A History of Mathematics From Mesopotamia to Modernity

Greeks,Practical andTheoretical 65

Archimedes (from whom Heron derived several other results).^5 If this is true, it shows Archimedes

as innovative in a more striking way than had been thought.

What Heron’s work shows, perhaps more than any other, is the existence of traditions which we

do not otherwise know about—and this is a warning against easy generalizations on the nature

of Greek mathematics. The quotation from Pappus does suggest that he was considered ‘special’ at

least in his interest in machines, but it is unclear how much should be built on that; many of the

machines which he describes, such as water-clocks, were certainly not of his invention. In parts

of theGeometrica(which may not be his, but can be seen as work by his ‘school’), he solves some

quadratic problems by the Babylonian recipe; and this is taken by some scholars as a reason to fit

his work into a genealogy which stretches (with notable gaps!) from the Babylonians through to

the Islamic algebraists of the ninth centuryce. However, he is also capable of advising intelligent

guesswork, and using what appear to be quite new methods. The following example is again unlike

our usual idea of Greek geometry.

In a right-angled triangle, the sum of the area and the perimeter is 280 feet;to separate the sides and find the area,

I proceed thus: Always look for the factors; now 280 can be factored into 2.140, 4.70, 5.56, 7.40, 8.35, 10.28, 14.20.

By inspection, we find 8 and 35 fulfil the requirements.

Note that the problem is not ‘well-posed’, by which I mean that it can have numerous solutions; as

an equation, it is (setting the short sides equal toaandb)

a+b+

√

a^2 +b^2 +

1

2

ab= 280

and of course the key point is that, for a ‘nice’ solutiona^2 +b^2 must be a square, itself a favourite

problem. It could be seen as an extension of the Babylonian problems which give you the side

of a square plus its length; but it is considerably harder and more geometrical. Heron’s solution

continues:

For take one-eighth of 280, getting 35 feet. Take 2 from 8, leaving 6 feet. Then 35 and 6 together make 41 feet.
Multiply this by itself, making 1681 feet. Now multiply 35 by 6, getting 210 feet. Multiply this by 8, getting 1680 feet.
Take thisawayfrom the 1680, leaving 1, whose square root is 1. Now take the 41 and subtract 1, leaving 40, whose
half is 20; this is the perpendicular, 20 feet. And again take 41 and add 1, getting 42 feet, of which the half is 21; and
let this be the base, 21 feet. And take 35 and subtract 6, leaving 29 feet. Now multiply the perpendicular and the base
together, [getting 420], of which the half is 210 feet; and the three sides comprising the perimeter amount to 70 feet;
add them to the area, getting 280 feet. (Geometrica, in Thomas 1939, pp. 503–9; Fauvel and Gray 5.C.2 (c))

This calculation is completely enigmatic as it stands—we can see that Heron’s answer is correct,
but what is he doing? Thomas’s translation gives a good explanation (due to Heath) related to the
formula of Appendix A. In fact, ifris the radius of the inscribed circle, ands=^12 (a+b+c)is half
the perimeter, then the area issr(see Appendix A) and so the sum of perimeter and area iss( 2 +r).
We can therefore ‘look for factors’, as Heron says, and guesss=35, 2+r=8.
One could go on to ask how exactly, without algebra, such a procedure could have been hit upon.
Such mathematics could be thought of, in the language of Høyrup (1994) as ‘subscientific’—short
on proof, although the numerical check is given; perhaps designed to display skill and virtuosity (or,
like a crossword, to keep the mind active) rather than to be of any use. And yet the relation to the
formula of Appendix A ties it in with ‘real’ geometry. Some time later than Heron (probably—the

5. In particular, he tended to use the ‘Archimedean approximation’ 3^17 forπ.