The price of curvature in Morley's triangle

Three trisectors and a triangle nobody expected drew Morley’s theorem: cut each angle of a flat triangle into three equal parts, and the trisectors nearest each side meet in the corners of an equilateral triangle, whatever the triangle. On a sphere the construction still makes sense — the trisectors are arcs of great circles, and they still meet in a small inner triangle — but the theorem fails. The inner triangle’s sides differ, by amounts far too small to see: within a tenth of a per cent for triangles whose angles add to well over 200°.

That essay measured the failure for three shapes and found it growing in proportion to the angle excess — the amount by which the angle sum exceeds 180°, which on a sphere of radius one is exactly the triangle’s area. It ended with the question this essay answers: what is the coefficient of that proportion, as a function of the triangle’s shape, and which shape makes it largest? That is a quantitative version of Morley’s theorem on curved surfaces — not “equilateral” but “equilateral to within so much per unit of enclosed curvature”.

Measuring the failure #

Two numbers describe a curved Morley triangle’s failure. The spread is how far its sides are from equal: the longest side divided by the shortest, minus one. The excess $E$ is how far the original triangle is from flat: its angle sum minus $\pi$, which on a unit sphere is its area and on a hyperbolic plane of curvature $-1$ is minus its area. A flat triangle has $E = 0$ and spread 0, by Morley’s theorem. A curved one has both positive.

To isolate the effect of shape from the effect of size, the figures take a flat triangle of chosen angles and carry it onto the curved surface by the exponential map at a point — each vertex is sent along the geodesic in its direction from the centre, for its distance — scaled down to a small size. As the size shrinks the curved triangle becomes indistinguishable from the flat one, both the excess and the spread tend to nought, and their ratio tends to a limit that depends only on the flat triangle’s angles. That limit is the coefficient $\kappa$. The figures compute the Morley triangle exactly at two small sizes, with all the construction done in three-dimensional coordinates — the sphere as the unit sphere, the hyperbolic plane as one sheet of a hyperboloid — and extrapolate the ratio to size nought. The extrapolation uses the fact that the ratio’s own error falls like the square of the size, so twice the size quadruples the error, and a combination of the two measurements cancels it: four times the smaller-size ratio minus the larger-size ratio, divided by three. The two sizes used are small enough for that to work and large enough for the spread to stand far above the rounding error, and where those two requirements conflict, for very thin triangles, the figures do not report a value.

The picture confirms the proportionality before it is used. For each of three shapes, triangles of thirteen sizes on the sphere and thirteen on the hyperbolic plane give points that lie on a V: spread rising in proportion to the excess on the right and to its absolute value on the left, with the same slope on both sides. The triangle with angles $80°, 70°, 30°$ has slope 0.00063; the obtuse one with $100°, 50°, 30°$ has 0.0018, three times larger; the equilateral one stays at nought at every size, on both surfaces, because three-fold symmetry forces its Morley triangle to be equilateral too.

Excess is enclosed curvature #

The excess is the right measure of size for this question because it is the only one the surface itself knows. On a unit sphere, Albert Girard’s theorem, published in 1629, says that a triangle’s area is exactly its angle sum minus $\pi$: a lune between two great circles meeting at angle $\theta$ covers $2\theta$ of the sphere’s $4\pi$, and the three lunes of a triangle cover the whole sphere once and the triangle and its opposite twice over, which gives the formula in two lines. On a sphere of radius $R$ the excess is the area divided by $R^2$, so it is the curvature times the area, the total curvature the triangle encloses. The same holds on the hyperbolic plane with the sign reversed, and on any surface the Gauss–Bonnet theorem makes the excess the integral of the curvature over the inside.

That is the sense in which flatness is a boundary case: the angle defects of a polyhedron add up to $720°$ because the total curvature of anything shaped like a sphere is $4\pi$, and a flat triangle is one that encloses none of it. Measuring the Morley failure per unit of excess asks how much the theorem breaks per unit of curvature enclosed, which is a question about the surface and not about the triangle’s size, and the answer, $\kappa$, depends only on the shape.

