Page:Squaring the circle a history of the problem (IA squaringcirclehi00hobsuoft).djvu/29

If the curve could be constructed, we should have a construction for the length $2a/\pi$, and thence one for $\pi$. It was at once seen that the construction of the curve itself involves the same difficulty as that of $\pi$.

The problem was considered by some of the Sophists, who made futile attempts to connect it with the discovery of "cyclical square numbers," i.e. such square numbers as end with the same cipher as the number itself, as for example 25 = 5², 36 = 6²; but the right path to a real treatment of the problem was discovered by Antiphon and further developed by Bryson, both of them contemporaries of Socrates (469—399 B.C.). Antiphon inscribed a square in the circle and passed on to an octagon, 16agon, &c., and thought that by proceeding far enough a polygon would be obtained of which the sides would be so small that they would coincide with the circle. Since a square can always be described so as to be equal to a rectilineal polygon, and a circle can be replaced by a polygon of equal area, the quadrature of the circle would be performed. That this procedure would give only an approximate solution he overlooked. The important improvement was introduced by Bryson of considering circumscribed as well as inscribed polygons; in this procedure he foreshadowed the notion of upper and lower limits in a limiting process. He thought that the area of the circle could be found by taking the mean of the areas of corresponding in- and circum-scribed polygons.

Hippocrates of Chios who lived in Athens in the second half of the fifth century B.C., and wrote the first text book on Geometry, was the first to give examples of curvilinear areas which admit of exact quadrature. These figures are the menisci or lunulae of Hippocrates.

If on the sides of a right-angled triangle $ACB$ semi-circles are described on the same side, the sum of

the areas of the two lunes $AEC$, $BDC/math>isequaltothatofthetriangle<math>ACB$. If the right-angled triangle is isosceles, the two lunes are equal, and each of them is half the area of the triangle. Thus the area of a lunula is found.

If $AC=CD=DB=$ radius $OA$ (see Fig. 3), the semi-circle $ACE$ is 1/4 of the semi-circle $ACDB$. We have now

$semicircleAB-3semicircleAC=ACDB-3.{\text{ meniscus }}ACE$,

and each of these expressions is $\textstyle {\frac {1}{4}}semicircleAB$ or half the circle on $AB$ as diameter. If then the meniscus $AEC$ were quadrable