344
HERON OF ALEXANDRIA
Quadratic equations solved in Heron.
We have already met with one such equation (in Metrica
HI. 4), namely x 2 — 14« + 46| = 0, the result only (x = 8-|)
being given. There are others in the Geometrica where the
process of solution is shown.
(1) Geometrica 24, 3 (Heib.). ‘Given a square such that the
sum of its area and perimeter is 896 feet: to separate the area
from the perimeter ’ : i.e. x‘ 2 + ‘ix= 896. Heron takes half of
4 and adds its square, completing the square on the left side.
(2) Geometrica 21, 9 and 24, 46 (Heib.) give one and the same
equation, Geom. 24, 47 another like it. ‘Given the sum of
the diameter, perimeter and area of a circle, to find each
of them/
The two equations are
\\di l + ^~d =212,
and ii d 2 + - 2 T 9 - d = 67\.
Our usual method is to begin by dividing by throughout,
so as to leave d' 1 as the first term. Heron’s is to multiply by
such a number as will leave a square as the first term. In this
case he multiplies by 154, giving 1 l 2 cZ 2 -h58 . lid = 212 . 154
or 67-|.154 as the case may be. Completing the square,
he obtains (11 d + 29) 2 = 32648 + 841 or 10395 + 841. Thus
11 ¿¿ + 29=: V(33489) or \/(11236), that is, 183 or 106.
Thus 11 d — 154 or 77, and cl = 14 or 7, as the case may be.
Indeterminate problems in the Geometrica.
Some very interesting indeterminate problems are now
included by Heiberg in the Geometrica? Two of them (chap.
24, 1-2) were included in the Geeponicus in Hultsch’s edition
(sections 78, 79^; the rest are new, having been found in the
Constantinople manuscript from which Scheme edited the
Metrica. As, however, these problems, to whatever period
they belong, are more akin to algebra than to mensuration,
they will be more properly described in a later chapter on
Algebra. ,
3 Heronis Alexandrini opera, voi. iv, p. 414. 28 sq.