A history of Greek mathematics - Wilbourhall.org

70 ARCHIMEDES
touches it at one point only. For, if possible, let the tangent
at P touch the spiral at another point Q.
Then, if we bisect
the angle POQ by OL meeting PQ in L and the spiral in R,
0P + 0Q=20R by the property of the spiral. But by
the property of the triangle (assumed, but easily proved)
OP+OQ > 2 0L, so that OL < OR, and some point of PQ
lies within the spiral. Hence PQ cuts the spiral, which is
contrary to the hypothesis.
Props. 16, 17 prove that the angle made by the tangent
at a point with the radius vector to that point is obtuse on the
'
forward ' side, and acute on the backward ' ' side, of the radius
vector.
Props. 18-20 give the fundamental proposition about the
tangent, that is to say, they give the length of the subtangent
at any point P (the distance between and the point of intersection
of the tangent with the perpendicular from to OP).
Archimedes always deals first with the first turn and then
with any subsequent turn, and with each complete turn before
parts or points of any particular turn. Thus he deals with
tangents in this order, (1) the tangent at A. the end of the first
turn, (2) the tangent at the end of the second and any subsequent
turn, (3)
the tangent at any intermediate point of the
first or any subsequent turn. We will take as illustrative
the case of the tangent at any intermediate point P of the first
turn (Prop. 20).
If OA be the initial line, P any point on the first turn, PT
the tangent at P and OT perpendicular to OP, then it is to be
proved that, if ASP be the circle through P with centre 0,
meeting PT in S, then
(subtangent OT) = (arc ASP).
I. If possible, let OT be greater than the arc ASP.
Measure off OU such that OU > arc ASP but < OT.
Then the ratio PO :
OU
is greater than the ratio P0 : OT,
i.e. greater than the ratio of %PS to the perpendicular from
on PS.
Therefore (Prop. 7) we can draw a straight line OQF meeting
TP produced in F, and the circle in Q, such that
FQ:PQ = P0:0U.