The five crescents that can be squared

A circle unrolled into a triangle found the area of a disc by cutting it into thin rings and laying them out straight: half the circumference times the radius, $\pi r^2$. The circle that will not square then proved the old suspicion that no ruler-and-compass construction turns that area into a square, because $\pi$ is transcendental. Between the two lies a question the Greeks asked first and answered only in part: if the whole circle cannot be squared, can any region bounded by circular arcs be?

The answer is yes, and the first example is one of the oldest surviving pieces of Greek mathematics. Hippocrates of Chios, around 440 BC, found a lune — a crescent between two circular arcs — whose area is exactly that of a triangle. This essay explains why it works, finds every lune of its kind that works, and shows why the list stops at five: the question turns out to be about which angles can be drawn, and for all but five ratios of arcs, the angle needed is a root of a polynomial that square roots cannot reach.

Hippocrates’ crescent #

Put a right isosceles triangle in a semicircle, its hypotenuse along the diameter. On one of its legs draw a second, smaller semicircle bulging outward. The crescent between the small semicircle and the arc of the large one is Hippocrates’ lune.

The argument is two lines, and both are about scaling. Areas of semicircles go as the square of their diameters. The leg is the hypotenuse divided by $\sqrt2$, so by Pythagoras its square is half the hypotenuse’s square, and the small semicircle has half the area of the large one — the same as the quarter-disc that the leg cuts off the large semicircle. Take the small semicircle, remove the part it shares with the large disc (a circular segment over the leg), and what remains is the lune; take the quarter-disc, remove the same segment, and what remains is the half-triangle. Equal areas less the same segment are equal. With the hypotenuse of length 2, the lune and the half-triangle both have area exactly $\tfrac12$, which the figure computes directly.

Nothing in the argument needed a value for $\pi$. That is the whole point: the area of a curved region came out free of $\pi$ because two quantities carrying $\pi$ were arranged to cancel. Hippocrates found two more lunes of the same kind, and his work was quoted by Simplicius a thousand years later from a lost history by Eudemus; it is the earliest piece of Greek geometry known in anything like its original form.

Where the π goes in general #

Every lune can be analysed the same way. A lune bounded by two arcs on a common chord is the region under the outer arc less the region under the inner one, and each of those is a circular segment: a sector of its circle less the triangle formed by the chord and the centre.

Let the outer circle have radius $r$ and its arc turn through the angle $2\alpha$ about its centre, and the inner circle radius $R$ with arc $2\beta$. The sectors have areas $r^2\alpha$ and $R^2\beta$, each a multiple of $\pi$ whenever the angles are rational fractions of a turn. So the lune’s area is free of $\pi$ exactly when the two sectors are equal:

$$r^2 \alpha = R^2 \beta.$$

The two arcs also share a chord, which ties the radii to the angles: half the chord is $r\sin\alpha$ and also $R\sin\beta$. Put $\alpha = m\theta$ and $\beta = n\theta$, so that the arcs turn through angles in the ratio $m : n$, and eliminate the radii. The condition becomes a single equation for $\theta$:

$$n \sin^2(m\theta) = m \sin^2(n\theta).$$

For each ratio there is one solution in the right range, and the lune built on it has the area of the polygon left over when the sectors cancel. The question of whether the lune can be squared — turned by ruler and compass into a square of the same area — is then the question of whether $\theta$ can be drawn. If it can, the arcs and centres can be drawn, the polygon can be drawn, and any polygon can be turned into a square by cutting and rearranging, which ruler and compass do.

An angle from a polynomial #

The angle $\theta$ is the root of an equation in sines, and sines of multiple angles are polynomials in the cosine. Writing $c = \cos\theta$, $\sin(k\theta)/\sin\theta$ is the Chebyshev polynomial $U_{k-1}(c)$, so the condition is

$$n\,U_{m-1}(c)^2 = m\,U_{n-1}(c)^2,$$

a polynomial equation in $c$ with whole-number coefficients, and in fact in $u = c^2$, since only even powers of $c$ occur. Every step is a square root proved what decides constructibility: a number can be drawn with ruler and compass exactly when it lies at the top of a tower of fields, each step a square root. For a root of a polynomial, that requires at least that the polynomial’s irreducible factor have degree a power of two, and in general that its Galois group be built from steps of size two.

So each ratio $m : n$ becomes a polynomial to examine.

