330 BROOKS — ORTHIC CURVES. [May 20, 



(i) x"" — aiX""-^ + a^A'"-' ^ . . . _ a2^_il'"-^ -|- a,~x'' = o, 



and 



(2) x"" — a'^x''-' + a',x''-' _+...— rA,_,I"-' + a',^^ = o. 



We now wish to show that the centroid of the orthocentric 7i^- 

 point in which these two curves intersect is the centre of the centre 

 circle of the pencil. If we rewrite (i) and (2) in terms of .t we get 



(i) {x — X,) {x — X,) ... (I — I-J = o, 



and 



(2) {x —x\) (x — x',) , . . (x — yj = o. 



If the j's refer to the elementary symmetrical functions of the 

 roots, we have 



Si = «2n-i, s\ = a',^_i, 1= I, 2, . . . /I— I. 



J-n = — (a-° — ai.Y°-^ + a.^x""-" . . . a„) a.,^-\ 



/, = — (a--^ — aVv°-^ + a^-- . . . a J a',-\ 



Now the eliminant of a- between these two equations is 



(^^ _ p^) (- _ -y . . . (I^ _ ?J. 



{x, — :v'i) (a;., — A-'^) . . . (X2 — A-'J. 



(^■n — X\) (I-, — I'^,) . . . (i, — A-'O = O. 



This is a function of degree ;^^ in x, and as a: occurs in s^ and /^ 

 alone, we need consider those terms alone in which the products j-^ 

 and /n appear. These are : 



.C — ^^C '^ a + 2 S^^^-S',' . . . //, 



or 



or, in terms of x, 



