Eighteen equilateral triangles

Morley’s theorem uses the trisectors nearest each side, and its proofs use that choice at every step. The first essay on the theorem ran a control on the other trisectors and found them failing: the triangle they cut out is not equilateral. That control used one alternative. There are many more, and when all of them are tried the picture changes completely.

A line through a vertex that trisects the angle there makes a third of the angle with one side. But the angle between two lines is only defined up to a half turn, and the lines through a vertex that make an angle with each side that is a third of the corner’s angle — modulo a half turn — come in three directions, not one: a third of the angle, a third plus $60°$, and a third plus $120°$. The first is the ordinary trisector; the other two trisect the angle together with the outside of the triangle, the exterior angles.

Label the choices $0$, $1$ and $2$ at each corner, so that each panel is named by three digits. Choice $000$ is Morley’s own triangle, the small one inside. Choice $111$ uses the second line at every corner and cuts out a large equilateral triangle enclosing the original; $222$ gives another, pointing the other way. And so on through all twenty-seven. Eighteen are equilateral. The nine that are not are the panels labelled $002$, $011$, $020$, $101$, $110$, $122$, $200$, $212$ and $221$ — every choice whose digits add to two, five or eight, and no other.

Why these lines rather than others? A line through a vertex does not know which side of it is the inside of the triangle. It meets the two sides at an angle, and it trisects the corner if those angles are a third and two thirds of the corner’s angle. But the two sides also bound the exterior angle, the one supplementary to the corner, and a line can trisect that instead — or trisect the full turn of $360°$ minus the corner, which is the reflex angle on the far side. Counted as lines rather than rays, all of these are the three directions at a third of the angle plus a multiple of sixty degrees. The generalised trisectors are not an arbitrary enlargement of the problem; they are every line through the vertex that has any claim to trisect something bounded by the two sides.

The rule of threes #

The search checks the rule on every panel, and it has a reason that the bookkeeping of the previous essay makes visible.

Write the chosen angle at each corner as $\alpha' = \alpha + 60°\,k_A$ and similarly $\beta'$ and $\gamma'$, where $k_A$, $k_B$, $k_C$ are the digits. Their sum is $60° + 60°(k_A + k_B + k_C)$. When the digits add to a multiple of three, the sum is $60°$ plus a multiple of $180°$; when they add to one more than a multiple of three, it is $120°$ plus a multiple of $180°$, which is $-60°$ modulo a half turn. And when they add to two more than a multiple of three, the three angles add to a multiple of $180°$.

Conway’s proof needs exactly one thing of the three angles: that they add to $60°$, so that the four angles round each corner of the middle piece make a full turn. Angles of lines are only defined up to a half turn, and adding to $-60°$ works just as well, with the middle triangle’s orientation reversed. What does not work is adding to $0°$ — then the pieces would need a middle angle of nothing, and the construction has no equilateral triangle to begin from. The equilateral triangles are exactly the choices for which the three generalised thirds add to $\pm 60°$ modulo a half turn, and the digit rule is that condition in disguise.

The same rule on another triangle #

The rule is a statement about every triangle, and a second shape shows it doing the same thing.

The shapes of the panels change — the obtuse corner throws some of the triangles far out and squeezes others — but the classification does not. The same nine labels fail and the same eighteen succeed. The figure checks every panel against the rule, on both triangles, and a panel that broke it would stop the drawing. The rule is independent of the shape because the digits add to a number that does not involve the angles at all: the condition is on $k_A + k_B + k_C$, and the angles enter only through the fixed sum $\alpha + \beta + \gamma = 60°$, which every triangle shares.

This is the kind of statement that a sweep over shapes, like the one the first essay ran for Morley’s own triangle, supports and does not prove, and here the proof is the angle count above. What the figures add is that the rule is sharp: the failing nine are not nearly equilateral, with ratios of longest to shortest side ranging up to nearly ten on the first triangle. There is no gradual degradation — the choice either satisfies the angle condition and is exactly equilateral, or does not and is not close.

Twenty-seven points, two triangles each #

The count of eighteen can be read another way. Each triangle is cut out by three points, one near each side, and the point near the side $BC$ depends only on the choices at $B$ and $C$: three lines from $B$, three from $C$, nine crossings. So there are nine candidate points near each side, twenty-seven in all, and each of the twenty-seven triangles picks one from each side.

