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

DIVISIONS OF FIGURES 339
two opposite sides which the required straight line cuts are
(a) parallel or (b) not parallel. In the first case (a) the
problem reduces to drawing a straight line through E intersecting
the parallel sides in points F, G such that BF+AG
is equal to a given length. In the second case (b) where
BG, AD are not parallel Heron supposes them to meet in H.
The angle at H is then given, and the area ABU. It is then
a question of cutting off from a triangle with vertex H a
triangle HFG of given area by a straight line drawn from E}
which is again a problem in Apollonius's Cutting-ojf of an
area. The auxiliary problem in case (a) is easily solved in
III. 16. Measure AH equal to the given length. Join BH
and bisect it at M. Then EM meets BG, AD in points such
that BF+ AG= the given length. For, by congruent triangles,
BF = GH.
The same problems are solved for the case of any polygon
in III. 14, 15.
A sphere is then divided (III. 17) into segments
such that their surfaces are in a given ratio, by means of
Archimedes, On the Sphere and Cylinder, II. 3, just as, in
III. 23, Prop. 4 of the same Book is used to divide a sphere
into segments having their volumes in a given ratio.
III. 18 is interesting because it recalls an ingenious proposition
in Euclid's book On Divisions. Heron's problem is
'
To divide a given circle into three equal parts by two straight
z 2
340 HERON OF ALEXANDRIA
lines ', and he observes that, ' as the problem is clearly not
rational, we shall, for practical convenience, make the division,
as exactly as possible, in the following
way.' AB is the side of an
equilateral triangle inscribed in the
circle. Let CD be the parallel
diameter, the centre of the circle,
and join A0, BO, AD, DB. Then
shall the segment ABD be very
nearly one-third of the circle.
For,
since AB is the side of an equilateral
triangle in the circle, the
sector OAEB is one-third of the
circle. And the triangle A OB forming part of the sector
is equal to the triangle ABB] therefore the segment AEB
rplus the triangle ABD is equal to one-third of the circle,
and the segment ABD only differs from this by the small
segment on BD as base, which may be neglected. Euclid's
proposition is to cut off one-third (or any fraction) of a circle
between two parallel chords (see vol. i, pp. 429-30).
III.
19 finds a point D within any triangle ABC such that
the triangles DBG, DOA, DAB are all equal ; and
then Heron
passes to the division of solid figures.
The solid figures divided in a given ratio (besides the
sphere) are the pyramid with base of any form (III. 20),
the cone (III. 21) and the frustum of a cone (III. 22), the
cutting planes being parallel to the base in each case. These
problems involve the extraction of the cube root of a number
which is in general not an exact cube, and the point of
interest is Heron's method of approximating to the cube root
in such a case. Take the case of the cone, and suppose that
the portion to be cut off at the top is to the rest of the cone as
m to n. We have to find the ratio in which the height or the
edge is cut by the plane parallel to the base which cuts
the cone in the given ratio. The volume of a cone being
^irc 2 h, where c is the radius of the base and h the height,
we have to find the height of the cone the volume of which
is .kirc 2 h, and, as the height hf is to the radius c f of
m + 7b
its base as h is to c, we have simply to find h! where