The coefficient over every shape #

A triangle’s shape is two numbers — three angles adding to $180°$ — and the hero lays all shapes out in a triangle of its own, each shape placed by its angles: the centre is the equilateral shape, each corner a degenerate triangle with one angle of $180°$, and each edge a degenerate triangle with one angle of nought. On a grid of 820 shapes, every angle at least $11.25°$, the coefficient is computed and shaded.

The pattern has the symmetry of the problem. Relabelling the corners of a triangle permutes its angles and its Morley sides together, so the map is unchanged by the six symmetries of the triangle of shapes, and the figure shows it: six-fold symmetric about the centre. On the grid, the centre is the only place where $\kappa$ vanishes. Moving away from it, the coefficient grows, slowly at first and then faster, and it is largest towards the corners, among obtuse shapes. On this grid its largest value is 0.00303, at angles of about $139°$, $23°$ and $19°$ — very nearly isosceles, with a wide obtuse apex.

The grid stops at $11.25°$ for a reason of arithmetic, not geometry. A triangle with an angle of a few degrees, carried onto the sphere at a size small enough for the first-order limit, encloses an excess so small that the spread — itself thousands of times smaller — disappears into the rounding error of double-precision arithmetic. The extrapolation then returns nonsense, and the figures leave those shapes out rather than shading them with it.

Along two lines of shapes #

Restricting to one-parameter families makes the coefficient a curve.

For isosceles triangles the Morley triangle is isosceles too, and the coefficient is a single curve through nought at the equilateral shape, apex $60°$. Below $60°$ — tall, narrow isosceles triangles — it rises to a small peak of 0.00045 at apex $35°$ and falls again. Above $60°$ — wide, flat ones — it rises much further, to 0.00302 at apex $137°$, and that peak is within a whisker of the global maximum. For right-angled triangles the coefficient is largest, 0.00139, at the isosceles right triangle, $90°, 45°, 45°$, which is where the right-angled family crosses the isosceles one.

The dashed curve, for the hyperbolic plane, lies exactly on the spherical one. That is not a coincidence of the shapes drawn: the failure is a first-order effect of curvature, so it is linear in the curvature, and a quantity linear in curvature has the same coefficient whether the curvature is positive or negative, with the excess changing sign alongside. The spread is a size, never negative, and that is why the V of the earlier figure is symmetric. The sphere and the hyperbolic plane break Morley’s theorem by exactly the same amount, to first order, for triangles of the same shape and the same enclosed curvature.

Which side is longest #

A spread says how unequal the Morley sides are, not which is longest, and the natural guess — the side facing the largest angle, or the smallest — turns out to be wrong both ways.

Of 406 shapes tested, the longest Morley side is the one nearest the largest angle in exactly one, and nearest the smallest angle in 140. In the rest it is the side nearest the middle angle. The picture shows the triangle of shapes divided into regions that meet at the centre, with the six-fold symmetry of the problem. Which side curvature lengthens depends on the shape in a way no simple ordering of the angles captures.

That is consistent with the structure of the first-order correction. To first order, each corner of the Morley triangle moves by an amount proportional to the excess in a direction set by the shape, and the change in each side is the difference of the displacements of its two ends projected along it. Three displacements and three projections give a pattern with more freedom than the ordering of three angles.

Why the failure is first order #

The flat theorem is exact, and the curved construction differs from the flat one only through curvature, so every quantity in it can be expanded in powers of the curvature. The zeroth-order term is the flat answer — an equilateral triangle. The first-order term is a correction proportional to the curvature times the square of the triangle’s size, which is the enclosed excess. There is no reason for that first-order correction to vanish, and the computation shows it does not, except where symmetry forces it.

Morley’s theorem is not alone in failing this way. Pythagoras’s theorem on a globe becomes $\cos c = \cos a \cos b$ for a right triangle on the unit sphere, and expanding both cosines to second order in the sides gives back $c^2 = a^2 + b^2$ with a correction of the next order, proportional to the triangle’s area times the curvature. Every exact theorem of plane geometry that involves lengths becomes, on a curved surface, the plane statement plus a term in the enclosed curvature, and the coefficient of that term is the first-order picture of how the theorem bends. What is unusual about Morley’s theorem is that its plane statement is an equality between three lengths that no construction obviously makes equal, so its curved correction has nothing to inherit from: the coefficient has to be computed rather than read off a formula.

