On Archimedes' Measurement of a Circle, Proposition 3

On Archimedes’ Measurement of a circle, Proposition 3

Mark Reeder February 2

1 10 The ratio of the circumference of any circle to its diameter is less than 3 7 but greater than 3 71 . Having related the area of a circle to its perimeter in Prop. 1, Archimedes next approximates the circle perimeter with circumscribed and inscribed regular polygons and then ﬁnds good rational estimates for these polygon perimeters, thereby approximating the ratio of circumference to diameter. The main geometric step is to see how the polygon perimeter changes when the number of sides is doubled. We will consider the circumscribed case. Let AC be a side of a regular circumscribing polygon, and let AD be a side of a regular polygon with the number of sides doubled.

θ O A θ B

To make Archimedes’ computation easier to follow, let

x = AC, y = AD, r = OA, c = OC, d = OD.

We want to express the new ratio y/r in terms of the old ratio x/r. But these numbers will be very small after a few subdivisions, so they will be difﬁcult to estimate. Instead, we will express r/y in terms of r/x. These are big numbers, which can be estimated by integers.

1 From Euclid VI.3, an angle bisector divides the opposite side in the same ratio as the other two sides of a triangle. Hence CD : DA = OC : OA. In our notation, this means x − y c x c + r r r c = , or = , or = + . y r y r y x x From Euclid I.47, we have r c r2 = 1 + , x x2 so that r r r r2 = + 1 + (1) y x x2 Thus, the new ratio r/y is expressed in terms of the old ratio r/x, as desired. We could then express the next ratio (after another doubling of sides) r/z in terms of the (now old) ratio r/y, and we could repeat this process until we got tired or ran out of sand. This recycling method, expressed in equation (1), is called a recursion formula.

Next comes the arithmetic work. Archimedes starts with a circumscribed hexagon. Here AOC is a ◦ 30 − 60 − 90 triangle, with angle ∠AOC = 30 , and r AO √ = = 3. x AC

D E 7.5° 15° 7.5° B A O

2 ◦ Then we subdivide, getting angle ∠AOD = 15 , and a new side y = AD with r r r r2 √ q √ √ = + 1 + = 3 + 1 + ( 3)2 = 2 + 3. y x x2

◦ Then we subdivide again, getting angle ∠AOE = 7.5 , and a new side z = AE, with s r r r2 √ q √ q √ q √ = + 1 + = 2 + 3 + 1 + (2 + 3)2 = 2 + 3 2 + 2 + 3 . z y y2

The numbers on the right side become more and more complicated, as we continue to subdivide. However, Archimedes never√ writes down such numbers, for he begins approximating√ immediately: instead of writing r/x = 3 etc. as we have done above, he approximates 3 very closely from below, in a curious way: r √ 265 = 3 > . (2) x 153 √ 265 This as a remarkably good approximation, for 3 = 1.73205..., while 153 = 1.73203.... With this approximation, Archimedes gets r √ 265 571 = 2 + 3 > 2 + = . y 153 153 Applying the recursion (1) to this approximation for r/y, he goes to the next step s √ r 571 5712 571 + 349450 > + 1 + = . (3) z 153 153 153

The next approximation is √ 1 349450 > 591 , (4) 8 which, when applied to (3) gives √ r 571 + 349450 1162 1 > > 8 . (5) z 153 153 Subdividing again to get a new side u = AF (the point F is between A and E and is not shown in the picture above), we get the new estimate  s  ! r 1 1 12 1 1 r 33 2334 1 > 1162 + 1532 + 1162 = 1162 + 1373943 > 4 , (6) u 153  8 8  153 8 64 153

3 using another approximation r 33 1 1373943 > 1172 . (7) 64 8 Subdividing one last time (pant pant) we get  s  r 1 1 12 > 2334 + 1532 + 2334 v 153  4 4  ! 1 1 r 1 = 2334 + 5472132 153 4 16 (8) 1 1 1 > 2334 + 2339 153 4 4 1 1 = 4673 , 153 2 using one last approximation r 1 1 5472132 > 2339 . (9) 16 4 In (8), v is one-half the side of a 96-gon. So we have perim. 96-gon 96 · 2v 96 · 153 1335 = < 1 = 3 + . diam.AB 2r 4673 2 9347 Remarkably, 7 · 1335 = 9345, so we get the weaker, but simpler-looking estimate circum. circle 1335 1335 1 < 3 + < 3 + = 3 + . diameter 9347 9345 7

