The circle and the number π

The last chapter ended with a jar wrapped in thread. Unwind the thread and lay it along the diameter: it fits three times, with a little tail left over. A mug gives the same, a wheel too, and so would the equator if you could wrap it. Exactly how many diameters fit into a circle? That question was asked by the scribes of Babylon and Egypt, by Archimedes, by Chinese astronomers, Indian poet-mathematicians, a Dutch fencing master and supercomputer programmers. The answer turned out to be a number that can’t be written as a terminating or a repeating decimal. We’ll follow the trail of this chase, and pick up the main theorems about the circle along the way.

Three and a bit

First let’s agree on the words we’ll use throughout the chapter.

The most direct way to measure a circle is to roll it along a ruler: in one turn a wheel covers exactly the length of its rim. Take a wheel of diameter $1$.

Ancient craftsmen measured the same way and got “three and a bit”. The Bible, in the First Book of Kings, describes a bronze basin for Solomon’s temple: “ten cubits from the one brim to the other… and a line of thirty cubits did compass it round about”. Here the ratio is exactly $3$. Babylonian scribes usually worked with three as well, but a tablet from Susa has $3\frac18 = 3.125$. The Egyptian Rhind papyrus (around 1550 BC) says to find the area of a round field of diameter $9$ as the area of a square of side $8$, and this recipe gives $\frac{256}{81} \approx 3.16$. The “bit” was measured by eye, sometimes more, sometimes less.

Before chasing the bit, we need to make sure it’s the same for every circle. It isn’t obvious: a small coin and the equator might have different ratios.

The idea: a circle is the limit of the polygons inscribed in it, and for polygons the ratio of perimeter to diameter can be seen from similarity.

Take two circles with a common centre $O$ and radii $\p1{r}$ and $\p2{R}$. Any two circles can be placed like this: moving them changes no lengths and no ratios.

Inscribe regular hexagons in both so that their vertices lie on the same rays from $O$. The rays cut each hexagon into six isosceles triangles with apex $O$.

Take one triangle from each. They share the angle at $O$, and their lateral sides are proportional: $\frac{\p1{r}}{\p2{R}}$ and $\frac{\p1{r}}{\p2{R}}$. By the similarity test with two sides and the included angle (chapter 17) the triangles are similar, and their bases, the sides of the hexagons, are in the ratio $\p1{r} : \p2{R}$. So the perimeters $P_6$ and $P'_6$ are in the same ratio, that is, $\frac{P_6}{2\p1{r}} = \frac{P'_6}{2\p2{R}}$. The same argument works for any number of sides $n$.

Keep doubling the number of sides: $12$, $24$, $48$… The perimeters of the inscribed polygons come arbitrarily close to the circumference; that’s how the length of a curve is defined, as the limit of the lengths of inscribed broken lines. For each $n$ the ratios of perimeter to diameter for the two circles are equal, so their limits are equal too: $\frac{C}{2\p1{r}} = \frac{C'}{2\p2{R}}$. This common number is called $\pi$. The word “limit” is used loosely here; chapter 25 will give it a precise meaning.

The circumference. The same number for every circle, $\pi \approx 3.14159$. The radius; $2r$ is the diameter, so equally $C = \pi d$. Example: in 1791 the metre was defined so that the Earth’s meridian through Paris, a circle through both poles, would be $40{,}000$ km long. So the Earth’s radius is $r = \frac{40{,}000}{2\pi} \approx 6366$ km; modern measurements give a mean radius of about $6371$ km.

A rope is stretched tight round the equator, then lengthened by $1$ metre and raised so that it’s the same height above the ground everywhere. How big is the gap?

Less than a millimetre About $16$ centimetres About a metre It depends on the size of the Earth

The rope’s length is $C = 2\pi r$. If $C$ grows by $1$, the radius grows by $\frac{1}{2\pi} \approx 0.159$ m, whatever $r$ is. You’d get the same for a ball or for the Sun. A cat could crawl through that gap.

Miletus, sixth century BC: the angle in a semicircle

Thales of Miletus, whom the Greeks considered the first philosopher, was by tradition the first not only to know geometric facts but to prove them. Diogenes Laertius, citing the writer Pamphila, reports that Thales was the first to inscribe a right triangle in a circle, and sacrificed an ox in his joy. Diogenes adds straight away that others attribute this to Pythagoras; it seems the ancient historians had one ox for all the theorems. In chapter 17 we saw that the circumcentre of a right triangle is the midpoint of the hypotenuse. Thales’ theorem states the converse.

The idea: cut the triangle with a radius into two isosceles triangles and add up the angles.