The proof by rotations that Alain Connes found in 1998 shows where the correction comes from. In the plane, the rotations about the three corners by twice a third of their angles compose to the identity once each has been applied three times, and their pairwise fixed points are the Morley corners; the equilateral conclusion is a statement about the group of rigid motions of the plane, in which rotations commute with translations up to a known twist. On the sphere the motions are rotations of space, the group is different, and the identity that forced the equilateral triangle holds only approximately, with an error proportional to the enclosed curvature — the same holonomy that makes a vector carried round a spherical triangle come back turned by its excess. In that reading the coefficient $\kappa$ measures how far Connes’s identity fails, translated into the lengths of Morley’s sides — an interpretation that a first-order expansion of his proof on the sphere would make exact.

The worst shape, drawn large #

The maximising shape is the one for which curvature does most damage per unit, and drawn large it is still almost perfect.

At this size the triangle’s angles are $142.1°$, $24.9°$ and $21.2°$, enclosing an excess of $8.1°$, and its Morley triangle has sides 0.07810, 0.07810 and 0.07806 on the unit sphere, a spread of 0.047 per cent. The first-order prediction is $0.00303 \times 8.1° \times \pi/180 = 0.043$ per cent, close to the measured value; the remainder is the second-order term that a triangle this large begins to show. Morley’s theorem on a sphere is wrong, and for every shape it is wrong by less than one part in three hundred of the enclosed excess.

What a tenth of a per cent means on the ground #

The numbers are small enough that a concrete scale helps. On the Earth, taken as a sphere of radius 6,371 kilometres, a triangle of three cities a thousand kilometres apart encloses about 433,000 square kilometres, an excess of about $0.61°$, or 0.0107 in radians. Its Morley triangle has sides of roughly 185 kilometres. With $\kappa$ at most 0.00303, the sides of that Morley triangle can differ by at most $0.00303 \times 0.0107$ of their length — about six metres in 185 kilometres. For a triangle near the equilateral shape, where $\kappa$ is far smaller, the difference is centimetres.

So the curvature of the Earth breaks Morley’s theorem by an amount a survey could detect but nobody drawing a map would ever see. That is the right scale for the band of a sphere that is a band of its cylinder and every other fact of spherical geometry that differs from the plane by a term in the enclosed area: the difference is real, exact, and second to everything a picture shows.

A coefficient computed, not derived #

The coefficient is computed, not derived. Every value of $\kappa$ comes from an exact construction at two small sizes and a Richardson extrapolation, checked by the agreement of sizes and by the vanishing at the equilateral shape, and it is accurate to about three significant figures. A formula for $\kappa$ in terms of the three angles would come from expanding the trisector intersections to first order in the curvature; that expansion is straightforward in principle and has not been carried out here.

Thin triangles are missing. For shapes with an angle below about $11°$ the computation loses its precision before reaching the first-order limit, so the grid stops there. Whether $\kappa$ stays bounded as a triangle degenerates, or grows without limit near the corners of the shape triangle, is not decided by these figures; the growth towards the obtuse corners suggests the second, and a formula would settle it.

First order only. The coefficient describes small triangles. For large ones the spread bends away from the straight line, and the V of the second figure is straight only near its point.

Still open: a formula, and the meaning of the triangle #

The two questions the earlier essay asked are now one question and a half. What is $\kappa$ as a function of the angles? The computed map says it is a smooth, symmetric function, nought at the centre, small everywhere on the grid and largest for wide obtuse shapes, with its maximum within a degree or two of the isosceles shape with apex $137°$. A closed form would give a quantitative Morley theorem: on any surface of curvature $K$, the Morley triangle of a small triangle of area $A$ is equilateral to within $\kappa \cdot |K| A$.

The other half is the question of what the curved triangle is. On the sphere it is nearly equilateral, and its failure is now measured shape by shape; but nobody has found a quantity — a combination of sides and angles, or an area — that the construction makes exactly equal at the three corners on every surface of constant curvature. The same question can be asked of the other seventeen of the eighteen equilateral triangles that trisectors cut out of a flat triangle, each of which presumably fails on a sphere at its own rate. If such a quantity exists, its first-order expansion must cancel exactly the pattern of longest and shortest sides that the fourth figure draws, and the irregular boundaries of that pattern suggest it is not a simple one.