The same question in space

The Reuleaux triangle is three circular arcs, each centred at the corner opposite it, each of radius the width. Equivalently it is the intersection of three discs of that radius centred at the corners of an equilateral triangle.

That second description generalises immediately. Take a regular tetrahedron of edge $s$, put a ball of radius $s$ at each of its four corners, and intersect. The result is a solid, it is called the Reuleaux tetrahedron, and the obvious guess about it is wrong.

Where the extra width comes from #

The failure is small and completely explicable, and the arithmetic is short enough to do.

Place the tetrahedron’s corners at $(1,1,1)$, $(1,-1,-1)$, $(-1,1,-1)$ and $(-1,-1,1)$, so each pair is $2\sqrt{2}$ apart and the edge is $s = 2\sqrt{2}$. The curved edge of the solid joining the first two corners lies where the spheres about the other two meet, so its points are at distance $s$ from both.

Look for such a point on the $x$-axis: $(x, 0, 0)$ at distance $s$ from $(-1,1,-1)$ gives $(x+1)^2 + 2 = 8$, so $x = \sqrt{6} - 1$. The opposite curved edge has its matching point at $1 - \sqrt{6}$, and the distance between them is $2(\sqrt{6} - 1)$, which is $s(\sqrt{6}-1)/\sqrt{2}$.

Divide by $s$ and the ratio is $1.0249$. The solid is two and a half per cent too wide across opposite edges, and it is exactly the width $s$ from a corner to the opposite curved face.

In the plane the same construction has no such defect, and the reason is a counting one. A Reuleaux triangle’s arcs meet at points; a Reuleaux tetrahedron’s spherical faces meet along curves, and a curve has room to bulge in a way a point does not. The failure is a fact about the dimension of the intersections rather than about tetrahedra.

Repairing it #

The bulging edges can be shaved, and the resulting bodies are Meissner’s, from 1911.

Take the six curved edges. Three of them meet at a corner; replace each of those three with a surface swept by rotating an arc, and the width becomes constant. There are two ways to choose the three edges — either three around a face or three around a corner — and both work, giving two Meissner solids that are not the same shape.

The construction is a repair rather than a derivation, which is worth noting. Nobody derived the Meissner bodies from a minimisation; they were built to fix a specific defect, and their status as candidates for the smallest volume is a conjecture made afterwards.

A body that is easy to check #

The spun Reuleaux triangle is the cleanest example of a constant-width solid, and it is worth understanding why it works, since the argument is one sentence and it explains why space is not always harder.

Spin any plane convex region about an axis of symmetry. A direction in space makes some angle with the axis, and the width in that direction is the width of the generating region in the corresponding planar direction — because the solid’s extreme points in that direction lie in the plane containing the axis and the direction. So a generating region of constant width gives a solid of constant width, provided the axis is one of the region’s own axes of symmetry.

The Reuleaux triangle has three such axes, each through a corner and the midpoint of the opposite arc. Spin about any of them and the result is a constant-width solid, and all three give the same shape.

The check the figure performs is the one that matters: the generating Reuleaux triangle’s width is measured off its drawn boundary in three hundred and sixty directions and comes out the same in all of them, and the solid’s volume and surface area are then computed by revolving that boundary as a chain of conical frusta — a construction that is exact for the polyline and converges as the polyline is refined. Blaschke’s relation is then checked against the two measured numbers rather than used to produce either.

That is the whole construction, and it means a constant-width solid can be produced without touching the difficulty the Reuleaux tetrahedron ran into. It is worth noticing what has been given up to get there: the spun triangle has a rotational symmetry the tetrahedron does not, and symmetry is what made the width argument one-dimensional. A body with no symmetry axis has to be checked direction by direction over a sphere, which is what the first figure does and what makes it expensive.

There is a converse worth stating too. Every plane convex region of constant width has axes of symmetry only if it happens to; a generic constant-width curve — the smooth ones built from several harmonics on the rung about support functions — has none, and cannot be spun into anything. So the construction is available for the Reuleaux polygons and for almost nothing else. What it does not produce is a small one: the spun triangle encloses $0.4495$ times the cube of the width, against the ball’s $0.5236$ and Meissner’s $0.4199$.

