No odd number of equal triangles

A rectangle made only of squares restricted the pieces of a dissection to one shape and found the answer decided by arithmetic: a rectangle can be squared exactly when its sides are in rational proportion. This essay restricts the pieces differently. They may be triangles of any shape at all, but they must all have the same area.

Cutting a square into two triangles of equal area is a single diagonal. Four is two diagonals. Six, eight, ten are no harder, and the construction is shown below. The question is what happens for odd numbers — whether a square can be cut into three triangles of equal area, or five, or seven.

Nobody could do it, and in 1965 Fred Richman asked whether it was possible at all. John Thomas proved in 1968 that it is impossible provided every corner of every triangle has rational coordinates with odd denominators. Paul Monsky removed that proviso in 1970. A square cannot be cut into an odd number of triangles of equal area — not by any arrangement, with any shapes, using any number of triangles.

Every even number is easy #

The even half of the answer takes one figure. Cut the square into $k$ equal vertical strips, and cut each strip along a diagonal. That gives $2k$ triangles, each of area $1/(2k)$, for every $k$.

So the theorem is really about parity — the same one-bit distinction that decides whether a shuffle can be undone by an even number of swaps, arriving from somewhere entirely different — and it is worth pausing on how strange that is. A dissection into an odd number of triangles is not hard to find — cut the square into three triangles by joining one corner to a point on the opposite side, and there are three triangles, of areas $\tfrac12$, $\tfrac14$, $\tfrac14$. The difficulty is only in making them equal. And there is no geometric reason visible in the picture why three equal ones should be harder than two: nothing about triangles, squares or angles seems to know the difference between odd and even. The strips are not the only way to reach an even count, either: a square cut into two equal rectangles, each cut into three equal triangles by joining a corner to points one third and two thirds of the way along the opposite side, gives six equal triangles of quite different shapes. Even counts are flexible and plentiful. Odd counts are not merely rare; there are none.

The proof confirms that nothing geometric does. It works by bringing in a completely different notion of size, in which being divisible by two makes a number small.

A second way to measure a number #

The ordinary absolute value measures how far a number is from zero on the line. The 2-adic absolute value measures how divisible it is by two. For a fraction $a/b$ in lowest terms, write it as $2^k \cdot (\text{odd}/\text{odd})$; then its 2-adic size is $|a/b|_2 = 2^{-k}$. So $4$ is small, of size $1/4$; $1/4$ is large, of size $4$; and every odd number, and every fraction with odd numerator and odd denominator, has size exactly $1$.

This is the size in which a series can converge to minus one: $1 + 2 + 4 + 8 + \cdots$ has terms growing smaller in this measure, and its sum is $-1$. It has one property the ordinary size lacks, and the proof needs it. The size of a sum is never more than the larger of the two sizes — and if the two sizes are different, the sum has exactly the larger one:

$$|x + y|_2 \le \max(|x|_2, |y|_2), \quad \text{with equality when } |x|_2 \ne |y|_2.$$

In words: in the 2-adic world, a small number added to a large one cannot change its size at all. A few examples fix the idea. The number $6 = 2 \cdot 3$ has size $1/2$; $3/4$ has size $4$; $5/12 = 5/(4 \cdot 3)$ has size $4$ as well; and $6 + 3/4 = 27/4$ has size $4$, the larger of the two, exactly as the rule says. The ordinary size of $27/4$ is dominated by the $6$; the 2-adic size is dominated by the $3/4$. Each measure sees a different term as the important one.

Three colours from two coordinates #

Now colour every point $(x, y)$ of the plane by comparing the 2-adic sizes of its coordinates. Colour it orange if both coordinates are small, $|x|_2 < 1$ and $|y|_2 < 1$. Otherwise, colour it blue if $x$ is at least as large as $y$, and green if $y$ is larger.

The picture looks like noise, and in a sense it is: the colouring has no continuity at all, and every small region of the square contains points of every colour. What it has instead are two exact algebraic properties.

The colours survive adding an orange point. If $(a, b)$ is orange and $(x, y)$ is any point, then $(x + a, y + b)$ has the same colour as $(x, y)$. Adding a small number to a large one does not change its size, and adding a small number to a small one keeps it small. So the colouring is unchanged by sliding the whole plane by any orange vector.

