The sums a triangulation cannot change

A total hung in a temple followed the Japanese theorem from a sangaku tablet to its proof. Cut a polygon inscribed in a circle into triangles by non-crossing diagonals, draw each triangle’s incircle, and add the radii. The total is the same for every way of cutting. That essay checked it on a hexagon’s fourteen triangulations and proved it through Carnot’s theorem on the distances from the circumcentre to the sides. It ended by asking which other quantities share the property.

This essay answers the question for a large family at once. In a triangle with circumradius $R$, the inradius is $r = R(\cos A + \cos B + \cos C - 1)$, so the Japanese theorem says that the sum of the cosines of all the angles of all the triangles does not depend on the triangulation. Replace the cosine with any function $f$ of an angle and the question becomes concrete: for which $f$ is the sum of $f$ over every angle of every triangle the same for all triangulations? The answer is a single condition on $f$, and its proof is the inscribed-angle theorem used once.

A sum over angles #

A triangulation of a convex polygon with $n$ vertices has $n - 2$ triangles and therefore $3(n - 2)$ angles. The number of triangulations is a Catalan number, the sequence one sequence, counting everything found counting a dozen other things: 14 for a hexagon, 42 for a heptagon, 132 for an octagon. For each triangulation and each function $f$, add $f(\text{angle})$ over all its angles. The sum depends on the polygon, on $f$ and possibly on the triangulation, and the question is when the last dependence vanishes.

Some cases are obvious. With $f(x) = x$ the sum is the total of all the angles, $(n - 2)\pi$, whatever the triangulation. With $f(x) = \tfrac12\sin 2x$ and the circle of radius one, the sum is the polygon’s area. A triangle inscribed in a unit circle has area $\tfrac12(\sin 2A + \sin 2B + \sin 2C)$, and the triangles of any triangulation tile the polygon. The Japanese theorem is the first case that is not obvious: $f(x) = \cos x$. The hero figure adds a fourth, $\cos 3x$, which nothing in the literature of sangaku suggests.

Eleven functions, forty-two triangulations #

Testing a function is mechanical: enumerate the 42 triangulations of a fixed cyclic heptagon, compute each sum, and measure the spread between the largest and smallest.

The result is a clean split. Six functions give spreads of $10^{-15}$ to $10^{-14}$, which is the arithmetic’s rounding error: $\cos x$, $\cos 3x$, $\cos 5x$, $\sin 2x$, $x$ and $(x - \pi/2)^3$. Five give spreads between $0.75$ and $4.7$: $\cos 2x$, $\sin x$, $\cos 4x$, $x^2$ and $\tan(x/2)$. There is nothing in between, no function that is nearly invariant. The six are exactly the functions for which $f(x) + f(\pi - x)$ is the same for every $x$, and the five are exactly those for which it is not. $\cos(\pi - x) = -\cos x$, so the cosine and its value at $\pi - x$ always add to nought. $\cos 3(\pi - x) = -\cos 3x$ likewise. But $\cos 2(\pi - x) = +\cos 2x$, so those add to $2\cos 2x$, which varies.

One flip, and the inscribed angles that cancel #

The proof needs two facts. The first is combinatorial: any triangulation can be turned into any other by a sequence of flips, each of which removes one diagonal, leaving a quadrilateral, and inserts the other diagonal of that quadrilateral. So a sum is the same for all triangulations as soon as no single flip changes it. The second fact is the theorem of an angle that does not move: angles inscribed in the same arc are equal. Neither fact alone is enough: the flips connect every triangulation to every other without saying anything about angles, and the inscribed-angle theorem relates angles without saying anything about how triangulations differ. Together they reduce a statement about every triangulation to a statement about one quadrilateral.

Cut a cyclic quadrilateral $ABCD$ by the diagonal $AC$. The two triangles have six angles. Two are the quadrilateral’s whole angles at $B$ and $D$, and four are pieces of its angles at $A$ and $C$: $\angle BAC$, $\angle CAD$, $\angle BCA$ and $\angle ACD$. Cut it instead by $BD$. Now the whole angles are at $A$ and $C$, and the four pieces are $\angle ABD$, $\angle DBC$, $\angle ADB$ and $\angle BDC$. Each piece on one side equals a piece on the other: $\angle BAC$ and $\angle BDC$ stand on the same chord $BC$, $\angle CAD$ and $\angle CBD$ on $CD$, and so on round the quadrilateral. The pieces cancel in pairs.

