Dudeney's four pieces are the fewest

Any two polygons of the same area can be cut into finitely many pieces that rearrange from one to the other. That is the Wallace–Bolyai–Gerwien theorem, and its proof is a construction: fan each polygon into triangles, turn each triangle into a rectangle, turn each rectangle into a square, and stack the squares. The construction works and is wasteful — it would cut a regular hexagon into thirty-nine pieces to make a square that five pieces can make — and the theorem says nothing about the fewest pieces any particular pair needs.

The most famous particular pair is an equilateral triangle and a square. Henry Dudeney published a four-piece dissection of one into the other in 1902, as a puzzle about a haberdasher cutting cloth, and in 1905 he exhibited it as a model of polished mahogany with brass hinges: the four pieces joined in a chain, so that the triangle swings closed into the square and back. The question that came with it — can it be done in three? — stayed open for a hundred and twenty-two years. It was settled in December 2024, by Erik Demaine, Tonan Kamata and Ryuhei Uehara: no dissection of an equilateral triangle into a square uses fewer than four pieces, even if the pieces may be turned over.

How the four pieces are cut #

Take the triangle with side 2, so that its area is $\sqrt3$ and the square of the same area has side $t = 3^{1/4} \approx 1.31607$. Mark the midpoints D of AB and E of BC. On the third side, find the point J whose distance from E is exactly $t$; the circle of radius $t$ about E crosses that side once, about a quarter of the way from A to C. Mark K on the same side one unit beyond J — half a side. Then drop perpendiculars from D and from K onto the line EJ, meeting it at L and at M.

The cuts are EJ, DL and KM, and they divide the triangle into four pieces: the quadrilateral ADLJ, the quadrilateral DBEL, the quadrilateral ECKM, and a fourth piece that is the right triangle JMK, with L lying on its side JM. The figure computes every point to machine precision and checks that the four areas — 0.40568, 0.68076, 0.39787 and 0.24775 — add to $\sqrt3$ exactly.

To make the square, leave piece 2 where it is, turn piece 1 half a turn about D, turn piece 3 half a turn about E, and turn piece 4 half a turn about K, carried round with piece 3. The four pieces then fill a square of side $t$ with no gap and no overlap: the figure computes the convex hull of the turned pieces, finds four sides of length 1.316074 each and an area of $\sqrt3$, and checks that no two pieces share any area.

A side drawn with compass and straightedge #

The construction asks for a length $t = 3^{1/4}$, and a puzzle about cutting cloth would be no use if that length could only be computed. It can be drawn. The square’s area is $\sqrt3$, the area of the triangle, and $\sqrt3$ is the product of two lengths already in the figure: half a side of the triangle, which is 1, and its height, which is $\sqrt3$. So $t$ is their geometric mean, $t = \sqrt{1 \cdot \sqrt3}$, and the geometric mean of two lengths is the oldest construction in the subject. Lay the two lengths end to end along a line, draw the semicircle on the whole as diameter, and raise a perpendicular at the point where they meet; its height up to the semicircle is the mean. The proof is that the two right triangles cut off by that perpendicular are similar, so the perpendicular is to one length as the other is to it.

Dudeney’s own description did exactly this, with the lengths taken from the triangle itself: extend the line from A to E — the triangle’s height, since E bisects the opposite side — by half a side, bisect the whole, draw the semicircle on it, and read off the mean on the perpendicular at E. A circle of that radius about E then marks J. Every point of the dissection is therefore reachable with compass and straightedge from the triangle’s corners, as it must be if the pieces are to be cut by a tailor with chalk and a string — and as every square root is, by the one rule that governs what can be drawn.

The pieces are far from equal. Of the triangle’s area, piece 2 takes 39.3 per cent, pieces 1 and 3 take 23.4 and 23.0, and the small triangle JMK takes 14.3. Nothing in the problem asks for equal pieces, and nothing in the proof of optimality depends on their sizes; the count is the only thing measured.

Why the lengths come out right #

The square is not a happy accident of the drawing. It is forced by a handful of equalities that the construction builds in, and the figure lays them out.

