564 
PROBLEMS AND SOLUTIONS. 
[485 
Solution by Professor Cayley. 
1. In the spherical triangle OPQ, whereof the sides OP, OQ, PQ are 0, (/>, ¡3 and 
cos ß — cos 0 cos </> 
the angle 0 is = a, the relation between these quantities is cos a = 
Sin 0 sin (f) 
hence treating a, ¡3 as constants, and 0, as variable angles connected by the fore 
going equation, it is required to show that we can find two fixed points S, H and 
a fixed line, such that taking M a variable point in this line and writing SM = r, 
HM = s, the relation between r and s (or equation of the fixed line in terms of 
r, s as coordinates of a point thereof) is obtained by substituting in the foregoing 
equation for 6 and c/> the values given by the two equations 
sin 6 = (r + s), sin <fi = (r — s), 
or as, for the sake of homogeneity, it will be more convenient to write these equations, 
TO sin 0 = (r + S), TO Sin (j) — (u — s). 
2. Suppose that the perpendicular distances of S, H from the fixed line are 
a and b, and that the distance between the feet of the two perpendiculars is 2c, then 
if x denote the distance of the point M from the midway point between the feet of 
the two perpendiculars, we have 
r = V {(c + xf + a 2 }, s = <v/{(c — xf + 6 2 }, 
and (a, b, c) being properly determined, the elimination of x from these equations 
should give between (r, s) a relation equivalent to that obtained by the elimination 
of (0, (f.>) from the before-mentioned equations. Or, what is the same thing, the 
elimination of (r, s, x) from the equations 
to sin 0 = r + s, m sin cf) = r — s t r = V{(c + x) 2 + a 2 }, s = \/{(c - ¿r) 2 + 6 2 } 
should give between (0, (f>) the relation 
cos ¡3 — cos 0 cos d> 
nna n —• L • 
cos a = 
that is, the last-mentioned equation should be obtained by the elimination of x from 
the equations 
to (sin 0 + sin cj)) = 2V |(c + x) 2 + a 2 }, to (sin 0 — sin c/>) = 2a/{(c — x) 2 + b 2 }. 
3. The equation in (0, cj)) may be written 
cos ß — cos a sin 0sincf) = cos 0 cos 0, 
or, squaring and reducing, 
that is 
sin 2 0 + sin 2 cf) = sin 2 ß + 2 cos a cos ß sin 0 sin + sin 2 a sin 2 0 sin 2 0,