Four equal quarters with two lines

One straight line can halve two shapes at once, wherever they lie on the page and whatever their shapes. That is the ham sandwich theorem in the plane, and its proof turns a line through every direction and watches a quantity change sign. The same idea answers a question about one shape and several lines: can a single shape be cut into four pieces of equal area by two lines? And if the lines are required to be perpendicular, like the arms of a cross?

Both answers are yes, for every shape. The first follows from the ham sandwich theorem in one step. The second needs a second turning argument, and it is the pattern for a sequence of questions about cutting one mass into equal pieces with several cuts — questions that are settled in two and three dimensions, false in five, and open in four.

A plate cut into quarters #

Take a plate shaped like the letter L. Two perpendicular lines, each of which halves the plate’s area, meet at a point; the question is whether the four pieces they cut can all be equal.

At an angle of 48.8° they are. The areas are not estimated by sampling: the plate is a polygon, each line is found as the exact position that halves its area, and each quadrant’s area is computed by clipping the polygon against two half-planes and applying the shoelace formula. All four come out at 25 per cent to eight decimal places. Finding the halving line in a given direction is itself a small search: slide a line of that direction across the plate, and the area on one side grows steadily from nothing to everything, so bisection on its position finds the one place where it is exactly half. Sixty halvings of the interval pin the line far more finely than any figure can show.

Two halving lines always divide a shape into two pairs of opposite pieces: the pieces on either side of the first line each hold half, so the two quadrants on one side sum to a half, and likewise for the second line. That leaves one degree of freedom. If one quadrant holds a quarter, then the quadrant beside it, sharing a half with it, holds a quarter too, and so do the others. The whole question is whether one quadrant can be made to hold exactly a quarter.

A sign that reverses every quarter-turn #

The answer comes from turning the cross. Let $\theta$ be the direction of the first line, with the second perpendicular to it, and let $q(\theta)$ be the share of area in the quadrant on the positive side of both.

Turn the cross through a right angle. The first line now lies where the second was, and the second where the first was, pointing the other way. The quadrant that was positive for both is now the one next to the original: positive for the old second line and negative for the old first. Its share is a half minus $q(\theta)$, since it and the original quadrant together make up one side of a halving line. So

$$q(\theta + 90°) - \tfrac14 = -\big(q(\theta) - \tfrac14\big).$$

The imbalance $q(\theta) - \tfrac14$ is a continuous function of the angle that becomes its own negative after a quarter-turn. If it is positive at some angle, it is negative a quarter-turn later, and in between it must pass through zero. For the L-shaped plate the crossing is at $48.8°$, and again at $138.8°$, which is the same pair of lines with their names exchanged. Nothing in the computation looked for the answer in the right place: the angle was scanned from zero, and the first sign change found it.

This is the same argument as the pair of opposite points on the Earth with equal temperatures, one dimension down: a quantity that changes sign under a symmetry of the circle must vanish somewhere on it. There the symmetry was a half-turn; here it is a quarter-turn, and the quantity changes sign under it.

Three shapes that are not convex #

Nothing in the argument used anything about the shape except that it has an area and that a line in any direction can be moved until it halves that area.

A crescent, whose centre of area lies outside it, and a shape of two lobes joined by a thin neck both have their quarterings, found the same way. For a non-convex shape a halving line can cross the boundary several times, and a quadrant can be several disconnected pieces; none of that matters to the argument, which only ever compares areas. For each of the three shapes the imbalance crosses zero once in each quarter-turn. There is no reason it must cross only once; for other shapes, several perpendicular crosses can quarter the same area.

Two lines at any angle #

Without the requirement that the lines be perpendicular the question is easier, and the answer is much larger. Pick any direction and a line in that direction halving the area. The two halves it leaves are two shapes, and the ham sandwich theorem gives a single line halving both of them. The two lines together cut four quarters.

