200 THE ELEMENTS DOWN TO PLATO’S TIME 
We also have 
r sin 6 = \ AB — r'sin 6' (1) 
In order that the lune may be squareable, we must have, in 
the first place, r 2$ _ T ,C ±Q\ 
Suppose that 6 = m 6', and it follows that 
r' — V m. r. 
Accordingly the area becomes 
-|r 2 (m sin 26' — sin 2 m 6') ; 
and it remains only to solve the equation (1) above, which 
becomes sin m Q' — v / m < s i n 
This reduces to a quadratic equation onty when m has one 
of the values o q 3 e 5 
'3* 
The solutions of Hippocrates correspond to the first three 
values of m. But the lune is squareable by ‘ plane ’ methods 
in the other two cases also, Clausen (1840) gave the last four 
cases of the problem as new 1 (it was not then known that 
Hippocrates had solved more than the first) ; but, according 
to M. Simon 2 , all five cases were given much earlier in 
a dissertation by Martin Johan Wallenius of Abo (Abveae, 
1766). As early as 1687 Tschirnhausen noted the existence 
of an infinite number of squareable portions of the first of 
Hippocrates’s lunes. Vieta 3 discussed the case in which m = 4, 
which of course leads to a cubic equation. 
(¡3) Reduction of the problem of doubling the cube to 
thè finding of two mean proportionals. 
We have already alluded to Hippocrates’s discovery of the 
reduction of the problem of duplicating the cube to that of 
finding two mean proportionals in continued proportion. That 
is, he discovered that, if 
a:x = x:y = y :b, 
then a ?J \x ?J — a:b. This shows that he could work with 
compound ratios, although for him the theory of proportion 
must still have been the incomplete, numerical, theory 
developed by the Pythagoreans. It has been suggested that 
1 Creile, xxi, 1840, pp. 375-6. 
2 GeschicJite der Math, im Altertum, p. 174. 
Yieta, Variorum de rebus mathematicis responsorum lib. viii, 1598.