Hogg. — On certain Tripolar Relations. 321 



X Y 7 



If P 1} P 2 be inverse points determined by the equations - — - : = — , 



(■ 11 1 n 



1 1 X.i — X 2 Yj — Y Zj — Z. 2 . , , 



then — = = r = - = kj — k.,, where k x and k 2 are the 



Z wi // 



roots of the equation * 2 [2 (aH 2 ) — 2 2 (6c cos Aww] 



- 2k a&c [2 (a cos A/)] + a 2 6V 2 = o. 



Hence the equation of the line bisecting perpendicularly the above pair 

 of inverse points is 



laa + mbfi + ncy = o. 



The two points whose pedal triangles are equilateral are determined by 

 the equations a 2 X = Z> 2 Y = c 2 Z. Hence these points are equidistant from 



the line — I- r + — = o, which is therefore perpendicular to the Brocard 

 a b c 



diameter 2 [a 2 (b 2 — c 2 ) X] = o, on which the points lie. 



The two points whose distances from A, B, C are proportional respec- 



X Y 7 



tively to a, b, c are given by the equations — ==- = —: they therefore 



(At (J C 



lie on Euler's line, 2[(& 2 — c 2 ) X] = o, and are equidistant from the line 

 a?a + b s fi + c 3 y = o. 



Let two pairs of inverse points be determined by the equations 



— =-- = - and — = -- = — ; the centre of the circle through them 



will be determined by the equations 



l x aa + mib/3 + n x cy = o 

 Uaa + mjbfi + n. 2 cy = o, 

 whence fla : 6/3 : cy = m^ — ra.yH, : w^ — n 2 l\ : Z^ — Z 2 m x , 

 and the equation of, the circle will be 



(to!» 2 — m 2 n 1 ) X + (^1^2 — n -A) Y" + (^m 2 - Z 2 ??i]) Z — o. 

 The two pairs of points will lie respectively on the diameters 

 (m, — iij) X -f (wj — Zj) Y + (Z x — m : ) Z = o 

 (ra 2 - ?i 2 ) X -f (n. 2 — L) Y + (Z 2 — ra 2 ) Z = o. 



We now proceed to find the equation in tripolar co-ordinates of the 

 inverse of the circle U = o with respect to S, the circum-circle of the 

 triangle ABC. 



U = ZX + mY + nZ - h 2 = o 



S = a cos AX + b cos BY + c cos CZ — abc = o. 



Let P be any point on U and Q its inverse. Let (X, Y, Z), (X 1 , Y 1 , Z 1 ) 

 be the tripolar co-ordinates of P and Q respectively, then 



X Y^ Z HP 2 



X I = Y 1 == Z 1 = ~W' 

 where H is the circum-centre. 



Hence HP 2 (ZX 1 + mY 1 + nZ 1 ) = /i 2 E 2 . 



11— Trans. 



