824 BROOKS — ORTHIC CURVES. '[May 20, 



and 



These are conjugate equations and so can be satisfied by the co- 

 ordinates of real points. Solving them, we get a pair of real points 





and 



(7^— VfTi^— 4^2 " ^1 — VtT^2 — 4^2 



• - 7 X-i - ' 



But further, we notice that the antipoints, x^, x^, and x.^, x^, of 

 these make the equation of the pencil vanish for all values of the 

 parameter. They are the remaining points of the orthocentric nine. 

 This leads to the theorem that : 



If five points of an orthocentric nine-point are on a circle^ of the 

 remaining four points two are real, two are imaginary ; and these 

 four form an orthocentric four-point. 



The centroid of the nine points is 



^^^ = i(A+ . . t,-\-2a,)=\<r,. 



This is the centre of the centre circle of the pencil. 



We can extend these results to the case of «^-points, 2n — i of 

 which lie on a circle. 



The pencil of orthic curves of order ;/ which go through 2;z — i 

 points of the unit circle is given by 



OC- _ (^^ _}_ /) ^^n-l _!_ (^^ ^ i^^^ ^^n-, _ _f_ . . . 



(^^20-2 + ^^2n-3) •^"""' — (<^2n-l + ^^2n-2) '^'"^ + ^2n^''^° = O. 



If we let (T^a-i = I J this becomes 



(x — t) (x^-' — a^x^-' + ^2^^-^ _-[-_.. a,_^) 

 + {xt— i) {x^-' —~<T,x--' + — . . . ^a-l) = o. 



Now since the coefficients of (x — /) and (xt — i) are conjugate 

 forms, there are n — i real points, in addition to the points /i, A 

 . . . . /2n-ij which are on all the curves of the pencil. Further, 



/ 