One relation instead of two #

In the plane, constant-width bodies have a fixed perimeter and a varying area, so there are two quantities and one is constant. In space there are three — width, surface area, volume — and the relation between them is exact.

Blaschke’s relation: every convex body of constant width $w$ in space satisfies

$$V = \frac{wS}{2} - \frac{\pi w^3}{3}.$$

The ball checks it immediately: $V = \pi w^3/6$ and $S = \pi w^2$, and the right-hand side is $\pi w^3/2 - \pi w^3/3$, which is $\pi w^3/6$. The spun triangle checks it too, with the volume and surface area measured independently off the revolved boundary.

The relation is worth a moment because it is not obvious and it is exact. It comes from the fact that a constant-width body and its reflection in the origin are the same shape up to translation, so the body’s Minkowski sum with its own reflection is a ball of radius $w$ — and expanding the volume of that sum in terms of mixed volumes gives the identity in three terms. Nothing about the body enters except its width, its surface area and its volume.

A consequence worth extracting: since $V$ is determined by $S$, and both are positive, the surface area of a constant-width body is at least $2\pi w^2/3$, attained when the volume is as small as it can be. Bounding one bounds the other, and every result about the minimum volume is equally a result about the minimum surface area — which is why the literature quotes sometimes one and sometimes the other and means the same thing.

So minimising the volume is the same problem as minimising the surface area, and the search has one unknown rather than two. That is a genuine simplification and it has not been enough, which is the most informative fact in this rung: even with the objective reduced to a single quantity tied to another by an identity, nobody can find the minimum.

The support function, one dimension up #

The planar rungs describe a shape by its support function on a circle, and the whole of their machinery is about that description. It survives into space and loses the feature that made it powerful.

A convex body in space has a support function $h$ on the sphere of directions, and constant width is again linear: $h(u) + h(-u) = w$ for every direction. So the constant-width bodies of a given width still form a convex set inside a linear space, and every argument that used only linearity carries over — Blaschke’s relation among them.

What does not carry over is the convexity condition. In the plane it is $h + h'' \ge 0$, a single number at each angle, and the extremal problem lands where that number reaches zero. In space convexity requires a matrix — the second derivative of $h$ along the sphere, plus $h$ times the identity — to be positive semi-definite, and a matrix can lose positivity in one direction while keeping it in another.

That is the difference between a corner and an edge. A planar extremal shape has isolated points where its curvature runs out. A spatial one has curves along which one principal curvature runs out and the other does not, and the Meissner solids are exactly that: their sharp features are edges, not vertices, and the repair replaces three of them.

So the machinery is the same and the constraint is one dimension richer, and the richness is precisely where the planar proof’s mechanism fails to generalise.

Why the spatial problem is harder #

The planar proof is delicate and it exists. Three things make the spatial version worse, and they are worth separating because only the third is really about dimension.

The family is larger. A planar constant-width shape is a support function on a circle with only odd harmonics; a spatial one is a support function on a sphere with an analogous condition, and the space of such functions is vastly bigger. An extremal argument has more extreme points to rule out.

The candidate is not unique. There are two Meissner solids of the same volume and different shapes. A minimisation problem with two answers is harder than one with a single answer, because uniqueness arguments — which is what most extremal proofs are — are unavailable from the start.

And the constraint is not local in the same way. In the plane, convexity is the single inequality $h + h'' \ge 0$; in space it is a condition on the Hessian of the support function being positive semi-definite, which is a matrix inequality and does not reduce to one number. The mechanism that made the planar answer land on the constraint boundary at three points has no clean analogue.

What is actually known in space #

A short list, because the gap between what is proved and what is believed is the point of this rung.

Proved: constant-width bodies in space exist and are plentiful; Blaschke’s relation holds for all of them; the ball is the largest; and the minimum volume is at least about $0.4098$ times the cube of the width, a bound from 2009.

Conjectured: the minimum is the Meissner solids’ $0.4199$. The gap between the proved lower bound and the conjectured value is under three per cent, which is close enough that a careful numerical search would be expected to find any counterexample, and none has.