For $2 : 1$ the polynomial is $2u - 1$, so $\cos^2\theta = \tfrac12$ and $\theta = 45°$: Hippocrates’ lune, though drawn above in a different position. For $3 : 1$ and $3 : 2$ it is a quadratic in $u$ — $8u^2 - 4u - 1$ and $16u^2 - 14u + 1$ — and a quadratic needs one square root. These are Hippocrates’ other two. For $4 : 1$ the polynomial in $\cos\theta$ is the irreducible cubic $4c^3 - 2c - 1$, and for $4 : 3$ the polynomial in $u$ is the irreducible cubic $48u^3 - 64u^2 + 20u - 1$; an irreducible cubic has degree three, not a power of two, and its roots cannot be drawn — the same obstruction that stops the cube from being doubled.

The quartics are the interesting rows. For $5 : 1$, $5 : 2$, $5 : 3$ and $5 : 4$ the polynomial in $u$ has degree four, a power of two, and degree alone decides nothing. What decides is the quartic’s group. Its resolvent cubic — the cubic whose roots are the three pairings of the quartic’s four roots, as three ways to pair four roots explained — has a rational root exactly when the group is small enough to be built from square roots. The figure computes each resolvent and searches its possible rational roots exhaustively. For $5 : 1$ and $5 : 3$ one exists, and the quartic is solvable by square roots. For $5 : 2$ and $5 : 4$ none does, the group contains a three-cycle, and a cube root would be needed.

Every ratio gives a polygon; only five give one that can be drawn #

There is a subtlety in the reasoning above that is easy to pass over. For every ratio $m : n$, the lune built at the angle $\theta$ that equalises the sectors has an area free of $\pi$: it equals the polygon formed by the chord’s ends and the two centres, whatever $m$ and $n$ are. The $4 : 1$ lune at $\theta = 27.792°$ has exactly the area of a certain quadrilateral, and so does the $5 : 2$ lune at $19.433°$. In that sense infinitely many lunes are “quadrable” — their areas are rectilinear.

What fails for those ratios is not the cancellation but the drawing. The quadrilateral that equals the $4 : 1$ lune has a vertex at a distance from the chord given by a root of an irreducible cubic, and no ruler-and-compass construction reaches that distance. So the polygon exists and has the right area, and it cannot be produced from the chord in finitely many steps. Squaring the lune would mean producing a square of that area, which would mean producing a length whose square is the area — and the area itself involves the cubic’s root. The classification is therefore about lengths, not areas: the five ratios are those for which every length in the figure is reached by square roots.

The distinction is the same one that separates the circle’s case from the lunes’. A circle’s area is $\pi r^2$, and no polygon whatever, constructible or not, has an area equal to it with the radius as the unit, because $\pi$ is not algebraic. A non-squarable lune’s area is algebraic and equals a perfectly definite polygon; it is merely a polygon that the instruments cannot reach. The lune lies between the two impossibilities that the circle that will not square and the cube that will not double describe, and for five ratios it escapes both.

Five angles in square roots #

A verdict of “constructible” is best backed by the construction itself. For each of the five ratios that pass, the figure writes $\cos^2\theta$ in square roots and checks it against the angle found numerically.

The first three are one square root deep: $\tfrac12$, $(1 + \sqrt3)/4$ and $(7 + \sqrt{33})/16$. The last two need a square root inside a square root: $(3 + \sqrt{5 + 4\sqrt5})/8$ for $5 : 1$, and a nested expression with $\sqrt{15}$ for $5 : 3$. Every square root in such an expression is one intersection of a circle with a line, so each expression is a construction written as a formula: draw the lengths, take the angle whose cosine squared is the result, draw the arcs, and the lune appears with its polygon beside it.

The two nested lunes are the reason the list was not completed in antiquity. Hippocrates’ three need only single square roots, which the Greeks handled freely. The other two were found by Martin Johan Wallenius in 1766, and the same two were later found again by Thomas Clausen in 1840, apparently without knowledge of Wallenius; Clausen conjectured that there were no more. The conjecture took another century.

Curved areas before the calculus #

Hippocrates’ lune matters historically for what it showed was possible. Before it, there was no evidence that any region bounded by curves could have its area found exactly; the circle resisted every attempt, and it would have been reasonable to suspect that curved boundaries simply make areas inexpressible. The lune proved otherwise, and it did so by a cancellation rather than a limit — no infinite process, no approximation, just two circles’ worth of $\pi$ arranged to annihilate each other.

The next great result of the same kind was Archimedes’ quadrature of the parabola two centuries later: a segment of a parabola has exactly four thirds the area of the triangle inscribed in it on the same base. That result needed an infinite process — the triangles added at each stage of the method of exhaustion form a geometric series with ratio a quarter — and it found a rectilinear area for a curve that is not circular at all. Between them, the two results map out the two ways a curved area can turn out to be polygonal: by cancellation of transcendental parts, as for the lunes, or by summation of an infinite rectilinear series, as for the parabola. The calculus later absorbed both, and made “the area under a curve” a single question; the lunes are the cases in which its answer contains no $\pi$ even though the curve is made of circles.