Let $AB$ be a diameter, $O$ the centre, and $C$ any other point of the circle. What is the angle $ACB$?

Draw the radius $OC$. The segments $OA$, $OB$ and $OC$ are radii, so all three are equal. So triangles $AOC$ and $BOC$ are isosceles.

In an isosceles triangle the base angles are equal (chapter 17). In triangle $AOC$ the angles at $A$ and at $C$ are equal; call them $\p1{\alpha}$. In triangle $BOC$ the angles at $B$ and at $C$ are equal; that’s $\p2{\beta}$. The angle $ACB$ is made up of the two angles at $C$: $\angle ACB = \p1{\alpha} + \p2{\beta}$.

The angles of triangle $ABC$ add up to $180^\circ$ (same chapter): at $A$ there’s $\p1{\alpha}$, at $B$ $\p2{\beta}$, at $C$ $\p1{\alpha} + \p2{\beta}$. We get $2\p1{\alpha} + 2\p2{\beta} = 180^\circ$.

So $\p1{\alpha} + \p2{\beta} = 90^\circ$, that is, $\angle ACB = 90^\circ$, wherever the point $C$ stands on the circle.

The converse is true too: if the angle $ACB$ is right, the point $C$ lies on the circle with diameter $AB$, as the chapter on triangles showed. So all the vertices of right angles standing on a segment $AB$ form a circle. A carpenter uses this to check a semicircular groove with a set square: if the groove is a true semicircle, the square touches the edge with its corner however it’s turned.

Alexandria, around 300 BC: Book III of the Elements

Euclid devoted the whole of Book III of the Elements to the circle. Two of its theorems, on the inscribed angle and on the tangent, are needed in almost every problem about circles.

The inscribed angle

The idea: draw a diameter from the vertex of the inscribed angle and reduce everything to isosceles triangles, as with Thales.

The inscribed angle $ACB$ and the central angle $AOB$ stand on the same arc $\p4{AB}$, the one without the point $C$.

Draw from $C$ the diameter $CD$ through the centre $O$.

Triangle $AOC$ is isosceles: $OA = OC$ as radii. So $\angle OAC = \angle OCA = \p1{\alpha}$. The angle $AOD$ is the exterior angle of this triangle at vertex $O$, and an exterior angle equals the sum of the two interior angles not adjacent to it (chapter 17): $\angle AOD = \p1{\alpha} + \p1{\alpha} = 2\p1{\alpha}$.

In the same way triangle $BOC$ is isosceles, $\angle OCB = \p2{\beta}$, and $\angle BOD = 2\p2{\beta}$.

If the centre $O$ lies inside angle $ACB$, then $\angle ACB = \p1{\alpha} + \p2{\beta}$, and $\angle AOB = 2\p1{\alpha} + 2\p2{\beta}$, twice as much. If the centre lies outside the angle, both angles are differences: $\angle ACB = |\p1{\alpha} - \p2{\beta}|$ and $\angle AOB = |2\p1{\alpha} - 2\p2{\beta}|$, and the ratio is the same. If the centre lies on an arm of the angle, one of $\alpha$, $\beta$ is zero. In every case $\angle AOB = 2\angle ACB$.

Three handy facts follow at once from the theorem.

The second fact is Thales’ theorem again: the arc on which an angle stands, with its vertex on the circle and arms through the ends of a diameter, is a semicircle, $180^\circ$, and the angle is half of it, $90^\circ$.

The quadrilateral $ABCD$ is inscribed in a circle, and angle $A$ is $75^\circ$. Find angle $C$ in degrees.

Angles $A$ and $C$ are opposite, and they stand on arcs that together make the whole circle. So $\angle A + \angle C = \frac{360^\circ}{2} = 180^\circ$, and $\angle C = 180^\circ - 75^\circ = 105^\circ$.

The tangent

The idea: in a right triangle the hypotenuse is longer than a leg, and the foot of a perpendicular is the nearest point.

A line $\p1{\ell}$ is drawn through the point $T$ of the circle, perpendicular to the radius $OT$.

Take any other point $X$ on $\p1{\ell}$. Triangle $OTX$ is right-angled, $OX$ is its hypotenuse and $OT$ a leg. The hypotenuse is longer than a leg (by Pythagoras’ theorem $OX^2 = OT^2 + TX^2 > OT^2$), so $OX > r$ and $X$ lies outside the circle. So $\p1{\ell}$ has one point in common with the circle, $T$, and $\p1{\ell}$ is a tangent.

Now the other way. Let a line $\p2{m}$ through $T$ be not perpendicular to $OT$; the second point changes its slope. Drop the perpendicular $OH$ to it from the centre. The point $H$ doesn’t coincide with $T$, or else $\p2{m}$ would be perpendicular to $OT$.

