A total hung in a temple

In the Edo period, from the seventeenth century to the nineteenth, Japanese mathematicians developed a tradition of geometry largely separate from Europe’s, and part of it was devotional. A problem solved, or a theorem found, might be painted on a wooden tablet — a sangaku — and hung under the eaves of a Shinto shrine or a Buddhist temple, as an offering and as a challenge to visitors. Hundreds survive. They are mostly about circles packed into triangles, squares and other circles, drawn in bright colours, and they often state a result without any proof.

One of them, dated to around 1800, records a fact that has since been called the Japanese theorem. Take a polygon whose corners all lie on one circle. Cut it into triangles by drawing diagonals that do not cross. In each triangle, draw the inscribed circle, the largest circle that fits inside it. Then the sum of the radii of those inscribed circles does not depend on how the polygon was cut.

The two cuttings in the figure share nothing. On the left, three diagonals fan out from a single corner at the bottom; on the right, they form a zigzag. The triangles have different shapes, their inscribed circles have different sizes, and no circle on one side corresponds to any circle on the other. The totals of the radii agree to six decimal places in the figure and to every digit of the arithmetic in the computation behind it. A visitor to the shrine was presumably expected to wonder why.

Fourteen cuttings, one total #

A hexagon can be cut into triangles by non-crossing diagonals in exactly fourteen ways, the fourth Catalan number, and the figure below checks all of them.

The dots form a perfectly flat line: all fourteen totals are $1.162227$. The open circles, for a hexagon that differs only in having one corner moved off the circle, scatter by more than ten per cent. The constancy is not a property of hexagons, or of triangulations, or of inscribed circles in general. It is a property of polygons inscribed in a circle, and of nothing else.

The fourteen triangulations are themselves a structure worth knowing. Any two of them are connected by a sequence of flips — remove one diagonal, leaving a quadrilateral, and replace it by the quadrilateral’s other diagonal — and the triangulations with flips between them form the corners and edges of a three-dimensional solid, the associahedron. That suggests where to look for a proof. If every flip leaves the total unchanged, every triangulation has the same total, because any one can be reached from any other by flips. And a flip only changes the triangles inside one quadrilateral. So the whole theorem reduces to the case of four points on a circle.

The quadrilateral, and a rectangle nobody asked for #

A quadrilateral with its corners on a circle can be cut along either diagonal, and the theorem says the two pairs of triangles have the same total of inradii.

The totals agree, as the flip argument needs. But the figure shows something extra that the theorem did not promise: the four centres of the inscribed circles — two from each cutting — form a rectangle. This is also on a sangaku, and it is the more surprising of the two facts, because a rectangle is a much stronger statement than an equality of sums. It says that the four incentres of the four triangles a cyclic quadrilateral’s diagonals make are arranged with perfect right angles, for every cyclic quadrilateral.

Why a rectangle appears is a story about arcs. The centre of the circle inscribed in a triangle — one of the classical centres, the average of the corners weighted by the opposite sides — lies on the bisector of each of its angles, and when the triangle is inscribed in a circle, the bisector of the angle at one corner passes through the midpoint of the opposite arc — the inscribed-angle theorem in its simplest use, since equal angles at the corner cut off equal arcs. The four triangles of a cyclic quadrilateral share their corners and arcs in pairs, so their incentres are each determined by a pair of arc midpoints, and carrying the angles through shows that the lines joining the incentres meet at right angles. The computation is long, the figure is the clearer statement, and it is not the shortest road to the equal sums. That goes through a theorem about a single triangle.

Carnot: three distances that add to R + r #

In 1803, Lazare Carnot — military engineer, revolutionary politician, and father of the physicist Sadi Carnot — published a theorem about one triangle and two circles, the one through its corners and the one inside it.

The distances from the circumcentre to the three sides add up to $R + r$, the circumradius plus the inradius — with one adjustment. When the triangle is obtuse, the circumcentre lies outside it, beyond the longest side, and the distance to that side must be counted as negative. The figure’s check uses that rule on 300 random triangles inscribed in a circle, of which 231 are obtuse — very close to the three in four that a random inscribed triangle is obtuse — and the identity holds to the last digit every time.

