106 Mr Brill, On the Generalization of [Feb. 24, 



If m 5 be the matrix corresponding to any fifth plane, we readily 

 obtain from equations (20) and (21) the result 



Lrn^h + Lm 2 m 5 + Lm 3 m 5 + Zm 4 ra s = 0. 



Hence if ^, % 2 , % 3 , % 4 denote the angles of the anharmonic 

 ratios corresponding to the respective sets of four planes consti- 

 tuted by the given plane, one of the faces of the tetrahedron, and 

 the two tangent planes to (18) through the line in which they 

 meet, we have 



P cos Xi + Q cos %2 + R c °s %s + 8 cos Xi = 0- 

 6. From equations (20) and (21) we have 



mJZrrh = (m 2 + m s + m 4 ) (Km 2 -f Km 3 + Km 4 ). 

 Expanding this we obtain 



( Wm,) 2 = ( Wm,} 2 + ( Wm 3 ) 2 + ( Wmtf + 2Lm s m 4 



+ 2Lm i m 2 + 2Lm 2 m 3 , 

 that is 



P 2 = Q 2 + R 2 + S 2 - 2RS cos a - 2SQ cos - 2QR cos 7. 



Similarly we should obtain 

 Q 2 = R 2 + S 2 + P- - 2SP cos i/r - 2PR cos <J> - 2RS cos a, 

 R* = S 2 + P 2 + Q 2 - 2PQ cos 6 - 2QS cos /3 - 2SP cos ^, 

 S 2 = P 2 + Q 2 + R 2 -2QR cos 7 - 2PP cos(j>-2PQ cos 0. 



By adding these four relations we obtain a symmetrical rela- 

 tion, which maybe obtained direct from equations (20) and (21) as 

 follows. We have 



(m x + m 2 + m 3 + m 4 ) (Knix + Km 2 + Km 3 + Km^) = 0, 



which on expansion becomes 



( W raj) 2 + ( W m 2 ) 2 + ( Wm 3 ) 2 + ( Wm^ 2 + 2Lm 3 m 4 + 2Lm i m 2 



+ 2Lm 2 m 3 + 2Lm x m 2 + 2Lm 1 m 3 + 2Lm 1 m i = 0. 



Interpreting this we have 



P 2 + Q 2 + R 2 + S 2 = 2RS cos a + 28Q cos /3 + 2QR cos 7 



+ 2PQ cos + 2PR cos </> + 2PS cos ^. 



We also have a set of equations of the form 



(w^ + w^) (Knii + Km 2 ) = (??i 3 + m 4 ) (iT??i 3 + Km^}. 



On expansion this gives 



p-2 + q 2 _ 2PQ cos 6 = R 2 + S 2 - 2RS cos a. 



