602 ME THOMAS STEPHENS DAVIES ON 



PROPOSITIONS XXVI., XLI. 



Let there be a regular polygon of m sides inscribed in a circle whose radius is p, 

 and let n be any number less than m ; and from any point Z in the circum- 

 ference of the circle, let lines be drawn to all the angular points of the polygon 

 then we shall have 



For in this case r= ; and the expression for Z A* 2 " takes the form 



= {2p 2 -2 2 cos (6-6k)}* 



Expanding the binomial factor by Lemma iii., and keeping in view that 

 Q v 6-t , . . . . B m are in arithmetical progression, having the common difference 



2-7T 



, we see that all the angular functions vanish, by Lemma iv. Whence the sum 

 is reduced as before to m times the term clear of the angles, and we thus have 



. 1 1.3.5 ---- (2-l) 2 * 



) - 2^ ' 1.2.3.... - ^T - 2 ? 



1.3.5 ____ (2)1-1} s 



1.2.3.... ;r~ 2 ? 



This is Proposition 41, as proposed by Dr STEWART. When n=2 it becomes 

 Prop. 26. 



CLASS III. Lines making equal A.ngles. 

 PROPOSITIONS XIV., XXXIV., XLV. 



Let there be m lines meeting in a point, and making all the angles round the point 

 equal, and let n be any number less than m : then if from any point Z, whose 

 distance from the common point of section 'is r, perpendiculars be drawn to all 

 the lines, Z A lf Z A 2 , . . . Z A m , we shall have 



!. m 1.3.5 ---- (2-l) j 

 2^ '1.2.3.... ^~ 



For simplification, take the line from Z to the given intersection as angular 

 origin, that intersection being the polar origin. Then O lt 6 2 } . , . . Q m of the 

 general equation are the angles made with this line by the perpendiculars to the 

 given lines. Also, in the equations of the given lines p l = p 2 = . . . = p m ; and 

 t , 2 , . . . . 6 m are in arithmetical progression, having the common difference 



- . Wherefore, the general form of the expression for Z A m 2 " is 



. 2n In 2 2 klT 



Z A m = r cos . 



