80 



Mr. Sylvester's Anahjtical Development 



Let 8 ^^~ ^l^,, then it is clear that O Q = O R, and the intersec- 

 tion of the two planes perpendicular to O Q, O R is therefore a line 

 perpendicular to the plane Q O R, and to the line which bisects 

 the angle Q O R. 



In fact if we draw Q T, R T perpen- 

 dicular to O Q, O R respectively in the 

 plane Q O R, the intersection in ques- 

 tion passes through Tand is perpendicu- 

 lar to O T ; also 



0T = 0Q.8ec(iR0Q) 

 to the first order of smallness. 



Now it is easj'^ to see (just as in page 57) that 



OQ 



ROP = 



and also 



smjM, 



QOP 



sin ju. 



.*. POP=QOPand .-. P O T is perpendicular to QO D. 



Hence the problem is reduced to finding I the intersection of two 

 lines T I, P I drawn in the same plane POT. 



Now because O T I, O P I are each right angles, a circle may 

 be made to pass through I, T, P, O. 



Hence the angle 



PI0=PT0 = ta„-1.0P2LZ0T 



= tan 



1 OlP X POR.co8^|u> 

 .'. tan P O I = sin i jtA . 



OPx 



= tan 



rfOP 



sin/u* 

 ^TOP 



cos I /A 



^1. 



d I, 



and O I = O P . sec POL 



Also the position of the plane P O I is known, and .*. the radius 

 is completely determined in magnitude and position. 



It may be worth while also to remark that the above construc- 

 tions enable us to form a series of equations between the magnitude 

 of the radius and its inclinations to the two prime perpendiculars. 



In fact, if we call ^t^ tt^, the two inclinations in question 



cos TT^ = cos P O I cos i, + sin P O I sin /^ . sin -i- 



cos IT// = cos P O I COS in -}- sin P O I sin <// 



sin i- 



