The History of Mathematics: A Brief Course

APOLLONIUS^309

FIGURE 22. The basis for solving the four-line locus problem.

lines from è to A and Ã meet the lines through Ã and A respectively, each parallel

to the tangent through the other endpoint, in points Ç and Æ respectively such that,

the following proportion holds: The ratio of the rectangle on AZ and ÃÇ to the

square on ΑÃ is the composite of the ratio of the square on EB to the square on

Β A and the ratio of the rectangle on A A and AT to the rectangle on AE and ET.

As we would write this relation,

AZ • TH : AT^2 :: EB^2 : ΒÄ^2 .ΑΑ • ÄÃ : AE • ET.

It is noteworthy that this theorem involves an expression that seldom occurs in

Greek mathematics: a composition of ratios involving squares. If we thought of it

in our terms, we would say that Apollonius was working in four dimensions. But he

wasn't. The composition of two ratios is possible only when both ratios are between

quantities of the same type, in this case areas. The proof given by Apollonius has

little in common with the way we would proceed nowadays, by multiplying and

dividing. Apollonius had to find an area C such that AZ • TH : C :: EB^2 : BA^2

and C : AT^2 :: AA • AT : ZE • ET. He could then "cancel" C in accordance with

the definition of compound ratio.

It is not at all obvious how this proposition makes it possible to solve the four-

line locus, and Apollonius does not fill in the details. We shall not attempt to do so,

either. To avoid excessive complexity, we merely state the four-line locus problem

and illustrate it. The data for the problem are four lines, which for definiteness we

suppose to intersect two at a time, and four given angles, one corresponding to each

line. The problem requires the locus of points Ñ such that if lines are drawn from

Ñ to the four lines, each making the corresponding angle with the given line (for

simplicity all shown as right angles in Fig. 23), the rectangle on two of the lines will

have a constant ratio to the rectangle on the other two. The solution is in general

a pair of conies.

The origin of this kind of problem may lie in the problem of two mean pro-

portionals, which was solved by drawing fixed reference lines and finding the loci

of points satisfying two conditions resembling this. The square on the line drawn

perpendicular to one reference line equals the rectangle on a fixed line and the line