Lay off on $\p2{m}$ beyond the point $H$ the segment $HT' = HT$. The right triangles $OHT$ and $OHT'$ are congruent by two legs: $OH$ is shared, and $HT = HT'$. So $OT' = OT = r$, and the point $T'$ lies on the circle too.

The line $\p2{m}$ has two points in common with the circle, $T$ and $T'$: it’s a secant, not a tangent. So the only possible tangent at the point $T$ is the line perpendicular to the radius $OT$, and it really is a tangent.

It follows, for example, that the points of contact of the incircle divide the sides of a triangle into pieces that are equal in pairs: from each vertex to the two neighbouring points of contact the distance is the same.

Crossing chords

Near the end of Book III Euclid proves a theorem that looks like a conjuring trick. Draw two chords through a point inside a disc, and the product of the pieces of one chord turns out equal to the product of the pieces of the other. A modern proof uses Pythagoras’ theorem and explains along the way what number is hiding behind this.

The idea: drop a perpendicular from the centre to the chord and apply Pythagoras’ theorem twice.

Through the point $P$, inside the disc or outside it, a line is drawn crossing the circle at points $A$ and $B$.

Let $M$ be the midpoint of the chord $AB$. The centre $O$ is equidistant from $A$ and $B$, so it lies on the perpendicular bisector of $AB$. So $OM \perp AB$.

The points $A$ and $B$ lie on the line on opposite sides of $M$, at distance $MA$. If $P$ is inside the disc, then $\p1{PA} = MA - PM$ and $\p2{PB} = MA + PM$; if outside, $\p1{PA} = PM - MA$ and $\p2{PB} = PM + MA$. In both cases $\p1{PA} \cdot \p2{PB} = |PM^2 - MA^2|$: a difference of squares.

Triangles $OMP$ and $OMA$ are right-angled, with the right angle at $M$. By Pythagoras’ theorem $PM^2 = OP^2 - OM^2$ and $MA^2 = r^2 - OM^2$. Subtract: $PM^2 - MA^2 = OP^2 - r^2$; the segment $OM$ has cancelled out.

We’ve got $\p1{PA} \cdot \p2{PB} = |OP^2 - r^2|$, and nothing on the right depends on the line. So for any other line through $P$, a chord or secant $CD$, the product is the same: $PA \cdot PB = PC \cdot PD$. When a secant turns into a tangent, the points $A$ and $B$ merge into the point of contact $T$, and we’re left with $PT^2 = OP^2 - r^2$, which is also clear from the right triangle $OTP$.

The classical proof does without Pythagoras: for chords $AB$ and $CD$ crossing at $P$, triangles $APC$ and $DPB$ are similar, because the angles at $C$ and at $B$ stand on the same arc $AD$ and so are equal, and the angles at $P$ are vertical. Similarity gives $\frac{PA}{PD} = \frac{PC}{PB}$, the same equation.

The chords $AB$ and $CD$ cross at the point $P$, with $AP = 3$, $PB = 8$, $CP = 4$. Find $PD$.

By the power of a point theorem $AP \cdot PB = CP \cdot PD$, that is, $3 \cdot 8 = 4 \cdot PD$, and $PD = 6$.

Syracuse, third century BC: π in a vice

The first person who didn’t guess $\pi$ but proved bounds for it was Archimedes. In a short book, Measurement of a Circle, he squeezed the circle between two polygons, one inscribed and one circumscribed, and kept doubling their number of sides: $6$, $12$, $24$, $48$, $96$. To compare a curve with broken lines, Archimedes needed an assumption, which he honestly set out at the start of his book On the Sphere and Cylinder.

The main thing follows from the assumption: the perimeter of the inscribed polygon is less than the circumference (each side is shorter than its arc), and the perimeter of the circumscribed one is greater (the two halves of its sides enclose the arc between the points of contact). It remains to compute the perimeters.

Archimedes knew neither sines nor decimals. He did each doubling with similar triangles and square roots, extracted by hand, with bounds erring on the safe side: for example, $\frac{265}{153} < \sqrt3 < \frac{1351}{780}$. For inscribed polygons the doubling step can be derived with Pythagoras’ theorem alone.

The idea: the new vertex is the midpoint of the arc, and we can reach it through two right triangles.

$AB = \p1{s}$ is a side of a regular $n$-gon inscribed in a circle of radius $1$.

