550 ON THE NINE-POINTS CIRCLE OE A SPHERICAL TRIANGLE. [964 
Thus p, q, r, p 1} q 1} denoting as above the cosines and sines of the half-sides 
of the triangle ABC, that is, the cosines and sines of the sides of the triangle 
A'B'C', we have 
tan B'F' = 
q — rp 
rp 1 
tan C'F' = 
r — pq 
№. ’ 
tan B'L' = , 
q + rp 
tan C'H = . 
r +pq 
First, for the points F', G', H', we have A'F', B'G', C'H', the perpendiculars from 
the angles on the opposite sides, meeting in a point O', the orthocentre of the 
triangle A'B'C': in fact, from the triangle A'B'F', right-angled at F', we have 
7-,/ fir 1 MTV TV . 1 cos 46 — COS 4c COS ia 
tan B'F = tan A B cos B = tan 4 c —.—= , 
smfcsmp 
_r x q — rp q — rp 
r r 1 p 1 rp 1 ’ 
as above, and similarly for the points G' and H'. 
I notice that we have 
sin B'F' = 
and thus 
sin B'F' = 
q — rp 
q — rp 
— rpf + r 2 pY} ’ + r 2 — !tpqr) ’ 
q-rp „.-p, _ r-pq 
\/(q 2 + r 2 — 2pqr) 
, sin C'F' = 
\J(q 2 + r 2 — 2pqr) ’ 
sin C'G' = 
r —pq 
\/(r 2 +p 2 — ¿pqr) ’ 
sin A'G' = 
p — qr 
\/(r 2 + p‘ J — 2pgr) ’ 
sin^ , FT / = - 77 — , , sin £'//' = - q - 2 rp 0 ; 
y \p +q — ¿pqr) y (p- + q 2 — ¿pqr) 
hence 
sin B'F'. sin tf'G'. sin A'H' = sin C'F'. sin ¿'G' . sin B'H', 
which (as is well known) is the condition for the intersection of the arcs A'F', B'G', C'H' 
in the orthocentre O'. 
But I say further that we have 
sin 2B'F' (= sin BF) = 
2 (q — rp) rp l 
q 2 + r 2 — 2pqr ’ 
and thence 
sin ¿C'F' (= sin OF) = 
2 (r — pg) gp : 
g 2 + r 2 — ¿pqr ’ 
sin BF. sin CG. sin AH = sin CF. sin AG. sin BH, 
and thus the arcs AF, BG, CH meet in a point which is obviously not the orthocentre 
of the triangle ABC.