Measurement of a Circle (Κύκλου μέτρησις)

Dimensio Circuli — Archimedes of Syracuse (ed. T. L. Heath, 1897, pp. 91–98; J. L. Heiberg, Vol. I, pp. 258–271)

Note

Edition Track (English): Public-domain translation from heath:1897 (pp. 91–98), verified page-by-page against 300 DPI scans of Heath (1897) and collated with Heiberg's critical Greek text (heiberg:1880, Vol. I, pp. 258–271) and Eutocius's arithmetic commentary (eutocius:in-archimedem). Switch tracks above for the Summary or Notes.

Proposition 1

[Folio Heath (1897), pp. 91–93 · Heiberg I, 258–261 | p. 91]
The area of any circle is equal to a right-angled triangle in which one of the sides about the right angle is equal to the radius, and the other to the circumference, of the circle.

Let ABCD be the given circle, and K the triangle described, having one of the sides about the right angle equal to the radius of the circle ABCD and the other equal to its circumference (ἴσος ἄρα ὁ κύκλος τῷ Ε τριγώνῳ, Heiberg I, 260).

Prop. 1: Circle = Right Triangle E Prop. 3: 6 → 12 → 24 → 48 → 96-gon Bounds O A r K L M Base LM = Circumference C = 2πr Area(Circle) = ½ r · C = Area(△KLM) Z C D (30°, 6-gon) F (15°, 12-gon) G (7½°, 24-gon) H (3¾°, 48→96) P 265/153 < √3 < 1351/780 3 + 10/71 < π < 3 + 1/7
Proposition 1 & Proposition 3 — Circle Rectification and 6-to-96-gon Inscribed/Circumscribed Bisection Bounds

Then, if the circle is not equal to K, it must be either greater or less, and we apply the double method of exhaustion:

  1. First Reductio — Suppose the circle greater than K: Inscribe a square ABCD, bisect the arcs AB, BC, CD, DA, then bisect (if necessary) the halves, and so on continually, until the sides of the inscribed regular polygon subtend segments whose sum is less than the excess of the area of the circle over K (Eucl. X. 1, Eucl. XII. 2; cf. euclid:elements). Thus the area of the inscribed polygon is greater than K.

    Let AE be any side of this polygon, and ON the perpendicular on AE from the centre O. Then ON is less than the radius of the circle, and therefore less than one of the sides about the right angle in K. Also the perimeter of the inscribed polygon is strictly less than the circumference of the circle (Sphere & Cyl. I, Assump. 1 — concave in the same direction), i.e. less than the other side about the right angle in K. Therefore the area of the polygon () is less than K; which is inconsistent with the hypothesis that the polygon is greater than K.

  2. Second Reductio — Suppose the circle less than K: Circumscribe a square, and let two adjacent sides, touching the circle at E and H, meet in T. Bisect the arcs between adjacent points of contact and draw tangents at the points of bisection. If A is the middle point of the arc EH, and FAKG is the tangent at A, the angle TAG is right; therefore and . It follows that the triangle FTG is greater than half the area TEAH bounded by the tangents and the arc. Continuing this bisection (Eucl. X. 1), we leave spaces between the circle and the circumscribed polygon whose sum is less than the excess of K over the circle; hence the circumscribed polygon is less than K.

    Yet the perpendicular OA from the centre on any side equals the radius of the circle, while the perimeter of the circumscribed polygon is strictly greater than the circumference of the circle (convexity postulate). Therefore the area of the circumscribed polygon is greater than the triangle K; which is impossible.

Since, then, the area of the circle is neither greater nor less than K, it is equal to it.

Proposition 2

[Folio Heath (1897), p. 93 · Heiberg I, 260–263 | p. 93]
The area of a circle is to the square on its diameter as to .

Let AB be the diameter of a circle, and let CGFE be a circumscribed square. Produce side FG so that and . Join AC, AD, AF.

Since and (Eucl. VI. 1), the whole triangle ACF is to triangleACG as . Now the circumscribed square CGFE is , so . Because approximates the circumference (established in Proposition 3) and AC equals the radius, triangleACF equals the circle by Proposition 1. Hence the circle bears to the square on its diameter the approximate ratio .

