Foundations

Approaching π

How Archimedes trapped the circle between two polygons and pinned down π to two decimal places, which doubling formula does it, and why the folded squares of Andreas Weiss rebuild exactly this method.

With the incommensurable lengths of Chapter 9 it is clear: not every ratio is a fraction. The most famous such ratio is that of circumference and diameter of a circle. Around 250 BC Archimedes of Syracuse pinned it down in his treatise Measurement of a Circle, with a method that remained the best until modern times.

The circle number

That the ratio is the same for all circles is because all circles are similar: a larger circle is an enlarged image of the smaller one, and circumference and diameter grow in the same proportion. Archimedes also proved that the area of a circle with radius $r$ equals that of a triangle with height $r$ and the circumference as base, that is $\pi r^2$. In what follows we calculate with the unit circle, the circle with radius 1, circumference $2\pi$ and area $\pi$.

The circle between two polygons

Archimedes took as a principle what one sees: a chord is shorter than its arc, and a polygon enclosing the circle has a larger circumference than the circle.

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A regular polygon inside the unit circle (blue) and one around it (red). Their half circumferences enclose π. Archimedes started with the hexagon and doubled four times up to the 96-gon.

Proof

The inscribed $n$-gon consists of $n$ isosceles triangles with legs 1 and the angle $\tfrac{2\pi}{n}$ at the apex. Each side is therefore $2 \sin\tfrac{\pi}{n}$ long, and the circumference is $2n \sin\tfrac{\pi}{n}$. For the circumscribed $n$-gon the triangles have height 1, each side is $2 \tan\tfrac{\pi}{n}$, the circumference $2n \tan\tfrac{\pi}{n}$. By the principle, the circumference of the circle $2\pi$ lies in between. Dividing by 2 gives the claim.

∎

For the hexagon this is $3 < \pi < 2\sqrt{3} \approx 3.46$. So the 3 before the decimal point comes from the hexagon: its circumference is exactly three times its diameter, as the page Simplexes, number theory and π remarks.

Doubling

Archimedes knew no trigonometric functions. He doubled the number of corners step by step with a calculation that in today’s notation looks like this:

Proof

With $x = \tfrac{\pi}{n}$, after taking reciprocals the first formula reads $\tfrac{1}{\tan(x/2)} = \tfrac{1}{\tan x} + \tfrac{1}{\sin x} = \tfrac{\cos x + 1}{\sin x}$, which is the well-known formula for the tangent of the half angle. For the second one calculates

$$\begin{aligned} a_{2n}\, b_n &= 2n \tan\tfrac{x}{2} \cdot n \cdot 2 \sin\tfrac{x}{2} \cos\tfrac{x}{2} \\ &= 4n^2 \sin^2\tfrac{x}{2} = b_{2n}^2. \end{aligned}$$

∎

Starting from the hexagon with $a_6 = 2\sqrt{3}$ and $b_6 = 3$, only the four basic operations and square roots are needed. After four doublings, at the 96-gon, Archimedes proved

$$3 + \frac{10}{71} \;<\; \pi \;<\; 3 + \frac{1}{7},$$

that is $3.1408 < \pi < 3.1429$. The upper value $\tfrac{22}{7}$ is still used as a rule of thumb today.

The folded squares

The page on the circle number reports an observation by Andreas Weiss: if on each side of a regular $n$-gon with side length 1 a unit square is folded inwards, the total area of the overlaps approaches $\pi$ as $n$ grows. This can be proven, and the proof shows why.

Vertices

From the 12-gon on, only neighbouring squares overlap, each time in a narrow kite at a corner.

Proof

From the 12-gon on only neighbouring squares overlap, as a geometric calculation shows (the checking script is in the project’s forschung folder). At a corner $B$ two sides meet at the exterior angle $\tfrac{2\pi}{n}$. There the two squares each have an edge of length 1 perpendicular to their side, and these edges also enclose the angle $\tfrac{2\pi}{n}$. The overlap is a kite made of two right triangles with leg 1 and angle $\tfrac{\pi}{n}$ at $B$, each with area $\tfrac{1}{2}\tan\tfrac{\pi}{n}$. So a kite has area $\tan\tfrac{\pi}{n}$, and $n$ corners give $n\tan\tfrac{\pi}{n}$.

The circumscribed $n$-gon of the unit circle consists of $n$ triangles with height 1 and base $2\tan\tfrac{\pi}{n}$, so it has the same area. It contains the circle, and the inscribed $2n$-gon with area $n \sin\tfrac{\pi}{n}$ lies inside the circle, so

$$n \sin\frac{\pi}{n} \;<\; \pi \;<\; n \tan\frac{\pi}{n}.$$

As $n$ grows both sides move together, since their ratio is $\cos\tfrac{\pi}{n} \to 1$.

∎

So the folded squares secretly rebuild the method of Archimedes. The number Andreas Weiss interpreted as an area is the area of the circumscribed polygon, and that it tends to $\pi$ is the theorem of Archimedes. There is hardly a finer example of an observation on this website coinciding with classical mathematics.

What a computer can check here

There is no Lean box for this chapter. Sine, tangent and the real numbers are not part of Lean’s core library but only of the large library Mathlib, which this section deliberately does not use. There one also finds formal proofs of bounds such as $3.14 < \pi < 3.15$, with a method related to Archimedes’ doubling.

That $\pi$ itself is not a fraction was proven by Johann Heinrich Lambert in 1761. That it cannot even be the solution of an equation with integer coefficients was proven by Ferdinand von Lindemann in 1882. With that it was also clear that squaring the circle with compass and straightedge is impossible, a question that had been open since antiquity.

Sources and further reading

Sources

• Archimedes, Measurement of a Circle, 3rd century BC
• Johann Heinrich Lambert, Mémoire sur quelques propriétés remarquables des quantités transcendantes circulaires et logarithmiques, Histoire de l’Académie royale des sciences et belles-lettres, Berlin 1761, printed 1768
• Ferdinand Lindemann, Über die Zahl π, Mathematische Annalen 20, 1882

Further reading

• Jörg Arndt and Christoph Haenel, Pi. Algorithmen, Computer, Arithmetik, Berlin 1998