166 Mr Brill, A new Geometrical Interpretation [Nov. 28, 
As was the case in Article 10 we see that these formulae may 
be derived, by aid of the theory of the polar triangle, from the 
formulae that can be deduced from the same analysis by means of 
Sir W. Hamilton’s method of interpretation ™. 
17. Suppose that we have three planes s,, s,, s,, and that 
s is the polar plane of the origin with respect to them. Then 
3s =s,+s,+s,, and if we draw a plane s, parallel to s, on the 
opposite side of the origin, and at one-third the distance from the 
origin, we shall have s,+s,+s,+s,=0. Further, since s passes 
through the point of concurrence of s,, s,, s,, we see that a plane 
through the origin parallel to s, will divide the perpendicular on s, 
from the said point of concurrence in the ratio 3:1. From sym- 
metry it is plain that similar statements will be true of the planes 
through the origin parallel to s,, s, and s,. Thus it is evident that 
the origin is the centroid of the tetrahedron ABCD formed by the 
four planes s,, ,, ,, 5,. Consequently, if p,, p,, p,, p, be the per- 
pendiculars from the origin on s,, s,, S,, 8,, we have 
p, ABCD = p,.ACDA =p,.ADAB=p,.AABD. 
From the equation s, +s,+8,+s,=0 we easily deduce that 
S.s.s.s.=—S.s,5,5.=S.s88,=—NS.s8 8,8 
2304 741 4°1°2 Ie2eSq 
Hence, adopting the notation of Article 12, we have 
(ABCD) 
1 —cos?(C, D) — cos’(D, B) — cos*(B, C) — 2 cos(C, D) cos(D, B) cos(B, C) 
2 (ACDAY’ 
~ 1—cos?(D, A) —cos*(A, C) — cos’(C, D) — 2 cos(D, A) cos(A, C) cos(C, D) 
(ADABY 
~ 1—cos’(A, B)— cos’(B, D) — cos*(D, A) — 2 cos(A, B) cos(B, D) cos(D, A) 
(AABCY’ 
~ 1—cos*(B, C) — cos?(C, A) — cos*(A, B) — 2 cos(B, C) cos(C, A) cos(A, B)’ 
it being understood that all the dihedral angles contained in these 
formulee are interior to the tetrahedron. This result has been 
given by Wolstenholme. 
18. For the sake of convenience we will write the denomi- 
nators of the four equal ratios of the preceding article as 7)’, 7)’, 
T?, 7,7, it being understood that T,, 7,, T,, 7’, are to be considered 
positive. Squaring the equation 
s,Ss,5,8, — $,S5,8,8, + $,Ss,8,5, — s,Ss,s,s, = 0, 
* Tait’s Quaternions, p. 57. 
