A history of Greek mathematics - Wilbourhall.org

66 ARCHIMEDES
points of intersection of a certain rectangular hyperbola
and a certain parabola. It is quite possible, however, that
such problems were in practice often solved by a mechanical
method, namely by placing a ruler, by trial, in the position of
the required line : for it is only necessary to place the ruler
so that it passes through the given point and then turn it
round that point as a pivot till the intercept becomes of the
given length. In Props. 6-9 we have a circle with centre 0,
a chord AB less than the diameter in it, OM the perpendicular
from on AB, BT the tangent at B, OT the straight line
through parallel to AB
;
B E
as the case may be, than the ratio BM :
:
is any ratio less or greater,
MO. Props. 6, 7
(Fig. 2) show that it is possible to draw a straight line OFP
Fig. 3.
:
meeting AB in F and the circle in P such that FP : PB=D E
(OP meeting AB in the case where D:E < BM:M0, and
meeting AB produced when D : E MO). In Props. 8, 9
> BM :
(Fig. 3) it is proved that it is possible to draw a straight line
OFP meeting AB in F, the circle in P and the tangent at B in
G, such that FP:BG = D:E (OP meeting AB itself in the case
where D : E < BM: MO, and meeting AB produced in the
case where D:E > BM : MO).
We will illustrate by the constructions in Props. 7, 8,
as it is these propositions which are actually cited later.
Prop. 7. If D : E is any ratio > BM : MO, it is required (Fig. 2)
to draw 0P F / / meeting the circle in P f and AB produced in
F' so that
F P / / :P B = / D:E.
Draw OT parallel to AB, and let the tangent to the circle at
B meet OT in T.