79] 
AND ON ANOTHER ALGEBRAICAL SYSTEM. 
469 
the result is easily shown to be 
l : m : 9 = A (B + H + 57 x ) : B (A + H + zr 2 ) : (H + OTj) (H + tz- 2 ) — AB. 
But the problem may be considered as the problem for two variables, analogous 
to that of determining the conic having a double contact with a given conic, and 
touching three conics each of them having a double contact with the given conic; 
and in this point of view I was led to the following solution. If we assume 
Bl — Hm = u, — HI + Am = v, 
Hence, writing K6 2 + ^ V = — s 2 , we have 
u = K6 + OTjS, v = KO + -3T 2 s, V + K 2 9 2 + Ks 2 = 0 ; 
and substituting these values of u, v in the last equation, 
A (KO + otjs) 2 + 2H(KO + Otis) (KO + '57 2 s) + B (K9 + vr 2 s) 2 + K 2 0 2 4- Ks 2 = 0, 
or reducing, 
K 2 0 2 (A + 2H + B +1) + 2K0s {(H + H) + (H + B) ot 2 } + s 2 (Anr-? + + Bvxg + K) = 0 ; 
whence 
[K6 (A + 2H + B +1) + s {(A + jH) + (H + B) S7 2 ]] 2 
= s 2 [{(yl + H) OTj + (H + B) ot 2 } 2 — (A + 2H + B +1) (Hw! 2 + 2ihr 1 sr 2 + B^ 2 2 + K)\ 
= S 2 {— K (vr 1 — '57 2 ) 2 — (A 57J 2 + 2 Htx x tb 2 + I?ct 2 2 ) — K (A + 2 H + B+ 1)}, 
= s 2 {— (A + K) 57j 2 — (B + K) Ts 2 — K (A + 2H + B + l) + 2 (K — H) vtivt»}, 
= s 2 {2'57 1 2 -5t 2 2 — K (A + 2H + B + 1) + 2 (K — H) ■5r 1 Br 2 }, 
= s 2 (vti'gtz + K — H) 2 :