430 Mr Burnside, On double-sixes. 



The equation connecting \ ^ is the same as that connecting 

 s, s'. Hence if the polygon whose sides lie on the two quadrics 

 is a hexagon, the circles must be such that the sides of a triangle 

 inscribed in the second circle touch the first. The conditions for 

 this are either 



(i) a' = R^- 2Rr, or (ii) a^ = R' + 2Rr. 



In the first case, 



^" :^2 ' ^^ — y:r~' 



giving A^ — 1 — ± B. 



In the second case, 



r^ 16R'^ 



^■' - r^ _ 4,Rr ' ^"~ r-"- ^Rr ' 



giving 



A^-1 = + AB. 



(a + b)(c + d) . ahcd + 1 



{a-b){c-d)' {a-b){c-d) 

 satisfy one of these relations, then obviously 

 (a — b)(c — d) abed + 1 



{a + b)(c + d)' (a + b){c + d) 



satisfy the other. In other words, if for a chosen pair of sets of 

 generators of the quadrics the polygon is a hexagon, then it is 

 also a hexagon for the other two sets. As pointed out above, 

 this is equivalent to establishing the existence of a double-six. 

 5. If the two quadrics are taken in the usual canonical form 

 ax^ + by^ + cz^ + df = 0, 

 a'x^ + b'y"" + ez" + d'f = 0, 

 it will be found that the above condition 



A'-1 = ±B, 



which ensures that a double-six lies on them, may be expressed 

 irrationally in the form 



1 1 1 ^ 



'^abo'd' + '^a'b'cd '^acb'd' + "Ja'c'bd ^/adb'c' + Va'd'ftc 



