The method Archimedes used for computing the perimeter of a circle, known as
Archimedes's Algorithm, consists, plainly enough, of computing the perimeters of
inscribed and escribed polygons with successively doubled sides. The usual starting
point is a hexagon, for which the perimeter is fairly easy to calculate. The really neat
math, however, is how one calculates the perimeters after that.
fig.1
In fig. 1 o is the center of the circle, lines ab and cd are perpendicular to ob, ca' is perpendicular to oa, and angle aob is equal to angle eob. Furthermore, it can be
shown that triangles a'bc and bfc are similar and isosceles. Triangles aa'c and oab are also similar triangles.
Since ca' and a'b are drawn tangent to the same circle and a' is their point of intersection they have the same length. Since they have the same length, if points a' and o were joined, the line joining them would bisect the angle aob. therefore, ca' is half of one side of an escribed polygon having twice as many sides as that of which ae is a side, and also one quarter of the sides covering an equal perimeter. Similarly, cb is half of the inscribed sides covering an equal perimeter.
Since aa'c is similar to oab, ca' is to aa' as ob is to oa. Since oc and ob are radii of the same circle they have the same length. But traingles cfo and aeo are also similar, so oc and ob are to ao as cf is to ae, that is, as the inner perimeter is to the outer perimeter.
But aa' is ab less a'b, that is ab less ca'. We can write the ratios out as
Becasue of the relationships to the perimeter established before, and using Pout, Pout', Pin, Pin' for first outer, second outer, first inner, second inner perimeters respectively, we can write this equation as
which, when solved for Pout gives
Using the similarities established at the beginnig we can see that
or sligtly re-arranged,
and actually repeating these calculations with the successive outputs is Archimedes' Algorithm.
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