Again 
of Edinburgh^ Session 1874-75. 
595 
Ps = sin r. 
V sm . 
therefore 
Pi 
/? _ y 
iU t>2 
'Pi 
)■ 
Sp 3 Y /3 p = 2 sin r sin r 2 sin r 3 
™ n ^ 1 2 3 sin sin p 2 sin jp 3 ’ 
Now 
therefore 
g <, sin r. sin r 9 sin r., Q 0 
VpiPiPz ~ 4-. — r ^ ba/?y . 
sm sin p sin p 3 
- Spipipt = 6 vol. of pyramid OP^Pg = 6Y 
- So j3y = 6 vol. of pyramid OABO = 6Y 
Y, = 4 
sin r, sm r„ sm r„ 
V. 
Also 
sm p x sm p 2 sm p 3 
SpYp x p 2 = 2 sin r sin r x sin r 2 . S<*Yy/? _ + Sff\ya 
sm p x sm p 2 sm p 3 
0 4 sin r sin r, sin r 9 a 0 
SpPi Pi = ^ ^ 1 e . ;vi<OT 8 Sa P7 1 
sm sm y? 2 sm jp 3 
4 sin r sin r t sin r 2 
sin _p A sin p 2 sin p 2 
V, 
calling pyramid OPP x P 2 = Y x , &c. 
Similarly we find the vols. of pyramids OPP 2 P 3 , &c., and arrive 
at this result — 
Y x sin r 3 + Y 2 sin r x + Y 3 sin r 2 = 3Y 4 sin r. 
Ex. (4.) To find the radius of Dr Hart's circle, i.e ., the circle 
which touches the inscribed circle and the three escribed circles. 
Let 7] be the vector of the pole of this circle, k its angular radius. 
Then since the circle touches all four circles, we must have, if z be 
its centre 
arc zPj = k + r L , zP 2 = k + r 2 , zP 3 = k + r 3 , zP = k - r . 
Hence 
S^pi = - cos ( k + r x ) = sin k sin r x - cos k cos r x 
S r)p. 2 = - cos (k -f r 2 ) = sin k sin r 2 - cos k cos r 2 
S>yp 3 = - cos (k 4- r 3 ) = sin k sin r 8 - cos k cos r 3 , 
4 ii 
VOL. VIII. 
