200 
ON SOME PROBLEMS OF ORTHOMORPHOSIS. 
[921 
and thence 
X, 2 + Y* = 
( 1 + kf P 2 + 1 - (1 + k 2 ) P 2 + k*P* _ 1 
k (1 + 2&P 2 + k 2 P 4 ) 
k’ 
that is, X 2 + Fj 2 — r = 0 as before. It is easy to see that the points A, 0, B of 
w 
figure 1 correspond to the points A, 0, B of figure 2, and hence that the area of 
the rectangle AOBLFMA of figure 1 corresponds to that of the semicircle AOBLFMA 
of figure 2. 
Returning to the equation X 1 + iY 1 = sn(X -ftF), if we write herein successively 
F a = \K' - /3, 
Y 1 = \K' + /3, sn i Y 1 = iQ 2 = sn i (%K r + /3), 
then we have 
iQi ■ iQ* = sn i {\K r - /3) sn i (\K‘ + P) = ~ p 
that is, 
QiQz = 
1 
k'' 
hence for q writing or Q 2 , we have in each case the same 
values of A and B, that is, we have the same bicircular quartic for two lines 
parallel to and equidistant from the line LM, but to one of these (viz. the line 
between LM and BA) there corresponds the half-perimeter lying within the semi 
circumference LFM, and to the other of them (viz. the line between LM and CD) 
there corresponds the half-perimeter lying without the semicircumference LFM. 
It may be shown in like manner that, to any line in figure 1 parallel to the 
axis OF, there corresponds in figure 2 a bicircular quartic of the like form 
(X* + F x 2 ) 2 - 2AX 2 -2BY 1 * + ^ = 0. 
13. Similarly in explanation of No. 5, observe that, starting from the equation 
x x + iyi = snl (x + iy) and writing snl;r=p, snl iy = i snl y = iq, we have 
p Vl — q 4 + iq Vl — p 4 
(Bi + vyi = ' 
1 —p 2 q 2 
p 
x, = - 
V1 — q 4 ^ _ q Vl — p 4 
i - pY * Vl ~ 1-pY"’ 
p 2 + q 2 
«i 2 + 2/i =fz 
pY 
that is, 
and thence 
writing x + y = we have 
cn x . . V l - p 2 „ 1 — p 2 , , . p 2 + f -, . 
sny = , that is, q=—===-, or o 2 = — , that is, =1, 
dn x V1 + p 2 1 + p 2 1 — p 2 ? 2 
and thus to the line x + y = there corresponds the circle x 2 + y 2 — 1 = 0. More 
precisely, to the line CD of figure 4, there corresponds the quarter-circumference CD