The new vertex of the $2n$-gon is the midpoint $M$ of the arc $AB$. The radius $OM$ bisects the angle $AOB$, and in the isosceles triangle $AOB$ the bisector of the apex angle is also the altitude and the median. So $OM$ is perpendicular to $AB$ and passes through the midpoint $K$ of the chord: $AK = KB = \frac{\p1{s}}{2}$.

In the right triangle $OKB$ the hypotenuse is $OB = 1$ and the leg $KB = \frac{s}{2}$. By Pythagoras’ theorem $\p3{OK} = \sqrt{1 - \frac{s^2}{4}}$.

The rest of the radius beyond the chord: $\p5{KM} = 1 - \p3{OK} = 1 - \sqrt{1 - \frac{s^2}{4}}$.

In the right triangle $BKM$: $\p2{s'}^2 = KB^2 + KM^2 = \frac{s^2}{4} + 1 - 2\sqrt{1 - \frac{s^2}{4}} + 1 - \frac{s^2}{4} = 2 - 2\sqrt{1 - \frac{s^2}{4}} = 2 - \sqrt{4 - s^2}$. Take the square root and we’re done.

The circumscribed polygon is the inscribed one scaled up by a factor of $\frac{1}{OK}$: its sides touch the circle, so the distance from the centre to a side is exactly $1$, while for the inscribed one it’s $OK$. So one formula gives both bounds. Archimedes took the doubling to $96$ sides and wrote the result in a form that has been quoted for twenty-two centuries.

The upper bound, $\frac{22}{7}$, has lived longest of all: people still use it for mental estimates. Legend has it that Archimedes died in 212 BC when the Romans took Syracuse: a soldier found him over drawings in the sand, and Archimedes supposedly cried out, “Do not disturb my circles!” It’s only a legend: Plutarch gives several different versions of his death.

Same place: the area of a disc

In the same book Archimedes proved that a disc has the same area as a right triangle with one leg equal to the radius and the other equal to the circumference. In our language, that’s the formula for the area of a disc. You can see it by slicing the disc like a pizza.

The idea: cut the disc into thin sectors and rearrange them into a shape almost indistinguishable from a rectangle.

Cut a disc of radius $r$ into $12$ equal sectors, wedges with their tip at the centre. Colour the upper half one colour and the lower half another.

Unfold each half into a “comb”: the wedges of the upper half stand tip down, those of the lower half tip up. The area doesn’t change: we’re only moving pieces.

Slide the combs into each other, teeth between teeth. We get a shape like a parallelogram. Its top edge is the arcs of the upper half, of total length half the circumference, $\p2{\pi r}$. The bottom edge is also $\pi r$. The sides and the height are roughly the radius $r$.

Slice more finely, into $36$ sectors. The edges get smoother and smoother, the sides more and more vertical, and the shape closer and closer to a $\pi r \times r$ rectangle. And its area is always exactly the area of the disc.

The area of the rectangle the shape approaches is $\pi r \cdot r = \pi r^2$. So $S = \pi r^2 = \frac12 \cdot \p2{2\pi r} \cdot r$. That last equation is Archimedes’ triangle with legs $2\pi r$ and $r$.

The word “approaches” here again hides a passage to a limit. Archimedes got round it more cleverly: he showed that the area of the disc can be neither larger nor smaller than the area of the triangle, since in either case a suitable inscribed or circumscribed polygon leads to a contradiction. Why area doesn’t change when pieces are rearranged, and what area is in the first place, is the subject of the next chapter, and the rigorous language for such arguments is provided by the integral.

The area of the disc. The same number as in the circumference. That’s no coincidence: the area is half the circumference times the radius. The radius, and squared: double the disc and its area grows fourfold. Example: a pizza of diameter $30$ cm has area $\pi \cdot 15^2 \approx 707$ cm², and one of diameter $40$ cm has $\pi \cdot 20^2 \approx 1257$ cm². The difference in diameter is a third, and in area almost $78\%$.

How many times larger in area is a pizza of diameter $40$ cm than one of diameter $30$ cm? Give the answer as a fraction.

The areas are in the ratio of the squares of the radii (or the diameters, it’s the same): $\left(\frac{40}{30}\right)^2 = \frac{16}{9} \approx 1.78$. The $\pi$ cancelled out.

China, third to fifth centuries: 355/113

In 263 Liu Hui, the commentator on the Nine Chapters, independently of Archimedes followed a similar path, only computing areas rather than perimeters, and got as far as a polygon with $3072$ sides. His result was $3.1416$. Two hundred years later Zu Chongzhi, an astronomer and inventor, went further still: $3.1415926 < \pi < 3.1415927$. Exactly how he computed it is unknown: his book hasn’t survived. Zu Chongzhi also left two fractions: the “rough ratio” $\frac{22}{7}$ and the “close ratio” $\frac{355}{113}$.

