132 A History ofMathematics
- It is easy to check that the vertices given are distant 2
√
2 from each other; which establishes
(1), since the faces must be equilateral triangles, and also answers (2). You can find the centre
of the sphere either by looking at the other half of the cube (the vertices are alternate vertices
of a cube), or by finding the centre of gravity, obviously(0, 0, 0). The radius is the length of the
line joining this to a vertex, that is,√
3, the diameter 2√
- Sos:d= 2
√
2:2
√
3 =
√
2
3 :1.
The ‘height of the triangular side’ is the height of an equilateral triangle of side 2√
2inour
model, that is,√
6 (using sin 60◦ =√
3 /2). The ratio of this todis now√
2:2=
√
1
2 :1.
The statement about area (= half times base times height) is ‘classical’. The volume is the
area of the base ( just found) times one third of the height, by the formula for the volume
of pyramids. To find the height of the pyramid, note that the three points which arenot
(−1,−1,− 1 )have centre of gravity(^13 ,^13 ,^13 ). The height is the length of the line which joins
this to(−1,−1,− 1 ), and it is easy to see that this is^23 .d. Hence al-K ̄ash ̄i’s ‘two-ninths of
the diameter’.
To prove it without using coordinates, look at Euclid XIII.13.