204 
ON A CORRESPONDENCE OF POINTS 
[512 
viz. this is 
= — a?yz — b 2 zx — c 2 xy + be cos A . x (x + y + z). 
Moreover, if A = twice the area of the triangle, then 
(A — /3 3 ) X — (a 2 — a s ) Y + a 2/ S 3 — a 3 /3 2 = Ax (x + y + z) = be sin A . x (cc + y + z) ; 
so that the equation becomes 
- a?yz — b 2 zx — c 2 xy + be sin A (cot A - cot L) x (x + y + z) = 0, 
or, what is the same thing, 
— a?yz - b 2 zx — c 2 xy + A (cot A — cot L) x (x + y + z) — 0, 
or, if we please, 
— a?yz — b 2 zx — c 2 xy + A (cot A — cot L) x = 0. 
Writing as before, 
— o?yz — b 2 zx — c~xy = U, x + y + z = fl, 
the equation is 
U + A (cot A — cot L) fix = 0; 
or forming the like equations of two other similar circles, we have the circles (B, C), 
(G, A), (A, B) containing the angles L, M, N respectively; and the equations are 
U + A (cot A — cot L ) fix = 0, 
U + A (cot B — cot M) fly = 0, 
U + A (cot G — cot N) flz = 0. 
Correspondence, A, B, G at P, and A', B', G' at P', subtending equal angles. 
5. Consider now the two figures A', B', C', subtending at P' the same angles 
L, M, N which A, B, G subtend at P; then we have 
mx +aoiA ~ cotL =°- n^+ cotA '- cotL =°> 
UKj+ cot 5 - cot M =°- + cot B ' - M = o, 
and thence 
UKz + c °t O - cot N =0, jj^y + cotV' —cotif=0; 
flAx + cotA= n'AV + cotA ’ 
DAy +cotB = + cot 
¿ +cot0= iiw + cot 0/5