282] 
ON A PARTICULAR CASE OF CASTILLON’s PROBLEM. 
437 
But to obtain equations involving only one of the sets (x, y, z), (x 1? y 1( z/ it is proper 
to write 
tan 2 / = tan (£ + g) tan £ = tan £ tan (£i — h), 
tan 2 g = tan (£ + /¿) tan rj = tan r/ 1 tan (£ — /), 
tan 2 h = tan (vj + /)tan £ = tan £ tanfo — gr); 
taking the first set of equations, we have 
x (z + m) 
1 — mz ’ 
z (y + l) 
therefore 
l 2 — rax 
m 2 (1 — nx) _ 
x + Pm ’ 
x + n 
therefore 
y + Z : 1 — Zy = m 2 + In + (l — m 2 n) x : n — lm 2 + (1 + lm?n) x ; 
therefore 
ft 2 [n — lm 2 + (1 + lm 2 n) x} (l 2 m + x) = (l 2 — mx) [m 2 + In + (l — ni 2 n) x}; 
therefore 
(l 2 m 2 + l 3 n — l 2 mn 3 + l 3 m 3 n 2 ) 
+ x (— n 3 + Zm 2 n 2 — l 2 mn 2 — l 3 m 3 n 3 — in 3 — Iran + l 3 — l 2 m 2 n) 
+ x 2 (— lm + m 3 n — n 2 — lm 2 n 3 ) = 0. 
Now l + m + n= Imn, and by means of this relation we find 
Coef. x° = l {lm 2 + Pn — n 2 (l + m+ n)+ m (l+ m + n) 2 }, 
Coef. x = — n 3 + mn (l + m + n) — In (l + m + n) — (l + m + n) 3 — m 3 
— Imn + l 3 — lm (l+m + n), 
or, reducing, 
| coef. x 3 = (m 3 + 2m 2 n — n 3 ) + l (3m 2 4- 2mn — n 2 ) + P (m + n) 
= (m + n) {(m 2 + mn — n 2 ) +1 (3m — n) + P] 
= (m + n) {(l + m) 2 + (l + m)(l + n) — (1 + n) 2 }, 
— \ coef. x = m 3 + m 2 n + mn 2 + n 3 + 21 (m 2 + 2mn + n 2 ) + 21 2 (m + n) 
= (m + n) {m 2 + n 2 + 21 (m + n) + №} 
= (m + w) {(Z + m) 2 + (Z + n) 2 }, 
Z coef. x 2 = (m 3 — 2m?i 2 — n 3 ) + Z (??r 2 — 2nm — 3n 2 ) — l 2 (m + n) 
= (m + w) {(m 2 — mn — n 2 ) + Z (m — 3?i) — Z 2 } 
= (m + w) {(Z + mf — (1 + m) (l + n) —(l + n) 2 };