Page:Squaring the circle a history of the problem (IA squaringcirclehi00hobsuoft).djvu/33

 indefinitely, then if the numbers which represent the perimeters of the successive polygons form a convergent sequence, of which the arithmetical limit is one and the same number for all sequences of polygons which satisfy the prescribed conditions, the circle has a length represented by this limit. It must be proved that this limit exists and is independent of the particular sequence employed, before we are entitled to regard the circle as rectifiable.

In his work, the measurement of a circle, Archimedes proves the following three theorems.

(1) 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.

(2) The area of the circle is to the square on its diameter as 11 to 14.

(3) The ratio of the circumference of any circle to its diameter is less than $$\textstyle 3\frac{1}{7}$$ but greater than $$\textstyle 3\frac{10}{71}$$.

It is clear that (2) must be regarded as entirely subordinate to (3). In order to estimate the accuracy of the statement in (3), we observe that

In order to form some idea of the wonderful power displayed by Archimedes in obtaining these results with the very limited means at his disposal, it is necessary to describe briefly the details of the method he employed.

His first theorem is established by using sequences of in- and circum-scribed polygons and a reductio ad absurdum, as in Euclid. 2, by the method already referred to above.

In order to establish the first part of (3), Archimedes considers a regular hexagon circumscribed to the circle.

In the figure, $$AC$$ is half one of the sides of this hexagon. Then

Bisecting the angle $$AOC$$, we obtain $$AD$$ half the side of a regular circumscribed 12agon. It is then shewn that $$\frac{OD}{DA} > \frac{591\frac{1}{8}}{153}$$. If $$OE$$ is the bisector of the angle $$DOA$$, $$AE$$ is half the side of a circumscribed 24agon, and it is then shewn that $$\frac{OE}{EA} > \frac{1172\frac{1}{8}}{153}$$ Next, bisecting $$EOA$$, we obtain $$AF$$ the half side of a 48agon, and it