The most four hinged rods can enclose

Join four rods end to end with hinges and the frame they make is floppy: push on one corner and the whole quadrilateral changes shape, with the same four side lengths, through a continuous family of positions. Its area changes as it flexes, from almost nothing when it collapses flat to some largest value in between. Which shape encloses the most?

The most area a fence can hold answered the version where the fence can bend freely: the circle. With straight rods of fixed lengths the answer is the next best thing. The largest quadrilateral is the one whose four corners lie on a circle, and its area has a formula in the side lengths alone, found by the Indian astronomer Brahmagupta in 628. The essay draws the flexing frame, measures its area at every shape, and finds both facts in the measurement before explaining them.

One degree of freedom #

A quadrilateral with given sides has one degree of freedom: fix the angle at one corner and the rest is determined, provided the frame can close up at all. The figures take the angle between the rods of length 3 and 6, place those two rods, measure the diagonal between their free ends, and complete the shape with a triangle on that diagonal whose other sides are 4 and 5 — on the far side, so that the quadrilateral is convex. Then the area is computed by the shoelace formula from the four corners. Sweeping the angle through twenty thousand values and keeping the largest area finds the tinted shape.

Its area is 18.97367, and its corners lie on a circle: the fourth corner sits on the circle through the other three to within the resolution of the sweep. The circle has radius 3.2873. Nothing in the search asked for a circle. The search only asked for the most area, and the circle appeared.

The shortfall from the circle #

The explanation is in a formula published by Carl Anton Bretschneider in 1842. For a quadrilateral with sides $a, b, c, d$, semiperimeter $s$ and opposite angles $A$ and $C$,

$$K^2 = (s - a)(s - b)(s - c)(s - d) - abcd\,\cos^2\!\frac{A + C}{2} .$$

The first term depends only on the sides. The second is never negative, and it vanishes exactly when $A + C = 180°$. Every measured shape in the figure sits on this curve. So the area is largest when two opposite angles are supplementary — and by the inscribed angle theorem, a quadrilateral has supplementary opposite angles exactly when its corners lie on a circle, since two inscribed angles standing on the two arcs between the same two points add up to half the full turn. At that shape the subtracted term is zero and

$$K = \sqrt{(s - a)(s - b)(s - c)(s - d)} .$$

For sides 3, 4, 5 and 6 the semiperimeter is 9 and the product is $6 \times 5 \times 4 \times 3 = 360$, so the area is $\sqrt{360}$. That is Brahmagupta’s formula. He stated it without proof and without the condition that the quadrilateral must be cyclic, which his commentators supplied; Bretschneider’s term measures exactly how much area is lost by a quadrilateral that is not.

Why flexing helps until the circle #

Bretschneider’s formula settles the question, but there is a direct argument that shows the circle appearing as the point where flexing stops paying. Split the quadrilateral by the diagonal joining the corners between sides $a, d$ and between sides $b, c$. Its area is

$$K = \tfrac12 ad\sin A + \tfrac12 bc\sin C,$$

and the diagonal is shared, so $a^2 + d^2 - 2ad\cos A = b^2 + c^2 - 2bc\cos C$. Flex the frame slightly: $A$ changes by a small amount and $C$ must change to keep the diagonal common, at the rate $ad\sin A\,dA = bc\sin C\,dC$. Substituting, the rate of change of area is

$$\frac{dK}{dA} = \frac{ad}{2}\cdot\frac{\sin(A + C)}{\sin C}.$$

This is positive while $A + C < 180°$, so opening the angle gains area, and negative beyond, so the area is largest exactly when $A + C = 180°$. No circle was assumed; the circle is where the derivative vanishes. The argument is the hinged-rod version of the variational argument for the circle as the best fence, where every small bulge of a non-circular curve can be moved to gain area.

The circle’s radius from the sides #

The radius of the circle is also fixed by the sides. Parameshvara, a fifteenth-century astronomer of the Kerala school, gave it as

$$R = \frac{1}{4K}\sqrt{(ab + cd)(ac + bd)(ad + bc)},$$

where $K$ is Brahmagupta’s area. For the frame with sides 3, 4, 5 and 6 that is $\sqrt{42 \times 39 \times 38}/(4\sqrt{360}) \approx 3.28729$, exactly the radius that solving the arc equation below gives, and the radius of the circle through three corners of the searched shape. The three factors under the root are, by Ptolemy’s theorem and its companions, products of the diagonals of the three quadrilaterals that the same sides make in their three cyclic orders, which is one more way of seeing that those orders share a circle.