The second fraction is astonishing. $\frac{355}{113} = 3.14159292\ldots$: six correct decimal places with a three-digit denominator. No fraction with a denominator below $16{,}604$ is closer to $\pi$. The secret lies in the continued fraction of $\pi$ from chapter 6: $\pi = [3; 7, 15, 1, 292, 1, 1, \dots]$. Stopping just before the big number $292$ gives an especially good approximation, and it’s exactly $\frac{355}{113}$. For about nine hundred years nobody knew $\pi$ more precisely than Zu Chongzhi.

Which is larger: $\frac{22}{7}$ or $\pi$?

$\frac{22}{7}$ $\pi$ They’re equal

$\frac{22}{7} = 3.142857\ldots$, and $\pi = 3.141592\ldots$ The difference is about $0.00126$, but it’s there. “Pi equals twenty-two sevenths” is the most tenacious misconception about this number.

Kerala, around 1400: π from an infinite sum

The next leap came from another direction. In southern India, in Kerala, the astronomer Madhava of Sangamagrama found that $\pi$ can be obtained without any polygons, from an infinite sum:

$$\frac{\pi}{4} = 1 - \frac13 + \frac15 - \frac17 + \frac19 - \dots$$

His works haven’t come down to us; we know of them from the writings of his followers, Nilakantha’s Tantrasangraha (1501) and Jyeshthadeva’s Yuktibhasha. In Europe Gottfried Leibniz found the same series independently in 1674, so it’s often called the Leibniz series. The series has great beauty and little use: after a thousand terms you get $3.1406$, only three correct digits. Madhava knew this and devised corrections that sped up the convergence; according to his followers, he found $\pi$ to eleven correct decimal places. Why the sum converges to exactly $\pi$ is shown in the chapter on series.

The polygons didn’t give up either. In 1424 Jamshid al-Kashi, working in Samarkand at the court of Ulugh Beg, took a polygon with $3 \cdot 2^{28}$ sides and computed $2\pi$ to sixteen places. In 1596 Ludolph van Ceulen, a fencing and mathematics teacher from Leiden, got twenty places, and by the end of his life thirty-five, going up to polygons with $2^{62}$ sides. The digits were carved on his gravestone, and in Germany $\pi$ was long called the “Ludolphine number”. The stone was lost, and a replica was put back in Leiden’s St Peter’s Church in 2000.

The era of polygons was ended by formulas like the one John Machin found in 1706: $\frac{\pi}{4} = 4\arctan\frac15 - \arctan\frac{1}{239}$. Its series converge so fast that Machin got a hundred places by hand. Formulas like this were used to compute $\pi$ until the middle of the twentieth century.

An angle measured by length

Where did the $\frac{\pi}{4}$ in a sum of fractions come from? The Madhava–Leibniz series is really about an angle: it equals the $45^\circ$ angle, whose tangent is one, measured not in degrees but by the length of an arc. In analysis angles are always measured this way. The degree is a convention of Babylonian origin: a full turn divided into $360$ parts. The natural measure of an angle is what fraction of the radius the arc it cuts out amounts to.

The idea of measuring an angle by an arc was stated clearly in 1714 by the English mathematician Roger Cotes, and the word “radian” appeared only in 1873, in exam questions set by James Thomson, the brother of Lord Kelvin. In radians the most important formulas about arcs become short.

The length of the arc. The radius of the circle. The central angle in radians. If it’s in degrees, multiply by $\frac{\pi}{180}$. The area of the sector, the part of the disc between two radii. With $\alpha = 2\pi$ we get $2\pi r$ and $\pi r^2$, as we should. Example: with $r = 6$ and an angle of $60^\circ = \frac{\pi}{3}$ the arc is $6 \cdot \frac{\pi}{3} = 2\pi \approx 6.28$, and the sector $\frac12 \cdot 36 \cdot \frac{\pi}{3} = 6\pi \approx 18.85$.

Both formulas are a proportion: the arc and the sector are the same fraction of the circle and the disc as the angle $\alpha$ is of a full turn $2\pi$. Why analysis needed radians becomes clear in the chapter on trigonometry and the chapter on the derivative: only in radians is the derivative of sine the cosine, with no extra factor.

How many radians are in an angle of $60^\circ$? You can write the answer using pi.

$180^\circ = \pi$ radians, so $60^\circ = \frac{60}{180}\pi = \frac{\pi}{3} \approx 1.047$.

