Mr Burnside, On double-sixes. 429 



hexagon. Then the existence of a double-six implies that when, 

 with the remaining set of generators on S and the remaining set 

 on S' the same construction is carried out, the polygon is again 

 in this case a hexagon. Conversely, a proof that this latter 

 polygon is a hexagon establishes the existence of a double-six, 



2. Any two arbitrary quadrics can, by a suitable projection 

 (which may be imaginary), be brought to the forms 



x'-y'- a?z' -I- hH'' = 0, 



cV - d^ -z^+t'' = 0. 

 A generator of a chosen system from each is given by 



X + y + \ {az -\-ht) = Q\ 



X{x — y) -\- az — ht = \ 

 and ex -\- dy + fi{z + t) = \ 



fi (ex — dy) + z — t = 0. 



The condition that these should meet is easily found to be 



XV + ^ (X' + At') + BXfi -1-1 = 0, 



(a + b)(c + d) r. . cibcd + 1 

 when . A = ) rr^ k , B = -4t 



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



The equations to the two other sets of generators, and the 

 corresponding conditions for them are obtained by writing — 6 

 and —diovb and d. It is to be noticed that writing — X for X 

 merely changes the sign of B. 



/ 1 — i^ 2^ \ 



3. The tangent at y^Y^r^, r j to x'^ + y'^ = r^ is 



x{\-t^)-\-yn = r{l + 1?). 

 If this meets {x + af + y^ = R^ in the point 





then \(R-a)-(R + a) t'^] (1 - t^) + ^Rtt' = r (1 4- ^') (1 4- P). 



Put t = ms, t' = ns'; 



then rnhv" (R-r + a) s'^s'^ -m?(R + r- a) s"" -n''(R + r + a) s'^ 



+ 4iRmnss' + R — r — a = 0. 

 If mV {R-r + a) = R-r — a, 



and ni'(R + r- a) = n'^(R + r + a), 



the equation is 



s^s'^ + A(s^ + s'2) + Bss +1 = 0, 



where A^=h- f- -, B^ = — 



{R-ry-a"' {R-ry-a" 



VOL. XV. PT. V. 28 