Shrinking one side gives Heron #

Let one side shrink. The largest quadrilateral with sides 5, 4, 3 and $d$ has area 13.311 when $d = 3$, 9.734 when $d = 1.5$, 7.250 when $d = 0.5$, and when $d$ reaches nothing the quadrilateral is a triangle and the formula becomes

$$K = \sqrt{s(s - a)(s - b)(s - c)},$$

Heron’s formula, which gives 6 for the 3-4-5 triangle. Heron of Alexandria proved it around AD 60, and it is a special case of Brahmagupta’s. A triangle needs no condition, because every triangle has a circumscribed circle; for four sides the circle is a genuine condition, met by exactly one shape of the hinged frame.

The shrinking also shows how the cyclic shape behaves as it degenerates. With the fourth side at 0.5 the quadrilateral is a triangle with one corner slightly cut off, still on a circle, and the circle converges to the triangle’s circumcircle as the side vanishes. So Brahmagupta’s formula is continuous in the sides in a strong sense: it passes smoothly into Heron’s without any special case, which is one reason to believe it is the right generalisation rather than one of several. The extended law of sines gives that circle’s size for a triangle, and for the quadrilateral the radius can be computed from the sides too.

The order of the sides does not matter #

Brahmagupta’s formula is symmetric in the four sides, so it cannot see their order. That is not an accident of algebra. A chord of length $x$ in a circle of radius $R$ cuts off an arc of fixed angle $2\arcsin(x/2R)$, and the triangle it makes with the centre has a fixed area. Rearranging the sides round the circle rearranges the arcs and these triangles without changing them, so the total angle is still a full turn, the radius is the same, and the area is the same. The figure places the four rods in all three essentially different cyclic orders on one circle, and all three areas agree to every digit shown.

So the hinged frames 3-4-5-6, 3-5-4-6 and 3-4-6-5 are quite different quadrilaterals, with different angles and different diagonals, but at their largest they enclose the same area on the same circle. That is the first sign that the area of a cyclic polygon depends on its sides as a set, not as a sequence.

The flexing frames themselves are different, though. A frame built in the order 3, 5, 4, 6 passes through a different family of shapes from one built 3, 4, 5, 6, and the two families share only their largest member’s area and circle. At any other angle the two frames enclose different areas. So the symmetry is special to the optimum: it is the circle, not the frame, that forgets the order. The same happens with any number of sides, which is why the area of the cyclic pentagon above is a symmetric function of its five sides even though there are twelve essentially different pentagons with those sides on that circle.

More rods: the circle without a formula #

For five or more hinged rods the largest enclosure is still the cyclic one. A polygon with given sides that is not cyclic can be improved: split it by a diagonal into pieces, and any piece that is not cyclic can be flexed to gain area while the others stay fixed. Only when every four consecutive corners are concyclic is no improvement possible, and then all the corners are on one circle. The problem was studied in the eighteenth century, by Gabriel Cramer among others, and the argument goes back to that period.

But the formula disappears. The radius of the circle is fixed by requiring the arcs cut off by the sides to add up to a full turn,

$$\sum_i 2\arcsin\frac{a_i}{2R} = 2\pi,$$

an equation with one arcsine for each side, solved here by bisection: radius 4.3461 and area 41.1744 for the pentagon, radius 3.9202 and area 36.5192 for the hexagon. For three and four sides the equation can be solved in closed form, which gives Heron and Brahmagupta. For five, David Robbins proved in 1994 that $16K^2$ is a root of a polynomial of degree 7 whose coefficients are polynomials in the squared sides, and the same holds for six; for more sides the degree grows, and Robbins conjectured a formula for that degree, which was later proved, by Maksym Fedorchuk and Igor Pak among others. The area is algebraic, but no expression in radicals like Brahmagupta’s exists in general.

Among all polygons with $n$ sides and a fixed perimeter, the regular one has the most area, which is the polygon version of the isoperimetric theorem. That follows from the hinged-rod result in two steps: with the side lengths fixed, the best polygon is cyclic, and among cyclic polygons with a fixed perimeter, equalising two adjacent sides gains area. The same reasoning on polygons of given diameter rather than given sides leads to quite different and much harder extremal shapes, because a diameter constraint does not behave like a hinge.

Brahmagupta’s verse #

Brahmagupta’s statement is a single verse of the Brāhmasphuṭasiddhānta, an astronomical treatise written in 628 at Bhinmal in what is now Rajasthan. It gives a rough area for any quadrilateral, the product of half the sums of opposite sides, and then the exact area as the square root of the product of the four terms obtained by subtracting each side from the semiperimeter. The verse does not say that the quadrilateral must be inscribed in a circle, and for a long time European historians took this as an error. Later commentators in India read the surrounding verses, which concern the diagonals and the circumscribed circle, as making the cyclic condition clear from context. The flexing frame shows what is at stake: for every other shape of the frame, the verse’s formula overstates the area, by exactly the term Bretschneider identified twelve centuries later.