Berlin, 1875: it rolls, but it isn’t a circle

Why are wheels round? The usual answer: a circle has the same width in every direction, so a board on rollers doesn’t rock. That’s true, but the circle has rivals. The German engineer Franz Reuleaux, a professor in Berlin and one of the founders of the theory of machines, analysed in a book of 1875 a shape that now bears his name.

The Reuleaux triangle will do as a roller but not as a wheel: a wheel’s axle must stay at one height, and its centre bobs up and down. But it found other work. It turns inside a square while touching all four sides, so a drill bit of this cross-section, if its centre is made to move along a special curve, cuts an almost square hole with slightly rounded corners; such bits were invented in the early twentieth century by the engineer Harry Watts. The British $20$p and $50$p coins are heptagons of constant width, built from arcs on the same principle: a vending machine that measures the coin’s width gets the same number however the coin is turned.

And here $\pi$ appears again. The boundary of the Reuleaux triangle is three arcs of $60^\circ$ each, that is, $\frac{\pi}{3}$ radians, of radius $w$. By the arc length formula the perimeter is $3 \cdot w \cdot \frac{\pi}{3} = \pi w$: exactly the same as for a circle of the same width.

We’ve just checked this for the Reuleaux triangle. In general the theorem was proved by the French mathematician Joseph Barbier in 1860, but an honest proof goes beyond this chapter: it needs Cauchy’s formula, by which the perimeter of a convex figure equals its mean width times $\pi$, and that’s proved with an integral. The idea is this: the width of a shape of constant width is $w$ in every direction, so its mean width is $w$ too. That’s why the widget above shows that in a full turn the triangle rolls a distance of $\pi w$, exactly like a circle.

The twentieth century: thirty-one pieces

To finish, a trap that catches almost everyone. Mark a few points on a circle and join every point to every other one with chords. The chords cut the disc into pieces. How many pieces do you get?

Start small. One point: no chords, the disc is whole, one piece. Two points: one chord, two pieces. Three points make a triangle and four pieces: the triangle itself and three crusts around it. Four points cut the disc into eight pieces, five into sixteen. There is one condition: the points are placed so that no three chords pass through the same point inside the circle. Why it matters will become clear shortly.

One, two, four, eight, sixteen: twice as many each time. Before you read on, place your bet.

How many pieces are there with six points on the circle, if no three chords meet at one point?

Thirty-one. If you wrote 32, don’t worry: that is exactly why this puzzle gets shown. Check it on the circle below by adding a sixth point.

Thirty-one

The circle below works like pencil and paper, except that it never loses count. It finds every crossing of the chords and every piece, even the tiny ones near the rim, and colours them so that neighbours don’t blend together. You can drag the points around the circle, add them and remove them.

With six points there are 31 pieces. Move the points however you like: as long as no three chords meet at one point, there are exactly 31, and a thirty-second never appears. After that the gap only widens. Seven points give 57 pieces instead of 64, eight give 99 instead of 128.

At the tenth point the sequence brushes against a power of two once more: $256 = 2^8$. That’s a coincidence; next come 386, 562, 794. The puzzle is known as Moser’s circle problem, after the Canadian mathematician Leo Moser.

A trap inside the trap

The condition about three chords isn’t pedantry. Put six points at the corners of a regular hexagon: the three long diagonals meet at the centre, and the count drops to 30. Where three chords would have crossed at three different points and fenced off a small triangle, there is now a single point and no triangle. Crossings that coincide always eat pieces (we’ll prove this in the next section). So 31 is the largest possible number of pieces for six points, and that is what the puzzle asks about.

Eight points sit at the corners of a regular octagon, and all the chords are drawn. How many pieces are there?

$128$, as the doubling promises Exactly $99$, as in the table Fewer than $99$ More than $99$

All four long diagonals of a regular octagon meet at its centre, and at eight more points three chords meet. Every such coincidence eats pieces, and 88 are left. The 99 in the table is the largest possible count: you get it only when no three chords pass through one point.

It’s the same trap as Euler’s formula from the introduction, only shorter: five agreements in a row, and a pattern that looks watertight breaks at the sixth step. Only an argument can confirm it “for every $n$”.

Where 31 comes from

To see why there are 31 pieces and not 32, we don’t need to count them one by one. We need to understand where pieces come from. Let’s draw the chords one at a time and watch how many pieces each one adds.

The pattern jumps out at once: a chord that crosses $k$ chords already drawn adds $k + 1$ pieces. If that always happens, the number of pieces is

$$\text{pieces} = 1 + \text{chords} + \text{crossings}:$$

the original disc, plus one for each chord, plus one for each crossing. But after the business with 32 we won’t take a pattern on trust. First we’ll count the terms, and then we’ll prove the formula always holds.