Carnot’s theorem is the reason for the Japanese theorem, and the deduction is short enough to give in full. Triangulate a cyclic polygon with $n$ corners into $n - 2$ triangles. Every triangle has the same circumcircle, the polygon’s circle, so the same $R$ and the same centre $O$. Write $d(s)$ for the signed distance from $O$ to a segment $s$, and $r_T$ for the inradius of a triangle $T$. Apply Carnot to each triangle and add:

$$\sum_{T} (R + r_T) = \sum_{T} \; \sum_{s \text{ a side of } T} d(s).$$

On the right, every side of every triangle is either a side of the polygon or a diagonal. Each diagonal is a side of exactly two triangles, one on each side of it, and $O$ is on the same side of the diagonal as one triangle’s third corner and on the opposite side from the other’s — so its signed distance is counted once positive and once negative, and cancels. What survives is the sum of the signed distances from $O$ to the polygon’s own sides, which does not depend on the triangulation at all. The left side is $(n - 2)R$ plus the total of the inradii. So the total of the inradii is a fixed quantity minus $(n - 2)R$, the same for every triangulation.

The proof turns a mysterious constant into a bookkeeping identity, and it shows exactly where the circle is used: every triangle must have the same circumcentre, so that the signed distances to a shared diagonal are measured from the same point and cancel. Move one corner off the circle and the triangles containing it acquire a different circumcentre; the diagonals’ distances no longer cancel, and the totals spread, exactly as the open circles in the second figure did.

What the total is, and where it goes as the polygon fills the circle #

Carnot’s bookkeeping does more than show the total is constant; it says what the constant is. For a cyclic polygon with $n$ corners on a circle of radius $R$, the total of the inradii of any triangulation is

$$\sum_{T} r_T = \sum_{s \text{ a side of the polygon}} d(s) \; - \; (n - 2)R.$$

For a regular polygon every side is at the same distance from the centre, $R\cos(\pi/n)$, so the total is $nR\cos(\pi/n) - (n - 2)R$. For the regular hexagon that is $R(3\sqrt3 - 4)$, about $1.196R$; for the irregular hexagon of the figures, whose sides are at different distances, it is the $1.162$ that every triangulation produced.

As the number of sides grows, $n\cos(\pi/n) - (n - 2) = 2 - n(1 - \cos(\pi/n))$, and since $1 - \cos(\pi/n)$ is about $\pi^2/2n^2$, the subtracted term shrinks like $\pi^2/2n$. So the total of the inradii of any triangulation of a regular polygon with many sides tends to $2R$, the diameter of the circle. A polygon with a thousand sides, cut into 998 triangles by any non-crossing diagonals whatever, has inscribed circles whose radii add up to within half a hundredth of the circle’s diameter — though most of those triangles are long slivers whose inscribed circles are tiny, and a fan from one corner and a balanced zigzag cut the polygon into completely different populations of triangles. The limit is a fact about the circle, reached through a polygon that fills it in the same way Archimedes reached the circle’s area.

Counting the cuttings #

The fourteen triangulations of a hexagon are part of a sequence that this subject has met before. A convex polygon with $n + 2$ corners can be triangulated in $C_n$ ways, where $C_n$ is the $n$-th Catalan number — $1, 2, 5, 14, 42, 132, \ldots$ — the same numbers that count balanced arrangements of brackets, mountain paths and binary trees. A heptagon has 42 triangulations, an octagon 132, and a polygon with twenty corners about 477 million. The Japanese theorem says that all of them, for a cyclic polygon, give the same total.

That is a large family for a single number to be constant across, and it shows the power of the flip argument. Checking 477 million triangulations one at a time would be hopeless; checking that a single flip preserves the total, and that flips connect everything, is a finite argument about one quadrilateral. The flip graph is connected — any triangulation can be turned into the fan from a single corner by flips, and so into any other — which is what makes the associahedron a single solid rather than a scattering of pieces. Invariance under a local move, plus connectedness under the move, is the shape of a great many proofs that something does not depend on choices, from the Euler characteristic to the invariants of knots, and the Japanese theorem is one of the most elementary instances.

Why the signs are the heart of it #

It is worth dwelling on the negative distances, because they are what makes Carnot’s theorem true for all triangles and they are easy to get wrong.

For an acute triangle the circumcentre is inside, all three distances are positive, and the theorem is a statement about three positive lengths. For a right triangle the circumcentre is the midpoint of the hypotenuse, the distance to the hypotenuse is nought, and the theorem says the other two distances — half of each leg — add to $R + r$, which can be checked with the familiar formula $r = (a + b - c)/2$ for the inradius of a right triangle. For an obtuse triangle the circumcentre is outside, and the long side separates it from the triangle. Counting that distance as positive gives the wrong answer, by twice that distance; counting it as negative restores the identity.

