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Knowing the result, Archimedes considered a secondary circle with the same area than the ellipse.
The radius of this circle is: Archimedes wanted to probe that the area of the ellipse is equal to the area of this secondary circle Archimedes considered an auxiliary polygon similar to P', the polygon inscribed in the auxiliary circle C'.
The relation between the areas of these two similar polygons is: Archimedes estarted his double 'reductio ad absurdum'.
"We can then inscribe in the secundary circle an equilateral polygon of 4n sides such that its area is greater than that of the ellipse. [cf. On the Sphere and Cylinder, I. 6.]" (Archimedes) Then
Then we can consider a similar polygon in the auxiliary circle and the corresponding polygon in the ellipse. "Supose that P' denotes the area of the polygon inscribed in the auxiliary circle, and P that of the polygon inscribed in the ellipse." (Archimedes) We already know that
Then "But this is imposible, because the later polygon is by hypothesis greater than the ellipse, and a fortiory greater than P.
Hence the secondary circle is not greater than the ellipse." (Archimedes)
In this case we inscribe in the ellipse a polygon P with 4n equal sides such that
Archimedes consider polygon P' inscribed to the auxiliary circle and a similar polygon inscribed in the secondary circle. As before which is imposible
This completes the double reductio ad absurdum proof. "Hence the secondary circle, being neither greater nor less than the ellipse, si equal to it; and the required result follows."(Archimedes)
"In esence, Archimedes has simply given a rigorous exhaustion proof of the intuitively evident fact that the area of the
ellipse is b/a times the area
REFERENCES
C.H. Edwards - The Historical Development of the Calculus (pag. 40-42) - Springer-Verlag New York Inc.
Archimedes - On Conoids and Spheroids -- The Works of Archimedes edited by T.L. Heath - Dover Publications, Inc.
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