Proposition 3

[Folio Heath (1897), pp. 93–98 · Heiberg I, 262–271 | p. 93]
The ratio of the circumference of any circle to its diameter is less than but greater than .

I. Proof of the Upper Bound ()

[Folio Heath (1897), pp. 93–96 · Heiberg I, 262–267 | p. 94]

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 (). Then:

  • , and .

  1. First Bisection (12-gon): Draw OD bisecting angleAOC. By the angle-bisector theorem (Eucl. VI. 3), , so , or . Hence:

    \[OA : AD > 571 : 153, \qquad OD^2 : AD^2 > (571^2 + 153^2) : 153^2 = 349450 : 23409,\]

    whence (Eucl. I. 47) .

  2. Second Bisection (24-gon): Let OE bisect angleAOD. By the same composition of ratios:

    \[OA : AE > 1162\frac{1}{8} : 153, \qquad OE : EA > 1172\frac{1}{8} : 153.\]

  3. Third Bisection (48-gon): Let OF bisect angleAOE. Then:

    \[OA : AF > 2334\frac{1}{4} : 153, \qquad OF : FA > 2339\frac{1}{4} : 153.\]

  4. Fourth Bisection (96-gon): Let OG bisect angleAOF. Then:

    \[OA : AG > 4673\frac{1}{2} : 153.\]

Since angleAOC, which is one-third of a right angle, has been bisected four times, angleAOG is of a right angle. Make angleAOHangleAOG on the opposite side of OA; then angleGOH is of a right angle, so GH is one side of a regular polygon of 96 sides circumscribed to the circle. Since and , the ratio of the diameter to the perimeter of the circumscribed 96-gon satisfies:

\[AB : P_{96}^{\text{circ}} > 4673\frac{1}{2} : (96 \times 153) = 4673\frac{1}{2} : 14688.\]

Now . A fortiori, the circumference of the circle—being strictly less than —is less than times the diameter.

II. Proof of the Lower Bound ()

[Folio Heath (1897), pp. 96–98 · Heiberg I, 266–271 | p. 96]

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; then angleACB in the semicircle is right (Eucl. III. 31), and:

  • , while .

  1. First Bisection (12-gon): Let AD bisect angleCAB, meeting BC in d and the circle in D. Join BD. By similar right triangles triangleADBsimtriangleACdsimtriangleBdD (Eucl. III. 21, Eucl. VI. 3):

    \[AD : DB = (AB + AC) : BC < 2911 : 780, \qquad AB : BD < 3013\frac{3}{4} : 780.\]

  2. Second Bisection (24-gon): Let AE bisect angleDAB, meeting the circle in E. Reducing by the factor after composition:

    \[AE : EB < 5924\frac{3}{4} : 780 = 1823 : 240, \qquad AB : BE < 1838\frac{9}{11} : 240.\]

  3. Third Bisection (48-gon): Let AF bisect angleEAB, meeting the circle in F. Reducing by the factor :

    \[AF : FB < 3661\frac{9}{11} : 240 = 1007 : 66, \qquad AB : BF < 1009\frac{1}{6} : 66.\]

  4. Fourth Bisection (96-gon): Let AG bisect angleFAB, meeting the circle in G. Then:

    \[AG : GB < 2016\frac{1}{6} : 66, \qquad AB : BG < 2017\frac{1}{4} : 66.\]

Inverting, the side BG of the inscribed regular 96-gon bears to the diameter AB a ratio greater than . Hence the perimeter of the inscribed 96-gon satisfies:

\[P_{96}^{\text{insc}} : AB > (96 \times 66) : 2017\frac{1}{4} = 6336 : 2017\frac{1}{4} = 3 + \frac{284\frac{1}{4}}{2017\frac{1}{4}} > 3\frac{10}{71},\]

since . A fortiori, the circumference of the circle—being strictly greater than —is greater than times the diameter.

Thus the ratio of the circumference to the diameter is less than and greater than .