So every direction of the first line gives a quartering — a whole circle of them — and the angle between the two lines varies continuously as the first one turns, mostly. Near 48° the second line swings round quickly through a right angle, and that crossing is the perpendicular quartering of the first figure; about 90° later it passes through a right angle again, the same cross with its arms exchanged. The perpendicular quarterings are the places where this one-parameter family happens to meet the extra condition. For a shape with a centre of symmetry, such as a rectangle or an ellipse, the picture is simpler: every line through the centre halves the area, so every quartering pair passes through the centre and only the angle between the lines has to be found; for a disc, every perpendicular cross through the centre works. For a lopsided shape like the L the crossing point moves as the first line turns, and the perpendicular cross is one position among them.

That view explains why perpendicularity costs one argument and no more. The family of quarterings is one-dimensional; asking for perpendicular lines is one condition; and a single turning argument supplies a solution. Asking for more — two lines that quarter the area and also bisect a second shape, say — would be a second condition on a one-dimensional family, and in general it cannot be guaranteed.

A second shape, a second sweep #

The two-lobed shape shows the sign change working on something far from symmetric.

The curve is lopsided and has a different amplitude, but it has the same antisymmetry — its value 90° later is its negative — so it has the same guaranteed zero. The antisymmetry is a property of the cross, not of the shape. Every shape inherits it.

Three planes and eight pieces #

The question has an obvious three-dimensional version: can three planes cut any solid into eight pieces of equal volume? Hugo Hadwiger proved in 1966 that they can. The proof is harder than the planar one, because a triple of planes has more freedom than a pair of lines and the conditions are more numerous, but it rests on the same kind of symmetry argument, now in the form of the Borsuk–Ulam theorem on a sphere: a continuous map from a sphere that respects the antipodal symmetry must hit zero.

The count of conditions against freedoms is what decides these questions. Cutting a mass in $d$ dimensions into $2^k$ equal pieces by $k$ hyperplanes imposes $2^k - 1$ conditions — the pieces’ volumes must be equal — on hyperplanes that have $d$ degrees of freedom each, $kd$ in all. For two lines in the plane that is three conditions on four freedoms, with one to spare, which is the circle of quarterings. For three planes in space it is seven conditions on nine freedoms. For $d$ hyperplanes in $d$ dimensions it is $2^d - 1$ conditions on $d^2$ freedoms, and for $d = 5$ the conditions, 31, outnumber the freedoms, 25.

The plane has a further result of the same kind, due to Robert and Ellen Buck in 1949: any shape can be cut into six pieces of equal area by three lines through a single point. The three lines have five freedoms between them — two for the common point and one angle for each line — and five conditions are needed to make six areas equal, so solutions are expected to be isolated, and a topological argument shows that at least one always exists. Four lines through a point cannot in general cut eight equal sectors: the count is seven conditions on six freedoms, and a shape made of a few small blobs placed awkwardly defeats every attempt. The counting is not a proof in either direction, but it is an accurate guide to which of these questions have positive answers.

Where it fails, and where nobody knows #

Branko Grünbaum asked in 1960 whether every mass in $d$ dimensions can be cut into $2^d$ equal pieces by $d$ hyperplanes. It is true for $d = 1$, trivially, for $d = 2$, by the argument above, and for $d = 3$, by Hadwiger’s theorem. David Avis showed in 1984 that it is false for $d \ge 5$: the counting of conditions against freedoms is not merely suggestive but decisive, because a mass concentrated along a curve of high enough degree — the moment curve — forces all the conditions to be met at once, and there are too many.

For $d = 4$ the counting does not decide — fifteen conditions against sixteen freedoms — and the question is open: nobody knows whether every mass in four-dimensional space can be cut into sixteen equal pieces by four hyperplanes. The general problem, of which this is one case — how many hyperplanes are needed to cut several masses into equal pieces all at once — has been answered for many combinations of dimension and number of masses by topological methods that generalise Borsuk–Ulam, and the case of four hyperplanes in four dimensions has resisted every one of them.

