TRIANGLE INSCRIBED IN A CIRCLE. 
77 
396] 
I find that in the function b-cAf + &c. the term in z 3 x is 
= a (4 a-b + 4a 2 c + 4a& 2 + 4 abc — 2 ¥c), 
whence in the function fr’c 2 # 4 + &c. the term in z 3 x is 
= z 3 xa (4a 2 6 + 4a 2 c + 2ab 2 + 4a6c - 2b-c), 
a portion whereof is = 4<z 3 x (a 3 b A a 3 c); and we thus obtain the numerical factor =4, 
and thence the identity 
6 2 c 2 ;» 4 + cWy* A ci-b 2 z i + &c. = 4bcyz (cy - bz)- + ... + &c. 
which equation I represent by 
A a? + By 3 + Cz 3 A 3 (Fy-z + Gz 3 x + Hx 2 y + Iyz 2 + Jzx 2 + Kxy 2 ) + 6 Lxyz = 0, 
or 
(A, B, G, F, G, H, I, J, K, L\x, y, zf = 0, 
viz. writing for shortness 
M =bc+ ca + ah, 
the values of the coefficients are as follows: 
^ = 3(6 + c) {be A M), F = - c (3a 2 + M), 
B = 3(c +a)(ca + M), G = -a(36 2 + M), 
G = 3 (a + b) (ab + M), H = — b (3c 2 + M), 
I =-b (3a 2 + M), 
J = — c (36 2 + M), 
K = — a (3c 2 + M), 
L=2(a + b + c)M — oabc. 
I remark that the cubic having a node at the point (1, 1, 1), the Hessian has 
at this point a node with the same tangents. The second derived functions for the 
Hessian are therefore at the node proportional to those of the cubic; it is easy to 
verify that we have in fact 
A + H + J — {b + c) M, L + F AI = — clM, 
K A B A F = (c A a) M, J + L + G = - bM, 
G AI AG =(a+ b)M, H a K A L — — cM, 
these values give also 
A AKAGA2LA2J A2H=0, 
HABAIA2FA 2La2K = 0, 
J + F aG A2I A2GA 2L =0, 
equations which merely express that the first derived functions vanish at the node. If, 
by these equations we express A, B, G in terms of the other coefficients, and substitute 
these values in the equation of the Hessian, this may be expressed in the form 
(;y-zf{ LxA(2Fa I Al)yA( Fa 21 Al)z] 
+ (z-xy{( GA2JAL)xA LyA(2GA JaL)z) 
A{x — y) 2 {(2 H A KaL)xA{Ha 2 K A L)y A L z) — 0, 
a form which puts in evidence the node (1, 1, 1).