73 
ARCHIMEDES 
Let OF'G meet the spiral in R'. 
Then, since PO = RO, we have, alter aando, 
F'R : RO = GP : 0 V 
> (arc PR): (arc ASP), a fortiori, 
whence F'O : RO < (arc ASR): (arc ASP) 
< OR'-.OP, 
so that F'O < OR'; which is impossible. 
Therefore OT is not less than the arc ASP. And it was 
proved not greater than the same arc. Therefore 
OT = (arc ASP). 
As particular cases (separately proved by Archimedes), if 
P be the extremity of the first turn and c r the circumference 
of the first circle, the subtangent = c 1 ; if P be the extremity 
of the second turn and c 2 the circumference of the ‘second 
circle’, the subtangent = 2c 2 ; and generally, if c n be the 
circumference of the rtth circle (the circle with the radius 
vector to the extremity of the nth turn as radius), the sub 
tangent to the tangent at the extremity of the wth turn = nc n . 
If P is a point on the nth turn, not the extremity, and the 
circle with 0 as centre and OP as radius cuts the initial line 
in K, while p is the circumference of the circle, the sub 
tangent to the tangent at P= {n—l)p + arc KP (measured 
‘ forward ’). x 
The remainder of the book (Props. 21-8) is devoted to 
finding the areas of portions of the spiral and its several 
turns cut off by the initial line or any two radii vectores. 
We will illustrate by the general case (Prop. 26), Take 
OB, OC, two bounding radii vectores, including an axfi BG 
of the spiral. With centre 0 and radius OC describe a circle. 
Divide the angle BOO into any number of equal parts by 
radii of this circle. The spiral meets these radii in points 
P, Q ... Y, Z such that the radii vectores OB, OP, OQ ... OZ, OC 
1 On the whole course of Archimedes’s proof of the property of the 
subtangent, see note in the Appendix.