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But (20) 
sin ACP A sin BCP Bey sin BAP c sin CAP 
Y AC Gime eae 7 “sin BAG sin CAG’ 
Therefore, putting 
usin ACF sin BOG Ww sin DAC sin BAG 
vy sin FOB sm AQOG’” v snBAD sin CAG’ 
uu+tvy + ww = 0. 
Let the great circle HF’ pass through the poles hkl, p qr, 
not being opposite extremities of a diameter of the sphere. 
Then uh+vk+wl=0, and up+vq¢+wr=0. 
These two equations are satisfied by making 
u=kr—lq, v=lp—hr, w=hq-—kp. 
The great circle passing through the poles hkl, pqr will be 
called a zone-circle, and will be denoted by the symbol u v w, 
or by any three integers in the same proportion. 
Condition that a pole may be in a zone-circle. 
22. It appears from (21) that when the zone-circle u v w 
passes through the pole wv w, we have 
uu + vy + ww = 0. 
Any integral values of u,v, w that satisfy this equation are 
the indices of a pole in the zone-circle uv w, and any integral 
values of u, v, w that satisfy it are the indices of a zone-circle 
passing through the pole wv w. 
The intersections of any two zone-curcles are poles. 
23. Let the zone-circles hk1, pqr intersect in the points 
_P, P’. Ifit be possible, let P be the poleuvw. Then (22) 
hu +kv+lw=0, and pu+qu+rw=0. 
