3« 



KANSAS UNIVERSITY QUAR'lERLV. 



-^ Reflex <()CB. 



<CAE:r=2 <APC=| Reflex <OCB. 



=^ Reflex <OCB — .V<APC. since <APC^<ACP 

 . •. <APC. 

 Cor. <COE'= 



Now suppose the line OAP to revolve about O, (any fixed point on 

 the circumference of a circle) and AP to remain equal to the radius 

 of the circle. Then it is asserted that if the locus of P be determined, 

 a given angle can be trisected by the aid of this curve. 



IHE LOCUS OK p. 



To find the polar equaiion of the locus of P, let O (fig. IV) be 

 the pole and OX, tangent to the circle at O, be the initial line; 

 let 0?z^/?, <P0Xi=:6', and R^-- Radius of the circle. 



Then the eiiuation of the circle, the locus of A, is 

 J?^.2 Rsin 8. 



Therefore, since AP=R, the equation of the locus of P is 

 (i) i?=2 Rsin ^+R==R (2 sin ^+i). 



To transform this equation from polar to rectangular co-ordinates, 



let O be the origin and OX the axis of abscissce. 



y y 



Put y?=i/y2_|_v2 and sin ^^^ -4r^^ 



Whence by substituting in (i j and reducing we get 



(2) x2+y2=.R (2y ±i'^;^^+7I) 



In order that (i) may represent the complete locus obtained by 

 using both the plus and minus values of the radical term in (2) the 

 equation must be written 



(3) i? = R(2 sin ^ n:!) 



The minus sign in both of these equations gives the locus of P', 

 where AP'=R. 



