172 Proceedings of the Royal Irish Academy. 



If the first equation in (2) be written 



1 1 £f^^ 1 



= 63 



%i — P Si — a \%i — a Si — y 



we see that geometrically it signifies that A^Bq is equal in length 



2ir . 



Co'Ao'y and is inclined to it at an angle — , or in fact that A^B^C,^ is 



o 



an equilateral triangle. 



Zi, Zi are therefore the origins round either of which ABC inverts 

 into an equilateral triangle. 



Fig. 2. 



Again, if we take the segment intercepted on a side of the triangle 

 ABC by the internal and external bisectors of the vertical angle, 

 and on it, as diameter, describe a circle, we have one of the circles 

 of Apollonius, and the three circles thus described have the points ^i, 

 Zi as their common points. These circles cut orthogonally the circum- 

 scribing circle of the triangle, and the points ^i, Z^^ are therefore inverse 

 points with respect to that circle. 



It is hardly necessary for me to indicate how the points Z^^ Z^ are 

 determined from the known equation (1), and in what follows I shall 

 therefore suppose them known in position, and from their positions 

 determine those of the points ABC. 



Solution of the Cubic. — Solving the equations (2), we obtain 

 for the complexes denoting Z^ and Zo the values 



Py + yaco + a^co- )8y + yaco^ + afSoi 



a + po) + yto^ a + p w + yo> 



and if we mark on the plane the points Zi, Z. defined by these quanti- 



