387 
pass through two opposite bisections, d 2 , d l2 , of the opposite 
edges C $3 and C 5 C 6 of the octahedron. 
The four axes, such as 0$ 7 , pass through the centres o l} o 7 
of the opposite and parallel faces, C 1 C$ S and C 6 C 5 C i of the 
octahedron, and are perpendicular to both of them. 
Owing to this property, the four axes 0$ 7 , &c., 0$ 6 , are 
called the octahedral axes of the cube. 
24. This property may be demonstrated as follows : — - 
Describe a square (fig. 27*, Plate IV.*), ACJ. \C 2 , haying 
each of its sides = 0 1 D 5 (fig. 27, Plate IV.). 
AC-JJfiz is evidently a fourth of the square 0 $ h 0 3 0 i} forming 
a face of the cube (fig. 27, Plate IV.). 
Draw the diagonals of the square C$ 2 , and AD l3 meeting 
in the point d v 0 $% bisected in d 1 will represent on a plane 
surface in (fig. 27*, Plate IV.*) the edge of the octahedron 
C l d l C 2 seen in perspective in (fig. 14, Plate II.). 
Produce D 1 C 1 and C 2 A (fig. 27*) to 0 1 and D 5 , making C l 0 1 
and AD 5 each = AD, a diagonal of the square Dfi-^AC^ 
Join make Ad b = Ad v Join C 1 d 5 and A0 V meeting in o x . 
Then C^d^ and Ao 1 0 1 (fig. 27*) represent on a plane surface 
the lines similarly shown in perspective in (fig. 14, Plate II.) 
25. To facilitate calculation we shall choose one of the sides 
of the square C X A C 2 D 1 as our unit. 
a/9 1 
Then AD X =\/ 2 and Ad x — Ad b = —y~ — -y- ■ 
By plane trigonometry tan Ad^C^ 
And angle Ad^— 54° 44' 8". 
Now (fig. 14, Plate II.) the lines C x d b and C 6 d 5 are both by 
construction perpendicular to the edge C 2 C 3 of the octahedron 
of two adjacent faces at the point d 5 . 
The angle 0$$^ therefore measures the inclination of these 
faces ; but this angle is evidently twice the angle Ad b C 1 (fig. 
27*, Plate IV.*). What is true with regard to the angle of 
inclination over the edge C^q is true by similarity and symmetry 
of construction of all the other edges of the octahedron. And 
therefore the angle of inclination of any two adjacent faces of 
the octahedron is 109° 28' 16". 
26. Again (fig. 27* Plate IV.*) tan A0 1 D 5 =A^= x / 2. 
but tan Ad $ ± = Therefore AO x D b = A d b C x ; 
also 0 1 AD b =90°—Ao 1 D b —90~Ad 5 C 1 ; consequently 
Ao x d 5 =90°, and the line Ao 1 is perpendicular to C x d b at the 
point o v 
By symmetry of construction the line 0 1 o 1 (fig. 14, Plate II.) 
AC 1 1 
A 
V2- 
