158 A History ofMathematics
Fig. 5Kepler’s picture of the circle, the straight line, and the infinitely small subdivisions.Theorem II.The area of a circle compared to the square of its diameter has the ratio approximately
11 to 14.Archimedes uses an indirect demonstration, which leads to a contradiction, about which many
authors have written much. But I think the meaning of it is this: [see Fig. 5]
The circumference of the circle BG has as many parts as it has points, say an infinite number; and
of these any two may be regarded as the base of an isosceles triangle of side AB—so that inside the
area of the circle there are to be found infinitely many triangles all coming together at a common
vertex A, the centre. Let the circumference then be extended to a straight line, and let the length BC
be equal to it [to the circumference], and AB perpendicular to BC. Then the bases of these infinitely
many triangles (or sectors) are imagined as lying on one straight line BC, arranged next to each
other. Let one little base [of the part of the circle] be BF, and let it be equal to CE [i.e. correspond to
CE on the straight line]; and join the points F, E, C to A.
Since the number of triangles ABF, AEC are the same on the straight line as in the circle, and the
bases BF, EC are equal, and they all have the same height BA, which is also the height of the sectors;
the triangles EAC, BAF will have the same area, and any one will equal a sector of the circle. And
since they all have their bases on the straight line BC, the triangle BAC, which is made up of all
of them, will be equal to all the sectors of the circle, that is to the area of the circle consisting of all of
them. This is what Archimedes wishes to prove by contradiction.
[The point is this. For Archimedes, you prove (say) that the areaaof BG is notgreater than^12 ×
(radius)×(circumference), by supposing itisgreater. Since the perimeters of circumscribed poly-
gons are as near as you like to the circumference of the circle, this means that for some such
polygonP,ais greater than^12 ×(radius)×(perimeter of P). But this is the area of P, which must be
greater than the areaa(see Fig. 6).]
Kepler is saying: if you choose the polygon to have an infinite number of infinitely small sides,
then it ‘is’ already equal to the circle, its perimeter is the circumference, and you can avoid the
tedious proof by contradiction. What could go wrong in this procedure?
The proof then ends by showing that the triangle BAC has area ‘nearly’ 11/14 times the square
on the diameter, using theorem I.]Appendix D
(From Galileo 1954, pp. 31–3)SALV. I take it for granted that you know which numbers are squares and which are not.