The corners of the unit square get the right colours for Sperner’s lemma. Along the bottom edge $y = 0$, so $y$ is as small as a number can be, and every point is orange or blue. Along the left edge only orange and green occur. Along the top edge $y = 1$ has size $1$, so no point is orange; it is blue or green. The right edge is blue or green as well, apart from the corners.

No line carries all three colours #

The key geometric fact about this very ungeometric colouring is that no straight line meets all three colours.

The proof is short. Slide the plane so that an orange point of the line is at the origin; colours do not change. Suppose now the line through the origin also contains a blue point $(x_1, y_1)$ and a green point $(x_2, y_2)$. Three points on a line through the origin satisfy $x_1 y_2 - x_2 y_1 = 0$. But blue means $|x_1|_2 \ge |y_1|_2$ and $|x_1|_2 \ge 1$, green means $|y_2|_2 > |x_2|_2$ and $|y_2|_2 \ge 1$, and multiplying, $|x_1 y_2|_2 > |x_2 y_1|_2$. By the rule for sums, $|x_1 y_2 - x_2 y_1|_2 = |x_1 y_2|_2 \ge 1$, so the expression is not zero, and the three points are not on a line after all. A line with no orange point is handled the same way with one of the other colours at the origin.

The same computation, applied to a triangle rather than a line, gives the second lemma. If a triangle has one corner of each colour, put the orange corner at the origin; its area is half of $|x_1 y_2 - x_2 y_1|$, and that expression has 2-adic size at least $1$. Halving it doubles the 2-adic size. Every triangle with three different colours at its corners has area of 2-adic size at least 2 — its area, as a fraction in lowest terms, has an even denominator.

Sperner’s lemma counts the three-coloured triangles #

The last ingredient is the counting lemma that the essay on fixed points used to prove that something always stays put, and that the rent-splitting argument used to divide a house. Cut a region into triangles, colour the corners with three colours following the boundary rules above, and the number of triangles whose corners carry all three colours is odd.

The proof is a count of doors. Call an edge a door if its ends are orange and blue. Count, for every triangle, how many of its three sides are doors, and add. A triangle with all three colours has exactly one door; any other triangle has none or two. So the total has the same parity as the number of three-coloured triangles.

Now count the same total by edges instead. A door inside the square is a side of two triangles and is counted twice. A door on the boundary is a side of one triangle and is counted once. So the total has the same parity as the number of boundary doors — and those occur only along the bottom edge, where the colour starts orange at $(0, 0)$, ends blue at $(1, 0)$, and switches between the two an odd number of times in between. The number of three-coloured triangles is therefore odd, and in particular it is not zero.

The random triangulations in the figure were built by inserting points with rational coordinates of many different denominators, and none of them was designed to have any particular property. Each was coloured and counted. Four hundred times the count came out odd, and 3,402 times a three-coloured triangle’s area came out with an even denominator. Neither outcome is luck; both are the lemmas, observed.

The contradiction, and where the odd number enters #

Now suppose the square were cut into $n$ triangles of equal area $1/n$, with $n$ odd. Refine nothing; just colour the corners of the triangles. Sperner’s lemma, in the version for dissections where corners can sit in the middle of other triangles’ edges, gives at least one three-coloured triangle. Its area has 2-adic size at least $2$. But its area is $1/n$, and with $n$ odd, $|1/n|_2 = 1$.

That is the contradiction, and the whole theorem is visible in it. An odd $n$ makes the equal areas 2-adically ordinary, of size exactly one. Monsky’s colouring forces some triangle to be 2-adically large. For even $n$ there is no conflict at all — $|1/n|_2$ is at least $2$ — and the equal-area cuts in the second figure exist.

Two details in that last step are worth making explicit, because each is a place where a first attempt goes wrong. The first is that a dissection need not be a triangulation: the corner of one triangle can sit in the middle of another triangle’s side, making a T-shaped junction, and the door-counting proof above assumed that every edge is shared by exactly two triangles or lies on the boundary. It extends anyway. A door along a long side that is broken into several shorter sides on the other side is still counted an even number of times from inside, because the colours along a line take at most two values and the switches between them pair up.

The second is that the equal-area hypothesis is used exactly once, at the very end, and only for one triangle. Monsky’s argument never needs to know the other triangles’ areas. It finds a single triangle whose area is 2-adically large and observes that $1/n$, for odd $n$, is not. That is why the proof is so robust: it would work equally well if all the triangles had areas of odd denominator, whatever those areas were.