What remains is the difference

$$f(B) + f(D) - f(A) - f(C).$$

Opposite angles of a cyclic quadrilateral add to $\pi$ — the inscribed-angle theorem again, since the two arcs they stand on make up the whole circle. So $D = \pi - B$ and $C = \pi - A$, and the difference is $[f(B) + f(\pi - B)] - [f(A) + f(\pi - A)]$. That is zero for every quadrilateral exactly when $f(x) + f(\pi - x)$ is the same for all $x$. The rule is proved in both directions. If $f(x) + f(\pi - x)$ is constant, no flip changes the sum. If not, a quadrilateral with angles chosen where it differs gives a flip that does, and that quadrilateral can be part of a larger cyclic polygon.

Carnot’s proof is the same proof #

The proof in a total hung in a temple went through Carnot’s theorem: in any triangle, the signed distances from the circumcentre to the three sides add to $R + r$. It looks like a different argument, and it is the same one. The signed distance from the circumcentre to a chord is $R$ times the cosine of the inscribed angle standing on that chord, measured from the side where the triangle lies. So Carnot’s theorem is the cosine formula for the inradius, written chord by chord instead of angle by angle.

Summed over a triangulation, each diagonal appears in two triangles, once on each side. Its two signed distances are $R\cos\theta$ and $R\cos(\pi - \theta)$, where $\theta$ and $\pi - \theta$ are the two inscribed angles standing on it, one from each side. These are supplementary because the two arcs on either side make up the whole circle. The two terms cancel, only the polygon’s own sides survive, and the total is fixed. That is the flip argument with the cancellation moved from the corners of a quadrilateral to the two sides of a chord, and it needs the same property of the cosine. For $\cos 3x$ the chord-by-chord version works identically: the diagonal contributes $\cos 3\theta + \cos 3(\pi - \theta) = 0$.

Seen this way, every invariant in the family is a sum over the chords of a cyclic polygon of a quantity that depends on the chord and on the side it is viewed from, odd under switching sides. The power of a point, which one number for every chord through a point attached to every chord through a fixed point, is a quantity of the same kind, attached to chords rather than to triangles.

Odd harmonics and even ones #

The criterion has a natural form for cosines of multiples of the angle.

$\cos k(\pi - x) = (-1)^k \cos kx$. For odd $k$ the two terms cancel, and for even $k$ they double. So every odd harmonic gives an invariant sum and every even one does not, on every cyclic polygon. The figure confirms it on forty random octagons, with spreads below $10^{-13}$ for $k = 1, 3, 5, 7$ and above 10 for $k = 2, 4, 6, 8$.

That gives the whole family a description. Write $y = x - \pi/2$, the angle measured from a right angle. The condition $f(x) + f(\pi - x) = c$ says that $f - c/2$ is an odd function of $y$. A sum of $f$ over the angles ignores the triangulation exactly when $f$, centred at a right angle, has only odd terms — $\sin$ of odd multiples of $y$, odd powers of $y$, or anything built from them. $\cos x = -\sin y$ is the simplest such function, which is why the Japanese theorem was found first.

Odd powers of the cosines #

The odd-function form of the criterion produces invariants that look nothing like the Japanese theorem. Since $\cos(\pi - x)^3 = -\cos^3 x$, the sum of the cubes of the cosines of all the angles is invariant. So is the sum of their fifth powers, and of any odd polynomial in the cosines. These are not new in disguise: $\cos^3 x = \tfrac14(\cos 3x + 3\cos x)$, so the sum of cubes is a combination of the sums of $\cos 3x$ and $\cos x$, both invariant already. The odd harmonics and the odd powers span the same family, the way a Fourier series and a power series describe the same odd functions in two different bases.

The even powers fail for the matching reason. The sum of the squared cosines is $\tfrac12\sum(1 + \cos 2x)$, which carries the non-invariant $\cos 2x$, and it varies from one triangulation to the next. So does every combination that contains an even harmonic with a nonzero coefficient. A proposed identity for cyclic polygons that involves a quadratic in the cosines, or the product $\cos A \cos B \cos C$ that appears in formulas for the orthocentre, cannot be true for all triangulations unless the even part cancels exactly. The criterion detects that in one line, without any figure.

Named quantities, sorted #

The criterion turns a list of familiar triangle quantities into a test, because most of them have formulas in the angles when the circumradius is one.

The inradius is $\cos A + \cos B + \cos C - 1$; its sum is invariant, which is the Japanese theorem. The squared distance from circumcentre to incentre is $OI^2 = 1 - 2r$ by Euler’s formula, linear in $r$, so its sum is invariant too. The same holds for the sum of the three exradii, $4 + r$, and for the area, $\tfrac12\sum \sin 2A$. The perimeter is $2(\sin A + \sin B + \sin C)$. Since $\sin(\pi - x) = \sin x$, that fails the criterion, and indeed the perimeters depend on which diagonals are drawn. The squared distance from circumcentre to orthocentre is $3 + 2\sum \cos 2A$, which fails because $\cos 2x$ does. And $r^2$ fails because squaring a sum of cosines is not a sum of odd functions.

