590 
Proceedings of the Royal Society 
x — — . 
sm p x 
Similarly 
_ sin r _ sin r 
^ sin p’ sin p 3 
Hence 
P = sin r ) • • (1.) 
\ sm Px sin p 2 sm pj 
To find the vectors of the poles of the escribed circles, let 
Pi Pa Pi be the vectors of Pj P 2 P 3 , the poles of the escribed circles 
opposite to ABC respectively. Then, as before, we may write 
p x - xa + y/3 + zy . 
Determining the scalars xyz as before, we have 
^ _ cos PjA' _ cos PjB' w _ cos PjC' 
X " cos AA' ’ y ~ ’ 2 ~ cos GO' ’ 
By geometry of the figure we see that 
I* ^'= 1 +^ PiB'=g-r, P,G '=|-r t . 
Hence 
Therefore 
_ _ sm r x _ sm r x _ sm r x 
sin p x 5 ^ sin p 2 5 2 sin p 3 ' 
p. = sm r x ( - PL- + J— + 
sin^ sm|) 2 sm p 3 / 
Similarly we find 
P -2 = sin r 2 ( PL- _ + -PL- ^ 
\smj}j sm p 2 sm py 
p 2 - sin r 3 ( — 1- ~J — — - PL- ) 
\sin pj sm p 3 sm pj 
( 2 .) 
(3.) 
(^) 
Coroll. : 
Pi | P-2 + Pi = a + P 7 = P 
sm r x sin r 2 sin r 3 sm p x sin p 2 sin p 3 sin r 
(a result which is useful further on). 
To find vector of the pole of the circumscribed circle, let cr be 
the vector of Q, the pole of the circumscribed circle. Then since 