Ptolemy’s diagonals #

The cyclic quadrilateral is determined by its sides, so its diagonals are too. Ptolemy’s theorem, which he used around AD 150 to compute his table of chords, says that the product of the diagonals equals the sum of the products of opposite sides, $pq = ac + bd$. The figure checks it on five quadrilaterals built on their circles, from the 3-4-5-6 frame with $pq = 39$ to the square of side 5 with $pq = 50$. Brahmagupta also gave the diagonals themselves, as square roots of rational expressions in the sides, and his formulas and Ptolemy’s agree.

Ptolemy’s theorem is the cyclic case of an inequality: for any four points, $pq \le ac + bd$, with equality exactly when they lie on a circle in that order. It is the same pattern as Bretschneider’s: a quantity that every quadrilateral has, with the circle as the extreme case. The power of a point, constant along every chord through a point, is a third face of the same circle geometry, and the triangulation sums of the Japanese theorem a fourth.

Bretschneider’s term is also a measure of how far from cyclic a quadrilateral is, and in that it resembles Bonnesen’s inequality, which bounds how far a closed curve is from a circle by how much area it falls short of the isoperimetric maximum. A quadrilateral whose area is within a small amount of Brahmagupta’s is necessarily close to cyclic, since $\cos^2((A + C)/2)$ must be small. Near-optimal means nearly the optimal shape, which is the stability that makes the extremal answer meaningful for real frames that cannot be set to a perfect angle.

Maximal area forces a circle #

The surprise is that maximising area produces a circle, and that the circle then makes everything else computable. The question asked about the flexing frame was purely about area; the answer brought in inscribed angles, a symmetric formula, invariance under reordering and Ptolemy’s identity, none of which the question mentioned. And the circle is not merely the best shape; it is the shape at which a derivative vanishes, because Bretschneider’s formula shows the area depends on the shape only through $\cos^2((A+C)/2)$, whose minimum is a smooth zero.

There is also a surprise in the history. Brahmagupta’s formula was stated in India nearly twelve centuries before Bretschneider explained exactly when it holds, and Robbins’s discovery that the pattern stops at four sides came another century and a half later, from a computer algebra calculation of what the pentagon’s area polynomial must be.

Twenty thousand angles, exactness from a formula #

The search for the largest shape samples twenty thousand angles, and its agreement with $\sqrt{360}$ is to four decimal places, not exact; the exactness comes from Bretschneider’s formula, which the figure checks on every sampled shape. The cyclic polygons with five and six sides are computed by solving the radius equation numerically, and the claim that they are the largest of their families is quoted, not searched. Robbins’s polynomials are quoted too, not computed: the figures show areas that are roots of degree-seven polynomials without exhibiting the polynomials.

The figures also assume the centre of the circle lies inside the polygon. For polygons with one side much longer than the rest the centre falls outside, and the radius equation changes sign on that side’s arc; the side sets drawn were chosen with the centre inside, which the computation checks.

Shapes swept, Brahmagupta and Robbins quoted #

Computed here: every shape of the hinged frame with sides 3, 4, 5, 6 over twenty thousand corner angles, its area, opposite angles and the circle through three corners; the agreement of every shape with Bretschneider’s formula; the largest area against $\sqrt{360}$; Heron’s limit; the three cyclic orders on one circle with equal areas; the radius and area of the cyclic pentagon and hexagon by bisection; and Ptolemy’s product on five cyclic quadrilaterals.

Quoted: Brahmagupta’s statement of 628 and his diagonal formulas; Heron’s and Ptolemy’s theorems; Bretschneider’s formula of 1842; Cramer’s problem and the maximality of cyclic polygons; Robbins’s polynomials and the work of Pak and of Fedorchuk and Pak.

Still open: the area polynomials of many sides #

Robbins’s polynomial for the pentagon has degree 7 in $16K^2$ and hundreds of terms; for the hexagon, more. For larger polygons the polynomials grow so fast that they have been computed only for a few further cases, though their degrees are known. No explicit description of the coefficients is known for general $n$, and computing the polynomial for many sides is beyond current methods. There is also a question about the other roots: a polynomial of degree 7 has seven roots, and Robbins observed that they correspond to the areas of the other polygons that close up with the same sides on a circle, including those that wind round the centre more than once or whose sides cross. Which roots are realised by which configurations, and how many real roots there are, is understood only for small cases.

The circle the frame finds #

Four hinged rods enclose the most area when their corners lie on a circle, and that area is $\sqrt{(s-a)(s-b)(s-c)(s-d)}$, Brahmagupta’s formula, because Bretschneider’s general formula subtracts a term that vanishes exactly when opposite angles add to 180°. The flexing frame with sides 3, 4, 5, 6 finds $\sqrt{360}$ by search, with its corners on a circle; Heron’s formula is the case of a vanishing side; the order of the sides changes the shape but not the area; and for five or six sides the circle still wins, with an area that is a root of a degree-seven polynomial rather than a formula.