679]
ON THE REGULAR SOLIDS.
271
each other in such wise that, taking two opposite points as poles, the relative situation
is as follows:
B
Longitudes.
1
—
?
3
0°,
120°,
240°
>
6
( 0°,
120°,
240°) + 22°
14',
6
(60°,
180°,
300°) + 22°
14',
3
60°,
180°,
300°
1
1
—
9
where the points B in the same horizontal line form a zone of points equidistant
from the point taken as the North Pole. Neglecting the 3+3 points B which lie
adjacent to the poles, the remaining 14 points B may be arranged as follows (/3 = 22 c 14'
as above):
B
Longitudes.
1
—
6
/?, 120 ° + (3, 240° + (3
-(3, 120 °-/3, 240°-/?,
6
60° + /3, 180° + /3, 300° + /3
60°-/?, 180 °~(3, 300°-/?.
1
—
And taking the two poles separately with each system of the remaining poles, we
have 2 systems each of 8 points B, which are, in fact, the summits of a cube
(hexahedron); each point B taken as North Pole thus belongs to two cubes; but
inasmuch as the cube has 8 summits, the number of the cubes thus obtained is
20 x 2 -i- 8, = 5; viz. the 20 points B form the summits of 5 cubes, each point B
of course belonging to 2 cubes.
It is to be added that, considering the 5 points B which form a face of the
dodecahedron, any diagonal BB of this dodecahedron is a side of a cube. We have
thus 12 x 5, =60, the number of the sides of the 5 cubes.
It is at once seen that the centres of the faces of a cube are points ©, and
that the mid-points of the sides of the cube are points <!>.
To each cube there corresponds of course an octahedron, the summits being
points ©, the centres of the faces points B, and the mid-points of the sides points
<I>; thus, for the five octahedra the summits are the 5x6,= 30, points ©; the
centres of the faces are 5x8,= 40, points B (each point B being thus a centre
of face for two octahedra), and the mid-points of the sides being the 5 x 12, = 60,
points <P.
Finally, considering the 8 points B which belong to a cube, we can, in four
different ways, select thereout 4 points B which are the summits of a tetrahedron;
