Page:Squaring the circle a history of the problem (IA squaringcirclehi00hobsuoft).djvu/28

The earlier Greek Mathematicians

It is to the Greek Mathematicians, the originators of Geometry as an abstract Science, that we owe the first systematic treatment of the problems of the quadrature and rectification of the circle. The oldest of the Greek Mathematicians, Thales of Miletus (640—548 B.C.) and Pythagoras of Samos (580—500 B.C.), probably introduced the Egyptian Geometry to the Greeks, but it is not known whether they dealt with the quadrature of the circle. According to Plutarch (in De exilio ), Anaxagoras of Clazomene (500—428 B.C.) employed his time during an incarceration in prison on Mathematical speculations, and constructed the quadrature of the circle. He probably made an approximate construction of an equal square, and was of opinion that he had obtained an exact solution. At all events, from this time the problem received continuous consideration.

About the year 420 B.C. Hippias of Elis invented a curve known as the τετραγωνίζουσα or Quadratrix, which is usually connected with the name of Dinostratus (second half of the fourth century) who studied the curve carefully, and who shewed that the use of the curve gives a construction for $\pi$.

This curve may be described as follows, using modern notation.

Let a point $Q$ starting at $A$ describe the circular quadrant $AB$ with uniform velocity, and let a point $R$

starting at $O$ describe the radius $OB$ with uniform velocity, and so that if $Q$ and $R$ start simultaneously they will reach the point $B$ simultaneously. Let the point $P$ be the intersection of $OQ$ with a line perpendicular to $OB$ drawn from $R$. The locus of $P$ is the quadratrix. Letting $\angle QOA=\theta$, and $OR=y$, the ratio $y/\theta$ is constant, and equal to $2a/\pi$, where $a$ denotes the radius of the circle. We have

$x=y\cot {\theta }$, or $x=y\cot {\frac {\pi y}{2a}}$,

the equation of the curve in rectangular coordinates. The curve will intersect the $x$ axis at the point

$x=\lim _{y=0}\left(y\cot {\frac {\pi y}{2a}}\right)=2a/\pi$.