unmarked straightedge. The obstruction is algebraic. Trisecting 60° would require constructing cos20°, and the triple-angle identity cos3θ = 4cos³θ − 3cosθ with 3θ = 60° gives 8x³ − 6x − 1 = 0 with x = cos20°. The rational root test leaves only ±1, ±½, ±¼ and ±⅛ as candidates, and none of them is a root, so the cubic is irreducible over the rationals and the field extension Q(cos20°) : Q has degree 3. Every compass-and-straightedge construction builds a tower of quadratic extensions, so any constructible number has degree a power of 2 over the rationals. Three is not a power of two, so cos20° is out of reach.

So the six trisectors in the animation are perfectly well-defined lines in the plane — they simply cannot be produced with the Greek toolkit. Widen the toolkit a little and they reappear: Archimedes' neusis construction trisects any angle using a straightedge with two marks on it, and paper folding does the same (the fold guaranteed by the Huzita–Hatori sixth axiom, used by Margherita Beloch in the 1930s, solves cubics and therefore trisects). The Morley figure is drawable; it just is not classically drawable.

This is the right way to read the theorem's reputation for being startling. Angle trisection is the textbook example of an operation that is badly behaved — impossible with the standard tools, tangled up with irreducible cubics, the thing cranks write letters about. And yet the object it produces, for every triangle without exception, is the most symmetric figure in plane geometry.

Proofs, and what the dragging demonstration does not do

First, the caveat the animation cannot state for itself. Dragging the outer triangle through a dozen shapes and watching three numbers stay equal is evidence, not proof. It checks finitely many configurations, at two decimal places, on a screen. Even agreement to sixteen digits across a million shapes would leave the general claim untouched. What the demonstration does well is something else: it shows that the result is not an artifact of one lucky diagram, which is exactly the worry a single static figure invites. The proof has to come from algebra.

The trigonometric proof is the calculation in the section above, carried to the end: law of sines in BPC, the triple-angle identity, then the law of cosines in BPR. It is short, fully rigorous, and completely unilluminating about why the answer is symmetric.

Conway's proof reverses the problem, which is why people remember it. Instead of trisecting a given triangle, write a = A⁄3, b = B⁄3, c = C⁄3, so that a + b + c = 60°. Now take seven triangles whose angles are written in terms of a, b, c and 60° — one of them equilateral, six of them arranged around it — scale them so that edges which must match do match, and show the seven fit together into a single triangle with angles A, B, C. Since any two triangles with the same angles are similar, the assembled figure is the original triangle up to scale, and the equilateral piece at its heart is the Morley triangle. Nothing is trisected anywhere in the argument; the trisection is built in from the start.

Alain Connes gave an algebraic proof in 1998, framed around the three rotations about A, B and C through twice the respective angles, whose composition is the identity. Working with the fixed points of their cube roots turns the theorem into an identity that makes sense over any commutative field in which 3 is invertible, which is a sharper statement than the Euclidean one. Richard Guy's “lighthouse theorem” (American Mathematical Monthly, 2007) derives Morley — along with several other results — from a statement about two families of evenly spaced rotating beams.

Frank Morley, 1899, and a decade of word of mouth

Frank Morley (1860–1937) was born in Woodbridge, Suffolk, read mathematics at Cambridge, and emigrated to the United States in 1887, teaching first at Haverford College and from 1900 at Johns Hopkins. He was president of the American Mathematical Society in 1919–20 and edited the American Journal of Mathematics for roughly two decades. He was also a formidable chess player who once beat the reigning world champion, Emanuel Lasker. His son Christopher Morley became a well-known novelist.

The trisector result was not what Morley was looking for. Around 1899 he was studying the cardioids inscribed in a triangle — curves tangent to all three sides — and the equilateral triangle fell out of that work as a special case of a far more general statement, published as On the metric geometry of the plane n-line in the Transactions of the American Mathematical Society in 1900. Buried in a general theory, in that form, it attracted no attention at all.

What spread instead was the trisector statement itself, passed around by word of mouth among Morley's friends in England and America. It reached print as a problem before it reached print as a theorem. The first published proof came in 1909, from M. T. Naraniengar, in Mathematical Questions and Solutions from the Educational Times; a further treatment by F. Glanville Taylor and W. F. Marr followed in the Proceedings of the Edinburgh Mathematical Society in 1914. By then it had acquired the name it still carries.

Eighteen Morley triangles, not one

The theorem as usually stated is a restriction of something larger. Use directed angles and the natural object at each vertex is not two rays but three: the lines through that vertex at A⁄3, A⁄3 + 60° and A⁄3 + 120°, which between them cover the trisectors of both the interior and the exterior angle. Three lines at each of three vertices is nine lines, and they meet in 27 points. Taylor and Marr showed in 1914 that those 27 points are the vertices of 18 equilateral triangles. The one in this animation, the one built from interior adjacent trisectors, is the first Morley triangle; the others are the second, third and so on, and every single one of them is equilateral.

The configuration also defines a family of triangle centers. The centroid of the first Morley triangle is the Morley center, catalogued as X(356) in Clark Kimberling's Encyclopedia of Triangle Centers. Triangle ABC and its first Morley triangle are moreover in perspective: the three lines joining corresponding vertices are concurrent, and that point of concurrency is the first Morley perspector, X(357). A figure that cannot be built with a compass nevertheless generates named, catalogued, perfectly ordinary points of triangle geometry.

One last thing the size formula makes precise. Because L = 8R·sin(A⁄3)·sin(B⁄3)·sin(C⁄3) depends on the angles only, two similar triangles have similar Morley figures — the construction commutes with scaling and rotation, as it must. The equilateral output, though, is the same shape no matter what goes in. That is the part that never stops being odd.

Five very different triangles, every row evaluated from L = 8R·sin(A⁄3)·sin(B⁄3)·sin(C⁄3). The third column is the Morley side measured in units of the circumradius, so the rows are directly comparable. Note how little it varies for ordinary triangles and how fast it collapses once one angle approaches 180° — but it is never anything other than exactly equilateral.

Triangle (angles A, B, C)	Trisected angles A⁄3, B⁄3, C⁄3	Morley side ÷ circumradius	Exact value or note
Equilateral — 60°, 60°, 60°	20°, 20°, 20°	0.320070	8sin³20° = 6sin20° − √3, the largest value any triangle can reach
Right isosceles — 90°, 45°, 45°	30°, 15°, 15°	0.267949	exactly 2 − √3
3-4-5 right — 90°, 53.1301°, 36.8699°	30°, 17.7100°, 12.2900°	0.259007	here R = 2.5, so the Morley side itself is 0.647518
Obtuse scalene — 95°, 52°, 33° (the shape the video opens on)	31.6667°, 17.3333°, 11°	0.238750	R = 2.319606 in the drawn figure, giving L = 0.5538
Near-degenerate sliver — 170°, 6°, 4°	56.6667°, 2°, 1.3333°	0.005428	about half a percent of the circumradius — tiny, and still exactly equilateral