There are as many chords as there are pairs of points. Each of the $n$ points can be joined to $n - 1$ others, but that counts every chord twice, once from each end, so there are $\frac{n(n-1)}{2}$ chords. With six points that’s $\frac{6 \cdot 5}{2} = 15$. The number of ways to choose $k$ things out of $n$ is written $\binom{n}{k}$ and read “$n$ choose $k$”; some books write $C(n, k)$ or ${}^nC_k$. So there are $\binom n2$ chords. We’ll look at these numbers closely in the chapter on combinatorics.

The crossings are more interesting. It turns out that every crossing corresponds to exactly one set of four points on the circle: the ends of the two chords that meet there. So there are as many crossings as sets of four: $\binom n4 = \frac{n(n-1)(n-2)(n-3)}{24}$. We divide by 24 because the same four points can be picked in order in $4 \cdot 3 \cdot 2 \cdot 1 = 24$ ways. Why fours and crossings match up one to one, the proof will explain.

The original disc: one piece while there are no chords. One piece for each chord. There are as many chords as pairs of points: $\binom n2 = \frac{n(n-1)}{2}$. One piece for each crossing. There are as many crossings as sets of four points on the circle: $\binom n4 = \frac{n(n-1)(n-2)(n-3)}{24}$. For $n = 5$: $1 + 10 + 5 = 16$. For $n = 6$: $1 + 15 + 15 = 31$. For $n = 7$: $1 + 21 + 35 = 57$. The formula holds when no three chords pass through one point; otherwise there are fewer pieces.

Let’s state the result as a theorem.

The trick is not to count the pieces of the finished picture, but to build it one chord at a time and watch how many pieces each chord adds.

Before any chords are drawn, the disc is a single piece. That is the term $\p1{1}$ in the formula. Draw the chords one at a time, in any order. Suppose the next chord crosses $k$ of those already drawn. The crossings on it are all different, because no three chords pass through one point. These $k$ points split the new chord into $k + 1$ segments. Each segment runs across one of the existing pieces from edge to edge and cuts it into exactly two. Why not three? The line through a chord meets the disc in exactly that chord, so every piece consists of the points of the disc lying on one particular side of each line drawn so far. Such a piece is convex: with any two of its points it contains the segment joining them. And a segment running across a convex shape from edge to edge splits it into exactly two parts, one on each side of the line. So the new chord adds $1 + k$ pieces: one for itself and one more for each crossing it creates. Draw all the chords this way. The “one for itself” terms add up to the total number of chords. A chord is a pair of points, and $n$ points have $\binom n2$ pairs. That is the term $\p2{\binom n2}$. All the $k$ terms together add up to the number of crossings. Each crossing was counted exactly once: when we drew the second of the two chords that meet there. When we drew the first, the crossing didn’t exist yet. It remains to count the crossings. Take four points in the order $P$, $Q$, $R$, $S$ around the circle. The chord $PR$ splits the disc in two, with $Q$ and $S$ on different sides, so the segment $QS$ must cross $PR$. But in the pairs $PQ$ and $RS$, and $PS$ and $QR$, both ends of one chord lie on the same side of the other, and the chords don’t meet. So of the three ways to split four points into two pairs, only the diagonals of the quadrilateral cross: every set of four gives exactly one crossing. Conversely, every crossing gives one set of four, the ends of its two chords. There are as many crossings as sets of four: $\p3{\binom n4}$. Add up the original piece, the pieces for the chords and the pieces for the crossings: $R(n) = \p1{1} + \p2{\binom n2} + \p3{\binom n4}$. With six points, $1 + 15 + 15 = 31$.

The condition about three chords isn’t in the theorem for decoration. Here is what happens when it’s broken.

Let’s rerun the proof of the theorem and see which term shrinks.

Three chords in general position cross pairwise at three different points and fence off a small triangle. Counting as in the proof: one piece, plus one for each chord and one for each crossing, $1 + 3 + 3 = 7$ in all. Slide the end $F$ of the third chord until the chord passes through the crossing of the other two. The triangle shrinks and finally disappears: all three chords now pass through one point. There are still three pairs of crossing chords, but only one crossing point. The third chord passes through a point that already exists and earns one unit for crossings, not two. That gives $1 + 3 + 2 = 6$ pieces, one fewer. The general case works the same way. Every step of the theorem’s proof still holds, except that a crossing on the new chord is counted once, however many chords pass through it. In the theorem every pair of crossing chords brought its own unit, and there are always $\binom n4$ such pairs, one per set of four points. If $m \ge 3$ chords meet at a point, their $\binom m2$ pairs bring only $m - 1$ units between them: the first of those chords gets nothing for that point, each later one gets one. Since $m - 1 < \binom m2$ for $m \ge 3$, there are fewer pieces. At the centre of the regular hexagon $m = 3$, and $\binom 32 - 2 = 1$ piece is lost: $31 - 1 = 30$.

