144 Professor Baker, On a proof of the theorem of a double six, etc, 



we may then take 



j) = a + k, p'^ a'+ k, m {b + p) = b + k, m' {b' + p') = b' + k, 



m' (c + />') = c + k, m (c' + p) = c' + k; 



these lead to m (b - c') = b - c' , m' (6'- c) = b'- c, and hence 



m = m'= 1, unless the lines d, d' have particular positions; whence 



p = p'=^ k. 



Thus we have 

 P = bB + c'C + p{B + C), P'= b'B'+ cC + p {B'+ C), etc., 

 which give, in particular, 



Q + Q'^ aA + a'A'+ cC + c'C'+ p{C+C'+A + A') 

 = -[bB + b'B'+ p{B + B')l 

 showing that the points B, B' , Q, Q' are coplanar. And similarly 

 the points C, C, R, R' are coplanar. 



§ 8. Now take a point 0, not in the original space of three 

 dimensions which we have considered ; with this original space the 

 point determines a four dimensional space containing both. 

 Therein take points A^, B-^, C^, A{, B{, C-l given by 



A^={a^p)A + 10, B:^^{b + p)B + mO, 

 A{ - (a' + p) A'+ I'O, B^ = (6'+ p) B'+ m'O, 

 Ci = (c + p) C + nO, 

 C/ = (c'+ p) C'+ n'O, 

 wherein I, m, n, V, m', n' are arbitrary numbers save for the single 

 relation 



Z + m + ?^ + Z'+ m'+ n'= 0. 

 Then we have 



^1 + 5i + Ci + A{+ B^+ C{= 0. 

 Take also Pj, Q-^, R-^, P{, Q{, R-l given by 

 Pi = P+(m + n')0, Qi = ^ + (w + r)0, Pi = P+(Z+m')0, 

 P/= P'+ (m'+ n) 0, Q^= Q'+ {n'+ I) 0, P/- P'+ {l'-\- m) 0; 

 then we have 



Pi = Pi + Ci', Q^ = C^ + A^, R,^A, + B^', 

 P^= Pi'+ Ci, Q^= Ci'+ A^, P/= A^+ Pi. 



The original figure in three dimensions is thus the projection 

 from of the figure now formed in four dimensions, and this is 

 exactly such a figure as that we considered originally. 



Geometrically, what is arbitrary in the four dimensional space 

 is the point 0, and five of the six points taken on the lines OA, OB, 

 00, OA', OB', OC. These being taken, it is no doubt possible to 

 complete the construction without use of the symbols. These seem 

 however to add to clearness. 



