Lecture 7 Archimedes on circles and spheres 

The Text: 'On the measurement of the Circle'
Archimedes of Syracuse, (-287 - -212) 
Proposition 1.
Source: Thomas 1. Heath (ed.), The Works of Archimedes Dover Edition, 1953, Cambridge University Press.

 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, K the triangle described.
 
 

Then, if the circle is not equal to K, it must be either greater or less. 

I. If possible, let the circle be greater than K.

 Inscribe a square ABCD, bisect the arcs AB, BC, CD, DA, then bisect (if necessary) the halves, and so on, until the sides of the inscribed polygon whose angular points are the points of division subtend segments whose sum is less than the excess of the area of the circle over K.

 Thus the area of the polygon is greater than K.

 Let AE be any side of it, and ON the perpendicular on AE from the centre 0.

 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 polygon is less than the circumference of the circle, so 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.

 Thus the area of the circle is not greater than K.

 II. If possible, let the circle be less than K.

 Circumscribe a square, and let two adjacent sides, touching the circle in E and H, meet in T. Bisect the arcs between adjacent points of contact and draw the tangents at the points of bisection. Let A be the middle point of the arc EH, and FAG the tangent at A.

 Then the angle TAG is a right angle.

 Therefore TG > GA > GH.

 It follows that the triangle FTG is greater than half the area TEAH.

 Similarly, if the arc AH be bisected and the tangent at the point of bisection be drawn, it will cut off from the area GAH more than one-half.

 Thus, by continuing the process, we shall ultimately arrive at a circumscribed polygon such that the spaces intercepted between it and the circle are together less than the excess of K over the area of the circle.

 Thus the area of the polygon wili be less than K.

 Now, since the perpendicular from 0 on any side of the polygon is equal to the radius of the circle, while the perimeter of the polygon is greater than the circumference of the circle, it follows that the area of the polygon is greater than the triangle K; which is impossible.

 Therefore, the area of the circle is not less than K. Since then the area of the circle is neither greater nor less than K, it is equal to K.

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Explanation

 Archimedes (-287- -212) is the most well-known of the Classical Greek mathematicians. Much of his work has survived in authentic translations. He was a master of the method of Exhaustion, though it appears many of his results were initially found by mechanical analogues.

  Proposition 1 appears to be a quadrature, as it finds a triangle equal to a circle, but it is not a construction or a computation, as it assumes you have a 'rectification of a circle' i.e. a line equal to the circumference. So we should regard it as a proof that the 'constant of proportionality' in the ratio of areas to radius squared is equal to the constant of proprtionality in the ratio of circumference to radius.

 It is a classical use of Exhaustion: ABCD is a given circle K is the right triangle with height the radius and length the circumference. 
[We would say area (1/2)r. 2 p  r = p r2]

  1. First case: Area circle < Area K: Inscribe in the circle a polygon P with Area K < Area P < Area circle. Now dissect P into triangles of height < the radius and sum of widths < circumference. Hence Area P < Area K. Contradiction.
  2. Second case: Area P > Area K. Circumscribe a polygon Q with Area K > Area Q > Area circle. Dissect Q into triangles of height the radius and sum of widths > circumference. Hence Area Q > Area K. Contradiction.
By trichotomy, Area K = Area circle.		 .

The Text: 'The surface area of a sphere' 
Proposition 33.
Source: Thomas 1. Heath (ed.), The Works of Archimedes Dover Edition, 1953, Cambridge University Press.

 The surface of any sphere is equal to four times the greatest circle in it.

 Let C be a circle equal to four times a great circle.

 Then if C is not equal to the surface of the sphere, it must either be less or greater.

 1. Suppose C less than the surface of the sphere.
 
 

It is then possible to find two lines,b, g, of which b is the greater, such that < (surface of sphere) : C [Prop. 2]. 

Take such lines, and let d be a mean proportional between them.

 Suppose similar regular polygons with 4n sides circumscribed about and inscribed in a great circle such that the ratio of their sides is less than the ratio b: d [Prop. 3].

 Let the polygons with the circle revolve together about a diameter common to all, describing solids of revolutions as before.

 Then (surface of outer solid) : (surface of inner solid) = (side of outer)2 : (side of inner)2 [Prop. 32] < b2 : d2, or b : g < (surface of sphere) : C, a fortiori.

 But this is impossible, since the surface of the circumscribed solid is greater than that of the sphere [Prop. 28], while the surface of the inscribed solid is less than C [Prop. 25].

 Therefore C is not less than the surface of the sphere.

 II. Suppose C greater than the surface of the sphere.

 Take lines b, g, of which b is the greater, such that: b : g < C : (surface of sphere).

 Circumscribe and inscribe to the great circle similar regular polygons, as before, such that their sides are in a ratio less than that of b to d, and suppose solids of revolution generated in the usual manner. Then, in this case, (surface of circumscribed solid) : (surface of inscribed solid) < C : (surface of sphere).

 But this is impossible, because the surface of the circumscribed solid is greater than C [Prop. 30], while the surface of the inscribed solid is less than that of the sphere [Prop. 23].

 Thus C is not greater than the surface of the sphere.

 Therefore, since it is neither greater nor less, C is equal to the surface of the sphere.

 .

Explanation

 Archimedes on surface area of a sphere

Archimedes proves that the area of a sphere is 4 x the area of a great circle of the sphere. 
[We would say surface area = 4.p r2]. 
He uses the proved results:

  1.  If Area A < Area B then there are lines b < c with b : c = A : B.
  2. There is a line d (the mean proportional ) with b : d = d : g . [We would say g2 = a b]. 
  3. If C is a circle and a < b are lines, there is an inscribed polygon P and a similar circumscribed polygon Q such that Area P : Area Q < a : b . 
  4. [We would say there are inscribed and circumscribed polygons whose areas are arbitrarily close]
  5. If a polygon is inscribed in a circle and the whole configuration revolved about an axis, forming a solid of revolution inside a sphere, then the surface area of the solid < surface area of sphere. Similarly if a polygon circumscribes the circe the surface area of the solid of revolution > surface of sphere.
  6. If two similar polygons are revolved round similar axes, the ratio of their surfaces is the ratio of their (linear dimension)2. Now he uses Method of Exhaustion to show that a circle of radius 4 x radius of sphere cannot have smaller or larger area. For if a it had smaller area, inscribe and circumscribe polygons to the great circle whose areas are intermediate and revolve to get a contradiction. Similarly get a contradiction if the circle has larger area.

    7.  
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Exercise

 Use the idea of Archimedes argument about the relationship of area of the circle to its circumference but with 'for all e there is a dinstead of the Method of Exhaustion. .

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 Last update: 5 June, 2000

 Author: Phill Schultz, schultz@maths.uwa.edu.au