How did Archimedes ﬁnd these approximations? According to John Wallis, Archimedes was as it were of set purpose to have covered up the traces of his investigation as if he had grudged posterity the secret of his method of inquiry while he wished to extort from them assent to his results. No one knows for sure how Archimedes found the curious (and very accurate) approximations (2), (4), (7), (9), but he may have used a method the ancient Greeks called Anthyphairesis, which we now call continued fractions. This is based on Euclid X.1: Two unequal magnitudes being set out, if from the greater there is subtracted a magnitude greater than its half, and from that which is left a magnitude

4 greater than its half, and if this process is repeated continually, then there will be left some magnitude less than the lesser magnitude set out. To use this, take a number x that you wish to approximate. Let a1 be the greatest integer ≤ x, and write the remainder as 1/x1. Next let a1 be the greatest integer ≤ x1 and write the remainder as 1/x2, etc. We get

1 x = a1 + x1 1 1 x1 = a2 + so x = a1 + x2 1 a2 + x2 1 1 x2 = a3 + so x = a1 + , etc. x3 1 a + 2 1 a3 + x3

Omitting the last xn in these fractions gives a sequence of better and better approximations to x by rational numbers that are alternately too big and too small.

x ≈ a1 too small 1 x ≈ a1 + too big a2 1 x ≈ a + , too small etc. 1 1 a2 + a3 As the Greeks may have known intuitively, these approximations are optimal, in the sense that each is closer to x than any other rational number with no larger denominator. However, to compute continued fractions as not as easy as it may appear: to produce the integers an, it seems that you need to have good approximations in hand already, so as to compute the√ xn’s. But watch what happens when we try this with x = 3. We have √ √ 1 3 = 1 + ( 3 − 1) = 1 + √ 3+1 √ √ 2 3 + 1 3 − 1 1 = 1 + = 1 + √ 2 2 3 + 1 √ √ 3 + 1 = 2 + ( 3 − 1).

5 √ Since we have arrived at a previous remainder, namely 3 − 1, the remainders will now repeat them- selves, and hence the an’s will just repeat as 1, 2, 1, 2 ... . So we get the approximations √ 3 ≈ 1 too small √ 1 3 ≈ 1 + too big 1 √ 1 3 ≈ 1 + too small 1 1 + 2 ... √ 1 265 3 ≈ 1 + = too small. 1 153 1 + 1 2 + 1 1 + 1 2 + 1 1 + 1 2 + 1 1 + 2 √ 265 Archimedes could have arrived at his approximation 3 > 153 by means of this long, but straightfor- ward application of Anthyphairesis. But some scholars think that a genius like Archimedes would have noticed that   265 1  1  = 5 +  . 153 3  1  5 + 10 Can you ﬁnd the number that gives this new continued fraction, and explain Archimedes’ trick?

6 √ 1 Some puzzling details. For the next approximation 349450 > 591 8 , the continued fractions are √ 349450 ≈ 591 too small √ 1 349450 ≈ 591 + too big 6 √ 1 1 349450 ≈ 591 + = 591 too small, 1 7 6 + 1 1 1 yet Archimedes uses 591 8 , which is not as good as 591 7 . The same thing happens with the next one. The third continued fraction approximation is r 33 1 1 1373943 ≈ 1172 + = 1172 too small, 64 1 7 6 + 1 1 1 but Archimedes uses 1172 8 . This one can be explained, however. A very simple way to get 1172 8 is to ﬁrst pull out the denominator of 64 and then use the ﬁrst continued fraction: r 33 1√ 1 9376 1 1 1373943 = 64 · 1373943 + 33 ≈ 9377 = + = 1172 . 64 8 8 8 8 8 The same method also gives the last approximation (9).

Final comment. The ﬁrst few continued fractions of π itself are π ≈ 3 too small 1 22 π ≈ 3 + = too big 7 7 1 333 π ≈ 3 + = too small, 1 106 7 + 15 1 355 π ≈ 3 + = too big. 1 113 7 + 1 15 + 1 So in effect, the ﬁrst half of Archimedes’ Proposition 3 is a computation of the second continued fraction of π. It is speculated that he knew further continued fractions. The fourth continued fraction 355 113 was found by the Chinese mathematician Zu Chongzhi (429-501 AD).