Fix a point near $BC$, which fixes the digits $k_B$ and $k_C$. The third digit $k_A$ can be any of three values, and exactly one of them makes the sum two more than a multiple of three. So every one of the twenty-seven points is a corner of exactly two equilateral triangles, and $27 \times 2 = 54$ corners, three per triangle, is eighteen triangles again. The configuration is as regular as it could be: no point is favoured, and each is shared between two of the equilateral triangles and one of the failures.

That regularity is the kind the search over every choice is good at exposing and bad at explaining. The count of two per point follows from the digit rule in a line, and the digit rule follows from the angle count; the search is what showed there was something to count.

Three directions for eighteen triangles #

Overlay all eighteen equilateral triangles and a further regularity appears.

Every side of every one of the eighteen points in one of just three directions, sixty degrees apart. The eighteen triangles are not scattered at random orientations; they are all copies of one equilateral triangle, scaled, moved, and some turned half round, with parallel sides. Many of them share entire sides, since the lines through the three vertices cross at a limited set of points: the eighteen triangles are drawn on only twenty-seven intersection points, and each point is a corner of several of them.

The common direction is simple to state. With the base $BC$ level, the sides make an angle of $(B - C)/3$ with it, modulo sixty degrees — for the $78°$, $54°$, $48°$ triangle, a third of $54° - 48°$, which is $2°$.

For the $100°$, $50°$, $30°$ triangle the tilt is a third of $20°$, which is $6.667°$, and again every one of the fifty-four sides respects it. The orientation of the whole family is determined by the base angles alone, and the apex angle, which fixes the triangle’s shape together with them, plays no part in the direction.

The tilt, measured on every shape #

The direction rule for Morley’s own triangle can be checked across all shapes, and it was stated without proof in the first essay on the theorem.

Across 91 shapes the measured tilt lands on $(B - C)/3$ exactly. An isosceles triangle, with equal base angles, has a Morley triangle with one side exactly level, which is what symmetry demands: the triangle is its own mirror image in the perpendicular bisector of the base, and so is its Morley triangle. For a general triangle the tilt is the amount by which that symmetry is broken, divided by three.

The rule can be recovered by an angle chase through Conway’s pieces, following the direction of the base round to the side of the middle triangle once by way of $B$ and once by way of $C$ and comparing; the figure’s job is the check, on every shape of the grid, that nothing was lost in the chase. What the chase makes plain is why only the base angles appear. The side of Morley’s triangle facing $A$ is built from pieces at $B$ and $C$, and the angle at $A$ enters only through the condition that the thirds add to sixty degrees, which fixes it once the other two are known.

The same rule holds for all eighteen triangles at once, which the overlays confirmed on two shapes. That is a strong statement: each generalised choice changes the angles by multiples of sixty degrees, and a change of sixty degrees in a line’s direction is invisible modulo sixty. The whole family inherits Morley’s own tilt because every member differs from it by turns the direction rule cannot see.

Where the eighteen come from #

The full statement — eighteen equilateral triangles, sides in three directions — was published by F. Glanville Taylor and W. L. Marr in 1914, and it is the natural setting of Morley’s own discovery. Frank Morley found the theorem in 1899 while studying cardioids tangent to the sides of a triangle. The centres of those cardioids lie on nine lines, falling into three sets of three parallel lines, and the lines cross in the corners of equilateral triangles. The trisectors are what the cardioids’ geometry picks out, and the famous triangle is the smallest and most central of the family.

That is why the theorem looked like a coincidence. Seen alone, the small triangle inside a triangle has no reason to be equilateral. Seen as one of eighteen, all with parallel sides, it is one member of a structure whose regularity is the point: the trisecting lines, taken all together, form a configuration with a threefold symmetry that the original triangle does not have, and the equilateral triangles are its visible trace.

The threefold symmetry has an algebraic source. Trisecting an angle is taking a cube root of a rotation, and a rotation has three cube roots — differing by turns of $120°$, which, for lines rather than rays, is $60°$ twice over. The three choices at each corner are the three cube roots, and the cube roots of unity that relate them are the source of the digit rule: the product of the three chosen roots is one of three possible values, and two of those values give equilateral triangles.