The cut EJ was chosen to have length $t$, and it becomes one side of the square. The other side is made by the two perpendiculars, DL from piece 1 and KM from pieces 3 and 4, placed end to end, and they are equal, each $t/2 = 0.65804$. Their equality is a consequence of D and K both being at the same distance — one unit — from where the half-turns carry them: D is the midpoint of AB, so turning half a turn about D sends A to B, and K is one unit from J along the side, so turning half a turn about K sends J to a point one unit back. The lengths JL and ME are equal too, 0.56309, so that when pieces 1 and 3 turn they exchange ends along the cut.

Underneath the equalities is a single idea, which the hinged-dissection essay called a strip: lay copies of the triangle, alternately turned, into an infinite strip, lay copies of the square into a strip of the same width, and superimpose the two at the right offset. Every cut of the dissection is a boundary in one strip or the other, and the pieces are the regions both strips agree on. The width of the square’s strip is $t$, the width of the triangle’s strip is fixed by its height, and the offset is the only free choice. Dudeney’s genius was to find the offset at which the strips’ boundaries create only four regions.

The swing #

The model Dudeney built was a chain: piece 1 hinged to piece 2 at D, piece 2 to piece 3 at E, piece 3 to piece 4 at K. Swung one way it is the triangle; swung the other it is the square.

A chain that works at its two ends need not work in between: pieces could collide halfway through the swing. The figure checks every pair of pieces for overlap at 361 positions through the half turn and finds none, provided each hinge turns in the right sense. Of the eight possible combinations of senses for the three hinges, only one passes; the other seven make pieces collide at some point of the swing, with overlaps as large as 0.39 of a unit of area. So the model is not merely a dissection with hinges added. The hinges and their directions are part of the design.

One dissection in a family #

The construction has one length fixed by the square: the cut EJ was set to $t$. Set it to any other length $w$ and the same recipe — midpoints, a point one unit beyond J, two perpendiculars, three half-turns — produces a rectangle of sides $w$ and $\sqrt3/w$, whenever the points land inside the triangle in the right order and the turned pieces do not collide.

The recipe works for $w$ from 1 to $\sqrt3$, so every rectangle whose sides differ by a factor of at most $\sqrt3$ is reached by four pieces of this shape, and the square sits in the middle of the range. Beyond the ends one of the recipe’s conditions fails: the point J would have to leave the side it is defined on, or the feet of the perpendiculars would fall off the cut, or the swing would make two pieces collide.

That three-piece dissections of the triangle into some rectangle exist is classical: cut across at half the height, cut the small top triangle down its altitude, and turn the two halves down beside the trapezoid, which is the slice-and-turn step of the general construction. The resulting rectangle is twice as long as it is high, outside the family above. So the three-piece question was never whether three pieces can turn the triangle into a rectangle. It was whether they can reach the square, which sits in the middle of the range where four pieces work easily.

The corners #

Every argument about the fewest pieces starts from bookkeeping, and the simplest bookkeeping is angles.

Each of the square’s four right angles must be built from piece corners adding to 90°, and each of the triangle’s three corners from piece corners adding to 60°. Each piece’s corners add to the right amount for its number of sides, and wherever a corner of one shape lies on a side of the other, the angles meeting there must add to 180°. In Dudeney’s dissection the four right angles of the square are the right angles at L in pieces 1 and 2 and at M in pieces 3 and 4. The triangle’s three corners do something prettier. Turning piece 1 half a turn about D carries A to B, and turning piece 3 half a turn about E carries C to B, so all three corners of the triangle land at the same point, and their three angles of 60° lie side by side to make a straight line along one side of the square.

That bookkeeping makes two pieces hopeless quickly: a single cut across the triangle has to supply all four right angles of the square and also gather the three 60° corners into straight lines, and a path with two ends has too few places to do both. Three pieces leave much more room, which is why the question lasted.