And unknown: whether the minimiser is unique up to symmetry, and whether there is any characterisation of the extremal bodies at all. In the plane the answer is a shape anybody can draw with a compass; in space the answer is believed to be a shape somebody built to fix a defect.

The three-per-cent gap deserves a comment, because it is the kind of gap that misleads. A lower bound within three per cent of a conjectured answer sounds like a problem nearly solved, and it is not: the bound comes from a general inequality that would not become an equality even if the conjecture were true, so closing the gap needs a different argument rather than a sharper version of the same one. A bound that is close is not the same as a bound that is nearly tight, and here the closeness is a coincidence of constants.

It is also worth saying what a numerical search can and cannot do. The family of constant-width bodies is infinite-dimensional, so a search must restrict to some finite parameterisation, and every such restriction has the problem the planar rung records: the minimiser may sit on a constraint boundary the parameterisation cannot reach. Numerical work in this area therefore confirms that nothing obvious beats Meissner and cannot do more.

The same failure, three times #

The break between two dimensions and three happens elsewhere on this site, and the three cases together make a pattern worth having.

Pick’s theorem has no version in space: a lattice polygon’s area is determined by two counts of dots, and a lattice polyhedron’s volume is not, because there is a family of tetrahedra with identical counts and every volume. The planar proof cuts the polygon into unimodular triangles and the spatial one cannot, because not every lattice tetrahedron subdivides that way.

Equal volume is not enough for scissors congruence: any two polygons of the same area can be cut into each other, and two polyhedra of the same volume generally cannot, because an invariant built from the dihedral angles obstructs it. The planar proof has no angles to worry about; a polygon’s angles are consumed by the cutting.

And constant width: the plane’s construction works and space’s does not, because spheres meet along curves.

Three failures, three different mechanisms, and one shared feature — in each case a planar argument’s key step involves objects meeting in something zero-dimensional, and in space the same objects meet in something one-dimensional with room to misbehave. Triangulations meet at vertices and tetrahedra along edges; polygons’ corners are points and polyhedra’s are angles along edges; discs meet at points and balls along circles.

That is not a theorem and it is a good heuristic. When a planar construction is being carried upward, find the step where two things meet, and check what they meet in.

What the pictures cannot show #

The Reuleaux tetrahedron’s widths are computed exactly and sampled over directions, so the reported maximum is the largest over the directions tried. It agrees with the closed form to three decimals, which is the accuracy the direction sampling allows.

The Meissner solids are not drawn. Their volume is quoted rather than computed, because a figure that revolves a boundary cannot build them — they are not solids of revolution, which is exactly why they are smaller than the one that is.

And the lower bound of $0.4098$ is stated with no argument. It comes from a substantial piece of convex geometry and is quoted here to locate the gap rather than to explain it.

Where the ladder goes next #

This rung closes the ladder: the family, the support function, the least area, and the same question one dimension up.

Named here as debts. The Meissner solids drawn properly, which needs a way to build a swept surface rather than a revolved one. And constant brightness, where the shadow’s area is held fixed instead of the width, and which has a completely different answer.

Sideways, the planar minimiser is the rung below, the linear condition on the support function is the rung below that, the construction that starts everything is the first rung, and the tetrahedron whose corners this uses is one of the five.

What is worth carrying away #

A construction that works in the plane is a construction, not a theorem, and the dimension it was drawn in is a hypothesis whether or not anybody wrote it down.

Intersecting discs at the corners of a triangle gives constant width. Intersecting balls at the corners of a tetrahedron gives two and a half per cent too much across opposite edges, and the reason is that spheres meet along curves where circles meet at points. Nothing in the plane construction announced that it was using the dimension.

The habit worth taking is to ask what a construction’s steps meet in. Every generalisation that fails between two and three dimensions fails at a step where two objects that met in a point now meet in a curve, and finding that step is faster than testing the conclusion.

The corollary is about repairs. The Meissner bodies fix the defect and are believed to be optimal, which is a coincidence rather than a consequence — the repair was designed to restore constant width, not to minimise anything. A patched construction that turns out to be extremal is a hint that the extremal problem is being solved by the constraint rather than by the objective, which is what happened in the plane, and is the most plausible reason to believe the conjecture.