One puzzle remains: why did the first five answers match the powers of two so neatly? It isn’t chance, and the explanation fits on one line.

Why the first five answers are powers of two

A power of two is the sum of a whole row of Pascal’s triangle: $2^{n-1} = \binom{n-1}{0} + \binom{n-1}{1} + \dots + \binom{n-1}{n-1}$, since each of $n - 1$ things can be taken or left, and $\binom{n-1}{k}$ counts the choices of exactly $k$ of them (details in the chapter on combinatorics). Moser’s answer is the sum of only the first five of these numbers. You can see this from Pascal’s rule $\binom{n-1}{k-1} + \binom{n-1}{k} = \binom nk$:

$$\binom{n-1}{0} + \underbrace{\binom{n-1}{1} + \binom{n-1}{2}}_{\binom n2} + \underbrace{\binom{n-1}{3} + \binom{n-1}{4}}_{\binom n4} = 1 + \binom n2 + \binom n4.$$

While $n \le 5$ the row has no other numbers, and the match is perfect. At $n = 6$ the last term $\binom 55 = 1$ drops out, hence $31 = 32 - 1$.

The same sum answers a question that sounds completely different: into how many pieces do $n - 1$ three-dimensional “planes” in general position cut four-dimensional space? In ordinary space the trap springs sooner: four flat cuts divide a watermelon into at most $1 + 4 + 6 + 4 = 15$ pieces, not 16. The proof uses the same trick of watching how many pieces each new cut adds.

There are eight points on a circle, and no three chords pass through one point. How many crossings do the chords have inside the circle?

As many as there are sets of four points: $\binom 84 = \frac{8 \cdot 7 \cdot 6 \cdot 5}{24} = 70$. As a check: $1 + 28 + 70 = 99$ pieces, as in the table.

What is the smallest number of points for which there are more than a thousand pieces?

Use the formula: $R(12) = 1 + 66 + 495 = 562$, $R(13) = 1 + 78 + 715 = 794$, $R(14) = 1 + 91 + 1001 = 1093$. Answer: 14 points. Doubling would have passed a thousand already at the eleventh point: $2^{10} = 1024$.

You can practise Moser’s formula in the trainer: it builds circles with different numbers of points and asks about chords, crossing points and pieces.

The twentieth and twenty-first centuries: trillions of digits

By the eighteenth century it was clear that the chase would never end. In the 1760s Johann Lambert proved that $\pi$ is irrational: its decimal expansion neither stops nor becomes periodic. In 1882 Ferdinand von Lindemann proved more: $\pi$ is transcendental, that is, it can’t be a root of any polynomial with integer coefficients. It follows that squaring the circle, constructing with compass and straightedge a square equal in area to a disc, is impossible: we discussed this in chapter 16, and the details are in chapter 43.

People didn’t stop computing because of that. In 1873 the Englishman William Shanks, after spending years on the calculations, published $707$ digits; in the mid-1940s it was discovered that only the first $527$ were right. In 1949 the ENIAC electronic computer calculated $2037$ digits in seventy hours, and nobody ever computed $\pi$ by hand again. Today records are measured in hundreds of trillions of digits, and they’re set to test computers and algorithms: one error in memory or in the program, and the digits part company with the checking run.

Is that much needed in practice? Engineers at NASA’s Jet Propulsion Laboratory have said that fifteen decimal places are enough for interplanetary navigation. About forty places would be enough to compute the circumference of a circle the size of the observable universe with an error smaller than a hydrogen atom. Everything else is sport, hardware testing and memory training. For memory there are mnemonics; the classic English one is attributed to the physicist James Jeans: “How I want a drink, alcoholic of course, after the heavy lectures involving quantum mechanics!” The letter counts of the words give $3.14159265358979$.

Finally, the trainer: inscribed and central angles, tangents, chords and secants, arc lengths and the areas of sectors.

Where next

We found the area of a disc by a trick: we sliced it into sectors and put together an almost-rectangle. But this trick hides questions that Archimedes got round rather than answered. Why doesn’t area change when a figure is cut up and rearranged? Can any figure be re-cut into a square of the same area, and if a polygon can, why can’t a disc? And what about solids: how much space does a ball take up, and why was Archimedes so proud of its volume that he asked for a sphere and a cylinder to be carved on his tombstone? Scissors, rearranging and the first limits: that’s chapter 20.