How did Archimedes accurately calculate π without modern technology or calculators?

Archimedes used mainly Euclidean geometry which he mastered perfectly, and his ingenious creative mind. He approximated the number π, (taken as the perimeter of a circle of diameter 1) from both sides, from below and above it, by calculating the lengths of sequence of pairs of regular polygons, inscribed and circumscribing that circle, the number of sides of each one of those regular polygons, is twice the number of sides of the previous polygon, starting with 6 sides (that is: 6, 62=12,122=24, 242=48, 482=96,…).

The initial step in Archimedes process is the calculations of the perimeters of the two hexagons, seen in the attached drawing:

It had been known long before Archimedes that the perimeter of the inscribed hexagon is

(recall that the radius r of the circle is taken to be 1/2), and the perimeter of the circumscribing hexagon is

All the steps in the rest of that process are inductive: The output of the n-th step is the pair of numbers:

and on the next (n+1)-th step Archimedes proved, purely geometrically, that by doubling the number of sides of both regular polygons, the resulting perimeters could be calculated by the following recursion formulas:

Archimedes, of course, could not write down these two formulas, as we do today, just describe the matter in words.

Observe that by the formula

the new perimeter can be interpreted as the harmonic mean of the perimeters of the two regular polygons calculated in the previous step, and by the second formula

the second new perimeter can be interpreted as the geometric mean of two perimeters, one is the perimeter of the first polygon just calculated in the present step and the other is the perimeter of the second polygon calculated in the previous step, and therefore, by the well-known property of harmonic and geometric means we may deduce that

and by induction it similarly follows that for every natural number n the next inequalities hold:

that is, as n is growing the two perimeters are getting closer to π from both sides, and the question to be asked here is if these approximation are really tending to π and how fast? The answers would come soon.

Archimedes conducted the first 5 steps up to the regular polygons with 96 sides. With the power of modern computers and the Hindu-Muslim decimal notation, we are able today to follow his calculations and deduce

and

and therefore we can write

Archimedes himself, without our modern technology, obtained:

which is amazingly close.

Suppose, now, we know the perimeters of the two regular polygons with m sides, inscribed (p) and circumscribed (q) in the circle with diameter 1. Let us focus at one of the m identical parts of these regular polygons, shown in the attached drawing

where O is the center of the circle, AB is one side of the inscribed polygon, CD is one side of circumscribed polygon, so that

Now, let N be the midpoint of the side CD, which is also the midpoint of the arc connecting A,B, M be the midpoint of the side *AB (notice that the point M lies inside the line interval ON* (WHY?), and let E,F be the intersection points of the two tangents to the circle at the points A,B. Then you may verify that AN,BN are two of the 2m sides of an inscribed regular polygon, and EF is one of the 2m sides of a circumscribed regular polygon. Hence, if p’,q’ are respectively the perimeters of the regular polygons with 2m sides, then

and

Also, the two right triangles OAE, ONE, (and OBF, ONF, too) are congruent, because they share the hypotenuse OE (or OF) and have equal legs OA, ON (and OB). Therefore

Now look at the two right triangles ONC,EAC, which share the inner angle near C, and therefore they are similar. Hence, by Thales theorem we get

On the other hand, the next two right triangles OMA,ONC are clearly similar and therefore, once again by Thales theorem we deduce

Hence one of the two recurrence relations obtained as follows:

In order to prove the second recurrence relation we use the identity between the next two angles

as alternating angels between two parallel lines. It follows at once that the next isosceles triangles

ABN, ANE

are similar triangles, from which we deduce, by Thales theorem, the equality between the next proportions

that is

which completes the proof of the validity of Archimedes process.

Archimedes didn’t continue beyond the calculations of the perimeters of the regular polygons with 96 sides, mainly because the lower and upper estimates he discovered were very close to each other, namely

that is, he understood that the error in replacing the exact (unknown) value of π by its upper estimate 22/7 is less the 0.002, which at that time, and centuries later, was considered negligible for all technical purposes.

However, today, with quite simple basic algebraic means, we can prove that the two Archimedes’ sequences are converging to the same limit, which is necessarily the number π, and get an estimate of how fast the they converge to π, and moreover, for any arbitrarily small positive error ε we can find the number of steps n required to calculate an Archimedes’ estimate of π with an error less than the given ε.

Instead of showing that the two Archimedes sequences

are converging to the same limit, we’ll show that the next sequence of ratios:

is converging to 2.

Indeed, from Archimedes recurrence relations we obtain the next single recurrence equation:

where the initial ratio is

The last recurrence equation implies the following results:

(i) For every natural number n we have

First, for n=1 it is true. Assume that it holds for some natural number n and deduce

and by induction this result follows at once.

(ii) For every natural number n we have

It follows from the previous result:

(iii) For every natural number n we deduce

Indeed, from the recurrence relation we get

and since

it follows

It readily yields

Now, we finally get to the point and since the sequence of the q’s is decreasing we obtain

and therefore

For example, if you wish to find the natural number n for which both estimates are at distance less than some ε>0 from π. we are looking for n such that

that is

This post will be followed soon by an algebraic improvement of Archimedes process of estimating π.