A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

MEASUREMENT OF A CIRCLE 51
sides continually doubled, beginning from a square, (b) by
circumscribing a similar set of regular polygons beginning
from a square, it being shown that, if the number of the
sides of these polygons be continually doubled, more than half
of the portion of the polygon outside the circle will be taken
away each time, so that we shall ultimately arrive at a circumscribed
polygon greater than the circle by a space less than
any assigned area.
Prop. 3, containing the arithmetical approximation to n, is
the most interesting. The method amounts to calculating
approximately the perimeter of two regular polygons of 96
sides, one of which is circumscribed, and the other inscribed,
to the circle ; and the calculation starts from a greater and
a lesser limit to the value of V 3, which Archimedes assumes
without remark as known, namely
265 s- a/Q ^ 1351
IT'S a
v
2a -* — ±
' 2a+
where a 2 is the nearest square number above or below a 2 ± b,
as the case may be. The use of the first part of this formula
by Heron, who made a number of such approximations, is
proved by a passage in his Metrica 1 , where a rule equivalent
to this is applied to \/720 ; the second part of the formula is
used by the Arabian Alkarkhi (eleventh century) who drew
from Greek sources, and one approximation in Heron may be
obtained in this way. 2 Another suggestion (that of Tannery
h
52 ARCHIMEDES
and Zeuthen) is that the successive solutions in integers of
the equations
x 2 —-Sy 3 y = 2 1 i
x 2 -3y<
= 2 -2)
may have been found in a similar way to those of the
equations x 2 — 2y
2
= ±1 given by Theon of Smyrna after
the Pythagoreans. The rest of the suggestions amount for the
most part to the use of the method of continued fractions
more or less disguised.
Applying the above formula, we easily find
2-i> V3 >2~§,
or | > \/3 > §.
Next, clearing of fractions, we consider 5 as an approximation
to V 3 . 3 2 or a/27, and we have
5 + T
2o > 3 ^3 > 6 + x
a
T ,
whence f| > V 3 > yy.
Clearing of
fractions again, and taking 26 as an approximation
to \/3 A5 2 or \/675, we have
which reduces to
26—& > 15^3 > 26-sV,
135
78 ;1 ^ J*\ ~> 265
q- > V 6 > y-g-j.
Archimedes first takes the case of the circumscribed polygon.
Let CA be the tangent at J. to a circular arc with centre 0.
Make the angle AOG equal to one-third of a right angle.
Bisect the angle AOG by OD, the angle AOD by OE, the
angle AOE by OF, and the angle AOFby OG. Produce GA
to AH, making AH equal to AG. The angle GOH is then
equal to the angle FOA which is ^th of a right angle, so
that GH is the side
96 sides.
of a circumscribed regular polygon with
1
Heron, Metrica, i. 8.
2
Stereom. ii, p. 184. 19, Hultsch; p. 154. 19, Heib. ^54 = 7^ = 7^
instead of 7 I
5
1
.
E 2
Now OA :AG[ = \/3:l] > 265:153, (1)
and OG :
CA = 2:1 = 306:153. (2)