SOLUTION FOR THREE TRACED FACES. 371 



SubstitutiDg (9) in (10) and reducing, 



=.{ 



(2y' — 2 xyp-^ — r^ + 1) (2 x" + 2 pxy — y- + 1) 

 — 4p-^ {x'—phf) 



(2x'-2pxy-T' + l)(2f + 2 xyp-' - ^^ ^ 1) .-,.. 



4p-'{x^-py) ' ^ J 



Cancelling 4 {x~—p'^y-), clearing of fractions, and changing to polar coordi- 

 nates, we have (since 2 x"^ — r^ = r"^ (2 cos'^cp — 1)= r'^ cos 2(p, and 2 ^/^ — r* 

 is similarly — r^ cos 2 c, and 2 :c?/ = f^ sin 2 <p), 



(r' cos 2 if -f 1)'^ —fv'' sin' 2 cp - f {1 - r' cos 2 cpj 



+ r* si)i- 2 ip = plx \ — ( — t'^ cos2 (f — r '- J 



(1 4- ''^ cos 2 (f -\- pr"^ sin 2 cr) — (1 + r- cos 2 c — pr' sin 2 c) 



(l - ,-' coa 2 c^ + ^ - J- 



Arranging now according to powers of r, cancelling and putting 1 for 

 sin- -f <ioi — 



•* iX-f) + 2 r cos 2 o (1 +/)-/ + 1 = 2 plz (r' - 1) 



or — 



.+2.eo.2.,-4±£:-^ + ;-E|;^;=0; (12) 



but since J9 = cot ^ , according to Chauvenet (Trig. Eq., 142) — 

 1 I p2 1 — ?9' 



-^^^ = 2csca, ^= -2co^a. (13) 



Dividing the numerator and denominator of the coefficients in (12) by p 

 and substituting from (13) we have — 



. ^ „ ^ 2 CSC a — 2 cot a -\- 2 k 



»■' + 2 ;■ cos 2 c. _2 eo( «-2l + ^^ eor« --2^^ = *^- 



From this, replacing I- by co^ 3, we have (1). 



For certain values of 0, equation (1) is much simplified, so that v and the 