The figure computes each quantity from the triangle’s vertices, not from these formulas, and the split matches the criterion exactly. It gives a quick way to read off whether any proposed generalisation of the Japanese theorem can be true. A quantity linear in the inradius always works. Anything involving the perimeter, the orthocentre or a square never does.

The flip graph #

The proof used the fact that flips connect every triangulation, and for the heptagon the connections can be drawn.

Every triangulation of a heptagon has four diagonals, and each can be flipped, so each point in the graph has exactly four neighbours. The graph is connected. It is the skeleton of the associahedron, the solid whose corners are triangulations, and its diameter for $n$-gons is $2n - 10$ for large $n$, a theorem of Daniel Sleator, Robert Tarjan and William Thurston settled completely by Lionel Pournin in 2014. The colours show $\sum \cos 2x$ changing along almost every edge. An invariant sum is constant along every edge, so it is constant on the whole graph.

The same structure appears in every side measured by one diameter, where Ptolemy’s relation among a cyclic quadrilateral’s sides and diagonals is the identity behind a flip. Quantities that are additive over a triangulation and unchanged by flips are the subject’s natural invariants. Angle sums are only one family of them.

Off the circle, almost nothing survives #

The circle is doing all the work, and moving a single vertex off it shows how much. On a convex polygon that is not inscribed in a circle, a flip still exchanges two triangulations of a quadrilateral. The inscribed-angle pairing of the small angles fails, because the four pieces no longer stand on common chords of a circle, and the opposite angles of the quadrilateral no longer add to $\pi$. Of the six invariant functions in the spread figure, only $f(x) = x$ survives on every convex polygon. The total of all the angles is $(n - 2)\pi$ regardless.

The area survives as a quantity, since triangles always tile the polygon, but not as the angle sum $\tfrac12\sum\sin 2x$. That formula assumed every triangle had circumradius one, and off the circle the triangles have different circumcircles. The Japanese theorem fails too, as a total hung in a temple showed by pushing one vertex of a hexagon off its circle and watching the fourteen totals spread. The invariants of this essay are invariants of cyclic polygons, and they are as special as the circle.

What the computation can and cannot show #

Each figure computes every triangulation exactly and reports spreads at the level of the arithmetic’s rounding. The split between $10^{-14}$ and order one is unambiguous. The criterion itself is proved by the flip argument, not by the figures, and the figures test it on particular polygons. A function could in principle pass on the heptagon and fail elsewhere. The proof says it cannot, because the condition involves only the quadrilateral’s angles, and every pair of supplementary angles occurs in some cyclic quadrilateral.

The figures also restrict attention to sums of a single function over individual angles. Quantities that depend on two angles of a triangle jointly, or on the triangle’s shape in ways not expressible angle by angle, need a different analysis. Some are invariant for other reasons; the area, for example, also equals a sum over the triangle’s edges of signed terms, and is invariant for that reason too.

Still open: invariants of other kinds #

The criterion settles which sums of an angle function are invariant. It does not settle which quantities of a triangle in general have triangulation-invariant sums. A quantity can be a function of all three angles at once, and the flip argument then gives a functional equation on pairs of triangles that share a diagonal. Its solutions include every sum of odd angle functions and every edge-additive quantity like the area. Whether there are others, and how to describe the whole space of solutions, is a question that has been studied for particular families and has no general answer.

The same question can be asked for polygons inscribed in a sphere or in the hyperbolic plane, where the inscribed-angle theorem fails in its simple form and the Japanese theorem needs modification. On those surfaces some versions hold, with the inradius replaced by a suitable function of it, and the full list of invariants is not known. The criterion here depends on opposite angles of a cyclic quadrilateral adding to $\pi$, and on the sphere they add to more.

A theorem that was one case of a rule #

The sangaku that recorded the Japanese theorem stated a single surprising equality. The cosine formula for the inradius turns it into a statement about a sum of cosines over angles. The inscribed-angle theorem, applied to one flip, reduces it to a property of the cosine: $\cos x + \cos(\pi - x) = 0$. Every function with that property, centred at a right angle and odd, gives a theorem of the same kind — $\cos 3x$ and $\cos 5x$ among them, and the area and the angle sum as the trivial members. The equality on the temple tablet was the first member of the family to be noticed, not a property special to the inradius. What made it the first is that the inradius is the one member with a picture: an incircle can be drawn inside each triangle and its radius measured with a ruler, while the sum of $\cos 3x$ has no circle to draw. The criterion explains the tablet’s theorem, and it also says that the theorem has many silent relatives, true on every cyclic polygon and visible only to someone who computes the sums.