A history of Greek mathematics - Wilbourhall.org

ON THE SPHERE AND CYLINDER, I 37
If $ is the surface of the cone, we have to prove that S= B.
For, if S is not equal to B, it must be either greater or less.
I. Suppose B < S.
Circumscribe a regular polygon about B, and inscribe a similar
polygon in it,
than S:B (Prop. 5).
such that the former has to the latter a ratio less
Describe about A a similar polygon and
set up from it a pyramid circumscribing the cone.
Then (polygon about A) : (polygon about B)
Therefore
= C 2 :E 2
= C:D
— (polygon about A) :
(surface of pyramid).
(surface of pyramid) = (polygon about B).
But (polygon about B) : (polygon in B) < S : B
therefore (surface of pyramid) : (polygou in B) < S :
But this is impossible, since (surface of pyramid) > S, while
(polygon in B) < B;
therefore B is not less than S.
II. Suppose B > S,
Circumscribe and inscribe similar regular polygons to B
such that the former has to the latter a ratio < B : S. Inscribe
in J. a similar polygon, and erect on A the inscribed pyramid.
Then (polygon in A) : (polygon in B) — C 2 : E
= C:D
2
> (polygon in A) : (surface of pyramid).
(The latter inference is clear, because the ratio of C:D is
greater than the ratio of the perpendiculars from the centre of
A and from the vertex of the pyramid respectively on any
side of the polygon in A ;
in other words, if /? < oc < \ir,
sincx
> sin/?.)
Therefore
(surface of pyramid) > (polygon in B).
But (polygon about B) :
whence (a fortiori)
(polygon in B) < B :
S,
(polygon about B) : (surface of pyramid) < B: S,
which is impossible, for (polygon about B) > B, while (surface
of pyramid) < #.
;
B.