Why a lower bound is a different kind of problem #

An upper bound on the number of pieces is a single object: exhibit a dissection, check that it works, and the count is settled from above. Dudeney’s took a page to describe and every check in this essay takes a fraction of a second. A lower bound is a statement about every possible dissection at once, and the possible dissections are not a list. The cuts may be straight or curved or wildly crooked; the pieces may be turned and flipped; and a piece of a three-piece dissection need not touch both outlines in any simple way. There is no grid to search and no finite family to enumerate until somebody finds a way to make one.

That is why the gap between four and three lasted so long while far harder-looking questions were settled. The essay on how many pieces a dissection needs made the point in general: for two pieces, whether a dissection exists can be decided by checking finitely many ways of matching the outlines, and three pieces is already outside any such method, because two cuts can meet inside the shape and the pieces need not each carry a whole chain of the original boundary. The 2024 proof is a way round exactly that difficulty for one pair of shapes. It finds the finitely many combinatorial patterns into which every three-piece candidate must fall, and then kills each pattern with arithmetic about lengths and angles.

How three pieces were ruled out #

Ruling out three pieces means ruling out every way of cutting a triangle into three pieces that could also make a square — not three particular shapes, but every shape the cuts could produce, with cuts that may be curved or crooked and pieces that may be rotated or turned over. The set of candidates is infinite, and no general method decides such questions for arbitrary pairs of shapes.

Demaine, Kamata and Uehara’s argument organises that infinite set into finitely many cases according to how the pieces’ boundaries are shared. Every stretch of a piece’s boundary lies on the outline of the triangle, on the outline of the square, on both, or on neither — and a stretch on neither is a cut, which must be matched against another piece’s cut in each assembly. Recording which stretches of which pieces meet which, in the triangle and in the square, gives a finite combinatorial object, which they call a matching diagram, and the lengths and angles along every matched stretch must agree in both shapes. With three pieces the diagrams are few enough to examine, and in each one the agreements it demands of lengths and angles cannot all hold at once. The argument is long and careful, and it does not reduce to a single invariant of the kind that settles the odd number of triangles or the rectangle made of squares.

What it settles is narrow and complete. Four pieces are necessary for an equilateral triangle and a square, whether or not pieces may be flipped. It does not settle the minimum for the hexagon and the square, or the pentagon and the square, or any other famous pair.

The construction checked, the bound quoted #

The figures verify the construction and the swing, not the lower bound. Everything drawn is a check on Dudeney’s dissection: that its pieces fill both shapes, that its swing has no collisions, that its lengths are right. None of it rules out a three-piece dissection; the 2024 proof does that, and it is a proof about shapes nobody drew.

The overlap check is at 361 positions. Two pieces could in principle graze each other between positions a quarter of a degree apart. The pieces are convex and move smoothly, and the smallest gap at the positions checked is comfortably larger than any such graze could be, but the check is a dense sample, not a continuous proof.

The family of rectangles uses Dudeney’s recipe only. Other four-piece dissections of the triangle into rectangles exist, and some reach rectangles outside the range drawn; the figure says which rectangles this one recipe reaches, not which rectangles four pieces can reach.

Still open: the fewest pieces for everything else #

Dudeney’s triangle and square is now the first non-trivial pair whose minimum is known exactly. For the regular hexagon and the square, the best known dissection has five pieces, and whether four could work is unknown. For the pentagon and the square the best known has six. For most pairs of regular polygons, the best known dissections come from strip constructions like Dudeney’s, found by hand and improved by computer search over the offset between strips, and almost none is known to be optimal. The method of the 2024 proof — classifying how boundaries can be shared and showing each case is contradictory — may extend to some of these, but the number of cases grows quickly with the number of pieces, and a proof that the hexagon needs five would have to rule out every four-piece candidate, a far larger family than three.

There is also the question that the hinged-dissection essay left: whether the fewest pieces for a hinged dissection can be more than the fewest for a free one. For the triangle and the square they agree, at four, and Dudeney’s chain shows it. Whether some pair of polygons can be cut into $n$ pieces freely but needs more than $n$ when the pieces must stay hinged together is open.