A relative: equilateral triangles on the sides #

Morley’s family is not the only way a triangle with no symmetry produces equilateral triangles, and the best-known alternative shows where the threefold regularity comes from. Build an equilateral triangle outward on each side of any triangle and join the three centres: the result is equilateral. That is Napoleon’s theorem, attributed to the emperor without much evidence and first published in the 1820s. Building the three triangles inward instead gives a second, smaller equilateral triangle, and the difference of the two areas is the area of the original.

Both Napoleon’s triangles and Morley’s are proved most cleanly in the complex plane, where multiplication is turning and three points form an equilateral triangle exactly when $X + \omega Y + \omega^2 Z = 0$ for a cube root of unity $\omega$. In Napoleon’s case the cube root enters because each built triangle turns a side by sixty degrees; in Morley’s, because each trisector is a cube root of a rotation. The resemblance is not superficial. Both theorems say that operations involving cube roots of rotations, applied round a triangle, land on the three points that the cube roots of unity single out — and the digit rule here is what arithmetic modulo three looks like when it is done with rotations.

The two families differ in what they need. Napoleon’s triangles are constructible with straightedge and compass, and the theorem could have been found by Euclid. Morley’s need angles cut in three, which the classical tools cannot do, and the family was found a century after Napoleon’s and two millennia after Euclid, alongside the nine-point circle and the other coincidences triangle geometry had already catalogued.

What the figures cannot show #

The twenty-seven panels are computed for two triangles, and the rule is checked on those two. The statement for every triangle rests on the angle count, and on Conway’s construction applied to angles that add to $\pm 60°$ modulo a half turn rather than to exactly $60°$ — a generalisation that needs care with orientations, since some of the generalised pieces are turned inside out, and which the figures do not display.

The overlays draw all eighteen triangles on one page, and at this scale several coincide along whole sides; the eye cannot count eighteen distinct outlines, and the figure’s own check, not the picture, is what establishes that eighteen are present. Nor do the figures show Morley’s cardioids, which are the construction the theorem came from and which would need a different kind of drawing.

And the direction figure reduces every angle to the range between $-30°$ and $30°$. The rule is a statement modulo sixty degrees, and the reduction is where it is read off; which of the three directions belongs to which side is a further piece of bookkeeping the figure discards.

The question it leaves: what else the trisectors know #

With eighteen equilateral triangles in hand, the natural question is what other structure the full configuration of trisectors carries — which further coincidences among its twenty-seven points and eighteen triangles are forced by the angles, and which merely hold for the shapes that happen to have been checked.

Triangle geometry answers such questions slowly and by accumulation. The encyclopedia of triangle centres maintained by Clark Kimberling lists tens of thousands of special points of a triangle, several of them defined through Morley’s configuration, each with its coordinates and the identities it is known to satisfy. Whether that list has an organising principle — a theory that would predict which coincidences hold, as the angle count predicted the eighteen here, rather than recording them one at a time — is open, and the modern view that many such coincidences are shadows of statements about cubic curves explains some of them and not others.

For Morley’s configuration specifically, one strand of the answer is the cardioids. Morley’s own framework treats the trisectors as the tangent directions of a family of curves, and in that setting the eighteen triangles and their parallel sides become statements about how three cardioids tangent to the three sides sit relative to each other. Whether every coincidence in the configuration has such a description, or whether some are genuinely separate facts, has not been settled by anyone.

One of eighteen #

The small equilateral triangle inside every triangle looked, on its own, like an accident with no cause. Allowing every trisector of every angle, inside and out, gives twenty-seven triangles, eighteen of them equilateral, all with sides in the same three directions, and a digit rule that picks them out.

The rule is an angle count: the three generalised thirds must add to sixty degrees, or its negative, modulo a half turn. The directions are a third of the difference of two base angles. Morley’s triangle is the member of the family nearest the centre, and its equality of sides is the family’s equality of sides seen through the one choice everyone happens to make first.

The lesson is a familiar one from elsewhere in mathematics. A result that looks isolated is often a single visible member of a family whose other members were excluded by a convention — here, the convention that an angle’s trisector lies inside the angle. Relax the convention, allow the lines that trisect the angle and its exterior together, and the isolated coincidence becomes a structure with a count, a rule and a direction. The surprise does not disappear; it moves from the small triangle to the fact that eighteen of twenty-seven choices succeed, and that the rule deciding which is arithmetic modulo three.