The sign rule is not a patch. Measured with signs, the distance from $O$ to a side is $R$ times the cosine of the opposite angle, which is negative exactly when that angle is obtuse; and Carnot’s theorem becomes the trigonometric identity $\cos A + \cos B + \cos C = 1 + r/R$, true for every triangle. That identity is the version usually proved, by the law of sines and a page of trigonometry, and its geometric reading — three signed perpendiculars from the circumcentre — is the one that makes the Japanese theorem fall out. In the triangulated polygon, a diagonal is the long side of the triangle on the far side of the centre and a short side of the triangle on the near side, which is why its two signed distances are always equal and opposite.

Off the circle, the totals come apart #

The second figure showed one hexagon pushed off its circle. The last figure pushes sixty of them by different amounts.

The disagreement is zero only at the left edge, where the corners are exactly on the circle, and it grows steadily from there, roughly in proportion to the perturbation. The sample is consistent with a converse: a convex polygon whose triangulations all give the same total of inradii must be inscribed in a circle. For quadrilaterals the converse is a theorem — if the two diagonals give equal totals, the quadrilateral is cyclic — and the flip argument suggests how it might extend, since equal totals for every flip would force each quadrilateral formed by two adjacent triangles to be cyclic, and a polygon in which every such quadrilateral is cyclic is cyclic.

So the Japanese theorem is not only a property of cyclic polygons; it characterises them. A polygon is inscribed in a circle exactly when the sum of its triangles’ inradii does not care how it is cut. That is a strange test for lying on a circle — it never mentions a circle, a centre or a radius — and it is the surprising connection this essay set out to find: a sum of radii of inscribed circles detects whether the polygon has a circumscribed one.

The sangaku tradition #

The Japanese temple geometry deserves more than an anecdote, because its style shaped what the theorem looks like. Sangaku problems typically gave a configuration of tangent circles and asked for one radius in terms of others; the answers were often stated as formulas without proof, sometimes with errors, and the methods behind them were transmitted in schools of mathematics through manuscripts and teaching. The tradition produced results that European mathematicians found independently, such as versions of Descartes’s theorem on mutually tangent circles and of the Malfatti problem, and some that were not found in Europe until much later.

The Japanese theorem appears on a tablet attributed to around 1800, stated for hexagons and cyclic quadrilaterals. Whether its makers had a proof of the general statement is not known; the published proofs in Europe and Japan date from later in the nineteenth century, and Carnot’s theorem, which gives the cleanest proof, was published in France in 1803, almost certainly unknown in Japan at the time. The theorem and its proof thus arrived on opposite sides of the world within a few years of each other, through different routes, and met only when historians compared them.

The flips the figures take on trust #

The flip argument is described and not drawn. The reduction from any polygon to quadrilaterals — any two triangulations are joined by flips, and a flip only changes one quadrilateral — is a combinatorial fact about triangulations that the figures assume. The fourteen-cutting figure checks every triangulation of one hexagon directly, which is evidence for that hexagon; that every cyclic polygon behaves the same way is the argument, not the picture.

The rectangle of incentres is measured, not explained. The quadrilateral figure checks that each of the four angles is a right angle to twelve places, and the text sketches why. The full proof, which goes through the midpoints of arcs and the fact that the bisector of an inscribed angle passes through the midpoint of the opposite arc, is longer than any figure here and is not given.

The converse is a sample. Sixty perturbed hexagons all showed unequal totals, and the disagreement grew with the perturbation. That a polygon with equal totals must be cyclic is a theorem for quadrilaterals and is strongly suggested for hexagons by the figure; the figure cannot exclude a special non-cyclic hexagon whose fourteen totals happen to agree.

Still open: which other sums do not care #

The Japanese theorem is one of a family of statements in which a sum over the pieces of a decomposition does not depend on the decomposition. The nine-point circle and the triangle’s other classical centres generate many quantities attached to a triangle, and for each one can ask whether its sum over the triangles of a cyclic polygon’s triangulation is invariant. For the inradius the answer is yes, by Carnot. For a few other quantities, built from the exradii and from distances between centres, invariance can be proved by the same kind of cancellation; for most quantities one might try, it fails.

Which quantities attached to a triangle have invariant sums over triangulations of cyclic polygons, and whether there is a single principle — beyond Carnot’s cancellation along diagonals — that produces all of them, is not settled. The question of whether a quantity’s invariance can always be traced to an identity like $\cos A + \cos B + \cos C = 1 + r/R$ is open in the sense that nobody has stated a theorem that says so.