8 Gee AND ADAMSON, Trisecting an Angle. 



BF or twice AB, and the angle ADB, or a, being equal 

 to one-third of the angle ABC, or /3. 



In this figure ^4Z> is obviously one of the trisectors 

 of the angle BAM, and also BF is one of the trisectors 

 of an angle DBK made equal to the angle ABC. 



IV. General method by the i?itersection of a conic and 



a circle. 



In an article on "The Trisection of a Given Angle," 7 " 

 L. Crawford states that " the general problem is solved 

 by considering the intersection of a conic with a circle, 

 the conic to pass through four points in the circle, three 

 being the angular points of an equilateral triangle. The 

 general equation is satisfied by a rectangular hyperbola, 

 also by a hyperbola of eccentricity two. Less simple 

 constructions are also possible with parabola and ellipse." 



Some of these methods will now be given. 



(i) Use of a rectangular hyperbola. 



The second method of Pappus already given is one 

 example. Its agreement with Crawford's statement may 

 be demonstrated by combining the equations of the 

 curves. Let A (Fig. 6) be taken as origin with axis of 

 x and y parallel to BC and BG ; then if AG — a and 

 BG = b, the equations of the hyperbola and the circle are : 



xy — ah 

 and 



(x-ay + (y-l>f = 4(<r + b' 2 )- 



The ordinates of the four points of intersection are found 

 by eliminating x and solving for y. One solution is 

 y=—b, and this corresponds to the intersection on BA 

 produced. The other three solutions are given by the 

 cubic equation : 



7 /'roc. Edinburgh Math. Soc, Vol. 16, p. 42. 



