Proposition 3

Archimedes

The ratio of the circumference of any circle to its diameter is less than 3 1/7 but greater than 3 10/71.

3\frac{10}{71} < \pi < 3\frac{1}{7}

PART I. Let AB be the diameter of any circle, O its centre, AC the tangent at A; and let the angle AOC be one-third of a right angle.

\begin{aligned} &\text{Then}\quad OA:AC\,[=\sqrt{3}:1] > 265:153 \quad (1)\\ &\text{and}\quad OC:CA\,[=2:1] = 306:153 \quad (2)\end{aligned}

First, draw OD bisecting the angle AOC and meeting AC in D.

\begin{aligned} &\text{Now}\quad CO:OA = CD:DA \qquad [\text{Eucl. VI. 3}]\\ &\text{so that}\quad CO+OA:CA = OA:AD\\ &\text{Therefore [by (1), (2)]}\quad OA:AD > 571:153 \quad (3)\\ &\text{Hence}\quad OD^2:AD^2 = (OA^2+AD^2):AD^2 > 349450:23409\\ &\text{so that}\quad OD:DA > 591\frac{1}{8}:153 \quad (4)\end{aligned}

T. L. Heath, The Works of Archimedes (Cambridge, 1897)

Secondly, let OE bisect the angle AOD, meeting AD in E.

\begin{aligned} &\text{so that}\quad DO+OA:DA = OA:AE\\ &\text{Therefore}\quad OA:AE > 1162\frac{1}{8}:153 \quad (5)\\ &\text{Hence}\quad OE^2:EA^2 > (1162\frac{1}{8})^2+153^2:153^2 > 1373943\frac{33}{64}:23409\\ &\text{Thus}\quad OE:EA > 1172\frac{1}{8}:153 \quad (6)\end{aligned}

Thirdly, let OF bisect the angle AOE and meet AE in F. We thus obtain the result that

\begin{aligned} &OA:AF\,[> (1162\frac{1}{8}+1172\frac{1}{8}):153] > 2334\frac{1}{4}:153 \quad (7)\\ &\text{Therefore}\quad OF:FA > 2339\frac{1}{4}:153 \quad (8)\end{aligned}

Fourthly, let OG bisect the angle AOF, meeting AF in G. We have then

OA:AG\,[> (2334\frac{1}{4}+2339\frac{1}{4}):153,\ \text{by (7), (8)}] > 4673\frac{1}{2}:153

Now the angle AOC, one-third of a right angle, has been bisected four times, so the angle AOG is 1/48 of a right angle. Make the angle AOH on the other side of OA equal to angle AOG, and let GA produced meet OH in H; then angle GOH is 1/24 of a right angle. Thus GH is one side of a regular polygon of 96 sides circumscribed to the given circle.

\begin{aligned} &\text{Since}\quad OA:AG > 4673\frac{1}{2}:153,\ \text{while}\ AB=2OA,\ GH=2AG,\\ &AB:(\text{perimeter of 96-gon})\,[> 4673\frac{1}{2}:153\times 96] > 4673\frac{1}{2}:14688\\ &\text{But}\quad \frac{14688}{4673\frac{1}{2}} = 3 + \frac{667\frac{1}{2}}{4673\frac{1}{2}} < 3\frac{1}{7}\end{aligned}

Therefore the circumference of the circle (being less than the perimeter of the polygon) is a fortiori less than 3 1/7 times the diameter AB.

PART II. Next let AB be the diameter of a circle, and let AC, meeting the circle in C, make the angle CAB equal to one-third of a right angle. Join BC.

\text{Then}\quad AC:CB\,[=\sqrt{3}:1] < 1351:780

First, let AD bisect the angle BAC and meet BC in d and the circle in D. Join BD. The triangles ADB, ACd, BDd are similar, whence

\begin{aligned} &BA+AC:BC = AD:DB\\ &\text{[But}\ AC:CB < 1351:780,\ \text{while}\ BA:BC = 1560:780]\\ &\text{Therefore}\quad AD:DB < 2911:780 \quad (1)\\ &\text{[Hence}\ AB^2:BD^2 < (2911^2+780^2):780^2 < 9082321:608400]\\ &\text{Thus}\quad AB:BD < 3013\frac{3}{4}:780 \quad (2)\end{aligned}

Secondly, let AE bisect the angle BAD, meeting the circle in E; and let BE be joined. Then we prove, in the same way as before, that

\begin{aligned} &AE:EB\,[= BA+AD:BD < (3013\frac{3}{4}+2911):780] < 5924\frac{3}{4}:780\\ &\quad < 5924\frac{3}{4}\times\frac{4}{13}:780\times\frac{4}{13} < 1823:240 \quad (3)\\ &\text{Therefore}\quad AB:BE < 1838\frac{9}{11}:240 \quad (4)\end{aligned}

Thirdly, let AF bisect the angle BAE, meeting the circle in F.

\begin{aligned} &AF:FB\,[= BA+AE:BE < 3661\frac{9}{11}:240] < 3661\frac{9}{11}\times\frac{11}{40}:240\times\frac{11}{40}\\ &\quad < 1007:66 \quad (5)\\ &\text{Therefore}\quad AB:BF < 1009\frac{1}{6}:66 \quad (6)\end{aligned}

Fourthly, let the angle BAF be bisected by AG meeting the circle in G.

\begin{aligned} &\text{Then}\quad AG:GB\,[= BA+AF:BF] < 2016\frac{1}{6}:66 \quad [\text{by (5), (6)}]\\ &\text{[And}\ AB^2:BG^2 < (2016\frac{1}{6})^2+66^2:66^2 < 4069284\frac{1}{36}:4356]\\ &\text{Therefore}\quad AB:BG < 2017\frac{1}{4}:66,\ \text{whence}\ BG:AB > 66:2017\frac{1}{4} \quad (7)\end{aligned}

Now the angle BAG, the result of the fourth bisection of the angle BAC, is one-forty-eighth of a right angle, so the angle subtended by BG at the centre is 1/24 of a right angle. Therefore BG is a side of a regular inscribed polygon of 96 sides.

\begin{aligned} &(\text{perimeter of 96-gon}):AB\,[> 96\times 66:2017\frac{1}{4}] > 6336:2017\frac{1}{4}\\ &\text{And}\quad \frac{6336}{2017\frac{1}{4}} > 3\frac{10}{71}\end{aligned}

Much more then is the circumference of the circle greater than 3 10/71 times the diameter.

\text{Thus the ratio of the circumference to the diameter is}\ < 3\frac{1}{7}\ \text{but}\ > 3\frac{10}{71}