The lunes also carried an early reputation they did not deserve. Aristotle cites a quadrature of the circle “by means of lunes” as an example of a false proof, and later commentators took this as an accusation that Hippocrates thought he had squared the circle. The fuller account, preserved from Eudemus’ history of geometry by Simplicius a thousand years later, credits Hippocrates with squaring three kinds of lune and then a lune together with a circle — a combination whose area could be found exactly although neither part’s could. If anyone drew the false conclusion, the error was in supposing that every lune could be squared. The classification of this essay is the precise version of what went wrong: five lunes can be squared, and the lune that would complete a circle quadrature is not among them.

Why there are no more #

Nikolai Chebotarev proved in 1934, and Anatoly Dorodnov completed in 1947, that among lunes whose two arcs have angles in a rational ratio, these five are the only ones that can be squared. The proof is the table’s method carried to every ratio: for each $m : n$ the angle $\theta$ is a root of the polynomial above, and showing that it is not constructible means showing that the polynomial’s relevant factor has a Galois group that is not built from steps of two. For small ratios that can be checked one at a time, as the figure does up to five; Chebotarev’s achievement was an argument covering every ratio at once, by analysing how the polynomials for general $m$ and $n$ factor and what their roots’ symmetries must be — the same kind of question that how a polynomial breaks modulo the primes answered statistically for single polynomials.

The classification also explains a striking feature of the table. The ratios that work are $2:1$, $3:1$, $3:2$, $5:1$ and $5:3$, and the obstruction for all others is the same one that blocks trisecting an angle and doubling a cube: a root of a polynomial whose structure requires an odd step. The lunes are, in this sense, the third of the classical construction problems to fall to Galois theory — and the only one of the three where the answer was “sometimes”.

What the figures do not decide #

The table goes only to ratios with $m \le 5$; the classification beyond that is the theorem, not the figure. For $m = 6$ the polynomial in $u$ has degree five and for $m = 7$ degree six, and checking that they are irreducible with non-constructible roots requires factoring them, which the figure does not attempt. Those rows would all read “no”, by Chebotarev and Dorodnov’s theorem. The figure’s own verdicts are exact where it gives them: the rational roots it strips are found by trying every candidate the rational root theorem allows, in whole-number arithmetic, and a quartic’s group is read off its resolvent cubic by the same exhaustive test, so no rounding enters any “yes” or “no” in the table.

Two restrictions in the statement deserve emphasis. The theorem concerns lunes whose arcs turn through angles in a rational ratio — commensurable lunes. For irrational ratios the sectors still cancel when $r^2\alpha = R^2\beta$, but then $\theta$ satisfies an equation that is not a polynomial, and whether any such lune can be squared is a different question, settled negatively only in special cases. And “squarable” here means squarable by ruler and compass in finitely many steps from the chord; a lune with constructible $\theta$ is squarable, and the figures construct the five angles, but they do not draw the full sequence of steps from chord to square.

Still open: lunes that are not commensurable #

The commensurable case is closed. The incommensurable case — lunes whose two arcs turn through angles with an irrational ratio — has received much less attention, and it is not known in general whether any of them can be squared. A squarable incommensurable lune would need an angle $\theta$ for which both $m\theta$ and $n\theta$, with $m/n$ irrational, produce a constructible configuration, which pushes the question into transcendental number theory, where results in the style of Lindemann’s theorem on $\pi$ settle particular cases; the general question, as far as the literature records it, remains open.

There is also a softer, historical question. Hippocrates’ lunes were long read as an attempt on the circle itself: if a lune could be squared, perhaps a circle could be written as lunes and polygons. Some ancient reports accuse him of claiming exactly that, through a figure in which a circle plus a lune equals a polygon — a figure that is correct but squares nothing, because the lune in it is not one of the squarable kind. Whether Hippocrates made the mistake or his reporters did is not settled by the surviving sources.

A cancellation with a short list #

The circle cannot be squared because $\pi$ is not an algebraic number at all. A lune escapes that obstruction by never asking for $\pi$: its two curved boundaries carry two multiples of $\pi$ and the lune is chosen so that they are equal and opposite. What is left is a question about one angle, and that question is algebraic — whether a root of a particular polynomial can be built from square roots.

Five times out of the infinitely many possible ratios, it can. Three were found by the first Greek geometer whose work survives, two more in the eighteenth century, and the proof that there are no others needed the deepest theorem about how polynomials factor. The crescents are a small chapter of geometry, and they contain in miniature its central story: an impossibility, a way round it, and a precise account of how far the way round goes.