A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

APPENDIX 559
But (arc ASP) = OT,by hypothesis ;
therefore it was necessary to prove, alter nando, that
(3) F'R :
(arc RP) < RO :
i.e.
OT, or PO : OT,
< PM:MO, where OM is perpendicular to SP.
560 APPENDIX
Next, for the point Q' on the ' forward ' side of the spiral
from P, suppose that in the figure of Prop. 9 or Prop. 7 (Fig. 2)
any radius OP of the circle meets A B produced in F, and
Similarly, in order to satisfy (2), it was necessary to
prove that
(4)
FQ:(sxgPQ) > PM:MO.
Now, as a matter of fact,
F'R :
but in the case of (4)
(3) is a fortiori satisfied if
(chord RP) < PM:MO
;
we cannot substitute the chord PQ for
the arc PQ, and we have to substitute PG\ where G' is the
point in which the tangent at P to
Fig. 1.
the circle meets OQ produced ; for
of course PG f > (arc PQ), so that (4)
is a fortiori satisfied if
FQ:PG'>PM:MO.
It is remarkable that Archimedes
uses for his proof of the'two cases Prop.
8 and Prop. 7 respectively, and makes
no use of Props 6 and 9, whereas
the above argument points precisely to the use of the figures
of the two latter propositions only.
For in the figure of Prop. 6 (Fig. 1), if OFP is any radius
cutting AB in F, and if PB produced cuts OT, the parallel to
AB through 0, in H, it is obvious, by parallels, that
PF : (chord PB) = OP :
PH.
Also PH becomes greater the farther P moves from B
towards A, so that the ratio PF :
while it is always less than OB :
PB diminishes continually,
BT (where BT is the tangent
at B and meets OH in T), i.e. always less than BM : MO.
Hence the relation (3) is always satisfied for any point R' of
the spiral on the backward ' ' side of P.
But (3) is equivalent to (1), from which it follows that F'R
is always less than RR', so that R f always lies on the side
of TP towards 0.
the tangent at B in G ;
Fig. 2.
parallel through to AB, in H, T.
and draw BPH, BGT meetmg 0T, the
Then PF:BG> FG: BG, since PF > FG,
> 0G :
GT, by parallels,
> OB :BT, a fortiori,
> BM:M0;
and obviously, as P moves away from B towards 0T, i.e.
moves away from B along BT, the ratio OG.GT increases
continually, while, as shown, PF:BG is always > BM:M0,
and, a fortiori,
That is,
PF:(8lycPB) > BM-.MO.
as G
(4) is always satisfied for any point Q' of the spiral
1
forward ' of P, so that (2) is also satisfied, and QQ' is always
less than QF.
It, will be observed that no vtvcri?, and nothing beyond
'
plane ' methods, is required in the above proof, and Pappus's
criticism of Archimedes's proof is therefore justified.
Let us now consider for a moment what Archimedes actually
does. In Prop. 8, which he uses to prove our proposition in
the ' backward' case (R', R, F'), he shows that, if P0 : 0V
is any ratio whatever less than P0 : 0T or PM : MO, we can
find points F' , G corresponding to any ratio P0 : 0V f where
0T < 0V < OF, i.e. we can find a point F' corresponding to
a ratio still nearer to P0 : 0T than P0 : OF is. This proves
that the ratio RF' : PG,
while it is always less than PM:M0,