Why the prime is two #

The number $2$ enters the proof in one place, and it is not the parity of $n$. It is the $\tfrac12$ in the formula for a triangle’s area. The expression $x_1 y_2 - x_2 y_1$ for a three-coloured triangle has 2-adic size at least one; halving it to get the area doubles the size, to at least two; and the equal areas $1/n$ have 2-adic size exactly one when $n$ is odd. Without the half, the argument would compare a size of at least one with a size of one and prove nothing.

So the theorem is, underneath, about the factor $\tfrac12$ that turns a determinant into an area — the same factor that makes a triangle half of a parallelogram and that makes lattice triangles have areas in half-integers. A square has area $1$, and cutting it into triangles means expressing $1$ as a sum of determinants each divided by two; for the halves to add up to $1$ with all terms equal to $1/n$, some $2$ has to be absorbed somewhere, and an odd $n$ has nowhere to absorb it.

This also explains the shape of the generalisations. For a regular polygon with $p$ sides, $p$ an odd prime, the corners have coordinates in a larger field built from cosines of multiples of $2\pi/p$, and a valuation extended to that field, attached to a prime above $p$, plays the role the 2-adic one plays here — which is why the counts that work for such polygons are multiples of the number of sides rather than even numbers.

A gap in the argument, and the axiom that fills it #

The figures used points with rational coordinates, and for those the 2-adic size is defined by counting factors of two. But a dissection of the square might put corners at irrational points — $\sqrt 2/3$, or $\pi/4$ — and there the size is not defined by anything so simple.

Monsky’s proof needs a 2-adic size on every real number, one that agrees with the usual one on the rationals and keeps the two properties above. Such a thing exists, by a theorem of Claude Chevalley on extending valuations from one field to a larger one. But the extension to all the real numbers cannot be written down: its existence depends on Zorn’s lemma, the working form of the axiom of choice. So the standard proof that no square can be cut into three equal triangles — a statement about finitely many points in the plane — passes through an object that cannot be exhibited.

Nobody believes the theorem is in any doubt, and there are ways around the dependence: for a given dissection only finitely many real numbers are involved, and a valuation on the field they generate can be built without choice. But the proof as usually given is a striking instance of a concrete, finite, elementary statement whose simplest known proof leans on a non-constructive axiom.

What the colourings cannot show #

The figures show the colouring on grids of rational points and on triangulations with rational corners. They cannot show the colouring of an irrational point, which exists only by the extension theorem, and they cannot show a dissection into an odd number of equal triangles failing, because there is nothing to draw. What they show are the two lemmas holding on every case tried, and Sperner’s parity holding on every triangulation tried. A check of that kind is worth having for a proof this indirect — it confirms that the colouring was defined with the right inequalities, strict where they must be strict — but it is evidence about the lemmas, not a substitute for them, and a single counterexample to either would have shown up as an even bar in the histogram.

Nor do the pictures give any sense of why the result is sharp in the way it is. It fails for other shapes in interesting ways. A regular pentagon, hexagon or any regular polygon with at least five sides can be cut into $m$ triangles of equal area exactly when $m$ is a multiple of its number of sides, as Elaine Kasimatis showed in 1989 using valuations for other primes. Every polygon with central symmetry behaves like the square — no odd number — which Monsky proved in 1990, answering a question of Sherman Stein. The colouring above is the prime $2$’s contribution; other shapes call on other primes.

Still open: how nearly equal an odd dissection can be #

The impossibility is exact, and it leaves a quantitative question: if the square is cut into an odd number $n$ of triangles, how close to equal can their areas be? The areas cannot all be $1/n$, but the difference between the largest and the smallest can be made positive and small.

It turns out it can be made astonishingly small. Bernd Schulze showed in 2011 that the spread can be made to shrink exponentially as $n$ grows, and computer searches by Jean-Philippe Labbé, Günter Rote and Günter Ziegler found dissections doing far better than that for every odd $n$ they could reach. Lower bounds are far weaker. The true rate at which the best odd dissections approach equality is not known, and neither is any structural description of the dissections that come closest. Like the count of pieces in an ordinary dissection, it is a question whose answers so far come from search, and whose lower bounds come from exactly the kind of 2-adic reasoning that proves the impossibility: an argument that some triangle’s area must be 2-adically large also bounds how close to $1/n$ it can be, and the bounds it gives are far from what the searches find.