Finding the cuts quickly #

The turning argument proves a quartering exists, and the figures find it by sampling angles and bisecting. For shapes given as finite sets of points rather than areas — a thousand points to be split into four groups of 250 by two lines — there are much faster methods. A line bisecting two point sets in the plane can be found in time proportional to the number of points, by an algorithm of Chi-Yuan Lo, Jiří Matoušek and William Steiger from 1994 that discards points which cannot affect the answer, a constant fraction at each round. Quartering one set by two lines reduces to that: halve the set in any direction, then bisect both halves at once.

The perpendicular version needs the rotation, and its fast algorithms search over the angle in the same spirit as the figure, but cleverly, using the fact that the imbalance changes only when the rotating lines pass through a point. For point sets, the continuous curve of the sweep figures becomes a step function with finitely many steps, and the zero of the continuous argument becomes a step at which the sign changes — found, like everything else here, by the reversal a quarter-turn forces.

Why the argument is fair division #

Cutting a mass into equal pieces with a few straight cuts is a model of dividing something fairly with simple instruments, and the family of results it belongs to is large. Two thieves can share a necklace with as few cuts as there are colours of bead; a labelled triangulation must contain an edge whose ends have opposite labels; colourings of Kneser graphs need as many colours as the topology forces. Each is the Borsuk–Ulam theorem in a different disguise, and each has a counting of conditions against freedoms behind it.

What the equipartition results add is geometry. The pieces must be cut by straight lines or flat planes, which is a severe restriction — a knife rather than a free choice of boundary — and the results say that, in low dimensions, straightness costs nothing: the cuts can still be chosen to make every piece equal. It is a statement of the kind that fair-division rules aim at by other means, obtained here purely from the symmetry of turning. There is a companion statement that asks for less and holds everywhere: every shape has a point such that every line through it leaves at least a third of the area on each side. That point, a relative of the centre points that Helly’s theorem produces, cannot in general be improved to one where every line halves the area, since only centrally symmetric shapes have such a point — which is why the quartering lines of an L-shaped plate do not meet at any fixed centre but at a point that depends on the direction.

What the clipping cannot show #

The areas in every figure are exact for the polygons drawn: the halving positions are found by bisection to many more digits than the figures print, and every quadrant’s area is a shoelace sum over a clipped polygon. The shapes are polygons, and the theorem is about any shape with an area; the polygons stand in for the general case, which the argument covers directly.

The sweeps are sampled at a finite number of angles, and a zero crossing between samples is found by bisection. Two zeros closer together than the sampling would be missed, and so the counts of perpendicular quarterings are counts of those found, not of those that exist. And the three-dimensional theorem, Avis’s counterexample and the open four-dimensional case are described, not computed: a solid cut by three planes into eight equal pieces is well within reach of the same method, but no computation touches the question in four dimensions, which is about every mass there is.

Still open: sixteen pieces in four dimensions #

Whether every mass in four-dimensional space can be divided into sixteen parts of equal measure by four hyperplanes is unknown. It is the only case of Grünbaum’s question left open, and the arithmetic makes it the tightest: one freedom to spare, as in the planar case, but in a space where the topological tools that settle the plane and three-dimensional space give no answer.

The planar case suggests why such questions resist. There, the one spare freedom was used by the turning argument, and the answer was a whole circle of solutions with the perpendicular ones inside it. In four dimensions the spare freedom is a single parameter again, but the configuration space of four hyperplanes is complicated enough that no one has found the symmetry that would force a solution to exist, or a mass for which none does.

What the turning buys #

The perpendicular cross has two arms, and turning it by a right angle swaps them. That swap reverses the sign of one quadrant’s excess over a quarter, and a quantity that reverses sign under a symmetry must vanish somewhere. It is the whole proof, and it quarters an L-shaped plate, a crescent and a pair of lobes without knowing anything about them. The same idea, dressed as the Borsuk–Ulam theorem, cuts a solid into eighths; counted against the freedoms it needs, it cannot cut a five-dimensional mass into thirty-seconds; and in four dimensions it has not yet been made to say either. Between the plate on the page and the open question in four dimensions there is only the same bookkeeping, done more carefully each time: how many things must be equal, how many ways the cuts can move, and whether a symmetry forces the two to meet.