318 
Proceedings of Royal Society of Edinburgh. [sess. 
or 
Pm-i = B + d m A - (A - a m B)p 1 , 
Qm — 1 R “I” (B 4" Ct m A jpj • 
These agree with (5), because m - 1 is even. And similarly we 
may prove the proposition when m is even. 
If now, in (2), we put n+ 1 for m, we have 
C + T)pi . ^ - 
p«+i=pi=i j -Cp lf n be evenj 
= ~ if n be odd, 
D + Cpj 
C and D being quaternions to be calculated (as above) from the data. 
The two cases require to be developed separately. 
Take first, the odd polygon: — 
then pf) 4- p i Cp 1 = C - D/q, 
or pfd 4- 8) 4- pfc 4- y)p 1 = c + y- (d + S)p 1 , 
if w^e exhibit the scalar and vector parts of the quaternions C and D. 
Cutting out the parts which cancel one another, and dividing by 2, 
this becomes 
dp Y 4- S Sp 1 4- /qSy/q - , c = 0, 
which, as p, is finite, divides itself at once into the two equations 
Sy/q + d = 0 , 
S 8p 1 — c = 0 . 
These planes intersect in a line which, by its intersections (if real) 
with the sphere, gives two possible positions of the first corner of 
the polygon. 
For the even polygon we have 
Pi-D - Pi C Pi = C 4- Dp T , 
or V /qS - pfiypi - y = 0 ; 
which may be written 
v. Pi (s-y w )=o. 
This equation gives 
pi = (* + y) ! (s + ~f) i 
