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 follow
ing way.’ HR is the side of an
equilateral triangle inscribed in the
circle. Let CD be the parallel
diameter, 0 the centre of the circle,
and join AO, BO, AD, DB. Then
shall the segment ABD be very
nearly one-third of the circle. For,
since AB is the side of an equi
lateral triangle in the circle, the
sector OAEB is one-third of the
circle. And the triangle AOB forming part of the sector
is equal to the triangle ADB; therefore the segment ABB
'plus 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 DBC, DC A, 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 he 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
TC 2 h, where c is the radius of the base and It the height,
we have to find the height of the cone the volume of which
. rc 2 h, and, as the height h' is to the radius c' of
its base as h is to c, we have simply to find h' where
